Chapter 3 Shigley's 9 Edition Solutions Manual
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Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times
Ans.
______________________________________________________________________________
1-8 C A = C B ,
1-7
10 + 0.8 P = 60 + 0.8 P 0.005 P 2
P 2 = 50/0.005
P = 100 parts Ans.
______________________________________________________________________________
1-9
Max. load = 1.10 P
Min. area = (0.95)2A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
nd
1.10
0.85 0.95
2
1.43
Ans.
______________________________________________________________________________
1-10 (a) X 1 + X 2 :
x1 x2 X 1 e1 X 2 e2
error e x1 x2 X 1 X 2
e1 e2
(b) X 1 X 2 :
Ans.
x1 x2 X 1 e1 X 2 e2
e x1 x2 X 1 X 2 e1 e2
( c) X 1 X 2 :
Ans.
x1 x2 X 1 e1 X 2 e2
e x1 x2 X 1 X 2 X 1e2 X 2 e1 e1e2
e
e
X 1e2 X 2 e1 X 1 X 2 1 2
X1 X 2
Ans.
Chapter 1 Solutions - Rev. B, Page 1/6
(d) X 1 /X 2 :
x1 X 1 e1 X 1 1 e1 X 1
x2 X 2 e2 X 2 1 e2 X 2
1
e2
e2
1
1
X2
X2
1 e1 X 1
e
e1
e2
e1
2
1
1
1
X1
X2
X1 X 2
1 e2 X 2
x
X
X e
e
Thus, e 1 1 1 1 2 Ans.
x2 X 2 X 2 X 1 X 2
______________________________________________________________________________
then
x 1 = 7 = 2.645 751 311 1
X 1 = 2.64
(3 correct digits)
x 2 = 8 = 2.828 427 124 7
(3 correct digits)
X 2 = 2.82
x 1 + x 2 = 5.474 178 435 8
e 1 = x 1 X 1 = 0.005 751 311 1
e 2 = x 2 X 2 = 0.008 427 124 7
e = e 1 + e 2 = 0.014 178 435 8
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks
X 1 = 2.65, X 2 = 2.83 (3 digit significant numbers)
(b)
e 1 = x 1 X 1 = 0.004 248 688 9
e 2 = x 2 X 2 = 0.001 572 875 3
e = e 1 + e 2 = 0.005 821 564 2
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.65 +2.83 0.001 572 875 3 = 5.474 178 435 8
Checks
______________________________________________________________________________
1-11 (a)
3
16 1000 25 10
S
1-12
2.5
d3
nd
Table A-17: d = 78 in Ans.
n
Factor of safety:
S
d 0.799 in
25 103
16 1000
7
8
3.29
Ans.
Ans.
3
______________________________________________________________________________
n
1-13 Eq. (1-5):
R = Ri = 0.98(0.96)0.94 = 0.88
i 1
Overall reliability = 88 percent
Ans.
______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 2/6
1-14
(a)
a = 1.500 0.001 in
b = 2.000 0.003 in
c = 3.000 0.004 in
d = 6.520 0.010 in
w d a b c = 6.520 1.5 2 3 = 0.020 in
tw tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020 0.018 in
Ans.
(b) From part (a), w min = 0.002 in. Thus, must add 0.008 in to d . Therefore,
d = 6.520 + 0.008 = 6.528 in
Ans.
______________________________________________________________________________
1-15 V = xyz, and x = a a, y = b b, z = c c,
V abc
V a a b b c c
abc bca acb abc abc bca cab abc
The higher order terms in are negligible. Thus,
V bca acb abc
and,
V bca acb abc a b c a b c
Ans.
V
abc
a
b
c
a
b
c
For the numerical values given, V 1.500 1.875 3.000 8.4375 in 3
V 0.002 0.003 0.004
0.00427
V
1.500 1.875 3.000
V 0.00427 8.4375 0.036 in 3
V = 8.438 0.036 in3 Ans.
______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 3/6
1-16
w max = 0.05 in, w min = 0.004 in
0.05 0.004
w=
0.027 in
2
Thus, w = 0.05 0.027 = 0.023 in, and then, w = 0.027 0.023 in.
w= a b c
0.027 a 0.042 1.5
a 1.569 in
tw =
Thus,
t
all
0.023 = t a + 0.002 + 0.005 t a = 0.016 in
a = 1.569 0.016 in
Ans.
______________________________________________________________________________
1-17
Do Di 2d 3.734 2 0.139 4.012 in
t Do tall 0.028 2 0.004 0.036 in
D o = 4.012 0.036 in
Ans.
______________________________________________________________________________
1-18 From O-Rings, Inc. (oringsusa.com), D i = 9.19 0.13 mm, d = 2.62 0.08 mm
Do Di 2d 9.19 2 2.62 14.43 mm
t Do tall 0.13 2 0.08 0.29 mm
D o = 14.43 0.29 mm
Ans.
______________________________________________________________________________
1-19 From O-Rings, Inc. (oringsusa.com), D i = 34.52 0.30 mm, d = 3.53 0.10 mm
Do Di 2d 34.52 2 3.53 41.58 mm
t Do tall 0.30 2 0.10 0.50 mm
D o = 41.58 0.50 mm
Ans.
______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 4/6
1-20
From O-Rings, Inc. (oringsusa.com), D i = 5.237 0.035 in, d = 0.103 0.003 in
Do Di 2d 5.237 2 0.103 5.443 in
t Do tall 0.035 2 0.003 0.041 in
D o = 5.443 0.041 in
Ans.
______________________________________________________________________________
1-21 From O-Rings, Inc. (oringsusa.com), D i = 1.100 0.012 in, d = 0.210 0.005 in
Do Di 2d 1.100 2 0.210 1.520 in
t Do tall 0.012 2 0.005 0.022 in
D o = 1.520 0.022 in
Ans.
______________________________________________________________________________
1-22 From Table A-2,
(a) = 150/6.89 = 21.8 kpsi
Ans.
(b) F = 2 /4.45 = 0.449 kip = 449 lbf
Ans.
(c) M = 150/0.113 = 1330 lbf in = 1.33 kip in
(d) A = 1500/ 25.42 = 2.33 in2
(e) I = 750/2.544 = 18.0 in4
(f) E = 145/6.89 = 21.0 Mpsi
(g) v = 75/1.61 = 46.6 mi/h
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
(h) V = 1000/946 = 1.06 qt
______________________________________________________________________________
1-23 From Table A-2,
(a) l = 5(0.305) = 1.53 m
(b) = 90(6.89) = 620 MPa
(c) p = 25(6.89) = 172 kPa
Ans.
Ans.
Ans.
Chapter 1 Solutions - Rev. B, Page 5/6
(d) Z =12(16.4) = 197 cm3
Ans.
(e) w = 0.208(175) = 36.4 N/m
Ans.
(f) = 0.001 89(25.4) = 0.0480 mm
(g) v = 1200(0.0051) = 6.12 m/s
Ans.
Ans.
(h) = 0.002 15(1) = 0.002 15 mm/mm
(i) V = 1830(25.43) = 30.0 (106) mm3
Ans.
Ans.
______________________________________________________________________________
1-24
(a) = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi
(b) = F /A = 9440/23.8 = 397 psi
Ans.
Ans.
(c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in
Ans.
(d) = Tl /GJ = 9740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95
Ans.
______________________________________________________________________________
1-25
(a) =F / wt = 1000/[25(5)] = 8 MPa
Ans.
(b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4
Ans.
(c) I = d4/64 = (25.4)4/64 = 20.4(103) mm4
Ans.
(d) =16T / d 3 = 16(25)103/[ (12.7)3] = 62.2 MPa Ans.
______________________________________________________________________________
1-26
(a) =F /A = 2 700/[ (0.750)2/4] = 6110 psi = 6.11 kpsi
Ans.
(b) = 32Fa/ d 3 = 32(180)31.5/[ (1.25)3] = 29 570 psi = 29.6 kpsi
Ans.
(c) Z = (d o 4 d i 4)/(32 d o ) = (1.504 1.004)/[32(1.50)] = 0.266 in3
Ans.
(d) k = (d 4G)/(8D 3 N) = 0.06254(11.3)106/[8(0.760)3 32] = 1.53 lbf/in
Ans.
______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 6/6
Chapter 2
2-1
From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: S ut = 380 (55) MPa (kpsi), S yt = 210 (30) Mpa (kpsi) Ans.
(b) SAE 1050 CD: S ut = 690 (100) MPa (kpsi), S yt = 580 (84) Mpa (kpsi) Ans.
(c) AISI 1141 Q&T at 540C (1000F): S ut = 896 (130) MPa (kpsi), S yt = 765 (111)
Mpa (kpsi) Ans.
(d) 2024-T4: S ut = 446 (64.8) MPa (kpsi), S yt = 296 (43.0) Mpa (kpsi) Ans.
(e) Ti-6Al-4V annealed: S ut = 900 (130) MPa (kpsi), S yt = 830 (120) Mpa (kpsi) Ans.
______________________________________________________________________________
2-2
(a) Maximize yield strength: Q&T at 425C (800F) Ans.
(b)Maximize elongation: Q&T at 650C (1200F) Ans.
______________________________________________________________________________
Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102
AISI 1018 CD steel: Tables A-20 and A-5
3
S y 370 10
47.4 kN m/kg
Ans.
76.5 102
2011-T6 aluminum: Tables A-22 and A-5
3
S y 169 10
62.3 kN m/kg
Ans.
26.6 102
Ti-6Al-4V titanium: Tables A-24c and A-5
830 103
Sy
187 kN m/kg
Ans.
43.4 102
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the
ultimate strength in tension
3
Sut 42.5 6.89 10
40.7 kN m/kg
Ans
70.6 102
______________________________________________________________________________
2-3
2-4
AISI 1018 CD steel: Table A-5
6
E 30.0 10
106 106 in Ans.
0.282
2011-T6 aluminum: Table A-5
6
E 10.4 10
106 106 in Ans.
0.098
Ti-6Al-6V titanium: Table A-5
Chapter 2 - Rev. D, Page 1/19
E
16.5 106
103 106 in
Ans.
0.160
No. 40 cast iron: Table A-5
6
E 14.5 10
55.8 106 in Ans.
0.260
______________________________________________________________________________
2-5
2G (1 v) E
From Table A-5
Steel: v
Aluminum:
v
30.0 2 11.5
2 11.5
v
Beryllium copper:
E 2G
2G
0.304
Ans.
10.4 2 3.90
0.333
2 3.90
v
2 7.0
0.286
Ans.
14.5 2 6.0
0.208 Ans.
2 6.0
______________________________________________________________________________
Gray cast iron:
2-6
v
18.0 2 7.0
Ans.
(a) A 0 = (0.503)2/4, = P i / A 0
For data in elastic range, = l / l 0 = l / 2
A
l l l0 l
For data in plastic range,
1 0 1
l0
l0
l0
A
On the next two pages, the data and plots are presented. Figure (a) shows the linear part of
the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the
complete range. Note: The exact value of A 0 is used without rounding off.
(b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans.
From Fig. (b) the equation for the dotted offset line is found to be
= 30.5(106) 61 000
(1)
The equation for the line between data points 8 and 9 is
= 7.60(105) + 42 900
(2)
Chapter 2 - Rev. D, Page 2/19
Solving Eqs. (1) and (2) simultaneously yields = 45.6 kpsi which is the 0.2 percent
offset yield strength. Thus, S y = 45.6 kpsi Ans.
The ultimate strength from Figure (c) is S u = 85.6 kpsi Ans.
The reduction in area is given by Eq. (2-12) is
R
A0 Af
A0
100
Data Point
Pi
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0
1000
2000
3000
4000
7000
8400
8800
9200
8800
9200
9100
13200
15200
17000
16400
14800
0.1987 0.1077
100 45.8 %
0.1987
0
0
0.0004
0.0006
0.001
0.0013
0.0023
0.0028
0.0036
0.0089
0.1984
0.1978
0.1963
0.1924
0.1875
0.1563
0.1307
0.1077
0.00020
0.00030
0.00050
0.00065
0.00115
0.00140
0.00180
0.00445
0.00158
0.00461
0.01229
0.03281
0.05980
0.27136
0.52037
0.84506
0
5032
10065
15097
20130
35227
42272
44285
46298
44285
46298
45795
66428
76492
85551
82531
74479
l, A i
Ans.
(a) Linear range
Chapter 2 - Rev. D, Page 3/19
(b) Offset yield
(c) Complete range
(c) The material is ductile since there is a large amount of deformation beyond yield.
(d) The closest material to the values of S y , S ut , and R is SAE 1045 HR with S y = 45 kpsi,
S ut = 82 kpsi, and R = 40 %. Ans.
______________________________________________________________________________
2-7
To plot
true
vs., the following equations are applied to the data.
P
true
A
Eq. (2-4)
Chapter 2 - Rev. D, Page 4/19
where
A0
(0.503)2
ln
l
l0
for 0 l 0.0028 in
ln
A0
A
for l 0.0028 in
0.1987 in 2
4
The results are summarized in the table below and plotted on the next page. The last 5
points of data are used to plot log vs log
The curve fit gives
m = 0.2306
log 0 = 5.1852 0 = 153.2 kpsi
Ans.
For 20% cold work, Eq. (2-14) and Eq. (2-17) give,
A = A 0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2
A0
0.1987
ln
0.2231
0.1590
A
Eq. (2-18): S y 0 m 153.2(0.2231)0.2306 108.4 kpsi
ln
Ans.
Eq. (2-19), with Su 85.6 from Prob. 2-6,
Su
Su
85.6
107 kpsi
1 W 1 0.2
P
L
0
1000
2000
3000
4000
7000
8400
8800
9200
9100
13200
15200
17000
16400
14800
0
0.0004
0.0006
0.001
0.0013
0.0023
0.0028
0.0036
0.0089
A
0.198 713
0.198 713
0.198 713
0.198 713
0.198 713
0.198 713
0.198 713
0.198 4
0.197 8
0.196 3
0.192 4
0.187 5
0.156 3
0.130 7
0.107 7
Ans.
true
log
0
0
0.000 2
5032.388 -3.699 01
0.000 3
10 064.78 -3.522 94
0.000 5
15 097.17 -3.301 14
0.000 65
20 129.55 -3.187 23
0.001 149 35 226.72 -2.939 55
0.001 399 42 272.06 -2.854 18
0.001 575 44 354.84 -2.802 61
0.004 604 46 511.63 -2.336 85
0.012 216 46 357.62 -1.913 05
0.032 284 68 607.07 -1.491 01
0.058 082 81 066.67 -1.235 96
0.240 083 108 765.20 -0.619 64
0.418 956 125 478.20 -0.377 83
0.612 511 137 418.80 -0.212 89
log true
3.701 774
4.002 804
4.178 895
4.303 834
4.546 872
4.626 053
4.646 941
4.667 562
4.666 121
4.836 369
4.908 842
5.036 49
5.098 568
5.138 046
Chapter 2 - Rev. D, Page 5/19
______________________________________________________________________________
2-8
Tangent modulus at = 0 is
E
5000 0
25 106 psi
3
0.2 10 0
Ans.
At = 20 kpsi
Chapter 2 - Rev. D, Page 6/19
26 19 103
E20
14.0 106
3
1.5 1 10
Ans.
psi
(10-3) (kpsi)
0
0
0.20
5
0.44
10
0.80
16
1.0
19
1.5
26
2.0
32
2.8
40
3.4
46
4.0
49
5.0
54
______________________________________________________________________________
2-9
W = 0.20,
(a) Before cold working: Annealed AISI 1018 steel. Table A-22, S y = 32 kpsi, S u = 49.5
kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.05
After cold working: Eq. (2-16), u = m = 0.25
A0
1
1
Eq. (2-14),
1.25
Ai 1 W 1 0.20
A
Eq. (2-17),
i ln 0 ln1.25 0.223 u
Ai
Eq. (2-18),
S y 0 im 90 0.223
Eq. (2-19),
Su
(b) Before:
Su 49.5
1.55
32
Sy
0.25
61.8 kpsi
Su
49.5
61.9 kpsi
1 W 1 0.20
After:
Ans. 93% increase
Ans. 25% increase
Su 61.9
1.00
S y 61.8
Ans.
Ans.
Ans.
Lost most of its ductility
______________________________________________________________________________
2-10 W = 0.20,
(a) Before cold working: AISI 1212 HR steel. Table A-22, S y = 28 kpsi, S u = 61.5 kpsi,
0 = 110 kpsi, m = 0.24, f = 0.85
After cold working: Eq. (2-16), u = m = 0.24
Chapter 2 - Rev. D, Page 7/19
Eq. (2-14),
Eq. (2-17),
A0
1
1
1.25
Ai 1 W 1 0.20
A
i ln 0 ln1.25 0.223 u
Ai
Eq. (2-18),
S y 0 im 110 0.223
Eq. (2-19),
Su
(b) Before:
Su 61.5
2.20
28
Sy
0.24
76.7 kpsi
Su
61.5
76.9 kpsi
1 W 1 0.20
After:
Ans. 174% increase
Ans. 25% increase
Su 76.9
1.00
S y 76.7
Ans.
Ans.
Ans.
Lost most of its ductility
______________________________________________________________________________
2-11 W = 0.20,
(a) Before cold working: 2024-T4 aluminum alloy. Table A-22, S y = 43.0 kpsi, S u =
64.8 kpsi, 0 = 100 kpsi, m = 0.15, f = 0.18
After cold working: Eq. (2-16), u = m = 0.15
A0
1
1
Eq. (2-14),
1.25
Ai 1 W 1 0.20
A
Eq. (2-17),
Ans.
i ln 0 ln1.25 0.223 f Material fractures.
Ai
______________________________________________________________________________
2-12 For H B = 275, Eq. (2-21), S u = 3.4(275) = 935 MPa
Ans.
______________________________________________________________________________
2-13 Gray cast iron, H B = 200.
Eq. (2-22),
S u = 0.23(200) 12.5 = 33.5 kpsi
Ans.
From Table A-24, this is probably ASTM No. 30 Gray cast iron
Ans.
______________________________________________________________________________
2-14 Eq. (2-21), 0.5H B = 100 H B = 200
Ans.
______________________________________________________________________________
Chapter 2 - Rev. D, Page 8/19
2-15 For the data given, converting H B to S u using Eq. (2-21)
S u (kpsi)
115
116
116
117
117.5
117.5
117.5
118
118
119.5
HB
230
232
232
234
235
235
235
236
236
239
S u = 1172
Su
S
N
u
S u 2 (kpsi)
13225
13456
13456
13689
13806.25
13806.25
13806.25
13924
13924
14280.25
S u 2 = 137373
1172
117.2 117 kpsi Ans.
10
Eq. (20-8),
10
S
2
u
NSu2
137373 10 117.2
s Su
1.27 kpsi
Ans.
N 1
9
______________________________________________________________________________
i 1
2
2-16 For the data given, converting H B to S u using Eq. (2-22)
HB
230
232
232
234
235
235
235
236
236
239
S u (kpsi)
40.4
40.86
40.86
41.32
41.55
41.55
41.55
41.78
41.78
42.47
S u = 414.12
S u 2 (kpsi)
1632.16
1669.54
1669.54
1707.342
1726.403
1726.403
1726.403
1745.568
1745.568
1803.701
S u 2 = 17152.63
Chapter 2 - Rev. D, Page 9/19
Su
S
u
N
414.12
41.4 kpsi Ans.
10
Eq. (20-8),
10
S
2
u
NSu2
17152.63 10 41.4
1.20
Ans.
N 1
9
______________________________________________________________________________
s Su
i 1
uR
2-17 (a)
2
45.52
34.5 in lbf / in 3
2(30)
Ans.
(b)
P
L
0
1000
2000
3000
4000
7000
8400
8800
9200
9100
13200
15200
17000
16400
14800
0
0.0004
0.0006
0.0010
0.0013
0.0023
0.0028
0.0036
0.0089
A
0.1963
0.1924
0.1875
0.1563
0.1307
0.1077
A0 / A – 1
= P/A 0
0.012 291
0.032 811
0.059 802
0.271 355
0.520 373
0.845 059
0
0.0002
0.0003
0.0005
0.000 65
0.001 15
0.0014
0.0018
0.004 45
0.012 291
0.032 811
0.059 802
0.271 355
0.520 373
0.845 059
0
5 032.39
10 064.78
15 097.17
20 129.55
35 226.72
42 272.06
44 285.02
46 297.97
45 794.73
66 427.53
76 492.30
85 550.60
82 531.17
74 479.35
From the figures on the next page,
5
1
uT Ai (43 000)(0.001 5) 45 000(0.004 45 0.001 5)
2
i 1
1
45 000 76 500 (0.059 8 0.004 45)
2
81 000 0.4 0.059 8 80 000 0.845 0.4
66.7 103 in lbf/in 3
Ans.
Chapter 2 - Rev. D, Page 10/19
Chapter 2 - Rev. D, Page 11/19
2-18, 2-19
These problems are for student research. No standard solutions are provided.
______________________________________________________________________________
2-20 Appropriate tables: Young’s modulus and Density (Table A-5)1020 HR and CD (Table A20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c)
Appropriate equations:
F
F
For diameter,
Sy
A / 4 d 2
d
4F
Sy
Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length,
Deflection/length = /L = F/(AE)
With F = 100 kips = 100(103) lbf,
Young's
Material Modulus
units
Mpsi
1020 HR
1020 CD
1040
4140
Al
Ti
Density
lbf/in^3
30
30
30
30
10.4
16.5
0.282
0.282
0.282
0.282
0.098
0.16
Yield
Weight/
Strength Cost/lbf Diameter length
kpsi
$/lbf
in
lbf/in
30
57
80
165
50
120
$0.27
$0.30
$0.35
$0.80
$1.10
$7.00
2.060
1.495
1.262
0.878
1.596
1.030
0.9400
0.4947
0.3525
0.1709
0.1960
0.1333
Cost/ Deflection/
length
length
$/in
in/in
$0.25
$0.15
$0.12
$0.14
$0.22
$0.93
1.000E‐03
1.900E‐03
2.667E‐03
5.500E‐03
4.808E‐03
7.273E‐03
The selected materials with minimum values are shaded in the table above.
Ans.
______________________________________________________________________________
2-21
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
would favor steel, cast iron, or maybe a less common ferrous material. The expectation
would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending
test could be done to check density or modulus of elasticity. The weight test is faster.
From the measured weight of 7.95 lbf, the unit weight is determined to be
w
W
7.95 lbf
0.281 lbf/in 3 0.28 lbf/in3
2
Al [ (1 in) / 4](36 in)
which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon
steel. Nickel steel and stainless steel have similar unit weights, but surface finish and
darker coloring do not favor their selection. To select a likely specification from Table
Chapter 2 - Rev. D, Page 12/19
A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength
of Su 0.5H B 0.5(200) 100 kpsi . Assuming the material is hot-rolled due to the
rough surface finish, appropriate choices from Table A-20 would be one of the higher
carbon steels, such as hot-rolled AISI 1050, 1060, or 1080.
Ans.
______________________________________________________________________________
2-22
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a
weight or bending test could be done to check density or modulus of elasticity. The
weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined
to be
W
2.9 lbf
w
0.103 lbf/in 3 0.10 lbf/in 3
Al [ (1 in)2 / 4](36 in)
which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5
for aluminum. No other materials come close to this unit weight, so the material is likely
aluminum. Ans.
______________________________________________________________________________
2-23
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To
further distinguish the material, either a weight or bending test could be done to check
density or modulus of elasticity. The weight test is faster. From the measured weight of
9 lbf, the unit weight is determined to be
w
W
9.0 lbf
0.318 lbf/in 3 0.32 lbf/in 3
2
Al [ (1 in) / 4](36 in)
which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5
for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain
additional insight. From the measured deflection and utilizing the deflection equation for
an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be
100 24
Fl 3
17.7 Mpsi
E
3Iy 3 (1) 4 64 (17 / 32)
3
which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi).
The conclusion is that the material is likely copper.
Ans.
______________________________________________________________________________
2-24 and 2-25 These problems are for student research. No standard solutions are provided.
______________________________________________________________________________
Chapter 2 - Rev. D, Page 13/19
2-26 For strength, = F/A = S A = F/S
For mass, m = Al = (F/S) l
Thus,
f 3 (M ) = /S , and maximize S/
( = 1)
In Fig. (2-19), draw lines parallel to S/
From the list of materials given, both aluminum alloy and high carbon heat treated
steel are good candidates, having greater potential than tungsten carbide or polycarbonate.
The higher strength aluminum alloys have a slightly greater potential. Other factors, such
as cost or availability, may dictate which to choose. Ans.
______________________________________________________________________________
2-27 For stiffness, k = AE/l A = kl/E
For mass, m = Al = (kl/E) l =kl2 /E
Thus,
f 3 (M) = /E , and maximize E/
( = 1)
In Fig. (2-16), draw lines parallel to E/
Chapter 2 - Rev. D, Page 14/19
From the list of materials given, tungsten carbide (WC) is best, closely followed by
aluminum alloys, and then followed by high carbon heat-treated steel. They are close
enough that other factors, like cost or availability, would likely dictate the best choice.
Polycarbonate polymer is clearly not a good choice compared to the other candidate
materials.
Ans.
______________________________________________________________________________
2-28 For strength,
= Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ].
The section modulus is strictly a function of the dimensions of the cross section and has
the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a
number. For example, for a circular cross section, C = 4
1
. Then, for strength, Eq.
(1) is
Fl
S
CA3/2
Fl
A
CS
2/3
(2)
Chapter 2 - Rev. D, Page 15/19
For mass,
Thus,
Fl
m Al
CS
2/3
F
l
C
2/3
l 5/3 2/3
S
f 3 (M) = /S 2/3, and maximize S 2/3/ ( = 2/3)
In Fig. (2-19), draw lines parallel to S 2/3/
From the list of materials given, a higher strength aluminum alloy has the greatest
potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly
not a good choice compared to the other candidate materials. .Ans.
______________________________________________________________________________
2-29 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular
section of height h and width b, where for a given scaled shape, h = cb, where c is a
Chapter 2 - Rev. D, Page 16/19
constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12
= cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant).
Thus, Eq. (2-27) becomes
1/2
kl 3
A
3CE
and Eq. (2-29) becomes
k 5/2
m Al
l 1/2
3C
E
1/2
Thus, minimize f3 M
E 1/ 2
, or maximize M
E1/2
. From Fig. (2-16)
From the list of materials given, aluminum alloys are clearly the best followed by steels
and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other
candidate materials. Ans.
______________________________________________________________________________
2-30 For stiffness, k = AE/l A = kl/E
For mass, m = Al = (kl/E) l =kl2 /E
Chapter 2 - Rev. D, Page 17/19
So,
f 3 (M) = /E, and maximize E/ . Thus, = 1. Ans.
______________________________________________________________________________
2-31 For strength, = F/A = S A = F/S
For mass, m = Al = (F/S) l
So, f 3 (M ) = /S, and maximize S/ . Thus, = 1. Ans.
______________________________________________________________________________
2-32 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular
section of height h and width b, where for a given scaled shape, h = cb, where c is a
constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12
= cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12.
Thus, Eq. (2-27) becomes
1/2
kl 3
A
3CE
and Eq. (2-29) becomes
k 5/2
m Al
l 1/2
3C
E
1/2
So, minimize f3 M
E1/2
, or maximize M
. Thus, = 1/2. Ans.
E 1/ 2
______________________________________________________________________________
2-33 For strength,
= Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ].
The section modulus is strictly a function of the dimensions of the cross section and has
the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a
number. For example, for a circular cross section, C = 4
1
. Then, for strength, Eq. (1)
is
Fl
S
CA3/2
For mass,
Fl
A
CS
Fl
m Al
CS
2/3
F
l
C
2/3
2/3
(2)
l 5/3 2/3
S
So, f 3 (M) = /S 2/3, and maximize S 2/3/. Thus, = 2/3. Ans.
______________________________________________________________________________
2-34 For stiffness, k=AE/l, or, A = kl/E.
Chapter 2 - Rev. D, Page 18/19
Thus, m = Al = (kl/E )l = kl 2 /E. Then, M = E / and = 1.
From Fig. 2-16, lines parallel to E / for ductile materials include steel, titanium,
molybdenum, aluminum alloys, and composites.
For strength, S = F/A, or, A = F/S.
Thus, m = Al = F/Sl = Fl /S. Then, M = S/ and = 1.
From Fig. 2-19, lines parallel to S/ give for ductile materials, steel, aluminum alloys,
nickel alloys, titanium, and composites.
Common to both stiffness and strength are steel, titanium, aluminum alloys, and
composites. Ans.
Chapter 2 - Rev. D, Page 19/19
Chapter 3
3-1
M o 0
18 RB 6(100) 0
RB 33.3 lbf Ans.
Fy 0
Ro RB 100 0
Ro 66.7 lbf Ans.
RC RB 33.3 lbf
Ans.
______________________________________________________________________________
3-2
Body AB:
Fx 0
RAx RBx
Fy 0
RAy RBy
M B 0
RAy (10) RAx (10) 0
RAx RAy
Body OAC:
M O 0
RAy (10) 100(30) 0
RAy 300 lbf
Ans.
Fx 0
ROx RAx 300 lbf
Fy 0
ROy RAy 100 0
ROy 200 lbf
Ans.
Ans.
______________________________________________________________________________
Chapter 3 - Rev. A, Page 1/100
3-3
0.8
1.39 kN Ans.
tan 30
0.8
RA
1.6 kN Ans.
sin 30
RO
______________________________________________________________________________
3-4
Step 1: Find R A & R E
4.5
7.794 m
tan 30
M A 0
h
9 RE 7.794(400 cos 30 )
4.5(400sin 30 ) 0
RE 400 N Ans.
F
x
F
y
0
RAx 400cos 30 0
RAx 346.4 N
0
RAy 400 400sin 30 0
RAy 200 N
RA 346.42 2002 400 N
Ans.
Step 2: Find components of R C on link 4 and R D
M
C
0
400(4.5) 7.794 1.9 RD 0
RD 305.4 N
F
F
Ans.
x
0
RCx 4 305.4 N
y
0
( RCy ) 4 400 N
Chapter 3 - Rev. A, Page 2/100
Step 3: Find components of R C on link 2
Fx 0
RCx 2 305.4 346.4 0
RCx 2 41 N
F
y
0
R
Cy 2
200 N
____________________________________________________________________________________________________________________
_
Chapter 3 - Rev. A, Page 3/100
3-5
M C 0
1500 R1 300(5) 1200(9) 0
R1 8.2 kN Ans.
Fy 0
8.2 9 5 R2 0 R2 5.8 kN
Ans.
M 1 8.2(300) 2460 N m Ans.
M 2 2460 0.8(900) 1740 N m Ans.
M 3 1740 5.8(300) 0 checks!
_____________________________________________________________________________
3-6
Fy 0
R0 500 40(6) 740 lbf Ans.
M 0 0
M 0 500(8) 40(6)(17) 8080 lbf in
Ans.
M 1 8080 740(8) 2160 lbf in Ans.
M 2 2160 240(6) 720 lbf in Ans.
1
M 3 720 (240)(6) 0 checks!
2
______________________________________________________________________________
Chapter 3 - Rev. A, Page 4/100
3-7
M B 0
2.2 R1 1(2) 1(4) 0
R1 0.91 kN Ans.
Fy 0
0.91 2 R2 4 0
R2 6.91 kN Ans.
M 1 0.91(1.2) 1.09 kN m
M 2 1.09 2.91(1) 4 kN m
M 3 4 4(1) 0 checks!
Ans.
Ans.
______________________________________________________________________________
3-8
Break at the hinge at B
Beam OB:
From symmetry,
R1 VB 200 lbf
Ans.
Beam BD:
M D 0
200(12) R2 (10) 40(10)(5) 0
R2 440 lbf Ans.
Fy 0
200 440 40(10) R3 0
R3 160 lbf Ans.
Chapter 3 - Rev. A, Page 5/100
M 1 200(4) 800 lbf in Ans.
M 2 800 200(4) 0 checks at hinge
M 3 800 200(6) 400 lbf in Ans.
1
M 4 400 (240)(6) 320 lbf in Ans.
2
1
M 5 320 (160)(4) 0 checks!
2
______________________________________________________________________________
3-9
q R1 x
1
9 x 300
0
1
5 x 1200
1
R2 x 1500
0
V R1 9 x 300 5 x 1200 R2 x 1500
1
1
0
M R1 x 9 x 300 5 x 1200 R2 x 1500
1
(1)
1
(2)
At x = 1500+ V = M = 0. Applying Eqs. (1) and (2),
R1 9 5 R2 0 R1 R2 14
1500 R1 9(1500 300) 5(1500 1200) 0
R2 14 8.2 5.8 kN
Ans.
R1 8.2 kN
Ans.
V 8.2 kN, M 8.2 x N m
0 x 300 :
300 x 1200 : V 8.2 9 0.8 kN
M 8.2 x 9( x 300) 0.8 x 2700 N m
1200 x 1500 : V 8.2 9 5 5.8 kN
M 8.2 x 9( x 300) 5( x 1200) 5.8 x 8700 N m
Plots of V and M are the same as in Prob. 3-5.
______________________________________________________________________________
Chapter 3 - Rev. A, Page 6/100
3-10
q R0 x
1
M0 x
V R0 M 0 x
1
2
500 x 8
1
0
40 x 14 40 x 20
0
1
500 x 8 40 x 14 40 x 20
1
2
M R0 x M 0 500 x 8 20 x 14 20 x 20
0
1
(1)
2
(2)
at x 20 in, V M 0, Eqs. (1) and (2) give
R0 500 40 20 14 0
R0 740 lbf
R0 (20) M 0 500(20 8) 20(20 14) 2 0
M 0 8080 lbf in
Ans.
Ans.
0 x 8 : V 740 lbf, M 740 x 8080 lbf in
8 x 14 : V 740 500 240 lbf
M 740 x 8080 500( x 8) 240 x 4080 lbf in
14 x 20 : V 740 500 40( x 14) 40 x 800 lbf
M 740 x 8080 500( x 8) 20( x 14) 2 20 x 2 800 x 8000 lbf in
Plots of V and M are the same as in Prob. 3-6.
______________________________________________________________________________
3-11
q R1 x
1
2 x 1.2
1
R2 x 2.2
0
0
1
4 x 3.2
V R1 2 x 1.2 R2 x 2.2 4 x 3.2
1
1
1
0
(1)
1
M R1 x 2 x 1.2 R2 x 2.2 4 x 3.2
at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2),
R1 2 R2 4 0
R1 R2 6
3.2 R1 2(2) R2 (1) 0 3.2 R1 R2 4
Solving Eqs. (3) and (4) simultaneously,
R 1 = -0.91 kN, R 2 = 6.91 kN Ans.
(2)
(3)
(4)
0 x 1.2 :
V 0.91 kN, M 0.91x kN m
1.2 x 2.2 : V 0.91 2 2.91 kN
M 0.91x 2( x 1.2) 2.91x 2.4 kN m
2.2 x 3.2 : V 0.91 2 6.91 4 kN
M 0.91x 2( x 1.2) 6.91( x 2.2) 4 x 12.8 kN m
Plots of V and M are the same as in Prob. 3-7.
______________________________________________________________________________
Chapter 3 - Rev. A, Page 7/100
3-12
q R1 x
1
400 x 4
1
0
R2 x 10
1
0
0
40 x 10 40 x 20 R3 x 20
0
1
1
V R1 400 x 4 R2 x 10 40 x 10 40 x 20 R3 x 20
1
1
2
0
2
M R1 x 400 x 4 R2 x 10 20 x 10 20 x 20 R3 x 20
R1 200 lbf
M 0 at x 8 in 8R1 400(8 4) 0
+
at x = 20 , V =M = 0. Applying Eqs. (1) and (2),
200 400 R2 40(10) R3 0
R2 R3 600
1
(1)
1
(2)
Ans.
200(20) 400(16) R2 (10) 20(10)2 0 R2 440 lbf
Ans.
R3 600 440 160 lbf
Ans.
0 x 4 : V 200 lbf, M 200 x lbf in
4 x 10 : V 200 400 200 lbf,
M 200 x 400( x 4) 200 x 1600 lbf in
10 x 20 : V 200 400 440 40( x 10) 640 40 x lbf
M 200 x 400( x 4) 440( x 10) 20 x 10 20 x 2 640 x 4800 lbf in
Plots of V and M are the same as in Prob. 3-8.
______________________________________________________________________________
2
3-13 Solution depends upon the beam selected.
______________________________________________________________________________
3-14
(a) Moment at center,
l 2a
xc
2
2
wl
l wl l
M c l 2a
a
2 2
2 2 4
At reaction, M r wa 2 2
a = 2.25, l = 10 in, w = 100 lbf/in
Mc
100(10) 10
2.25 125 lbf in
2 4
Mr
100 2.252
253 lbf in
2
(b) Optimal occurs when M c M r
Ans.
Chapter 3 - Rev. A, Page 8/100
2
wl l
wa
a
a 2 al 0.25l 2 0
2 4
2
Taking the positive root
1
l
l l 2 4 0.25l 2
2 1 0.207 l
2
2
for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in
M min 100 2 2.07 2 214 lbf in
a
Ans.
______________________________________________________________________________
3-15
(a)
20 10
5 kpsi
2
20 10
CD
15 kpsi
2
C
R 152 82 17 kpsi
1 5 17 22 kpsi
2 5 17 12 kpsi
1
8
p tan 1 14.04 cw
2
15
1 R 17 kpsi
s 45 14.04 30.96 ccw
(b)
9 16
12.5 kpsi
2
16 9
CD
3.5 kpsi
2
C
R 52 3.52 6.10 kpsi
1 12.5 6.1 18.6 kpsi
2 12.5 6.1 6.4 kpsi
1
5
tan 1
27.5 ccw
2
3.5
1 R 6.10 kpsi
p
s 45 27.5 17.5 cw
Chapter 3 - Rev. A, Page 9/100
(c)
24 10
17 kpsi
2
24 10
CD
7 kpsi
2
C
R 7 2 6 2 9.22 kpsi
1 17 9.22 26.22 kpsi
2 17 9.22 7.78 kpsi
1
7
p 90 tan 1 69.7 ccw
2
6
1 R 9.22 kpsi
s 69.7 45 24.7 ccw
(d)
12 22
5 kpsi
2
12 22
CD
17 kpsi
2
C
R 172 122 20.81 kpsi
1 5 20.81 25.81 kpsi
2 5 20.81 15.81 kpsi
1
17
p 90 tan 1 72.39 cw
2
12
Chapter 3 - Rev. A, Page 10/100
1 R 20.81 kpsi
s 72.39 45 27.39 cw
______________________________________________________________________________
Chapter 3 - Rev. A, Page 11/100
3-16
(a)
8 7
0.5 MPa
2
87
CD
7.5 MPa
2
C
R 7.52 62 9.60 MPa
1 9.60 0.5 9.10 MPa
2 0.5 9.6 10.1 Mpa
1
7.5
p 90 tan 1
70.67 cw
2
6
1 R 9.60 MPa
s 70.67 45 25.67 cw
(b)
96
1.5 MPa
2
96
CD
7.5 MPa
2
C
R 7.52 32 8.078 MPa
1 1.5 8.078 9.58 MPa
2 1.5 8.078 6.58 MPa
1
3
10.9 cw
2
7.5
1 R 8.078 MPa
p tan 1
s 45 10.9 34.1 ccw
Chapter 3 - Rev. A, Page 12/100
(c)
12 4
4 MPa
2
12 4
CD
8 MPa
2
C
R 82 7 2 10.63 MPa
1 4 10.63 14.63 MPa
2 4 10.63 6.63 MPa
1
8
p 90 tan 1 69.4 ccw
2
7
1 R 10.63 MPa
s 69.4 45 24.4 ccw
(d)
65
0.5 MPa
2
65
CD
5.5 MPa
2
C
R 5.52 82 9.71 MPa
1 0.5 9.71 10.21 MPa
2 0.5 9.71 9.21 MPa
1
8
27.75 ccw
2
5.5
1 R 9.71 MPa
p tan 1
s 45 27.75 17.25 cw
______________________________________________________________________________
Chapter 3 - Rev. A, Page 13/100
3-17
(a)
12 6
9 kpsi
2
12 6
CD
3 kpsi
2
C
R 32 42 5 kpsi
1 5 9 14 kpsi
2 9 5 4 kpsi
1
4
tan 1 26.6 ccw
3
2
1 R 5 kpsi
p
s 45 26.6 18.4 ccw
(b)
30 10
10 kpsi
2
30 10
CD
20 kpsi
2
C
R 202 102 22.36 kpsi
1 10 22.36 32.36 kpsi
2 10 22.36 12.36 kpsi
1
10
p tan 1 13.28 ccw
2
20
1 R 22.36 kpsi
s 45 13.28 31.72 cw
Chapter 3 - Rev. A, Page 14/100
(c)
10 18
4 kpsi
2
10 18
CD
14 kpsi
2
C
R 142 92 16.64 kpsi
1 4 16.64 20.64 kpsi
2 4 16.64 12.64 kpsi
1
14
p 90 tan 1 73.63 cw
2
9
1 R 16.64 kpsi
s 73.63 45 28.63 cw
(d)
9 19
14 kpsi
2
19 9
CD
5 kpsi
2
C
R 52 82 9.434 kpsi
1 14 9.43 23.43 kpsi
2 14 9.43 4.57 kpsi
1
5
p 90 tan 1 61.0 cw
2
8
1 R 9.34 kpsi
s 61 45 16 cw
______________________________________________________________________________
Chapter 3 - Rev. A, Page 15/100
3-18
(a)
80 30
55 MPa
2
80 30
CD
25 MPa
2
C
R 252 202 32.02 MPa
1 0 MPa
2 55 32.02 22.98 23.0 MPa
3 55 32.0 87.0 MPa
1 2
23
11.5 MPa,
2
2 3 32.0 MPa,
1 3
87
43.5 MPa
2
(b)
30 60
15 MPa
2
60 30
CD
45 MPa
2
C
R 452 302 54.1 MPa
1 15 54.1 39.1 MPa
2 0 MPa
3 15 54.1 69.1 MPa
39.1 69.1
54.1 MPa
2
39.1
19.6 MPa
1 2
2
69.1
34.6 MPa
2 3
2
1 3
Chapter 3 - Rev. A, Page 16/100
(c)
40 0
20 MPa
2
40 0
20 MPa
CD
2
C
R 202 202 28.3 MPa
1 20 28.3 48.3 MPa
2 20 28.3 8.3 MPa
3 z 30 MPa
1 3
48.3 30
39.1 MPa,
2
1 2 28.3 MPa,
2 3
30 8.3
10.9 MPa
2
(d)
50
25 MPa
2
50
CD
25 MPa
2
C
R 252 302 39.1 MPa
1 25 39.1 64.1 MPa
2 25 39.1 14.1 MPa
3 z 20 MPa
64.1 20
20 14.1
42.1 MPa, 1 2 39.1 MPa, 2 3
2.95 MPa
2
2
______________________________________________________________________________
1 3
3-19
(a)
Since there are no shear stresses on the
stress element, the stress element
already represents principal stresses.
1 x 10 kpsi
2 0 kpsi
3 y 4 kpsi
10 (4)
7 kpsi
2
10
1 2 5 kpsi
2
0 (4)
2 3
2 kpsi
2
1 3
Chapter 3 - Rev. A, Page 17/100
(b)
0 10
5 kpsi
2
10 0
CD
5 kpsi
2
C
R 52 42 6.40 kpsi
1 5 6.40 11.40 kpsi
2 0 kpsi, 3 5 6.40 1.40 kpsi
1 3 R 6.40 kpsi,
1 2
11.40
5.70 kpsi,
2
3
1.40
0.70 kpsi
2
(c)
2 8
5 kpsi
2
82
CD
3 kpsi
2
C
R 32 42 5 kpsi
1 5 5 0 kpsi, 2 0 kpsi
3 5 5 10 kpsi
1 3
10
5 kpsi,
2
1 2 0 kpsi,
2 3 5 kpsi
(d)
10 30
10 kpsi
2
10 30
CD
20 kpsi
2
C
R 202 102 22.36 kpsi
1 10 22.36 12.36 kpsi
2 0 kpsi
3 10 22.36 32.36 kpsi
12.36
32.36
6.18 kpsi, 2 3
16.18 kpsi
2
2
______________________________________________________________________________
1 3 22.36 kpsi,
1 2
Chapter 3 - Rev. A, Page 18/100
3-20
From Eq. (3-15),
3 (6 18 12) 2 6(18) (6)(12) 18(12) 92 62 (15) 2
6(18)(12) 2(9)(6)(15) (6)(6)2 18(15)2 (12)(9)2 0
3 594 3186 0
Roots are: 21.04, 5.67, –26.71 kpsi Ans.
21.04 5.67
7.69 kpsi
1 2
2
5.67 26.71
16.19 kpsi
2 3
2
21.04 26.71
23.88 kpsi
Ans.
max 1 3
2
_____________________________________________________________________________
3-21
From Eq. (3-15)
3 (20 0 20) 2 20(0) 20(20) 0(20) 402 20 2 02
2
2
20(0)(20) 2(40) 20 2 (0) 20 20 2 0(0)2 20(40)2 0
3 40 2 2 000 48 000 0
Roots are: 60, 20, –40 kpsi
60 20
20 kpsi
2
20 40
30 kpsi
2 3
2
60 40
50 kpsi
max 1 3
2
Ans.
1 2
Ans.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 19/100
3-22
From Eq. (3-15)
2
2
3 (10 40 40) 2 10(40) 10(40) 40(40) 202 40 20
10(40)(40) 2(20)(40)(20) 10(40) 2 40(20) 2 40(20) 2 0
3 90 2 0
Roots are: 90, 0, 0 MPa
Ans.
2 3 0
1 2 1 3 max
90
45 MPa
2
Ans.
_____________________________________________________________________________
3-23
F
15000
33 950 psi 34.0 kpsi
A 4 0.752
60
FL
L
33 950
0.0679 in
AE
E
30 106
0.0679
1130 106 1130
1
60
L
From Table A-5, v = 0.292
2 v1 0.292(1130) 330
d 2 d 330 10
6
Ans.
Ans.
Ans.
Ans.
(0.75) 248 10 in
6
Ans.
_____________________________________________________________________________
3-24
F
3000
6790 psi 6.79 kpsi Ans.
A 4 0.752
60
FL
L
6790
0.0392 in
AE
E
10.4 106
0.0392
653 106 653
60
L
From Table A-5, v = 0.333
1
2 v1 0.333(653) 217
Ans.
Ans.
Ans.
d 2 d 217 106 (0.75) 163 10 6 in
Ans.
Chapter 3 - Rev. A, Page 20/100
_____________________________________________________________________________
3-25
d 0.0001d
0.0001
2
d
d
From Table A-5, v = 0.326, E = 119 GPa
0.0001
1 2
306.7 106
v
0.326
FL
F
and , so
AE
A
E
=
1 E 306.7 106 (119) 109 36.5 MPa
L
F A 36.5 10
0.03
2
25 800 N 25.8 kN
Ans.
4
S y = 70 MPa > , so elastic deformation assumption is valid.
_____________________________________________________________________________
6
3-26
FL
L
8(12)
20 000
0.185 in
AE
E
10.4 106
Ans.
_____________________________________________________________________________
3-27
FL
L
3
140 106
0.00586 m 5.86 mm
AE
E
71.7 109
Ans.
_____________________________________________________________________________
3-28
FL
L
10(12)
15 000
0.173 in
AE
E
10.4 10 6
Ans.
_____________________________________________________________________________
3-29
With z 0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
x
E x
E y
v
1
1 v
v 1
E x vE y
1 v
2
E x v y
1 v2
Likewise,
Chapter 3 - Rev. A, Page 21/100
y
E y x
1 v2
From Table A-5, E = 207 GPa and ν = 0.292. Thus,
E x v y
x
207 109 0.0019 0.292 0.000 72
1 v
1 0.292
9
207 10 0.000 72 0.292 0.0019
2
y
2
10 382 MPa
6
Ans.
10 37.4 MPa
6
Ans.
1 0.292
_____________________________________________________________________________
3-30
2
With z 0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
x
E x
v
E y
1
1
v
v
1
E x vE y
1 v2
E x v y
1 v2
Likewise,
y
E y x
1 v2
From Table A-5, E = 71.7 GPa and ν = 0.333. Thus,
E x v y 71.7 109 0.0019 0.333 0.000 72
x
106 134 MPa Ans.
2
2
1 v
1 0.333
9
71.7 10 0.000 72 0.333 0.0019
y
106 7.04 MPa Ans.
2
1 0.333
_____________________________________________________________________________
3-31
c
(a) R1 F
l
6M
2
bh
F
(b) m m
F
ac
F
l
6 ac
bh 2l
Ans.
F
F
bh 2 l
6 ac
2
bm b hm h lm l1 1( s)( s ) 2 ( s ) 2
s
( s)( s)
am a cm c
M max R1a
Ans.
For equal stress, the model load varies by the square of the scale factor.
_____________________________________________________________________________
3-32
Chapter 3 - Rev. A, Page 22/100
R1
(a)
(b)
wl
,
2
M max
x l /2
6M
6 wl 2 3Wl
bh 2 bh 2 8
4bh 2
w l l wl 2
l
2 2 2
8
2
4 bh
W
Ans.
3 l
Wm ( m / )(bm / b)(hm / h) 2 1( s)( s ) 2
s2
W
lm / l
s
Ans.
wmlm
wm s 2
s2
s Ans.
wl
w
s
For equal stress, the model load w varies linearly with the scale factor.
_____________________________________________________________________________
3-33
(a)
Can solve by iteration or derive
equations for the general case. Find
maximum moment under wheel W3 .
WT W at centroid of W’s
l x3 d3
RA
WT
l
Under wheel 3,
M 3 RA x3 W1a13 W2 a23
For maximum,
l x3 d3 W
dM 3
W
0 l d3 2 x3 T
dx3
l
Substitute into M M 3
l d3
x W1a13 W2 a23
T 3
l
x3
l d3
2
2
WT W1a13 W2 a23
4l
This means the midpoint of d 3 intersects the midpoint of the beam.
For wheel i,
l di
xi
,
2
Mi
l di
4l
2
i 1
WT W j a ji
j 1
Note for wheel 1: W j a ji 0
WT 104.4,
W1 W2 W3 W4
104.4
26.1 kips
4
476
(1200 238) 2
238 in, M 1
(104.4) 20 128 kip in
2
4(1200)
Wheel 2: d 2 238 84 154 in
Wheel 1: d1
Chapter 3 - Rev. A, Page 23/100
M2
(1200 154) 2
(104.4) 26.1(84) 21 605 kip in M max
4(1200)
Ans.
Check if all of the wheels are on the rail.
(b)
xmax 600 77 523 in Ans.
(c)
See above sketch.
(d)
Inner axles
_____________________________________________________________________________
3-34
(a) Let a = total area of entire envelope
Let b = area of side notch
A a 2b 40(3)(25) 25 34 2150 mm2
1
1
3
3
40 75 34 25
12
12
6
4
I 1.36 10 mm
Ans.
I I a 2Ib
Dimensions in mm.
(b)
Aa 0.375(1.875) 0.703 125 in 2
Ab 0.375(1.75) 0.656 25 in 2
A 2(0.703125) 0.656 25 2.0625 in 2
2(0.703 125)(0.9375) 0.656 25(0.6875)
0.858 in Ans.
2.0625
0.375(1.875)3
0.206 in 4
Ia
12
1.75(0.375)3
0.007 69 in 4
Ib
12
I1 2 0.206 0.703 125(0.0795) 2 0.00769 0.656 25(0.1705) 2 0.448 in 4
y
Ans.
(c)
Use two negative areas.
Aa 625 mm 2 , Ab 5625 mm 2 , Ac 10 000 mm 2
A 10 000 5625 625 3750 mm 2 ;
Chapter 3 - Rev. A, Page 24/100
ya 6.25 mm, yb 50 mm, yc 50 mm
10 000(50) 5625(50) 625(6.25)
57.29 mm
3750
c1 100 57.29 42.71 mm Ans.
y
Ans.
50(12.5)3
8138 mm 4
12
75(75)3
Ib
2.637 106 mm 4
12
100(100)3
Ic
8.333 106 in 4
12
2
2
I1 8.333 106 10 000(7.29) 2 2.637 106 5625 7.29 8138 625 57.29 6.25
Ia
I1 4.29 106 in 4
Ans.
(d)
Aa 4 0.875 3.5 in 2
Ab 2.5 0.875 2.1875 in 2
A Aa Ab 5.6875 in 2
2.9375 3.5 1.25(2.1875)
2.288 in Ans.
5.6875
1
1
3
2
3
2
I (4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.25
12
12
4
I 5.20 in
Ans.
_____________________________________________________________________________
y
3-35
1
(20)(40)3 1.067 105 mm 4
12
A 20(40) 800 mm 2
M max is at A. At the bottom of the section,
Mc 450 000(20)
max
84.3 MPa Ans.
I
1.067 105
I
Due to V, max is between A and B at y = 0.
3 V 3 3000
max
5.63 MPa Ans.
2 A 2 800
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 25/100
3-36
1
(1)(2)3 0.6667 in 4
12
A 1(2) 2 in 2
I
M o 0
8RA 100(8)(12) 0
RA 1200 lbf
Ro 1200 100(8) 400 lbf
M max is at A. At the top of the beam,
max
Mc 3200(0.5)
2400 psi
I
0.6667
Ans.
Due to V, max is at A, at y = 0.
3 V 3 800
600 psi Ans.
2 A 2 2
_____________________________________________________________________________
max
3-37
1
(0.75)(2)3 0.5 in 4
12
A (0.75)(2) 1.5 in 2
I
M A 0
15 RB 1000(20) 0
RB 1333.3 lbf
RA 3000 1333.3 1000 2666.7 lbf
M max is at B. At the top of the beam,
max
Mc 5000(1)
10 000 psi
I
0.5
Ans.
Due to V, max is between B and C at y = 0.
3 V 3 1000
1000 psi Ans.
2 A 2 1.5
_____________________________________________________________________________
max
Chapter 3 - Rev. A, Page 26/100
3-38
I
d4
(50)4
306.796 103 mm 4
64
64
2
d
(50)2
A
1963 mm 2
4
4
M B 0
6(300)(150) 200 RA 0
RA 1350 kN
RB 6(300) 1350 450 kN
max
M max is at A. At the top,
Due to V, max is at A, at y = 0.
4 V 4 750
2
0.509 kN/mm 509 MPa Ans.
3 A 3 1963
_____________________________________________________________________________
max
3-39
8 I
wl 2
wl 2 c
max
w max
8
8I
cl 2
(a) l 48 in; Table A-8, I 0.537 in 4
M max
w
8 12 103 0.537
1 482
22.38 lbf/in
Ans.
(b) l 60 in, I 1 12 2 33 1 12 1.625 2.6253 2.051 in 4
w
8 12 103 2.051
1.5 602
36.5 lbf/in
Ans.
(c) l 60 in; Table A-6, I 2 0.703 1.406 in 4
y = 0.717 in, c max = 1.783 in
8 12 103 1.406
w
21.0 lbf/in Ans.
1.783 602
(d) l 60 in, Table A-7, I 2.07 in 4
w
8 12 103 2.07
1.5 602
36.8 lbf/in
Ans.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 27/100
Mc
I
3-40
I 0.54 3.068 103 in 4 , A 0.52 0.1963 in 2
64
4
Model
500(0.5) 500(0.75 / 2)
218.75 lbf in
2
2
Mc 218.75(0.25)
I
3.068 10 3
(c)
M
17 825 psi 17.8 kpsi
max
Ans.
4 V 4 500
3400 psi 3.4 kpsi
3 A 3 0.1963
Ans.
Model (d)
M 500(0.625) 312.5 lbf in
Mc 312.5(0.25)
I
3.068 103
25 464 psi 25.5 kpsi
max
Ans.
4 V 4 500
3400 psi 3.4 kpsi
3 A 3 0.1963
Ans.
Model
M 500(0.4375) 218.75 lbf in
(e)
Mc 218.75(0.25)
I
3.068 103
17 825 psi 17.8 kpsi
max
Ans.
4 V 4 500
3400 psi 3.4 kpsi
3 A 3 0.1963
Ans.
_____________________________________________________________________________
3-41
Chapter 3 - Rev. A, Page 28/100
I
12 1018 mm , A 12 113.1 mm
64
4
4
4
2
2
Model (c)
2000(6) 2000(9)
M
15 000 N mm
2
2
Mc 15 000(6)
I
1018
88.4 N/mm 2 88.4 MPa Ans.
max
4V
3A
4 2000
2
23.6 N/mm 23.6 MPa
3 113.1
Model (d)
M 2000(12) 24 000 N mm
Mc 24 000(6)
1018
I
141.5 N/mm 2 141.5 MPa Ans.
4 V 4 2000
2
max
23.6 N/mm 23.6 MPa
3 A 3 113.1
Ans.
Ans.
Model (e)
M 2000(7.5) 15000 N mm
Mc 15000(6)
1018
I
88.4 N/mm2 88.4 MPa Ans.
4 V 4 2000
2
max
23.6 N/mm 23.6 MPa
3 A 3 113.1
Ans.
_____________________________________________________________________________
Mc M d / 2 32 M
3-42 (a)
d 4 / 64 d 3
I
Chapter 3 - Rev. A, Page 29/100
d
(b)
d
(c)
d
3
32 M
3
32(218.75)
0.420 in Ans.
(30 000)
V
V
A d2 / 4
4V
4(500)
0.206 in
(15 000)
Ans.
4V 4
V
3 A 3 d 2 / 4
4 4V
3
4 4(500)
0.238 in
3 (15 000)
Ans.
_____________________________________________________________________________
3-43
p1 p2
1
x l terms for x l a
a
p
p
1
2
2
V F p1 x l 1
x l terms for x l a
2a
p
p p2
2
3
M Fx 1 x l 1
x l terms for x l a
2
6a
q F x
1
0
p1 x l
At x (l a ) , V M 0, terms for x > l + a = 0
p1 p2 2
2F
a 0 p1 p2
a
2a
2
pa
p p2 3
6 F (l a)
F (l a) 1 1
a 0
2 p1 p2
a2
2
6a
F p1a
From (1) and (2)
From similar triangles
p1
2F
(3l 2a),
a2
b
a
p2 p1 p2
p2
2F
(3l a )
a2
ap2
b
p1 p2
(1)
(2)
(3)
(4)
Chapter 3 - Rev. A, Page 30/100
M max occurs where V = 0
xmax l a 2b
p1
p p2
(a 2b) 2 1
(a 2b)3
2
6a
p
p p2
( a 2b)3
Fl F ( a 2b) 1 ( a 2b) 2 1
2
6a
Normally M max = Fl
M max F (l a 2b)
The fractional increase in the magnitude is
F (a 2b) p1 2 (a 2b) 2 p1 p2 6a (a 2b)3
Fl
(5)
For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in
(3)
(4)
p1
2(1500)
3 1.5 2(1.2) 14 375 lbf/in
1.22
p2
2(1500)
3 1.5 1.2 11 875 lbf/in
1.22
b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in
Substituting into (5) yields
= 0.036 89 or 3.7% higher than -Fl
_____________________________________________________________________________
3-44
Chapter 3 - Rev. A, Page 31/100
300(30) 40
1800 6900 lbf
2
30
300(30) 10
R2
1800 3900 lbf
2
30
3900
a
13 in
300
R1
M B = 1800(10) = 18 000 lbfin
M x = 27 in = (1/2)3900(13) = 25 350 lbfin
0.5(3) 2.5(3)
1.5 in
6
1
I1 (3)(13 ) 0.25 in 4
12
1
I 2 (1)(33 ) 2.25 in 4
12
Applying the parallel-axis theorem,
I z 0.25 3(1.5 0.5) 2 2.25 3(2.5 1.5) 2 8.5 in 4
y
18000(1.5)
3176 psi
8.5
18000(2.5)
At x 10 in, y 2.5 in, x
5294 psi
8.5
(a)
25350(1.5)
At x 27 in, y 1.5 in, x
4474 psi
8.5
25350(2.5)
At x 27 in, y 2.5 in, x
7456 psi
8.5
Max tension 5294 psi
Ans.
Ans.
Max compression 7456 psi
At x 10 in, y 1.5 in, x
(b)
The maximum shear stress due to V is at B, at the neutral axis.
Vmax 5100 lbf
Q yA 1.25(2.5)(1) 3.125 in 3
VQ 5100(3.125)
Ans.
1875 psi
max V
Ib
8.5(1)
(c)
There are three potentially critical locations for the maximum shear stress, all at x
= 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where
Chapter 3 - Rev. A, Page 32/100
the transverse shear is maximum, or (iii) in the web just above the flange where bending
stress and shear stress are in their largest combination.
For (i):
The maximum bending stress was previously found to be 7456 psi, and the shear
stress is zero. From Mohr’s circle,
7456
max max
3728 psi
2
2
For (ii):
The bending stress is zero, and the transverse shear stress was found previously to be
1875 psi. Thus, max = 1875 psi.
For (iii):
The bending stress at y = – 0.5 in is
18000( 0.5)
x
1059 psi
8.5
The transverse shear stress is
Q yA (1)(3)(1) 3.0 in 3
VQ 5100(3.0)
1800 psi
Ib
8.5(1)
From Mohr’s circle,
2
1059
2
max
1800 1876 psi
2
The critical location is at x = 27 in, at the top surface, where max = 3728 psi.
Ans.
_____________________________________________________________________________
3-45
(a) L = 10 in. Element A:
My
(1000)(10)(0.5)
103 101.9 kpsi
4
( / 64)(1)
I
VQ
A
, Q 0 A 0
Ib
A
2
max
2
101.9
2
A A2
(0) 50.9 kpsi
2
2
Ans.
Element B:
B
My
, y0
I
4r
Q yA
3
B 0
2
3
4 0.5
r 4r
1/12 in 3
6
6
2
3
Chapter 3 - Rev. A, Page 33/100
VQ (1000)(1/12)
103 1.698 kpsi
4
Ib ( / 64)(1) (1)
B
0
2
max 1.6982 1.698 kpsi
2
Ans.
Element C:
C
My
(1000)(10)(0.25) 3
10 50.93 kpsi
( / 64)(1)4
I
r
r
r
y1
y1
y1
Q ydA y (2 x) dy y 2 r 2 y 2 dy
2 2
r y2
3
3/2
r
y1
2
r2 r2
3
3/ 2
r 2 y12
3/ 2
2 2
r y12
3
For C, y 1 = r /2 =0.25 in
Q
3/ 2
3/2
2
0.52 0.252 0.05413 in3
3
b 2 x 2 r 2 y12 2 0.52 0.252 0.866 in
C
VQ
(1000)(0.05413)
103 1.273 kpsi
Ib ( / 64)(1) 4 (0.866)
max
50.93
2
(1.273) 25.50 kpsi
2
2
Ans.
(b) Neglecting transverse shear stress:
Element A: Since the transverse shear stress at point A is zero, there is no change.
max 50.9 kpsi Ans.
% error 0% Ans.
Element B: Since the only stress at point B is transverse shear stress, neglecting
the transverse shear stress ignores the entire stress.
2
0
max 0 psi Ans.
2
1.698 0
% error
* (100) 100% Ans.
1.698
Chapter 3 - Rev. A, Page 34/100
Element C:
2
max
50.93
25.47 kpsi
2
Ans.
25.50 25.47
% error
*(100) 0.12% Ans.
25.50
(c) Repeating the process with different beam lengths produces the results in the table.
L = 10 in
A
B
C
L = 4 in
A
B
C
L = 1 in
A
B
C
L = 0.1in
A
B
C
Bending
stress,
kpsi)
Transverse
shear stress,
kpsi)
Max shear
stress,
max kpsi)
Max shear
stress,
neglecting
max kpsi)
% error
102
0
50.9
0
1.70
1.27
50.9
1.70
25.50
50.9
0
25.47
0
100
0.12
40.7
0
20.4
0
1.70
1.27
20.4
1.70
10.26
20.4
0
10.19
0
100
0.77
10.2
0
5.09
0
1.70
1.27
5.09
1.70
2.85
5.09
0
2.55
0
100
10.6
1.02
0
0.509
0
1.70
1.27
0.509
1.70
1.30
0.509
0
0.255
0
100
80.4
Discussion:
The transverse shear stress is only significant in determining the critical stress element as
the length of the cantilever beam becomes smaller. As this length decreases, bending
stress reduces greatly and transverse shear stress stays the same. This causes the critical
element location to go from being at point A, on the surface, to point B, in the center. The
maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in,
where it is at point B at the center. When the critical stress element is at point A, there is
no error from neglecting transverse shear stress, since it is zero at that location.
Neglecting the transverse shear stress has extreme significance at the stress element at the
center at point B, but that location is probably only of practical significance for very short
beam lengths.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 35/100
3-46
c
F
l
c
M Fx 0 x a
l
6M 6 c l Fx
2
bh
bh 2
R1
h
6 Fcx
lb max
0 xa
Ans.
_____________________________________________________________________________
3-47
c
From Problem 3-46, R1 F V , 0 x a
l
3 V 3 (c / l ) F
3 Fc
max
h
2 bh 2 bh
2 lb max
Ans.
6 Fcx
.
lb max
Sub in x = e and equate to h above.
3 Fc
6 Fce
2 lb max
lb max
From Problem 3-46, h( x)
3 Fc max
Ans.
2
8 lb max
_____________________________________________________________________________
e
3-48
(a)
x-z plane
M O 0 1.5(0.5) 2(1.5) sin(30 )(2.25) R2 z (3)
R2 z 1.375 kN Ans.
Fz 0 R1 z 1.5 2(1.5) sin(30 ) 1.375
R1z 1.625 kN Ans.
x-y plane
M O 0 2(1.5) cos(30 )(2.25) R2 y (3)
R2 y 1.949 kN
Ans.
Fy 0 R1 y 2(1.5) cos(30 ) 1.949
R1 y 0.6491 kN
Ans.
Chapter 3 - Rev. A, Page 36/100
(b)
(c) The transverse shear and bending moments for most points of interest can readily be
taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are
parabolic, and are obtained by integrating the linear expressions for shear. For
convenience, use a coordinate shift of x = x – 1.5. Then, for 0 < x < 1.5,
Vz x 0.125
My
x
V dx
z
2
2
0.125x C
At x 0, M y C 0.9375 M y 0.5 x 0.125 x 0.9375
2
1.949
x 0.6491 1.732 x 0.6491
1.125
1.732
2
Mz
x 0.6491x C
2
2
At x 0, M z C 0.9737 M z 0.8662 x 0.125 x 0.9375
By programming these bending moment equations, we can find M y , M z , and their vector
combination at any point along the beam. The maximum combined bending moment is
found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key
locations on the shear and bending moment diagrams.
Vy
x (m)
0
0.5
1.5
1.625
1.875
3
V z (kN)
–1.625
–1.625
–0.1250
0
0.2500
1.375
V y (kN)
0.6491
0.6491
0.6491
0.4327
0
–1.949
Mz
My
V (kN) (kNm) (kNm)
1.750
0
0
1.750 –0.8125 0.3246
0.6610 0.9375 0.9737
0.4327 –0.9453 1.041
0.2500 –0.9141 1.095
2.385
0
0
M
(kNm)
0
0.8749
1.352
1.406
1.427
0
Chapter 3 - Rev. A, Page 37/100
(d) The bending stress is obtained from Eq. (3-27),
M z yA M y zA
x
Iz
Iy
The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34
(a), where distances from the neutral axes for both bending moments will be maximum.
At A, for M z , y A = –37.5 mm, and for M y , z A = –20 mm.
40(75)3 34(25)3
Iz
1.36(106 ) mm 4 1.36(106 ) m 4
12
12
3
25(40) 25(6)3
2.67(105 ) mm 4 2.67(107 ) m 4
Iy 2
12
12
It is apparent the maximum bending moment, and thus the maximum stress, will be in the
parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the
bending moment equations previously derived, the maximum tensile bending stress is
found at x = 1.77 m, where M y = – 0.9408 kN·m, M z = 1.075 kN·m, and x = 100.1 MPa.
Ans.
_____________________________________________________________________________
3-49
(a) x-z plane
3
600
(10) M Oy
M O 0 (1000)(4)
5
2
M Oy 1842.6 lbf in Ans.
3
600
Fz 0 ROz (1000)
5
2
ROz 175.7 lbf Ans.
x-y plane
4
600
(10) M Oz
M O 0 (1000)(4)
5
2
M Oz 7442.5 lbf in Ans.
4
600
Fy 0 ROy (1000)
5
2
ROy 1224.3 lbf Ans.
Chapter 3 - Rev. A, Page 38/100
(b)
(
(c)
1/2
V ( x) Vy ( x) 2 Vz ( x) 2
1/ 2
M ( x) M y ( x ) 2 M z ( x ) 2
x (m)
0
4
10
V z (kN)
–175.7
–175.7
424.3
V y (kN)
1224.3
1224.3
424.3
V (kN)
1237
1237
600
M y (kNm) M z (kNm) M (kNm)
–1842.6
–7442.6
7667
–2545.4
–2545.4
3600
0
0
0
(d) The maximum tensile bending stress will be at the outer corner of the cross section in
the positive y, negative z quadrant, where y = 1.5 in and z = –1 in.
2(3)3 (1.625)(2.625)3
Iz
2.051 in 4
12
12
3
3(2) (2.625)(1.625)3
Iy
1.601 in 4
12
12
At x = 0, using Eq. (3-27),
M y M z
x z y
Iz
Iy
(7442.6)(1.5) (1842.6)(1)
x
6594 psi
2.051
1.601
Check at x = 4 in,
(2545.4)(1.5) ( 2545.4)( 1)
x
2706 psi
2.051
1.601
The critical location is at x = 0, where x = 6594 psi. Ans.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 39/100
3-50
The area within the wall median line, A m , is
Square: Am (b t )2 . From Eq. (3-45)
Tsq 2 Am t all 2(b t ) 2 t all
Round: Am (b t ) 2 / 4
Trd 2 (b t ) 2 t all / 4
Ratio of Torques
Tsq
2(b t )2 t all
4
1.27
2
Trd (b t ) t all / 2
Twist per unit length from Eq. (3-46) is
1
TLm
2 A t L
L
L
m all2 m all m C m
2
4GAmt
4GAm t
2G Am
Am
Square:
sq C
Round:
rd C
4(b t )
(b t )2
(b t )
2
(b t ) / 4
C
4(b t )
(b t )2
Ratio equals 1. Twists are the same.
_____________________________________________________________________________
3-51
(a) The area enclosed by the section median line is A m = (1 0.0625)2 = 0.8789 in2 and
the length of the section median line is L m = 4(1 0.0625) = 3.75 in. From Eq. (3-45),
T 2 Amt 2(0.8789)(0.0625)(12 000) 1318 lbf in
From Eq. (3-46),
1l
Ans.
(1318)(3.75) 36
TLml
0.0801 rad 4.59 Ans.
2
6
2
4GAmt 4 11.5 10 (0.8789) 0.0625
(b) The radius at the median line is r m = 0.125 + (0.5)(0.0625) = 0.15625 in. The area enclosed
by the section median line is A m = (1 0.0625)2 – 4(0.15625)2 + 4(π /4)(0.15625)2 = 0.8579
in2. The length of the section median line is L m = 4[1 – 0.0625 – 2(0.15625)] + 2π(0.15625) =
3.482 in.
Chapter 3 - Rev. A, Page 40/100
From Eq. (3-45),
T 2 Amt 2(0.8579)(0.0625)(12 000) 1287 lbf in
Ans.
From Eq. (3-46),
(1287)(3.482) 36
TLml
1l
0.0762 rad 4.37 Ans.
2
6
2
4GAmt 4 11.5 10 (0.8579) 0.0625
_____________________________________________________________________________
3-52
1
3Ti
GLi ci3
T T1 T2 T3
Ti
1G
3
1GLi ci3
3
Li ci3
3
Ans.
i 1
From Eq. (3-47), G 1 c
G and 1 are constant, therefore the largest shear stress occurs when c is a maximum.
max G1cmax
Ans.
_____________________________________________________________________________
3-53
(b) Solve part (b) first since the twist is needed for part (a).
max allow 12 6.89 82.7 MPa
1
max
Gcmax
0.348 rad/m
79.3 10 (0.003)
82.7 106
9
(a)
T1
T2
T3
1GL1c13
3
2GL2 c23
3
3GL3c33
Ans.
0.348(79.3) 109 (0.020)(0.0023 )
3
0.348(79.3) 109 (0.030)(0.0033 )
3
9
0.348(79.3) 10 (0)(03 )
1.47 N m
Ans.
7.45 N m
Ans.
0 Ans.
3
3
T T1 T2 T3 1.47 7.45 0 8.92 N m Ans.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 41/100
3-54
(b) Solve part (b) first since the twist is needed for part (a).
12 000
1 max
8.35 103 rad/in Ans.
6
Gcmax 11.5 10 (0.125)
(a)
T1
T2
1GL1c13
3
2GL2 c23
8.35 103 11.5 106 0.75 0.06253
3
3
8.35 10 11.5 106 1 0.1253
5.86 lbf in
Ans.
62.52 lbf in Ans.
3
3
6
3
3GL3c33 8.35 10 11.5 10 0.625 0.0625
4.88 lbf in Ans.
T3
3
3
T T1 T2 T3 5.86 62.52 4.88 73.3 lbf in Ans.
_____________________________________________________________________________
3
3-55
(b) Solve part (b) first since the twist is needed for part (a).
max allow 12 6.89 82.7 MPa
1
max
Gcmax
0.348 rad/m
79.3 10 (0.003)
82.7 106
9
(a)
T1
T2
T3
1GL1c13
3
2GL2 c23
3
0.348(79.3) 109 (0.020)(0.0023 )
3GL3c33
Ans.
3
0.348(79.3) 109 (0.030)(0.0033 )
3
9
0.348(79.3) 10 (0.025)(0.0023 )
1.47 N m
7.45 N m
Ans.
Ans.
1.84 N m
Ans.
3
3
T T1 T2 T3 1.47 7.45 1.84 10.8 N m Ans.
_____________________________________________________________________________
3-56
(a) From Eq. (3-40), with two 2-mm strips,
80 106 0.030 0.0022
T
3.08 N m
3 1.8 / (b / c)
3 1.8 / 0.030 / 0.002
max bc 2
Tmax 2(3.08) 6.16 N m
Ans.
Chapter 3 - Rev. A, Page 42/100
From the table on p. 102, with b/c = 30/2 = 15, and has a value between 0.313 and 0.333.
From Eq. (3-40),
1
0.321
3 1.8 / (30 / 2)
From Eq. (3-41),
Tl
3.08(0.3)
0.151 rad
3
bc G 0.321 0.030 0.0023 79.3 109
kt
T
6.16
40.8 N m
0.151
Ans.
Ans.
From Eq. (3-40), with a single 4-mm strip,
80 106 0.030 0.0042
11.9 N m
Tmax
3 1.8 / (b / c)
3 1.8 / 0.030 / 0.004
max bc 2
Ans.
Interpolating from the table on p. 102, with b/c = 30/4 = 7.5,
7.5 6
(0.307 0.299) 0.299 0.305
86
From Eq. (3-41)
Tl
11.9(0.3)
0.0769 rad
3
bc G 0.305 0.030 0.0043 79.3 109
kt
T
11.9
155 N m
0.0769
Ans.
Ans.
(b) From Eq. (3-47), with two 2-mm strips,
2
6
Lc 2 0.030 0.002 80 10
3.20 N m
T
3
3
Tmax 2(3.20) 6.40 N m Ans.
3Tl
3(3.20)(0.3)
3
0.151 rad
Lc G 0.030 0.0023 79.3 109
kt T 6.40 0.151 42.4 N m
Ans.
Ans.
From Eq. (3-47), with a single 4-mm strip,
Tmax
2
6
Lc 2 0.030 0.004 80 10
12.8 N m
3
3
Ans.
Chapter 3 - Rev. A, Page 43/100
3Tl
3(12.8)(0.3)
0.0757 rad
3
Lc G 0.030 0.0043 79.3 109
kt T 12.8 0.0757 169 N m
Ans.
Ans.
The results for the spring constants when using Eq. (3-47) are slightly larger than when using
Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal
infinity). The spring constants when considering one solid strip are significantly larger (almost
four times larger) than when considering two thin strips because two thin strips would be able
to slip along the center plane.
_____________________________________________________________________________
3-57
(a) Obtain the torque from the given power and speed using Eq. (3-44).
H
(40 000)
T 9.55 9.55
152.8 N m
n
2500
Tr 16T
max 3
J d
13
16 152.8
0.0223 m 22.3 mm
70 106
H
(40000)
1528 N m
(b) T 9.55 9.55
n
250
13
16T
d
max
Ans.
13
16(1528)
0.0481 m 48.1 mm Ans.
d
70 106
_____________________________________________________________________________
3-58
(a) Obtain the torque from the given power and speed using Eq. (3-42).
63025H 63025(50)
T
1261 lbf in
n
2500
Tr 16T
max 3
J d
13
13
16 1261
16T
d
0.685 in
max
(20 000)
63025H 63025(50)
12610 lbf in
(b) T
n
250
Ans.
13
16(12 610)
d
1.48 in Ans.
(20 000)
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 44/100
3-59
max
16T
d3
T
max d 3
50 106 0.033
265 N m
16
16
Tn 265(2000)
55.5 103 W 55.5 kW Ans.
Eq. (3-44), H
9.55
9.55
_____________________________________________________________________________
3-60
16T
3 T d 3 110 106 0.0203 173 N m
16
16
d
0.0204 79.3 109 15
4
Tl
d G
180
l
JG
32T
32(173)
l 1.89 m Ans.
_____________________________________________________________________________
3-61
16T
T d 3 30 000 0.753 2485 lbf in
3
16
16
d
Tl
32Tl
32(2485)(24)
0.167 rad 9.57 Ans.
4
4
6
JG d G 0.75 11.5 10
_____________________________________________________________________________
3-62
(a) Tsolid
J max do4 max
r
16do
Thollow
J max (d o4 di4 ) max
r
16d o
364
Tsolid Thollow
di4
%T
(100%) 4 (100%)
(100%) 65.6%
Tsolid
do
404
(b) Wsolid kdo2 ,
Whollow k do 2 di 2
Ans.
362
Wsolid Whollow
di2
%W
(100%) 2 (100%)
(100%) 81.0%
Wsolid
do
402
Ans.
_____________________________________________________________________________
3-63
4
4
J max d 4 max
J max d xd max
Thollow
(a) Tsolid
r
r
16d
16d
4
T
T
( xd )
%T solid hollow (100%)
(100%) x 4 (100%) Ans.
4
Tsolid
d
Chapter 3 - Rev. A, Page 45/100
Whollow k d 2 xd
(b) Wsolid kd 2
2
xd (100%) x 2 (100%)
Wsolid Whollow
(100%)
Wsolid
d2
2
%W
Ans.
Plot %T and %W versus x.
The value of greatest difference in percent reduction of weight and torque is 25% and
occurs at x 2 2 .
_____________________________________________________________________________
3-64
Tc
(a)
J
120 10
4200 d 2
6
4
32 d 4 0.70d
2.8149 104
d
3
13
2.8149 104
6.17 102 m 61.7 mm
d
6
120(10 )
From Table A-17, the next preferred size is d = 80 mm.
Ans.
d i = 0.7d = 56 mm. The next preferred size smaller is d i = 50 mm Ans.
(b)
4200 d i 2
4200 0.050 2
Tc
30.8 MPa
Ans.
4
J 32 d 4 d 32 0.080 4 0.050 4
i
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 46/100
3-65
T 9.55
H
(1500)
9.55
1433 N m
n
10
13
16 1433
0.045 m 45 mm
=
80 106
From Table A-17, select 50 mm. Ans.
16 2 1433
(a) start
117 106 Pa 117 MPa Ans.
3
0.050
16T
3
dC
13
16T
dC
(b) Design activity
_____________________________________________________________________________
3-66
T
63 025 H 63 025(1)
7880 lbf in
n
8
16T
3
dC
13
16T
dC
13
16 7880
=
15 000
1.39 in
From Table A-17, select 1.40 in.
Ans.
_____________________________________________________________________________
3-67
For a square cross section with side length b, and a circular section with diameter d,
Asquare Acircular b 2 d 2
b
d
4
2
From Eq. (3-40) with b = c,
3
T
1.8 T 1.8 T 2
T
(4.8) 6.896 3
max square 2 3
3 3
3
bc
b/c b
1 d
d
For the circular cross section,
16T
T
max circular 3 5.093 3
d
d
T
max square 6.896 d 3
1.354
max circular 5.093 T
d3
The shear stress in the square cross section is 35.4% greater.
Ans.
(b) For the square cross section, from the table on p. 102, β = 0.141. From Eq. (3-41),
Chapter 3 - Rev. A, Page 47/100
square
Tl
Tl
4
3
bc G b G
Tl
4
0.141
d G
2
11.50
Tl
d 4G
For the circular cross section,
Tl
Tl
Tl
rd
10.19 4
4
GJ G d 32
d G
Tl
sq 11.50 d 4G
1.129
rd 10.19 Tl
4
d G
The angle of twist in the square cross section is 12.9% greater. Ans.
_____________________________________________________________________________
3-68 (a)
T1 0.15T2
T 0 (500 75)(4) T
2
1700 4.25T2 0
T1 0.15 400 60 lbf
(b)
M
O
T1 5 1700 T2 0.15T2 5
T2 400 lbf
Ans.
Ans.
0 575(10) 460(28) RC (40)
RC 178.25 178 lbf
F 0R
O
Ans.
575 460 178.25
RO 293.25 lbf
Ans.
(c)
Chapter 3 - Rev. A, Page 48/100
(d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbf·in. Since the
shaft rotates, each stress element will experience both positive and negative bending
stress as it moves from tension to compression. The torque transmitted through the shaft
from A to B is T = (500 75)(4) = 1700 lbf·in. For a stress element on the outer surface
where the bending stress and the torsional stress are both maximum,
Mc 32 M 32 2932.5
15 294 psi = 15.3 kpsi Ans.
I
d3
(1.25)3
Tr 16T 16(1700)
Ans.
3
4433 psi = 4.43 kpsi
J d
(1.25)3
(e)
2
x
2
2
15.3
2
15.3
x xy
1, 2
4.43
2
2
2
2
1 16.5 kpsi
Ans.
2 1.19 kpsi
Ans.
2
2
2
2
15.3
max x xy
Ans.
4.43 8.84 kpsi
2
2
_____________________________________________________________________________
3-69 (a)
T2 0.15T1
T 0 1800 270 (200) T
2
306 103 106.25T1 0
T1 (125) 306 103 125 0.15T1 T1
T1 2880 N Ans.
T2 0.15 2880 432 N Ans.
(b)
M
O
0 3312(230) RC (510) 2070(810)
RC 1794 N Ans.
F
y
0 RO 3312 1794 2070
RO 3036 N Ans.
(c)
Chapter 3 - Rev. A, Page 49/100
(d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. Since the
shaft rotates, each stress element will experience both positive and negative bending
stress as it moves from tension to compression. The torque transmitted through the shaft
from A to B is T = (1800 270)(0.200) = 306 N·m. For a stress element on the outer
surface where the bending stress and the torsional stress are both maximum,
Mc 32 M 32 698.3
263 103 Pa 263 MPa Ans.
3
3
I
d
(0.030)
Tr 16T
16(306)
Ans.
3
57.7 106 Pa 57.7MPa
3
J d
(0.030)
(e)
2
x
2
2
263
2
263
x xy
1, 2
57.7
2
2
2
2
1 275 MPa
Ans.
2 12.1 MPa
Ans.
2
2
2
2
263
max x xy
Ans.
57.7 144 MPa
2
2
_____________________________________________________________________________
3-70
(a)
T2 0.15T1
T 0 300 50 (4) T
T1 (3) 1000 0.15T1 T1 (3)
1000 2.55T1 0
T1 392.16 lbf Ans.
2
T2 0.15 392.16 58.82 lbf Ans.
(b)
Chapter 3 - Rev. A, Page 50/100
M
Oy
0 450.98(16) RC z (22)
RC z 327.99 lbf Ans.
F
z
0 RO z 450.98 327.99
RO z 122.99 lbf Ans.
M
Oz
0 350(8) RC y (22)
RC y 127.27 lbf Ans.
F
y
0 RO y 350 127.27
RO y 222.73 lbf Ans.
Chapter 3 - Rev. A, Page 51/100
(c)
(d) Combine the bending moments from both planes at A and B to find the critical
location.
M A (983.92) 2 (1781.84) 2 2035 lbf in
M B (1967.84) 2 (763.65) 2 2111 lbf in
The critical location is at B. The torque transmitted through the shaft from A to B is T =
(300 50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending
stress and the torsional stress are both maximum,
Mc 32 M 32 2111
21 502 psi = 21.5 kpsi Ans.
I
d3
(1)3
Tr 16T 16(1000)
Ans.
3
5093 psi = 5.09 kpsi
J d
(1)3
(e)
2
x
2
2
21.5
2
21.5
x xy
5.09
2
2
2
2
1 22.6 kpsi
Ans.
1, 2
2 1.14 kpsi
2
Ans.
2
2
2
21.5
max x xy
Ans.
5.09 11.9 kpsi
2
2
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 52/100
3-71
(a)
T2 0.15T1
T 0 300 45 (125) T
2
31 875 127.5T1 0
T1 (150) 31 875 0.15T1 T1 (150)
T1 250 N mm Ans.
T2 0.15 250 37.5 N mm Ans.
(b)
M
Oy
0 345sin 45o (300) 287.5(700) RC z (850)
RC z 150.7 N
F
z
Ans.
0 RO z 345 cos 45o 287.5 150.7
RO z 107.2 N
M
Oz
Ans.
0 345sin 45o (300) RC y (850)
RC y 86.10 N
F
y
Ans.
0 RO y 345cos 45o 86.10
RO y 157.9 N
Ans.
(c)
(
d
)
F
r
o
m
t
h
e
b
e
n
ding moment diagrams, it is clear that the critical location is at A where both planes have
the maximum bending moment. Combining the bending moments from the two planes,
M
47.37 32.16
2
2
57.26 N m
Chapter 3 - Rev. A, Page 53/100
The torque transmitted through the shaft from A to B is T = (300 45)(0.125) = 31.88
N·m. For a stress element on the outer surface where the bending stress and the torsional
stress are both maximum,
Mc 32 M 32 57.26
72.9 106 Pa 72.9 MPa
Ans.
3
3
I
d
(0.020)
Tr 16T 16(31.88)
Ans.
3
20.3 106 Pa 20.3 MPa
3
J d
(0.020)
(e)
2
x
2
2
72.9
2
72.9
x xy
1, 2
20.3
2
2
2
2
1 78.2 MPa
Ans.
2 5.27 MPa
Ans.
2
2
2
2
72.9
max x xy
Ans.
20.3 41.7 MPa
2
2
_____________________________________________________________________________
3-72
(a)
T 0 300(cos 20º )(10) FB (cos 20º )(4)
FB 750 lbf Ans.
(b)
M
Oz
0 300(cos 20º )(16) 750(sin 20º )(39) RC y (30)
RC y 183 lbf Ans.
F
y
0 RO y 300(cos 20º ) 183 750(sin 20º )
RO y 208 lbf Ans.
M
Oy
0 300(sin 20º )(16) RC z (30) 750(cos 20º )(39)
RC z 861 lbf Ans.
F
z
0 RO z 300(sin 20º ) 861 750(cos 20º )
RO z 259 lbf
Ans.
Chapter 3 - Rev. A, Page 54/100
(c)
(d) Combine the bending moments from both planes at A and C to find the critical
location.
M A (3336) 2 (4149) 2 5324 lbf in
M C (2308) 2 (6343) 2 6750 lbf in
The critical location is at C. The torque transmitted through the shaft from A to B is
T 300 cos 20º 10 2819 lbf in . For a stress element on the outer surface where the
bending stress and the torsional stress are both maximum,
Mc 32 M 32 6750
35 203 psi = 35.2 kpsi
Ans.
I
d3
(1.25)3
Tr 16T 16(2819)
Ans.
3
7351 psi = 7.35 kpsi
J d
(1.25)3
(e)
2
x
2
2
35.2
2
35.2
x xy
7.35
2
2
2
2
1 36.7 kpsi
Ans.
1, 2
2 1.47 kpsi
2
Ans.
2
2
2
35.2
max x xy
Ans.
7.35 19.1 kpsi
2
2
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 55/100
3-73
(a)
T 0 11 000(cos 20º )(300) F (cos 25º )(150)
B
FB 22 810 N
(b)
M
Oz
0 11 000(sin 20º )(400) 22 810(sin 25º )(750) RC y (1050)
RC y 8319 N
F
y
Ans.
0 RO y 11000(sin 20º ) 22 810 sin(25º ) 8319
RO y 5083 N
M
Ans.
Oy
Ans.
0 11 000(cos 20º )(400) 22 810(cos 25º )(750) RC z (1050)
RC z 10 830 N Ans.
F
z
0 RO z 11 000(cos 20º ) 22 810(cos 25º ) 10 830
RO z 494 N
Ans.
(c)
(d) From the bending moment diagrams, it is clear that the critical location is at B where
both planes have the maximum bending moment. Combining the bending moments from
the two planes,
M
2496 3249
2
2
4097 N m
The torque transmitted through the shaft from A to B is
T 11 000 cos 20º 0.3 3101 N m .
For a stress element on the outer surface where the bending stress and the torsional stress
are both maximum,
Chapter 3 - Rev. A, Page 56/100
Mc 32 M 32 4097
333.9 106 Pa 333.9 MPa
Ans.
3
3
I
d
(0.050)
Tr 16T 16(3101)
Ans.
3
126.3 106 Pa 126.3 MPa
3
J d
(0.050)
(e)
2
x
2
2
333.9
2
333.9
x xy
1, 2
126.3
2
2
2
2
1 376 MPa
Ans.
2 42.4 MPa
Ans.
2
2
2
2
333.9
max x xy
Ans.
126.3 209 MPa
2
2
_____________________________________________________________________________
3-74
(a)
M D z 6.13Cx 3.8(92.8) 3.88(362.8) 0
C x 287.2 lbf
Ans.
M C z 6.13Dx 2.33(92.8) 3.88(362.8) 0
Dx 194.4 lbf
3.8
(808) 500.9 lbf
6.13
2.33
0 Dz
(808) 307.1 lbf
6.13
M D x 0
M C x
Ans.
Cz
Ans.
Ans.
(b) For DQC, let x, y, z correspond to the original y, x, z axes.
Chapter 3 - Rev. A, Page 57/100
(c) The critical stress element is just to the right of Q, where the bending moment in both
planes is maximum, and where the torsional and axial loads exist.
T 808(3.88) 3135 lbf in
M 669.22 1167 2 1345 lbf in
16T 16(3135)
3
11 070 psi Ans.
d
1.133
b
32 M
32(1345)
9495 psi
3
d
1.133
Ans.
a
F
362.8
362 psi
A
( / 4) 1.132
Ans.
(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
max 9495 362 9857 psi
2
9857
2
11 070 12 118 psi 12.1 kpsi
2
max
Ans.
2
9857
9857
2
1, 2
11 070
2
2
1 7189 psi 7.19 kpsi Ans.
2 17 046 psi 17.0 kpsi
Ans.
_____________________________________________________________________________
3-75
(a)
M D z 0
6.13Cx 3.8(46.6) 3.88(140) 0
C x 117.5 lbf
Ans.
M C z 0
6.13Dx 2.33(46.6) 3.88(140) 0
Dx 70.9 lbf
3.8
(406) 251.7 lbf
6.13
2.33
0 Dz
(406) 154.3 lbf
6.13
M D x 0
M C x
Ans.
Cz
Ans.
Ans.
Chapter 3 - Rev. A, Page 58/100
(b) For DQC, let x, y, z correspond to the original y, x, z axes.
(c) The critical stress element is just to the right of Q, where the bending moment in both
planes is maximum, and where the torsional and axial loads exist.
T 406(3.88) 1575 lbf in
M 273.82 586.32 647.1 lbf in
16T 16(1575)
3
8021 psi Ans.
d
13
b
32 M
32(647.1)
6591 psi
3
d
13
a
F
140
178.3 psi
A
( / 4) 12
Ans.
Ans.
(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
max 6591 178.3 6769 psi
2
max
6769
2
8021 8706 psi 8.71 kpsi
2
Ans.
2
1, 2
6769
6769
2
8021
2
2
Chapter 3 - Rev. A, Page 59/100
1 5321 psi 5.32 kpsi Ans.
2 12 090 psi 12.1 kpsi Ans.
_____________________________________________________________________________
3-76
M B z 5.62(362.8) 1.3(92.8) 3 Ay 0
Ay 639.4 lbf
Ans.
M A z 2.62(362.8) 1.3(92.8) 3By 0
B y 276.6 lbf
5.62
(808) 1513.7 lbf
3
2.62
0 Bz
(808) 705.7 lbf
3
M B y 0
M A y
Ans.
Az
Ans.
Ans.
(b)
(c) The critical stress element is just to the left of A, where the bending moment in both
planes is maximum, and where the torsional and axial loads exist.
Chapter 3 - Rev. A, Page 60/100
T 808(1.3) 1050 lbf in
16(1050)
7847 psi Ans.
0.883
M (829.8) 2 (2117) 2 2274 lbf in
32M
32(2274)
b
33 990 psi
3
d
0.883
a
F
92.8
153 psi
A
( / 4) 0.882
Ans.
Ans.
(d) The critical stress will occur when the bending stress and axial stress are both in
compression.
max 33 990 153 34143 psi
2
max
34 143
2
7847 18 789 psi 18.8 kpsi
2
Ans.
2
34143
34143
2
1, 2
7847
2
2
1 1717 psi 1.72 kpsi Ans.
2 35 860 psi 35.9 kpsi
Ans.
_____________________________________________________________________________
3-77
Ft
T
100
1600 N
c / 2 0.125 / 2
Fn 1600 tan 20 582.4 N
TC Ft b 2 1600 0.250 2 200 N m
P
TC
200
2667 N
a 2 0.150 2
M A z 0
450 RDy 582.4(325) 2667(75) 0
RDy 865.1 N
M A y 0 450 RDz 1600(325) RDz 1156 N
Fy 0 RAy 865.1 582.4 2667 RAy 2384 N
Fz 0 RAz 1156 1600 RAz 444 N
Chapter 3 - Rev. A, Page 61/100
AB The maximum bending moment will either be at B or C. If this is not obvious, sketch
the shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B AB RA2 y RA2z 0.075 23842 4442 181.9 N m
M C CD RD2 y RD2 z 0.125 865.12 11562 180.5 N m
The stresses at B and C are almost identical, but the maximum stresses occur at B.
32M B 32(181.9)
B
68.6 106 Pa 68.6 MPa
3
3
d
0.030
B
max
Ans.
16TB
16(200)
37.7 106 Pa 37.7 MPa
3
3
d
0.030
2
B
2
68.6
68.6
2
B B2
37.7 85.3 MPa
2
2
2
2
2
Ans.
2
68.6
2
max B B2
37.7 51.0 MPa Ans.
2
2
_____________________________________________________________________________
3-78
Ft
T
100
1600 N
c / 2 0.125 / 2
Fn 1600 tan 20 582.4 N
TC Ft b 2 1600 0.250 2 200 N m
P
TC
200
2667 N
a 2 0.150 2
RDy 420.6 N
M A z 0 450 RDy 582.4(325)
M A y 0 450 RDz 1600(325) 2667(75) RDz 711.1 N
Fy 0 RAy 420.6 582.4
Fz 0 RAz 711.1 1600 2667
RAy 161.8 N
RAz 1778 N
Chapter 3 - Rev. A, Page 62/100
The maximum bending moment will either be at B or C. If this is not obvious, sketch
shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B AB RA2 y RA2z 0.075 161.82 1778 133.9 N m
2
M C CD RD2 y RD2 z 0.125 420.62 711.12 103.3 N m
The maximum stresses occur at B. Ans.
32M B 32(133.9)
B
50.5 106 Pa 50.5 MPa
3
3
d
0.030
B
16TB
16(200)
37.7 106 Pa 37.7 MPa
3
3
d
0.030
max
2
B
2
50.5
50.5
2
B B2
37.7 70.6 MPa
2
2
2
2
2
Ans.
2
50.5
2
max B B2
37.7 45.4 MPa
2
2
Ans.
_____________________________________________________________________________
3-79
T
900
180 lbf
c / 2 10 / 2
Fn 180 tan 20 65.5 lbf
Ft
TC Ft b 2 180 5 2 450 lbf in
P
TC
450
150 lbf
a 2 6 2
M A z 0 20 RDy 65.5(14) 150(4) RDy 75.9 lbf
RDz 126 lbf
M A y 0 20 RDz 180(14)
Fy 0 RAy 75.9 65.5 150
Fz 0 RAz 126 180
RAy 140 lbf
RAz 54.0 lbf
Chapter 3 - Rev. A, Page 63/100
The maximum bending moment will either be at B or C. If this is not obvious, sketch
shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B AB RA2 y RA2z 4 1402 542 600 lbf in
M C CD RD2 y RD2 z 6 75.92 1262 883 lbf in
The maximum stresses occur at C.
Ans.
C
32M C
32(883)
3460 psi
3
d
1.3753
C
16TC
16(450)
882 psi
d 3 1.3753
max
2
C
2
3460
3460
2
C C2
882 3670 psi
2
2
2
2
2
Ans.
2
3460
2
max C C2
882 1940 psi
2
2
Ans.
_____________________________________________________________________________
3-80
(a) Rod AB experiences constant torsion throughout its length, and maximum bending
moment at the wall. Both torsional shear stress and bending stress will be maximum on
the outer surface. The transverse shear will be very small compared to bending and
torsion, due to the reasonably high length to diameter ratio, so it will not dominate the
determination of the critical location. The critical stress element will be at the wall, at
either the top (compression) or the bottom (tension) on the y axis. We will select the
bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
Mc M d / 2 32M 32 8 200
x
16 297 psi 16.3 kpsi
3
d 4 / 64 d 3
I
1
xz
Tr T d / 2 16T 16 5 200
5093 psi 5.09 kpsi
3
J d 4 / 32 d 3
1
Chapter 3 - Rev. A, Page 64/100
(c)
2
x
2
16.3
2
2
16.3
x xz
1, 2
5.09
2
2
2
2
1 17.8 kpsi
Ans.
2 1.46 kpsi
Ans.
2
2
2
2
16.3
max x xz
Ans.
5.09 9.61 kpsi
2
2
_____________________________________________________________________________
3-81
(a) Rod AB experiences constant torsion throughout its length, and maximum bending
moments at the wall in both planes of bending. Both torsional shear stress and bending
stress will be maximum on the outer surface. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be on the outer surface at the wall, with its critical location determined by the plane
of the combined bending moments.
M y = – (100)(8) = – 800 lbf·in
M z = (175)(8) = 1400 lbf·in
M tot M y2 M z2
800
2
14002 1612 lbf in
My
1 800
tan
29.7º
1400
Mz
The combined bending moment vector is at an angle of 29.7º CCW from the z axis. The
critical bending stress location, and thus the critical stress element, will be ±90º from this
vector, as shown. There are two equally critical stress elements, one in tension (119.7º
CCW from the z axis) and the other in compression (60.3º CW from the z axis). We’ll
continue the analysis with the element in tension.
(b) Transverse shear is zero at the critical stress elements on the outer surfaces.
M c M d / 2 32 M tot 32 1612
x tot tot 4
16 420 psi 16.4 kpsi
3
d / 64
d3
I
1
= tan 1
Tr T d / 2 16T 16 5 175
4456 psi 4.46 kpsi
3
J d 4 / 32 d 3
1
Chapter 3 - Rev. A, Page 65/100
(c)
2
x
2
16.4
2
16.4
x 2
1, 2
4.46
2
2
2
2
1 17.5 kpsi
Ans.
2 1.13 kpsi
Ans.
2
2
2
16.4
max x 2
Ans.
4.46 9.33 kpsi
2
2
_____________________________________________________________________________
3-82
(a) Rod AB experiences constant torsion and constant axial tension throughout its length,
and maximum bending moments at the wall from both planes of bending. Both torsional
shear stress and bending stress will be maximum on the outer surface. The transverse
shear will be very small compared to bending and torsion, due to the reasonably high
length to diameter ratio, so it will not dominate the determination of the critical location.
The critical stress element will be on the outer surface at the wall, with its critical
location determined by the plane of the combined bending moments.
M y = – (100)(8) – (75)(5) = – 1175 lbf·in
M z = (–200)(8) = –1600 lbf·in
M tot M y2 M z2
1175 1600
2
2
1985 lbf in
My
1 1175
tan
36.3º
1600
Mz
The combined bending moment vector is at an angle of 36.3º CW from the negative z
axis. The critical bending stress location will be ±90º from this vector, as shown. Since
there is an axial stress in tension, the critical stress element will be where the bending is
also in tension. The critical stress element is therefore on the outer surface at the wall, at
an angle of 36.3º CW from the y axis.
(b) Transverse shear is zero at the critical stress element on the outer surface.
M c M d / 2 32M tot 32 1985
x ,bend tot tot 4
20 220 psi 20.2 kpsi
3
d / 64
d3
I
1
= tan 1
x ,axial
Fx
Fx
75
95.5 psi 0.1 kpsi , which is essentially negligible
2
A d / 4 12 / 4
x x ,axial x ,bend 20 220 95.5 20 316 psi 20.3 kpsi
Tr 16T 16 5 200
5093 psi 5.09 kpsi
3
J d3
1
Chapter 3 - Rev. A, Page 66/100
(c)
2
x
2
20.3
2
20.3
x 2
1, 2
5.09
2
2
2
2
Ans.
1 21.5 kpsi
2 1.20 kpsi
2
Ans.
2
2
20.3
Ans.
max x 2
5.09 11.4 kpsi
2
2
_____________________________________________________________________________
3-83
T = (2)(200) = 400 lbf·in
The maximum shear stress due to torsion occurs in the middle of the longest side of the
rectangular cross section. From the table on p. 102, with b/c = 1.5/0.25 = 6, = 0.299.
From Eq. (3-40),
T
400
max
14 270 psi 14.3 kpsi Ans.
2
bc 0.299 1.5 0.252
____________________________________________________________________________
3-84
(a) The cross section at A will experience bending, torsion, and transverse shear. Both
torsional shear stress and bending stress will be maximum on the outer surface. The
transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be at either the top (compression) or the
bottom (tension) on the y axis. We’ll select the bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
Mc M d / 2 32M 32 11 250
x
28 011 psi 28.0 kpsi
3
d 4 / 64 d 3
I
1
xz
Tr T d / 2 16T 16 12 250
15 279 psi 15.3 kpsi
3
J d 4 / 32 d 3
1
Chapter 3 - Rev. A, Page 67/100
(c)
2
x
2
28.0
2
2
28.0
x xz
1, 2
15.3
2
2
2
2
Ans.
1 34.7 kpsi
2 6.7 kpsi
Ans.
2
2
2
2
28.0
Ans.
max x xz
15.3 20.7 kpsi
2
2
____________________________________________________________________________
3-85
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be maximum on the outer surface. The
transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be on the outer surface, with its critical
location determined by the plane of the combined bending moments.
M y = (300)(12) = 3600 lbf·in
M z = (250)(11) = 2750 lbf·in
M tot M y2 M z2
3600 2750
2
M
z
My
= tan 1
2
4530 lbf in
2750
tan 1
37.4º
3600
The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 37.4º CCW from the z axis.
(b)
M c M d / 2 32M tot 32 4530
x ,bend tot tot 4
46 142 psi 46.1 kpsi
3
d / 64
d3
I
1
x ,axial
Fx
Fx
300
382 psi 0.382 kpsi
2
A d / 4 12 / 4
x x ,axial x ,bend 46 142 382 46 524 psi 46.5 kpsi
Tr 16T 16 12 250
15 279 psi 15.3 kpsi
3
J d3
1
Chapter 3 - Rev. A, Page 68/100
(c)
2
x
2
46.5
2
46.5
x 2
1, 2
15.3
2
2
2
2
Ans.
1 51.1 kpsi
2 4.58 kpsi
Ans.
2
max
2
2
46.5
x 2
15.3 27.8 kpsi
2
2
Ans.
____________________________________________________________________________
3-86
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be maximum on the outer surface. The
transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be on the outer surface, with its critical
location determined by the plane of the combined bending moments.
M y = (300)(12) – (–100)(11) = 4700 lbf·in
M z = (250)(11) = 2750 lbf·in
M tot M y2 M z2
4700 2750
2
M
z
My
= tan 1
2
5445 lbf in
2750
tan 1
30.3º
4700
The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 30.3º CCW from the z axis.
(b)
M c M d / 2 32M tot 32 5445
x ,bend tot tot 4
55 462 psi 55.5 kpsi
3
d / 64
d3
I
1
Chapter 3 - Rev. A, Page 69/100
x ,axial
Fx
Fx
300
382 psi 0.382 kpsi
2
A d / 4 12 / 4
x x ,axial x ,bend 55 462 382 55 844 psi 55.8 kpsi
Tr 16T 16 12 250
15 279 psi 15.3 kpsi
3
J d3
1
(c)
2
x
2
55.8
2
55.8
x 2
15.3
2
2
2
2
Ans.
1 59.7 kpsi
1, 2
2 3.92 kpsi
Ans.
2
max
2
2
55.8
x 2
15.3 31.8 kpsi
2
2
Ans.
____________________________________________________________________________
3-87
(a) The cross section at A will experience bending, torsion, and transverse shear. Both
torsional shear stress and bending stress will be maximum on the outer surface, where the
stress concentration will also be applicable. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select
the bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
r / d 0.125 /1 0.125
D / d 1.5 /1 1.5
Fig. A-15-8
K t ,torsion 1.39
K t ,bend 1.59
x K t ,bend
Fig. A-15-9
32 11 250
Mc
32M
K t ,bend
(1.59)
44 538 psi 44.5 kpsi
3
3
I
d
1
xz K t ,torsion
16 12 250
Tr
16T
K t ,torsion
(1.39)
21 238 psi 21.2 kpsi
3
3
J
d
1
Chapter 3 - Rev. A, Page 70/100
(c)
2
x
2
44.5
2
2
44.5
x xz
21.2
2
2
2
2
Ans.
1 53.0 kpsi
1, 2
2 8.48 kpsi
Ans.
2
2
2
2
44.5
Ans.
max x xz
21.2 30.7 kpsi
2
2
____________________________________________________________________________
3-88
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be maximum on the outer surface,
where the stress concentration will also be applicable. The transverse shear will be very
small compared to bending and torsion, due to the reasonably high length to diameter
ratio, so it will not dominate the determination of the critical location. The critical stress
element will be on the outer surface, with its critical location determined by the plane of
the combined bending moments.
M y = (300)(12) = 3600 lbf·in
M z = (250)(11) = 2750 lbf·in
M tot M y2 M z2
3600 2750
2
M
z
My
= tan 1
2
4530 lbf in
2750
tan 1
37.4º
3600
The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 37.4º CCW from the z axis.
(b)
r / d 0.125 /1 0.125
D / d 1.5 /1 1.5
Fig. A-15-7
K t ,axial 1.75
K t ,torsion 1.39
Fig. A-15-8
K t ,bend 1.59
Fig. A-15-9
Chapter 3 - Rev. A, Page 71/100
x ,bend K t ,bend
32 4530
Mc
32M
K t ,bend
(1.59)
73 366 psi 73.4 kpsi
3
3
I
d
1
x ,axial K t ,axial
Fx
300
1.75
668 psi 0.668 kpsi
2
A
1 / 4
x x ,axial x ,bend 73 366 668 74 034 psi 74.0 kpsi
Kt ,torsion
16 12 250
Tr
16T
Kt ,torsion
(1.39)
21 238 psi 21.2 kpsi
3
3
J
d
1
(c)
2
x
2
74.0
2
74.0
1, 2 x 2
21.2
2
2
2
2
Ans.
1 79.6 kpsi
2 5.64 kpsi
Ans.
2
max
2
2
74.0
x 2
21.2 42.6 kpsi
2
2
Ans.
____________________________________________________________________________
3-89
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be maximum on the outer surface,
where the stress concentration is also applicable. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be on the outer surface, with its critical location determined by the plane of the
combined bending moments.
M y = (300)(12) – (–100)(11) = 4700 lbf·in
M z = (250)(11) = 2750 lbf·in
M tot M y2 M z2
4700 2750
2
M
z
My
= tan 1
2
5445 lbf in
2750
tan 1
30.3º
4700
Chapter 3 - Rev. A, Page 72/100
The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 30.3º CCW from the z axis.
(b)
r / d 0.125 /1 0.125
D / d 1.5 /1 1.5
Fig. A-15-7
K t ,axial 1.75
K t ,torsion 1.39
Fig. A-15-8
K t ,bend 1.59
Fig. A-15-9
x ,bend K t ,bend
32 5445
Mc
32M
K t ,bend
(1.59)
88185 psi 88.2 kpsi
3
3
I
d
1
x ,axial K t ,axial
Fx
300
1.75
668 psi 0.668 kpsi
2
A
1 / 4
x x ,axial x ,bend 88185 668 88 853 psi 88.9 kpsi
Kt ,torsion
16 12 250
Tr
16T
Kt ,torsion
(1.39)
21 238 psi 21.2 kpsi
3
3
J
d
1
(c)
2
x
2
88.9
2
88.9
x 2
21.2
2
2
2
2
Ans.
1 93.7 kpsi
1, 2
2 4.80 kpsi
2
Ans.
2
2
88.9
Ans.
max x 2
21.2 49.2 kpsi
2
2
____________________________________________________________________________
3-90
(a) M = F(p / 4), c = p / 4, I = bh3 / 12, b = d r n t , h = p / 2
Chapter 3 - Rev. A, Page 73/100
b
F p / 4 p / 4
Mc
Fp 2
3
I
bh3 /12
16 d r nt p / 2 /12
6F
Ans.
d r nt p
F
F
4F
(b) a 2
2
dr / 4
dr
A
b
Ans.
Tr T d r / 2 16T
Ans.
J
d r4 / 32 d r3
(c) The bending stress causes compression in the x direction. The axial stress causes
compression in the y direction. The torsional stress shears across the y face in the negative z
direction.
t
(d) Analyze the stress element from part (c) using the equations developed in parts (a) and (b).
d r d p 1.5 0.25 1.25 in
x b
6 1500
6F
4584 psi = 4.584 kpsi
d r nt p
1.25 2 0.25
y a
4 1500
4F
=
= 1222 psi = 1.222 kpsi
d r2
1.252
yz t
16 235
16T
=
= 612.8 psi = 0.6128 kpsi
3
dr
1.253
Use Eq. (3-15) for the three-dimensional stress element.
2
2
3 4.584 1.222 2 4.584 1.222 0.6128 4.584 0.6128 0
5.806 5.226 1.721 0
3
2
The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are
1 = 0.2543 kpsi, 2 = –1.476 kpsi, and 3 = – 4.584 kpsi.
Ans.
From Eq. (3-16), the principal shear stresses are
Chapter 3 - Rev. A, Page 74/100
0.2543 1.476
0.8652 kpsi
Ans.
2
2
3 1.476 4.584
2/3 2
1.554 kpsi
Ans.
2
2
0.2543 4.584
1/3 1 3
2.419 kpsi
Ans.
2
2
____________________________________________________________________________
1/2
3-91
1 2
As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = r i .
Therefore, from Eq. (3-50)
ri2 pi ro2
1
ro2 ri2 ri2
r2 r2
pi o2 i2 Ans.
r r
o i
t ,max
ri2 pi ro2
1 pi Ans.
ro2 ri2 ri2
______________________________________________________________________________
r ,max
3-92
If p i = 0, Eq. (3-49) becomes
t
po ro2 ri 2 ro2 po / r 2
ro2 ri 2
po ro2 ri 2
1
ro2 ri 2 r 2
The maximum tangential stress occurs at r = r i . So
t ,max
2 po ro2
Ans.
ro2 ri 2
For σ r , we have
r
po ro2 ri 2 ro2 po / r 2
ro2 ri 2
po ro2 ri 2
2 2 2 1
ro ri r
So σ r = 0 at r = r i . Thus at r = r o
po ro2 ri 2 ro2
r ,max 2 2 2 po Ans.
ro ri ro
______________________________________________________________________________
Chapter 3 - Rev. A, Page 75/100
3-93
The force due to the pressure on half of the sphere is resisted by the stress that is
distributed around the center plane of the sphere. All planes are the same, so
p / 4 di2 pdi
( t )av 1 2
di t
4t
Ans.
The radial stress on the inner surface of the shell is, 3 = p Ans.
______________________________________________________________________________
3-94
σt > σl > σr
τ max = (σ t − σ r )/2 at r = r i
max
ro2 pi
1 ri 2 pi ro2 ri 2 pi ro2
2 2 1 2 2 2 1 2 2 2
2 ro ri ri ro ri ri ro ri
ro2 ri 2
32 2.752
(10 000) 1597 psi Ans.
max
ro 2
32
______________________________________________________________________________
3-95
pi
σt > σl > σr
τ max = (σ t − σ r )/2 at r = r i
r2 p r2
r2 p
1 r 2 p r 2 r 2 p r 2
max 2i i 2 1 o2 2i i 2 1 o2 2i i 2 o2 2o i 2
2 ro ri ri ro ri ri ro ri ri ro ri
ri ro
( max pi )
max
100
(25 4)106
91.7 mm
25 106
t ro ri 100 91.7 8.3 mm Ans.
______________________________________________________________________________
3-96
σt > σl > σr
τ max = (σ t − σ r )/2 at r = r i
ri 2 pi ro2
ro2 pi
1 ri 2 pi ro2 ri 2 pi ro2
1
1
2 ro2 ri 2 ri 2 ro2 ri 2 ri 2 ro2 ri 2 ri 2 ro2 ri 2
max
42 (500)
2
4129 psi Ans.
4 3.752
______________________________________________________________________________
3-97
From Eq. (3-49) with p i = 0,
Chapter 3 - Rev. A, Page 76/100
t
ro2 po ri 2
1
ro2 ri 2 r 2
ro2 po ri 2
1
ro2 ri 2 r 2
σ t > σ l > σ r , and since σ t and σ r are negative,
τ max = (σ r − σ t )/2 at r = r o
r
ro2 po ri 2 ri 2 po
1 ro2 po ri 2 ro2 po ri 2
1
1
2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2
max
ro2 ri 2
32 2.752
(10 000) 1900 psi Ans.
max
ri 2
2.752
______________________________________________________________________________
3-98
po
From Eq. (3-49) with p i = 0,
r2 p r2
t 2o o 2 1 i 2
ro ri r
r
ro2 po ri 2
1
ro2 ri 2 r 2
σ t > σ l > σ r, and since σ t and σ r are negative,
τ max = (σ r − σ t )/2 at r = r o
ro2 po ri 2 ri 2 po
1 ro2 po ri 2 ro2 po ri 2
1
1
2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2
max
ri ro
max
( max po )
100
25 106
25 4 106
92.8 mm
t ro ri 100 92.8 7.2 mm Ans.
______________________________________________________________________________
3-99
From Eq. (3-49) with p i = 0,
t
ro2 po ri 2
1
ro2 ri 2 r 2
ro2 po ri 2
r 2 2 1 2
ro ri r
σ t > σ l > σ r , and since σ t and σ r are negative,
τ max = (σ r − σ t )/2 at r = r o
Chapter 3 - Rev. A, Page 77/100
ro2 po ri 2 ri 2 po
1 ro2 po ri 2 ro2 po ri 2
1
1
2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2 ro 2 ro2 ri 2
max
3.752 (500)
3629 psi Ans.
42 3.752
______________________________________________________________________________
3-100 From Table A-20, S y =490 MPa
From Eq. (3-49) with p i = 0,
ro2 po ri 2
t 2 2 1 2
ro ri r
Maximum will occur at r = r i
0.8(490) 25 19
(r r )
2ro2 po
82.8 MPa Ans.
t ,max 2 2 po t ,max o2 i
2ro
2(252 )
ro ri
______________________________________________________________________________
2
2
2
2
3-101 From Table A-20, S y = 71 kpsi
From Eq. (3-49) with p i = 0,
t
ro2 po ri 2
1
ro2 ri 2 r 2
Maximum will occur at r = r i
t ,max ro ri
0.8(71) 1 0.75
2r 2 p
12.4 kpsi Ans.
t ,max 2 o o2 po
2
2ro
2(12 )
ro ri
______________________________________________________________________________
2
2
2
2
3-102 From Table A-20, S y =490 MPa
From Eq. (3-50)
t
ri 2 pi ro2
1
ro2 ri 2 r 2
Maximum will occur at r = r i
t ,max
2
2
ri 2 pi ro2 pi ro ri
2 2 1 2
ro ri ri
ro2 ri 2
pi
t ,max (ro2 ri 2 )
0.8(490) (252 192 )
105 MPa
Ans.
ro2 ri 2
(252 192 )
______________________________________________________________________________
Chapter 3 - Rev. A, Page 78/100
3-103 From Table A-20, S y =71 MPa
From Eq. (3-50)
ri 2 pi ro2
t 2 2 1 2
ro ri r
Maximum will occur at r = r i
ri 2 pi ro2 pi (ro2 ri 2 )
t ,max 2 2 1 2
ro ri ri
ro2 ri 2
pi
t ,max (ro2 ri 2 )
0.8(71) (12 0.752 ) 15.9 ksi
Ans.
ro2 ri 2
(12 0.752 )
______________________________________________________________________________
3-104 The longitudinal stress will be due to the weight of the vessel above the maximum stress
point. From Table A-5, the unit weight of steel is s = 0.282 lbf/in3. The area of the wall
is
A wall = ( /4)(3602 358.52) = 846. 5 in2
The volume of the wall and dome are
V wall = A wall h = 846.5 (720) = 609.5 (103) in3
V dome = (2 /3)(1803 179.253) = 152.0 (103) in3
The weight of the structure on the wall area at the tank bottom is
W = s V total = 0.282(609.5 +152.0) (103) = 214.7(103) lbf
214.7 103
W
l
254 psi
846.5
Awall
The maximum pressure will occur at the bottom of the tank, p i = water h. From Eq. (3-50)
with r ri
ro2 ri 2
ri 2 pi ro2
p
1
i 2
2
ro2 ri 2 ri 2
ro ri
1 ft 2 1802 179.252
62.4(55)
5708 5710 psi Ans.
2
2
2
144
in
180
179.25
2
2
1 ft 2
r p r
r 2i i 2 1 o2 pi 62.4(55)
23.8 psi Ans.
2
ro ri ri
144 in
Note: These stresses are very idealized as the floor of the tank will restrict the values
calculated.
t
Chapter 3 - Rev. A, Page 79/100
Since 1 2 3 , 1 = t = 5708 psi, 2 = r = 24 psi and 3 = l = 254 psi.
From Eq. (3-16),
5708 254
2981 2980 psi
2
5708 24
1 2
2866 2870 psi
Ans.
2
24 254
2 3
115 psi
2
______________________________________________________________________________
1 3
3-105 Stresses from additional pressure are,
Eq. (3-51),
50 179.252
5963 psi
l 50psi
1802 179.252
( r ) 50 psi = 50 psi
Eq. (3-50)
1802 179.252
11 975 psi
t 50psi 50 2
180 179.252
Adding these to the stresses found in Prob. 3-104 gives
t = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans.
r = 23.8 50 = 73.8 psi Ans.
l = 254 + 5963 = 5709 psi
Ans.
Note: These stresses are very idealized as the floor of the tank will restrict the values
calculated.
From Eq. (3-16)
17 683 73.8
1 3
8879 psi
2
17 683 5709
1 2
5987 psi
Ans.
2
5709 23.8
2 3
2866 psi
2
______________________________________________________________________________
3-106 Since σ t and σ r are both positive and σ t > σ r
max t max 2
From Eq. (3-55), t is maximum at r = r i = 0.3125 in. The term
Chapter 3 - Rev. A, Page 80/100
0.282 2 5000 3 0.292
82.42 lbf/in
60
8
386
0.31252 2.752 1 3(0.292)
2
2
2
82.42 0.3125 2.75
0.3125
3 0.292
0.31252
1260 psi
2
3
8
2
t max
max
1260
630 psi
2
Ans.
r 2r 2
Radial stress:
r k ri2 ro2 i 2o r 2
r
Maxima:
r 2r 2
d r
k 2 i 3o 2r 0 r ri ro 0.3125(2.75) 0.927 in
dr
r
0.31252 2.752
2
2
2
0.927
r max 82.42 0.3125 2.75
0.927 2
490 psi Ans.
______________________________________________________________________________
3-107 = 2 (2000)/60 = 209.4 rad/s, = 3320 20 kg/m3, = 0.24, r i = 0.01 m, r o = 0.125 m
Using Eq. (3-55)
2
3 0.24
2
2 1 3(0.24)
0.012 (10)6
0.01 (0.125) (0.125)
3 0.24
8
1.85 MPa Ans.
______________________________________________________________________________
t 3320(209.4)2
3-108 = 2 (12 000)/60 = 1256.6 rad/s,
5 /16
6.749 104 lbf s 2 / in 4
2
2
386 1 16 4 5 0.75
The maximum shear stress occurs at bore where max = t /2. From Eq. (3-55)
1 3(0.20)
2 3 0.20
2
2
2
( t ) max 6.749(104 ) 1256.6
(0.375) 2
0.375 2.5 2.5
3 0.20
8
5360 psi
Chapter 3 - Rev. A, Page 81/100
max = 5360 / 2 = 2680 psi
Ans.
______________________________________________________________________________
3-109 = 2 (3500)/60 = 366.5 rad/s,
mass of blade = m = V = (0.282 / 386) [1.25(30)(0.125)] = 3.425(103) lbfs2/in
F = (m/2) 2r
= [3.425(103)/2]( 366.52)(7.5)
= 1725 lbf
A nom = (1.25 0.5)(1/8) = 0.093 75 in2
nom = F/ A nom = 1725/0.093 75 = 18 400 psi Ans.
Note: Stress concentration Fig. A-15-1 gives K t = 2.25 which increases σ max and fatigue.
______________________________________________________________________________
3-110 = 0.292, E = 207 GPa, r i = 0, R = 25 mm, r o = 50 mm
Eq. (3-57),
207(109 )
p
2(0.025)3
(0.052 0.0252 )(0.0252 0) 9
3
10 3.105(10 )
2
(0.05 0)
where p is in MPa and is in mm.
(1)
Maximum interference,
1
max [50.042 50.000] 0.021 mm Ans.
2
Minimum interference,
1
min [50.026 50.025] 0.0005 mm Ans.
2
From Eq. (1)
p max = 3.105(103)(0.021) = 65.2 MPa
Ans.
p min = 3.105(103)(0.0005) = 1.55 MPa Ans.
______________________________________________________________________________
3-111 = 0.292, E = 30 Mpsi, r i = 0, R = 1 in, r o = 2 in
Eq. (3-57),
30(106 ) (22 12 )(12 0)
7
p
1.125(10 )
3
2
2(1 )
(2 0)
(1)
where p is in psi and is in inches.
Maximum interference,
Chapter 3 - Rev. A, Page 82/100
1
2
Minimum interference,
max [2.0016 2.0000] 0.0008 in
1
2
min [2.0010 2.0010] 0
Ans.
Ans.
From Eq. (1),
p max = 1.125(107)(0.0008) = 9 000 psi
Ans.
p min = 1.125(107)(0) = 0 Ans.
______________________________________________________________________________
3-112 = 0.292, E = 207 GPa, r i = 0, R = 25 mm, r o = 50 mm
Eq. (3-57),
207(109 ) (0.052 0.0252 )(0.0252 0) 9
3
10 3.105(10 )
3
2
2(0.025)
(0.05 0)
where p is in MPa and is in mm.
p
(1)
Maximum interference,
1
max [50.059 50.000] 0.0295 mm Ans.
2
Minimum interference,
1
min [50.043 50.025] 0.009 mm Ans.
2
From Eq. (1)
p max = 3.105(103)(0.0295) = 91.6 MPa
Ans.
p min = 3.105(103)(0.009) = 27.9 MPa Ans.
______________________________________________________________________________
3-113 = 0.292, E = 30 Mpsi, r i = 0, R = 1 in, r o = 2 in
Eq. (3-57),
30(106 ) (22 12 )(12 0)
7
p
1.125(10 )
3
2
2(1 )
(2 0)
(1)
where p is in psi and is in inches.
Maximum interference,
1
max [2.0023 2.0000] 0.00115 in
2
Minimum interference,
Ans.
Chapter 3 - Rev. A, Page 83/100
1
2
min [2.0017 2.0010] 0.00035
Ans.
From Eq. (1),
p max = 1.125(107)(0.00115) = 12 940 psi
p min = 1.125(107)(0.00035) = 3 938
Ans.
Ans.
______________________________________________________________________________
3-114 = 0.292, E = 207 GPa, r i = 0, R = 25 mm, r o = 50 mm
Eq. (3-57),
207(109 )
2(0.025)3
(0.052 0.0252 )(0.0252 0) 9
3
10 3.105(10 )
2
(0.05 0)
where p is in MPa and is in mm.
p
(1)
Maximum interference,
1
max [50.086 50.000] 0.043 mm Ans.
2
Minimum interference,
1
min [50.070 50.025] 0.0225 mm Ans.
2
From Eq. (1)
p max = 3.105(103)(0.043) = 134 MPa
Ans.
p min = 3.105(103)(0.0225) = 69.9 MPa Ans.
______________________________________________________________________________
3-115 = 0.292, E = 30 Mpsi, r i = 0, R = 1 in, r o = 2 in
Eq. (3-57),
30(106 ) (22 12 )(12 0)
7
p
1.125(10 )
3
2
2(1 )
(2 0)
(1)
where p is in psi and is in inches.
Maximum interference,
1
max [2.0034 2.0000] 0.0017 in
2
Minimum interference,
1
2
min [2.0028 2.0010] 0.0009
Ans.
Ans.
From Eq. (1),
Chapter 3 - Rev. A, Page 84/100
p max = 1.125(107)(0.0017) = 19 130 psi
Ans.
p min = 1.125(107)(0.0009) = 10 130 Ans.
______________________________________________________________________________
3-116 From Table A-5, E i = E o = 30 Mpsi, i o r i = 0, R = 1 in, r o = 1.5 in
1
The radial interference is 2.002 2.000 0.001 in Ans.
2
Eq. (3-57),
2
2
2
2
E ro R R ri
p
2 R3
ro2 ri 2
8333 psi 83.3 kpsi
30 10 0.001 1.5 1 1 0
2 1
1.5 0
6
2
3
2
2
2
Ans.
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
R2 r 2
12 02
( t )i r R p 2 i 2 (8333) 2 2 8333 psi 8.33 kpsi Ans.
R ri
1 0
ro2 R 2
1.52 12
(8333)
21 670 psi 21.7 kpsi Ans.
rR
ro2 R 2
1.52 12
______________________________________________________________________________
( t )o
p
3-117 From Table A-5, E i = 30 Mpsi, E o =14.5 Mpsi, i o
r i = 0, R = 1 in, r o = 1.5 in
1
The radial interference is 2.002 2.000 0.001 in Ans.
2
Eq. (3-56),
p
p
1 r 2 R2
1
R o2
o
2
Ei
Eo ro R
R 2 ri 2
2 2 i
R ri
0.001
1.52 12
12 02
1
1
1
0.211
0.292
2
2
2
2
6
6
1
0
30
10
14.5 10 1.5 1
4599 psi
Ans.
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
R2 r 2
12 02
( t )i r R p 2 i 2 (4599) 2 2 4599 psi Ans.
R ri
1 0
Chapter 3 - Rev. A, Page 85/100
ro2 R 2
1.52 12
( t )o r R p 2
(4599)
11 960 psi Ans.
ro R 2
1.52 12
______________________________________________________________________________
3-118 From Table A-5, E i = E o = 30 Mpsi, i o r i = 0, R = 0.5 in, r o = 1 in
The minimum and maximum radial interferences are
1
min 1.002 1.002 0.000 in Ans.
2
1
max 1.003 1.001 0.001 in Ans.
2
Since the minimum interference is zero, the minimum pressure and tangential stresses are
zero.
Ans.
The maximum pressure is obtained from Eq. (3-57).
2
2
2
2
E ro R R ri
p 3
ro2 ri 2
2R
p
30 106 0.001 12 0.52 0.52 0
2 0.53
12 0
22 500 psi
Ans
The maximum tangential stresses at the interface for the inner and outer members are
given by Eqs. (3-58) and (3-59), respectively.
R2 r 2
0.52 02
( t )i r R p 2 i 2 (22 500) 2 2 22 500 psi Ans.
R ri
0.5 0
ro2 R 2
12 0.52
( t )o r R p 2
(22 500) 2
37 500 psi Ans.
ro R 2
1 0.52
______________________________________________________________________________
3-119 From Table A-5, E i = 10.4 Mpsi, E o =30 Mpsi, i o
r i = 0, R = 1 in, r o = 1.5 in
The minimum and maximum radial interferences are
1
min [2.003 2.002] 0.0005 in Ans.
2
1
max [2.006 2.000] 0.003 in Ans.
2
Eq. (3-56),
Chapter 3 - Rev. A, Page 86/100
p
p
1 r 2 R2
1
R o2
o
2
Ei
Eo ro R
R 2 ri 2
2 2 i
R ri
1
1.52 12
12 02
1
1
0.292
0.333
2
2
2
2
6
6
1
0
10.4
10
30 10 1.5 1
p 6.229 10 psi
p 6.229 10
p 6.229 10
6
6
min
min
6
max
max
Ans.
6.229 106 0.0005 3114.6 psi 3.11 kpsi
6.229 106 0.003 18 687 psi 18.7 kpsi
Ans.
Ans.
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
Minimum interference:
R2 r 2
12 02
( t )i min pmin 2 i2 (3.11) 2 2 3.11 kpsi Ans.
R ri
1 0
( t )o
min
pmin
ro2 R 2
1.52 12
(3.11)
8.09 kpsi Ans.
ro2 R 2
1.52 12
Maximum interference:
R 2 ri 2
12 02
( t )i max pmax 2 2 (18.7) 2 2 18.7 kpsi Ans.
R ri
1 0
ro2 R 2
1.52 12
( t )o max pmax 2
(18.7) 2 2 48.6 kpsi Ans.
ro R 2
1.5 1
______________________________________________________________________________
3-120 d 20 mm, ri 37.5 mm, ro 57.5 mm
From Table 3-4, for R = 10 mm,
rc 37.5 10 47.5 mm
rn
102
2 47.5 47.5 10
2
2
46.96772 mm
e rc rn 47.5 46.96772 0.53228 mm
ci rn ri 46.9677 37.5 9.4677 mm
co ro rn 57.5 46.9677 10.5323 mm
A d 2 / 4 (20)2 / 4 314.16 mm 2
M Frc 4000(47.5) 190 000 N mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
Chapter 3 - Rev. A, Page 87/100
F Mci
4000
190 000(9.4677)
300 MPa
A Aeri 314.16 314.16(0.53228)(37.5)
i
Ans.
F Mco
4000
190 000(10.5323)
195 MPa Ans.
A Aero 314.16 314.16(0.53228)(57.5)
______________________________________________________________________________
o
3-121 d 0.75 in, ri 1.25 in, ro 2.0 in
From Table 3-4, for R = 0.375 in,
rc 1.25 0.375 1.625 in
rn
0.3752
2 1.625 1.6252 0.3752
1.60307 in
e rc rn 1.625 1.60307 0.02193 in
ci rn ri 1.60307 1.25 0.35307 in
co ro rn 2.0 1.60307 0.39693 in
A d 2 / 4 (0.75) 2 / 4 0.44179 in 2
M Frc 750(1.625) 1218.8 lbf in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
F Mci
750
1218.8(0.35307)
37 230 psi 37.2 kpsi
A Aeri 0.44179 0.44179(0.02193)(1.25)
i
Ans.
F Mco
750
1218.8(0.39693)
23 269 psi 23.3 kpsi Ans.
A Aero 0.44179 0.44179(0.02193)(2.0)
______________________________________________________________________________
o
3-122 d 6 mm, ri 10 mm, ro 16 mm
From Table 3-4, for R = 3 mm,
rc 10 3 13 mm
rn
32
2 13 132 32
12.82456 mm
e rc rn 13 12.82456 0.17544 mm
ci rn ri 12.82456 10 2.82456 mm
co ro rn 16 12.82456 3.17544 mm
A d 2 / 4 (6)2 / 4 28.2743 mm 2
M Frc 300(13) 3900 N mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
Chapter 3 - Rev. A, Page 88/100
F Mci
300
3900(2.82456)
233 MPa
A Aeri 28.2743 28.2743(0.17544)(10)
i
Ans.
F Mco
300
3900(3.17544)
145 MPa Ans.
A Aero 28.2743 28.2743(0.17544)(16)
______________________________________________________________________________
o
3-123
d 6 mm, ri 10 mm, ro 16 mm
From Table 3-4, for R = 3 mm,
rc 10 3 13 mm
rn
32
2 13 132 32
12.82456 mm
e rc rn 13 12.82456 0.17544 mm
ci rn ri 12.82456 10 2.82456 mm
co ro rn 16 12.82456 3.17544 mm
A d 2 / 4 (6)2 / 4 28.2743 mm 2
The angle of the line of radius centers is
Rd /2
1 10 6 / 2
sin 1
sin
30
Rd R
10 6 10
M F R d / 2 sin 300 10 6 / 2 sin 30 1950 N mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
F sin Mci 300sin 30
1950(2.82456)
116 MPa
A
Aeri
28.2743
28.2743(0.17544)(10)
i
Ans.
F sin Mco 300sin 30
1950(3.17544)
72.7 MPa Ans.
A
Aero
28.2743
28.2743(0.17544)(16)
Note that the shear stress due to the shear force is zero at the surface.
______________________________________________________________________________
o
3-124 d 0.25 in, ri 0.5 in, ro 0.75 in
From Table 3-4, for R = 0.125 in,
rc 0.5 0.125 0.625 in
rn
0.1252
2 0.625 0.6252 0.1252
0.618686 in
e rc rn 0.625 0.618686 0.006314 in
ci rn ri 0.618686 0.5 0.118686 in
co ro rn 0.75 0.618686 0.131314 in
Chapter 3 - Rev. A, Page 89/100
A d 2 / 4 (0.25) 2 / 4 0.049087 in 2
M Frc 75(0.625) 46.875 lbf in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
i
F Mci
75
46.875(0.118686)
37 428 psi 37.4 kpsi
A Aeri 0.049087 0.049087(0.006314)(0.5)
Ans.
F Mco
75
46.875(0.131314)
24 952 psi 25.0 kpsi Ans.
A Aero 0.049087 0.049087(0.006314)(0.75)
______________________________________________________________________________
o
3-125 d 0.25 in, ri 0.5 in, ro 0.75 in
From Table 3-4, for R = 0.125 in,
rc 0.5 0.125 0.625 in
rn
0.1252
2 0.625 0.6252 0.1252
0.618686 in
e rc rn 0.625 0.618686 0.006314 in
ci rn ri 0.618686 0.5 0.118686 in
co ro rn 0.75 0.618686 0.131314 in
A d 2 / 4 (0.25) 2 / 4 0.049087 in 2
The angle of the line of radius centers is
Rd /2
1 0.5 0.25 / 2
sin 1
sin
30
R
d
R
0.5
0.25
0.5
M F R d / 2 sin 75 0.5 0.25 / 2 sin 30 23.44 lbf in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
i
F sin Mci 75sin 30
23.44(0.118686)
18 716 psi 18.7 kpsi
A
Aeri 0.049087 0.049087(0.006314)(0.5)
Ans.
F sin Mco 75sin 30
23.44(0.131314)
12 478 psi 12.5 kpsi Ans.
A
Aero 0.049087 0.049087(0.006314)(0.75)
Note that the shear stress due to the shear force is zero at the surface.
______________________________________________________________________________
o
3-126
(a)
3(4) 0.5(0.1094) 8021 psi 8.02 kpsi
Mc
I
(0.75) 0.10943 /12
Ans.
(b) r i = 0.125 in, r o = r i + h = 0.125 + 0.1094 = 0.2344 in
From Table 3-4,
Chapter 3 - Rev. A, Page 90/100
rc 0.125 (0.5)(0.1094) 0.1797 in
0.1094
0.174006 in
ln(0.2344 / 0.125)
e rc rn 0.1797 0.174006 0.005694 in
rn
ci rn ri 0.174006 0.125 0.049006 in
co ro rn 0.2344 0.174006 0.060394 in
A bh 0.75(0.1094) 0.08205 in 2
M 3(4) 12 lbf in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
i
Mci
12(0.049006)
10 070 psi 10.1 kpsi
Aeri 0.08205(0.005694)(0.125)
o
Mco
12(0.060394)
6618 psi 6.62 kpsi
0.08205(0.005694)(0.2344)
Aero
i 10.1
1.26
8.02
6.62
Ko o
0.825
8.02
(c) K i
Ans.
Ans.
Ans.
Ans.
______________________________________________________________________________
3-127
(a)
3(4)0.5(0.1406) 4856 psi 4.86 kpsi
Mc
I
(0.75) 0.14063 /12
Ans.
(b) r i = 0.125 in, r o = r i + h = 0.125 + 0.1406 = 0.2656 in
From Table 3-4,
rc 0.125 (0.5)(0.1406) 0.1953 in
0.1406
0.186552 in
ln(0.2656 / 0.125)
e rc rn 0.1953 0.186552 0.008748 in
rn
ci rn ri 0.186552 0.125 0.061552 in
co ro rn 0.2656 0.186552 0.079048 in
A bh 0.75(0.1406) 0.10545 in 2
M 3(4) 12 lbf in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
Chapter 3 - Rev. A, Page 91/100
i
Mci
12(0.061552)
6406 psi 6.41 kpsi
Aeri 0.10545(0.008748)(0.125)
o
Ans.
Mco
12(0.079048)
3872 psi 3.87 kpsi
0.10545(0.008748)(0.2656)
Aero
Ans.
i 6.41
Ans.
1.32
4.86
3.87
Ko o
0.80
Ans.
4.86
______________________________________________________________________________
(c) K i
3-128
(a)
3(4)0.5(0.1094) 8021 psi 8.02 kpsi
Mc
I
(0.75) 0.10943 /12
Ans.
(b) r i = 0.25 in, r o = r i + h = 0.25 + 0.1094 = 0.3594 in
From Table 3-4,
rc 0.25 (0.5)(0.1094) 0.3047 in
0.1094
0.301398 in
ln(0.3594 / 0.25)
e rc rn 0.3047 0.301398 0.003302 in
rn
ci rn ri 0.301398 0.25 0.051398 in
co ro rn 0.3594 0.301398 0.058002 in
A bh 0.75(0.1094) 0.08205 in 2
M 3(4) 12 lbf in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
i
Mci
12(0.051398)
9106 psi 9.11 kpsi
Aeri 0.08205(0.003302)(0.25)
o
Ans.
Mco
12(0.058002)
7148 psi 7.15 kpsi
0.08205(0.003302)(0.3594)
Aero
Ans.
i 9.11
Ans.
1.14
8.02
7.15
Ko o
0.89
Ans.
8.02
______________________________________________________________________________
(c) K i
3-129 r i = 25 mm, r o = r i + h = 25 + 87 = 112 mm, r c = 25 + 87/2 = 68.5 mm
The radius of the neutral axis is found from Eq. (3-63), given below.
Chapter 3 - Rev. A, Page 92/100
rn
A
dA / r
For a rectangular area with constant width b, the denominator is
ro bdr
r
ri r b ln roi
Applying this equation over each of the four rectangular areas,
dA
45
54.5
92 112
9 ln 31 ln
31 ln
9 ln
16.3769
r
45
25
82.5 92
A 2 20(9) 31(9.5) 949 mm2
rn
A
dA / r
949
57.9475 mm
16.3769
e rc rn 68.5 57.9475 10.5525 mm
ci rn ri 57.9475 25 32.9475 mm
co ro rn 112 57.9475 54.0525 mm
M = 150F 2 = 150(3.2) = 480 kN·mm
We need to find the forces transmitted through the section in order to determine the axial
stress. It is not immediately obvious which plane should be used for resolving the axial
versus shear directions. It is convenient to use the plane containing the reaction force at
the bushing, which assumes its contribution resolves entirely into shear force. To find the
angle of this plane, find the resultant of F 1 and F 2 .
Fx F1x F2 x 2.4 cos 60 3.2 cos 0 4.40 kN
Fy F1 y F2 y 2.4sin 60 3.2sin 0 2.08 kN
F 4.402 2.082
12
4.87 kN
This is the pin force on the lever which acts in a direction
tan 1
Fy
Fx
tan 1
2.08
25.3
4.40
On the surface 25.3° from the horizontal, find the internal forces in the tangential and
normal directions. Resolving F 1 into components,
Ft 2.4 cos 60 25.3 1.97 kN
Fn 2.4sin 60 25.3 1.37 kN
The transverse shear stress is zero at the inner and outer surfaces. Using Eq. (3-65) for
the bending stress, and combining with the axial stress due to F n ,
Chapter 3 - Rev. A, Page 93/100
i
Fn Mci 1370 3200 150 (32.9475)
64.6 MPa
A Aeri 949
949(10.5525)(25)
Ans.
Fn Mco 1370 3200 150 (54.0525)
21.7 MPa Ans.
A Aero 949
949(10.5525)(112)
______________________________________________________________________________
o
3-130 r i = 2 in, r o = r i + h = 2 + 4 = 6 in, rc 2 0.5(4) 4 in
A (6 2 0.75)(0.75) 2.4375 in 2
Similar to Prob. 3-129,
dA
3.625
6
0.75ln
0.75ln
0.682 920 in
r
2
4.375
A
2.4375
rn
3.56923 in
0.682
920
(
/
)
dA
r
e rc rn 4 3.56923 0.43077 in
ci rn ri 3.56923 2 1.56923 in
co ro rn 6 3.56923 2.43077 in
M Frc 6000(4) 24 000 lbf in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
i
F Mci
6000
24 000(1.56923)
20 396 psi 20.4 kpsi
A Aeri 2.4375 2.4375(0.43077)(2)
Ans.
F Mco
6000
24 000(2.43077)
6 799 psi 6.80 kpsi Ans.
A Aero 2.4375 2.4375(0.43077)(6)
______________________________________________________________________________
o
3-131
r i = 12 in, r o = r i + h = 12 + 3 = 15 in, r c = 12 + 3/2 = 13.5 in
I a 3b (1.53 )(0.75) 1.988 in 4
4
4
A ab (1.5)(0.75) 3.534
M 20(3 1.5) 90 kip in
Since the radius is large compared to the cross section, assume Eq. 3-67 is applicable for
the bending stress. Combining the bending stress and the axial stress,
F Mci rc
20
90(1.5)(13.5)
82.1 kpsi Ans.
A
Iri
3.534 (1.988)(12)
F Mc r
20
90(1.5)(13.5)
o o c
55.5 kpsi Ans.
A
Iro
3.534
1.988(15)
i
______________________________________________________________________________
Chapter 3 - Rev. A, Page 94/100
3-132
r i = 1.25 in, r o = r i + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in
r c = (r i + r o ) / 2 = (1.25 + 3.25)/2 = 2.25 in Ans.
ro
dA
For outer rectangle,
b ln
ri
r
A
r2
, A r2
For circle,
O
2
2
dA r
2
r
r
r
c
c
O
O
dA
2 (rc rc2 r 2 )
r O
Combine the integrals subtracting the circle from the rectangle
dA
3.25
1.25ln
2 2.25 2.252 0.52 0.840 904 in
r
1.25
A 1.25(2) (0.52 ) 1.714 60 in 2
Ans.
A
1.714 60
rn
2.0390 in Ans.
(dA / r ) 0.840 904
e rc rn 2.25 2.0390 0.2110 in Ans.
ci rn ri 2.0390 1.25 0.7890 in
co ro rn 3.25 2.0390 1.2110 in
M 2000(4.5 1.25 0.5 0.5) 13 500 lbf in
F Mci
2000
13 500(0.7890)
i
20 720 psi = 20.7 kpsi Ans.
A Aeri 1.7146 1.7146(0.2110)(1.25)
F Mco
2000
13 500(1.2110)
o
12 738 psi 12.7 kpsi Ans.
A Aero 1.7146 1.7146(0.2110)(3.25)
______________________________________________________________________________
3-133 From Eq. (3-68),
2 1 2 E
1 3 3
F
2 1 d
8
13
a KF 1 3
Use 0.292, F in newtons, E in N/mm2 and d in mm, then
1/3
3 [(1 0.2922 ) / 207 000]
K
1/ 30
8
0.03685
From Eq. (3-69),
pmax
3F
3F
3F 1/3
3F 1/3
352 F 1/3 MPa
2 a 2 2 ( KF 1/3 ) 2 2 K 2 2 (0.03685)2
Chapter 3 - Rev. A, Page 95/100
From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is
equal to – p max .
max z pmax 352 F 1/3 MPa
Ans.
From Fig. 3-37,
max 0.3 pmax 106 F 1/3 MPa
Ans.
______________________________________________________________________________
3-134 From Eq. (3-68),
2
2
3F 1 1 E1 1 2 E2
a
1 d1 1 d 2
8
3
3 10 1 0.292
a
8
3
2
207 000 1 0.333 71 700 0.0990 mm
2
1 25 1 40
From Eq. (3-69),
3 10
3F
pmax
487.2 MPa
2
2 a
2 0.09902
From Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a.
z 0.48a 0.48 0.0990 0.0475 mm
Ans.
The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.
1 2 487.2 1 0.48 tan 1 1/ 0.48 1 0.333
3
1
101.3 MPa
2 1 0.482
487.2
396.0 MPa
1 0.482
From Eq. (3-72),
101.3 396.0 147.4 MPa Ans.
max 1 3
2
2
Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow
for calculating and plotting the stresses versus the depth for specific values of . For =
0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with max
= 0.3025 p max .
Chapter 3 - Rev. A, Page 96/100
This gives max = 0.3025 p max = (0.3025)(487.2) = 147.38 MPa. Even though the depth
assumption was a little off, it did not have significant effect on the the maximum shear
stress.
______________________________________________________________________________
3-135 From the solution to Prob. 3-134, a = 0.0990 mm and p max = 487.2 MPa. Assuming
applicability of Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a =
0.0475 mm. Ans.
The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.
1 2 487.2 1 0.48 tan 1 1/ 0.48 1 0.292
3
1
92.09 MPa
2 1 0.482
487.2
396.0 MPa
1 0.482
From Eq. (3-72),
92.09 396.0 152.0 MPa Ans.
max 1 3
2
2
Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow
for calculating and plotting the stresses versus the depth for specific values of . For =
0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with max =
0.3119 p max .
______________________________________________________________________________
3-136 From Eq. (3-68),
2 1 2 E
F
3
a3
8 1 d1 1 d 2
2
3 20 2 1 0.292 207 000
0.1258 mm
a3
1 30 1
8
From Eq. (3-69),
3 20
3F
pmax
603.4 MPa
2
2 a
2 0.12582
From Fig. 3-37, the maximum shear stress occurs at a depth of
z 0.48a 0.48 0.1258 0.0604 mm
Ans.
Also from Fig. 3-37, the maximum shear stress is
max 0.3 pmax 0.3(603.4) 181 MPa
Ans.
Chapter 3 - Rev. A, Page 97/100
______________________________________________________________________________
3-137 Aluminum Plate-Ball interface: From Eq. (3-68),
2
2
3F 1 1 E1 1 2 E2
a3
1 d1 1 d 2
8
3F 1 0.292
a
8
2
3
30 10 1 0.333 10.4 10
6
2
6
3.517 103 F 1/3 in
1 11
From Eq. (3-69),
3F
3F
3.860 104 F 1/3 psi
pmax
2
2
3
1/3
2 a
2 3.517 10 F
By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference
in the maximum shear stress for the plate and the ball will be due to poisson’s ratio in Eq.
(3-70). The larger poisson’s ratio will create the greater maximum shear stress, so the
aluminum plate will be the critical element in this interface. Applying the equations for
the aluminum plate,
1 3.86 104 F 1/3 1 0.48 tan 1 1/ 0.48 1 0.333
3
3.86 104 F 1/3
1 0.48
From Eq. (3-72),
max
1 3
2
1
8025 F 1/3 psi
2
2 1 0.48
3.137 104 F 1/3 psi
8025F 3.137 10 F
1.167 10 F
2
1/3
4
1/3
4
1/3
2
Comparing this stress to the allowable stress, and solving for F,
psi
3
20 000
5.03 lbf
F
4
1.167 10
Table-Ball interface: From Eq. (3-68),
3F 1 0.292
a3
8
2
30 10 1 0.211 14.5 10
6
1 11
2
6
3.306 103 F 1/3 in
From Eq. (3-69),
Chapter 3 - Rev. A, Page 98/100
3F
3F
4.369 104 F 1/3 psi
2
2
3
1/3
2 a
2 3.306 10 F
The steel ball has a higher poisson’s ratio than the cast iron table, so it will dominate.
pmax
1 4.369 104 F 1/3 1 0.48 tan 1 1/ 0.48 1 0.292
3
4.369 104 F 1/3
1 0.48
From Eq. (3-72),
max
1 3
2
1
8258 F 1/3 psi
2
2 1 0.48
3.551104 F 1/3 psi
8258F 3.55110 F
1.363 10 F
2
1/3
4
1/3
4
1/3
2
Comparing this stress to the allowable stress, and solving for F,
psi
3
20 000
3.16 lbf
F
4
1.363 10
The steel ball is critical, with F = 3.16 lbf.
Ans.
______________________________________________________________________________
3-138 v 1 = 0.333, E 1 = 10.4 Mpsi, l = 2 in, d 1 = 1.25 in, v 2 = 0.211, E 2 = 14.5 Mpsi, d 2 = –12
in.
With b = K c F1/2
2 1 0.3332 10.4 106 1 0.2112 14.5 106
Kc
(2)
1/1.25 1/12
12
2.336 104
By examination of Eqs. (3-75), (3-76), and (3-77, it can be seen that the only difference
in the maximum shear stress for the two materials will be due to poisson’s ratio in Eq. (375). The larger poisson’s ratio will create the greater maximum shear stress, so the
aluminum roller will be the critical element in this interface. Instead of applying these
equations, we will assume the poisson’s ratio for aluminum of 0.333 is close enough to
0.3 to make Fig. 3-39 applicable.
max 0.3 pmax
pmax
4000
13 300 psi
0.3
From Eq. (3-74), p max = 2F / (bl ), so we have
Chapter 3 - Rev. A, Page 99/100
2F
2F 1 2
lK c F 1 2 lK c
pmax
So,
lK c pmax
F
2
2
(2)(2.336) 104 (13 300)
2
95.3 lbf Ans.
______________________________________________________________________________
2
3-139
v = 0.292, E = 30 Mpsi, l = 0.75 in, d 1 = 2(0.47) = 0.94 in, d 2 = 2(0.62) = 1.24 in.
Eq. (3-73):
2(40) 2 1 0.2922 30 106
b
(0.75)
1/ 0.94 1/1.24
12
1.052 103 in
Eq. (3-74):
pmax
2 40
2F
32 275 psi 32.3 kpsi
bl
1.052 103 0.75
Ans.
From Fig. 3-39,
max 0.3 pmax 0.3(32 275)=9682.5 psi 9.68 kpsi
Ans.
______________________________________________________________________________
3-140 Use Eqs. (3-73) through (3-77).
1/ 2
2 F (1 12 ) / E1 (1 v22 ) / E2
b
(1 / d1 ) (1 / d 2 )
l
1/2
2(600) (1 0.2922 ) / (30(106 )) (1 0.2922 ) / (30(106 ))
1/ 5 1/
(2)
b 0.007 631 in
pmax
2F
2(600)
25 028 psi
bl (0.007 631)(2)
Chapter 3 - Rev. A, Page 100/100
z2 z
x 2 pmax 1 2 2 0.292 25 028
b
b
Ans.
7102 psi 7.10 kpsi
1 0.7862 0.786
z2
1 2 0.7862
1 2 2
z
b
2 25 028
2 0.786
y pmax
1 0.7862
b
z2
1 2
b
4 646 psi 4.65 kpsi Ans.
pmax
25 028
z
19 677 psi 19.7 kpsi
Ans.
2
z
1 0.7862
1 2
b
z 4 646 19 677
Ans.
max y
7 516 psi 7.52 kpsi
2
2
______________________________________________________________________________
3-141 Use Eqs. (3-73) through (3-77).
2 F 1 12 E1 1 2 2 E2
b
l
1/ d1 1/ d 2
12
2(2000) 1 0.2922 207 103 1 0.2112 100 103
(40)
1/150 1 /
b 0.2583 mm
12
pmax
2F
2(2000)
123.2 MPa
bl (0.2583)(40)
z2 z
x 2 pmax 1 2 2 0.292 123.2 1 0.7862 0.786
b
b
Ans.
35.0 MPa
z2
1 2 0.7862
1 2 2
z
b
y pmax
2
123.2
2 0.786
2
2
b
z
1 0.786
1
b2
Ans.
22.9 MPa
pmax
123.2
z
96.9 MPa
Ans.
2
z
1 0.7862
1 2
b
Chapter 3 - Rev. A, Page 101/100
max
y z
2
22.9 96.9
37.0 MPa
2
Ans.
______________________________________________________________________________
3-142 Note to the Instructor: The first printing incorrectly had a width w = 1.25 mm instead of
w = 1.25 in. The solution presented here reflects the correction which will be made in
subsequent printings.
Use Eqs. (3-73) through (3-77).
2 F 1 12 E1 1 2 2 E2
b
l
1/ d1 1/ d 2
12
2(250) 1 0.2112 14.5 106 1 0.2112 14.5 106
(1.25)
1/ 3 1/
b 0.007 095 in
12
2F
2(250)
17 946 psi
bl (0.007 095)(1.25)
z2 z
x 2 pmax 1 2 2 0.21117 946
b
b
Ans.
3 680 psi 3.68 kpsi
pmax
1 0.7862 0.786
z2
1 2 0.7862
1 2 2
z
b
2
17 946
2 0.786
y pmax
1 0.7862
b
z2
1 2
b
3 332 psi 3.33 kpsi Ans.
pmax
17 946
z
14 109 psi 14.1 kpsi
Ans.
2
z
1 0.7862
1 2
b
z 3 332 14 109
Ans.
max y
5 389 psi 5.39 kpsi
2
2
______________________________________________________________________________
Chapter 3 - Rev. A, Page 102/100
Chapter 4
For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For a cantilever, k l = F/ , = F/k l . For
the assembly, k = F/y, or, y = F/k = l +
Thus
F Fl 2 F
y
k
kT kl
Solving for k
kk
1
k 2
2l T
Ans.
l
1 kl l kT
kT kl
______________________________________________________________________________
4-1
For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For each cantilever, k l = F/ l , l =
F/k l , and, L = F/k L . For the assembly, k = F/y, or, y = F/k = l + l + L .
Thus
F Fl 2 F F
y
k
kT kl k L
Solving for k
k L kl kT
1
k 2
Ans.
2
l
1 1 kl k Ll kT k L kT kl
kT kl k L
______________________________________________________________________________
4-2
4-3
(a) For a torsion bar, k =T/ =GJ/l.
Two springs in parallel, with J =d i 4/32,
and d 1 = d 1 = d,
J G J G d4 d4
k 1 2 G 1 2
x
l x 32 x l x
1
1
Gd 4
32
x lx
Deflection equation,
Ans.
(1)
T1 x T2 l x
JG
JG
T2 l x
(2)
x
From statics, T 1 + T 2 = T = 1500. Substitute Eq. (2)
results in
T1
Chapter 4 - Rev B, Page 1/81
lx
T2
T2 1500
x
T2 1500
x
l
Ans.
(3)
lx
Ans. (4)
l
1
1
3
(b) From Eq. (1), k 0.54 11.5 106
28.2 10 lbf in/rad
32
5 10 5
10 5
From Eq. (4), T1 1500
750 lbf in Ans.
10
5
From Eq. (3), T2 1500 750 lbf in
Ans.
10
16Ti 16 1500
From either section,
30.6 103 psi 30.6 kpsi
Ans.
3
3
di
0.5
T1 1500
Substitute into Eq. (2) resulting in
Ans.
______________________________________________________________________________
4-4
Deflection to be the same as Prob. 4-3 where T 1 = 750 lbfin, l 1 = l / 2 = 5 in, and d 1 = 0.5
in
1=2=
T1 4
32
Or,
4
1
d G
T2 6
32
4
2
d G
32
T1 15 103 d14
0.5 G
4
4T1 6T2
4 60 103
4
d1
d2
(1)
(2)
T2 10 103 d 24
Equal stress, 1 2
750 5
(3)
16T1 16T2
T
T
13 23
3
3
d1 d 2
d1 d 2
(4)
Divide Eq. (4) by the first two equations of Eq.(1) results in
T1
T2
3
d1
d3
2 d 2 1.5d1 (5)
4T1 4T2
d14
d 24
Statics, T 1 + T 2 = 1500
(6)
Substitute in Eqs. (2) and (3), with Eq. (5) gives
15 103 d14 10 103 1.5d1 1500
4
Solving for d 1 and substituting it back into Eq. (5) gives
d 1 = 0.388 8 in, d 2 = 0.583 2 in
Ans.
Chapter 4 - Rev B, Page 2/81
From Eqs. (2) and (3),
T 1 = 15(103)(0.388 8)4 = 343 lbfin
T 2 = 10(103)(0.583 2)4 = 1 157 lbfin
Ans.
Ans.
Deflection of T is
1
343 4
T1l1
0.053 18 rad
J1G / 32 0.388 84 11.5 106
Spring constant is
k
T
The stress in d 1 is
1
16 343
16T1
29.7 103 psi 29.7 kpsi
3
d1 0.388 8 3
Ans.
The stress in d 1 is
2
16 1 157
16T2
29.7 103 psi 29.7 kpsi
3
3
d 2 0.583 2
Ans.
1
1500
28.2 103 lbf in
0.053 18
Ans.
______________________________________________________________________________
4-5
(a) Let the radii of the straight sections be r 1 = d 1 /2 and r 2 = d 2 /2. Let the angle of the
taper be where tan = (r 2 r 1 )/2. Thus, the radius in the taper as a function of x is
r = r 1 + x tan , and the area is A = (r 1 + x tan )2. The deflection of the tapered portion
is
l
l
F
F
dx
F
1
dx
2
AE
E 0 r1 x tan
E r1 x tan tan
0
1 1
F 1
F
1
E r1 tan tan r1 l tan E tan r1 r2
F
r2 r1
F
l tan
Fl
E tan r1r2
E tan r1r2
r1r2 E
4 Fl
d1d 2 E
l
0
Ans.
(b) For section 1,
1
Fl
4 Fl
4(1000)(2)
3.40(104 ) in
2
2
6
AE d1 E (0.5 )(30)(10 )
For the tapered section,
4 Fl
4
1000(2)
2.26(104 ) in
d1d 2 E (0.5)(0.75)(30)(106 )
For section 2,
Ans.
Ans.
Chapter 4 - Rev B, Page 3/81
Fl
4 Fl
4(1000)(2)
1.51(104 ) in Ans.
2
2
6
AE d1 E (0.75 )(30)(10 )
______________________________________________________________________________
2
4-6
(a) Let the radii of the straight sections be r 1 = d 1 /2 and r 2 = d 2 /2. Let the angle of the
taper be where tan = (r 2 r 1 )/2. Thus, the radius in the taper as a function of x is
r = r 1 + x tan , and the polar second area moment is J = ( /2) (r 1 + x tan )4. The
angular deflection of the tapered portion is
l
l
T
2T
dx
1 2T
1
dx
4
3 G r1 x tan 3 tan
G 0 r1 x tan
GJ
0
l
0
2
2 T 1
1
T 1 1
3
3
3 G r1 tan tan r1 l tan 3 G tan r13 r23
2
2
r23 r13
2
2 T l r23 r13
2 Tl r1 r1r2 r2
T
3 G tan r13r23
3 G r2 r1 r13r23
3 G
r13r23
2
2
32 Tl d1 d1d 2 d 2
3 G
d13d 23
Ans.
(b) The deflections, in degrees, are
For section 1,
Tl 180 32Tl 180
32(1500)(2) 180
2.44 deg
4
GJ d1 G (0.54 )11.5(106 )
1
Ans.
For the tapered section,
32 Tl (d12 d1d 2 d 2 2 ) 180
3
Gd13d 23
2
2
32 (1500)(2) 0.5 (0.5)(0.75) 0.75 180
1.14 deg
3
11.5(106 )(0.53 )(.753 )
Ans.
For section 2,
Tl 180 32Tl 180
32(1500)(2) 180
Ans.
0.481 deg
4
GJ d 2 G (0.754 )11.5(106 )
______________________________________________________________________________
2
4-7
The area and the elastic modulus remain constant, however the force changes with respect
to x. From Table A-5 the unit weight of steel is = 0.282 lbf/in3, and the elastic modulus is
E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top).
F = (A)(lx)
Chapter 4 - Rev B, Page 4/81
Fdx w l
l 2 0.282 500(12)
1
(l x)dx lx x 2
0.169 in
AE E 0
E
2 0 2E
2(30)106
2
l
c
l
o
From the weight at the bottom of the cable,
4(5000) 500(12)
4Wl
Wl
5.093 in
(0.52 )30(106 )
AE d 2 E
c W 0.169 5.093 5.262 in
Ans.
W
The percentage of total elongation due to the cable’s own weight
0.169
(100) 3.21%
Ans.
5.262
______________________________________________________________________________
4-8
F y = 0 = R 1 F R 1 = F
M A = 0 = M 1 Fa M 1 = Fa
V AB = F, M AB =F (x a ), V BC = M BC = 0
Section AB:
1
F x2
F
x
a
dx
AB
ax C1
EI
EI 2
AB = 0 at x = 0 C 1 = 0
y AB
(1)
F x2
F x3
x2
ax dx
a C2
EI 2
EI 6
2
y AB = 0 at x = 0 C 2 = 0
Fx 2
y AB
x 3a
6 EI
(2)
Ans.
Section BC:
BC
1
EI
0 dx 0 C
3
From Eq. (1), at x = a (with C 1 = 0),
Fa 2
2 EI
Fa 2
Fa 2
dx
x C4
2 EI
2 EI
F a2
Fa 2
= C 3 . Thus,
a
(
a
)
2 EI
EI 2
BC
yBC
(3)
Chapter 4 - Rev B, Page 5/81
From Eq. (2), at x = a (with C 2 = 0), y
Fa 2
Fa 3
a C4
2 EI
3EI
yBC
C4
F a3
a2
Fa 3
. Thus, from Eq. (3)
a
EI 6
2
3EI
Fa 3
6 EI
Fa 2
Fa3 Fa 2
x
a 3x
2 EI
6 EI 6 EI
Substitute into Eq. (3)
Ans.
The maximum deflection occurs at x= l,
Fa 2
Ans.
a 3l
6 EI
______________________________________________________________________________
ymax
4-9
M C = 0 = F (l /2) R 1 l R 1 = F /2
F y = 0 = F /2 + R 2 F R 2 = F /2
Break at 0 x l /2:
V AB = R 1 = F /2,
M AB = R 1 x = Fx /2
Break at l /2 x l :
V BC = R 1 F = R 2 = F /2,
M BC = R 1 x F ( x l / 2) = F (l x) /2
Section AB:
AB
1
EI
Fx
F x2
C1
dx
2
EI 4
2
l
F
2 C 0
From symmetry, AB = 0 at x = l /2
1
4 EI
AB
F x 2 Fl 2
F
4 x2 l 2
EI 4 16 EI 16 EI
y AB
F
16 EI
C1
Fl 2
. Thus,
16 EI
(1)
F 4x 3 2
4 x l dx 16 EI 3 l x C2
2
2
Chapter 4 - Rev B, Page 6/81
y AB = 0 at x = 0
y AB
C 2 = 0, and,
Fx
4 x 2 3l 2
48 EI
(2)
y BC is not given, because with symmetry, Eq. (2) can be used in this region. The
maximum deflection occurs at x =l /2,
ymax
l
F
2
Fl 3
2 l
2
4 3l
48EI 2
48EI
Ans.
______________________________________________________________________________
4-10 From Table A-6, for each angle, I 1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4
From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam
3 with w = 1 N/mm and l = 3000 mm.
Fa 2
wl4
(a 3l )
6 EI
8EI
2500(2000) 2
(1)(3000)4
2000
3(3000)
6(207)103 (4.14)106
8(207)(103 )(4.14)(106 )
25.4 mm
Ans.
ymax
M O Fa ( wl 2 / 2)
= 2500(2000) [1(30002)/2] = 9.5(106) Nmm
From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom
surface is y = 29.0 100= 71 mm. The maximum stress is compressive at the bottom of
the beam at the wall. This stress is
My
9.5(106 )(71)
163 MPa Ans.
max
I
4.14(106 )
______________________________________________________________________________
Chapter 4 - Rev B, Page 7/81
4-11
14
10
(450) (300) 465 lbf
20
20
6
10
(450) (300) 285 lbf
RC
20
20
RO
M 1 = 465(6)12 = 33.48(103) lbfin
M 2 = 33.48(103) +15(4)12
= 34.20(103) lbfin
M max
34.2
Z 2.28 in 3
15
Z
Z
For deflections, use beams 5 and 6 of Table A-9
2
F1a[l (l / 2)] l
l F2l 3
2
y x 10ft
a 2l
6 EIl
2 48 EI
2
max
450(72)(120)
300(2403 )
2
2
2
0.5
120 72 240 48(30)(106 ) I
6(30)(106 ) I (240)
I 12.60 in 4 I / 2 6.30 in 4
Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) =
6.00 in3
12.60 1
ymidspan
0.421 in
14.98 2
34.2
max
5.70 kpsi
6.00
______________________________________________________________________________
4-12
I
(1.54 ) 0.2485 in 4
64
From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and
l = 39 in
Fba 2
wa
y
[a b 2 l 2 ]
(2la 2 a 3 l 3 )
6 EIl
24 EI
yA
340(24)15
152 242 392
6
6(30)10 (0.2485)39
(150 /12)(15)
2(39)(152 ) 153 393 0.0978 in
24(30)106 (0.2485)
Ans.
At x = l /2 = 19.5 in
Chapter 4 - Rev B, Page 8/81
y
2
2
3
Fa[l (l / 2)] l
l w(l / 2) l l
2
3
a
2
l
2
l
l
6 EIl
2 24 EI 2 2
2
y
340(15)(19.5)
19.52 152 392
6(30)(106 )(0.2485)(39)
(150 /12)(19.5)
2(39)(19.52 ) 19.53 393 0.1027 in
6
24(30)(10 )(0.2485)
Ans.
0.1027 0.0978
(100) 5.01% Ans.
0.0978
______________________________________________________________________________
% difference
4-13 I
1
(6)(323 ) 16.384 103 mm 4
12
From Table A-9-10, beam 10
Fa 2
yC
(l a )
3EI
Fax 2
y AB
l x2
6 EIl
dy AB
Fa 2
(l 3 x 2 )
dx
6 EIl
dy AB
A
dx
Fal 2 Fal
A
6 EIl 6 EI
Fa 2l
yO A a
6 EI
At x = 0,
With both loads,
Fa 2l Fa 2
yO
(l a )
6 EI 3EI
Fa 2
400(3002 )
(3l 2a )
3(500) 2(300) 3.72 mm Ans.
6 EI
6(207)103 (16.384)103
At midspan,
2
2 Fa(l / 2) 2 l 3 Fal 2
3
400(300)(5002 )
1.11 mm Ans.
yE
l
6 EIl
24 207 103 16.384 103
2 24 EI
_____________________________________________________________________________
4-14 I (24 1.54 ) 0.5369 in 4
64
Chapter 4 - Rev B, Page 9/81
From Table A-5, E = 10.4 Mpsi
From Table A-9, beams 1 and 2, by superposition
3
2
200 4(12)
300 2(12)
FB l 3 FA a 2
yB
(a 3l )
2(12) 3(4)(12)
3EI 6 EI
3(10.4)106 (0.5369) 6(10.4)106 (0.5369)
yB 1.94 in Ans.
______________________________________________________________________________
4-15 From Table A-7, I = 2(1.85) = 3.70 in4
From Table A-5, E = 30.0 Mpsi
From Table A-9, beams 1 and 3, by superposition
5 2(5 /12) (60 ) 0.182 in Ans.
Fl 3 ( w wc )l 4
150(603 )
yA
3EI
8 EI
3(30)106 (3.70)
8(30)106 (3.70)
______________________________________________________________________________
4
4-16 I
d
64
From Table A-5, E 207(103 ) MPa
From Table A-9, beams 5 and 9, with F C = F A = F, by superposition
F l3
Fa
1
FB l 3 2 Fa(4a 2 3l 2 )
yB B
(4a 2 3l 2 ) I
48EI 24 EI
48 EyB
1
I
550(10003 ) 2 375 (250) 4(2502 ) 3(10002 )
3
48(207)10 2
4
53.624 103 mm4
d
4
64
I
4
64
(53.624)103 32.3 mm
Ans.
______________________________________________________________________________
4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by
superposition
yBC
MA 3
Fax 2
x 3lx 2 2l 2 x
l x2
6 EIl
6 EIl
1
M A x 3 3lx 2 2l 2 x Fax l 2 x 2
Ans.
6 EIl
d M
F (x l)
A x3 3lx 2 2l 2 x ( x l )
[( x l )2 a(3x l )]
6 EI
x l
dx 6 EIl
y AB
M Al
F (x l)
(x l)
[( x l ) 2 a (3 x l )]
6 EI
6 EI
Chapter 4 - Rev B, Page 10/81
(x l)
Ans.
M Al F ( x l ) 2 a (3 x l )
6 EI
______________________________________________________________________________
4-18 Note to the instructor: Beams with discontinuous loading are better solved using
singularity functions. This eliminates matching the slopes and displacements at the
discontinuity as is done in this solution.
a
wa
M C 0 R1l wa l a 2 R1 2l 2l a Ans.
wa
wa 2
Fy 0 2l 2l a R2 wa R2 2l Ans.
wa
w
VAB R1 wx =
2l a wx = 2l a x a 2 Ans.
2l
2l
2
wa
VBC R2
Ans.
2l
w
x2
M AB VAB dx 2l ax a 2 x C1
2l
2
wx
2al a 2 lx Ans.
M AB 0 at x 0 C1 0
M AB
2l
wa 2
wa 2
M BC VBC dx
dx
x C2
2l
2l
wa 2
wa 2
M BC 0 at x l C2
M BC
(l x)
Ans.
2
2l
M
1 wx
1 w 2 1 2 2 1 3
AB AB dx
2al a 2 lx dx
alx a x lx C3
3
EI
EI 2l
EI 2l
2
1 w 2 1 2 2 1 3
alx a x lx C3 dx
2
3
EI 2l
1 w 1 3 1 2 3 1 4
alx a x lx C3 x C4
6
12
EI 2l 3
y AB AB dx
y AB 0 at x 0 C4 0
BC
M
1
BC dx
EI
EI
wa 2
1 wa 2
1 2
2l (l x) dx EI 2l lx 2 x C5
AB BC at x a
1
EI
1 2
w 2 1 4 1 3
1 wa 2
wa 3
ala
a
la
C
la
a
C
C
C5
3
5
3
2l
EI 2l
2
3
2
6
(1)
Chapter 4 - Rev B, Page 11/81
1 wa 2
1 2
1 wa 2 1 2 1 3
lx
x
C
dx
5
lx x C5 x C6
EI 2l
2
EI 2l 2
6
2 2
wa l
0 at x l C6
C5l
6
1 wa 2 1 2 1 3 1 3
lx x l C5 ( x l )
EI 2l 2
6
3
yBC BC dx
yBC
yBC
y AB yBC at x a
1 5 1 4
w 1
wa 2 1 2 1 3 1 3
3
ala
a
la
C
a
3
la a l C5 (a l )
2l 3
6
12
2l 2
6
3
wa 2
C3 a
(2)
3la 2 4l 3 C5 (a l )
24l
wa 2
Substituting (1) into (2) yields C5
a 2 4l 2 . Substituting this back into (2) gives
24l
2
wa
4al a 2 4l 2 . Thus,
C3
24l
w
y AB
4alx3 2a 2 x3 lx 4 4a3lx a 4 x 4a 2l 2 x
24 EIl
wx
2
y AB
Ans.
2ax 2 (2l a ) lx 3 a 2 2l a
24 EIl
w
yBC
6a 2lx 2 2a 2 x 3 a 4 x 4a 2l 2 x a 4l
Ans.
24 EIl
This result is sufficient for y BC . However, this can be shown to be equivalent to
w
w
yBC
4alx 3 2a 2 x 3 lx 4 4a 2l 2 x 4a 3lx a 4 x
( x a)4
24 EIl
24 EI
w
yBC y AB
( x a)4
Ans.
24 EI
by expanding this or by solving the problem using singularity functions.
______________________________________________________________________________
4-19
The beam can be broken up into a uniform load w downward from points A to C and a
uniform load upward from points A to B.
wx
wx
2
2
2bx 2 (2l b) lx 3 b 2 2l b
2ax 2 (2l a) lx 3 a 2 2l a
y AB
24 EIl
24 EIl
wx
2
2
2bx 2 (2l b) b 2 2l b 2ax 2 (2l a ) a 2 2l a
24 EIl
w
2
2bx 3 (2l b) lx 4 b 2 x 2l b
24 EIl
yBC
4alx 3 2a 2 x 3 lx 4 4a 2l 2 x 4a 3lx a 4 x l ( x a ) 4
Ans.
Ans.
Chapter 4 - Rev B, Page 12/81
w
4blx 3 2b 2 x 3 lx 4 4b 2l 2 x 4b3lx b 4 x l ( x b) 4
24 EIl
w
4alx 3 2a 2 x3 lx 4 4a 2l 2 x 4a 3lx a 4 x l ( x a) 4
24 EIl
w
( x b) 4 ( x a ) 4 y AB
Ans.
24 EI
______________________________________________________________________________
yCD
4-20
Note to the instructor: See the note in the solution for Problem 4-18.
wa 2
wa
0
F
R
wa RB
2l a Ans.
y
B
2l
2l
For region BC, isolate right-hand element of length (l + a x)
wa 2
,
VAB RA
VBC w l a x
Ans.
2l
wa 2
w
2
M AB RA x
x,
M BC l a x
Ans.
2l
2
wa 2 2
EI AB M AB dx
x C1
4l
wa 2 3
EIy AB
x C1 x C2
12l
wa 2 3
y AB = 0 at x = 0 C 2 = 0 EIy AB
x C1 x
12l
w a 2l
y AB = 0 at x = l C1
12
wa 2 3 wa 2l
wa 2 x 2
wa 2 x 2
2
EIy AB
x
x
l x y AB
l x 2 Ans.
12l
12
12l
12 EIl
w
3
EI BC M BC dx l a x C3
6
w
4
EIyBC l a x C3 x C4
24
wa 4
wa 4
(1)
y BC = 0 at x = l
C3l C4 0 C4
C3l
24
24
wa 2l wa 2l wa 3
wa 2
AB = BC at x = l
C3 C3
l a
4
12
6
6
wa 2 2
a 4l l a . Substitute back into y BC
Substitute C 3 into Eq. (1) gives C4
24
1 w
wa 2
wa 4 wa 2l
4
yBC
l
a
x
x
l
a
l a
EI 24
6
24
6
w
4
l a x 4a 2 l x l a a 4
24 EI
Ans.
Chapter 4 - Rev B, Page 13/81
Table A-9, beam 7,
w l 100(10)
R1 R2
500 lbf
2
2
100 x
wx
2(10) x 2 x 3 103
y AB
2lx 2 x 3 l 3
6
24 EI
24 30 10 0.05
4-21
2.7778 106 x 20 x 2 x 3 1000
Slope:
AB
d y AB
w
6lx 2 4 x3 l 3
24 EI
dx
w
wl 3
2
3
3
At x = l, AB x l
6l l 4l l 24EI
24 EI
100 103
w l3
yBC AB x l x l
x l
x 10 2.7778 103 x 10
6
24 EI
24(30)10 (0.05)
From Prob. 4-20,
2
wa 2 100 4
RA
80 lbf
2l
2(10)
RB
100 4
wa
2l a
2(10) 4 480 lbf
2l
2(10)
100 42 x
wa 2 x 2
2
l
x
10 2 x 2 8.8889 106 x 100 x 2
6
12 EIl
12 30 10 0.05
w
4
l a x 4a 2 l x l a a 4
24 EI
y AB
yBC
100
10 4 x 4 4 42 10 x 10 4 44
6
24 30 10 0.05
2.7778 106 14 x 896 x 9216
Superposition,
RA 500 80 420 lbf
RB 500 480 980 lbf
4
y AB 2.7778 10 6 x 20 x 2 x 3 1000 8.8889 10 6 x 100 x 2
Ans.
Ans.
yBC 2.7778 103 x 10 2.7778 106 14 x 896 x 9216
Ans.
The deflection equations can be simplified further. However, they are sufficient for
plotting.
Using a spreadsheet,
4
x
y
0
0.5
1
1.5
2
2.5
3
3.5
0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027
x
y
4
4.5
5
5.5
6
6.5
7
7.5
-0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268
Chapter 4 - Rev B, Page 14/81
x
y
x
y
8
8.5
9
9.5
10
10.5
11
11.5
-0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036
12
0.001244
12.5
0.001419
13
0.001575
13.5
14
0.001722 0.001867
______________________________________________________________________________
4-22
(a) Useful relations
k
F 48 EI
3
y
l
3
kl 3 1800 36
I
0.05832 in 4
6
48 E 48(30)10
From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or,
h
4
6 I 4 6(0.05832)
0.514 in
5
5
h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically
changes the spring rate, so changing the base will make finding a close solution easier.
Trial and error was applied to find the combination of values from Table A-17 that
yielded the closet desired spring rate.
h (in)
1/2
1/2
1/2
9/16
9/16
b (in)
5
5½
5¾
5
4
b/h
10
11
11.5
8.89
7.11
k (lbf/in)
1608
1768
1849
2289
1831
Chapter 4 - Rev B, Page 15/81
h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is
still reasonably close to 10.
(b) I 5.5(0.5)3 /12 0.05729 in 4
4 I 4(60)103 (0.05729)
Mc ( Fl / 4)c
F
1528 lbf
I
I
lc
36 (0.25)
(1528) 36
Fl 3
0.864 in Ans.
y
48 EI 48(30)106 (0.05729)
______________________________________________________________________________
3
4-23 From the solutions to Prob. 3-68, T1 60 lbf and T2 400 lbf
I
d4
64
(1.25)4
64
0.1198 in 4
From Table A-9, beam 6,
Fb x
Fb x
z A 1 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 )
6 EIl
6 EIl
x 10in
(575)(30)(10)
102 302 402
6
6(30)10 (0.1198)(40)
460(12)(10)
102 122 402 0.0332 in
6(30)106 (0.1198)(40)
Ans.
dz
d Fb x
Fb x
1 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 )
6 EIl
dx x 10in
x 10in
dx 6 EIl
Fb
Fb
1 1 (3 x 2 b12 l 2 ) 2 2 (3x 2 b2 2 l 2 )
6 EIl
6 EIl
x 10in
A y
(575)(30)
3 102 302 402
6
6(30)10 (0.1198)(40)
460(12)
3 102 122 402
6(30)106 (0.1198)(40)
6.02(104 ) rad
Ans.
______________________________________________________________________________
4-24 From the solutions to Prob. 3-69, T1 2880 N and T2 432 N
I
d4
64
(30)4
64
39.76 103 mm 4
Chapter 4 - Rev B, Page 16/81
The load in between the supports supplies an angle to the overhanging end of the beam.
That angle is found by taking the derivative of the deflection from that load. From Table
A-9, beams 6 (subscript 1) and 10 (subscript 2),
y A BC
C
a2 beam 6 y A beam10
(1)
d F1a1 l x 2
Fa
x a12 2lx 1 1 6lx 3x 2 a12 2l 2
x l
x l 6 EIl
dx 6 EIl
BC C
F1a1 2
l a12
6 EIl
Equation (1) is thus
F1a1 2
Fa2
l a12 a2 2 2 (l a2 )
6 EIl
3EI
2070(3002 )
3312(230)
2
2
510
230
300
3(207)103 (39.76)103 510 300
6(207)103 (39.76)103 (510)
7.99 mm
Ans.
yA
The slope at A, relative to the z axis is
A z
F1a1 2
d F (x l)
( x l )2 a2 (3 x l )
(l a12 ) 2
6 EIl
x l a2
dx 6 EI
F1a1 2
F
l a12 2 3( x l ) 2 3a2 ( x l ) a2 (3 x l )
x l a2
6 EIl
6 EI
Fa
F
1 1 (l 2 a12 ) 2 3a2 2 2la2
6 EIl
6 EI
3312(230)
5102 2302
6(207)103 (39.76)103 (510)
2070
3(3002 ) 2(510)(300)
3
3
6(207)10 (39.76)10
0.0304 rad
Ans.
______________________________________________________________________________
4-25 From the solutions to Prob. 3-70, T1 392.16 lbf and T2 58.82 lbf
I
d4
64
From Table A-9, beam 6,
(1) 4
64
0.049 09 in 4
Chapter 4 - Rev B, Page 17/81
( 350)(14)(8)
Fb x
82 14 2 22 2 0.0452 in Ans.
y A 1 1 x 2 b12 l 2
6
6 EIl
x 8in 6(30)10 (0.049 09)(22)
( 450.98)(6)(8)
F b x
z A 2 2 ( x 2 b2 2 l 2 )
82 6 2 22 2 0.0428 in Ans.
6
6
EIl
6(30)10
(0.049
09)(22)
x 8in
The displacement magnitude is y A2 z A2 0.04522 0.04282 0.0622 in
Ans.
d y
d Fb x
Fb
1 1 x 2 b12 l 2
1 1 (3a12 b12 l 2 )
x a1 6 EIl
d x x a1 dx 6 EIl
A z
( 350)(14)
3 82 14 2 22 2 0.00242 rad
6
6(30)10 (0.04909)(22)
Ans.
dz
d F b x
Fb
2 2 ( x 2 b22 l 2 )
2 2 3a12 b22 l 2
x a1 6 EIl
dx 6 EIl
d x x a1
A y
(450.98)(6)
3 82 62 222 0.00356 rad
6
6(30)10 (0.04909)(22)
Ans.
The slope magnitude is A 0.002422 0.00356 0.00430 rad Ans.
2
______________________________________________________________________________
4-26 From the solutions to Prob. 3-71, T1 250 N and T2 37.5 N
I
d4
64
(20) 4
64
7 854 mm 4
345sin 45o (550)(300) 3002 5502 8502
F b x
y A 1 y 1 ( x 2 b12 l 2 )
6(207)103 (7 854)(850)
6 EIl
x 300mm
1.60 mm Ans.
Fb x
F b x
z A 1z 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 )
6 EIl
6 EIl
x 300mm
345cos 45 (550)(300)
o
6(207)103 (7 854)(850)
300
2
5502 8502
287.5(150)(300)
3002 1502 8502 0.650 mm
3
6(207)10 (7 854)(850)
Ans.
The displacement magnitude is y A2 z A2 1.602 0.650 1.73 mm
2
Ans.
Chapter 4 - Rev B, Page 18/81
F1 y b1
d y
d F1 y b1 x 2 2 2
(3a12 b12 l 2 )
x b1 l
d x x a1 dx 6 EIl
x a1 6 EIl
A z
345sin 45o (550)
3 3002 5502 8502 0.00243 rad
6(207)10 (7 854)(850)
Ans.
3
dz
d F b x
Fb x
1z 1 x 2 b12 l 2 2 2 x 2 b2 2 l 2
6 EIl
x a1
dx 6 EIl
d x x a1
A y
F1z b1
Fb
3a12 b12 l 2 2 2 3a12 b22 l 2
6 EIl
6 EIl
o
345cos 45 (550) 3 3002 5502 8502
6(207)103 (7 854)(850)
287.5(150)
3 3002 1502 8502 1.91 104 rad
6(207)103 (7 854)(850)
Ans.
The slope magnitude is A 0.002432 0.0001912 0.00244 rad Ans.
______________________________________________________________________________
4-27 From the solutions to Prob. 3-72, FB 750 lbf
I
d4
64
(1.25) 4
64
0.1198 in 4
From Table A-9, beams 6 (subscript 1) and 10 (subscript 2)
F1 y b1 x 2 2 2 F2 y a2 x 2
yA
x b1 l
l x2
6 EIl
6 EIl
x 16in
300 cos 20 (14)(16)
o
16
6(30)10 (0.119 8)(30)
2
6
0.0805 in
14 30
2
2
750sin 20 (9)(16)
o
30
6(30)10 (0.119 8)(30)
2
162
2
162
6
Ans.
F ax
F b x
z A 1z 1 x 2 b12 l 2 2 z 2 l 2 x 2
6 EIl
6 EIl
x 16in
300sin 20 (14)(16)
o
6(30)106 (0.119 8)(30)
Ans.
0.1169 in
16
2
14 30
2
2
750 cos 20 (9)(16)
o
6(30)106 (0.119 8)(30)
30
The displacement magnitude is y A2 z A2 0.08052 0.1169 0.142 in
2
Ans.
Chapter 4 - Rev B, Page 19/81
d F1 y b1 x 2
F2 y a2 x 2
d y
x b12 l 2
l x2
6 EIl
d x x a1 dx 6 EIl
x a1
A z
3a b l 6EIl l 3a
6 EIl
300 cos 20 (14) 3 16 14 30
6(30)10 (0.119 8)(30)
F1 y b1
2
1
2
1
2
F2 y a2
2
2
1
o
2
2
2
6
750sin 20 (9)
o
302 3 162 8.06 105 rad
6(30)10 (0.119 8)(30)
6
Ans.
dz
d F b x
F ax
1z 1 x 2 b12 l 2 2 z 2 l 2 x 2
6 EIl
x a1
dx 6 EIl
d x x a1
A y
F1z b1
F a
3a12 b12 l 2 2 z 2 l 2 3a12
6 EIl
6 EIl
o
750 cos 20o (9)
300sin 20 (14)
2
2
2
302 3 162
3 16 14 30
6
6
6(30)10 (0.119 8)(30)
6(30)10 (0.119 8)(30)
0.00115 rad
Ans.
The slope magnitude is A 8.06 105 0.001152 0.00115 rad Ans.
______________________________________________________________________________
2
4-28 From the solutions to Prob. 3-73, F B = 22.8 (103) N
4
d 4 50
306.8 103 mm 4
I
64
64
From Table A-9, beam 6,
F bx
F b x
y A 1 y 1 ( x 2 b12 l 2 ) 2 y 2 ( x 2 b2 2 l 2 )
6 EIl
6 EIl
x 400mm
11103 sin 20o (650)(400)
4002 6502 10502
3
3
6(207)10 (306.8)10 (1050)
22.8 103 sin 25o (300)(400)
4002 3002 10502
3
3
6(207)10 (306.8)10 (1050)
3.735 mm
Ans.
Chapter 4 - Rev B, Page 20/81
F bx
F b x
z A 1z 1 ( x 2 b12 l 2 ) 2 z 2 ( x 2 b2 2 l 2 )
6 EIl
6 EIl
x 400mm
11103 cos 20o (650)(400)
4002 6502 10502
3
6(207)10 (306.8)103 (1050)
22.8 103 cos 25o (300)(400)
4002 3002 10502 1.791 mm
6(207)103 (306.8)103 (1050)
The displacement magnitude is y A2 z A2
3.735
2
1.7912 4.14 mm
Ans.
Ans.
d y
d F b x
F bx
1z 1 x 2 b12 l 2 2 z 2 x 2 b2 2 l 2
6 EIl
x a1
d x x a1 dx 6 EIl
A z
3a b l 6EIl 3a b l
6 EIl
1110 sin 20 (650)
3 400 650
6(207)10 (306.8)10 (1050)
F1 y b1
2
1
3
2
1
F2 y b2
2
2
1
2
2
o
2
3
2
3
2
10502
22.8 103 sin 25o (300)
3 4002 3002 10502
6(207)103 (306.8)103 (1050)
0.00507 rad
Ans.
dz
d F b x
F bx
1z 1 x 2 b12 l 2 2 z 2 x 2 b2 2 l 2
6 EIl
x a1
dx 6 EIl
d x x a1
A y
F1z b1
F b
3a12 b12 l 2 2 z 2 3a12 b22 l 2
6 EIl
6 EIl
11103 cos 20o (650)
3 4002 6502 10502
3
3
6(207)10 (306.8)10 (1050)
22.8 103 cos 25o (300)
3 4002 3002 10502
3
3
6(207)10 (306.8)10 (1050)
0.00489 rad
Ans.
The slope magnitude is A
0.00507 0.00489
2
2
0.00704 rad Ans.
______________________________________________________________________________
4-29 From the solutions to Prob. 3-68, T 1 = 60 lbf and T 2 = 400 lbf , and Prob. 4-23, I = 0.119 8
in4. From Table A-9, beam 6,
Chapter 4 - Rev B, Page 21/81
dz
d F1z b1 x 2
F bx
x b12 l 2 2 z 2 x 2 b22 l 2
6 EIl
x 0
dx 6 EIl
d x x 0
F b
F b
575(30)
1z 1 b12 l 2 2 z 2 b2 2 l 2
302 402
6 EIl
6 EIl
6(30)106 (0.119 8)(40)
460(12)
122 402 0.00468 rad
Ans.
6
6(30)10 (0.119 8)(40)
O y
d F1z a1 l x 2
F a l x 2
dz
x a12 2lx 2 z 2
x a22 2lx
6 EIl
d x x l
x l
dx 6 EIl
F a
F a
1z 1 6lx 2l 2 3 x 2 a12 2 z 2 6lx 2l 2 3 x 2 a22
6 EIl
6 EIl
x l
C y
F1z a1 2
F a
l a12 2 z 2 l 2 a22
6 EIl
6 EIl
2
460(28) 402 282
575(10) 40 102
0.00219 rad
Ans.
6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40)
______________________________________________________________________________
4-30 From the solutions to Prob. 3-69, T 1 = 2 880 N and T 2 = 432 N, and Prob. 4-24, I = 39.76
(103) mm4. From Table A-9, beams 6 and 10
d y
d Fb x
Fa x
O z 1 1 ( x 2 b12 l 2 ) 2 2 (l 2 x 2 )
6 EIl
x 0
d x x 0 dx 6 EIl
Fa
Fb
Fal
Fb
1 1 (3x 2 b12 l 2 ) 2 2 (l 2 3x 2 ) 1 1 (b12 l 2 ) 2 2
6 EIl
6 EI
6 EIl
x 0 6 EIl
2 070(300)(510)
3 312(280)
2802 5102
3
3
6(207)103 (39.76)103
6(207)10 (39.76)10 (510)
0.0131 rad
Ans.
d y
d F1a1 (l x) 2
Fa x
( x a12 2lx) 2 2 (l 2 x 2 )
6 EIl
x l
d x x l dx 6 EIl
C z
Fa
Fa
Fal
Fa
1 1 (6lx 2l 2 3 x 2 a12 ) 2 2 (l 2 3 x 2 ) 1 1 (l 2 a12 ) 2 2
6 EIl
3EI
6 EIl
x l 6 EIl
2 070(300)(510)
3 312(230)
(5102 2302 )
3
3
6(207)10 (39.76)10 (510)
3(207)103 (39.76)103
0.0191 rad
Ans.
______________________________________________________________________________
4-31 From the solutions to Prob. 3-70, T 1 = 392.19 lbf and T 2 = 58.82 lbf , and Prob. 4-25, I =
0.0490 9 in4. From Table A-9, beam 6
Chapter 4 - Rev B, Page 22/81
F1 y b1 2 2
d y
d F1 y b1 x 2
x b12 l 2
(b1 l )
d x x 0 dx 6 EIl
x 0 6 EIl
350(14)
142 222 0.00726 rad
Ans.
6
6(30)10 (0.04909)(22)
O z
dz
d F2 z b2 x 2
F b
x b22 l 2 2 z 2 b22 l 2
6 EIl
x 0
dx 6 EIl
d x x 0
450.98(6)
6 2 22 2
6
6(30)10 (0.0490 9)(22)
Ans.
0.00624 rad
O y
The slope magnitude is O 0.007262 0.00624 0.00957 rad Ans.
2
d F1 y a1 (l x) 2
d y
x a12 2lx
d x x l dx 6 EIl
x l
F1 a1
F1 a1
y 6lx 2l 2 3x 2 a12 y (l 2 a12 )
6 EIl
x l 6 EIl
C z
350(8)
222 82 0.00605 rad
6
6(30)10 (0.0491)(22)
Ans.
dz
d F2 z a2 (l x ) 2
x a22 2lx
x l
dx 6 EIl
d x x l
C y
F a
F a
2 z 2 6lx 2l 2 3 x 2 a22 2 z 2 l 2 a22
6 EIl
6 EIl
x l
450.98(16)
222 162 0.00846 rad
6(30)106 (0.0490 9)(22)
The slope magnitude is C
0.00605
2
Ans.
0.008462 0.0104 rad Ans.
______________________________________________________________________________
4-32 From the solutions to Prob. 3-71, T 1 =250 N and T 1 =37.5 N, and Prob. 4-26, I = 7 854
mm4. From Table A-9, beam 6
d F1 y b1 x 2
F1 y b1 2 2
d y
x b12 l 2
(b1 l )
d x x 0 dx 6 EIl
x 0 6 EIl
O z
345sin 45o (550)
5502 8502 0.00680 rad
3
6(207)10 (7 854)(850)
Ans.
Chapter 4 - Rev B, Page 23/81
dz
d F1z b1 x 2 2 2 F2 z b2 x 2 2 2
x b1 l 6EIl x b2 l
dx 6 EIl
x 0
d x x 0
O y
345cos 45o (550)
F1z b1 2 2 F2 z b2 2 2
b1 l
b2 l
5502 8502
3
6 EIl
6 EIl
6(207)10 (7 854)(850)
287.5(150)
1502 8502 0.00316 rad Ans.
6(207)103 (7 854)(850)
The slope magnitude is O 0.00680 2 0.00316 2 0.00750 rad Ans.
d F1 y a1 (l x) 2
F1 y a1
d y
x a12 2lx
6lx 2l 2 3x 2 a12
d x x l dx 6 EIl
x l 6 EIl
x l
C z
345sin 45o (300)
(l 2 a12 )
8502 3002 0.00558 rad
3
6 EIl
6(207)10 (7 854)(850)
F1 y a1
Ans.
dz
d F1z a1 (l x) 2
F a (l x) 2
x a12 2lx 2 z 2
x a22 2lx
6 EIl
x l
dx 6 EIl
d x x l
C y
345cos 45o (300)
F1z a1 2
F2 z a2 2
2
2
l a1 6EIl l a2 6(207)103 (7 854)(850) 8502 3002
6 EIl
287.5(700)
8502 7002 6.04 105 rad
Ans.
3
6(207)10 (7 854)(850)
The slope magnitude is C
0.00558
2
6.04 105 0.00558 rad Ans.
2
________________________________________________________________________
4-33 From the solutions to Prob. 3-72, F B = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From
Table A-9, beams 6 and 10
d F b x
F ax
d y
O z 1 y 1 x 2 b12 l 2 2 y 2 l 2 x 2
6 EIl
d x x 0 dx 6 EIl
x 0
F a
F b
F al
F b
1 y 1 3x 2 b12 l 2 2 y 2 l 2 3 x 2 1 y 1 b12 l 2 2 y 2
6 EIl
6 EI
6 EIl
x 0 6 EIl
300 cos 20o (14)
750sin 20o (9)(30)
2
2
14 30 6(30)106 (0.119 8) 0.00751 rad
6(30)106 (0.119 8)(30)
Ans.
Chapter 4 - Rev B, Page 24/81
dz
d F1z b1 x 2
F ax
x b12 l 2 2 z 2 l 2 x 2
6 EIl
x 0
dx 6 EIl
d x x 0
O y
F a
F b
F al
F b
1z 1 3x 2 b12 l 2 2 z 2 l 2 3 x 2 1z 1 b12 l 2 2 z 2
6 EIl
6 EIl
6 EI
6 EIl
x 0
750 cos 20o (9)(30)
300sin 20o (14)
2
2
14 30 6(30)106 (0.119 8) 0.0104 rad
6(30)106 (0.119 8)(30)
Ans.
The slope magnitude is O 0.007512 0.0104 2 0.0128 rad Ans.
F ax
dy
d F1 y a1 (l x) 2
x a12 2lx 2 y 2 l 2 x 2
6 EIl
dx x l dx 6 EIl
x l
F a
F a
F al
F a
1 y 1 6lx 2l 2 3x 2 a12 2 y 2 l 2 3x 2 1 y 1 (l 2 a12 ) 2 y 2
6 EIl
3EI
6 EIl
x l 6 EIl
C z
300 cos 20o (16)
750sin 20o (9)(30)
2
2
30
16 3(30)106 (0.119 8) 0.0109 rad
6(30)106 (0.119 8)(30)
Ans.
dz
d F1z a1 (l x) 2
F ax
x a12 2lx 2 z 2 l 2 x 2
6 EIl
x l
dx 6 EIl
d x x l
C y
F a
F a
F al
F a
1z 1 6lx 2l 2 3 x 2 a12 2 z 2 l 2 3 x 2 1z 1 l 2 a12 2 z 2
6 EIl
6 EIl
3EI
6 EIl
x l
750 cos 20o (9)(30)
300sin 20o (16)
2
2
30 16 3(30)106 (0.119 8) 0.0193 rad
6(30)106 (0.119 8)(30)
The slope magnitude is C
0.0109 0.0193
2
2
Ans.
0.0222 rad Ans.
______________________________________________________________________________
4-34 From the solutions to Prob. 3-73, F B = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4.
From Table A-9, beam 6
d F1 y b1 x 2 2 2 F2 y b2 x 2 2 2
d y
x b1 l 6EIl x b2 l
d x x 0 dx 6 EIl
x 0
O z
11103 sin 20o (650)
6502 10502
b l
b l
3
3
6 EIl
6 EIl
6(207)10 (306.8)10 (1050)
F1 y b1
2
1
2
F2 y b2
2
2
2
22.8 103 sin 25o (300)
3002 10502 0.0115 rad
6(207)103 (306.8)103 (1050)
Ans.
Chapter 4 - Rev B, Page 25/81
dz
d F1z b1 x 2
F bx
x b12 l 2 2 z 2 x 2 b22 l 2
6 EIl
x 0
dx 6 EIl
d x x 0
F b
F b
1z 1 b12 l 2 2 z 2 b22 l 2
6 EIl
6 EIl
11103 cos 20o (650)
6502 10502
3
3
6(207)10 (306.8)10 (1050)
O y
22.8 103 cos 25o (300)
3002 10502 0.00427 rad
6(207)103 (306.8)103 (1050)
The slope magnitude is O
0.0115 0.00427
2
2
Ans.
0.0123 rad Ans.
F2 y a2 (l x) 2
d y
d F1 y a1 (l x) 2
x a12 2lx
x a22 2lx
6 EIl
d x x l dx 6 EIl
x l
F2 y a2
F1 y a1
(6lx 2l 2 3 x 2 a12 )
6lx 2l 2 3x 2 a22
6 EIl
6 EIl
x l
C z
11103 sin 20o (400)
10502 4002
l a
l a
3
3
6(207)10 (306.8)10 (1050)
6 EIl
6 EIl
F1 y a1
2
2
1
F2 y a2
2
2
2
22.8 103 sin 25o (750)
10502 7502 0.0133 rad
6(207)103 (306.8)103 (1050)
Ans.
dz
d F1z a1 (l x) 2
F a (l x) 2
x a12 2lx 2 z 2
x a22 2lx
6 EIl
x l
dx 6 EIl
d x x l
C y
F a
F a
1z 1 6lx 2l 2 3 x 2 a12 2 z 2 6lx 2l 2 3 x 2 a22
6 EIl
6 EIl
x l
11103 cos 20o (400)
F1z a1 2
F2 z a2 2
2
2
l a1 6 EIl l a2 6(207)10
10502 4002
3
6 EIl
(306.8)103 (1050)
22.8 103 cos 25o (750)
10502 7502 0.0112 rad
3
3
6(207)10 (306.8)10 (1050)
Ans.
The slope magnitude is C 0.01332 0.01122 0.0174 rad Ans.
______________________________________________________________________________
4-35 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing
at O where ( O ) y = 0.00468 rad.
Since is inversely proportional to I,
Chapter 4 - Rev B, Page 26/81
new I new = old I old
4
I new = d new
/64 = old I old / new
1/4
or,
d new
64 old
I old
new
The absolute sign is used as the old slope may be negative.
1/4
64 0.00468
d new
0.119 8 1.82 in
Ans.
0.00105
______________________________________________________________________________
4-36 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the
bearing at C where ( C ) y = 0.0191 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
64 old
I old
new
1/ 4
64 0.0191
d new
39.76 103 62.0 mm
Ans.
0.00105
______________________________________________________________________________
4-37 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
64 old
I old
new
1/4
64 0.0104
d new
0.0491 1.77 in
Ans.
0.00105
______________________________________________________________________________
4-38 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad.
See the solution to Prob. 4-35 for the development of the equation
Chapter 4 - Rev B, Page 27/81
1/4
d new
64 old
I old
new
1/4
64 0.00750
d new
7 854 32.7 mm
Ans.
0.00105
______________________________________________________________________________
4-39 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-33, I = 0.119 8 in4, and the maximum slope = 0.0222 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
64 old
I old
new
1/4
64 0.0222
d new
0.119 8 2.68 in
Ans.
0.00105
______________________________________________________________________________
4-40 The required new slope in radians is new = 0.06( /180) = 0.00105 rad.
In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
64 old
I old
new
1/4
64 0.0174
d new
306.8 103 100.9 mm
Ans.
0.00105
______________________________________________________________________________
4-41 I AB = 14/64 = 0.04909 in4, J AB = 2 I AB = 0.09818 in4, I BC = (0.25)(1.5)3/12 = 0.07031 in4,
I CD = (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 102, b/c = 1.5/0.25 = 6 = 0.299.
The deflection can be broken down into several parts
1. The vertical deflection of B due to force and moment acting on B (y 1 ).
2. The vertical deflection due to the slope at B, B1 , due to the force and moment acting on
B (y 2 = CD B1 = 2 B1 ).
Chapter 4 - Rev B, Page 28/81
3. The vertical deflection due to the rotation at B, B2 , due to the torsion acting at B (y 3 =
BC B1 = 5 B1 ).
4. The vertical deflection of C due to the force acting on C (y 4 ).
5. The rotation at C, C , due to the torsion acting at C (y 3 = CD C = 2 C ).
6. The vertical deflection of D due to the force acting on D (y 5 ).
1. From Table A-9, beams 1 and 4 with F = 200 lbf and M B = 2(200) = 400 lbfin
200 63
400 62
y1
0.01467 in
3 30 106 0.04909 2 30 106 0.04909
2. From Table A-9, beams 1 and 4
d Fx 2
M B x 2
M x
Fx
B1
x 3l
3x 6l B
EI x l
2 EI x l 6 EI
dx 6 EI
6
l
200 6 2 400 0.004074 rad
Fl 2M B
6
2 EI
2 30 10 0.04909
y 2 = 2(0.004072) = 0.00815 in
3. The torsion at B is T B = 5(200) = 1000 lbfin. From Eq. (4-5)
TL
1000 6
0.005314 rad
B2
6
JG AB 0.09818 11.5 10
y 3 = 5(0.005314) = 0.02657 in
4. For bending of BC, from Table A-9, beam 1
y4
200 53
3 30 106 0.07031
0.00395 in
5. For twist of BC, from Eq. (3-41), p. 102, with T = 2(200) = 400 lbfin
C
400 5
0.299 1.5 0.253 11.5 106
0.02482 rad
y 5 = 2(0.02482) = 0.04964 in
6. For bending of CD, from Table A-9, beam 1
y6
200 23
3 30 106 0.01553
0.00114 in
Chapter 4 - Rev B, Page 29/81
Summing the deflections results in
6
yD yi 0.01467 0.00815 0.02657 0.00395 0.04964 0.00114 0.1041 in Ans.
i 1
This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71.
______________________________________________________________________________
4-42 The deflection of D in the x direction due to F z is from:
1. The deflection due to the slope at B, B1 , due to the force and moment acting on B (x 1 =
BC B1 = 5 B1 ).
2. The deflection due to the moment acting on C (x 2 ).
1. For AB, I AB = 14/64 = 0.04909 in4. From Table A-9, beams 1 and 4
2
M B x 2
M x
d Fx
Fx
x
l
3
3x 6l B
EI x l
2 EI x l 6 EI
dx 6 EI
B1
6
l
100 6 2 200 0.002037 rad
Fl 2M B
6
2 EI
2 30 10 0.04909
x 1 = 5( 0.002037) = 0.01019 in
2. For BC, I BC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4
x2
2 100 5
M Cl 2
0.04267 in
2 EI
2 30 106 0.001953
The deflection of D in the x direction due to F x is from:
3. The elongation of AB due to the tension. For AB, the area is A = 12/4 = 0.7854 in2
150 6
Fl
x3
3.82 105 in
6
AE
0.7854
30
10
AB
4. The deflection due to the slope at B, B2 , due to the moment acting on B (x 1 = BC B2 =
5 B2 ). With I AB = 0.04907 in4,
B2
5 150 6
M Bl
0.003056 rad
EI
30 106 0.04909
Chapter 4 - Rev B, Page 30/81
x 4 = 5( 0.003056) = 0.01528 in
5. The deflection at C due to the bending force acting on C. With I BC = 0.001953 in4
150 53
Fl 3
x5
0.10667 in
3 30 106 0.001953
3EI BC
6. The elongation of CD due to the tension. For CD, the area is A = (0.752)/4 = 0.4418
in2
150 2
Fl
x6
2.26 105 in
6
AE CD 0.4418 30 10
Summing the deflections results in
6
xD xi 0.01019 0.04267 3.82 105
i 1
0.01528 0.10667 2.26 105 0.1749 in Ans.
______________________________________________________________________________
4-43
J OA = J BC = (1.54)/32 = 0.4970 in4, J AB = (14)/32 = 0.09817 in4, I AB = (14)/64 =
0.04909 in4, and I CD = (0.754)/64 = 0.01553 in4.
T lOA l AB lBC
Tl
Tl
Tl
GJ OA GJ AB GJ BC G J OA J AB J BC
250(12) 2
9
2
0.0260 rad
6
11.5(10 ) 0.4970 0.09817 0.4970
Simplified
Tl
250(12)(13)
s
GJ 11.5 106 0.09817
Ans.
s 0.0345 rad
Ans.
Simplified is 0.0345/0.0260 = 1.33 times greater Ans.
yD
Fy lOC 3
3EI AB
s lCD
Fy lCD 3
3EI CD
250 133
3(30)106 0.04909
0.0345(12)
250 123
3(30)106 0.01553
yD 0.847 in
Ans.
______________________________________________________________________________
4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches
Chapter 4 - Rev B, Page 31/81
y
3000 /12 x 2 25 x 2 x3 25 12 3
wx
2lx 2 x3 l 3
24 EI
24 30 106 485
7.159 1010 x 27 106 600 x 2 x3
Ans.
The maximum height occurs at x = 25(12)/2 = 150 in
ymax 7.159 1010 150 27 106 600 1502 1503 1.812 in
Ans.
______________________________________________________________________________
4-45 From Table A-9-6,
Fbx 2
x b2 l 2
6 EIl
Fb 3
yL
x b2 x l 2 x
6 EIl
dyL
Fb
3x2 b2 l 2
dx 6 EIl
yL
dyL
dx
Let
dyL
dx
Fb b 2 l 2
6 EIl
x 0
and set I
x 0
dL
d L4
64
. Thus,
32 Fb b 2 l 2
1/ 4
Ans.
3 El
For the other end view, observe beam 6 of Table A-9 from the back of the page, noting
that a and b interchange as do x and –x
dR
32 Fa l 2 a 2
3 El
1/4
Ans.
For a uniform diameter shaft the necessary diameter is the larger of d L and d R .
______________________________________________________________________________
4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the
solution for d L of Prob. 4-45,
Chapter 4 - Rev B, Page 32/81
32nFb l 2 b 2
d
3 El
1/4
d
4
32(1.28)(3000)(200) 3002 2002
3 (207)103 (300)(0.001)
d 38.1 mm
I
38.14
Ans.
103.4 103 mm 4
64
From Table A-9, beam 6, the maximum deflection will occur in BC where dy BC /dx = 0
d Fa l x 2
x a 2 2lx 0 3 x 2 6lx a 2 2l 2 0
dx 6 EIl
3x 2 6 300 x 1002 2 3002 0 x 2 600 x 63333 0
x
1
600 600 2 4(1)63 333 463.3, 136.7 mm
2
x = 136.7 mm is acceptable.
Fa l x 2
ymax
x a 2 2lx
6 EIl
x 136.7 mm
3 103 100 300 136.7
136.7 2 1002 2 300 136.7 0.0678 mm
6 207 10 103.4 10 300
3
3
Ans.
______________________________________________________________________________
4-47 I = (1.254)/64 = 0.1198 in4. From Table A-9, beam 6
2
F a (l x) 2
F b x
( x a12 2lx) 2 2 ( x 2 b2 2 l 2 )
1 1
6 EIl
6 EIl
2
2
150(5)(20 8)
2
2
8 5 2(20)(8)
6
6(30)10
0.1198
(20)
250(10)(8)
2
2
2
8
10
20
6
6(30)10 0.1198 (20)
2
1/2
Ans.
0.0120 in
______________________________________________________________________________
Chapter 4 - Rev B, Page 33/81
4-48 I = (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9.
For F 1 = 150 lbf:
0x5
150 15 x
Fb x
y 1 1 x 2 b12 l 2
x 2 152 202
6
6 EIl
6 30 10 0.1198 20
5.217 106 x x 2 175
5 x 20
y
(1)
F1a1 l x 2
150 5 20 x
x 2 52 2 20 x
x a12 2lx
6
6 EIl
6 30 10 0.1198 20
1.739 106 20 x x 2 40 x 25
(2)
For F 2 = 250 lbf:
0 x 10
250 10 x
Fb x
z 2 2 x 2 b22 l 2
x 2 102 202
6
6 EIl
6 30 10 0.1198 20
5.797 106 x x 2 300
(3)
10 x 20
F a l x 2
250 10 20 x
x 2 102 2 20 x
z 2 2
x a22 2lx
6 EIl
6 30 106 0.1198 20
5.797 106 20 x x 2 40 x 100
(4)
Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too
numerous to tabulate here but the plot is shown below, where the maximum deflection of
= 0.01255 in occurs at x = 9.9 in. Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 34/81
4-49 The larger slope will occur at the left end.
From Table A-9, beam 8
MBx 2
( x 3a 2 6al 2l 2 )
6 EIl
dy AB M B
(3 x 2 3a 2 6al 2l 2 )
6 EIl
dx
y AB
With I = d 4/64, the slope at the left bearing is
dy AB
dx
A
x0
MB
(3a 2 6al 2l 2 )
4
6 E d / 64 l
Solving for d
32M B
32(1000)
3(42 ) 6(4)(10) 2 102
d4
3a 2 6al 2l 2 4
6
3 E Al
3 (30)10 (0.002)(10)
0.461 in
Ans.
______________________________________________________________________________
4-50 From Table A-5, E = 10.4 Mpsi
M O = 0 = 18 F BC 6(100) F BC = 33.33 lbf
The cross sectional area of rod BC is A = (0.52)/4 = 0.1963 in2.
The deflection at point B will be equal to the elongation of the rod BC.
33.33(12)
FL
6.79 105 in
yB
6
AE BC 0.1963 30 10
Ans.
______________________________________________________________________________
4-51 M O = 0 = 6 F AC 11(100)
F AC = 183.3 lbf
The deflection at point A in the negative y direction is equal to the elongation of the rod
AC. From Table A-5, E s = 30 Mpsi.
183.3 12
FL
yA
3.735 104 in
2
6
0.5 / 4 30 10
AE AC
By similar triangles the deflection at B due to the elongation of the rod AC is
y A y B1
6
18
yB1 3 y A 3(3.735)104 0.00112 in
From Table A-5, E a = 10.4 Mpsi
The bar can then be treated as a simply supported beam with an overhang AB. From Table
A-9, beam 10
Chapter 4 - Rev B, Page 35/81
dy
Fa 2
Fa 2
d F (x l)
2
yB 2 BD BC
l
a
x
l
a
x
l
(
)
7
(
)
(3
)
(l a)
x l a 3EI
dx 6 EI
dx x l a 3EI
Fa 2
F
Fa 2
7 Fa
2
3( x l ) 3a ( x l ) a (3 x l ) |x l a
(l a )
(2l 3a)
(l a)
7
3EI
6 EI
3EI
6 EI
100 52
7 100 5
(6 5)
2(6) 3(5)
3(10.4)106 0.25(23 ) / 12
6(10.4)106 0.25(23 ) / 12
0.01438 in
Ans.
y B = y B1 + y B2 = 0.00112 0.01438 = 0.0155 in
______________________________________________________________________________
4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa.
FlOC l AB 2
Fl AC l AB 2
Fl AB 3
Fl AB 3
Tl
Tl
yB
l
l
AB
AB
3EI AB G dOC 4 / 32 G d AC 4 / 32 3E d34 / 64
GJ OC
GJ AC
l AC
32 Fl AB 2 lOC
2l AB
4
4
4
GdOC Gd AC 3Ed AB
The spring rate is k = F/ y B . Thus
32l AB 2
k
lOC
l AC
2l AB
4
4
4
GdOC Gd AC 3Ed AB
1
32 2002
2 200
200
200
3
4
3
4
3
4
79.3
10
18
79.3
10
12
3
207
10
8
1
8.10 N/mm
Ans.
_____________________________________________________________________________
4-53 For the beam deflection, use beam 5 of Table A-9.
F
R1 R2
2
F
F
1
, and 2
2k1
2k 2
y AB 1
1 2
l
x
Fx
(4 x 2 3l 3 )
48EI
1 k2 k1
x
y AB F
x
(4 x 2 3l 3 )
48EI
2k1 2k1k2l
Ans.
Chapter 4 - Rev B, Page 36/81
For BC, since Table A-9 does not have an equation (because of symmetry) an equation
will need to be developed as the problem is no longer symmetric. This can be done easily
using beam 6 of Table A-9 with a = l /2
F l / 2 l x 2 l 2
F Fk2 Fk1
yBC
x
x 2lx
2k1
2k1k 2l
EIl
4
1 k2 k1
l x 4 x 2 l 2 8lx Ans.
F
x
48 EI
2k1 2k1k2l
______________________________________________________________________________
4-54
Fa
F
, and R2 (l a )
l
l
Fa
F
1
, and 2
(l a )
lk1
lk2
R1
y AB 1
1 2
l
x
Fax 2
(l x 2 )
6 EIl
a
x
ax 2
y AB F
(l x 2 )
Ans.
k a k1 l a
2 2
6 EIl
k1l k1k2l
F (x l)
( x l ) 2 a(3 x l )
yBC 1 1 2 x
l
6 EI
a
x
(x l)
( x l ) 2 a(3 x l )
yBC F
Ans.
k a k1 l a
2 2
6 EI
k1l k1k2l
______________________________________________________________________________
4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB
(Table A-9, beam 6)
y AB
Fbx 2
x b2 l 2
6 EIl
dy AB
Fb
3x 2 b2 l 2 0
dx
6 EIl
x
l 2 b2
l2
, xmax
0.577l
3
3
3x 2 b2 l 2 0
Ans.
For x l/2, xmin l 0.577l 0.423l Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 37/81
4-56
M O 1(3000)(1500) 2500(2000)
9.5 106 N·mm
RO 1(3000) 2500 5 500 N
From Prob. 4-10,
I 4.14(106 ) mm 4
x2
1
M 9.5 10 5500 x 2500 x - 2000
2
dy
x3
2
EI
9.5 106 x 2 750 x 2 1250 x 2000 C1
dx
6
6
dy
0 at x 0 C1 0
dx
dy
x3
2
EI
9.5 106 x 2 750 x 2 1250 x 2000
dx
6
x4
3
6
2
3
EIy 4.75 10 x 916.67 x 416.67 x 2000 C2
24
y 0 at x 0 C2 0 , and therefore
y
1
3
114 106 x 2 22 103 x 3 x 4 10 103 x 2000
24 EI
yB
1
114 106 30002 22 103 30003
24 207 103 4.14 106
3
30004 10 103 3000 2000
25.4 mm
Ans.
M O = 9.5 (106) Nm. The maximum stress is compressive at the bottom of the beam where
y = 29.0 100 = 71 mm
9.5 106 (71)
My
163 106 Pa 163MPa Ans.
max
6
I
4.14(10 )
The solutions are the same as Prob. 4-10.
______________________________________________________________________________
4-57 See Prob. 4-11 for reactions: R O = 465 lbf and R C = 285 lbf. Using lbf and inch units
Chapter 4 - Rev B, Page 38/81
M = 465 x 450 x 721 300 x 1201
dy
2
2
EI
232.5 x 2 225 x 72 150 x 120 C1
dx
EIy = 77.5 x3 75 x 723 50 x 1203 C 1 x
y = 0 at x = 0 C 2 = 0
y = 0 at x = 240 in
0 = 77.5(2403) 75(240 72)3 50(240 120)3 + C 1 x
and,
EIy = 77.5 x3 75 x 723 50 x 1203 2.622(106) x
C 1 = 2.622(106) lbfin2
Substituting y = 0.5 in at x = 120 in gives
30(106) I ( 0.5) = 77.5 (1203) 75(120 72)3 50(120 120)3 2.622(106)(120)
I = 12.60 in4
Select two 5 in 6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4
12.60 1
Ans.
0.421 in
14.98 2
The maximum moment occurs at x = 120 in where M max = 34.2(103) lbfin
ymidspan
Mc 34.2(103 )(2.5)
max
5 710 psi O.K.
I
14.98
The solutions are the same as Prob. 4-17.
______________________________________________________________________________
4-58 I = (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in.
1
24
RO 12.5 39 (340) 453.0 lbf
2
39
12.5 2
1
M 453.0 x
x 340 x 15
2
dy
12.5 3
2
EI
226.5 x 2
x 170 x 15 C1
6
dx
3
3
EIy 75.5 x 0.5208 x 4 56.67 x 15 C1 x C2
y 0 at x 0 C2 0
y 0 at x 39 in
y
C1 6.385(10 4 ) lbf in 2 Thus,
1
3
75.5 x 3 0.5208 x 4 56.67 x 15 6.385 104 x
EI
Evaluating at x = 15 in,
Chapter 4 - Rev B, Page 39/81
1
75.5 153 0.5208 154 56.67 15 15 3 6.385 104 (15)
30(10 )(0.2485)
0.0978 in Ans.
yA
6
1
75.5 19.53 0.5208 19.54 56.67 19.5 15 3 6.385 104 (19.5)
30(10 )(0.2485)
0.1027 in Ans.
ymidspan
6
5 % difference
Ans.
The solutions are the same as Prob. 4-12.
______________________________________________________________________________
3 14 100
7 14 100
4-59 I = 0.05 in4, RA
420 lbf and RB
980 lbf
10
10
M = 420 x 50 x2 + 980 x 10 1
EI
dy
2
210 x 2 16.667 x 3 490 x 10 C1
dx
3
EIy 70 x 3 4.167 x 4 163.3 x 10 C1 x C2
y = 0 at x = 0 C 2 = 0
y = 0 at x = 10 in C 1 = 2 833 lbfin2. Thus,
y
1
70 x 3 4.167 x 4 163.3 x 10 3 2833 x
6
30 10 0.05
3
6.667 107 70 x 3 4.167 x 4 163.3 x 10 2833 x
Ans.
The tabular results and plot are exactly the same as Prob. 4-21.
______________________________________________________________________________
4-60 R A = R B = 400 N, and I = 6(323) /12 = 16 384 mm4.
First half of beam,
M = 400 x + 400 x 300 1
dy
2
EI
200 x 2 200 x 300 C1
dx
From symmetry, dy/dx = 0 at x = 550 mm
0 = 200(5502) + 200(550 – 300) 2 + C 1
C 1 = 48(106) N·mm2
EIy = 66.67 x3 + 66.67 x 300 3 + 48(106) x + C 2
Chapter 4 - Rev B, Page 40/81
y = 0 at x = 300 mm
C 2 = 12.60(109) N·mm3.
The term (EI)1 = [207(103)16 384] 1 = 2.949 (1010 ) Thus
y = 2.949 (1010) [ 66.67 x3 + 66.67 x 300 3 + 48(106) x 12.60(109)]
y O = 3.72 mm Ans.
y x = 550 mm =2.949 (1010) [ 66.67 (5503) + 66.67 (550 300)3
+ 48(106) 550 12.60(109)] = 1.11 mm Ans.
The solutions are the same as Prob. 4-13.
______________________________________________________________________________
4-61
1
M A Fa
l
1
M A 0 M A R2l F (l a) R2 l Fl Fa M A
M
B
0 R1l Fa M A
M R1 x M A R2 x l
R1
1
dy 1
1
2
R1 x 2 M A x R2 x l C1
dx 2
2
1
1
1
3
EIy R1 x 3 M A x 2 R2 x l C1 x C2
6
2
6
EI
y = 0 at x = 0
C2 = 0
y = 0 at x = l
1
1
C1 R1l 2 M Al . Thus,
6
2
EIy
y
1
1
1
1
3
1
R1 x 3 M A x 2 R2 x l R1l 2 M Al x
6
2
6
2
6
1
3
M A Fa x3 3M A x 2l Fl Fa M A x l Fal 2 2M Al 2 x
6 EIl
Ans.
In regions,
1
M A Fa x 3 3M A x 2l Fal 2 2 M Al 2 x
6 EIl
x
M A x 2 3lx 2l 2 Fa l 2 x 2
Ans.
6 EIl
y AB
Chapter 4 - Rev B, Page 41/81
1
3
M A Fa x3 3M A x 2l Fl Fa M A x l Fal 2 2M Al 2 x
6 EIl
1
3
3
M A x3 3 x 2l x l 2 xl 2 F ax3 l a x l axl 2
6 EIl
1
2
M A x l l 2 Fl x l x l a 3 x l
6 EIl
yBC
x l
6 EI
M l F x l a 3x l
2
A
Ans.
The solutions reduce to the same as Prob. 4-17.
______________________________________________________________________________
w b a
1
R1
4-62 M D 0 R1l w b a l b b a
2l b a
2
2l
w
w
2
2
M R1 x
xa
xb
2
2
dy 1
w
w
3
3
EI
R1 x 2
xa
x b C1
6
6
dx 2
1
w
w
4
4
EIy R1 x 3
xa
x b C1 x C2
6
24
24
y = 0 at x = 0
y = 0 at x = l
C2 = 0
1 1
w
w
4
4
C1 R1l 3 l a l b
l 6
24
24
y
1
EI
1 w b a
w
w
4
2l b a x3 x a x b
2l
24
24
6
4
1 1 w b a
w
w
4
4
x
2l b a l 3 l a l b
l 6
2l
24
24
w
4
2 b a 2l b a x 3 l x a l x b
24 EIl
4
4
4
x 2 b a 2l b a l 2 l a l b
Ans.
The above answer is sufficient. In regions,
Chapter 4 - Rev B, Page 42/81
w
4
4
2 b a 2l b a x3 x 2 b a 2l b a l 2 l a l b
24 EIl
wx
4
4
2 b a 2l b a x 2 2 b a 2l b a l 2 l a l b
24 EIl
y AB
yBC
w
4
2 b a 2l b a x3 l x a
24 EIl
4
4
x 2 b a 2l b a l 2 l a l b
yCD
w
4
4
2 b a 2l b a x 3 l x a l x b
24 EIl
4
4
x 2 b a 2l b a l 2 l a l b
These equations can be shown to be equivalent to the results found in Prob. 4-19.
______________________________________________________________________________
4-63 I 1 = (1.3754)/64 = 0.1755 in4, I 2 = (1.754)/64 = 0.4604 in4,
R 1 = 0.5(180)(10) = 900 lbf
Since the loading and geometry are symmetric, we will only write the equations for half
the beam
2
For 0 x 8 in M 900 x 90 x 3
At x = 3, M = 2700 lbfin
Writing an equation for M / I, as seen in the figure,
the magnitude and slope reduce since I 2 > I 1 .
To reduce the magnitude at x = 3 in, we add the
term, 2700(1/I 1 1/ I 2 ) x 3 0. The slope of 900 at x = 3 in is also reduced. We
account for this with a ramp function, x 31 . Thus,
1 1
1 1
M 900 x
90
0
1
2700 x 3 900 x 3
x 3
I
I1
I2
I1 I 2
I1 I 2
0
1
5128 x 9520 x 3 3173 x 3 195.5 x 3
E
2
2
dy
1
2
3
2564 x 2 9520 x 3 1587 x 3 65.17 x 3 C1
dx
Boundary Condition:
dy
0 at x 8 in
dx
Chapter 4 - Rev B, Page 43/81
0 2564 8 9520 8 3 1587 8 3 65.17 8 3 C1
2
2
3
C 1 = 68.67 (103) lbf/in2
2
3
4
Ey 854.7 x 3 4760 x 3 529 x 3 16.29 x 3 68.67(103 ) x C2
y = 0 at x = 0
C2 = 0
Thus, for 0 x 8 in
1
2
3
4
y
854.7 x3 4760 x 3 529 x 3 16.29 x 3 68.7(103 ) x
6
30(10 )
Ans.
Using a spreadsheet, the following graph represents the deflection equation found above
The maximum is ymax 0.0102 in at x 8 in Ans.
______________________________________________________________________________
4-64 The force and moment reactions at the left support
are F and Fl respectively. The bending moment
equation is
M = Fx Fl
Plots for M and M /I are shown.
M /I can be expressed using singularity functions
M
F
Fl Fl
l
x
x
I
2 I1
2 I1 4 I1
2
0
F
l
x
2 I1
2
1
Chapter 4 - Rev B, Page 44/81
where the step down and increase in slope at x = l /2 are given by the last two terms.
Integrate
dy F 2 Fl
Fl
l
E
x
x
x
dx 4 I1
2 I1
4 I1
2
dy/dx = 0 at x = 0 C 1 = 0
1
F
l
x
4 I1
2
2
C1
2
3
F 3 Fl 2 Fl
l
F
l
Ey
x
x
x
x
C2
12 I1
4 I1
8 I1
2
12 I1
2
y = 0 at x = 0 C 2 = 0
2
3
l
F 3
l
2
2 x
y
2 x 6lx 3l x
24 EI1
2
2
3
2
F l
5 Fl 3
l
y x l /2
Ans.
2 6l 3l (0) 2(0)
24 EI1 2
96 EI1
2
2
3
F
l
3Fl 3
3
l
2
y x l
Ans.
2 l 6l l 3l l 2 x
24 EI1
2
16 EI1
2
The answers are identical to Ex. 4-10.
______________________________________________________________________________
4-65 Place a dummy force, Q, at the center. The reaction, R 1 = wl / 2 + Q / 2
M x
wx2
wl Q
M x
Q 2
2
2 2
Integrating for half the beam and doubling the results
1
ymax 2
EI
l /2
0
M
M
Q
2
dx
Q 0 EI
l /2
0
wl
wx2 x
x
2
2 dx
2
Note, after differentiating with respect to Q, it can be set to zero
l/2
w x 3l x 4
5w
ymax
x
l
x
dx
Ans.
0
2 EI 3
4 0
384 EI
______________________________________________________________________________
w
2 EI
l /2
2
4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at
the free end at positive to the left
M
wx2
x
M Qx
Q
2
Chapter 4 - Rev B, Page 45/81
l
l
1 l M
1 wx 2
w
ymax M
dx
x
dx
x 3 dx
EI
Q
EI
EI
2
2
Q 0
0
0
0
wl 4
Ans.
8 EI
______________________________________________________________________________
4-67 From Table A-7, I 1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4
First treat the end force as a variable, F.
Adding weight of channels of 2(5)/12 =
0.833 lbf/in. Using the variable x as
shown in the figure
M F x
5.833 2
x F x 2.917 x 2
2
M
x
F
A
1
EI
60
0
M
M
1
dx
EI
F
60
0
( F x 2.917 x 2 )( x ) d x
(150 / 3)(603 ) (2.917 / 4)(604 )
0.182 in in the direction of the 150 lbf force
30(106 )(3.70)
y A 0.182 in Ans.
______________________________________________________________________________
4-68 The energy includes torsion in AC, torsion in CO, and bending in AB.
Neglecting transverse shear in AB
M Fx,
M
x
F
In AC and CO,
T
T Fl AB ,
l AB
F
The total energy is
T 2l
T 2l
U
2GJ AC 2GJ CO
l AB
0
M2
dx
2 EI AB
The deflection at the tip is
Chapter 4 - Rev B, Page 46/81
U Tl AC T TlCO T
F GJ AC F GJ CO F
l AB
0
Tl l
Tl l
M M
1
dx AC AB CO AB
EI 3 F
GJ AC
GJ CO
EI AB
l AB
Fx dx
2
0
2
2
3
3
Tl AC l AB TlCO l AB Fl AB
Fl AC l AB
FlCO l AB
Fl AB
4
4
4
GJ AC
GJ CO 3EI AB G d AC
/ 32 G dCO
/ 32 3E d AB
/ 64
2
l AC
l
32 Fl AB
2l AB
CO4
4
4
Gd AC GdCO 3Ed AB
k
F
2
32l AB
l AC
l
2l AB
CO4
4
4
Gd AC Gd CO 3Ed AB
1
1
2 200
200
200
8.10 N/mm Ans.
32 2002 79.3 103 184 79.3 103 124 3 207 103 84
______________________________________________________________________________
4-69 I 1 = (1.3754)/64 = 0.1755 in4, I 2 = (1.754)/64 = 0.4604 in4
Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in
R1
1
1
(10)180 Q 900 0.5Q
2
2
For 0 x 3 in M 900 0.5Q x
M
0.5 x
Q
M
0.5 x
Q
For 3 x 13 in M 900 0.5Q x 90( x 3)2
By symmetry it is equivalent to use twice the integral from 0 to 8
3
8
8 M M
1
1
2
2
2
dx
900 x dx
900 x 90 x 3 x dx
EI 2 3
0 EI Q Q 0 EI1 0
3
300 x 3
1
EI1 0 EI 2
8
1 4
9 2
3
3
300 x 90( 4 x 2 x 2 x )
3
120.2 103
8100
1
8100
3
3
145.5 10 25.3110
30 106 0.1755 30 106 0.4604
EI1 EI 2
0.0102 in
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 47/81
4-70 I = (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (0.52)/4 = 0.1963 in2.
Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB.
Resolve the force F into components in the x and y directions obtaining 0.6 F in the
horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates
strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force
creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the
dummy variable x to originate at the end where the loads are applied on each segment,
0.6 F: AB
OA
M
0.6 x
F
M 0.6 F x
M 4.2 F
Fa 0.6 F
M
4.2
F
Fa
0.6
F
0.8 F: AB
M 0.8 F x
M
0.8 x
F
OA
M 0.8 F x
M
0.8 x
F
T
5.6
F
Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection
of B in the direction of F is*
T 5.6 F
B F
U Fa L Fa TL T 1
M
M
dx
F AE OA F JG OA F EI
F
0.6 15 15
5.6 15 15
0.6
5.6
6
0.1963 30 10
6.136 103 11.5 106
7
15
15 4.22
15
2
x
d
x
dx
0.6
30 106 3.068 103 0
30 106 3.068 103 0
7
15
15
15
2
2
0.8 x d x
0.8 x d x
6
6
3
3
30 10 3.068 10 0
30 10 3.068 10 0
1.38 105 0.1000 6.71103 0.0431 0.0119 0.1173
0.279 in
Ans.
Chapter 4 - Rev B, Page 48/81
*Note. This is not the actual deflection of point B. For this, dummy forces must be placed
on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and
then substitute the values of F x = 9 lbf, F y = 12 lbf, and F z = 0. This can be done
separately and then use superposition. The actual deflections of B are
B = 0.0831 i 0.2862 j 0.00770 k in
From this, the deflection of B in the direction of F is
B F 0.6 0.0831 0.8 0.2862 0.279 in
which agrees with our result.
______________________________________________________________________________
4-71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending.
I AB = (14)/64 = 0.04909 in4, J AB = 2 I AB = 0.09818 in4, I BC = 0.25(1.53)/12 = 0.07031 in4,
I CD = (0.754)/64 = 0.01553 in4.
For the torsion of bar BC, Eq. (3-41) is in the form of =TL/(JG), where the equivalent of
J is J eq = bc 3. With b/c = 1.5/0.25 = 6, J BC = bc 3 = 0.299(1.5)0.253 = 7.008 (103) in4.
Use the dummy variable x to originate at the end where the loads are applied on each
segment,
M
AB: Bending M F x 2 F
x 2
F
T
Torsion T 5 F
5
F
M
BC: Bending M F x
x
F
T
Torsion T 2 F
2
F
M
CD: Bending M F x
x
F
U
Tl T
1
M
M
dx
F
JG F
EI
F
6
5F 6
2F 5
1
2
5
2
F x 2 d x
6
3
6
6
0.09818 11.5 10
7.008 10 11.5 10
30 10 0.04909 0
D
5
2
1
1
F x 2d x
F x 2d x
6
6
30 10 0.07031 0
30 10 0.01553 0
1.329 104 F 2.482 104 F 1.141104 F 1.98 105 F 5.72 106 F
5.207 10 4 F 5.207 104 200 0.104 in
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 49/81
4-72 A AB = (12)/4 = 0.7854 in2, I AB = (14)/64 = 0.04909 in4, I BC = 1.5 (0.253)/12 = 1.953
(103) in4, A CD = (0.752)/4 = 0.4418 in2, I AB = (0.754)/64 = 0.01553 in4. For ( D ) x let
F = F x = 150 lbf and F z = 100 lbf . Use the dummy variable x to originate at the end
where the loads are applied on each segment,
M y
M y Fz x
0
CD:
F
Fa
Fa F
1
F
M y
M y F x 2 Fz
x
BC:
F
Fa
Fa Fz
0
F
M y
5
M y 5F 2 Fz Fz x
AB:
F
Fa
Fa F
1
F
5
Fa
U
FL
1
D x
F x 2 Fz x d x
F AE CD F EI BC 0
1
EI AB
F 2
0.4418 30 10
6
6
5F 2 F
z
0
1
FL Fa
Fz x 5 d x
AE AB F
1
3
F
5 Fz 52
3
30 10 1.953 10 3
6
F 6
F
1
25 F 6 10 Fz 6 z 62 5
1
6
2
30 10 0.04909
0.7854 30 10
6
1.509 107 F 7.112 104 F 4.267 104 Fz 1.019 104 F
1.019 104 Fz 2.546 107 F 8.135 104 F 5.286 104 Fz
Substituting F = F x = 150 lbf and F z = 100 lbf gives
D x 8.135 104 150 5.286 104 100 0.1749 in
Ans.
______________________________________________________________________________
4-73 I OA = I BC = (1.54)/64 = 0.2485 in4, J OA = J BC = 2 I OA = 0.4970 in4, I AB = (14)/64 =
0.04909 in4, J AB = 2 I AB = 0.09818 in4, I CD = (0.754)/64 = 0.01553 in4
Let F y = F, and use the dummy variable x to originate at the end where the loads are
applied on each segment,
Chapter 4 - Rev B, Page 50/81
OC:
M Fx
M
x,
F
DC:
M Fx
M
x
F
T 12 F
T
12
F
U
1
M
TL T
M
dx
EI
F
F
JG OC F
The terms involving the torsion and bending moments in OC must be split up because of
the changing second-area moments.
D y
D y
2
12 F 4
12 F 9
1
12
12
F x 2d x
6
6
6
0.4970 11.5 10
0.09818 11.5 10
30 10 0.2485 0
11
13
12
1
1
1
F x 2d x
F x 2d x
F x 2d x
6
6
6
30 10 0.04909 2
30 10 0.2485 11
30 10 0.01553 0
1.008 104 F 1.148 103 F 3.58 107 F
2.994 104 F 3.872 105 F 1.2363 103 F
2.824 103 F 2.824 103 250 0.706 in
Ans.
For the simplified shaft OC,
B y
12 F 13
13
12
1
1
12
F x 2d x
F x 2d x
6
6
6
0.09818 11.510
30 10 0.04909 0
30 10 0.01553 0
1.6580 103 F 4.973 104 F 1.2363 103 F 3.392 103 F 3.392 103 250
0.848 in
Ans.
Simplified is 0.848/0.706 = 1.20 times greater Ans.
______________________________________________________________________________
4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is
R C = Q + (6/18)100 = Q + 33.33
This is the axial force in member BC. Isolating the beam, we find that the moment is not a
function of Q, and thus does not contribute to the strain energy. Thus, only energy in the
member BC needs to be considered. Let the axial force in BC be F, where
Chapter 4 - Rev B, Page 51/81
F Q 33.33
F
1
Q
FL F
0 33.3312
1 6.79 105 in Ans.
2
6
AE
Q
BC
Q 0 0.5 / 4 30 10
Q 0
______________________________________________________________________________
B
4-75
U
Q
I OB = 0.25(23)/12 = 0.1667 in4
A AC = (0.52)/4 = 0.1963 in2
M O = 0 = 6 R C 11(100) 18 Q
R C = 3Q + 183.3
M A = 0 = 6 R O 5(100) 12 Q
R O = 2Q + 83.33
Bending in OB.
BD: Bending in BD is only due to Q which when set to zero after differentiation
gives no contribution.
AD: Using the variable x as shown in the figure above
M
7 x
Q
OA: Using the variable x as shown in the figure above
M 100 x Q 7 x
M 2Q 83.33 x
Axial in AC:
F 3Q 183.3
M
2 x
Q
F
3
Q
Chapter 4 - Rev B, Page 52/81
U
FL F
1
M
B
M Q dx
Q Q 0 AE Q Q 0 EI
Q 0
183.3 12
1
3
6
0.1963 30 10
EI
5
6
100 x 7 x d x 2 83.33 x dx
2
0
0
5
6
2
1.12110
100
x
7
x
d
x
166.7
x
dx
0
10.4 106 0.1667 0
1
3
1.121103 5.768 107 100 129.2 166.7 72 0.0155 in
Ans.
______________________________________________________________________________
4-76
There is no bending in AB. Using the variable, rotating counterclockwise from B
M
R sin
P
Fr
cos
P
F
sin
P
M PR sin
Fr P cos
F P sin
MF
2 PR sin 2
P
A 6(4) 24 mm 2 , ro 40 12 (6) 43 mm, ri 40 12 (6) 37 mm,
From Table 3-4, p.121, for a rectangular cross section
6
rn
39.92489 mm
ln(43 / 37)
From Eq. (4-33), the eccentricity is e = R r n =40 39.92489 = 0.07511 mm
From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa
From Table 4-1, C = 1.2
From Eq. (4-38)
M M
1 MF
2
2 F R F
2
2 CFr R Fr
d
d
d
0 AeE
0 AE
0 AE
0
AG P
P
P
P
d
2
P R sin
2 PR sin
2 2 PR sin
2 CPR cos
d
d
d
d
0
0
0
0
AeE
AE
AE
AG
EC
PR R
(10)(40) 40
(207 103 )(1.2)
1
2
1
2
4 AE e
G 4(24)(207 103 ) 0.07511
79.3 103
0.0338 mm
Ans.
______________________________________________________________________________
2
2
2
2
Chapter 4 - Rev B, Page 53/81
4-77
Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q
which when set to zero after differentiation gives no contribution. For section BC use the
variable, rotating counterclockwise from B
M
M PR sin Q R R sin
R 1 sin
Q
Fr
cos
Fr P Q cos
Q
F
F P Q sin
sin
Q
MF PR sin QR 1 sin P Q sin
MF
PR sin 2 PR sin 1 sin 2QR sin 1 sin
Q
But after differentiation, we can set Q = 0. Thus,
MF
PR sin 1 2sin
Q
A 6(4) 24 mm 2 , ro 40 12 (6) 43 mm, ri 40 12 (6) 37 mm,
From Table 3-4, p.121, for a rectangular cross section
6
rn
39.92489 mm
ln(43 / 37)
From Eq. (4-33), the eccentricity is e = R r n =40 39.92489 = 0.07511 mm
From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa
From Table 4-1, C = 1.2
From Eq. (4-38)
M M
1 MF
2
2 F R F
2
2 CFr R Fr
d
d
d
d
0 AeE Q
0 AE Q
0 AE
0
Q
AG Q
PR 2 2
PR 2 2
PR 2
sin
1
sin
sin
sin 1 2sin d
d
d
AeE 0
AE 0
AE 0
CPR 2
cos 2 d
0
AG
2
CE
R
PR PR
PR CPR PR
1
2
1 2
4 G
4 AeE 4 AE 4
AE 4 AG AE 4 e
3
1.2 207 10
40
1
2
4 79.3 103
4 0.07511
0.0766 mm
Ans.
______________________________________________________________________________
10 40
24 207 103
Chapter 4 - Rev B, Page 54/81
4-78
Note to the Instructor. The cross section
shown in the first printing is incorrect and the
solution presented here reflects the correction
which will be made in subsequent printings.
The corrected cross section should appear as
shown in this figure. We apologize for any
inconvenience.
A = 3(2.25) 2.25(1.5) = 3.375 in2
(1 1.5)(3)(2.25) (1 0.75 1.125)(1.5)(2.25)
R
2.125 in
3.375
Section is equivalent to the “T” section of Table 3-4, p. 121,
2.25(0.75) 0.75(2.25)
1.7960 in
2.25ln[(1 0.75) /1] 0.75ln[(1 3) / (1 0.75)]
e R rn 2.125 1.7960 0.329 in
rn
For the straight section
1
I z (2.25) 33 2.25(3)(1.5 1.125) 2
12
2
1
2.25
3
(1.5) 2.25 1.5(2.25) 0.75
1.125
2
12
2.689 in 4
For 0 x 4 in
M Fx
M
x,
F
V F
V
1
F
For /2
Fr F cos
Fr
cos ,
F
F F sin
F
sin
F
M
(4 2.125sin )
F
MF
MF F (4 2.125sin ) F sin
2 F (4 2.365sin ) sin
F
M F (4 2.125sin )
Chapter 4 - Rev B, Page 55/81
Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the
curved part, integrating from 0 to π/2, and double the results
/2
2 1 4 2
F (4)(1)
(4 2.125sin ) 2
Fx
dx
F
d
E I 0
3.375(G / E ) 0
3.375(0.329)
/2 2 F (4 2.125sin ) sin
F sin 2 (2.125)
d
d
0
0
3.375
3.375
2
/2 (1) F cos (2.125)
d
0
3.375(G / E )
/2
Substitute
I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi
2 6700 43
4
1
16 17(1) 4.516
6
30 10 3 2.689 3.375(11.5 / 30) 3.375(0.329) 2
4
2.125
2
2.125
4 1 2.125
3.375 4 3.375
4 3.375 11.5 / 30 4
Ans.
0.0226 in
______________________________________________________________________________
4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to , and double the results
M
M FR 1 cos
R 1 cos
F
Fr
Fr F sin
sin
F
F
cos
F F cos
F
MF F 2 Rcos 1 cos
MF
2 FRcos 1 cos
F
From Eq. (4-38),
FR 2
FR
2
(1 cos ) 2 d
cos 2 d
0
0
AeE
AE
2 FR
1.2 FR 2
cos 1 cos d
sin d
AE 0
AG 0
2 FR 3 R 3
E
0.6
AE 2 e 2
G
A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4,
p. 121,
Chapter 4 - Rev B, Page 56/81
h
4.5
34.95173 mm
ro
37.25
ln
ln
32.75
ri
and e = R r n = 35 34.95173 = 0.04827 mm. Thus,
2 F 35 3
35
3
207
0.6
0.08583F
3
13.5 207 10 2 0.04827 2
79.3
1
where F is in N. For = 1 mm, F
11.65 N
Ans.
0.08583
Note: The first term in the equation for dominates and this is from the bending moment.
Try Eq. (4-41), and compare the results.
______________________________________________________________________________
4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal
reaction, to applied at B, subject to the constraint ( B ) H 0.
rn
M
M
R sin
H
FR
(1 cos ) HR sin
2
0
2
By symmetry, we may consider only half of the wire form and use twice the strain energy
Eq. (4-41) then becomes,
( B ) H
/2
0
U
2
H EI
/2
M
M H Rd 0
0
FR
2 (1 cos ) HR sin ( R sin ) R d 0
F F
F 30
H 0 H
9.55 N
2 4
4
Ans.
Reaction at A is the same where H goes to the left. Substituting H into the moment
equation we get,
M
FR
(1 cos ) 2sin
2
R
M
[ (1 cos ) 2sin ]
F 2
0
2
Chapter 4 - Rev B, Page 57/81
U
2 M
2 /2 FR 2
P
[ (1 cos ) 2sin ]2 R d
M
Rd
2
0
P
EI F
EI
4
3
/2
FR
( 2 2 cos 2 4sin 2 2 2 cos 4 sin 4 sin cos ) d
2
0
2 EI
FR 3 2
2 4 2 2 4 2
2
2 EI 2
4
4
(30)(403 )
(3 2 8 4) FR 3 (3 2 8 4)
0.224 mm Ans.
8
8
EI
207 103 24 / 64
______________________________________________________________________________
4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the
curved beam portion. The shear and axial components will be negligible compared to
bending.
Place a fictitious force Q pointing to the left at point A.
M
M PR sin Q( R sin l )
R sin l
Q
Note that the strain energy in the straight portion is zero since there is no real force in that
section.
From Eq. (4-41),
/2
0
PR 2
EI
1 M
M
EI Q
/2
0
1
Rd
Q 0 EI
R sin 2 l sin d
/2
0
PR sin R sin l Rd
PR 2
1(52 )
R
l
(5) 4
6
4
EI 4
30 10 0.125 / 64 4
0.551 in
Ans.
______________________________________________________________________________
4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
M AB
x
P
Straight portion:
M AB Px
Curved portion:
M BC P R (1 cos ) l
M BC
R (1 cos ) l
P
From Eq. (4-41) with the addition of the bending strain energy in the straight portion of
the wire,
Chapter 4 - Rev B, Page 58/81
/2 1
M BC
M AB
1
M AB
dx
M BC
Rd
0 EI
0
P
EI
P
P l 2
PR /2
2
x dx
R(1 cos ) l d
EI 0
EI 0
3
Pl
PR /2 2
R (1 2 cos cos 2 ) 2 Rl (1 cos ) l 2 d
0
3EI EI
Pl 3 PR /2 2
R cos 2 2 R 2 2 Rl cos ( R l ) 2 d
3EI EI 0
Pl 3 PR 2
R 2 R 2 2 Rl ( R l ) 2
2
3EI EI 4
l
l3 3
2
2
3 4 R R 2 R 2 Rl 2 R( R l )
3
4 3
1
2
2
(5 ) 5 2(5 ) 2(5)(4) 5 5 4
6
4
2
30 10 0.125 / 64 3 4
P
EI
0.850 in Ans.
______________________________________________________________________________
4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
Place a dummy force, Q, at A vertically downward. The only load in the straight section is
the axial force, Q. Since this will be zero, there is no contribution.
In the curved section
M
R 1 cos
Q
M PR sin QR 1 cos
From Eq. (4-41)
/2 1 M
1
M
Rd
0
EI Q
Q 0 EI
PR 3
EI
/2
0
/2
0
PR sin R 1 cos Rd
PR 3 1 PR 3
sin sin cos d
1
EI 2 2 EI
1 53
2 30 106 0.1254 / 64
0.174 in
Ans.
______________________________________________________________________________
4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
Chapter 4 - Rev B, Page 59/81
Place a dummy force, Q, at A vertically downward. The load in the straight section is the
axial force, Q, whereas the bending moment is only a function of P and is not a function
of Q. When setting Q = 0, there is no axial or bending contribution.
In the curved section
M
R sin
Q
M P R 1 cos l QR sin
From Eq. (4-41)
/2
0
1 M
1
M
Rd
EI Q
Q 0 EI
PR 2
EI
/2
0
P R 1 cos l R sin Rd
/2
R sin R sin cos l sin d
0
1 52
PR 2
1
PR 2
R
l
R
R 2l
EI
2
2 EI
5 2 4 0.452 in
2 30 106 0.1254 / 64
Since the deflection is negative, is in the opposite direction of Q. Thus the deflection is
0.452 in
Ans.
______________________________________________________________________________
4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is
tension
FAB
1
FAB F
F
For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no
bending in section DE. For section BCD, let be counterclockwise originating at D
M
M FR sin
R sin
0
F
Using Eqs. (4-29) and (4-41)
Fl
F
1
M
Fl
FR3
AB
sin 2 d
1 0
M
Rd
0
EI F
AE
EI
AE AB F
403
Fl FR3 F l R3
9.81
80
AE 2 EI
E A 2 I 207 103 22 / 4 2 24 / 64
6.067 mm
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 60/81
4-86 A OA = 2(0.25) = 0.5 in2,
I OAB = 0.25(23)/12 = 0.1667 in4,
I AC = (0.54)/64 = 3.068 (10-3) in4
Applying a force F at point B, using
statics, the reaction forces at O and C
are as shown.
OA: Axial FOA 3F
FOA
3
F
M OA
2 x
F
Bending M OA 2 Fx
M AB
x
F
M AB F x
AB: Bending
AC: Isolating the upper curved section
M AC
3R sin cos 1
F
10
20
1
1
Fl FOA
2
4
Fx
dx
F x 2d x
AE
F
EI
EI
OAB 0
OAB 0
OA
M AC 3FR sin cos 1
9 FR3
EI AC
/2
sin cos 1
2
d
0
4 F 103
F 203
3F 10
3
0.5 10.4 106
3 10.4 106 0.1667 3 10.4 106 0.1667
9 F 103
30 10 3.068 10
6
/2
3
sin
2
2sin cos 2sin cos 2 2 cos 1 d
0
1.731105 F 7.691104 F 1.538 103 F 0.09778 F 1 2 2
4
2
4
0.0162 F 0.0162 100 1.62 in
Ans.
_____________________________________________________________________________
4-87 A OA = 2(0.25) = 0.5 in2,
I OAB = 0.25(23)/12 = 0.1667 in4,
I AC = (0.54)/64 = 3.068 (10-3) in4
Applying a vertical dummy force, Q, at A,
from statics the reactions are as shown. The
dummy force is transmitted through section
Chapter 4 - Rev B, Page 61/81
OA and member AC.
FOA
1
Q
OA: FOA 3F Q
M AC 3F Q R sin 3F Q R 1 cos
AC:
M AC
R sin cos 1
Q
Fl FOA 1 /2
M AC
Rd
M AC
Q
AE OA Q EI AC 0
Q 0
3FlOA
3FR 3
AE OA EI AC
3 100 10
10.4 10 0.5
6
/2
sin cos 1
2
d
0
3 100 103
1 2 2 0.462 in
4
2
30 10 3.068 10 4
6
3
Ans.
______________________________________________________________________________
4-88
I = (64)/64 = 63.62 mm4
0 /2
M
R sin
F
T
R (1 cos )
T FR (1 cos )
F
According to Castigliano’s theorem, a positive
U/ F will yield a deflection of A in the negative y direction. Thus the deflection in the
positive y direction is
M FR sin
( A ) y
U
1
F
EI
/2
0
F ( R sin ) 2 R d
1
GJ
/2
0
F [ R (1 cos )]2 R d
Integrating and substituting J 2 I and G E / 2 1
FR 3
3
4 (1 ) 4 2 4 8 (3 8) 4 EI
(250)(80)3
[4 8 (3 8)(0.29)]
12.5 mm Ans.
4(200)103 63.62
______________________________________________________________________________
FR 3
( A ) y
EI
4-89 The force applied to the copper and steel wire assembly is
Fc Fs 400 lbf
(1)
Since the deflections are equal, c s
Chapter 4 - Rev B, Page 62/81
Fl Fl
AE c AE s
Fc l
Fs l
2
6
3( / 4)(0.1019) (17.2)10
( / 4)(0.1055) 2 (30)106
Yields, Fc 1.6046 Fs . Substituting this into Eq. (1) gives
1.604 Fs Fs 2.6046Fs 400 Fs 153.6 lbf
Fc 1.6046 Fs 246.5 lbf
F
246.5
c c
10 075 psi 10.1 kpsi Ans.
Ac 3( / 4)(0.1019) 2
F
153.6
s s
17 571 psi 17.6 kpsi Ans.
As ( / 4)(0.10552 )
153.6(100)(12)
Fl
0.703 in
Ans.
2
6
AE s ( / 4)(0.1055) (30)10
______________________________________________________________________________
4-90
(a)
b 0.75(65) 48.8 kpsi
Bolt stress
Ans.
Fb 6 b Ab 6(48.8) (0.52 ) 57.5 kips
4
F
57.43
c b
13.9 kpsi Ans.
Ac ( / 4)(5.52 52 )
Total bolt force
Cylinder stress
(b) Force from pressure
P
F x = 0
P b + P c = 9.82
D2
4
p
(52 )
4
(500) 9817 lbf 9.82 kip
(1)
Since c b ,
Pc l
Pbl
2
2
( / 4)(5.5 5 ) E 6( / 4)(0.52 ) E
P c = 3.5 P b
(2)
Substituting this into Eq. (1)
P b + 3.5 P b = 4.5 P b = 9.82 P b = 2.182 kip. From Eq. (2), P c = 7.638 kip
Using the results of (a) above, the total bolt and cylinder stresses are
2.182
b 48.8
50.7 kpsi Ans.
6( / 4)(0.52 )
Chapter 4 - Rev B, Page 63/81
7.638
12.0 kpsi Ans.
( / 4)(5.52 52 )
______________________________________________________________________________
c 13.9
4-91
(1)
Tc + Ts = T
c = s
Tc l
Tl
s
JG c JG s
Substitute this into Eq. (1)
JG c
T T T
JG s s s
Ts
Tc
JG c
T
JG s s
(2)
JG s
T
JG s JG c
The percentage of the total torque carried by the shell is
% Torque
100 JG s
JG s JG c
Ans.
______________________________________________________________________________
4-92 R O + R B = W
(1)
OA = AB
Fl
Fl
AE OA AE AB
400 RO 600 RB
3
RO RB
AE
AE
2
Substitute this unto Eq. (1)
3
RB RB 4
2
RB 1.6 kN
(2)
Ans.
3
RO 1.6 2.4 kN Ans.
2
2 400(400)
Fl
0.0223 mm Ans.
A
3
AE OA 10(60)(71.7)(10 )
______________________________________________________________________________
From Eq. (2)
4-93 See figure in Prob. 4-92 solution.
Procedure 1:
1. Let R B be the redundant reaction.
Chapter 4 - Rev B, Page 64/81
2. Statics. R O + R B = 4 000 N
3. Deflection of point B. B
R O = 4 000 R B
(1)
RB 600 RB 4000 400
0
AE
AE
(2)
4. From Eq. (2), AE cancels and R B = 1 600 N Ans.
and from Eq. (1), R O = 4 000 1 600 = 2 400 N Ans.
2 400(400)
Fl
0.0223 mm Ans.
3
AE OA 10(60)(71.7)(10 )
______________________________________________________________________________
A
4-94 (a) Without the right-hand wall the deflection of point C would be
5 103 8
2 103 5
Fl
C
AE / 4 0.752 10.4 106 / 4 0.52 10.4 106
0.01360 in 0.005 in Hits wall
Ans.
(b) Let R C be the reaction of the wall at C acting to the left (). Thus, the deflection of
point C is now
5 103 RC 8
2 103 RC 5
C
/ 4 0.752 10.4 106 / 4 0.52 10.4 106
0.01360
4 RC
5
8
2 0.005
6
2
10.4 10 0.75 0.5
or,
0.01360 4.190 10 6 RC 0.005
RC 2 053 lbf 2.05 kip Ans.
Statics. Considering +, 5 000 R A 2 053 = 0 R A = 2 947 lbf = 2.95 kip Ans.
Deflection. AB is 2 947 lbf in tension. Thus
RA 8
2 947 8
5.13 103 in Ans.
2
6
AAB E / 4 0.75 10.4 10
______________________________________________________________________________
B AB
4-95 Since OA = AB ,
TOA (4) TAB (6)
JG
JG
3
TOA TAB
2
(1)
Chapter 4 - Rev B, Page 65/81
Statics. T OA + T AB = 200
(2)
Substitute Eq. (1) into Eq. (2),
3
5
TAB TAB TAB 200
2
2
From Eq. (1)
TAB 80 lbf in
Ans.
3
3
TOA TAB 80 120 lbf in
Ans.
2
2
80 6
180
A
0.3900 Ans.
4
6
/ 32 0.5 11.5 10
max
16T
d3
AB
16 80
0.53
OA
16 120
0.53
4890 psi 4.89 kpsi
3260 psi 3.26 kpsi
Ans.
Ans.
______________________________________________________________________________
4-96 Since OA = AB ,
TOA (4)
TAB (6)
4
/ 32 0.5 G / 32 0.754 G
Statics. T OA + T AB = 200
TOA 0.2963 TAB
TAB 154.3 lbf in
TOA 0.2963TAB 0.2963 154.3 45.7 lbf in
A
154.3 6
180
0.1480
4
6
/ 32 0.75 11.5 10
max
(1)
(2)
Substitute Eq. (1) into Eq. (2),
0.2963TAB TAB 1.2963TAB 200
From Eq. (1)
16T
d3
AB
OA
16 154.3
0.753
16 45.7
0.53
Ans.
Ans.
Ans.
1862 psi 1.86 kpsi
1862 psi 1.86 kpsi
Ans.
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 66/81
4-97 Procedure 1.
1. Arbitrarily, choose R C as a redundant reaction.
2. Statics. F x = 0,
12(103) 6(103) R O R C = 0
(1)
R O = 6(103) R C
3. The deflection of point C.
12(103 ) 6(103 ) RC (20) 6(103 ) RC (10) R (15)
C
C
0
AE
AE
AE
4. The deflection equation simplifies to
45 R C + 60(103) = 0 R C = 1 333 lbf 1.33 kip Ans.
R O = 6(103) 1 333 = 4 667 lbf 4.67 kip Ans.
From Eq. (1),
F AB = F B + R C = 6 +1.333 = 7.333 kips compression
FAB 7.333
14.7 kpsi Ans.
A
(0.5)(1)
Deflection of A. Since OA is in tension,
Rl
4 667(20)
A OA O OA
0.00622 in Ans.
AE
(0.5)(1)(30)106
______________________________________________________________________________
AB
4-98 Procedure 1.
1. Choose R B as redundant reaction.
2. Statics. R C = wl R B
(1)
1
(2)
M C wl 2 RB l a
2
3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9,
R l a w l a
4l l a l a 2 6l 2 0
yB B
3EI
24 EI
3
2
4. Solving for R B .
w 2
2
RB
6l 4l l a l a
8 l a
w
3l 2 2al a 2
8 l a
Ans.
Substituting this into Eqs. (1) and (2) gives
Chapter 4 - Rev B, Page 67/81
RC wl RB
w
5l 2 10al a 2
8 l a
Ans.
1 2
w
Ans.
wl RB l a l 2 2al a 2
2
8
______________________________________________________________________________
MC
4-99 See figure in Prob. 4-98 solution.
Procedure 1.
1. Choose R B as redundant reaction.
2. Statics. R C = wl R B
(1)
1
(2)
M C wl 2 RB l a
2
3. Deflection equation for point B. Let the variable x start at point A and to the right. Using
singularity functions, the bending moment as a function of x is
1
M w x 2 RB x a
2
1
M
xa
RB
1
U
1
M
yB
M
dx
RB EI 0 RB
l
1
1
1 1 2
w x 2 0 dx
w x RB x a x a dx 0
EI 0 2
EI a 2
l
l
or,
1 1
a
3
3
R
w l 4 a 4 l 3 a 3 B l a a a 0
2 4
3
3
Solving for R B gives
RB
w
8 l a
3
w
2
2
3 l 4 a 4 4 a l 3 a 3
8 l a 3l 2al a
Ans.
From Eqs. (1) and (2)
RC wl RB
MC
w
5l 2 10al a 2
8 l a
1 2
w
wl RB l a l 2 2al a 2
2
8
Ans.
Ans.
Chapter 4 - Rev B, Page 68/81
______________________________________________________________________________
4-100 Note: When setting up the equations for this problem, no rounding of numbers was
made. It turns out that the deflection equation is very sensitive to rounding.
Procedure 2.
1. Statics.
R 1 + R 2 = wl
1 2
wl
2
2. Bending moment equation.
R2l M 1
(1)
(2)
1
M R1 x w x 2 M 1
2
dy 1
1
EI
R1 x 2 w x 3 M 1 x C1
6
dx 2
1
1
1
EIy R1 x 3 w x 4 M 1 x 2 C1 x C2
6
24
2
(3)
(4)
EI = 30(106)(0.85) = 25.5(106) lbfin2.
3. Boundary condition 1. At x = 0, y = R 1 /k 1 = R 1 /[1.5(106)]. Substitute into Eq. (4)
with value of EI yields C 2 = 17 R 1 .
Boundary condition 2. At x = 0, dy /dx = M 1 /k 2 = M 1 /[2.5(106)]. Substitute into
Eq.
(3) with value of EI yields C 1 = 10.2 M 1 .
Boundary condition 3. At x = l, y = R 2 /k 3 = R 1 /[2.0(106)]. Substitute into Eq. (4)
with value of EI yields
1 3 1
1
R1l wl 4 M 1l 2 10.2 M 1l 17 R1
(5)
6
24
2
Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in,
are
12.75 R2
1
0
1
24
1
0
2287 12.75 532.8
R1 1
3
R2 12 10
M 1 576
Solving, the simultaneous equations yields
R 1 = 554.59 lbf, R 2 = 445.41.59 lbf, M 1 = 1310.1 lbfin
Ans.
For the deflection at x = l /2 = 12 in, Eq. (4) gives
Chapter 4 - Rev B, Page 69/81
y x 12in
1
1 500 4 1
1
12 1310.1122
554.59 123
6
6
24
12
2
25.5 10
10.2 1310.112 17 554.59
5.51103 in
Ans.
______________________________________________________________________________
4-101 Cable area, A
4
(0.52 ) 0.1963 in 2
Procedure 2.
1. Statics. R A + F BE + F DF = 5(103)
(1)
3 F DF + F BE = 10(103)
(2)
2. Bending moment equation.
1
M RA x FBE x 16 5000 x 32
1
dy 1
1
2
2
RA x 2 FBE x 16 2500 x 32 C1
dx 2
2
1
1
2500
3
3
EIy RA x3 FBE x 16
x 32 C1 x C2
6
6
3
EI
3. B.C. 1: At x = 0, y = 0
(3)
(4)
C2 = 0
B.C. 2: At x = 16 in,
FBE (38)
Fl
6.453(106 ) FBE
yB
6
0.1963(30)10
AE BE
Substituting into Eq. (4) and evaluating at x = 16 in
1
EIyB 30(106 )(1.2)( 6.453)(106 ) FBE RA 163 C1 (16)
6
(5)
Simplifying gives 682.7 R A + 232.3 F BE + 16 C 1 = 0
B.C. 2: At x = 48 in,
FDF (38)
Fl
6.453(106 ) FDF
yD
6
0.1963(30)10
AE
DF
Substituting into Eq. (4) and evaluating at x = 48 in,
1
1
2500
(48 32)3 48C1
RA 483 FBE (48 16)3
6
6
3
18 432 R A + 5 461 F BE + 232.3 F DF + 48 C 1 = 3.413(106)
EIyD 232.3FDF
Simplifying gives
(6)
Chapter 4 - Rev B, Page 70/81
Equations (1), (2), (5) and (6) in matrix form are
1
1
0 RA 5000
1
F 10 000
0
1
3
0
BE
0
682.7 232.3
0
16 FDF
6
3.413
10
C
18
432
5
461
232.3
48
1
Solve simultaneously or use software. The results are
R A = 970.5 lbf, F BE = 3956 lbf, F DF = 2015 lbf, and C 1 = 16 020 lbfin2.
3956
2015
BE
20.2 kpsi, DF
10.3 kpsi Ans.
0.1963
0.1963
EI = 30(106)(1.2) = 36(106) lbfin2
y
1
2500
3
3
970.5 3 3956
x
x 16
x 32 16 020 x
6
6
6
3
36 10
161.8 x 659.3 x 16
1
36 106
B: x = 16 in,
3
3
3
833.3 x 32 16 020 x
yB
1
161.8 163 16 020 16 0.0255 in
6
36 10
yC
1
161.8 323 659.3 32 16 3 16 020 32
6
36 10
Ans.
C: x = 32 in,
D: x = 48 in,
yD
0.0865 in
Ans.
1
161.8 483 659.3 48 16 3 833.3 48 32 3 16 020 48
6
36 10
0.0131 in Ans.
______________________________________________________________________________
4-102 Beam: EI = 207(103)21(103)
= 4.347(109) Nmm2.
Rods: A = ( /4)82 = 50.27 mm2.
Procedure 2.
1. Statics.
Chapter 4 - Rev B, Page 71/81
R C + F BE F DF = 2 000
(1)
R C + 2F BE = 6 000
(2)
2. Bending moment equation.
M = 2 000 x + F BE x 75 1 + R C x 150 1
dy
1
1
2
2
1000 x 2 FBE x 75 RC x 150 C1
dx
2
2
1000 3 1
1
3
3
EIy
x FBE x 75 RC x 150 C1 x C2
3
6
6
(3)
EI
(4)
3. B.C 1. At x = 75 mm,
FBE 50
Fl
4.805 106 FBE
yB
3
50.27 207 10
AE BE
Substituting into Eq. (4) at x = 75 mm,
4.347 109 4.805 106 FBE
1000
753 C1 75 C2
3
Simplifying gives
20.89 103 FBE 75C1 C2 140.6 106
(5)
B.C 2. At x = 150 mm, y = 0. From Eq. (4),
or,
1000
1
3
1503 FBE 150 75 C1 150 C2 0
3
6
70.31103 FBE 150C1 C2 1.125 109
(6)
B.C 3. At x = 225 mm,
FDF 65
Fl
6.246 106 FDF
yD
3
AE DF 50.27 207 10
Substituting into Eq. (4) at x = 225 mm,
Chapter 4 - Rev B, Page 72/81
4.347 109 6.246 106 FDF
1000
1
3
2253 FBE 225 75
3
6
1
3
RC 225 150 C1 225 C2
6
Simplifying gives
70.31103 RC 562.5 103 FBE 27.15 103 FDF 225C1 C2 3.797 109
(7)
Equations (1), (2), (5), (6), and (7) in matrix form are
2 103
1
1
1
0 0
RC
3
1
2
0
0
0
6 10
FBE
0
20.89 103
0
75 1 F 140.6 106
DF
3
9
0
70.3110
0
150 1 C1
1.125 10
3
3
3
C2
9
70.3110 562.5 10 27.15 10 225 1
3.797 10
Solve simultaneously or use software. The results are
Ans.
R C = 2378 N, F BE = 4189 N, F DF = 189.2 N
7
2
8
3
and C 1 = 1.036 (10 ) Nmm , C 2 = 7.243 (10 ) Nmm .
The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF = 189/50.27= 3.8 MPa Ans.
The deflections are
From Eq. (4) y A
1
7.243 108 0.167 mm
9
4.347 10
Ans.
For points B and D use the axial deflection equations*.
4189 50
Fl
yB
0.0201 mm
50.27 207 103
AE BE
Ans.
189 65
Fl
yD
Ans.
1.18 103 mm
3
AE DF 50.27 207 10
*Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for
calculating the very small deflections, especially for point D.
______________________________________________________________________________
4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have
Chapter 4 - Rev B, Page 73/81
U
0
M A
The bending moment at an angle to the x axis is
FR
M
M MA
1
1 cos
2
M A
The rotation at A is
/2
U
1
M
M
Rd 0
A
M A EI 0
M A
Thus,
1
EI
/2
M
A
0
FR
1 cos 1 Rd 0
2
FR FR
0
MA
2 2
2
or,
FR 2
1
2
Substituting this into the equation for M gives
FR
2
M
cos (1)
2
The maximum occurs at B where = /2
MA
M max M B
FR
Ans.
(b) Assume B is supported on a knife edge. The deflection of point D is U/ F. We will
deal with the quarter-ring segment and multiply the results by 4. From Eq. (1)
M R
2
cos
F 2
Thus,
U
4
D
F EI
/2
0
M
FR3
M
Rd
F
EI
/2
0
2
2
FR3 2
cos
d
EI 4
3
FR
2 8 Ans.
4 EI
______________________________________________________________________________
4-104
C 2 EI
Pcr
l2
I
D4 d 4
D4
1 K
64
64
2
4
C E D
Pcr 2
1 K 4
l
64
4
where K
d
D
Chapter 4 - Rev B, Page 74/81
1/ 4
64 Pcr l 2
D 3
Ans.
4
CE 1 K
______________________________________________________________________________
D 2 1 K 2 ,
I
D 4 1 K 4
D 4 1 K 2 1 K 2 , where K = d / D.
4
64
64
The radius of gyration, k, is given by
I D2
2
k
1 K 2
A 16
From Eq. (4-46)
S y2l 2
S y2l 2
Pcr
S
S
y
y
4 2 k 2CE
4 2 D 2 / 16 1 K 2 CE
/ 4 D 2 1 K 2
4-105 A
4 Pcr D 1 K
2
2
S
y
D 2 1 K 2 S y 4 Pcr
4S y2l 2 D 2 1 K 2
2 D 2 1 K 2 CE
4S y2l 2 1 K 2
1 K 2 CE
4 S y2l 2 1 K 2
4 Pcr
D
2
2
2
S y 1 K 1 K CE 1 K S y
1/2
1/2
S yl 2
Pcr
2
Ans.
2
2
2
S y 1 K CE 1 K
______________________________________________________________________________
0.9
4-106 (a) M A 0, (0.75)(800)
0.92 0.52
FBO (0.5) 0 FBO 1373 N
Using n d = 4, design for F cr = n d F BO = 4(1373) = 5492 N
l 0.92 0.52 1.03 m, S y 165 MPa
In-plane:
1/ 2
1/ 2
bh3 / 12
I
k
A
bh
l
1.03
142.7
k 0.007218
0.2887 h 0.2887(0.025) 0.007 218 m,
C 1.0
1/ 2
2
9
l 2 (207)(10 )
6
k 1 165(10 )
157.4
Chapter 4 - Rev B, Page 75/81
Since (l / k )1 (l / k ) use Johnson formula.
Try 25 mm x 12 mm,
2
165 106
1
6
Pcr 0.025(0.012) 165 10
(142.7)
29.1 kN
9
2
1(207)10
This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane.
Out-of-plane: k 0.2887(0.012) 0.003 464 in,
l
1.03
297.3
k 0.003 464
Since (l / k )1 (l / k ) use Euler equation.
Pcr 0.025(0.012)
C 1.2
1.2 2 207 109
8321 N
297.32
This is greater than the design load of 5492 N found earlier. It is also significantly less
than the in-plane P cr found earlier, so the out-of-plane condition will dominate. Iterate
the process to find the minimum h that gives P cr greater than the design load.
With h = 0.010, P cr = 4815 N (too small)
h = 0.011, P cr = 6409 N (acceptable)
Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans.
P
1373
10.4 106 Pa 10.4 MPa
dh
0.012(0.011)
No, bearing stress is not significant. Ans.
______________________________________________________________________________
(b) b
4-107 This is an open-ended design problem with no one distinct solution.
______________________________________________________________________________
F = 1500( /4)22 = 4712 lbf. From Table A-20, S y = 37.5 kpsi
P cr = n d F = 2.5(4712) = 11 780 lbf
4-108
(a) Assume Euler with C = 1
1/4
64 11790 502
64 Pcr l 2
d 3
3
6
CE
1
30
10
Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in
Pcr l 2
I d
64
C 2 E
4
1/ 4
1.193 in
Chapter 4 - Rev B, Page 76/81
l
50
160
k 0.3125
2 2 (1)30 106
126
37.5 103
2
6
4
30 10 / 64 1.25
14194 lbf
Pcr
502
1/2
1/2
2
l 2 CE
k 1 S y
Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory.
use Euler
Ans.
1/4
64 11780 162
0.675 in, so use d = 0.750 in
d 3
6
1 30 10
k = 0.750/4 = 0.1875 in
l
16
85.33
use Johnson
k 0.1875
(b)
2
37.5 103
1
3
Pcr 0.750 37.5 10
85.33
12 748 lbf
6
4
2
1 30 10
2
Use d = 0.75 in.
( c)
n( a )
14194
3.01
4 712
Ans.
12 748
2.71
Ans.
4 712
______________________________________________________________________________
n(b )
4-109 From Table A-20, S y = 180 MPa
4F sin = 2 943
735.8
sin
In range of operation, F is maximum when = 15
735.8
Fmax
2843 N per bar
sin15o
F
P cr = n d F max = 3.50 (2 843) = 9 951 N
l = 350 mm, h = 30 mm
Chapter 4 - Rev B, Page 77/81
Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm
l
350
242.6
k 1.443
2
9
l 2 1.4 207 10
180 106
k 1
Pcr A
C 2 E
l / k
2
5(30)
1/ 2
178.3
1.4 2 207 103
242.6
2
use Euler
7 290 N
Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm
350
l
202.1
k 1.732
1.4 2 207 103
C 2 E
12 605 N
6(30)
Pcr A
2
2
202.1
l / k
O.K. Use 25 6 mm bars Ans. The factor of safety is
12 605
4.43 Ans.
2843
______________________________________________________________________________
n
4-110 P = 1 500 + 9 000 = 10 500 lbf
Ans.
M A = 10 500 (4.5/2) 9 000 (4.5) +M = 0
M = 16 874 lbfin
e = M / P = 16 874/10 500 = 1.607 in Ans.
From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq.
(4-55)
k2
I 2.059
0.953 in 2
A 2.160
P ec
10500 1.607 3 / 2
1
17157 psi 17.16 kpsi Ans.
1 2
A k
2.160
0.953
______________________________________________________________________________
c
4-111 This is a design problem which has no single distinct solution.
______________________________________________________________________________
Chapter 4 - Rev B, Page 78/81
4-112 Loss of potential energy of weight = W (h + )
1
Increase in potential energy of spring = k 2
2
1
W (h + ) = k 2
2
2
W
2
W
or, 2
h 0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields
k
k
2 0.6 1.2 = 0
Taking the positive root (see discussion on p. 192)
1
max 0.6 (0.6) 2 4(1.2) 1.436 in
2
Ans.
Ans.
F max = k max = 100 (1.436) = 143.6 lbf
______________________________________________________________________________
4-113 The drop of weight W 1 converts potential energy, W 1 h, to kinetic energy
1 W1 2
v1 .
2 g
Equating these provides the velocity of W 1 at impact with W 2 .
1 W1 2
v1
v1 2 gh
(1)
2 g
Since the collision is inelastic, momentum is conserved. That is, (m 1 + m 2 ) v 2 = m 1 v 1 ,
where v 2 is the velocity of W 1 + W 2 after impact. Thus
W1h
W1 W2
W
v2 1 v1
g
g
v2
W1
W1
v1
2 gh
W1 W2
W1 W2
(2)
The kinetic and potential energies of W 1 + W 2 are then converted to potential energy of
the spring. Thus,
1 W1 W2 2
1
v2 W1 W2 k 2
2
g
2
Substituting in Eq. (1) and rearranging results in
W1 W2
W12 h
2
2
0
(3)
k
W1 W2 k
Solving for the positive root (see discussion on p. 192)
2
2
1 W1 W2
W12 h
W1 W2
8
2
4
2
k
k
W1 W2 k
(4)
Chapter 4 - Rev B, Page 79/81
W 1 = 40 N, W 2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm.
1
2
40 400
402 200
40 400
29.06 mm
4
8
32
40 400 32
32
2
2
Ans.
F max = k = 32(29.06) = 930 N Ans.
______________________________________________________________________________
1
4-114 The initial potential energy of the k 1 spring is V i = k1a 2 . The movement of the weight
2
1
1
2
W the distance y gives a final potential of V f = k1 a y k2 y 2 . Equating the two
2
2
energies give
1 2 1
1
2
k1a k1 a y k2 y 2
2
2
2
Simplifying gives
k1 k2 y 2 2ak1 y 0
2k1a
. Without damping the weight will vibrate between
k1 k2
2k1a
Ans.
these two limits. The maximum displacement is thus y max =
k1 k2
With W = 5 lbf, k 1 = 10 lbf/in, k 2 = 20 lbf/in, and a = 0.25 in
This has two roots, y = 0,
2 0.25 10
0.1667 in
Ans.
10 20
______________________________________________________________________________
ymax
Chapter 4 - Rev B, Page 80/81
Chapter 6
6-1
Eq. (2-21):
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Sut 3.4 H B 3.4(300) 1020 MPa
Se 0.5Sut 0.5(1020) 510 MPa
a 1.58, b 0.085
ka aSutb 1.58(1020) 0.085 0.877
Eq. (6-20):
kb 1.24d 0.107 1.24(10) 0.107 0.969
Eq. (6-18):
Se ka kb Se (0.877)(0.969)(510) 433 MPa Ans.
______________________________________________________________________________
6-2
(a) Table A-20:
Eq. (6-8):
S ut = 80 kpsi
Se 0.5(80) 40 kpsi
Ans.
(b) Table A-20:
Eq. (6-8):
S ut = 90 kpsi
Se 0.5(90) 45 kpsi
Ans.
(c) Aluminum has no endurance limit. Ans.
(d) Eq. (6-8):
S ut > 200 kpsi, Se 100 kpsi
Ans.
______________________________________________________________________________
6-3
Sut 120 kpsi, rev 70 kpsi
Fig. 6-18:
f 0.82
Eq. (6-8):
Se Se 0.5(120) 60 kpsi
Eq. (6-14):
( f Sut ) 2 0.82(120)
a
161.4 kpsi
Se
60
Eq. (6-15):
f Sut
1
b log
3
Se
2
1/ b
1
0.82(120)
log
0.0716
3
60
1
70 0.0716
116 700 cycles Ans.
Eq. (6-16):
N rev
161.4
a
______________________________________________________________________________
6-4
Sut 1600 MPa, rev 900 MPa
Fig. 6-18:
S ut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77.
Eq. (6-8):
S ut > 1400 MPa, so S e = 700 MPa
Eq. (6-14):
( f Sut ) 2 0.77(1600)
a
2168.3 MPa
700
Se
2
Chapter 6 - Rev. A, Page 1/66
Eq. (6-15):
f Sut
1
b log
3
Se
1
0.77(1600)
log
0.081838
3
700
1
1/ b
rev
900 0.081838
46 400 cycles Ans.
Eq. (6-16):
N
2168.3
a
______________________________________________________________________________
6-5
Sut 230 kpsi, N 150 000 cycles
Fig. 6-18, point is off the graph, so estimate:
f = 0.77
Eq. (6-8):
S ut > 200 kpsi, so Se Se 100 kpsi
Eq. (6-14):
( f Sut ) 2 0.77(230)
a
313.6 kpsi
Se
100
Eq. (6-15):
f Sut
1
b log
3
Se
Eq. (6-13):
S f aN b 313.6(150 000)0.08274 117.0 kpsi
2
1
0.77(230)
log
0.08274
3
100
Ans.
______________________________________________________________________________
6-6
Sut 1100 MPa = 160 kpsi
Fig. 6-18:
f = 0.79
Eq. (6-8):
Se Se 0.5(1100) 550 MPa
Eq. (6-14):
( f Sut ) 2 0.79(1100)
a
1373 MPa
Se
550
Eq. (6-15):
f Sut
1
b log
3
Se
Eq. (6-13):
S f aN b 1373(150 000)0.06622 624 MPa
2
1
0.79(1100)
log
0.06622
3
550
Ans.
______________________________________________________________________________
6-7
Sut 150 kpsi, S yt 135 kpsi, N 500 cycles
Fig. 6-18:
f = 0.798
From Fig. 6-10, we note that below 103 cycles on the S-N diagram constitutes the lowcycle region, in which Eq. (6-17) is applicable.
Chapter 6 - Rev. A, Page 2/66
Eq. (6-17):
log 0.798 /3
S f Sut N log f /3 150 500
122 kpsi
Ans.
The testing should be done at a completely reversed stress of 122 kpsi, which is below
the yield strength, so it is possible. Ans.
______________________________________________________________________________
6-8
The general equation for a line on a log S f - log N scale is S f = aNb, which is Eq. (6-13).
By taking the log of both sides, we can get the equation of the line in slope-intercept
form.
log S f b log N log a
Substitute the two known points to solve for unknowns a and b. Substituting point (1,
S ut ),
log Sut b log(1) log a
From which a Sut . Substituting point (103 , f Sut ) and a Sut
log f Sut b log103 log Sut
From which b 1/ 3 log f
S f Sut N (log f )/3
1 N 103
______________________________________________________________________________
6-9
Read from graph: 103 ,90 and (106 ,50). From S aN b
log S1 log a b log N1
log S 2 log a b log N 2
From which
log a
log S1 log N 2 log S 2 log N1
log N 2 / N1
log 90 log106 log 50 log103
log106 /103
2.2095
a 10log a 102.2095 162.0 kpsi
log 50 / 90
b
0.0851
3
( S f ) ax 162 N 0.0851 103 N 106 in kpsi
Ans.
Chapter 6 - Rev. A, Page 3/66
Check:
( S f ) ax 3 162(103 ) 0.0851 90 kpsi
10
( S f ) ax 6 162(106 ) 0.0851 50 kpsi
10
The end points agree.
______________________________________________________________________________
6-10
d = 1.5 in, S ut = 110 kpsi
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Se 0.5(110) 55 kpsi
a = 2.70, b = 0.265
k a aSut b 2.70(110) 0.265 0.777
Since the loading situation is not specified, we’ll assume rotating bending or torsion so
Eq. (6-20) is applicable. This would be the worst case.
kb 0.879d 0.107 0.879(1.5) 0.107 0.842
Eq. (6-18): Se ka kb Se 0.777(0.842)(55) 36.0 kpsi
Ans.
______________________________________________________________________________
6-11
For AISI 4340 as-forged steel,
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Eq. (6-20):
S e = 100 kpsi
a = 39.9, b = 0.995
k a = 39.9(260)0.995 = 0.158
0.107
0.75
0.907
kb
0.30
Each of the other modifying factors is unity.
S e = 0.158(0.907)(100) = 14.3 kpsi
For AISI 1040:
Se 0.5(113) 56.5 kpsi
ka 39.9(113) 0.995 0.362
kb 0.907 (same as 4340)
Each of the other modifying factors is unity
Se 0.362(0.907)(56.5) 18.6 kpsi
Not only is AISI 1040 steel a contender, it has a superior endurance strength.
______________________________________________________________________________
Chapter 6 - Rev. A, Page 4/66
6-12
D = 1 in, d = 0.8 in, T = 1800 lbfin, f = 0.9, and from Table A-20 for AISI 1020 CD,
S ut = 68 kpsi, and S y = 57 kpsi.
r 0.1
D
1
(a) Fig. A-15-15:
0.125,
1.25, K ts 1.40
d 0.8
d 0.8
Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and
(6-35b). We’ll use the equations.
a 0.190 2.51103 68 1.35 105 68 2.67 108 683 0.07335
2
1
qs
a
r
1
Eq. (6-32):
1
0.812
0.07335
1
0.1
K fs = 1 + q s (K ts 1) = 1 + 0.812(1.40 1) = 1.32
For a purely reversing torque of T = 1800 lbfin,
a K fs
Tr K fs 16T 1.32(16)(1800)
23 635 psi 23.6 kpsi
J
d3
(0.8)3
Eq. (6-8):
Se 0.5(68) 34 kpsi
Eq. (6-19):
k a = 2.70(68)0.265 = 0.883
Eq. (6-20):
k b = 0.879(0.8)0.107 = 0.900
Eq. (6-26):
k c = 0.59
Eq. (6-18) (labeling for shear):
S se = 0.883(0.900)(0.59)(34) = 15.9 kpsi
For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear.
Eq. (6-54):
S su = 0.67 S ut = 0.67(68) = 45.6 kpsi
Adjusting the fatigue strength equations for shear,
2
2
f S su
0.9(45.6)
Eq. (6-14):
a
105.9 kpsi
S se
15.9
Eq. (6-15):
f S su
1
1
0.9(45.6)
b log
log
0.137 27
3
3
15.9
S se
Eq. (6-16):
b 23.3 0.137 27
61.7 103 cycles
N a
a 105.9
1
1
Ans.
Chapter 6 - Rev. A, Page 5/66
(b) For an operating temperature of 750 F, the temperature modification factor,
from Table 6-4 is k d = 0.90.
S se = 0.883(0.900)(0.59)(0.9)(34) = 14.3 kpsi
f Ssu
a
S se
2
0.9(45.6)
f S su
1
b log
3
S se
2
14.3
117.8 kpsi
1
0.9(45.6)
log
0.152 62
3
14.3
1
1
b 23.3 0.152 62
40.9 103 cycles
N a
Ans.
117.8
a
______________________________________________________________________________
6-13
L 0.6 m, Fa 2 kN, n 1.5, N 104 cycles, Sut 770 MPa, S y 420 MPa (Table A-20)
First evaluate the fatigue strength.
Se 0.5(770) 385 MPa
ka 57.7(770) 0.718 0.488
Since the size is not yet known, assume a
typical value of k b = 0.85 and check later.
All other modifiers are equal to one.
Eq. (6-18):
S e = 0.488(0.85)(385) = 160 MPa
In kpsi, S ut = 770/6.89 = 112 kpsi
Fig. 6-18:
Eq. (6-14):
Eq. (6-15):
Eq. (6-13):
f = 0.83
2
2
f Sut
0.83(770)
a
2553 MPa
Se
160
f Sut
1
1
0.83(770)
b log
log
0.2005
3
3
160
Se
S f aN b 2553(104 )0.2005 403 MPa
Now evaluate the stress.
M max (2000 N)(0.6 m) 1200 N m
a max
Mc M b / 2 6 M 6 1200 7200
3 Pa, with b in m.
I
b(b3 ) /12 b3
b3
b
Compare strength to stress and solve for the necessary b.
Chapter 6 - Rev. A, Page 6/66
n
Sf
a
403 106
7200 / b3
1.5
b = 0.0299 m Select b = 30 mm.
Since the size factor was guessed, go back and check it now.
1/2
Eq. (6-25):
d e 0.808 hb 0.808b 0.808 30 24.24 mm
0.107
24.2
Eq. (6-20):
0.88
kb
7.62
Our guess of 0.85 was slightly conservative, so we will accept the result of
b = 30 mm.
Ans.
Checking yield,
max
7200
106 267 MPa
3
0.030
420
1.57
max 267
______________________________________________________________________________
ny
6-14
Sy
Given: w =2.5 in, t = 3/8 in, d = 0.5 in, n d = 2. From Table A-20, for AISI 1020 CD,
S ut = 68 kpsi and S y = 57 kpsi.
Eq. (6-8):
Se 0.5(68) 34 kpsi
Table 6-2:
Eq. (6-21):
Eq. (6-26):
ka 2.70(68) 0.265 0.88
k b = 1 (axial loading)
k c = 0.85
Eq. (6-18):
S e = 0.88(1)(0.85)(34) = 25.4 kpsi
Table A-15-1: d / w 0.5 / 2.5 0.2, K t 2.5
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). The relatively large radius is off the graph of Fig. 6-20, so we’ll assume the
curves continue according to the same trend and use the equations to estimate the notch
sensitivity.
a 0.246 3.08 103 68 1.51105 68 2.67 108 683 0.09799
2
1
0.836
a 1 0.09799
1
0.25
r
K f 1 q ( K t 1) 1 0.836(2.5 1) 2.25
q
Eq. (6-32):
1
Chapter 6 - Rev. A, Page 7/66
Fa
2.25 Fa
=
3Fa
A (3 / 8)(2.5 0.5)
a K f
Since a finite life was not mentioned, we’ll assume infinite life is desired, so the
completely reversed stress must stay below the endurance limit.
25.4
2
a 3Fa
Fa 4.23 kips Ans.
______________________________________________________________________________
nf
6-15
Se
Given: D 2 in, d 1.8 in, r 0.1 in, M max 25 000 lbf in, M min 0.
From Table A-20, for AISI 1095 HR, S ut = 120 kpsi and S y = 66 kpsi.
Eq. (6-8):
Se 0.5Sut 0.5 120 60 kpsi
Eq. (6-19):
Eq. (6-24):
k a aSutb 2.70(120) 0.265 0.76
Eq. (6-20):
kb 0.879 d e 0.107 0.879(0.666) 0.107 0.92
Eq. (6-26):
kc 1
Eq. (6-18):
Se ka kb kc Se (0.76)(0.92)(1)(60) 42.0 kpsi
d e 0.370d 0.370(1.8) 0.666 in
Fig. A-15-14: D / d 2 / 1.8 1.11,
r / d 0.1 / 1.8 0.056
Kt 2.1
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We’ll use the equations.
a 0.246 3.08 103 120 1.51105 120 2.67 108 1203 0.04770
2
1
q
1
Eq. (6-32):
a
r
1
0.87
0.04770
1
0.1
K f 1 q ( K t 1) 1 0.87(2.1 1) 1.96
I ( / 64)d 4 ( / 64)(1.8)4 0.5153 in 4
Mc 25 000(1.8 / 2)
43 664 psi 43.7 kpsi
I
0.5153
0
max
min
Chapter 6 - Rev. A, Page 8/66
m K f
Eq. (6-36):
a K f
max min
2
max min
2
1.96
43.7 0 42.8 kpsi
1.96
2
43.7 0
2
42.8 kpsi
1 a m 42.8 42.8
nf
Se Sut 42.0 120
Eq. (6-46):
n f 0.73
Ans.
A factor of safety less than unity indicates a finite life.
Check for yielding. It is not necessary to include the stress concentration for static
yielding of a ductile material.
66
1.51 Ans.
max 43.7
______________________________________________________________________________
ny
6-16
Sy
From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at
the left bearing and 3.9 kN at the right bearing. The critical location will be at the
shoulder fillet between the 35 mm and the 50 mm diameters, where the bending moment
is large, the diameter is smaller, and the stress concentration exists. The bending moment
at this point is M = 2.1(200) = 420 kN·mm. With a rotating shaft, the bending stress will
be completely reversed.
rev
Mc 420 (35 / 2)
0.09978 kN/mm 2 99.8 MPa
4
I
( / 64)(35)
This stress is far below the yield strength of 390 MPa, so yielding is not predicted. Find
the stress concentration factor for the fatigue analysis.
Fig. A-15-9: r/d = 3/35 = 0.086, D/d = 50/35 = 1.43, K t =1.7
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We’ll use the equations, with S ut = 470 MPa = 68.2 kpsi and r = 3 mm = 0.118
in.
a 0.246 3.08 103 68.2 1.51105 68.2 2.67 108 68.2 0.09771
2
1
0.78
a 1 0.09771
1
0.118
r
K f 1 q ( K t 1) 1 0.78(1.7 1) 1.55
q
Eq. (6-32):
1
3
Chapter 6 - Rev. A, Page 9/66
Eq. (6-8):
S e' 0.5Sut 0.5(470) 235 MPa
Eq. (6-19):
k a aSutb 4.51(470) 0.265 0.88
Eq. (6-24):
Eq. (6-26):
kb 1.24d 0.107 1.24(35) 0.107 0.85
Eq. (6-18):
S e k a kb kc S e' (0.88)(0.85)(1)(235) 176 MPa
kc 1
176
Ans.
1.14 Infinite life is predicted.
K f rev 1.55 99.8
______________________________________________________________________________
nf
6-17
Se
From a free-body diagram analysis, the
bearing reaction forces are found to be R A =
2000 lbf and R B = 1500 lbf. The shear-force
and bending-moment diagrams are shown.
The critical location will be at the shoulder
fillet between the 1-5/8 in and the 1-7/8 in
diameters, where the bending moment is
large, the diameter is smaller, and the stress
concentration exists.
M = 16 000 – 500 (2.5) = 14 750 lbf · in
With a rotating shaft, the bending stress will
be completely reversed.
Mc 14 750(1.625 / 2)
35.0 kpsi
I
( / 64)(1.625) 4
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
rev
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, K t =1.95
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We will use the equations.
a 0.246 3.08 103 85 1.51105 85 2.67 108 85 0.07690
2
1
q
1
0.76 .
0.07690
1
0.0625
Eq. (6-32):
a
r
K f 1 q ( K t 1) 1 0.76(1.95 1) 1.72
Eq. (6-8):
S e' 0.5Sut 0.5(85) 42.5 kpsi
1
3
Chapter 6 - Rev. A, Page 10/66
Eq. (6-19):
k a aSutb 2.70(85) 0.265 0.832
Eq. (6-20):
Eq. (6-26):
kb 0.879d 0.107 0.879(1.625) 0.107 0.835
Eq. (6-18):
S e k a kb kc S e' (0.832)(0.835)(1)(42.5) 29.5 kpsi
kc 1
29.5
Ans.
0.49
K f rev 1.72 35.0
Infinite life is not predicted. Use the S-N diagram to estimate the life.
nf
Se
Fig. 6-18: f = 0.867
Eq. (6-14):
Eq. (6-15):
f Sut
a
Se
2
0.867(85)
2
29.5
184.1
f Sut
1
1
0.867(85)
b log
log
0.1325
3
3
29.5
Se
1
1
K f rev b (1.72)(35.0) 0.1325
Eq. (6-16):
N
4611 cycles
a 184.1
N = 4600 cycles
Ans.
______________________________________________________________________________
6-18
From a free-body diagram analysis, the
bearing reaction forces are found to be R A =
1600 lbf and R B = 2000 lbf. The shear-force
and bending-moment diagrams are shown.
The critical location will be at the shoulder
fillet between the 1-5/8 in and the 1-7/8 in
diameters, where the bending moment is
large, the diameter is smaller, and the stress
concentration exists.
M = 12 800 + 400 (2.5) = 13 800 lbf · in
With a rotating shaft, the bending stress will
be completely reversed.
Mc 13 800(1.625 / 2)
rev
32.8 kpsi
I
( / 64)(1.625) 4
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, K t =1.95
Chapter 6 - Rev. A, Page 11/66
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We will use the equations
a 0.246 3.08 103 85 1.51105 85 2.67 108 85 0.07690
2
Eq. (6-32):
1
0.76
0.07690
a 1
1
0.0625
r
K f 1 q ( K t 1) 1 0.76(1.95 1) 1.72
Eq. (6-8):
S e' 0.5Sut 0.5(85) 42.5 kpsi
Eq. (6-19):
k a aSutb 2.70(85) 0.265 0.832
Eq. (6-20):
Eq. (6-26):
kb 0.879d 0.107 0.879(1.625) 0.107 0.835
Eq. (6-18):
S e k a kb kc S e' (0.832)(0.835)(1)(42.5) 29.5 kpsi
q
1
3
kc 1
29.5
Ans.
0.52
K f rev 1.72 32.8
Infinite life is not predicted. Use the S-N diagram to estimate the life.
Fig. 6-18:
f = 0.867
2
2
f Sut
0.867(85)
Eq. (6-14):
a
184.1
Se
29.5
Se
nf
Eq. (6-15):
f Sut
1
b log
3
Se
1
1
0.867(85)
log
0.1325
3
29.5
1
K f rev b (1.72)(32.8) 0.1325
Eq. (6-16):
N
7527 cycles
a 184.1
N = 7500 cycles
Ans.
______________________________________________________________________________
6-19
Table A-20: Sut 120 kpsi, S y 66 kpsi
N = (950 rev/min)(10 hr)(60 min/hr) = 570 000 cycles
One approach is to guess a diameter and solve the problem as an iterative analysis
problem. Alternatively, we can estimate the few modifying parameters that are dependent
on the diameter and solve the stress equation for the diameter, then iterate to check the
estimates. We’ll use the second approach since it should require only one iteration, since
the estimates on the modifying parameters should be pretty close.
Chapter 6 - Rev. A, Page 12/66
First, we’ll evaluate the stress. From a free-body diagram analysis, the reaction forces at
the bearings are R 1 = 2 kips and R 2 = 6 kips. The critical stress location is in the middle
of the span at the shoulder, where the bending moment is high, the shaft diameter is
smaller, and a stress concentration factor exists. If the critical location is not obvious,
prepare a complete bending moment diagram and evaluate at any potentially critical
locations. Evaluating at the critical shoulder,
M 2 kip 10 in 20 kip in
rev
Mc M d / 2 32 M 32 20 203.7
kpsi
I
d 4 / 64 d 3
d3
d3
Now we’ll get the notch sensitivity and stress concentration factor. The notch sensitivity
depends on the fillet radius, which depends on the unknown diameter. For now, we’ll
estimate a value for q = 0.85 from observation of Fig. 6-20, and check it later.
Fig. A-15-9: D / d 1.4d / d 1.4,
Eq. (6-32):
r / d 0.1d / d 0.1,
Kt 1.65
K f 1 q ( K t 1) 1 0.85(1.65 1) 1.55
Now we will evaluate the fatigue strength.
Se' 0.5(120) 60 kpsi
ka 2.70(120) 0.265 0.76
Since the diameter is not yet known, assume a typical value of k b = 0.85 and check later.
All other modifiers are equal to one.
S e = (0.76)(0.85)(60) = 38.8 kpsi
Determine the desired fatigue strength from the S-N diagram.
Fig. 6-18:
Eq. (6-14):
f = 0.82
2
2
f Sut
0.82(120)
a
249.6
Se
38.8
Eq. (6-15):
f Sut
1
b log
3
Se
1
0.82(120)
log
0.1347
3
38.8
Eq. (6-13):
S f aN b 249.6(570 000)0.1347 41.9 kpsi
Compare strength to stress and solve for the necessary d.
Chapter 6 - Rev. A, Page 13/66
nf
Sf
K f rev
41.9
1.6
1.55 203.7 / d 3
d = 2.29 in
Since the size factor and notch sensitivity were guessed, go back and check them now.
Eq. (6-20):
kb 0.91d 0.157 0.91 2.29
0.157
0.80
Our guess of 0.85 was conservative. From Fig. 6-20 with r = d/10 = 0.229 in, we are off
the graph, but it appears our guess for q is low. Assuming the trend of the graph
continues, we’ll choose q = 0.91 and iterate the problem with the new values of k b and q.
Intermediate results are S e = 36.5 kpsi, S f = 39.6 kpsi, and K f = 1.59. This gives
nf
Sf
K f rev
39.6
1.6
1.59 203.7 / d 3
d = 2.36 in
Ans.
A quick check of k b and q show that our estimates are still reasonable for this diameter.
______________________________________________________________________________
6-20
Se 40 kpsi, S y 60 kpsi, Sut 80 kpsi, m 15 kpsi, a 25 kpsi, m a 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
252 3 0
25.00 kpsi
1/2
2
02 3 15
2
2
max
3 max
max
1/2
25.98 kpsi
1/2
2
2
a m 3 a m
1/2
252 3 152 36.06 kpsi
S
60
ny y
1.66 Ans.
max
36.06
(a) Modified Goodman, Table 6-6
nf
1
1.05
(25.00 / 40) (25.98 / 80)
Ans.
(b) Gerber, Table 6-7
2
2
2(25.98)(40)
1 80 25.00
nf
1.31
1 1
2 25.98 40
80(25.00)
Ans.
Chapter 6 - Rev. A, Page 14/66
(c) ASME-Elliptic, Table 6-8
1
1.32 Ans.
(25.00 / 40) (25.98 / 60) 2
______________________________________________________________________________
nf
6-21
2
Se 40 kpsi, S y 60 kpsi, Sut 80 kpsi, m 20 kpsi, a 10 kpsi, m a 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/ 2
m m2 3 m2
1/2
1/2
2
102 3 0
10.00 kpsi
1/2
2
02 3 20
2
2
max
3 max
max
1/2
34.64 kpsi
1/2
2
2
a m 3 a m
1/2
102 3 202 36.06 kpsi
S
60
ny y
1.66 Ans.
max
36.06
(a) Modified Goodman, Table 6-6
1
nf
1.46
(10.00 / 40) (34.64 / 80)
(b) Gerber, Table 6-7
Ans.
2
2
2(34.64)(40)
1 80 10.00
nf
1.74
1 1
2 34.64 40
80(10.00)
Ans.
(c) ASME-Elliptic, Table 6-8
1
1.59 Ans.
(10.00 / 40) (34.64 / 60) 2
______________________________________________________________________________
nf
6-22
2
S e 40 kpsi, S y 60 kpsi, Sut 80 kpsi, a 10 kpsi, m 15 kpsi, a 12 kpsi, m 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/ 2
2
122 3 10
1/ 2
2
02 3 15
21.07 kpsi
25.98 kpsi
Chapter 6 - Rev. A, Page 15/66
2
2
max
3 max
max
1/ 2
1/ 2
2
2
a m 3 a m
1/ 2
2
2
12 0 3 10 15 44.93 kpsi
S
60
ny y
1.34 Ans.
max
44.93
(a) Modified Goodman, Table 6-6
nf
1
1.17
(21.07 / 40) (25.98 / 80)
Ans.
(b) Gerber, Table 6-7
2
2
2(25.98)(40)
1 80 21.07
nf
1.47
1 1
2 25.98 40
80(21.07)
(c) ASME-Elliptic, Table 6-8
Ans.
1
1.47 Ans.
(21.07 / 40) (25.98 / 60) 2
______________________________________________________________________________
nf
6-23
2
S e 40 kpsi, S y 60 kpsi, Sut 80 kpsi, a 30 kpsi, m a a 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
02 3 30
51.96 kpsi
0 kpsi
2
2
max
3 max
max
1/ 2
1/ 2
2
2
a m 3 a m
1/ 2
2
3 30 51.96 kpsi
Sy
60
ny
1.15 Ans.
max
51.96
(a) Modified Goodman, Table 6-6
nf
1
0.77
(51.96 / 40)
Ans.
(b) Gerber criterion of Table 6-7 is only valid for m > 0; therefore use Eq. (6-47).
Chapter 6 - Rev. A, Page 16/66
nf
a
Se
1
nf
Se
40
0.77
a 51.96
Ans.
(c) ASME-Elliptic, Table 6-8
1
0.77
(51.96 / 40) 2
nf
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. Since ' m = 0,
the stress state is completely reversed and the S-N diagram is applicable for ' a .
Fig. 6-18: f = 0.875
2
( f Sut ) 2 0.875(80)
Eq. (6-14):
a
122.5
Se
40
Eq. (6-15):
f Sut
1
1
0.875(80)
b log
log
0.08101
3
3
40
Se
1
1/ b
51.96 0.08101
39 600 cycles Ans.
Eq. (6-16):
N rev
122.5
a
______________________________________________________________________________
6-24
S e 40 kpsi, S y 60 kpsi, Sut 80 kpsi, a 15 kpsi, m 15 kpsi, m a 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
02 3 15
1/ 2
2
152 3 0
2
2
max
max
3 max
1/ 2
25.98 kpsi
15.00 kpsi
1/ 2
2
2
a m 3 a m
2 1/2
15 3 15 30.00 kpsi
S
60
ny y
2.00 Ans.
max
30
2
(a) Modified Goodman, Table 6-6
nf
1
1.19
(25.98 / 40) (15.00 / 80)
Ans.
(b) Gerber, Table 6-7
2
2
2(15.00)(40)
1 80 25.98
nf
1.43
1 1
2 15.00 40
80(25.98)
Ans.
Chapter 6 - Rev. A, Page 17/66
(c) ASME-Elliptic, Table 6-8
1
1.44 Ans.
(25.98 / 40) (15.00 / 60) 2
______________________________________________________________________________
nf
6-25
2
Given: Fmax 28 kN, Fmin 28 kN . From Table A-20, for AISI 1040
CD, Sut 590 MPa, S y 490 MPa,
Check for yielding
F
28 000
max max
147.4 N/mm2 147.4 MPa
A
10(25 6)
490
3.32 Ans.
max 147.4
Determine the fatigue factor of safety based on infinite life
ny
Sy
Eq. (6-8):
S e' 0.5(590) 295 MPa
Eq. (6-19):
Eq. (6-21):
Eq. (6-26):
k a aSutb 4.51(590) 0.265 0.832
Eq. (6-18):
S e k a kb kc S e' (0.832)(1)(0.85)(295) 208.6 MPa
kb 1 (axial)
kc 0.85
Fig. 6-20:
q = 0.83
Fig. A-15-1: d / w 0.24, K t 2.44
K f 1 q ( K t 1) 1 0.83(2.44 1) 2.20
a K f
28 000 28 000
Fmax Fmin
2.2
324.2 MPa
2A
2(10)(25 6)
Fmax Fmin
0
2A
1 a m 324.2
0
n f Se Sut 208.6 590
m K f
n f 0.64
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. Since m = 0,
the stress state is completely reversed and the S-N diagram is applicable for a .
S ut = 590/6.89 = 85.6 kpsi
Fig. 6-18:
f = 0.87
Chapter 6 - Rev. A, Page 18/66
Eq. (6-14):
( f Sut )2 0.87(590)
a
1263
Se
208.6
Eq. (6-15):
f Sut
1
1
0.87(590)
b log
log
0.1304
3
3
208.6
Se
2
1/ b
1
324.2 0.1304
33 812 cycles
Eq. (6-16):
N rev
1263
a
N = 34 000 cycles
Ans.
________________________________________________________________________
6-26
Sut 590 MPa, S y 490 MPa, Fmax 28 kN, Fmin 12 kN
Check for yielding
Fmax
28 000
147.4 N/mm2 147.4 MPa
A
10(25 6)
S
490
ny y
3.32 Ans.
max 147.4
max
Determine the fatigue factor of safety based on infinite life.
From Prob. 6-25: S e 208.6 MPa, K f 2.2
a K f
28 000 12 000
Fmax Fmin
2.2
92.63 MPa
2A
2(10)(25 6)
m K f
28 000 12 000
Fmax Fmin
2.2
231.6 MPa
2A
2(10)(25 6)
Modified Goodman criteria:
1 a m 92.63 231.6
n f Se Sut 208.6 590
n f 1.20
Ans.
Gerber criteria:
2
2
2 m Se
1 Sut a
nf
1 1
2 m Se
Sut a
2
2
2(231.6)(208.6)
1 590 92.63
1 1
2 231.6 208.6
590(92.63)
n f 1.49
Ans.
Chapter 6 - Rev. A, Page 19/66
ASME-Elliptic criteria:
1
1
2
2
( a / Se ) ( m / S y )
(92.63 / 208.6) (231.6 / 490)2
nf
2
= 1.54
Ans.
The results are consistent with Fig. 6-27, where for a mean stress that is about half of the
yield strength, the Modified Goodman line should predict failure significantly before the
other two.
______________________________________________________________________________
6-27
Sut 590 MPa, S y 490 MPa
(a) Fmax 28 kN, Fmin 0 kN
Check for yielding
max
ny
Fmax
28 000
147.4 N/mm2 147.4 MPa
A
10(25 6)
Sy
max
490
3.32
147.4
Ans.
From Prob. 6-25: S e 208.6 MPa, K f 2.2
a K f
Fmax Fmin
28 000 0
2.2
162.1 MPa
2A
2(10)(25 6)
m K f
28 000 0
Fmax Fmin
2.2
162.1 MPa
2A
2(10)(25 6)
1 a m 162.1 162.1
nf
Se Sut 208.6 590
n f 0.95
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
Fig. 6-18:
a
162.1
223.5 MPa
1 ( m / Sut ) 1 (162.1/ 590)
f = 0.87
( f Sut )2 0.87(590)
a
1263
208.6
Se
2
Eq. (6-14):
Chapter 6 - Rev. A, Page 20/66
Eq. (6-15):
f Sut
1
b log
3
Se
Eq. (6-16):
N rev
a
1/ b
1
0.87(590)
log
0.1304
3
208.6
1
223.5 0.1304
586 000 cycles
1263
Ans.
(b) Fmax 28 kN, Fmin 12 kN
The maximum load is the same as in part (a), so
max 147.4 MPa
n y 3.32 Ans.
Factor of safety based on infinite life:
Fmax Fmin
28 000 12 000
2.2
92.63 MPa
2A
2(10)(25 6)
a K f
28 000 12 000
Fmax Fmin
2.2
231.6 MPa
2A
2(10)(25 6)
1 a m 92.63 231.6
n f Se Sut 208.6 590
m K f
n f 1.20
Ans.
(c) Fmax 12 kN, Fmin 28 kN
The compressive load is the largest, so check it for yielding.
min
ny
Fmin
28 000
147.4 MPa
A 10(25 6)
S yc
min
490
3.32
147.4
Ans.
Factor of safety based on infinite life:
For m < 0,
a K f
12 000 28 000
Fmax Fmin
2.2
231.6 MPa
2A
2(10)(25 6)
m K f
12 000 28 000
Fmax Fmin
2.2
92.63 MPa
2A
2(10)(25
6)
nf
Se
a
208.6
0.90
231.6
Ans.
Chapter 6 - Rev. A, Page 21/66
Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative
mean stress, we shall assume the equivalent completely reversed stress is the same as the
actual alternating stress. Get a and b from part (a).
1
1/ b
231.6 0.1304
Eq. (6-16):
N rev
Ans.
446 000 cycles
1263
a
______________________________________________________________________________
6-28
Eq. (2-21): S ut = 0.5(400) = 200 kpsi
Eq. (6-8):
S e' 0.5(200) 100 kpsi
Eq. (6-19):
Eq. (6-25):
ka aSutb 14.4(200) 0.718 0.321
Eq. (6-20):
kb 0.879d e 0.107 0.879(0.1388) 0.107 1.09
de 0.37d 0.37(0.375) 0.1388 in
Since we have used the equivalent diameter method to get the size factor, and in doing so
introduced greater uncertainties, we will choose not to use a size factor greater than one.
Let k b = 1.
Eq. (6-18):
Se (0.321)(1)(100) 32.1 kpsi
40 20
40 20
Fa
10 lb
Fm
30 lb
2
2
32 M a 32(10)(12)
a
23.18 kpsi
d3
(0.375)3
32 M m 32(30)(12)
m
69.54 kpsi
d3
(0.375)3
(a) Modified Goodman criterion
1 a m 23.18 69.54
nf
Se Sut
32.1
200
n f 0.94
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
a
23.18
35.54 kpsi
1 ( m / Sut ) 1 (69.54 / 200)
f = 0.775
rev
Fig. 6-18:
( f Sut ) 2 0.775(200)
a
748.4
32.1
Se
2
Eq. (6-14):
Chapter 6 - Rev. A, Page 22/66
Eq. (6-15):
f Sut
1
b log
3
Se
1/ b
Eq. (6-16):
rev
N
a
1
0.775(200)
log
0.228
3
32.1
1
35.54 0.228
637 000 cycles
748.4
Ans.
(b) Gerber criterion, Table 6-7
2
2
2 m Se
1 Sut a
1 1
nf
2 m Se
Sut a
2
2
2(69.54)(32.1)
1 200 23.18
1 1
2 69.54 32.1
200(23.18)
Ans.
1.16 Infinite life is predicted
______________________________________________________________________________
6-29
E 207.0 GPa
1
(a) I (20)(43 ) 106.7 mm 4
12
Fl 3
3EIy
y
F 3
3EI
l
9
3(207)(10 )(106.7)(10 12 )(2)(10 3 )
Fmin
48.3 N
1403 (109 )
Fmax
3(207)(109 )(106.7)(10 12 )(6)(10 3 )
144.9 N
1403 (10 9 )
Ans.
Ans.
(b) Get the fatigue strength information.
Eq. (2-21):
S ut = =3.4H B = 3.4(490) = 1666 MPa
From problem statement: S y = 0.9S ut = 0.9(1666) = 1499 MPa
Eq. (6-8):
Se 700 MPa
Eq. (6-19):
Eq. (6-25):
Eq. (6-20):
Eq. (6-18):
k a = 1.58(1666)-0.085 = 0.84
d e = 0.808[20(4)]1/2 = 7.23 mm
k b = 1.24(7.23)-0.107 = 1.00
S e = 0.84(1)(700) = 588 MPa
This is a relatively thick curved beam, so
use the method in Sect. 3-18 to find the
stresses. The maximum bending moment
will be to the centroid of the section as
shown.
Chapter 6 - Rev. A, Page 23/66
M = 142F N·mm, A = 4(20) = 80 mm2, h = 4 mm, r i = 4 mm, r o = r i + h = 8 mm,
r c = r i + h/2 = 6 mm
Table 3-4:
rn
h
4
5.7708 mm
ln(ro / ri ) ln(8 / 4)
e rc rn 6 5.7708 0.2292 mm
ci rn ri 5.7708 4 1.7708 mm
co ro rn 8 5.7708 2.2292 mm
Get the stresses at the inner and outer surfaces from Eq. (3-65) with the axial stresses
added. The signs have been set to account for tension and compression as appropriate.
i
o
Mci F
(142 F )(1.7708) F
3.441F MPa
Aeri A
80(0.2292)(4) 80
Mco F (142 F )(2.2292) F
2.145F MPa
Aero A
80(0.2292)(8) 80
( i ) min 3.441(144.9) 498.6 MPa
( i ) max 3.441(48.3) 166.2 MPa
( o )min 2.145(48.3) 103.6 MPa
( o )max 2.145(144.9) 310.8 MPa
( i ) a
166.2 498.6
2
166.2 MPa
166.2 498.6
332.4 MPa
2
310.8 103.6
( o ) a
103.6 MPa
2
310.8 103.6
( o ) m
207.2 MPa
2
To check for yielding, we note that the largest stress is –498.6 MPa (compression) on the
inner radius. This is considerably less than the estimated yield strength of 1499 MPa, so
yielding is not predicted.
( i ) m
Check for fatigue on both inner and outer radii since one has a compressive mean stress
and the other has a tensile mean stress.
Inner radius:
S
588
Since m < 0, n f e
3.54
a 166.2
Chapter 6 - Rev. A, Page 24/66
Outer radius:
Since m > 0, we will use the Modified Goodman line.
103.6 207.2
1/ n f a m
Se Sut
588 1666
n f 3.33
Infinite life is predicted at both inner and outer radii. Ans.
______________________________________________________________________________
6-30
From Table A-20, for AISI 1018 CD, Sut 64 kpsi, S y 54 kpsi
Eq. (6-8):
Se' 0.5(64) 32 kpsi
Eq. (6-19):
Eq. (6-20):
Eq. (6-26):
Eq. (6-18):
k a 2.70(64) 0.265 0.897
kb 1 (axial)
kc 0.85
Se (0.897)(1)(0.85)(32) 24.4 kpsi
Fillet:
Fig. A-15-5: D / d 3.5 / 3 1.17, r / d 0.25 / 3 0.083, K t 1.85
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f 1 q ( K t 1) 1 0.85(1.85 1) 1.72
max
Fmax
5
3.33 kpsi
w2 h 3.0(0.5)
min
16
10.67 kpsi
3.0(0.5)
a K f
max min
2
max min
2
m K f
ny
Sy
min
1.72
3.33 (10.67)
12.0 kpsi
2
3.33 (10.67)
6.31 kpsi
1.72
2
54
5.06
10.67
Does not yield.
Since the midrange stress is negative,
nf
Se
a
24.4
2.03
12.0
Chapter 6 - Rev. A, Page 25/66
Hole:
Fig. A-15-1: d / w1 0.4 / 3.5 0.11 K t 2.68
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f 1 0.85(2.68 1) 2.43
max
Fmax
5
3.226 kpsi
h w1 d 0.5(3.5 0.4)
min
Fmin
16
10.32 kpsi
h w1 d 0.5(3.5 0.4)
a K f
max min
2
2.43
max min
2
m K f
ny
Sy
min
3.226 (10.32)
16.5 kpsi
2
3.226 (10.32)
8.62 kpsi
2.43
2
54
5.23
10.32
does not yield
Since the midrange stress is negative,
nf
Se
a
24.4
1.48
16.5
Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of
safety of n f 1.48. Ans.
______________________________________________________________________________
6-31
Sut 64 kpsi, S y 54 kpsi
Eq. (6-8):
Se' 0.5(64) 32 kpsi
Eq. (6-19):
k a 2.70(64) 0.265 0.897
Eq. (6-20):
Eq. (6-26):
Eq. (6-18):
kb 1 (axial)
kc 0.85
Se (0.897)(1)(0.85)(32) 24.4 kpsi
Fillet:
Fig. A-15-5: D / d 2.5 /1.5 1.67, r / d 0.25 /1.5 0.17, K t 2.1
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f 1 q ( K t 1) 1 0.85(2.1 1) 1.94
Chapter 6 - Rev. A, Page 26/66
max
Fmax
16
21.3 kpsi
w2 h 1.5(0.5)
4
5.33 kpsi
1.5(0.5)
min
21.3 ( 5.33)
1.94
25.8 kpsi
a K f max
2
2
min
max min
21.3 ( 5.33)
1.94
15.5 kpsi
2
2
m K f
ny
Sy
max
54
2.54
21.3
Does not yield.
Using Modified Goodman criteria,
1 a m 25.8 15.5
nf
Se Sut 24.4 64
n f 0.77
Hole:
Fig. A-15-1: d / w1 0.4 / 2.5 0.16 K t 2.55
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f 1 0.85(2.55 1) 2.32
max
Fmax
16
15.2 kpsi
h w1 d 0.5(2.5 0.4)
min
Fmin
4
3.81 kpsi
h w1 d 0.5(2.5 0.4)
max min
15.2 (3.81)
2.32
22.1 kpsi
2
2
min
15.2 (3.81)
m K f max
2.32
13.2 kpsi
2
2
S
54
ny y
3.55 Does not yield.
max 15.2
a K f
Using Modified Goodman criteria
1 a m 22.1 13.2
nf
Se Sut 24.4 64
n f 0.90
Chapter 6 - Rev. A, Page 27/66
Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor
of safety of n f 0.77
Ans.
______________________________________________________________________________
6-32
Sut 64 kpsi, S y 54 kpsi
From Prob. 6-30, the fatigue factor of safety at the hole is n f = 1.48. To match this at the
fillet,
S
S
24.4
nf e a e
16.5 kpsi
n f 1.48
a
where S e is unchanged from Prob. 6-30. The only aspect of a that is affected by the
fillet radius is the fatigue stress concentration factor. Obtaining a in terms of K f ,
a K f
max min
2
Kf
3.33 (10.67)
7.00 K f
2
Equating to the desired stress, and solving for K f ,
a 7.00 K f 16.5
K f 2.36
Assume since we are expecting to get a smaller fillet radius than the original, that q will
be back on the graph of Fig. 6-20, so we’ll estimate q = 0.8.
K f 1 0.80( K t 1) 2.36
K t 2.7
From Fig. A-15-5, with D / d = 3.5/3 = 1.17 and K t = 2.6, find r / d. Choosing r / d =
0.03, and with d = w 2 = 3.0,
r 0.03w2 0.03 3.0 0.09 in
At this small radius, our estimate for q is too high. From Fig. 6-20, with r = 0.09, q
should be about 0.75. Iterating, we get K t = 2.8. This is at a difficult range on Fig. A-155 to read the graph with any confidence, but we’ll estimate r / d = 0.02, giving r = 0.06
in. This is a very rough estimate, but it clearly demonstrates that the fillet radius can be
relatively sharp to match the fatigue factor of safety of the hole.
Ans.
______________________________________________________________________________
6-33
S y 60 kpsi, Sut 110 kpsi
Inner fiber where rc 3 / 4 in
3
3
ro
0.84375
4 16(2)
3 3
ri
0.65625
4 32
Table 3-4, p. 121,
Chapter 6 - Rev. A, Page 28/66
rn
h
3 / 16
0.74608 in
ro
0.84375
ln
ln
0.65625
ri
e rc rn 0.75 0.74608 0.00392 in
ci rn ri 0.74608 0.65625 0.08983
3 3
A 0.035156 in 2
16 16
Eq. (3-65), p. 119,
i
Mci
T (0.08983)
993.3T
Aeri (0.035156)(0.00392)(0.65625)
where T is in lbf·in and i is in psi.
1
2
a 496.7T
m (993.3)T 496.7T
Eq. (6-8):
Se' 0.5 110 55 kpsi
Eq. (6-19):
ka 2.70(110) 0.265 0.777
Eq. (6-25):
d e 0.808 3 /16 3 /16
Eq. (6-20):
kb 0.879 0.1515
Eq. (6-19):
Se (0.777)(1)(55) 42.7 kpsi
1/2
0.107
0.1515 in
1.08 (round to 1)
For a compressive midrange component, a S e / n f . Thus,
42.7
3
T 28.7 lbf in
0.4967T
Outer fiber where rc 2.5 in
3
2.59375
32
3
ri 2.5
2.40625
32
3 / 16
rn
2.49883
2.59375
ln
2.40625
e 2.5 2.49883 0.00117 in
co 2.59375 2.49883 0.09492 in
ro 2.5
Chapter 6 - Rev. A, Page 29/66
o
Mco
T (0.09492)
889.7T psi
Aero (0.035156)(0.00117)(2.59375)
1
2
(a) Using Eq. (6-46), for modified Goodman, we have
a m 1
Se Sut n
m a (889.7T ) 444.9T psi
0.4449T 0.4449T 1
42.7
110
3
T 23.0 lbf in
Ans.
(b) Gerber, Eq. (6-47), at the outer fiber,
2
n a n m
1
Se Sut
2
3(0.4449T ) 3(0.4449T )
1
42.7
110
T 28.2 lbf in Ans.
(c) To guard against yield, use T of part (b) and the inner stress.
60
2.14 Ans.
i 0.9933(28.2)
______________________________________________________________________________
ny
6-34
Sy
From Prob. 6-33, S e 42.7 kpsi, S y 60 kpsi, and Sut 110 kpsi
(a) Assuming the beam is straight,
max
Goodman:
Mc M h / 2 6M
6T
2
910.2T
3
I
bh /12 bh
(3 /16)3
0.4551T 0.4551T 1
42.7
110
3
T 22.5 lbf in
Ans.
2
(b) Gerber:
3(0.4551T ) 3(0.4551T )
1
42.7
110
T 27.6 lbf in Ans.
Chapter 6 - Rev. A, Page 30/66
60
2.39 Ans.
max 0.9102(27.6)
______________________________________________________________________________
6-35 K f ,bend 1.4, K f ,axial 1.1, K f ,tors 2.0, S y 300 MPa, Sut 400 MPa, Se 200 MPa
(c)
ny
Sy
Bending:
m 0, a 60 MPa
Axial:
m 20 MPa, a 0
Torsion:
m 25 MPa, a 25 MPa
Eqs. (6-55) and (6-56):
a
1.4(60) 0
m
0 1.1(20)
2
2
3 2.0(25) 120.6 MPa
2
3 2.0(25) 89.35 MPa
2
Using Modified Goodman, Eq. (6-46),
1 a m 120.6 89.35
200
400
nf
Se Sut
n f 1.21
Ans.
a m ,
Check for yielding, using the conservative max
300
1.43 Ans.
a m 120.6 89.35
______________________________________________________________________________
ny
6-36
Sy
K f ,bend 1.4, K f ,tors 2.0, S y 300 MPa, Sut 400 MPa, Se 200 MPa
Bending: max 150 MPa, min 40 MPa, m 55 MPa, a 95 MPa
Torsion: m 90 MPa, a 9 MPa
Eqs. (6-55) and (6-56):
a
1.4(95)
m
1.4(55)
2
2
3 2.0(9) 136.6 MPa
2
3 2.0(90) 321.1 MPa
2
Using Modified Goodman,
1 a m 136.6 321.1
200
400
n f Se Sut
n f 0.67
Ans.
a m ,
Check for yielding, using the conservative max
Chapter 6 - Rev. A, Page 31/66
300
0.66 Ans.
a m 136.6 321.1
Since the conservative yield check indicates yielding, we will check more carefully with
obtained directly from the maximum stresses, using the distortion energy
with max
failure theory, without stress concentrations. Note that this is exactly the method used for
static failure in Ch. 5.
ny
max
Sy
max
3 max
2
2
150
3 90 9 227.8 MPa
2
2
Sy
300
1.32 Ans.
max
227.8
Since yielding is not predicted, and infinite life is not predicted, we would like to
estimate a life from the S-N diagram. First, find an equivalent completely reversed stress
(See Ex. 6-12).
ny
a
136.6
692.5 MPa
1 ( m / Sut ) 1 (321.1/ 400)
This stress is much higher than the ultimate strength, rendering it impractical for the S-N
diagram. We must conclude that the stresses from the combination loading, when
increased by the stress concentration factors, produce such a high midrange stress that the
equivalent completely reversed stress method is not practical to use. Without testing, we
are unable to predict a life.
______________________________________________________________________________
rev
6-37
Table A-20: S ut 64 kpsi, S y 54 kpsi
From Prob. 3-68, the critical stress element experiences = 15.3 kpsi and = 4.43 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 15.3 kpsi, m = 0 kpsi, a = 0 kpsi, m = 4.43 kpsi. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
15.32 3 0
15.3 kpsi
1/2
2
0 2 3 4.43
2
2
max
3 max
max
1/ 2
7.67 kpsi
1/2
2
15.32 3 4.43
17.11 kpsi
Check for yielding, using the distortion energy failure theory.
S
54
ny y
3.16
max
17.11
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5 64 32 kpsi
Chapter 6 - Rev. A, Page 32/66
Eq. (6-19):
ka 2.70(64) 0.265 0.90
Eq. (6-20):
kb 0.879(1.25) 0.107 0.86
Eq. (6-18):
Se 0.90(0.86)(32) 24.8 kpsi
Using Modified Goodman,
1 a m 15.3 7.67
nf
Se Sut 24.8 64
n f 1.36 Ans.
______________________________________________________________________________
6-38
Table A-20: S ut 440 MPa, S y 370 MPa
From Prob. 3-69, the critical stress element experiences = 263 MPa and = 57.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 263 MPa, m = 0, a = 0 MPa, m = 57.7 MPa. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
a a2 3 a2
1/ 2
m m2 3 m2
1/ 2
1/ 2
2
2632 3 0
263 MPa
1/ 2
2
02 3 57.7
2
2
max
3 max
max
1/ 2
99.9 MPa
1/2
2
2632 3 57.7
281 MPa
Check for yielding, using the distortion energy failure theory.
Sy
370
1.32
max
281
Obtain the modifying factors and endurance limit.
ny
Eq. (6-8):
Se 0.5 440 220 MPa
Eq. (6-19):
ka 4.51(440) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 1.24(30) 0.107 0.86
Se 0.90(0.86)(220) 170 MPa
Using Modified Goodman,
1 a m 263 99.9
nf
Se Sut 170 440
n f 0.56
Infinite life is not predicted.
Ans.
______________________________________________________________________________
Chapter 6 - Rev. A, Page 33/66
6-39
Table A-20: S ut 64 kpsi, S y 54 kpsi
From Prob. 3-70, the critical stress element experiences = 21.5 kpsi and = 5.09 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 21.5 kpsi, m = 0 kpsi, a = 0 kpsi, m = 5.09 kpsi. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
21.52 3 0
1/ 2
2
0 2 3 5.09
2
2
max
3 max
max
1/ 2
21.5 kpsi
1/ 2
8.82 kpsi
2
21.52 3 5.09
1/ 2
23.24 kpsi
Check for yielding, using the distortion energy failure theory.
ny
Sy
54
2.32
max
23.24
Obtain the modifying factors and endurance limit.
ka 2.70(64) 0.265 0.90
kb 0.879(1) 0.107 0.88
Se 0.90(0.88)(0.5)(64) 25.3 kpsi
Using Modified Goodman,
1 a m 21.5 8.82
nf
Se Sut 25.3 64
n f 1.01 Ans.
______________________________________________________________________________
6-40
Table A-20: S ut 440 MPa, S y 370 MPa
From Prob. 3-71, the critical stress element experiences = 72.9 MPa and = 20.3 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 72.9 MPa, m = 0 MPa, a = 0 MPa, m = 20.3 MPa. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
72.92 3 0
1/2
2
0 2 3 20.3
2
2
max
3 max
max
1/ 2
72.9 MPa
35.2 MPa
1/2
2
72.92 3 20.3
80.9 MPa
Check for yielding, using the distortion energy failure theory.
Chapter 6 - Rev. A, Page 34/66
ny
Sy
370
4.57
max
80.9
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5 440 220 MPa
Eq. (6-19):
ka 4.51(440) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 1.24(20) 0.107 0.90
Se 0.90(0.90)(220) 178.2 MPa
Using Modified Goodman,
1 a m
72.9 35.2
nf
Se Sut 178.2 440
n f 2.04 Ans.
______________________________________________________________________________
6-41
Table A-20: S ut 64 kpsi, S y 54 kpsi
From Prob. 3-72, the critical stress element experiences = 35.2 kpsi and = 7.35 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 35.2 kpsi, m = 0 kpsi, a = 0 kpsi, m = 7.35 kpsi. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
35.22 3 0
1/ 2
1/2
2
02 3 7.35
2
2
max
3 max
max
1/2
35.2 kpsi
12.7 kpsi
2
35.22 3 7.35
1/ 2
37.4 kpsi
Check for yielding, using the distortion energy failure theory.
ny
Sy
54
1.44
max
37.4
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
ka 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879(1.25) 0.107 0.86
Se 0.90(0.86)(32) 24.8 kpsi
Chapter 6 - Rev. A, Page 35/66
Using Modified Goodman,
1 a m 35.2 12.7
nf
Se Sut 24.8 64
n f 0.62 Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-42
Table A-20: S ut 440 MPa, S y 370 MPa
From Prob. 3-73, the critical stress element experiences = 333.9 MPa and = 126.3
MPa. The bending is completely reversed due to the rotation, and the torsion is steady,
giving a = 333.9 MPa, m = 0 MPa, a = 0 MPa, m = 126.3 MPa. Obtain von Mises
stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
333.9 2 3 0
1/ 2
2
0 2 3 126.3
2
2
max
3 max
max
1/ 2
1/2
333.9 MPa
218.8 MPa
2
333.9 2 3 126.3
1/ 2
399.2 MPa
Check for yielding, using the distortion energy failure theory.
ny
Sy
370
0.93
max
399.2
The sample fails by yielding, infinite life is not predicted.
Ans.
The fatigue analysis will be continued only to obtain the requested fatigue factor of
safety, though the yielding failure will dictate the life.
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(440) 220 MPa
Eq. (6-19):
ka 4.51(440) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 1.24(50) 0.107 0.82
Se 0.90(0.82)(220) 162.4 MPa
Using Modified Goodman,
1 a m 333.9 218.8
nf
Se Sut 162.4 440
n f 0.39
Infinite life is not predicted.
Ans.
______________________________________________________________________________
Chapter 6 - Rev. A, Page 36/66
6-43
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-74, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
a ,bend 9.495 kpsi,
a ,axial 0 kpsi,
a 0 kpsi,
m ,bend 0 kpsi
m,axial 0.362 kpsi
m 11.07 kpsi
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
2
9.495 3 0
1/ 2
9.495 kpsi
2
2
0.362 3 11.07
2
2
max
3 max
max
1/ 2
1/ 2
19.18 kpsi
2
2
9.495 0.362 3 11.07
1/ 2
21.56 kpsi
Check for yielding, using the distortion energy failure theory.
S
54
ny y
2.50
max
21.56
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
ka 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879(1.13) 0.107 0.87
Se 0.90(0.87)(32) 25.1 kpsi
Using Modified Goodman,
1 a m 9.495 19.18
nf
Se Sut
25.1
64
n f 1.47 Ans.
______________________________________________________________________________
6-44
Table A-20: S ut 64 kpsi, S y 54 kpsi
From Prob. 3-76, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
a ,bend 33.99 kpsi, m ,bend 0 kpsi
a ,axial 0 kpsi,
a 0 kpsi,
m,axial 0.153 kpsi
m 7.847 kpsi
Chapter 6 - Rev. A, Page 37/66
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
2
33.99 3 0
1/2
33.99 kpsi
2
2
0.153 3 7.847
2
2
max
3 max
max
1/ 2
1/ 2
13.59 kpsi
2
2
33.99 0.153 3 7.847
1/2
36.75 kpsi
Check for yielding, using the distortion energy failure theory.
ny
Sy
54
1.47
max
36.75
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
ka 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879(0.88) 0.107 0.89
Se 0.90(0.89)(32) 25.6 kpsi
Using Modified Goodman,
1 a m 33.99 13.59
nf
Se Sut
25.6
64
n f 0.65
Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-45
Table A-20: S ut 440 MPa, S y 370 MPa
From Prob. 3-77, the critical stress element experiences = 68.6 MPa and = 37.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 68.6 MPa, m = 0 MPa, a = 0 MPa, m = 37.7 MPa. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
2
68.6 2 3 0
1/ 2
2
0 2 3 37.7
2
2
max
3 max
max
1/ 2
1/ 2
68.6 MPa
65.3 MPa
2 1/ 2
68.6 2 3 37.7 94.7 MPa
Check for yielding, using the distortion energy failure theory.
ny
Sy
370
3.91
max
94.7
Chapter 6 - Rev. A, Page 38/66
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(440) 220 MPa
Eq. (6-19):
ka 4.51(440) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 1.24(30) 0.107 0.86
Se 0.90(0.86)(220) 170 MPa
Using Modified Goodman,
1 a m 68.6 65.3
nf
Se Sut 170 440
n f 1.81 Ans.
______________________________________________________________________________
6-46
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-79, the critical stress element experiences = 3.46 kpsi and = 0.882 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 3.46 kpsi, m = 0, a = 0 kpsi, m = 0.882 kpsi. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
a a2 3 a2
1/2
m m2 3 m2
1/2
1/2
2
3.462 3 0
3.46 kpsi
1/2
2
02 3 0.882
2
2
max
3 max
max
1/ 2
1.53 kpsi
1/ 2
2
3.462 3 0.882
3.78 kpsi
Check for yielding, using the distortion energy failure theory.
ny
Sy
54
14.3
max
3.78
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
ka 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879(1.375) 0.107 0.85
Se 0.90(0.85)(32) 24.5 kpsi
Using Modified Goodman,
Chapter 6 - Rev. A, Page 39/66
1 a m 3.46 1.53
nf
S e Sut 24.5 64
n f 6.06
Ans.
______________________________________________________________________________
6-47
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-80, the critical stress element experiences = 16.3 kpsi and = 5.09 kpsi.
Since the load is applied and released repeatedly, this gives max = 16.3 kpsi, min = 0
kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently, m = a = 8.15 kpsi, m = a = 2.55
kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a 0.246 3.08 103 64 1.51105 64 2.67 108 64 0.10373
2
1
1
0.75
0.10373
a 1
1
0.1
r
K f 1 q ( K t 1) 1 0.75(1.5 1) 1.38
q
Eq. (6-32):
3
For torsion, from Eqs. (6-34) and (6-35b),
a 0.190 2.51103 64 1.35 105 64 2.67 108 64 0.07800
2
1
1
0.80
a 1 0.07800
1
0.1
r
K fs 1 qs ( K ts 1) 1 0.80(2.1 1) 1.88
q
Eq. (6-32):
3
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
a 1.38 8.15 3 1.88 2.55
2
2 1/ 2
13.98 kpsi
m a 13.98 kpsi
a m ,
Check for yielding, using the conservative max
Sy
54
ny
1.93
a m 13.98 13.98
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
k a aSutb 2.70(64) 0.265 0.90
Chapter 6 - Rev. A, Page 40/66
Eq. (6-24):
de 0.370d 0.370 1 0.370 in
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.98
Se (0.90)(0.98)(32) 28.2 kpsi
Using Modified Goodman,
1 a m 13.98 13.98
nf
S e Sut
28.2
64
n f 1.40
Ans.
______________________________________________________________________________
6-48
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-81, the critical stress element experiences = 16.4 kpsi and = 4.46 kpsi.
Since the load is applied and released repeatedly, this gives max = 16.4 kpsi, min = 0
kpsi, max = 4.46 kpsi, min = 0 kpsi. Consequently, m = a = 8.20 kpsi, m = a = 2.23
kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a 0.246 3.08 103 64 1.51105 64 2.67 108 64 0.10373
2
1
1
0.75
a 1 0.10373
1
0.1
r
K f 1 q ( K t 1) 1 0.75(1.5 1) 1.38
q
Eq. (6-32):
3
For torsion, from Eqs. (6-34) and (6-35b),
a 0.190 2.51103 64 1.35 105 64 2.67 108 64 0.07800
2
1
1
0.80
a 1 0.07800
1
0.1
r
K fs 1 qs ( K ts 1) 1 0.80(2.1 1) 1.88
q
Eq. (6-32):
3
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
a 1.38 8.20 3 1.88 2.23
2
2 1/ 2
13.45 kpsi
m a 13.45 kpsi
a m ,
Check for yielding, using the conservative max
Chapter 6 - Rev. A, Page 41/66
ny
Sy
a m
54
2.01
13.45 13.45
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
Eq. (6-24):
k a aSutb 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.98
d e 0.370d 0.370(1) 0.370 in
Se (0.90)(0.98)(32) 28.2 kpsi
Using Modified Goodman,
1 a m 13.45 13.45
nf
S e Sut
28.2
64
n f 1.46
Ans.
______________________________________________________________________________
6-49
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-82, the critical stress element experiences repeatedly applied bending,
axial, and torsional stresses of x,bend = 20.2 kpsi, x,axial = 0.1 kpsi, and = 5.09 kpsi..
Since the axial stress is practically negligible compared to the bending stress, we will
simply combine the two and not treat the axial stress separately for stress concentration
factor and load factor. This gives max = 20.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min =
0 kpsi. Consequently, m = a = 10.15 kpsi, m = a = 2.55 kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a 0.246 3.08 103 64 1.51105 64 2.67 108 64 0.10373
2
1
1
0.75
a 1 0.10373
1
0.1
r
K f 1 q ( K t 1) 1 0.75(1.5 1) 1.38
q
Eq. (6-32):
3
For torsion, from Eqs. (6-34) and (6-35b),
a 0.190 2.51103 64 1.35 105 64 2.67 108 64 0.07800
2
1
q
1
a
r
3
1
0.80
0.07800
1
0.1
Chapter 6 - Rev. A, Page 42/66
Eq. (6-32):
K fs 1 qs ( K ts 1) 1 0.80(2.1 1) 1.88
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
a 1.38 10.15 3 1.88 2.55
2
2 1/ 2
16.28 kpsi
m a 16.28 kpsi
a m ,
Check for yielding, using the conservative max
Sy
54
ny
1.66
a m 16.28 16.28
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
Eq. (6-24):
k a aSutb 2.70(64) 0.265 0.90
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.98
d e 0.370d 0.370(1) 0.370 in
Se (0.90)(0.98)(32) 28.2 kpsi
Using Modified Goodman,
1 a m 16.28 16.28
nf
S e Sut
28.2
64
n f 1.20
Ans.
____________________________________________________________________________
6-50
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-83, the critical stress element on the neutral axis in the middle of the
longest side of the rectangular cross section experiences a repeatedly applied shear stress
of max = 14.3 kpsi, min = 0 kpsi. Thus, m = a = 7.15 kpsi. Since the stress is entirely
shear, it is convenient to check for yielding using the standard Maximum Shear Stress
theory.
S / 2 54 / 2
ny y
1.89
max
14.3
Find the modifiers and endurance limit.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
k a aSutb 2.70(64) 0.265 0.90
Chapter 6 - Rev. A, Page 43/66
The size factor for a torsionally loaded rectangular cross section is not readily available.
Following the procedure on p. 289, we need an equivalent diameter based on the 95
percent stress area. However, the stress situation in this case is nonlinear, as described on
p. 102. Noting that the maximum stress occurs at the middle of the longest side, or with a
radius from the center of the cross section equal to half of the shortest side, we will
simply choose an equivalent diameter equal to the length of the shortest side.
de 0.25 in
Eq. (6-20):
kb 0.879de 0.107 0.879(0.25)0.107 1.02
We will round down to k b = 1.
Eq. (6-26):
Eq. (6-18):
kc 0.59
S se 0.9(1)(0.59)(32) 17.0 kpsi
Since the stress is entirely shear, we choose to use a load factor k c = 0.59, and convert the
ultimate strength to a shear value rather than using the combination loading method of
Sec. 6-14. From Eq. (6-54), S su = 0.67S u = 0.67 (64) = 42.9 kpsi.
Using Modified Goodman,
1
1
1.70 Ans.
( a / S se ) ( m / S su ) (7.15 / 17.0) (7.15 / 42.9)
______________________________________________________________________________
nf
6-51
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-84, the critical stress element experiences = 28.0 kpsi and = 15.3 kpsi.
Since the load is applied and released repeatedly, this gives max = 28.0 kpsi, min = 0
kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently, m = a = 14.0 kpsi, m = a = 7.65
kpsi. From Table A-15-8 and A-15-9,
D / d 1.5 / 1 1.5,
r / d 0.125 / 1 0.125
K t ,bend 1.60,
K t ,tors 1.39
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: q bend = 0.78, q tors = 0.82
Eq. (6-32):
K f ,bend 1 qbend Kt ,bend 1 1 0.78 1.60 1 1.47
K f ,tors 1 qtors Kt ,tors 1 1 0.82 1.39 1 1.32
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
Chapter 6 - Rev. A, Page 44/66
a 1.47 14.0 3 1.32 7.65
2
2 1/2
27.0 kpsi
m a 27.0 kpsi
a m ,
Check for yielding, using the conservative max
Sy
54
ny
1.00
a m 27.0 27.0
Since stress concentrations are included in this quick yield check, the low factor of safety
is acceptable.
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
k a aSutb 2.70(64) 0.265 0.897
Eq. (6-24):
de 0.370d 0.370 1 0.370 in
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.978
Se (0.897)(0.978)(0.5)(64) 28.1 kpsi
Using Modified Goodman,
1 a m 27.0 27.0
nf
Se Sut 28.1 64
n f 0.72
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
a
27.0
46.7 kpsi
1 ( m / Sut ) 1 (27.0 / 64)
Fig. 6-18:
f = 0.9
Eq. (6-14):
( f Sut ) 2 0.9(64)
a
118.07
Se
28.1
Eq. (6-15):
f Sut
1
1
0.9(64)
b log
log
0.1039
3
3
28.1
Se
2
1/ b
1
46.7 0.1039
7534 cycles 7500 cycles
Eq. (6-16):
N rev
Ans.
118.07
a
______________________________________________________________________________
6-52
Table A-20: Sut 64 kpsi, S y 54 kpsi
Chapter 6 - Rev. A, Page 45/66
From Prob. 3-85, the critical stress element experiences x,bend = 46.1 kpsi, x,axial = 0.382
kpsi and = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to
demonstrate the process. Since the load is applied and released repeatedly, this gives
max,bend = 46.1 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max =
15.3 kpsi, min = 0 kpsi. Consequently, m,bend = a,bend = 23.05 kpsi, m,axial = a,axial =
0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9,
D / d 1.5 / 1 1.5,
K t ,bend 1.60,
r / d 0.125 / 1 0.125
K t ,tors 1.39,
K t ,axial 1.75
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: q bend = q axial =0.78, q tors = 0.82
Eq. (6-32):
K f ,bend 1 qbend Kt ,bend 1 1 0.78 1.60 1 1.47
K f ,axial 1 qaxial Kt ,axial 1 1 0.78 1.75 1 1.59
K f ,tors 1 qtors Kt ,tors 1 1 0.82 1.39 1 1.32
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1/2
2
0.191
2
a 1.47 23.05 1.59
3 1.32 7.65
0.85
m 1.47 23.05 1.59 0.191 3 1.32 7.65
2
2 1/ 2
38.45 kpsi
38.40 kpsi
a m ,
Check for yielding, using the conservative max
Sy
54
ny
0.70
a m 38.45 38.40
Since the conservative yield check indicates yielding, we will check more carefully with
obtained directly from the maximum stresses, using the distortion energy
with max
failure theory, without stress concentrations. Note that this is exactly the method used for
static failure in Ch. 5.
max
ny
max,axial 3 max
2
max,bend
Sy
54
1.01
max
53.5
2
46.1 0.382
2
3 15.3 53.5 kpsi
2
Ans.
This shows that yielding is imminent, and further analysis of fatigue life should not be
interpreted as a guarantee of more than one cycle of life.
Chapter 6 - Rev. A, Page 46/66
Eq. (6-8):
Se 0.5(64) 32 kpsi
Eq. (6-19):
k a aSutb 2.70(64) 0.265 0.897
Eq. (6-24):
de 0.370d 0.370 1 0.370 in
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.978
Se (0.897)(0.978)(0.5)(64) 28.1 kpsi
Using Modified Goodman,
1 a m 38.45 38.40
nf
Se Sut
28.1
64
n f 0.51
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
a
38.45
96.1 kpsi
1 ( m / Sut ) 1 (38.40 / 64)
This stress is much higher than the ultimate strength, rendering it impractical for the S-N
diagram. We must conclude that the fluctuating stresses from the combination loading,
when increased by the stress concentration factors, are so far from the Goodman line that
the equivalent completely reversed stress method is not practical to use. Without testing,
we are unable to predict a life.
______________________________________________________________________________
6-53
Table A-20: Sut 64 kpsi, S y 54 kpsi
From Prob. 3-86, the critical stress element experiences x,bend = 55.5 kpsi, x,axial = 0.382
kpsi and = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to
demonstrate the process. Since the load is applied and released repeatedly, this gives
max,bend = 55.5 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max =
15.3 kpsi, min = 0 kpsi. Consequently, m,bend = a,bend = 27.75 kpsi, m,axial = a,axial =
0.191 kpsi, m = a = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9,
D / d 1.5 / 1 1.5,
r / d 0.125 / 1 0.125
K t ,bend 1.60,
K t ,tors 1.39,
K t ,axial 1.75
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: q bend = q axial =0.78, q tors = 0.82
Eq. (6-32):
K f ,bend 1 qbend Kt ,bend 1 1 0.78 1.60 1 1.47
K f ,axial 1 qaxial Kt ,axial 1 1 0.78 1.75 1 1.59
K f ,tors 1 qtors Kt ,tors 1 1 0.82 1.39 1 1.32
Chapter 6 - Rev. A, Page 47/66
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1/2
2
0.191
2
a 1.47 27.75 1.59
3
1.32
7.65
0.85
m 1.47 27.75 1.59 0.191 3 1.32 7.65
2
2 1/ 2
44.71 kpsi
44.66 kpsi
Since these stresses are relatively high compared to the yield strength, we will go ahead
and check for yielding using the distortion energy failure theory.
max
ny
max,axial 3 max
2
max,bend
Sy
54
0.87
max
61.8
2
55.5 0.382
2
3 15.3 61.8 kpsi
2
Ans.
This shows that yielding is predicted. Further analysis of fatigue life is just to be able to
report the fatigue factor of safety, though the life will be dictated by the static yielding
failure, i.e. N = 1/2 cycle.
Ans.
Eq. (6-8):
Se 0.5 64 32 kpsi
Eq. (6-19):
k a aSutb 2.70(64) 0.265 0.897
Eq. (6-24):
de 0.370d 0.370 1 0.370 in
Eq. (6-20):
Eq. (6-18):
kb 0.879 d e 0.107 0.879(0.370) 0.107 0.978
Se (0.897)(0.978)(0.5)(64) 28.1 kpsi
Using Modified Goodman,
1 a m 44.71 44.66
nf
Se Sut
28.1
64
n f 0.44
Ans.
______________________________________________________________________________
6-54 From Table A-20, for AISI 1040 CD, S ut = 85 kpsi and S y = 71 kpsi. From the solution to
Prob. 6-17 we find the completely reversed stress at the critical shoulder fillet to be rev =
35.0 kpsi, producing a = 35.0 kpsi and m = 0 kpsi. This problem adds a steady torque
which creates torsional stresses of
m
Tr 2500 1.625 / 2
2967 psi 2.97 kpsi, a 0 kpsi
J
1.6254 / 32
From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15,
K t,bend =1.95, K t,tors =1.60
Chapter 6 - Rev. A, Page 48/66
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: q bend = 0.76, q tors = 0.81
Eq. (6-32):
K f ,bend 1 qbend Kt ,bend 1 1 0.76 1.95 1 1.72
K f ,tors 1 qtors Kt ,tors 1 1 0.811.60 1 1.49
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1.72 0 3 1.49 2.97
a 1.72 35.0 3 1.49 0
2 1/ 2
2
2 1/ 2
2
m
60.2 kpsi
7.66 kpsi
a m ,
Check for yielding, using the conservative max
Sy
71
1.05
ny
a m 60.2 7.66
From the solution to Prob. 6-17, S e = 29.5 kpsi. Using Modified Goodman,
1 a m 60.2 7.66
nf
Se Sut 29.5 85
n f 0.47
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
Fig. 6-18:
Eq. (6-14):
a
60.2
66.2 kpsi
1 ( m / Sut ) 1 (7.66 / 85)
f = 0.867
2
2
f Sut
0.867(85)
a
184.1
Se
29.5
Eq. (6-15):
f Sut
1
b log
3
Se
Eq. (6-16):
N rev
a
1/ b
1
0.867(85)
log
0.1325
3
29.5
1
66.2 0.1325
2251 cycles
184.1
N = 2300 cycles
Ans.
______________________________________________________________________________
Chapter 6 - Rev. A, Page 49/66
6-55
From the solution to Prob. 6-18 we find the completely reversed stress at the critical
shoulder fillet to be rev = 32.8 kpsi, producing a = 32.8 kpsi and m = 0 kpsi. This
problem adds a steady torque which creates torsional stresses of
m
Tr 2200 1.625 / 2
2611 psi 2.61 kpsi, a 0 kpsi
J
1.6254 / 32
From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15,
K t,bend =1.95, K t,tors =1.60
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: q bend = 0.76, q tors = 0.81
Eq. (6-32):
K f ,bend 1 qbend Kt ,bend 1 1 0.76 1.95 1 1.72
K f ,tors 1 qtors Kt ,tors 1 1 0.811.60 1 1.49
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1.72 0 3 1.49 2.61
a 1.72 32.8 3 1.49 0
2
2
m
2 1/ 2
2 1/ 2
56.4 kpsi
6.74 kpsi
a m ,
Check for yielding, using the conservative max
Sy
71
1.12
ny
a m 56.4 6.74
From the solution to Prob. 6-18, S e = 29.5 kpsi. Using Modified Goodman,
1 a m 56.4 6.74
nf
Se Sut 29.5 85
n f 0.50
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
Fig. 6-18:
a
56.4
61.3 kpsi
1 ( m / Sut ) 1 (6.74 / 85)
f = 0.867
Chapter 6 - Rev. A, Page 50/66
f Sut
a
Eq. (6-14):
Se
2
0.867(85)
29.5
f Sut
1
b log
3
Se
Eq. (6-15):
1/ b
Eq. (6-16): N rev
a
2
184.1
1
0.867(85)
log
0.1325
3
29.5
1
61.3 0.1325
4022 cycles
184.1
N = 4000 cycles
Ans.
______________________________________________________________________________
6-56
Sut 55 kpsi, S y 30 kpsi, K ts 1.6, L 2 ft, Fmin 150 lbf , Fmax 500 lbf
Eqs. (6-34) and (6-35b), or Fig. 6-21: q s = 0.80
Eq. (6-32): K fs 1 qs Kts 1 1 0.80 1.6 1 1.48
Tmax 500(2) 1000 lbf in, Tmin 150(2) 300 lbf in
max
min
m
16 K fsTmax
d
3
16 K fsTmin
d
max min
3
2
a max min
2
16(1.48)(1000)
11 251 psi 11.25 kpsi
(0.875)3
16(1.48)(300)
3375 psi 3.38 kpsi
(0.875)3
11.25 3.38
7.32 kpsi
2
11.25 3.38
3.94 kpsi
2
Since the stress is entirely shear, it is convenient to check for yielding using the standard
Maximum Shear Stress theory.
S y / 2 30 / 2
1.33
ny
max 11.25
Find the modifiers and endurance limit.
Se 0.5(55) 27.5 kpsi
Eq. (6-8):
Eq. (6-19):
Eq. (6-24):
k a 14.4(55) 0.718 0.81
Eq. (6-20):
Eq. (6-26):
Eq. (6-18):
kb 0.879(0.324) 0.107 0.99
d e 0.370(0.875) 0.324 in
kc 0.59
S se 0.81(0.99)(0.59)(27.5) 13.0 kpsi
Chapter 6 - Rev. A, Page 51/66
Since the stress is entirely shear, we will use a load factor k c = 0.59, and convert the
ultimate strength to a shear value rather than using the combination loading method of
Sec. 6-14. From Eq. (6-54), S su = 0.67S u = 0.67 (55) = 36.9 kpsi.
(a) Modified Goodman, Table 6-6
nf
1
1
1.99
( a / S se ) ( m / S su ) (3.94 / 13.0) (7.32 / 36.9)
Ans.
(b) Gerber, Table 6-7
2
2
2 m S se
1 S su a
nf
1 1
2 m S se
Ssu a
2
2
2(7.32)(13.0)
1 36.9 3.94
1 1
2 7.32 13.0
36.9(3.94)
n f 2.49 Ans.
______________________________________________________________________________
6-57
Sut 145 kpsi, S y 120 kpsi
From Eqs. (6-34) and (6-35a), or Fig. 6-20, with a notch radius of 0.1 in, q = 0.9. Thus,
with K t = 3 from the problem statement,
K f 1 q ( K t 1) 1 0.9(3 1) 2.80
4 P 2.80(4)( P)
2.476 P
d2
(1.2) 2
1
m a (2.476 P) 1.238 P
2
f P D d 0.3P 6 1.2
Tmax
0.54 P
4
4
max K f
From Eqs. (6-34) and (6-35b), or Fig. 6-21, with a notch radius of 0.1 in, qs 0.92. Thus,
with K ts = 1.8 from the problem statement,
K fs 1 qs ( K ts 1) 1 0.92(1.8 1) 1.74
16 K fsT
16(1.74)(0.54 P)
2.769 P
d
(1.2)3
2.769 P
a m max
1.385 P
2
2
Eqs. (6-55) and (6-56):
max
3
Chapter 6 - Rev. A, Page 52/66
a [( a / 0.85)2 3 a2 ]1/2 [(1.238P / 0.85) 2 3(1.385P) 2 ]1/2 2.81P
m [ m 2 3 m2 ]1/2 [(1.238P) 2 3(1.385 P) 2 ]1/2 2.70 P
Eq. (6-8):
Se 0.5(145) 72.5 kpsi
Eq. (6-19):
ka 2.70(145)0.265 0.722
Eq. (6-20):
Eq. (6-18):
kb 0.879(1.2) 0.107 0.862
Se (0.722)(0.862)(72.5) 45.12 kpsi
Modified Goodman:
1 a m 2.81P 2.70 P 1
n f Se Sut 45.12
145
3
P 4.12 kips
Ans.
120
5.29 Ans.
a m (2.81)(4.12) (2.70)(4.12)
______________________________________________________________________________
Yield (conservative): ny
6-58
Sy
From Prob. 6-57, K f 2.80, K f s 1.74, S e 45.12 kpsi
4 Pmax
4(18)
2.80
44.56 kpsi
2
d
(1.22 )
4P
4(4.5)
min K f min2 2.80
11.14 kpsi
d
(1.2) 2
Dd
6 1.2
Tmax f Pmax
0.3(18)
9.72 kip in
4
4
Dd
6 1.2
Tmin f Pmin
0.3(4.5)
2.43 kip in
4
4
16Tmax
16(9.72)
max K f s
1.74
49.85 kpsi
3
d
(1.2)3
16Tmin
16(2.43)
min K f s
1.74
12.46 kpsi
3
d
(1.2)3
44.56 (11.14)
a
16.71 kpsi
2
44.56 (11.14)
m
27.85 kpsi
2
49.85 12.46
a
18.70 kpsi
2
49.85 12.46
m
31.16 kpsi
2
max K f
Chapter 6 - Rev. A, Page 53/66
Eqs. (6-55) and (6-56):
a [( a / 0.85)2 3 a2 ]1/2 [(16.71/ 0.85) 2 3(18.70) 2 ]1/2 37.89 kpsi
m [ m 2 3 m2 ]1/2 [(27.85) 2 3(31.16)2 ]1/2 60.73 kpsi
Modified Goodman:
1 a m 37.89 60.73
n f Se Sut 45.12 145
n f = 0.79
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
a
37.89
65.2 kpsi
1 ( m / Sut ) 1 (60.73 / 145)
Fig. 6-18:
f = 0.8
Eq. (6-14):
f Sut
a
2
0.8(145)
2
45.12
Se
298.2
Eq. (6-15):
f Sut
1
1
0.8(145)
b log
log
0.1367
3
3
45.12
Se
Eq. (6-16):
N rev
a
1/ b
1
65.2 0.1367
67 607 cycles
298.2
N = 67 600 cycles
Ans.
______________________________________________________________________________
6-59
For AISI 1020 CD, From Table A-20, S y = 390 MPa, S ut = 470 MPa. Given: S e = 175
MPa.
360 160
360 160
First Loading:
260 MPa,
100 MPa
m 1
a 1
2
2
Goodman:
a e1
a 1
1 m 1 / Sut
100
223.8 MPa Se finite life
1 260 / 470
Chapter 6 - Rev. A, Page 54/66
0.9 470
a
1022.5 MPa
175
0.9 470
1
b log
0.127 767
3
175
2
1/0.127 767
223.8
N
145 920 cycles
1022.5
320 200
320 200
Second loading: m 2
60 MPa,
260 MPa
a 2
2
2
a e 2
(a) Miner’s method:
n1 n2
1
N1 N 2
260
298.0 MPa
1 60 / 470
298.0
N2
1022.5
1/0.127767
15 520 cycles
n2
80 000
1
145 920 15 520
n2 7000 cycles Ans.
(b) Manson’s method: The number of cycles remaining after the first loading
N remaining =145 920 80 000 = 65 920 cycles
Two data points: 0.9(470) MPa, 103 cycles
223.8 MPa, 65 920 cycles
a2 103
0.9 470
b
223.8
a2 65 920 2
b2
1.8901 0.015170 2
b
log1.8901
0.151 997
log 0.015170
223.8
1208.7 MPa
a2
0.151 997
65 920
b2
1/ 0.151 997
298.0
10 000 cycles
n2
Ans.
1208.7
______________________________________________________________________________
6-60
Given: S e = 50 kpsi, S ut = 140 kpsi, f =0.8. Using Miner’s method,
Chapter 6 - Rev. A, Page 55/66
0.8 140
a
250.88 kpsi
50
0.8 140
1
b log
0.116 749
3
50
1/ 0.116 749
95
1 95 kpsi,
N1
4100 cycles
250.88
2
1/ 0.116 749
2 80 kpsi,
80
N2
250.88
3 65 kpsi,
65
N3
250.88
17 850 cycles
1/ 0.116 749
105 700 cycles
0.2 N 0.5 N
0.3N
1 N 12 600 cycles Ans.
4100 17 850 105 700
______________________________________________________________________________
6-61
Given: S ut = 530 MPa, S e = 210 MPa, and f = 0.9.
(a) Miner’s method
0.9 530
a
1083.47 MPa
210
0.9 530
1
b log
0.118 766
3
210
2
1/ 0.118 766
350
1 350 MPa, N1
13 550 cycles
1083.47
1/ 0.118 766
260
2 260 MPa, N 2
165 600 cycles
1083.47
225
1/ 0.118 766
3 225 MPa, N 3
1083.47
559 400 cycles
n1 n2 n3
1
N1 N 2 N 3
n3
5000
50 000
184 100 cycles
13 550 165 600 559 400
Ans.
(b) Manson’s method:
The life remaining after the first series of cycling is N R1 = 13 550 5000 = 8550
cycles. The two data points required to define S e,1 are [0.9(530), 103] and (350, 8550).
Chapter 6 - Rev. A, Page 56/66
a2 103
0.9 530
b
350
a2 8550 2
b2
b2
a2
1.3629 0.11696 2
b
log 1.362 9
0.144 280
log 0.116 96
350
8550
0.144 280
1292.3 MPa
1/0.144 280
260
67 090 cycles
N2
1292.3
N R 2 67 090 50 000 17 090 cycles
a3 103
0.9 530
b
260
a3 17 090 3
b3
b3
1.834 6 0.058 514 2
b
log 1.834 6
260
0.213 785, a3
2088.7 MPa
0.213 785
log 0.058 514
17 090
1/0.213 785
225
N3
33 610 cycles Ans.
2088.7
______________________________________________________________________________
6-62
Given: S e = 45 kpsi, S ut = 85 kpsi, f = 0.86, and a = 35 kpsi and m = 30 kpsi for 12
(103) cycles.
a
35
Gerber equivalent reversing stress: rev
39.98 kpsi
2
2
1 m / Sut 1 30 / 85
(a) Miner’s method: rev < S e . According to the method, this means that the endurance
limit has not been reduced and the new endurance limit is S e = 45 kpsi. Ans.
(b) Manson’s method: Again, rev < S e . According to the method, this means that the
material has not been damaged and the endurance limit has not been reduced. Thus,
the new endurance limit is S e = 45 kpsi. Ans.
______________________________________________________________________________
6-63
Given: S e = 45 kpsi, S ut = 85 kpsi, f = 0.86, and a = 35 kpsi and m = 30 kpsi for 12
(103) cycles.
a
35
Goodman equivalent reversing stress: rev
54.09 kpsi
1 m / Sut 1 30 / 85
Initial cycling
Chapter 6 - Rev. A, Page 57/66
0.86 85
a
116.00 kpsi
45
0.86 85
1
b log
0.070 235
3
45
2
1/ 0.070 235
1 54.09 kpsi,
54.09
N1
116.00
52 190 cycles
(a) Miner’s method (see discussion on p. 325): The number of remaining cycles at 54.09
kpsi is N remaining = 52 190 12 000 = 40 190 cycles. The new coefficients are b = b,
and a =S f /Nb = 54.09/(40 190) 0.070 235 = 113.89 kpsi. The new endurance limit is
Se,1 aN eb 113.89 106
0.070 235
43.2 kpsi
Ans.
(b) Manson’s method (see discussion on p. 326): The number of remaining cycles at
54.09 kpsi is N remaining = 52 190 12 000 = 40 190 cycles. At 103 cycles,
S f = 0.86(85) = 73.1 kpsi. The new coefficients are
b = [log(73.1/54.09)]/log(103/40 190) = 0.081 540 and a = 1 / (N remaining ) b =
54.09/(40 190) 0.081 540 = 128.39 kpsi. The new endurance limit is
Se,1 aN eb 128.39 106
0.081 540
41.6 kpsi
Ans.
______________________________________________________________________________
6-64
Given S ut =1030LN(1, 0.0508) MPa
From Table 6-10:
a = 1.58, b = 0.086, C = 0.120
0.086
Eq. (6-72) and Table 6-10): k a 1.58 1030
LN 1, 0.120 0.870LN 1, 0.120
From Prob. 6-1:
k b = 0.97
Eqs. (6-70) and (6-71):
0.138)]
S e = [0.870LN(1, 0.120)] (0.97) [0.506(1030)LN(1,
S e 0.870 (0.97)(0.506)(1030) = 440 MPa
and,
C Se (0.122 + 0.1382)1/2 = 0.183
S e =440LN(1, 0.183) MPa
Ans.
______________________________________________________________________________
Chapter 6 - Rev. A, Page 58/66
6-65
A Priori Decisions:
• Material and condition: 1020 CD, S ut = 68 LN(1, 0.28), and
S y = 57 LN(1, 0.058) kpsi
• Reliability goal: R = 0.99 (z = 2.326, Table A-10)
• Function:
Critical location—hole
• Variabilities:
Cka 0.058
Ckc 0.125
CSe 0.138
CSe Cka2 Ckc2 CS2e
1/2
(0.0582 0.1252 0.1382 )1/2 0.195
CKf 0.10
CFa 0.20
C a (0.102 0.202 )1/2 0.234
Cn
CSe2 C2a
0.1952 0.2342
0.297
1 C2a
1 0.2342
Resulting in a design factor n f of,
Eq. (6-59): n f exp[(2.326) ln(1 0.297 2 ) ln 1 0.297 2 ] 2.05
• Decision: Set n f = 2.05
Now proceed deterministically using the mean values:
Table 6-10:
k a 2.67 68
Eq. (6-21):
kb = 1
Table 6-11:
kc 1.23 68
Eq. (6-70):
Se 0.506 68 34.4 kpsi
Eq. (6-71):
Se 0.873 1 0.886 34.4 26.6 kpsi
0.265
0.873
0.0778
0.886
From Prob. 6-14, K f = 2.26. Thus,
Chapter 6 - Rev. A, Page 59/66
a K f
t
Decision:
Fa
Fa
F
S
Kf
Kf a e
A
t 2.5 0.5
2t n f
n f K f Fa
2Se
Use t =
in
3
8
2.05 2.26 3.8
0.331 in
2 26.6
Ans.
______________________________________________________________________________
6-66
Rotation is presumed. M and S ut are given as deterministic, but notice that is not;
therefore, a reliability estimation can be made.
From Eq. (6-70): S e = 0.506(780)LN(1, 0.138) = 394.7 LN(1, 0.138)
Table 6-13: k a = 4.45(780) 0.265LN(1, 0.058) = 0.762 LN(1, 0.058)
Based on d = 32 6 = 26 mm, Eq. (6-20) gives
26
kb
7.62
0.107
0.877
Conservatism is not necessary
S e 0.762LN 1, 0.058 (0.877)(394.7) LN(1, 0.138)
Se 263.8 MPa
CSe (0.0582 0.1382 )1/2 0.150
S e 263.8LN(1, 0.150) MPa
Fig. A-15-14: D/d = 32/26 = 1.23, r/d = 3/26 = 0.115. Thus, K t 1.75, and Eq. (6-78)
and Table 6-15 gives
Kt
1.75
1.64
Kf
2 1.75 1 104 / 780
2 K t 1 a
1
1
1.75
Kt
3
r
From Table 6-15, C Kf = 0.15. Thus,
K
f
= 1.64LN(1, 0.15)
The bending stress is
Kf
32(160)
32M
1.64LN(1, 0.15)
3
3
d
(0.026)
152 106 LN(1, 0.15) Pa 152LN(1, 0.15) MPa
From Eq. (5-43), p. 250,
Chapter 6 - Rev. A, Page 60/66
1 C2
ln S
1 CS2
z
ln 1 CS2 1 C2
1 0.15 / 1 0.15
2.61
ln 1 0.15 1 0.15
ln 263.8 / 152
2
2
2
2
From Table A-10, p f = 0.004 53. Thus, R = 1 0.004 53 = 0.995
Ans.
Note: The correlation method uses only the mean of S ut ; its variability is already
included in the 0.138. When a deterministic load, in this case M, is used in a reliability
estimate, engineers state, “For a Design Load of M, the reliability is 0.995.” They are, in
fact, referring to a Deterministic Design Load.
______________________________________________________________________________
6-67
For completely reversed torsion, k a and k b of Prob. 6-66 apply, but k c must also be
considered. Sut = 780/6.89 = 113 kpsi
k c = 0.328(113)0.125LN(1, 0.125) = 0.592LN(1, 0.125)
Eq. 6-74:
Note 0.590 is close to 0.577.
S e k a kbk cSe
0.762[LN(1, 0.058)](0.877)[0.592LN(1, 0.125)][394.7LN(1, 0.138)]
Se 0.762(0.877)(0.592)(394.7) 156.2 MPa
CSe (0.0582 0.1252 0.1382 )1/2 0.195
S e 156.2LN(1, 0.195) MPa
Fig. A-15-15: D/d = 1.23, r/d = 0.115, then K ts 1.40. From Eq. (6-78) and
Table 7-8
K ts
1.40
1.34
K fs
2 1.40 1 104 / 780
2 K ts 1 a
1
1
1.40
K ts
3
r
From Table 6-15, C Kf = 0.15. Thus,
K
fs
= 1.34LN(1, 0.15)
The torsional stress is
K fs
16 160
16T
LN
1.34
(1,
0.15)
3
d3
0.026
62.1106 LN(1, 0.15) Pa 62.1LN(1, 0.15) MPa
Chapter 6 - Rev. A, Page 61/66
From Eq. (5-43), p. 250,
ln (156.2 / 62.1) (1 0.152 ) / (1 0.1952 )
3.75
z
2
2
ln[(1 0.195 )(1 0.15 )]
From Table A-10, p f = 0.000 09
R = 1 p f = 1 0.000 09 = 0.999 91
Ans.
For a design with completely-reversed torsion of 160 N · m, the reliability is 0.999 91.
The improvement over bending comes from a smaller stress-concentration factor in
torsion. See the note at the end of the solution of Prob. 6-66 for the reason for the
phraseology.
______________________________________________________________________________
6-68
Given: S ut = 58 kpsi.
Eq. (6-70):
S e = 0.506(76) LN(1, 0.138) = 38.5 LN(1, 0.138) kpsi
Table 6-13:
k a = 14.5(76) 0.719 LN(1, 0.11) = 0.644 LN(1, 0.11)
Eq. (6-24):
d e = 0.370(1.5) = 0.555 in
Eq. (6-20):
Eq. (6-70):
k b = (0.555/0.3)0.107 = 0.936
S e = [0.644 LN(1, 0.11)](0.936)[38.5 LN(1, 0.138)]
Se 0.644 0.936 38.5 23.2 kpsi
C Se = (0.112 + 0.1382)1/2 = 0.176
S e =23.2 LN(1, 0.176) kpsi
Table A-16: d/D = 0, a/D = (3/16)/1.5 = 0.125, A = 0.80 K t = 2.20.
From Eqs. (6-78) and (6-79) and Table 6-15
Chapter 6 - Rev. A, Page 62/66
Kf
2.20LN(1, 0.10)
1.83LN(1, 0.10)
2 2.20 1 5 / 76
1
2.20
0.125
Table A-16:
Z net
AD3
32
(0.80)(1.53 )
32
0.265 in 3
M
1.5
1.83LN(1, 0.10)
Z net
0.265
10.4LN (1, 0.10) kpsi
10.4 kpsi
C 0.10
Kf
Eq. (5-43), p. 250:
ln (23.2 /10.4) (1 0.102 ) / (1 0.1762 )
3.94
z
2
2
ln[(1 0.176 )(1 0.10 )]
Table A-10: p f = 0.000 041 5 R = 1 p f = 1 0.000 041 5 = 0.999 96 Ans.
______________________________________________________________________________
6-69
From Prob. 6-68:
S e = 23.2 LN(1, 0.138) kpsi
k a = 0.644LN(1, 0.11)
k b = 0.936
Eq. (6-74):
k c = 0.328(76)0.125LN(1, 0.125) = 0.564 LN(1, 0.125)
Eq. (6-71): S e = [0.644LN(1, 0.11)](0.936)[ 0.564 LN(1, 0.125)][ 23.2 LN(1, 0.138)]
Se 0.644 0.936 0.564 23.2 7.89 kpsi
C Se = (0.112 +0.1252 + 0.1383)1/2 = 0.216
Table A-16: d/D = 0, a/D = (3/16)/1.5 = 0.125, A = 0.89, K ts = 1.64
From Eqs. (6-78) and(7-79), and Table 6-15
Kfs
1.64LN (1, 0.10)
1.40LN (1, 0.10)
2 1.64 1 5 / 76
1
1.64
3 / 32
Chapter 6 - Rev. A, Page 63/66
Table A-16:
AD 4 (0.89)(1.54 )
0.4423 in 4
J net
32
32
TD
2(1.5)
a K f s a 1.40[LN (1, 0.10)]
4.75LN (1, 0.10) kpsi
2 J net
2 0.4423
From Eq. (6-57):
z
ln(7.89 / 4.75) (1 0.102 ) / (1 0.2162 )
ln[(1 0.102 )(1 0.2162 )]
2.08
Table A-10, p f = 0.0188,
R = 1 p f = 1 0.0188 = 0.981 Ans.
______________________________________________________________________________
6-70
This is a very important task for the student to attempt before starting Part 3. It illustrates
the drawback of the deterministic factor of safety method. It also identifies the a priori
decisions and their consequences.
The range of force fluctuation in Prob. 6-30 is 16 to + 5 kip, or 21 kip. Let the
repeatedly-applied F a be 10.5 kip. The stochastic properties of this heat of AISI 1018 CD
are given in the problem statement.
Function
Axial
Fatigue load
Consequences
F a = 10.5 kip
C Fa = 0
C kc = 0.125
Overall reliability R ≥ 0.998; z = 3.09
with twin fillets
C Kf = 0.11
R 0.998 0.999
Cold rolled or machined
C ka = 0.058
surfaces
Ambient temperature
C kd = 0
Use correlation method
C 0.138
Stress amplitude
C Kf = 0.11
C a = 0.11
Significant strength S e
CSe (0.0582 0.1252 0.1382 )1/ 2 0.195
Choose the mean design factor which will meet the reliability goal. From Eq. (6-88)
Cn
0.1952 0.112
0.223
1 0.112
n exp (3.09) ln(1 0.2232 ) ln 1 0.2232
n 2.02
Chapter 6 - Rev. A, Page 64/66
In Prob. 6-30, it was found that the hole was the significant location that controlled the
analysis. Thus,
S
a e
n
S
Fa
S
a e K f
e
n
h w1 d n
We need to determine S e
ka 2.67 Sut-0.265 2.67(64)-0.265 0.887
kb = 1
kc 1.23Sut 0.0778 1.23(64) 0.0778 0.890
kd ke 1
S e 0.887(1)(0.890)(1)(1)(0.506)(64) 25.6 kpsi
From the solution to Prob. 6-30, the stress concentration factor at the hole is K t = 2.68.
From Eq. (6-78) and Table 6-15
Kf
2.68
2.20
2 2.68 1 5 / 64
1
2.68
0.2
2.20(2.02)(10.5)
0.588 Ans.
w1 d Se 3.5 0.4 (25.6)
______________________________________________________________________________
h
6-71
K f nFa
Fa 1200 lbf
Sut 80 kpsi
(a) Strength
k a = 2.67(80) 0265LN(1, 0.058) = 0.836 LN(1, 0.058)
kb = 1
k c = 1.23(80) 0.0778LN(1, 0.125) = 0.875 LN(1, 0.125)
Chapter 6 - Rev. A, Page 65/66
Se 0.506(80)LN(1, 0.138) 40.5LN(1, 0.138) kpsi
S e 0.836LN (1, 0.058) (1) 0.875LN(1, 0.125) 40.5LN(1, 0.138)
Se 0.836(1)(0.875)(40.5) 29.6 kpsi
CSe (0.0582 0.1252 0.1382 )1/2 0.195
Stress: Fig. A-15-1; d/w = 0.75/1.5 = 0.5, K t = 2.18. From Eqs. (6-78), (6-79) and
Table 6-15
2.18LN(1, 0.10)
1.96LN(1, 0.10)
2 2.18 1 5 / 80
1
2.18
0.375
Fa
, C 0.10
a K f
( w d )t
Kf
a
K f Fa
( w d )t
1.96(1.2)
12.54 kpsi
(1.5 0.75)(0.25)
S a Se 29.6 kpsi
ln ( S a / a ) 1 C2 1 CS2
z
2
2
ln 1 C 1 CS
1 0.10 / 1 0.195
3.9
ln 1 0.10 1 0.195
ln 29.6 /12.48
2
2
From Table A-20, p f = 4.81(10 5)
2
2
R = 1 4.81(10 5) = 0.999 955
Ans.
(b) All computer programs will differ in detail.
______________________________________________________________________________
6-72 to 6-78 Computer programs are very useful for automating specific tasks in the design
process. All computer programs will differ in detail.
Chapter 6 - Rev. A, Page 66/66
Chapter 7
7-1
(a) DE-Gerber, Eq. (7-10):
A 4 K f M a 3 K fsTa 4 (2.2)(70) 3 (1.8)(45) 338.4 N m
2
2
2
2
B 4 K f M m 3 K fsTm 4 (2.2)(55) 3 (1.8)(35) 265.5 N m
2
2
2
6
8(2)(338.4) 2(265.5) 210 10
d
1
1
6
6
210 10 338.4 700 10
3
d = 25.85 (10 ) m = 25.85 mm Ans.
2
2 1/2
1/3
(b) DE-elliptic, Eq. (7-12) can be shown to be
1/3
2
2
338.4
265.5
16(2)
2
2
6
6
210 10
560 10
3
d = 25.77 (10 ) m = 25.77 mm Ans.
16n
d
1/3
A
B
Se2 S y2
2
2
(c) DE-Soderberg, Eq. (7-14) can be shown to be
1/3
16(2) 338.4
16n A B
265.5
d
6
210 10
560 106
Se S y
d = 27.70 (103) m = 27.70 mm Ans.
1/3
(d) DE-Goodman: Eq. (7-8) can be shown to be
1/3
16(2) 338.4
265.5
6
210 106
700
10
3
d = 27.27 (10 ) m = 27.27 mm Ans.
________________________________________________________________________
Criterion
d (mm)
Compared to DE-Gerber
DE-Gerber
25.85
DE-Elliptic
25.77
0.31% Lower
Less conservative
DE-Soderberg
27.70
7.2% Higher
More conservative
DE-Goodman
27.27
5.5% Higher
More conservative
______________________________________________________________________________
1/3
16n A B
d
Se Sut
7-2
This problem has to be done by successive trials, since S e is a function of shaft size. The
material is SAE 2340 for which S ut = 175 kpsi, S y = 160 kpsi, and H B ≥ 370.
Chapter 7 - Rev. A, Page 1/45
Eq. (6-19), p. 287:
k a 2.70(175) 0.265 0.69
Trial #1: Choose d r = 0.75 in
Eq. (6-20), p. 288:
kb 0.879(0.75) 0.107 0.91
Eq. (6-8), p.282:
Se 0.5Sut 0.5 175 87.5 kpsi
S e = 0.69 (0.91)(87.5) = 54.9 kpsi
Eq. (6-18), p. 287:
d r d 2r 0.75 D 2 D / 20 0.65D
d
0.75
D r
1.15 in
0.65 0.65
D 1.15
r
0.058 in
20 20
Fig. A-15-14:
d d r 2r 0.75 2(0.058) 0.808 in
d 0.808
1.08
dr
0.75
r 0.058
0.077
dr
0.75
K t = 1.9
Fig. 6-20, p. 295:
r = 0.058 in, q = 0.90
Eq. (6-32), p. 295:
K f = 1 + 0.90 (1.9 – 1) = 1.81
Fig. A-15-15:
K ts = 1.5
Fig. 6-21, p. 296:
r = 0.058 in, q s = 0.92
Eq. (6-32), p. 295:
K fs = 1 + 0.92 (1.5 – 1) = 1.46
We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as d r , and
M m = T a = 0,
1/3
2
2 1/2
1.46(400)
16(2.5) 1.81(600)
dr
4
3
3
160 103
54.9 10
d r = 0.799 in
Trial #2: Choose d r = 0.799 in.
kb 0.879(0.799) 0.107 0.90
S e = 0.69 (0.90)(0.5)(175) = 54.3 kpsi
d
0.799
D r
1.23 in
0.65 0.65
r = D / 20 = 1.23/20 = 0.062 in
Chapter 7 - Rev. A, Page 2/45
Figs. A-15-14 and A-15-15:
d d r 2r 0.799 2(0.062) 0.923 in
d 0.923
1.16
d r 0.799
r 0.062
0.078
d r 0.799
With these ratios only slightly different from the previous iteration, we are at the limit of
readability of the figures. We will keep the same values as before.
K t 1.9, K ts 1.5, q 0.90, qs 0.92
K f 1.81, K fs 1.46
Using Eq. (7-12) produces d r = 0.802 in. Further iteration produces no change. With
d r = 0.802 in,
0.802
D
1.23 in
0.65
d 0.75(1.23) 0.92 in
A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In
nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans.
______________________________________________________________________________
7-3
F cos 20(d / 2) = T A , F = 2 T A / ( d cos 20) = 2(340) / (0.150 cos 20) = 4824 N.
The maximum bending moment will be at point C, with M C = 4824(0.100) = 482.4 N·m.
Due to the rotation, the bending is completely reversed, while the torsion is constant.
Thus, M a = 482.4 N·m, T m = 340 N·m, M m = T a = 0.
For sharp fillet radii at the shoulders, from Table 7-1, K t = 2.7, and K ts = 2.2. Examining
Figs. 6-20 and 6-21 (pp. 295 and 296 respectively) with Sut 560 MPa, conservatively
estimate q = 0.8 and qs 0.9. These estimates can be checked once a specific fillet radius
is determined.
Eq. (6-32):
K f 1 0.8(2.7 1) 2.4
K fs 1 0.9(2.2 1) 2.1
(a) We will choose to include fatigue stress concentration factors even for the static
analysis to avoid localized yielding.
1/2
Eq. (7-15):
max
2
32 K f M a 2
16 K fsTm
3
3
3
d
d
Chapter 7 - Rev. A, Page 3/45
Eq. (7-16):
Sy
d 3S y
n
max
16
4 K M 2 3 K T 2
f
a
fs m
1/ 2
Solving for d,
1/3
16n
1/2
4( K f M a ) 2 3( K fsTa ) 2
d
S y
16(2.5)
420 106
4 (2.4)(482.4) 3 (2.1)(340)
d = 0.0430 m = 43.0 mm
(b)
2 1/2
2
1/3
Ans.
k a 4.51(560) 0.265 0.84
Assume k b = 0.85 for now. Check later once a diameter is known.
S e = 0.84(0.85)(0.5)(560) = 200 MPa
Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with M m Ta 0.
2
2.1(340)
16(2.5) 2.4(482.4)
3
d
4
6
420 106
200 10
0.0534 m 53.4 mm
2 1/ 2
1/3
With this diameter, we can refine our estimates for k b and q.
Eq. (6-20):
kb 1.51d 0.157 1.51 53.4
0.157
0.81
Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (53.4) = 1.07 mm.
Fig. (6-20):
Fig. (6-21):
q = 0.72
q s = 0.77
Iterating with these new estimates,
Eq. (6-32):
Eq. (6-18):
Eq. (7-12):
K f = 1 + 0.72 (2.7 – 1) = 2.2
K fs = 1 + 0.77 (2.2 – 1) = 1.9
S e = 0.84(0.81)(0.5)(560) = 191 MPa
d = 53 mm
Ans.
Further iteration does not change the results.
_____________________________________________________________________________
Chapter 7 - Rev. A, Page 4/45
7-4
We have a design task of identifying bending moment and torsion diagrams which are
preliminary to an industrial roller shaft design. Let point C represent the center of the
span of the roller.
FCy 30(8) 240 lbf
FCz 0.4(240) 96 lbf
T FCz (2) 96(2) 192 lbf in
T 192
FBz
128 lbf
1.5 1.5
FBy FBz tan 20 128 tan 20 46.6 lbf
(a) xy-plane
M O 240(5.75) FAy (11.5) 46.6(14.25) 0
240(5.75) 46.6(14.25)
FAy
62.3 lbf
11.5
M A FOy (11.5) 46.6(2.75) 240(5.75) 0
240(5.75) 46.6(2.75)
FOy
131.1 lbf
11.5
Bending moment diagram:
xz-plane
Chapter 7 - Rev. A, Page 5/45
M O 0 96(5.75) FAz (11.5) 128(14.25)
96(5.75) 128(14.25)
FAz
206.6 lbf
11.5
M A 0 FOz (11.5) 128(2.75) 96(5.75)
96(5.75) 128(2.75)
FOz
17.4 lbf
11.5
Bending moment diagram:
M C 100 2 ( 754) 2 761 lbf in
M A ( 128) 2 ( 352) 2 375 lbf in
Torque: The torque is constant from C to B, with a magnitude previously obtained of 192
lbf·in.
(b) xy-plane
2
2
M xy 131.1x 15 x 1.75 15 x 9.75 62.3 x 11.5
1
Bending moment diagram:
Chapter 7 - Rev. A, Page 6/45
M max = –516 lbf · in and occurs at 6.12 in.
M C 131.1(5.75) 15(5.75 1.75) 2 514 lbf in
This is reduced from 754 lbf · in found in part (a). The maximum occurs
at x 6.12 in rather than C, but it is close enough.
xz-plane
2
2
M xz 17.4 x 6 x 1.75 6 x 9.75 206.6 x 11.5
1
Bending moment diagram:
Let M net M xy2 M xz2
Plot M net (x), 1.75 ≤ x ≤ 11.5 in
M max = 516 lbf · in at x = 6.25 in
Torque: The torque rises from 0 to 192 lbf·in linearly across the roller, then is constant to
B. Ans.
______________________________________________________________________________
This is a design problem, which can have many acceptable designs. See the solution for
Prob. 7-17 for an example of the design process.
______________________________________________________________________________
7-5
Chapter 7 - Rev. A, Page 7/45
7-6
If students have access to finite element or beam analysis software, have them model the
shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in
diameter sections will not affect the deflection results much, so model the 1 in diameter
as 1.25 in. Also, ignore the step in AB.
From Prob. 7-4, integrate M xy and M xz .
xy plane, with dy/dx = y'
131.1 2
62.3
3
3
2
x 5 x 1.75 5 x 9.75
x 11.5 C1
2
2
131.1 3 5
5
62.3
4
4
3
EIy
x x 1.75 x 9.75
x 11.5 C1 x C2
6
4
4
6
EIy
(1)
y 0 at x 0
From (1),
C2 0
y 0 at x 11.5 C1 1908.4 lbf in 3
x = 0:
EIy' = 1908.4
x = 11.5:
EIy' = –2153.1
xz plane (treating z )
17.4 2
206.6
3
3
2
x 2 x 1.75 2 x 9.75
x 11.5 C3
2
2
17.4 3 1
1
206.6
4
4
3
EIz
x x 1.75 x 9.75
x 11.5 C3 x C4
6
2
2
6
EIz
(2)
z 0 at x 0
C4 0
From (2),
z 0 at x 11.5 C3 8.975 lbf in 3
x = 0:
EIz' = 8.975
x = 11.5:
EIz' = –683.5
At O:
EI 1908.42 8.9752 1908.4 lbf in 3
Chapter 7 - Rev. A, Page 8/45
EI (2153.1) 2 ( 683.5) 2 2259.0 lbf in 3 (dictates size)
At A:
2259
0.000 628 rad
30 10 / 64 1.254
n
0.001
1.59
0.000 628
6
At gear mesh, B
xy plane
With I I1 in section OCA,
yA 2153.1/ EI1
Since y' B/A is a cantilever, from Table A-9-1, with I I 2 in section AB
Fx( x 2l ) 46.6
yB / A
(2.75)[2.75 2(2.75)] 176.2 / EI 2
2 EI 2
2 EI 2
2153.1
176.2
yB yA yB / A
30 106 / 64 1.254 30 106 / 64 0.8754
= –0.000 803 rad (magnitude greater than 0.0005 rad)
xz plane
128 2.752
484
z A
683.5
,
EI1
z B
683.5
0.000 751 rad
30 10 / 64 1.254
30 106 / 64 0.8754
z B / A
6
2 EI 2
EI 2
484
B (0.000 803) 2 ( 0.000 751) 2 0.00110 rad
Crowned teeth must be used.
Finite element results:
O 5.47(104 ) rad
A 7.09(104 ) rad
B 1.10(103 ) rad
Error in simplified model
3.0%
11.4%
0.0%
Chapter 7 - Rev. A, Page 9/45
The simplified model yielded reasonable results.
Strength
Sut 72 kpsi, S y 39.5 kpsi
At the shoulder at A, x 10.75 in. From Prob. 7-4,
M xy 209.3 lbf in, M xz 293.0 lbf in, T 192 lbf in
M ( 209.3) 2 ( 293) 2 360.0 lbf in
Se 0.5(72) 36 kpsi
k a 2.70(72) 0.265 0.869
0.107
1
kb
0.879
0.3
kc k d ke k f 1
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Se 0.869(0.879)(36) 27.5 kpsi
D / d = 1.25, r / d = 0.03
K ts = 1.8
K t = 2.3
q = 0.65
q s = 0.70
K f 1 0.65(2.3 1) 1.85
K fs 1 0.70(1.8 1) 1.56
Using DE-ASME Elliptic, Eq. (7-11) with M m Ta 0,
1/2
2
2
1.56(192)
1
16 1.85(360)
3
4
n 13 27 500
39 500
n = 3.91
Perform a similar analysis at the profile keyway under the gear.
The main problem with the design is the undersized shaft overhang with excessive slope
at the gear. The use of crowned-teeth in the gears will eliminate this problem.
______________________________________________________________________________
7-7 through 7-16
These are design problems, which can have many acceptable designs. See the solution for
Prob. 7-17 for an example of the design process.
______________________________________________________________________________
7-17
(a) One possible shaft layout is shown in part (e). Both bearings and the gear will be
located against shoulders. The gear and the motor will transmit the torque through the
Chapter 7 - Rev. A, Page 10/45
keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the
shaft in the housing, while the right bearing will float in the housing.
(b) From summing moments around the shaft axis, the tangential transmitted load
through the gear will be
Wt T / (d / 2) 2500 / (4 / 2) 1250 lbf
The radial component of gear force is related by the pressure angle.
Wr Wt tan 1250 tan 20 455 lbf
W Wr2 Wt 2
1/2
4552 12502
1/2
1330 lbf
Reactions RA and RB , and the load W are all in the same plane. From force and moment
balance,
RA 1330(2 /11) 242 lbf
RB 1330(9 /11) 1088 lbf
M max RA (9) 242(9) 2178 lbf in
Shear force, bending moment, and torque diagrams can now be obtained.
(c) Potential critical locations occur at each stress concentration (shoulders and keyways).
To be thorough, the stress at each potentially critical location should be evaluated. For
Chapter 7 - Rev. A, Page 11/45
now, we will choose the most likely critical location, by observation of the loading
situation, to be in the keyway for the gear. At this point there is a large stress
concentration, a large bending moment, and the torque is present. The other locations
either have small bending moments, or no torque. The stress concentration for the
keyway is highest at the ends. For simplicity, and to be conservative, we will use the
maximum bending moment, even though it will have dropped off a little at the end of the
keyway.
(d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is
completely reversed and the torque is steady.
M a 2178 lbf in Tm 2500 lbf in M m Ta 0
From Table 7-1, estimate stress concentrations for the end-milled keyseat to be K t = 2.14
and K ts = 3.0. For the relatively low strength steel specified (AISI 1020 CD), roughly
estimate notch sensitivities of q = 0.75 and q s = 0.80, obtained by observation of Figs. 620 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02 (p.
373), and a shaft diameter of up to 3 inches.
Eq. (6-32):
K f 1 0.75(2.14 1) 1.9
K fs 1 0.8(3.0 1) 2.6
Eq. (6-19):
ka 2.70(68) 0.265 0.883
For estimating kb , guess d 2 in.
Eq. (6-20)
Eq. (6-18)
kb (2 / 0.3) 0.107 0.816
Se 0.883(0.816)(0.5)(68) 24.5 kpsi
Selecting the DE-Goodman criteria for a conservative first design,
1/3
Eq. (7-8):
1/2
2 1/2
3 K T 2
16n 4 K f M a
fs
m
d
Se
Sut
1/3
1/2
2 1/2
3 2.6 2500 2
4
1.9
2178
16(1.5)
d
24
500
68
000
d 1.57 in
Ans.
With this diameter, the estimates for notch sensitivity and size factor were conservative,
but close enough for a first iteration until deflections are checked. Check yielding with
this diameter.
Chapter 7 - Rev. A, Page 12/45
1/2
Eq. (7-15):
max
2
32 K f M a 2
16 K fsTm
3
3
3
d
d
2
32(1.9)(2178) 2
16(2.6)(2500)
max
3
3
3
(1.57)
(1.57)
57 / 18.4 3.1 Ans.
n y S y / max
1/ 2
18389 psi 18.4 kpsi
(e) Now estimate other diameters to provide typical shoulder supports for the gear and
bearings (p. 372). Also, estimate the gear and bearing widths.
(f) Entering this shaft geometry into beam analysis software (or Finite Element software),
the following deflections are determined:
Left bearing slope:
0.000 532 rad
Right bearing slope:
0.000 850 rad
Gear slope:
0.000 545 rad
Right end of shaft slope:
0.000 850 rad
Gear deflection:
0.001 45 in
Right end of shaft deflection:
0.005 10 in
Comparing these deflections to the recommendations in Table 7-2, everything is within
typical range except the gear slope is a little high for an uncrowned gear.
(g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad.
Since all other deflections are acceptable, we will target an increase in diameter only for
the long section between the left bearing and the gear. Increasing this diameter from the
proposed 1.56 in to 1.75 in, produces a gear slope of 0.000 401 rad. All other
deflections are improved as well.
______________________________________________________________________________
Chapter 7 - Rev. A, Page 13/45
7-18
(a) Use the distortion-energy elliptic failure locus. The torque and moment loadings on
the shaft are shown in the solution to Prob. 7-17.
Candidate critical locations for strength:
Left seat keyway
Right bearing shoulder
Right keyway
Table A-20 for 1030 HR: Sut 68 kpsi, S y 37.5 kpsi, H B 137
Eq. (6-8):
Se 0.5(68) 34.0 kpsi
Eq. (6-19):
ka 2.70(68) 0.265 0.883
kc k d k e 1
Left keyway
See Table 7-1 for keyway stress concentration factors,
Kt 2.14
Profile keyway
Kts 3.0
For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities.
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
q 0.51
qs 0.57
K fs 1 qs ( K ts 1) 1 0.57(3.0 1) 2.1
K f 1 0.51(2.14 1) 1.6
0.107
Eq. (6-20):
Eq. (6-18):
1.875
kb
0.822
0.30
Se 0.883(0.822)(34.0) 24.7 kpsi
1
Eq. (7-11):
2 2
1.6(2178) 2
2.1(2500)
1
16
3
4
n f (1.8753 ) 24 700
37 500
n f = 3.5
Ans.
Right bearing shoulder
The text does not give minimum and maximum shoulder diameters for 03-series bearings
(roller). Use D = 1.75 in.
r 0.030
D 1.75
0.019,
1.11
d 1.574
d 1.574
Fig. A-15-9:
Fig. A-15-8:
Kt 2.4
K ts 1.6
Chapter 7 - Rev. A, Page 14/45
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
q 0.65
qs 0.70
K f 1 0.65(2.4 1) 1.91
K fs 1 0.70(1.6 1) 1.42
0.453
M 2178
493 lbf in
2
1/ 2
Eq. (7-11):
2
1.91(493) 2
1.42(2500)
1
16
4
3
37 500
n f (1.5743 ) 24 700
n f = 4.2
Ans.
Right keyway
Use the same stress concentration factors as for the left keyway. There is no bending
moment, thus Eq. (7-11) reduces to:
1 16 3K fsTm 16 3(2.1)(2500)
nf
d 3S y
1.53 (37 500)
n f = 2.7
Ans.
Yielding
Check for yielding at the left keyway, where the completely reversed bending is
maximum, and the steady torque is present. Using Eq. (7-15), with M m = T a = 0,
1/2
max
2
32 K f M a 2
16 K fsTm
3
3
3
d
d
1/2
2
2
16 2.1 2500
32
1.6
2178
3
1.875 3
1.875 3
8791 psi 8.79 kpsi
S
37.5
ny y
4.3
Ans.
max
8.79
Check in smaller diameter at right end of shaft where only steady torsion exists.
1/2
max
16 K fsTm 2
3
3
d
1/2
16 2.1 2500 2
3
3
1.5
13 722 psi 13.7 kpsi
Chapter 7 - Rev. A, Page 15/45
ny
Sy
37.5
2.7
max
13.7
Ans.
(b) One could take pains to model this shaft exactly, using finite element software.
However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875
in. The reductions in diameter at the bearings will change the results insignificantly. Use
E = 30 Mpsi for steel.
To the left of the load, from Table A-9, case 6, p. 1015,
AB
d y AB
Fb
1449(2)(3 x 2 2 2 112 )
(3 x 2 b 2 l 2 )
dx
6 EIl
6(30)(106 )( / 64)(1.8754 )(11)
2.4124(10 6 )(3 x 2 117)
At x = 0 in: 2.823(104 ) rad
At x = 9 in: 3.040(104 ) rad
To the right of the load, from Table A-9, case 6, p. 1015,
BC
d yBC
Fa
3 x 2 6 xl 2l 2 a 2
dx
6 EIl
At x = l = 11 in:
Fa 2
1449(9)(112 92 )
l a2
4.342(10 4 ) rad
6 EIl
6(30)(106 )( / 64)(1.8754 )(11)
Obtain allowable slopes from Table 7-2.
Left bearing:
n fs
Allowable slope
0.001
3.5
Actual slope
0.000 282 3
n fs
0.0008
1.8
0.000 434 2
Ans.
Right bearing:
Ans.
Gear mesh slope:
Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the
slope on the next shaft, we know that it will need to have a larger diameter and be stiffer.
At the moment we can say
0.0005
1.6
Ans.
0.000 304
______________________________________________________________________________
n fs
Chapter 7 - Rev. A, Page 16/45
7-19
The most likely critical locations for fatigue are at locations where the bending moment is
high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-72, indicate decreasing
moments in both planes to the left of A and to the right of C, with combined values at A
and C of M A = 5324 lbf·in and M C = 6750 lbf·in. The torque is constant between A and
B, with T = 2819 lbf·in. The most likely critical locations are at the stress concentrations
near A and C. The two shoulders near A can be eliminated since the shoulders near C
have the same geometry but a higher bending moment. We will consider the following
potentially critical locations:
keyway at A
shoulder to the left of C
shoulder to the right of C
Table A-20:
Eq. (6-8):
S ut = 64 kpsi, S y = 54 kpsi
Se 0.5(64) 32.0 kpsi
Eq. (6-19):
ka 2.70(64) 0.265 0.897
kc k d k e 1
Keyway at A
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in,
r = 0.02d = 0.035 in.
Table 7-1:
K t = 2.14, K ts = 3.0
Fig. 6-20:
q = 0.65
Fig. 6-21:
q s = 0.71
Eq. (6-32):
K f 1 q K t 1 1 0.65(2.14 1) 1.7
K fs 1 qs ( K ts 1) 1 0.71(3.0 1) 2.4
0.107
Eq. (6-20):
Eq. (6-18):
1.75
kb
0.828
0.30
Se 0.897(0.828)(32) 23.8 kpsi
Chapter 7 - Rev. A, Page 17/45
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eq. (7-9) with M m = T a = 0,
A 4 K f M a 4 1.7 5324 18102 lbf in 18.10 kip in
2
2
B 3 K fsTm 3 2.4 2819 11 718 lbf in 11.72 kip in
2
2
2 1/2
2
BS
e
1
1
ASut
2 1/2
8 18.10
2
11.72
23.8
1 1
1.753 23.8 18.10 64
1
8A
n d 3 Se
n = 1.3
Shoulder to the left of C
r / d = 0.0625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43
Fig. A-15-9:
Fig. A-15-8:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
K t = 2.2
K ts = 1.8
q = 0.71
q s = 0.76
K f 1 q K t 1 1 0.71(2.2 1) 1.9
K fs 1 qs ( K ts 1) 1 0.76(1.8 1) 1.6
0.107
Eq. (6-20):
Eq. (6-18):
1.75
kb
0.828
0.30
Se 0.897(0.828)(32) 23.8 kpsi
For convenience, we will use the full value of the bending moment at C, even though it
will be slightly less at the shoulder. Using Eq. (7-9) with M m = T a = 0,
A 4 K f M a 4 1.9 6750 25 650 lbf in 25.65 kip in
2
2
B 3 K fsTm 3 1.6 2819 7812 lbf in 7.812 kip in
2
2
2 1/2
1
8 A 2 BSe
1 1
n d 3 Se ASut
2 1/2
8 25.65
2 7.812 23.8
1 1
1.753 23.8 25.65 64
Chapter 7 - Rev. A, Page 18/45
n = 0.96
Shoulder to the right of C
r / d = 0.0625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35
Fig. A-15-9:
Fig. A-15-8:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
K t = 2.0
K ts = 1.7
q = 0.71
q s = 0.76
K f 1 q K t 1 1 0.71(2.0 1) 1.7
K fs 1 qs ( K ts 1) 1 0.76(1.7 1) 1.5
0.107
1.3
Eq. (6-20):
kb
0.855
0.30
Eq. (6-18):
Se 0.897(0.855)(32) 24.5 kpsi
For convenience, we will use the full value of the bending moment at C, even though it
will be slightly less at the shoulder. Using Eq. (7-9) with M m = T a = 0,
A 4 K f M a 4 1.7 6750 22 950 lbf in 22.95 kip in
2
2
B 3 K fsTm 3 1.5 2819 7324 lbf in 7.324 kip in
2
2
2 1/2
1
8 A 2 BSe
1 1
n d 3 Se ASut
2 1/2
2 7.324 24.5
1 1
1.33 24.5 22.95 64
n = 0.45
8 22.95
The critical location is at the shoulder to the right of C, where n = 0.45 and finite life is
predicted. Ans.
Though not explicitly called for in the problem statement, a static check for yielding is
especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with
M m = T a = 0,
1/2
max
2
32 K f M a 2
16 K fsTm
3
3
3
d
d
1/2
2
2
16 1.5 2819
32
1.7
6750
3
3
3
1.3
1.3
55 845 psi 55.8 kpsi
Chapter 7 - Rev. A, Page 19/45
n
Sy
54
0.97
max
55.8
This indicates localized yielding is predicted at the stress-concentration, though after
localized cold-working it may not be a problem. The finite fatigue life is still likely to be
the failure mode that will dictate whether this shaft is acceptable.
It is interesting to note the impact of stress concentration on the acceptability of the
proposed design. This problem is linked with several previous problems (see Table 1-1,
p. 24) in which the shaft was considered to have a constant diameter of 1.25 in. In each of
the previous problems, the 1.25 in diameter was more than adequate for deflection, static,
and fatigue considerations. In this problem, even though practically the entire shaft has
diameters larger than 1.25 in, the stress concentrations significantly reduce the
anticipated fatigue life.
______________________________________________________________________________
7-20
For a shaft with significantly varying diameters over its length, we will choose to use
shaft analysis software or finite element software to calculate the deflections. Entering
the geometry from the shaft as defined in Prob. 7-19, and the loading as defined in Prob.
3-72, the following deflection magnitudes are determined:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope Deflection
(rad)
(in)
0.00640 0.00000
0.00434 0.00000
0.00260 0.04839
0.01078 0.07517
Comparing these values to the recommended limits in Table 7-2, we find that they are all
out of the desired range. This is not unexpected since the stress analysis of Prob. 7-19
also indicated the shaft is undersized for infinite life. The slope at the right gear is the
most excessive, so we will attempt to increase all diameters to bring it into compliance.
Using Eq. (7-18) at the right gear,
d new nd dy / dx old
d old
slope all
1/4
(1)(0.01078)
0.0005
1/4
2.15
Multiplying all diameters by 2.15, we obtain the following deflections:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope
(rad)
0.00030
0.00020
0.00012
0.00050
Deflection
(in)
0.00000
0.00000
0.00225
0.00350
Chapter 7 - Rev. A, Page 20/45
This brings the slope at the right gear just to the limit for an uncrowned gear, and all
other slopes well below the recommended limits. For the gear deflections, the values are
below recommended limits as long as the diametral pitch is less than 20.
______________________________________________________________________________
7-21
The most likely critical locations for fatigue are at locations where the bending moment is
high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-73, indicate both
planes have a maximum bending moment at B. At this location, the combined bending
moment from both planes is M = 4097 N·m, and the torque is T = 3101 N·m. The
shoulder to the right of B will be eliminated since its diameter is only slightly smaller,
and there is no torque. Comparing the shoulder to the left of B with the keyway at B, the
primary difference between the two is the stress concentration, since they both have
essentially the same bending moment, torque, and size. We will check the stress
concentration factors for both to determine which is critical.
Table A-20:
S ut = 440 MPa, S y = 370 MPa
Keyway at A
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 50 mm,
r = 0.02d = 1 mm.
Table 7-1:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
K t = 2.14, K ts = 3.0
q = 0.66
q s = 0.72
K f 1 q K t 1 1 0.66(2.14 1) 1.8
K fs 1 qs ( K ts 1) 1 0.72(3.0 1) 2.4
Shoulder to the left of B
r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5
Fig. A-15-9:
K t = 2.2
Chapter 7 - Rev. A, Page 21/45
Fig. A-15-8: K ts = 1.8
Fig. 6-20:
q = 0.73
Fig. 6-21:
q s = 0.78
Eq. (6-32):
K f 1 q K t 1 1 0.73(2.2 1) 1.9
K fs 1 qs ( K ts 1) 1 0.78(1.8 1) 1.6
Examination of the stress concentration factors indicates the keyway will be the critical
location.
Eq. (6-8):
Se 0.5(440) 220 MPa
Eq. (6-19):
ka 4.51(440) 0.265 0.899
0.107
Eq. (6-20):
Eq. (6-18):
50
kb
0.818
7.62
kc k d k e 1
Se 0.899(0.818)(220) 162 MPa
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations. Using Eq. (7-9) with M m = T a = 0,
A 4 K f M a 4 1.8 4097 14 750 N m
2
2
B 3 K fsTm 3 2.4 3101 12 890 N m
2
2
2 1/2
1
8 A 2 BSe
1 1
n d 3 Se ASut
2 1/2
6
2
12
890
162
1
0
8 14 750
1 1
3
6
6
0.050 162 10 14 750 440 10
n = 0.25
Infinite life is not predicted.
Ans.
Though not explicitly called for in the problem statement, a static check for yielding is
especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with
M m = T a = 0,
1/2
max
2
32 K f M a 2
16 K fsTm
3
3
3
d
d
1/2
2
2
16 2.4 3101
32
1.8
4097
3
0.050 3
0.050 3
7.98 108 Pa 798 MPa
Chapter 7 - Rev. A, Page 22/45
n
Sy
370
0.46
max
798
This indicates localized yielding is predicted at the stress-concentration. Even without
the stress concentration effects, the static factor of safety turns out to be 0.93. Static
failure is predicted, rendering this proposed shaft design unacceptable.
This problem is linked with several previous problems (see Table 1-1, p. 24) in which the
shaft was considered to have a constant diameter of 50 mm. The results here are
consistent with the previous problems, in which the 50 mm diameter was found to
slightly undersized for static, and significantly undersized for fatigue. Though in the
current problem much of the shaft has larger than 50 mm diameter, the added
contribution of stress concentration limits the fatigue life.
______________________________________________________________________________
7-22
For a shaft with significantly varying diameters over its length, we will choose to use
shaft analysis software or finite element software to calculate the deflections. Entering
the geometry from the shaft as defined in Prob. 7-21, and the loading as defined in Prob.
3-73, the following deflection magnitudes are determined:
Location
Slope
(rad)
0.01445
0.01843
0.00358
0.00366
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Deflection
(mm)
0.000
0.000
3.761
3.676
Comparing these values to the recommended limits in Table 7-2, we find that they are all
well out of the desired range. This is not unexpected since the stress analysis in Prob.
7-21 also indicated the shaft is undersized for infinite life. The transverse deflection at
the left gear is the most excessive, so we will attempt to increase all diameters to bring it
into compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable
deflection of y all = 0.01 in = 0.254 mm,
d new nd yold
d old
yall
1/4
(1)(3.761)
0.254
1/4
1.96
Multiplying all diameters by 2, we obtain the following deflections:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope
(rad)
0.00090
0.00115
0.00022
0.00023
Deflection
(mm)
0.000
0.000
0.235
0.230
Chapter 7 - Rev. A, Page 23/45
This brings the deflection at the gears just within the limit for a spur gear (assuming P <
10 teeth/in), and all other deflections well below the recommended limits.
______________________________________________________________________________
7-23
(a) Label the approximate locations of the effective centers of the bearings as A and B,
the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one
gear, we can combine the radial and tangential gear forces into a single resultant force
with an accompanying torque, and handle the statics problem in a single plane. From
statics, the resultant reactions at the bearings can be found to be R A = 209.9 lbf and R B =
464.5 lbf. The bending moment and torque diagrams are shown, with the maximum
bending moment at D of M D = 209.9(6.98) = 1459 lbf·in and a torque transmitted from D
to C of T = 633 (8/2) = 2532 lbf·in. Due to the shaft rotation, the bending stress on any
stress element will be completely reversed, while the torsional stress will be steady.
Since we do not have any information about the fan, we will ignore any axial load that it
would introduce. It would not likely contribute much compared to the bending anyway.
Potentially critical locations are identified as follows:
Keyway at C, where the torque is high, the diameter is small, and the keyway creates
a stress concentration.
Chapter 7 - Rev. A, Page 24/45
Keyway at D, where the bending moment is maximum, the torque is high, and the
keyway creates a stress concentration.
Groove at E, where the diameter is smaller than at D, the bending moment is still
high, and the groove creates a stress concentration. There is no torque here, though.
Shoulder at F, where the diameter is smaller than at D or E, the bending moment is
still moderate, and the shoulder creates a stress concentration. There is no torque
here, though.
The shoulder to the left of D can be eliminated since the change in diameter is very
slight, so that the stress concentration will undoubtedly be much less than at D.
Table A-20:
Eq. (6-8):
S ut = 68 kpsi, S y = 57 kpsi
Se 0.5(68) 34.0 kpsi
Eq. (6-19):
ka 2.70(68) 0.265 0.883
Keyway at C
Since there is only steady torsion here, only a static check needs to be performed. We’ll
use the maximum shear stress theory.
Eq. (5-3):
Tr 2532 1.00 / 2
12.9 kpsi
J
1.004 / 32
ny
Sy / 2
57 / 2
2.21
12.9
Keyway at D
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in,
r = 0.02d = 0.035 in.
Table 7-1:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
K t = 2.14, K ts = 3.0
q = 0.66
q s = 0.72
K f 1 q K t 1 1 0.66(2.14 1) 1.8
K fs 1 qs ( K ts 1) 1 0.72(3.0 1) 2.4
0.107
Eq. (6-20):
Eq. (6-18):
1.75
kb
0.828
0.30
Se 0.883(0.828)(34.0) 24.9 kpsi
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eq. (7-9) with M m = T a = 0,
Chapter 7 - Rev. A, Page 25/45
A 4 K f M a 4 1.8 1459 5252 lbf in 5.252 kip in
2
2
B 3 K fsTm 3 2.4 2532 10 525 lbf in 10.53 kip in
2
2
2 1/2
2 BSe
1 1
ASut
2 1/2
8 5.252
2 10.53 24.9
1 1
1.753 24.9 5.252 68
n = 3.59
Ans.
1
8A
n d 3 Se
Groove at E
We will assume Figs. A-15-14 is applicable since the 2 in diameter to the right of the
groove is relatively narrow and will likely not allow the stress flow to fully develop. (See
Fig.7-9 for the stress flow concept.)
r / d = 0.1 / 1.55 = 0.065,
D / d = 1.75 / 1.55 = 1.13
Fig. A-15-14: K t = 2.1
Fig. 6-20:
q = 0.76
Eq. (6-32):
K f 1 q K t 1 1 0.76(2.1 1) 1.8
0.107
Eq. (6-20):
Eq. (6-18):
1.55
kb
0.839
0.30
Se 0.883(0.839)(34) 25.2 kpsi
Using Eq. (7-9) with M m = T a = T m = 0,
A 4 K f M a 4 1.8 1115 4122 lbf in 4.122 kip in
2
2
B=0
2 1/2
1
8 A 2 BSe
1 1
n d 3 Se ASut
1/2
8 4.122
1 0 2
1
1.553 25.2
n = 4.47
Ans.
Shoulder at F
Fig. A-15-9:
Fig. 6-20:
r / d = 0.125 / 1.40 = 0.089,
K t = 1.7
q = 0.78
D / d = 2.0 / 1.40 = 1.43
Chapter 7 - Rev. A, Page 26/45
Eq. (6-32):
K f 1 q K t 1 1 0.78(1.7 1) 1.5
0.107
Eq. (6-20):
Eq. (6-18):
1.40
kb
0.848
0.30
Se 0.883(0.848)(34) 25.5 kpsi
Using Eq. (7-9) with M m = T a = T m = 0,
A 4 K f M a 4 1.5 845 2535 lbf in 2.535 kip in
2
2
B=0
2 1/2
2 BSe
1 1
ASut
8 2.535
2 1/2
1 1 0
3
1.40 25.5
1
8A
n d 3 Se
n = 5.42
Ans.
(b)
The deflection will not be much affected by the details of fillet radii, grooves, and
keyways, so these can be ignored. Also, the slight diameter changes, as well as the
narrow 2.0 in diameter section, can be neglected. We will model the shaft with the
following three sections:
Section Diameter Length
(in)
(in)
1
1.00
2.90
2
1.70
7.77
3
1.40
2.20
The deflection problem can readily (though tediously) be solved with singularity
functions. For examples, see Ex. 4-7, p. 159, or the solution to Prob. 7-24. Alternatively,
shaft analysis software or finite element software may be used. Using any of the
methods, the results should be as follows:
Location
Left bearing A
Right bearing B
Fan C
Gear D
Slope
Deflection
(rad)
(in)
0.000290 0.000000
0.000400 0.000000
0.000290 0.000404
0.000146 0.000928
Chapter 7 - Rev. A, Page 27/45
Comparing these values to the recommended limits in Table 7-2, we find that they are all
within the recommended range.
______________________________________________________________________________
7-24
Shaft analysis software or finite element software can be utilized if available. Here we
will demonstrate how the problem can be simplified and solved using singularity
functions.
Deflection: First we will ignore the steps near the bearings where the bending moments
are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10
mm. The full bending stresses will not develop at the outer fibers so full stiffness will not
develop either. Thus, ignore this step and let the diameter be 45 mm.
Statics: Left support: R1 7(315 140) / 315 3.889 kN
Right support: R2 7(140) / 315 3.111 kN
Determine the bending moment at each step.
100
140
210
275
315
x(mm)
0
40
M(N · m) 0 155.56 388.89 544.44 326.67 124.44 0
I 35 = (/64)(0.0354) = 7.366(10-8) m4, I 40 = 1.257(10-7) m4, I 45 = 2.013(10-7) m4
Plot M/I as a function of x.
x(m)
0
0.04
0.04
0.1
0.1
0.14
0.14
0.21
0.21
0.275
0.275
0.315
M/I (109 N/m3)
0
2.112
1.2375
3.094
1.932
2.705
2.705
1.623
2.6
0.99
1.6894
0
Step
Slope
52.8
Slope
–0.8745
30.942
–21.86
–1.162
19.325
–11.617
0
–15.457
–34.78
0.977
-24.769
-9.312
0.6994
-42.235
-17.47
Chapter 7 - Rev. A, Page 28/45
The steps and the change of slopes are evaluated in the table. From these, the function
M/I can be generated:
0
1
0
M / I 52.8 x 0.8745 x 0.04 21.86 x 0.04 1.162 x 0.1
1
1
0
11.617 x 0.1 34.78 x 0.14 0.977 x 0.21
1
9.312 x 0.21 0.6994 x 0.275
0
1
17.47 x 0.275 109
Integrate twice:
E
dy
1
2
1
26.4 x 2 0.8745 x 0.04 10.93 x 0.04 1.162 x 0.1
dx
1
2
2
5.81 x 0.1 17.39 x 0.14 0.977 x 0.21
2
1
4.655 x 0.21 0.6994 x 0.275 8.735 x 0.275
2
C1 109 (1)
2
3
Ey 8.8 x 3 0.4373 x 0.04 3.643 x 0.04 0.581 x 0.1
3
3
2
1.937 x 0.1 5.797 x 0.14 0.4885 x 0.21
3
1.552 x 0.21 0.3497 x 0.275
2
2.912 x 0.275
3
2
C1x C2 109
Boundary conditions: y = 0 at x = 0 yields C 2 = 0;
y = 0 at x = 0.315 m yields C 1 = –0.295 25 N/m2.
Equation (1) with C 1 = –0.295 25 provides the slopes at the bearings and gear. The
following table gives the results in the second column. The third column gives the results
from a similar finite element model. The fourth column gives the results of a full model
which models the 35 and 55 mm diameter steps.
x (mm)
0
140
315
(rad)
–0.001 4260
–0.000 1466
0.001 3120
F.E. Model
–0.001 4270
–0.000 1467
0.001 3280
Full F.E. Model
–0.001 4160
–0.000 1646
0.001 3150
Chapter 7 - Rev. A, Page 29/45
The main discrepancy between the results is at the gear location (x = 140 mm). The larger
value in the full model is caused by the stiffer 55 mm diameter step. As was stated
earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere
between the full model and the simplified model which in any event is a small value. As
expected, modeling the 30 mm dia. as 35 mm does not affect the results much.
It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load
has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the
maximum load should be F max = (0.001/0.001 426)7 = 4.91 kN. With a design factor this
would be reduced further.
To increase the stiffness of the shaft, apply Eq. (7-18) to the most offending deflection (at
x = 0) to determine a multiplier to be used for all diameters.
d new nd dy / dx old
d old
slope all
1/ 4
(1)(0.0014260)
0.001
1/4
1.093
Form a table:
20.00 30.00 35.00 40.00 45.00 55.00
Old d, mm
New ideal d, mm
21.86 32.79 38.26 43.72 49.19 60.12
Rounded up d, mm 22.00 34.00 40.00 44.00 50.00 62.00
Repeating the full finite element model results in
= –9.30 10-4 rad
x = 0:
x = 140 mm: = –1.09 10-4 rad
x = 315 mm: = 8.65 10-4 rad
This is well within our goal. Have the students try a goal of 0.0005 rad at the gears.
Strength: Due to stress concentrations and reduced shaft diameters, there are a number of
locations to look at. A table of nominal stresses is given below. Note that torsion is only
to the right of the 7 kN load. Using = 32M/(d3) and = 16T/(d3),
x (mm)
(MPa)
(MPa)
(MPa)
0 15
40
100
110
140
210
275
300
330
0 22.0 37.0 61.9 47.8 60.9 52.0 39.6 17.6
0
0 0
0
0
0
6
8.5 12.7 20.2 68.1
0 22.0 37.0 61.9 47.8 61.8 53.1 45.3 39.2 118.0
Table A-20 for AISI 1020 CD steel: S ut = 470 MPa, S y = 390 MPa
At x = 210 mm:
Eq. (6-19):
k a 4.51(470) 0.265 0.883
Chapter 7 - Rev. A, Page 30/45
Eq. (6-20):
Eq. (6-18):
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
kb (40 / 7.62) 0.107 0.837
S e = 0.883 (0.837)(0.5)(470) = 174 MPa
D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05
K ts = 1.4
K t = 1.9
q = 0.75
q s = 0.79
K f = 1 + 0.75(1.9 –1) = 1.68
K f s = 1 + 0.79(1.4 – 1) = 1.32
Choosing DE-ASME Elliptic to inherently include the yield check, from Eq. (7-11), with
M m = T a = 0,
1/2
2
2
1.32(107)
1
16
1.68(326.67)
3
4
6
n 0.043 174 106
390
10
n 1.98
At x = 330 mm:
The von Mises stress is the highest but it comes from the steady torque only.
D / d = 30 / 20 = 1.5, r / d = 2 / 20 = 0.1
Fig. A-15-9: K ts = 1.42
Fig. 6-21:
q s = 0.79
Eq. (6-32):
K f s = 1 + 0.79(1.42 – 1) = 1.33
Eq. (7-11):
1.33(107)
1
16
3
n 3
390 106
n = 2.49
Note that since there is only a steady torque, Eq. (7-11) reduces to essentially the
equivalent of the distortion energy failure theory.
Check the other locations.
If worse-case is at x = 210 mm, the changes discussed for the slope criterion will
improve the strength issue.
______________________________________________________________________________
7-25 and 7-26 With these design tasks each student will travel different paths and almost all
details will differ. The important points are
The student gets a blank piece of paper, a statement of function, and some constraints
– explicit and implied. At this point in the course, this is a good experience.
It is a good preparation for the capstone design course.
Chapter 7 - Rev. A, Page 31/45
The adequacy of their design must be demonstrated and possibly include a designer’s
notebook.
Many of the fundaments of the course, based on this text and this course, are useful.
The student will find them useful and notice that he/she is doing it.
Don’t let the students create a time sink for themselves. Tell them how far you want
them to go.
______________________________________________________________________________
7-27
This task was once given as a final exam problem. This problem is a learning experience.
Following the task statement, the following guidance was added.
Take the first half hour, resisting the temptation of putting pencil to paper, and decide
what the problem really is.
Take another twenty minutes to list several possible remedies.
Pick one, and show your instructor how you would implement it.
The students’ initial reaction is that he/she does not know much from the problem
statement. Then, slowly the realization sets in that they do know some important things
that the designer did not. They knew how it failed, where it failed, and that the design
wasn’t good enough; it was close, though.
Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and
the problem may not be solved.
To many students’ credit, they chose to keep the shaft geometry, and selected a new
material to realize about twice the Brinell hardness.
______________________________________________________________________________
7-28
In Eq. (7-22) set
d4
d2
I
, A
4
64
to obtain
2
d gE
l 4
(1)
or
d
4l 2
gE
2
(2)
(a) From Eq. (1) and Table A-5
2
9
0.025 9.81(207)(10 )
883 rad/s
76.5 103
0.6 4
Ans.
Chapter 7 - Rev. A, Page 32/45
(b) From Eq. (1), we observe that the critical speed is linearly proportional to the
diameter. Thus, to double the critical speed, we should double the diameter to d = 50
mm.
Ans.
(c) From Eq. (2),
l
2 d
gE
4 l
Since d / l is the same regardless of the scale,
l constant 0.6(883) 529.8
529.8
1766 rad/s Ans.
0.3
Thus the first critical speed doubles.
______________________________________________________________________________
7-29
From Prob. 7-28, 883 rad/s
A 4.909 10 4 m 2 ,
E 207(109 ) Pa,
I 1.917 10 8 m 4 ,
7.65 104 N/m 3
w A l 4.909 10 4 7.65 10 4 (0.6) 22.53 N
One element:
Eq. (7-24):
0.3(0.3) 0.62 0.32 0.32
(0.6)
y w 22.53(1.134) 10 2.555 10 m
y 6.528 10
w y 22.53(2.555) 10 5.756 10
w y 22.53(6.528) 10 1.47110
11
9
6(207) 10 (1.917) 10
8
1.134 106 m/N
6
1
5
1 11
10
2
1
5
2
4
10
8
5.756 104
w y
9.81
620 rad/s
1 g
w y 2
1.471 108
(30% low)
Two elements:
Chapter 7 - Rev. A, Page 33/45
11 22
12 21
6.379 10 m/N
(0.6)
0.45(0.15) 0.62 0.452 0.152
9
6(207) 10 (1.917) 10
8
7
0.15(0.15)(0.62 0.152 0.152 )
4.961 107 m/N
8
9
6(207) 10 (1.917) 10 (0.6)
y1 y2 w111 w212 11.265(6.379) 10 7 11.265(4.961) 10 7 1.277 105 m
y y 1.632 10
2
1
2
2
10
2(11.265)(1.632) 10 3.677 10
w y 2(11.265)(1.277) 10 5 2.877 10 4
10
w y 2
2.877 104
1 9.81
3.677 109
9
876 rad/s
(0.8% low)
Three elements:
11 33
0.5(0.1) 0.62 0.52 0.12
9
6(207) 10 (1.917) 10
8
0.3(0.3) 0.62 0.32 0.32
22
6(207) 10 (1.917) 10
9
12 32
8
(0.6)
(0.6)
6(207) 10 (1.917) 10
0.1(0.1) 0.62 0.12 0.12
8
3.500 107 m/N
1.134 106 m/N
0.3(0.1) 0.62 0.32 0.12
9
(0.6)
5.460 107 m/N
(0.6)
y 7.51 3.500 10 5.460 10 2.380 10 8.516 10
y 7.51 5.460 10 1.134 10 5.460 10 1.672 10
y 7.51 2.380 10 5.460 10 3.500 10 8.516 10
w y 7.51 8.516 10 1.672 10 8.516 10 2.535 10
w y 7.518.516 10 1.672 10 8.516 10 3.189 10
13
9
6(207) 10 (1.917) 10
2.380 107 m/N
8
7
7
7
6
7
6
7
5
7
7
7
6
1
2
3
6
2
5
6
2
6
5
2
4
6
2
9
Chapter 7 - Rev. A, Page 34/45
2.535 104
1 9.81
3.189 109
883 rad/s
The result is the same as in Prob. 7-28. The point was to show that convergence is rapid
using a static deflection beam equation. The method works because:
If a deflection curve is chosen which meets the boundary conditions of momentfree and deflection-free ends, as in this problem, the strain energy is not very
sensitive to the equation used.
Since the static bending equation is available, and meets the moment-free and
deflection-free ends, it works.
______________________________________________________________________________
7-30
(a) For two bodies, Eq. (7-26) is
(m111 1/ 2 )
m212
m1 21
( m2 22 1/ 2 )
0
Expanding the determinant yields,
2
1
1
2 (m111 m2 22 ) 2 m1m2 (11 22 12 21 ) 0
1
(1)
Eq. (1) has two roots 1 / 12 and 1 / 22 . Thus
1
1 1
1
2 2 2 2 0
1 2
or,
2
2
1 1 1 1
1 1
2 2 2 2 2 0
1 2 1 2
(2)
Equate the third terms of Eqs. (1) and (2), which must be identical.
1 1
2
1
2
2
m1m2 (11 22 12 21 )
1
22
12 m1m2 (11 22 12 21 )
and it follows that
Chapter 7 - Rev. A, Page 35/45
2
g2
1
1 w1w2 (11 22 12 21 )
Ans.
(b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w 1 = 35 lbf,
w 2 = 55 lbf) is 1 = 124.8 rad/s. From part (a), using influence coefficients,
1
3862
466 rad/s Ans.
124.8 35(55) 2.061(3.534) 2.2342 108
______________________________________________________________________________
2
7-31
In Eq. (7-22), for 1 , the term I / A appears. For a hollow uniform diameter shaft,
1
2
2
2
2
do4 di4 / 64
I
1 d o di d o di 1
d o2 di2
2
2
2
2
A
16
d
d
4
do di / 4
o
i
This means that when a solid shaft is hollowed out, the critical speed increases beyond
that of the solid shaft of the same size. By how much?
(1/ 4) do2 di2
(1/ 4) d o2
d
1 i
do
2
The possible values of di are 0 di d o , so the range of the critical speeds is
1 1 0 to about 1 1 1
or from 1 to 2 1. Ans.
______________________________________________________________________________
7-32
All steps will be modeled using singularity functions with a spreadsheet. Programming
both loads will enable the user to first set the left load to 1, the right load to 0 and
calculate 11 and 21 . Then set the left load to 0 and the right to 1 to get 12 and 22 . The
spreadsheet shows the 11 and 21 calculation. A table for M / I vs. x is easy to make.
First, draw the bending-moment diagram as shown with the data.
x
M
x
M
0
0
1
2
3
0.875 1.75 1.625
9
10
11
0.875 0.75 0.625
12
0.5
4
1.5
5
6
7
1.375 1.25 1.125
13
14
15
0.375 0.25 0.125
8
1
16
0
Chapter 7 - Rev. A, Page 36/45
The second-area moments are:
0 x 1 in and 15 x 16 in, I1 2 4 / 64 0.7854 in 4
1 x 9 in , I 2 2.4724 / 64 1.833 in 4
9 x 15 in , I 3 2.7634 / 64 2.861 in 4
Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot
x
M/I
x
M/I
0
0
1
1
2
2
3
4
5
6
7
8
1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456
9
9
10
11
12
13
14
14
15
15
0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592
16
0
From this diagram, one can see where changes in value (steps) and slope occur. Using a
spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in,
M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step
change of 0.4774 1.1141 = 0.6367 lbf/in3. A slope change also occurs at at x = 1 in.
Chapter 7 - Rev. A, Page 37/45
The slope for 0 x 1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547
0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774 1.1141 = 0.6367 lbf/in2. Following
this approach, a table is made of all the changes. The table shown indicates the column
letters and row numbers for the spreadsheet.
A
x
1a
1b
2
2
9a
9b
14
14
15a
15b
16
1
2
3
4
5
6
7
8
9
10
11
12
B
M
0.875
0.875
1.75
1.75
0.875
0.875
0.25
0.25
0.125
0.125
0
C
M/I
1.114085
0.477358
0.954716
0.954716
0.477358
0.305854
0.087387
0.087387
0.043693
0.159155
0.000000
D
step
0.000000
-0.636727
0.000000
0.000000
0.000000
-0.171504
0.000000
0.000000
0.000000
0.115461
0.000000
E
Slope
1.114085
0.477358
0.477358
-0.068194
-0.068194
-0.043693
-0.043693
-0.043693
-0.043693
-0.159155
-0.159155
F
Slope
0.000000
-0.636727
0.000000
-0.545552
0.000000
0.024501
0.000000
0.000000
0.000000
-0.115461
0.000000
The equation for M / I in terms of the spreadsheet cell locations is:
0
1
M / I E2 ( x) D3 x 1 F3 x 1 F5 x 2
0
1
1
0
D7 x 9 F7 x 9 D11 x 15 F11 x 15
1
Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary
conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C 1 = 4.906
lbf/in and C 2 = 0). Substitution back into the deflection equation at x = 2 and 14 in
provides the ’s. The results are: 11 = 2.917(10–7) and 12 = 1.627(10–7). Repeat for
F 1 = 0 and F 2 = 1, resulting in 21 = 1.627(10–7) and 22 = 2.231(10–7). This can be
verified by finite element analysis.
y1
y2
y12
wy
18(2.917)(107 ) 32(1.627)(107 ) 1.046(105 )
18(1.627)(107 ) 32(2.231)(107 ) 1.007(105 )
1.093(1010 ), y22 1.014(1010 )
5.105(104 ), w y 2 5.212(109 )
Neglecting the shaft, Eq. (7-23) gives
1 386
5.105(104 )
6149 rad/s or 58 720 rev/min
5.212(109 )
Ans.
Chapter 7 - Rev. A, Page 38/45
Without the loads, we will model the shaft using 2 elements, one between 0 x 9 in,
and one between 0 x 16 in. As an approximation, we will place their weights at
x = 9/2 = 4.5 in, and x = 9 + (16 9)/2 = 12.5 in. From Table A-5, the weight density of
steel is = 0.282 lbf/in3. The weight of the left element is
w1 d 2l 0.282 22 1 2.4722 8 11.7 lbf
4
4
The right element is
w2 0.282 2.7632 6 22 1 11.0 lbf
4
The spreadsheet can be easily modified to give
11 9.605 10 7 , 12 21 5.718 10 7 , 22 5.472 10 7
y1 1.753 10 5 , y2 1.27110 5
y12 3.072 10 10 , y22 1.615 10 10
w y 3.449 10 , w y
4
2
5.371109
3.449 104
4980 rad/s
1 386
9
5.37110
A finite element model of the exact shaft gives 1 = 5340 rad/s. The simple model is
6.8% low.
Combination: Using Dunkerley’s equation, Eq. (7-32):
1
1
1
1 3870 rad/s Ans.
2
6149 49802
______________________________________________________________________________
2
1
7-33
We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows K ts as a decreasing monotonic as a function of
a/D. All is not what it seems. Let us change the basis for data presentation to the full
section rather than the net section.
K ts 0 K ts 0
Chapter 7 - Rev. A, Page 39/45
K ts
32T
32T
K ts
3
3
AD
D
Therefore
K ts
K ts
A
Form a table:
K ts has the following attributes:
It exhibits a minimum;
It changes little over a wide range;
Its minimum is a stationary point minimum at a / D 0.100;
Our knowledge of the minima location is
0.075 (a / D ) 0.125
We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft
diameter, for greatest shaft capacity. However, it is not catastrophic if one forgets the
rule.
______________________________________________________________________________
7-34
From the solution to Prob. 3-72, the torque to be transmitted through the key from the
gear to the shaft is T = 2819 lbf·in. From Prob. 7-19, the nominal shaft diameter
supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a
1.00 in shaft diameter. The force applied to the key is
F
T
2819
5638 lbf
r 1.00 / 2
Selecting 1020 CD steel for the key, with S y = 57 kpsi, and using the distortion-energy
theory, S sy = 0.577 S y = (0.577)(57) = 32.9 kpsi.
Failure by shear across the key:
Chapter 7 - Rev. A, Page 40/45
F F
A tl
S
S
n sy sy
F / tl
l
1.1 5638
nF
0.754 in
tS sy 0.25 32 900
Failure by crushing:
Sy
Sy
F
F
n
A t / 2 l
2 F / tl
l
2 Fn 2 5638 1.1
0.870 in
tS y 0.25 57 103
Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans.
______________________________________________________________________________
7-35
From the solution to Prob. 3-73, the torque to be transmitted through the key from the
gear to the shaft is T = 3101 N·m. From Prob. 7-21, the nominal shaft diameter
supporting the gear is 50 mm. To determine an appropriate key size for the shaft
diameter, we can either convert to inches and use Table 7-6, or we can look up standard
metric key sizes from the internet or a machine design handbook. It turns out that the
recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement
specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6
as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent
to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to
standard sizes.
The force applied to the key is
F
T
3101
124 103 N
r 0.050 / 2
Selecting 1020 CD steel for the key, with S y = 390 MPa, and using the distortion-energy
theory, S sy = 0.577 S y = 0.577(390) = 225 MPa.
Failure by shear across the key:
F F
A tl
n
S sy
S sy
F / tl
1.1124 103
nF
l
0.0433 m 43.3 mm
tS sy 0.014 225 106
Failure by crushing:
Chapter 7 - Rev. A, Page 41/45
n
Sy
F
F
A t / 2 l
Sy
2 F / tl
3
2 Fn 2 124 10 1.1
0.0500 m 50.0 mm
l
tS y 0.014 390 106
Select 14 mm square key, 50 mm long, 1020 CD steel. Ans.
______________________________________________________________________________
7-36
Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is
designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and
d = 0.011 mm. From Table A-12, the fundamental deviation is F = 0 mm.
Hole:
Eq. (7-36):
D max = D + D = 15 + 0.018 = 15.018 mm
D min = D = 15.000 mm
Ans.
Ans.
Shaft:
Eq. (7-37):
Ans.
d max = d + F = 15.000 + 0 = 15.000 mm
Ans.
d min = d + F – d = 15.000 + 0 – 0.011 = 14.989 mm
______________________________________________________________________________
7-37 Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as
H7/s6. From Table A-13, the tolerance grades are D = 0.0010 in and d = 0.0006 in.
From Table A-14, the fundamental deviation is F = 0.0017 in.
Hole:
Eq. (7-36):
D max = D + D = 1.75 + 0.0010 = 1.7510 in
D min = D = 1.7500 in
Ans.
Ans.
Shaft:
Eq. (7-38):
Ans.
d min = d + F = 1.75 + 0.0017 = 1.7517 in
d max = d + F + d = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans.
______________________________________________________________________________
7-38
Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6.
From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. From
Table A-12, the fundamental deviation is F = –0.009 mm.
Hole:
Eq. (7-36):
D max = D + D = 45 + 0.025 = 45.025 mm
D min = D = 45.000 mm
Ans.
Ans.
Shaft:
Eq. (7-37):
Ans.
d max = d + F = 45.000 + (–0.009) = 44.991 mm
Ans.
d min = d + F – d = 45.000 + (–0.009) – 0.016 = 44.975 mm
______________________________________________________________________________
Chapter 7 - Rev. A, Page 42/45
7-39
Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as
H8/f7. From Table A-13, the tolerance grades are D = 0.0015 in and d = 0.0010 in.
From Table A-14, the fundamental deviation is F = –0.0010 in.
Hole:
Eq. (7-36):
D max = D + D = 1.250 + 0.0015 = 1.2515 in
D min = D = 1.2500 in
Ans.
Ans.
Shaft:
Eq. (7-37):
Ans.
d max = d + F = 1.250 + (–0.0010) = 1.2490 in
Ans.
d min = d + F – d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in
______________________________________________________________________________
7-40
Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is
designated as H7/p6. From Table A-11, the tolerance grades are D = 0.025 mm and
d = 0.016 mm. From Table A-12, the fundamental deviation is F = 0.026 mm.
Hole:
Eq. (7-36):
D max = D + D = 35 + 0.025 = 35.025 mm
D min = D = 35.000 mm
The bearing bore specifications are within the hole specifications for a locational
interference fit. Now find the necessary shaft sizes.
Shaft:
Eq. (7-38):
Ans.
d min = d + F = 35 + 0.026 = 35.026 mm
Ans.
d max = d + F + d = 35 + 0.026 + 0.016 = 35.042 mm
______________________________________________________________________________
7-41
Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is
designated as H7/p6. From Table A-13, the tolerance grades are D = 0.0010 in and
d = 0.0006 in. From Table A-14, the fundamental deviation is F = 0.0010 in.
Hole:
Eq. (7-36):
D max = D + D = 1.5000 + 0.0010 = 1.5010 in
D min = D = 1.5000 in
The bearing bore specifications exactly match the requirements for a locational
interference fit. Now check the shaft.
Shaft:
Eq. (7-38):
d min = d + F = 1.5000 + 0.0010 = 1.5010 in
d max = d + F + d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in
Chapter 7 - Rev. A, Page 43/45
The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of
1.5016 in, and therefore does not meet the specifications for the locational interference
fit. Ans.
______________________________________________________________________________
7-42
(a) Basic size is D = d = 35 mm.
Table 7-9:
H7/s6 is specified for medium drive fit.
Table A-11: Tolerance grades are D = 0.025 mm and d = 0.016 mm.
Table A-12: Fundamental deviation is F 0.043 mm.
Eq. (7-36):
D max = D + D = 35 + 0.025 = 35.025 mm
D min = D = 35.000 mm
Ans.
Eq. (7-38):
d min = d + F = 35 + 0.043 = 35.043 mm
Ans.
d max = d + F + d = 35 + 0.043 + 0.016 = 35.059 mm
(b)
Eq. (7-42):
Eq. (7-43):
Eq. (7-40):
min d min Dmax 35.043 35.025 0.018 mm
max d max Dmin 35.059 35.000 0.059 mm
pmax
2
2
2
2
E max d o d d d i
d o2 di2
2d 3
207 109 0.059 602 352 352 0
602 0
2 353
pmin
Ans.
2
2
2
2
E min d o d d di
2d 3
d o2 di2
207 109 0.018 602 352 352 0
602 0
2 353
115 MPa
35.1 MPa
Ans.
(c) For the shaft:
Eq. (7-44):
t ,shaft p 115 MPa
Eq. (7-46):
r ,shaft p 115 MPa
Eq. (5-13):
12 1 2 22
1/ 2
( 115) 2 ( 115)( 115) ( 115) 2
n S y / 390 / 115 3.4
1/2
115 MPa
Ans.
For the hub:
602 352
d o2 d 2
234 MPa
115
2
2
do2 d 2
60 35
Eq. (7-45):
t ,hub p
Eq. (7-46):
r ,hub p 115 MPa
Chapter 7 - Rev. A, Page 44/45
Eq. (5-13):
12 1 2 22
1/ 2
(234) 2 (234)( 115) ( 115) 2
n S y / 600 / 308 1.9
1/ 2
308 MPa
Ans.
(d) A value for the static coefficient of friction for steel to steel can be obtained online or
from a physics textbook as approximately f = 0.8.
Eq. (7-49)
T ( / 2) f pmin ld 2
( / 2)(0.8)(35.1) 106 (0.050)(0.035) 2 2700 N m
Ans.
______________________________________________________________________________
Chapter 7 - Rev. A, Page 45/45
Chapter 8
Note to the Instructor for Probs. 8-41 to 8-44. These problems, as well as many others in this
chapter are best implemented using a spreadsheet.
8-1
(a) Thread depth= 2.5 mm Ans.
Width = 2.5 mm Ans.
d m = 25 - 1.25 - 1.25 = 22.5 mm
d r = 25 - 5 = 20 mm
l = p = 5 mm Ans.
(b) Thread depth = 2.5 mm Ans.
Width at pitch line = 2.5 mm Ans.
d m = 22.5 mm
d r = 20 mm
l = p = 5 mm Ans.
______________________________________________________________________________
8-2
From Table 8-1,
d r d 1.226 869 p
d m d 0.649 519 p
d 1.226 869 p d 0.649 519 p
d 0.938 194 p
d
2
At
d 2
(d 0.938 194 p)2 Ans.
4
4
______________________________________________________________________________
8-3
From Eq. (c) of Sec. 8-2,
tan f
1 f tan
Pd
Fd m tan f
TR R m
2
2 1 f tan
T
Fl / (2 ) 1 f tan
1 f tan
e 0
tan
TR
Fd m / 2 tan f
tan f
PR F
Ans.
Chap. 8 Solutions - Rev. A, Page 1/69
Using f = 0.08, form a table and plot the efficiency curve.
e
, deg.
0
0
0
0.678
20
0.796
30
0.838
40
0.8517
45
0.8519
______________________________________________________________________________
8-4
Given F = 5 kN, l = 5 mm, and d m = d p/2 = 25 5/2 = 22.5 mm, the torque required to
raise the load is found using Eqs. (8-1) and (8-6)
TR
5 22.5 5 0.09 22.5 5 0.06 45
15.85 N m
2
2
22.5 0.09 5
Ans.
The torque required to lower the load, from Eqs. (8-2) and (8-6) is
TL
5 22.5 0.09 22.5 5 5 0.06 45
7.83 N m
2
2
22.5 0.09 5
Ans.
Since T L is positive, the thread is self-locking. From Eq.(8-4) the efficiency is
5 5
Ans.
0.251
2 15.85
______________________________________________________________________________
e
Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom
segment of the screws must be in compression. Whereas, tension specimens and their
grips must be in tension. Both screws must be of the same-hand threads.
______________________________________________________________________________
8-5
8-6
Screws rotate at an angular rate of
n
1720
28.67 rev/min
60
Chap. 8 Solutions - Rev. A, Page 2/69
(a) The lead is 0.25 in, so the linear speed of the press head is
V = 28.67(0.25) = 7.17 in/min Ans.
(b) F = 2500 lbf/screw
d m 2 0.25 / 2 1.875 in
sec 1 / cos(29o / 2) 1.033
Eq. (8-5):
TR
2500(1.875) 0.25 (0.05)(1.875)(1.033)
221.0 lbf · in
2
(1.875) 0.05(0.25)(1.033)
Eq. (8-6):
Tc 2500(0.08)(3.5 / 2) 350 lbf · in
Ttotal 350 221.0 571 lbf · in/screw
571(2)
Tmotor
20.04 lbf · in
60(0.95)
Tn
20.04(1720)
H
0.547 hp Ans.
63 025
63 025
______________________________________________________________________________
8-7
Note to the Instructor: The statement for this problem in the first printing of this edition
was vague regarding the effective handle length. For the printings to follow the statement
“The overall length is 4.25 in.” will be replaced by “ A force will be applied to the handle
at a radius of 3 12 in from the screw centerline.” We apologize if this has caused any
inconvenience.
L 3.5 in
T 3.5F
3
3
M L F 3.5 F 3.125F
8
8
S y 41 kpsi
32M
32(3.125) F
Sy
41 000
3
d
(0.1875)3
F 8.49 lbf
T 3.5(8.49) 29.7 lbf · in Ans.
(b) Eq. (8-5), 2 = 60 , l = 1/10 = 0.1 in, f = 0.15, sec = 1.155, p = 0.1 in
Chap. 8 Solutions - Rev. A, Page 3/69
3
0.649 519 0.1 0.6850 in
4
F (0.6850) 0.1 (0.15)(0.6850)(1.155)
TR clamp
2
(0.6850) 0.15(0.1)(1.155)
TR 0.075 86Fclamp
TR
29.7
Fclamp
392 lbf Ans.
0.075 86 0.075 86
dm
(c) The column has one end fixed and the other end pivoted. Base the decision on the
mean diameter column. Input: C = 1.2, D = 0.685 in, A = (0.6852)/4 = 0.369 in2, S y = 41
kpsi, E = 30(106) psi, L = 6 in, k = D/4 =0.171 25 in, L/k = 35.04. From Eq. (4-45),
2 2 1.2 30 106
131.7
41 000
From Eq. (4-46), the limiting clamping force for buckling is
1/2
1/2
2
l 2 CE
k 1 S y
2
Sy l 1
Fclamp Pc r A S y
2 k CE
2
41103
1
3
0.369 4110
35.04
14.6 103 lbf
6
2
1.2 30 10
Ans
(d) This is a subject for class discussion.
______________________________________________________________________________
8-8
T = 8(3.5) = 28 lbf in
dm
3 1
0.6667 in
4 12
l =
1
= 0.1667 in,
6
=
290
= 14.50,
2
sec 14.50 = 1.033
From Eqs. (8-5) and (8-6)
Ttotal
0.6667 F
2
0.1667 0.15 0.6667 1.033 0.15 1 F
0.1542 F
2
0.6667 0.15 0.1667 1.033
28
182 lbf
Ans.
0.1542
_____________________________________________________________________________
F
Chap. 8 Solutions - Rev. A, Page 4/69
8-9
d m = 1.5 0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in
From Eq. (8-1) and Eq. (8-6)
2.2 103 (1.375) 0.5 (0.10)(1.375) 2.2 103 (0.15)(2.25)
TR
(1.375) 0.10(0.5)
2
2
330 371 701 lbf · in
Since n = V/l = 2/0.5 = 4 rev/s = 240 rev/min
so the power is
701 240
Tn
2.67 hp
Ans.
63 025
63 025
______________________________________________________________________________
H
8-10 d m = 40 4 = 36 mm, l = p = 8 mm
From Eqs. (8-1) and (8-6)
T
H
T
F
36 F 8 (0.14)(36) 0.09(100) F
2 (36) 0.14(8)
2
(3.831 4.5) F 8.33F N · m (F in kN)
2 n 2 (1) 2 rad/s
T
3000
H
477 N · m
2
477
57.3 kN Ans.
8.33
Fl
57.3(8)
0.153 Ans.
2 T
2 (477)
______________________________________________________________________________
e
8-11 (a) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding
up,
L = 45 mm
Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(14) +6 = 34 mm
From Table 8-7, l d = L L T = 45 34 = 11 mm, l t = l l d = 2(15) 11 = 19 mm,
A d = (142) / 4 = 153.9 mm2. From Table 8-1, A t = 115 mm2. From Eq. (8-17)
Chap. 8 Solutions - Rev. A, Page 5/69
kb
153.9 115 207
Ad At E
874.6 MN/m
Ad lt At ld 153.9 19 115 11
Ans.
(c) From Eq. (8-22), with l = 2(15) = 30 mm
km
0.5774 207 14
0.5774 Ed
3 116.5 MN/m
0.5774l 0.5d
0.5774 30 0.5 14
2 ln 5
2 ln 5
0.5774l 2.5d
0.5774 30 2.5 14
Ans.
8-12 (a) Table A-31, nut height H = 12.8 mm. Table A-33, washer thickness t = 3.5 mm. Thus,
the grip is l = 2(15) + 3.5 = 33.5 mm. L ≥ l + H = 33.5 + 12.8 = 46.3 mm. Rounding up
L = 50 mm
Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(14) +6 = 34 mm
From Table 8-7, l d = L L T = 50 34 = 16 mm, l t = l l d = 33.5 16 = 17.5 mm,
A d = (142) / 4 = 153.9 mm2. From Table 8-1, A t = 115 mm2. From Eq. (8-17)
kb
153.9 115 207
Ad At E
808.2 MN/m
Ad lt At ld 153.9 17.5 115 16
Ans.
(c)
From Eq. (8-22)
km
0.5774 207 14
0.5774 Ed
2 969 MN/m
0.5774l 0.5d
0.5774 33.5 0.5 14
2 ln 5
2 ln 5
0.5774l 2.5d
0.5774 33.5 2.5 14
Ans.
______________________________________________________________________________
Chap. 8 Solutions - Rev. A, Page 6/69
8-13 (a) Table 8-7, l = h + d /2 = 15 + 14/2 = 22 mm. L ≥ h + 1.5d = 36 mm. Rounding up
L = 40 mm
Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(14) +6 = 34 mm
From Table 8-7, l d = L L T = 40 34 = 6 mm, l t = l l d = 22 6 = 16 mm
A d = (142) / 4 = 153.9 mm2. From Table 8-1, A t = 115 mm2. From Eq. (8-17)
kb
153.9 115 207
Ad At E
1 162.2 MN/m
Ad lt At ld 153.9 16 115 6
Ans.
(c) From Eq. (8-22), with l = 22 mm
km
0.5774 207 14
0.5774 Ed
3 624.4 MN/m
0.5774l 0.5d
0.5774 22 0.5 14
2 ln 5
2 ln 5
0.5774l 2.5d
0.5774 22 2.5 14
Ans.
______________________________________________________________________________
8-14 (a) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in.
Rounding up, L = 3.5 in
Ans.
(b) From Eq. (8-13), L T = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, l d = L L T = 3.5 1.25 = 2.25 in, l t = l l d = 3 2.25 = 0.75 in
A d = (0.52)/4 = 0.1963 in2. From Table 8-2, A t = 0.1419 in2. From Eq. (8-17)
kb
0.1963 0.1419 30
Ad At E
1.79 Mlbf/in
Ad lt At ld 0.1963 0.75 0.1419 2.25
Ans.
Chap. 8 Solutions - Rev. A, Page 7/69
(c)
Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
k1
0.5774 30 0.5
1.155 1.5 0.75 0.5 0.75 0.5
ln
1.155 1.5 0.75 0.5 0.75 0.5
22.65 Mlbf/in
Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30 = 1.905 in, E = 30
Mpsi. Eq. (8-20) k 2 = 210.7 Mlbf/in
Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)
k 3 = 12.27 Mlbf/in
From Eq. (8-18)
k m = (1/k 1 + 1/k 2 +1/k 3 )1 = (1/22.65 + 1/210.7 + 1/12.27)1 = 7.67 Mlbf/in
Ans.
8-15 (a) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.19
in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in.
Rounding up, L = 3.75 in
Ans.
(b) From Eq. (8-13), L T = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, l d = L L T = 3.75 1.25 = 2.5 in, l t = l l d = 3.19 2.5 = 0.69 in
A d = (0.52)/4 = 0.1963 in2. From Table 8-2, A t = 0.1419 in2. From Eq. (8-17)
Chap. 8 Solutions - Rev. A, Page 8/69
kb
0.1963 0.1419 30
Ad At E
1.705 Mlbf/in
Ad lt At ld 0.1963 0.69 0.1419 2.5
Ans.
(c)
Each steel washer frustum: t = 0.095 in, d = 0.531 in (Table A-32), D = 0.75 in, E = 30
Mpsi. From Eq. (8-20)
k1
0.5774 30 0.531
1.155 0.095 0.75 0.531 0.75 0.531
ln
1.155 0.095 0.75 0.531 0.75 0.531
89.20 Mlbf/in
Top plate, top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in,
E = 30 Mpsi. Eq. (8-20) k 2 = 28.99 Mlbf/in
Top plate, lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.860 + 2(1) tan 30 = 2.015 in,
E = 30 Mpsi. Eq. (8-20) k 3 = 234.08 Mlbf/in
Cast iron: t = 1 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30 = 0.860 in, E = 14.5 Mpsi
(Table 8-8). Eq. (8-20) k 4 = 15.99 Mlbf/in
From Eq. (8-18)
k m = (2/k 1 + 1/k 2 +1/k 3 +1/k 4 )1 = (2/89.20 + 1/28.99 + 1/234.08 + 1/15.99)1
= 8.08 Mlbf/in Ans.
______________________________________________________________________________
8-16 (a) From Table 8-7, l = h + d /2 = 2 + 0.5/2 = 2.25 in.
L ≥ h + 1.5 d = 2 + 1.5(0.5) = 2.75 in Ans.
(b) From Table 8-7, L T = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
Chap. 8 Solutions - Rev. A, Page 9/69
l d = L L T = 2.75 1.25 = 1.5 in, l t = l l d = 2.25 1.5 = 0.75 in
A d = (0.52)/4 = 0.1963 in2. From Table 8-2, A t = 0.1419 in2. From Eq. (8-17)
kb
0.1963 0.1419 30
Ad At E
2.321 Mlbf/in
Ad lt At ld 0.1963 0.75 0.1419 1.5
Ans.
(c)
Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
k1
0.5774 30 0.5
1.155 1.125 0.75 0.5 0.75 0.5
ln
1.155 1.125 0.75 0.5 0.75 0.5
24.48 Mlbf/in
Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30 = 1.039 in, E =
30 Mpsi. Eq. (8-20) k 2 = 49.36 Mlbf/in
Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20)
k 3 = 23.49 Mlbf/in
From Eq. (8-18)
k m = (1/k 1 + 1/k 2 +1/k 3 )1 = (1/24.48 + 1/49.36 + 1/23.49)1 = 9.645 Mlbf/in Ans.
______________________________________________________________________________
8-17 a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in
Ans.
Chap. 8 Solutions - Rev. A, Page 10/69
(b) From Eq. (8-13), L T = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, l d = L L T = 4.75 1.25 = 3.5 in, l t = l l d = 4.19 3.5 = 0.69 in
A d = (0.52)/4 = 0.1963 in2. From Table 8-2, A t = 0.1419 in2. From Eq. (8-17)
kb
0.1963 0.1419 30
Ad At E
1.322 Mlbf/in
Ad lt At ld 0.1963 0.69 0.1419 3.5
Ans.
(c)
Upper and lower halves are the same. For the upper half,
Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20)
k1
0.5774 30 0.531
1.155 0.095 0.75 0.531 0.75 0.531
ln
1.155 0.095 0.75 0.531 0.75 0.531
89.20 Mlbf/in
Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30 = 0.860 in, and E = 10.3
Mpsi. Eq. (8-20) k 2 = 9.24 Mlbf/in
For the top half, km = (1/k 1 + 1/k 2 )1 = (1/89.20 + 1/9.24)1 = 8.373 Mlbf/in
Since the bottom half is the same, the overall stiffness is given by
k m = (1/ km + 1/ km )1 = km /2 = 8.373/2 = 4.19 Mlbf/in
Ans
______________________________________________________________________________
8-18 (a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in
Ans.
Chap. 8 Solutions - Rev. A, Page 11/69
(b) From Eq. (8-13), L T = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, l d = L L T = 4.75 1.25 = 3.5 in, l t = l l d = 4.19 3.5 = 0.69 in
A d = (0.52)/4 = 0.1963 in2. From Table 8-2, A t = 0.1419 in2. From Eq. (8-17)
kb
0.1963 0.1419 30
Ad At E
1.322 Mlbf/in
Ad lt At ld 0.1963 0.69 0.1419 3.5
Ans.
(c)
Upper aluminum frustum: t = [4 + 2(0.095)] /2 = 2.095 in, d = 0.5 in, D = 0.75 in, and
E = 10.3 Mpsi. From Eq. (8-20)
0.5774 10.3 0.5
k1
7.23 Mlbf/in
1.155 2.095 0.75 0.5 0.75 0.5
ln
1.155 2.095 0.75 0.5 0.75 0.5
Lower aluminum frustum: t = 4 2.095 = 1.905 in, d = 0.5 in,
D = 0.75 +4(0.095) tan 30 = 0.969 in, and E = 10.3 Mpsi. Eq. (8-20) k 2 = 11.34
Mlbf/in
Steel washers frustum: t = 2(0.095) = 0.190 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi.
Eq. (8-20) k 3 = 53.91 Mlbf/in
From Eq. (8-18)
k m = (1/k 1 + 1/k 2 +1/k 3 )1 = (1/7.23 + 1/11.34 + 1/53.91)1 = 4.08 Mlbf/in Ans.
______________________________________________________________________________
8-19 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 50 + 8.4 = 58.4 mm.
Chap. 8 Solutions - Rev. A, Page 12/69
Rounding up, L = 60 mm
Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(10) + 6 = 26 mm, l d = L L T = 60 26 =
34 mm, l t = l l = 50 34 = 16 mm. A d = (102) / 4 = 78.54 mm2. From Table 8-1,
A t = 58 mm2. From Eq. (8-17)
kb
78.54 58.0 207
Ad At E
292.1 MN/m
Ad lt At ld 78.54 16 58.0 34
Ans.
(c)
Upper and lower frustums are the same. For the upper half,
Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa.
From Eq. (8-20)
0.5774 7110
k1
1576 MN/m
1.155 10 15 10 15 10
ln
1.155 10 15 10 15 10
Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E = 207
GPa. From Eq. (8-20)
k2
0.5774 207 10
1.155 15 26.55 10 26.55 10
ln
1.155 15 26.55 10 26.55 10
11 440 MN/m
For the top half, km = (1/k 1 + 1/k 2 )1 = (1/1576 + 1/11 440)1 = 1385 MN/m
Chap. 8 Solutions - Rev. A, Page 13/69
Since the bottom half is the same, the overall stiffness is given by
k m = (1/ km + 1/ km )1 = km /2 = 1385/2 = 692.5 MN/m
Ans.
8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm.
Rounding up, L = 70 mm
Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(10) + 6 = 26 mm, l d = L L T = 70 26 =
44 mm, l t = l l d = 60 44 = 16 mm. A d = (102) / 4 = 78.54 mm2. From Table 8-1,
A t = 58 mm2. From Eq. (8-17)
kb
78.54 58.0 207
Ad At E
247.6 MN/m
Ad lt At ld 78.54 16 58.0 44
Ans.
(c)
Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq.
(8-20)
Chap. 8 Solutions - Rev. A, Page 14/69
k1
0.5774 10.3 71
1.155 2.095 15 10 15 10
ln
1.155 2.095 15 10 15 10
1576 MN/m
Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq.
(8-20) k 2 = 1 201 MN/m
Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E =
207 GPa. Eq. (8-20) k 3 = 9 781 MN/m
Lower steel frustum: t = 10 mm, d = 10 mm, D = 15 + 2(20) tan 30 = 38.09 mm, and E =
207 GPa. Eq. (8-20) k 4 = 29 070 MN/m
From Eq. (8-18)
k m = (1/k 1 + 1/k 2 +1/k 3 +1/k 4 )1 = (1/1 576 + 1/1 201 + 1/9 781 +1/29 070)1
= 623.5 MN/m Ans.
______________________________________________________________________________
8-21 (a) From Table 8-7, l = h + d /2 = 10 + 30 + 10/2 = 45 mm. L ≥ h + 1.5 d =
10 + 30 + 1.5(10) = 55 mm Ans.
(b) From Eq. (8-14), L T = 2d + 6 = 2(10) + 6 = 26 mm, l d = L L T = 55 26 =
29 mm, l t = l l d = 45 29 = 16 mm. A d = (102) / 4 = 78.54 mm2. From Table 8-1,
A t = 58 mm2. From Eq. (8-17)
78.54 58.0 207
Ad At E
kb
320.9 MN/m
Ans.
Ad lt At ld 78.54 16 58.0 29
(c)
Chap. 8 Solutions - Rev. A, Page 15/69
Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq.
(8-20)
0.5774 10.3 71
k1
1576 MN/m
1.155 2.095 15 10 15 10
ln
1.155 2.095 15 10 15 10
Lower aluminum frustum: t = 5 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq.
(8-20) k 2 = 2 300 MN/m
Top steel frustum: t = 12.5 mm, d = 10 mm, D = 15 + 2(10) tan 30 = 26.55 mm, and E =
207 GPa. Eq. (8-20) k 3 = 12 759 MN/m
Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30 = 20.77 mm, and E
= 207 GPa. Eq. (8-20) k 4 = 6 806 MN/m
From Eq. (8-18)
k m = (1/k 1 + 1/k 2 +1/k 3 +1/k 4 )1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)1
= 772.4 MN/m Ans.
______________________________________________________________________________
kb
8-22 Equation (f ), p. 436: C
kb k m
Eq. (8-17):
Eq. (8-22):
kb
Ad At E
Ad lt At ld
km
0.5774 207 d
0.5774 40 0.5d
2 ln 5
0.5774 40 2.5d
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are mm, mm2, MN/m):
d
10
12
14
16
20
24
30
At
58
84.3
115
157
245
353
561
Ad
78.53982
113.0973
153.938
201.0619
314.1593
452.3893
706.8583
H
8.4
10.8
12.8
14.8
18
21.5
25.6
L>
48.4
50.8
52.8
54.8
58
61.5
65.6
L
50
55
55
55
60
65
70
LT
26
30
34
38
46
54
66
Chap. 8 Solutions - Rev. A, Page 16/69
d
10
12
14
16
20
24
30
l
40
40
40
40
40
40
40
ld
24
25
21
17
14
11
4
lt
16
15
19
23
26
29
36
kb
356.0129
518.8172
686.2578
895.9182
1373.719
1944.24
2964.343
km
1751.566
2235.192
2761.721
3330.796
4595.515
6027.684
8487.533
C
0.16892
0.188386
0.199032
0.211966
0.230133
0.243886
0.258852
The 14 mm would probably be ok, but to satisfy the question, use a 16 mm bolt Ans.
_____________________________________________________________________________
kb
8-23 Equation (f ), p. 436: C
kb k m
Ad At E
Eq. (8-17):
kb
Ad lt At ld
For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in:
k1
0.5774 30 d
1.155 1.5 0.5d 2.5d
ln
1.155 1.5 2.5d 0.5d
0.5774 30 d
1.733 0.5d
ln 5
1.733 2.5d
Lower steel frustum, with D = 1.5d + 2(1) tan 30 = 1.5d + 1.155, and t = 0.5 in:
0.5774 30 d
k2
1.733 0.5d 2.5d 1.155
ln
1.733 2.5d 0.5d 1.155
Chap. 8 Solutions - Rev. A, Page 17/69
For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d, and t = 1 in:
k3
0.5774 14.5 d
1.155 0.5d
ln 5
1.155 2.5d
k m = (1/k 1 +1/k 2 +1/k 3 )1
Overall,
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in):
d
0.375
0.4375
0.5
0.5625
0.625
0.75
0.875
At
0.0775
0.1063
0.1419
0.182
0.226
0.334
0.462
d
0.375
0.4375
0.5
0.5625
0.625
0.75
0.875
ld
2.5
2.375
2.25
2.125
2.25
2
1.75
Use a
9
16
Ad
H
L>
0.110447 0.328125 3.328125
0.15033 0.375
3.375
0.19635 0.4375 3.4375
0.248505 0.484375 3.484375
0.306796 0.546875 3.546875
0.441786 0.640625 3.640625
0.60132
0.75
3.75
lt
0.5
0.625
0.75
0.875
0.75
1
1.25
L
3.5
3.5
3.5
3.5
3.75
3.75
3.75
LT
1
1.125
1.25
1.375
1.5
1.75
2
l
3
3
3
3
3
3
3
k1
k2
k3
kb
km
C
1.031389 15.94599 178.7801 8.461979 5.362481 0.161309
1.383882 19.21506 194.465 10.30557 6.484256 0.175884
1.791626 22.65332 210.6084 12.26874 7.668728 0.189383
2.245705 26.25931 227.2109 14.35052 8.915294 0.20121
2.816255 30.03179 244.2728 16.55009 10.22344 0.215976
3.988786 38.07191 279.7762 21.29991 13.02271 0.234476
5.341985 46.7663 317.1203 26.51374 16.06359 0.24956
12 UNC 3.5 in long bolt
Ans.
______________________________________________________________________________
8-24 Equation (f ), p. 436:
Eq. (8-17):
kb
C
kb
kb k m
Ad At E
Ad lt At ld
Chap. 8 Solutions - Rev. A, Page 18/69
Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2:
k1
0.5774 10.3 d
1.155 l / 2 0.5d
ln 5
1.155 l / 2 2.5d
Middle frustum, with E = 10.3 Mpsi, D = 1.5d + 2(l 0.5) tan 30, and t = 0.5 l /2
k2
0.5774 10.3 d
1.155 0.5 0.5l 0.5d 2 l 0.5 tan 30 2.5d 2 l 0.5 tan 30
ln
1.155 0.5 0.5l 2.5d 2 l 0.5 tan 30 0.5d 2 l 0.5 tan 30
0
0
0
0
Lower frustum, with E = 30Mpsi, D = 1.5 d, t = l 0.5
k3
0.5774 30 d
1.155 l 0.5 0.5d
ln 5
1.155 l 0.5 2.5d
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in)
Chap. 8 Solutions - Rev. A, Page 19/69
Size
1
2
3
4
5
6
8
10
d
0.073
0.086
0.099
0.112
0.125
0.138
0.164
0.19
At
0.00263
0.0037
0.00487
0.00604
0.00796
0.00909
0.014
0.0175
Ad
0.004185
0.005809
0.007698
0.009852
0.012272
0.014957
0.021124
0.028353
L>
0.6095
0.629
0.6485
0.668
0.6875
0.707
0.746
0.785
L
0.75
0.75
0.75
0.75
0.75
0.75
0.75
1
LT
0.396
0.422
0.448
0.474
0.5
0.526
0.578
0.63
l
0.5365
0.543
0.5495
0.556
0.5625
0.569
0.582
0.595
ld
0.354
0.328
0.302
0.276
0.25
0.224
0.172
0.37
Size
1
2
3
4
5
6
8
10
d
0.073
0.086
0.099
0.112
0.125
0.138
0.164
0.19
lt
0.1825
0.215
0.2475
0.28
0.3125
0.345
0.41
0.225
kb
0.194841
0.261839
0.333134
0.403377
0.503097
0.566787
0.801537
1.15799
k1
1.084468
1.321595
1.570439
1.830494
2.101297
2.382414
2.974009
3.602349
k2
1.954599
2.449694
2.993366
3.587564
4.234381
4.936066
6.513824
8.342138
k3
7.09432
8.357692
9.621064
10.88444
12.14781
13.41118
15.93792
18.46467
km
0.635049
0.778497
0.930427
1.090613
1.258846
1.434931
1.809923
2.214214
C
0.23478
0.251687
0.263647
0.27
0.285535
0.28315
0.306931
0.343393
The lowest coarse series screw is a 164 UNC 0.75 in long up to a 632 UNC 0.75 in
long. Ans.
______________________________________________________________________________
8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa
k1
0.5774 207 14
1.155 20 21 14 21 14
ln
1.155 20 21 14 21 14
k m = (1/k 1 + 1/k 1 )1 = k 1 /2 = 5523/2 = 2762 MN/m
5523 MN/m
Ans.
From Eq. (8-22) with l = 40 mm
km
0.5774 207 14
0.5774 40 0.5 14
2 ln 5
0.5774 40 2.5 14
2762 MN/m
Ans.
which agrees with the earlier calculation.
Chap. 8 Solutions - Rev. A, Page 20/69
For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73
k m = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m
Ans.
This is 2.9% higher than the earlier calculations.
______________________________________________________________________________
8-26 (a) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31).
L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in.
Table 8-7, L T = 2d + 0.5 = 2(0.75) + 0.5 = 2 in, l d = L L T = 10.75 2 = 8.75 in,
l t = l l d = 10 8.75 = 1.25 in
A d = (0.752)/4 = 0.4418 in2, A t = 0.373 in2 (Table 8-2)
Eq. (8-17),
0.4418 0.373 30
Ad At E
kb
1.296 Mlbf/in
Ans.
Ad lt At ld 0.4418 1.25 0.373 8.75
Eq. (4-4), p. 149,
2
2
Am Em / 4 1.125 0.75 30
km
1.657 Mlbf/in
Ans.
l
10
Eq. (f), p. 436, C = k b /(k b + k m ) = 1.296/(1.296 + 1.657) = 0.439
Ans.
(b)
Let: N t = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then,
= b + m = Nt p = Nt / N
(1)
But, b = F i / k b , and, m = F i / k m . Substituting these into Eq. (1) and solving for F i gives
Chap. 8 Solutions - Rev. A, Page 21/69
Fi
kb k m N t
kb k m N
2
1.296 1.657 106 1 / 3
15 150 lbf
Ans.
1.296 1.657 16
______________________________________________________________________________
8-27 Proof for the turn-of-nut equation is given in the solution of Prob. 8-26, Eq. (2), where
N t = / 360.
The relationship between the turn-of-nut method and the torque-wrench method is as
follows.
k km
Nt b
Fi N
kb k m
T KFd
i
(turn-of-nut)
(torque-wrench)
Eliminate F i
k km NT
Nt b
Ans.
360
kbkm Kd
______________________________________________________________________________
8-28 (a) From Ex. 8-4, F i = 14.4 kip, k b = 5.21(106) lbf/in, k m = 8.95(106) lbf/in
Eq. (8-27):
T = kF i d = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans.
From Prob. 8-27,
5.21 8.95
k km
Nt b
(14.4)(103 )11
Fi N
6
k
k
5.21
8.95
10
b m
0.0481 turns 17.3
Ans.
Bolt group is (1.5) / (5/8) = 2.4 diameters. Answer is much lower than RB&W
recommendations.
______________________________________________________________________________
8-29 C = k b / (k b + k m ) = 3/(3+12) = 0.2, P = P total / N = 80/6 = 13.33 kips/bolt
Table 8-2, A t = 0.141 9 in2; Table 8-9, S p = 120 kpsi; Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(0.141 9)(120) = 12.77 kips
(a) From Eq. (8-28), the factor of safety for yielding is
120 0.141 9
1.10
CP Fi 0.2 13.33 12.77
(b) From Eq. (8-29), the overload factor is
np
S p At
Ans.
Chap. 8 Solutions - Rev. A, Page 22/69
nL
S p At Fi
CP
120 0.141 9 12.77
1.60
0.2 13.33
Ans.
(c) From Eq. (803), the joint separation factor of safety is
Fi
12.77
1.20
Ans.
P 1 C 13.33 1 0.2
______________________________________________________________________________
n0
1/2 13 UNC Grade 8 bolt, K = 0.20
(a) Proof strength, Table 8-9, S p = 120 kpsi
Table 8-2, A t = 0.141 9 in2
Maximum, F i = S p A t = 120(0.141 9) = 17.0 kips
Ans.
(b) From Prob. 8-29, C = 0.2, P = 13.33 kips
Joint separation, Eq. (8-30) with n 0 = 1
Minimum F i = P (1 C) = 13.33(1 0.2) = 10.66 kips Ans.
(c) Fi = (17.0 + 10.66)/2 = 13.8 kips
Eq. (8-27), T = KF i d = 0.2(13.8)103(0.5)/12 = 115 lbf ft
Ans.
______________________________________________________________________________
8-30
8-31 (a) Table 8-1, A t = 20.1 mm2. Table 8-11, S p = 380 MPa.
Eq. (8-31), F i = 0.75 F p = 0.75 A t S p = 0.75(20.1)380(103) = 5.73 kN
Eq. (f ), p. 436,
C
kb
1
0.278
kb k m 1 2.6
Eq. (8-28) with n p = 1,
3
S A Fi 0.25S p At 0.25 20.1 380 10
P p t
6.869 kN
C
C
0.278
P total = NP = 8(6.869) = 55.0 kN Ans.
(b) Eq. (8-30) with n 0 = 1,
F
5.73
P i
7.94 kN
1 C 1 0.278
P total = NP = 8(7.94) = 63.5 kN Ans. Bolt stress would exceed proof strength
______________________________________________________________________________
8-32 (a) Table 8-2, A t = 0.141 9 in2. Table 8-9, S p = 120 kpsi.
Eq. (8-31), F i = 0.75 F p = 0.75 A t S p = 0.75(0.141 9)120 = 12.77 kips
Eq. (f ), p. 436,
C
kb
4
0.25
kb km 4 12
Chap. 8 Solutions - Rev. A, Page 23/69
Eq. (8-28) with n p = 1,
S p At Fi 0.25 NS p At
Ptotal N
C
C
80 0.25
P C
N total
4.70
0.25S p At 0.25 120 0.141 9
Round to N = 5 bolts
Ans.
(b) Eq. (8-30) with n 0 = 1,
F
Ptotal N i
1 C
Ptotal 1 C 80 1 0.25
4.70
12.77
Fi
Round to N = 5 bolts Ans.
______________________________________________________________________________
N
8-33 Bolts: From Table A-31, the nut height is H = 10.8 mm. L ≥ l +H = 40 + 10.8 = 50.8
mm. Although Table A-17 indicates to go to 60 mm, 55 mm is readily available
Round up to L = 55 mm
Ans.
Eq. (8-14): L T = 2d + 6 = 2(12) + 6 = 30 mm
Table 8-7: l d = L L T = 55 30 = 25 mm, l t = l l d = 40 25 = 15 mm
A d = (122)/4 = 113.1 mm2, Table 8-1: A t = 84.3 mm2
Eq. (8-17):
kb
113.1 84.3 207
Ad At E
518.8 MN/m
Ad lt At ld 113.115 84.3 25
Members: Steel cyl. head: t = 20 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20),
k1
0.5774 207 12
1.155 20 18 12 18 12
ln
1.155 20 18 12 18 12
4470 MN/m
Cast iron: t = 20 mm, d = 12 mm, D = 18 mm, E = 100 GPa (from
Table 8-8). The only difference from k 1 is the material
k 2 = (100/207)(4470) = 2159 MN/m
Eq. (8-18): k m = (1/4470 + 1/2159)1 = 1456 MN/m
Chap. 8 Solutions - Rev. A, Page 24/69
C = k b / (k b + k m ) = 518.8/(518.8+1456) = 0.263
Table 8-11: S p = 650 MPa
Assume non-permanent connection. Eqs. (8-31) and (8-32)
F i = 0.75 A t S p = 0.75(84.3)(650)103 = 41.1 kN
The total external load is P total = p g A c , where A c is the diameter of the cylinder which is
100 mm. The external load per bolt is P = P total /N. Thus
P = [6 (1002)/4](103)/10 = 4.712 kN/bolt
Yielding factor of safety, Eq. (8-28):
np
S p At
CP Fi
650 84.3103
0.263 4.712 41.10
1.29
Ans.
Overload factor of safety, Eq. (8-29):
nL
S p At Fi
CP
650 84.3103 41.10
11.1
0.263 4.712
Ans.
Separation factor of safety, Eq. (8-30):
Fi
41.10
11.8
Ans.
P 1 C 4.712 1 0.263
______________________________________________________________________________
n0
8-34
Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in.
L ≥ l + H = 1.125 + 7/16 = 1.563 in.
Round up to L = 1.75 in
Ans.
Eq. (8-13): L T = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in
Table 8-7: l d = L L T = 1.75 1.25 = 0.5 in, l t = l l d = 1.125 0.5 = 0.625 in
A d = (0.52)/4 = 0.196 3 in2, Table 8-2: A t = 0.141 9 in2
Eq. (8-17):
kb
0.196 3 0.141 9 30
Ad At E
4.316 Mlbf/in
Ad lt At ld 0.196 3 0.625 0.141 9 0.5
Chap. 8 Solutions - Rev. A, Page 25/69
Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20),
k1
0.5774 30 0.5
1.155 0.5 0.75 0.5 0.75 0.5
ln
1.155 0.5 0.75 0.5 0.75 0.5
33.30 Mlbf/in
Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 =
0.5625 in.
Upper frustum, t = 0.5625 0.5 = 0.0625 in, d = 0.5 in,
D = 0.75 + 2(0.5) tan 30 = 1.327 in, E = 14.5 Mpsi (from Table 8-8)
Eq. (8-20) k 2 = 292.7 Mlbf/in
Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi
Eq. (8-20) k 3 = 15.26 Mlbf/in
Eq. (8-18): k m = (1/33.30 + 1/292.7 + 1/15.26)1 = 10.10 Mlbf/in
C = k b / (k b + k m ) = 4.316/(4.316+10.10) = 0.299
Table 8-9: S p = 85 kpsi
Assume non-permanent connection. Eqs. (8-31) and (8-32)
F i = 0.75 A t S p = 0.75(0.141 9)(85) = 9.05 kips
The total external load is P total = p g A c , where A c is the diameter of the cylinder which is
3.5 in. The external load per bolt is P = P total /N. Thus
P = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28):
np
S p At
CP Fi
85 0.141 9
1.27
0.299 1.443 9.05
Ans.
Overload factor of safety, Eq. (8-29):
nL
S p At Fi
CP
85 0.141 9 9.05
6.98
0.299 1.443
Ans.
Separation factor of safety, Eq. (8-30):
Chap. 8 Solutions - Rev. A, Page 26/69
Fi
9.05
8.95
Ans.
P 1 C 1.443 1 0.299
______________________________________________________________________________
n0
8-35 Bolts: Grip: l = 20 + 25 = 45 mm. From Table A-31, the nut height is H = 8.4 mm.
L ≥ l +H = 45 + 8.4 = 53.4 mm. Although Table A-17 indicates to go to 60 mm, 55 mm is
readily available
Round up to L = 55 mm
Ans.
Eq. (8-14): L T = 2d + 6 = 2(10) + 6 = 26 mm
Table 8-7: l d = L L T = 55 26 = 29 mm, l t = l l d = 45 29 = 16 mm
A d = (102)/4 = 78.5 mm2, Table 8-1: A t = 58.0 mm2
Eq. (8-17):
kb
78.5 58.0 207
Ad At E
320.8 MN/m
Ad lt At ld 78.5 16 58.0 29
Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20),
k1
0.5774 207 10
1.155 20 15 10 15 10
ln
1.155 20 15 10 15 10
3503 MN/m
Cast iron: Has two frusta. Midpoint of complete joint is at (20 + 25)/2 = 22.5 mm
Upper frustum, t = 22.5 20 = 2.5 mm, d = 10 mm,
D = 15 + 2(20) tan 30 = 38.09 mm, E = 100 GPa (from Table 8-8),
Eq. (8-20) k 2 = 45 880 MN/m
Lower frustum, t = 22.5 mm, d = 10 mm, D = 15 mm, E = 100 GPa
Eq. (8-20) k 3 = 1632 MN/m
Eq. (8-18): k m = (1/3503 + 1/45 880 + 1/1632)1 = 1087 MN/m
C = k b / (k b + k m ) = 320.8/(320.8+1087) = 0.228
Table 8-11: S p = 830 MPa
Assume non-permanent connection. Eqs. (8-31) and (8-32)
F i = 0.75 A t S p = 0.75(58.0)(830)103 = 36.1 kN
Chap. 8 Solutions - Rev. A, Page 27/69
The total external load is P total = p g A c , where A c is the diameter of the cylinder which is
0.8 m. The external load per bolt is P = P total /N. Thus
P = [550 (0.82)/4]/36 = 7.679 kN/bolt
Yielding factor of safety, Eq. (8-28):
np
S p At
CP Fi
830 58.0 103
0.228 7.679 36.1
1.27
Ans.
Overload factor of safety, Eq. (8-29):
nL
S p At Fi
CP
830 58.0 103 36.1
6.88 Ans.
0.228 7.679
Separation factor of safety, Eq. (8-30):
Fi
36.1
6.09
Ans.
P 1 C 7.679 1 0.228
______________________________________________________________________________
n0
8-36 Bolts: Grip, l = 3/8 + 1/2 = 0.875 in. From Table A-31, the nut height is H = 3/8 in.
L ≥ l + H = 0.875 + 3/8 = 1.25 in.
Let L = 1.25 in Ans.
Eq. (8-13): L T = 2d + 0.25 = 2(7/16) + 0.25 = 1.125 in
Table 8-7: l d = L L T = 1.25 1.125 = 0.125 in, l t = l l d = 0.875 0.125 =
0.75 in
A d = (7/16)2/4 = 0.150 3 in2, Table 8-2: A t = 0.106 3 in2
Eq. (8-17),
kb
0.150 3 0.106 3 30
Ad At E
3.804 Mlbf/in
Ad lt At ld 0.150 3 0.75 0.106 3 0.125
Members: Steel cyl. head: t = 0.375 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi. Eq.
(8-20),
Chap. 8 Solutions - Rev. A, Page 28/69
k1
0.5774 30 0.4375
1.155 0.375 0.65625 0.4375 0.65625 0.4375
ln
1.155 0.375 0.65625 0.4375 0.65625 0.4375
31.40 Mlbf/in
Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 =
0.4375 in.
Upper frustum, t = 0.4375 0.375 = 0.0625 in, d = 0.4375 in,
D = 0.65625 + 2(0.375) tan 30 = 1.089 in, E = 14.5 Mpsi (from Table
8-8)
Eq. (8-20) k 2 = 195.5 Mlbf/in
Lower frustum, t = 0.4375 in, d = 0.4375 in, D = 0.65625 in, E = 14.5
Mpsi
Eq. (8-20) k 3 = 14.08 Mlbf/in
Eq. (8-18): k m = (1/31.40 + 1/195.5 + 1/14.08)1 = 9.261 Mlbf/in
C = k b / (k b + k m ) = 3.804/(3.804 + 9.261) = 0.291
Table 8-9: S p = 120 kpsi
Assume non-permanent connection. Eqs. (8-31) and (8-32)
F i = 0.75 A t S p = 0.75(0.106 3)(120) = 9.57 kips
The total external load is P total = p g A c , where A c is the diameter of the cylinder which is
3.25 in. The external load per bolt is P = P total /N. Thus
P = [1 200 (3.252)/4](103)/8 = 1.244 kips/bolt
Yielding factor of safety, Eq. (8-28):
np
S p At
CP Fi
120 0.106 3
1.28
0.2911.244 9.57
Ans.
Overload factor of safety, Eq. (8-29):
nL
S p At Fi
CP
120 0.106 3 9.57
8.80 Ans.
0.2911.244
Separation factor of safety, Eq. (8-30):
Chap. 8 Solutions - Rev. A, Page 29/69
n0
Fi
9.57
10.9
P 1 C 1.244 1 0.291
Ans.
______________________________________________________________________________
8-37 From Table 8-7, h = t 1 = 20 mm
For t 2 > d, l = h + d /2 = 20 + 12/2 = 26 mm
L ≥ h + 1.5 d = 20 + 1.5(12) = 38 mm. Round up to L = 40 mm
L T = 2d + 6 = 2(12) + 6 = 30 mm
l d = L L T = 40 20 = 10 mm
l t = l l d = 26 10 = 16 mm
From Table 8-1, A t = 84.3 mm2. A d = (122)/4 = 113.1 mm2
Eq. (8-17),
113.1 84.3 207
Ad At E
kb
744.0 MN/m
Ad lt At ld 113.116 84.3 10
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 13 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20)
k1
0.5774 207 12
1.155 13 18 12 18 12
ln
1.155 13 18 12 18 12
5 316 MN/m
Middle frusta, steel: t = 20 13 = 7 mm, d = 12 mm, D = 18 + 2(13 7) tan 30 = 24.93
mm, E = 207 GPa. Eq. (8-20) k 2 = 15 660 MN/m
Lower frusta, cast iron: t = 26 20 = 6 mm, d = 12 mm, D = 18 mm, E = 100 GPa (see
Table 8-8). Eq. (8-20) k 3 = 3 887 MN/m
Eq. (8-18),
k m = (1/5 316 + 1/15 660 + 1/3 887)1 = 1 964 MN/m
C = k b / (k b + k m ) = 744.0/(744.0 + 1 964) = 0.275
Table 8-11: S p = 650 MPa. From Prob. 8-33, P = 4.712 kN. Assume a non-permanent
connection. Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(84.3)(650)103 = 41.1 kN
Yielding factor of safety, Eq. (8-28)
650 84.3103
1.29
np
CP Fi 0.275 4.712 41.1
S p At
Ans.
Overload factor of safety, Eq. (8-29)
Chap. 8 Solutions - Rev. A, Page 30/69
nL
S p At Fi
CP
650 84.3103 41.1
0.275 4.712
10.7
Ans.
Separation factor of safety, Eq. (8-30)
Fi
41.1
12.0
Ans.
P 1 C 4.712 1 0.275
______________________________________________________________________________
n0
8-38 From Table 8-7, h = t 1 = 0.5 in
For t 2 > d, l = h + d /2 = 0.5 + 0.5/2 = 0.75 in
L ≥ h + 1.5 d = 0.5 + 1.5(0.5) = 1.25 in. Let L = 1.25 in
L T = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in. All threaded.
From Table 8-1, A t = 0.141 9 in2. The bolt stiffness is k b = A t E / l = 0.141 9(30)/0.75 =
5.676 Mlbf/in
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi
0.5774 30 0.5
38.45 Mlbf/in
k1
1.155 0.375 0.75 0.5 0.75 0.5
ln
1.155 0.375 0.75 0.5 0.75 0.5
Middle frusta, steel: t = 0.5 0.375 = 0.125 in, d = 0.5 in,
D = 0.75 + 2(0.75 0.5) tan 30 = 1.039 in, E = 30 Mpsi.
Eq. (8-20) k 2 = 184.3 Mlbf/in
Lower frusta, cast iron: t = 0.75 0.5 = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi.
Eq. (8-20) k 3 = 23.49 Mlbf/in
Eq. (8-18),
k m = (1/38.45 + 1/184.3 + 1/23.49)1 = 13.51 Mlbf/in
C = k b / (k b + k m ) = 5.676 / (5.676 + 13.51) = 0.296
Table 8-9, S p = 85 kpsi. From Prob. 8-34, P = 1.443 kips/bolt. Assume a non-permanent
connection. Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(0.141 9)(85) = 9.05 kips
Yielding factor of safety, Eq. (8-28)
np
S p At
CP Fi
85 0.141 9
1.27
0.296 1.443 9.05
Ans.
Overload factor of safety, Eq. (8-29)
Chap. 8 Solutions - Rev. A, Page 31/69
nL
S p At Fi
CP
85 0.141 9 9.05
7.05
0.296 1.443
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.05
8.91
Ans.
P 1 C 1.443 1 0.296
______________________________________________________________________________
n0
8-39 From Table 8-7, h = t 1 = 20 mm
For t 2 > d, l = h + d /2 = 20 + 10/2 = 25 mm
L ≥ h + 1.5 d = 20 + 1.5(10) = 35 mm. Let L = 35 mm
L T = 2d + 6 = 2(10) + 6 = 26 mm
l d = L L T = 35 26 = 9 mm
l t = l l d = 25 9 = 16 mm
From Table 8-1, A t = 58.0 mm2. A d = (102)/4 = 78.5 mm2
Eq. (8-17),
78.5 58.0 207
Ad At E
530.1 MN/m
kb
Ad lt At ld 78.5 16 58.0 9
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 12.5 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20)
0.5774 207 10
4 163 MN/m
k1
1.155 12.5 15 10 15 10
ln
1.155 12.5 15 10 15 10
Middle frusta, steel: t = 20 12.5 = 7.5 mm, d = 10 mm, D = 15 + 2(12.5 7.5) tan 30 =
20.77 mm, E = 207 GPa. Eq. (8-20) k 2 = 10 975 MN/m
Lower frusta, cast iron: t = 25 20 = 5 mm, d = 10 mm, D = 15 mm, E = 100 GPa (see
Table 8-8). Eq. (8-20) k 3 = 3 239 MN/m
Eq. (8-18),
k m = (1/4 163 + 1/10 975 + 1/3 239)1 = 1 562 MN/m
C = k b / (k b + k m ) = 530.1/(530.1 + 1 562) = 0.253
Table 8-11: S p = 830 MPa. From Prob. 8-35, P = 7.679 kN/bolt. Assume a non-permanent
connection. Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(58.0)(830)103 = 36.1 kN
Yielding factor of safety, Eq. (8-28)
Chap. 8 Solutions - Rev. A, Page 32/69
np
S p At
CP Fi
830 58.0 103
0.253 7.679 36.1
1.27
Ans.
Overload factor of safety, Eq. (8-29)
nL
S p At Fi
CP
830 58.0 103 36.1
6.20
0.253 7.679
Ans.
Separation factor of safety, Eq. (8-30)
Fi
36.1
6.29
Ans.
P 1 C 7.679 1 0.253
______________________________________________________________________________
n0
8-40 From Table 8-7, h = t 1 = 0.375 in
For t 2 > d, l = h + d /2 = 0.375 + 0.4375/2 = 0.59375 in
L ≥ h + 1.5 d = 0.375 + 1.5(0.4375) = 1.031 in. Round up to L = 1.25 in
L T = 2d + 0.25 = 2(0.4375) + 0.25 = 1.125 in
l d = L L T = 1.25 1.125 = 0.125
l t = l l d = 0.59375 0.125 = 0.46875 in
A d = (7/16)2/4 = 0.150 3 in2, Table 8-2: A t = 0.106 3 in2
Eq. (8-17),
0.150 3 0.106 3 30
Ad At E
5.724 Mlbf/in
Ad lt At ld 0.150 3 0.46875 0.106 3 0.125
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi
kb
k1
0.5774 30 0.4375
1.155 0.296875 0.656255 0.4375 0.75 0.656255
ln
1.155 0.296875 0.75 0.656255 0.75 0.656255
35.52 Mlbf/in
Middle frusta, steel: t = 0.375 0.296875 = 0.078125 in, d = 0.4375 in,
D = 0.65625 + 2(0.59375 0.375) tan 30 = 0.9088 in, E = 30 Mpsi.
Eq. (8-20) k 2 = 215.8 Mlbf/in
Lower frusta, cast iron: t = 0.59375 0.375 = 0.21875 in, d = 0.4375 in, D = 0.65625 in,
E = 14.5 Mpsi. Eq. (8-20) k 3 = 20.55 Mlbf/in
Eq. (8-18),
k m = (1/35.52 + 1/215.8 + 1/20.55)1 = 12.28 Mlbf/in
C = k b / (k b + k m ) = 5.724/(5.724 + 12.28) = 0.318
Chap. 8 Solutions - Rev. A, Page 33/69
Table 8-9, S p = 120 kpsi. From Prob. 8-34, P = 1.244 kips/bolt. Assume a non-permanent
connection. Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(0.106 3)(120) = 9.57 kips
Yielding factor of safety, Eq. (8-28)
np
S p At
CP Fi
120 0.106 3
1.28
0.318 1.244 9.57
Ans.
Overload factor of safety, Eq. (8-29)
nL
S p At Fi
CP
120 0.106 3 9.57
8.05
0.318 1.244
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.57
11.3
Ans.
P 1 C 1.244 1 0.318
______________________________________________________________________________
n0
8-41 This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on members,
and combining using Eq. (8-18), yields k m = 1 141 MN/m (see Prob. 8-33 for method of
calculation.
2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is
rounded up from the calculation of l + H = 40 + 8.4 = 48.4 mm to 50 mm. Next,
calculations are made for L T = 2(10) + 6 = 26 mm, l d = 50 26 = 24 mm, l t = 40 24 =
16 mm. From step 1, A d = (102)/4 = 78.54 mm2. Next, from Table 8-1, A t = 78.54 mm2.
From Eq. (8-17), k b = 356 MN/m. Finally, from Eq. (e), p. 421, C = 0.238.
3. From Prob. 8-33, the bolt circle diameter is E = 200 mm. Substituting this for D b in Eq.
(8-34), the number of bolts are
N
Db
200
4d
4 10
Rounding this up gives N = 16.
15.7
4. Next, select a grade bolt. Based on the solution to Prob. 8-33, the strength of ISO 9.8
was so high to give very large factors of safety for overload and separation. Try ISO 4.6
Chap. 8 Solutions - Rev. A, Page 34/69
with S p = 225 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, F i =
9.79 kN.
5. The external load requirement per bolt is P = 1.15 p g A c /N, where from Prob 8-33, p g =
6 MPa, and A c = (1002)/4. This gives P = 3.39 kN/bolt.
6. Using Eqs. (8-28) to (8-30) yield n p = 1.23, n L = 4.05, and n 0 = 3.79.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. The results for four bolt sizes are shown below. The dimension of each
term is consistent with the example given above.
d
8
10
12
14
km
854
1141
1456
1950
H
6.8
8.4
10.8
12.8
L
50
50
55
55
LT
22
26
30
34
ld
28
24
25
21
lt
12
16
15
19
d
8
10
12
14
C
0.215
0.238
0.263
0.276
N
20
16
13*
12
Sp
225
225
225
225
Fi
6.18
9.79
14.23
19.41
P
2.71
3.39
4.17
4.52
np
1.22
1.23
1.24
1.25
Ad
At
kb
50.26 36.6 233.9
78.54 58
356
113.1 84.3 518.8
153.9 115 686.3
nL
3.53
4.05
4.33
5.19
n0
2.90
3.79
4.63
5.94
*Rounded down from13.08997, so spacing is slightly greater than four diameters.
Any one of the solutions is acceptable. A decision-maker might be cost such as
N cost/bolt, and/or N cost per hole, etc.
________________________________________________________________________
8-42 This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 0.5 in. Using Eq. (8-20) on three frusta
(see Prob. 8-34 solution), and combining using Eq. (8-19), yields k m = 10.10 Mlbf/in.
2. Look up the nut height in Table A-31. For the example, H = 0.4375 in. From this, L is
rounded up from the calculation of l + H = 1.125 + 0.4375 = 1.5625 in to 1.75 in. Next,
calculations are made for L T = 2(0.5) + 0.25 = 1.25 in, l d = 1.75 1.25 = 0.5 in, l t = 1.125
0.5 = 0.625 in. From step 1, A d = (0.52)/4 = 0.1963 in2. Next, from Table 8-1, A t =
0.141 9 in2. From Eq. (8-17), k b = 4.316 Mlbf/in. Finally, from Eq. (e), p. 421, C = 0.299.
3. From Prob. 8-34, the bolt circle diameter is E = 6 in. Substituting this for D b in Eq. (834), for the number of bolts
Chap. 8 Solutions - Rev. A, Page 35/69
N
Db
6
4d
4 0.5
Rounding this up gives N = 10.
9.425
4. Next, select a grade bolt. Based on the solution to Prob. 8-34, the strength of SAE grade
5 was adequate. Use this with S p = 85 kpsi. From Eqs. (8-31) and (8-32) for a nonpermanent connection, F i = 9.046 kips.
5. The external load requirement per bolt is P = 1.15 p g A c /N, where from Prob 8-34,
p g = 1 500 psi, and A c = (3.52)/4 . This gives P = 1.660 kips/bolt.
6. Using Eqs. (8-28) to (8-30) yield n p = 1.26, n L = 6.07, and n 0 = 7.78.
d
0.375
0.4375
0.5
0.5625
km
6.75
9.17
10.10
11.98
H
0.3281
0.375
0.4375
0.4844
d
0.375
0.4375
0.5
0.5625
C
0.261
0.273
0.299
0.308
N
13
11
10
9
L
LT
ld
1.5
1
0.5
1.5 1.125 0.375
1.75 1.25 0.5
1.75 1.375 0.375
Sp
85
85
85
85
Fi
4.941
6.777
9.046
11.6
P
1.277
1.509
1.660
1.844
lt
0.625
0.75
0.625
0.75
np
1.25
1.26
1.26
1.27
Ad
At
0.1104 0.0775
0.1503 0.1063
0.1963 0.1419
0.2485 0.182
nL
4.95
5.48
6.07
6.81
kb
2.383
3.141
4.316
5.329
n0
5.24
6.18
7.78
9.09
Any one of the solutions is acceptable. A decision-maker might be cost such as
N cost/bolt, and/or N cost per hole, etc.
________________________________________________________________________
8-43 This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on three frusta
(see Prob. 8-35 solution), and combining using Eq. (8-19), yields k m = 1 087 MN/m.
2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is
rounded up from the calculation of l + H = 45 + 8.4 = 53.4 mm to 55 mm. Next,
calculations are made for L T = 2(10) + 6 = 26 mm, l d = 55 26 = 29 mm, l t = 45 29 =
16 mm. From step 1, A d = (102)/4 = 78.54 mm2. Next, from Table 8-1, A t = 58.0 mm2.
From Eq. (8-17), k b = 320.9 MN/m. Finally, from Eq. (e), p. 421, C = 0.228.
3. From Prob. 8-35, the bolt circle diameter is E = 1000 mm. Substituting this for D b in
Eq. (8-34), for the number of bolts
Chap. 8 Solutions - Rev. A, Page 36/69
N
Db
1000
78.5
4d
4 10
Rounding this up gives N = 79. A rather large number, since the bolt circle diameter, E is
so large. Try larger bolts.
4. Next, select a grade bolt. Based on the solution to Prob. 8-35, the strength of ISO 9.8
was so high to give very large factors of safety for overload and separation. Try ISO 5.8
with S p = 380 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, F i =
16.53 kN.
5. The external load requirement per bolt is P = 1.15 p g A c /N, where from Prob 8-35, p g
= 0.550 MPa, and A c = (8002)/4 . This gives P = 4.024 kN/bolt.
6. Using Eqs. (8-28) to (8-30) yield n p = 1.26, n L = 6.01, and n 0 = 5.32.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. The results for three bolt sizes are shown below. The dimension of
each term is consistent with the example given above.
d
10
20
36
km
1087
3055
6725
H
8.4
18
31
L
55
65
80
d
10
20
36
C
0.228
0.308
0.361
N
79
40
22
Sp
380
380
380
LT
26
46
78
ld
29
19
2
lt
16
26
43
Fi
P
np
16.53 4.024 1.26
69.83 7.948 1.29
232.8 14.45 1.3
Ad
78.54
314.2
1018
At
58
245
817
nL
6.01
9.5
14.9
n0
5.32
12.7
25.2
kb
320.9
1242
3791
A large range is presented here. Any one of the solutions is acceptable. A decision-maker
might be cost such as N cost/bolt, and/or N cost per hole, etc.
________________________________________________________________________
8-44 This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 0.375 in. Using Eq. (8-20) on three
frusta (see Prob. 8-36 solution), and combining using Eq. (8-19), yields k m = 7.42 Mlbf/in.
2. Look up the nut height in Table A-31. For the example, H = 0.3281 in. From this,
L ≥ l + H = 0.875 + 0.3281 = 1.2031 in. Rounding up, L = 1.25. Next, calculations are
made for L T = 2(0.375) + 0.25 = 1 in, l d = 1.25 1 = 0.25 in, l t = 0.875 0.25 = 0.625 in.
Chap. 8 Solutions - Rev. A, Page 37/69
From step 1, A d = (0.3752)/4 = 0.1104 in2. Next, from Table 8-1, A t = 0.0775 in2. From
Eq. (8-17), k b = 2.905 Mlbf/in. Finally, from Eq. (e), p. 421, C = 0.263.
3. From Prob. 8-36, the bolt circle diameter is E = 6 in. Substituting this for D b in Eq. (834), for the number of bolts
N
Db
6
4d
4 0.375
Rounding this up gives N = 13.
12.6
4. Next, select a grade bolt. Based on the solution to Prob. 8-36, the strength of SAE grade
8 seemed high for overload and separation. Try SAE grade 5 with S p = 85 kpsi. From Eqs.
(8-31) and (8-32) for a non-permanent connection, F i = 4.941 kips.
5. The external load requirement per bolt is P = 1.15 p g A c /N, where from Prob 8-34,
p g = 1 200 psi, and A c = (3.252)/4. This gives P = 0.881 kips/bolt.
6. Using Eqs. (8-28) to (8-30) yield n p = 1.27, n L = 6.65, and n 0 = 7.81.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. For this solution we only looked at one bolt size, 83 16 , but evaluated
changing the bolt grade. The results for four bolt grades are shown below. The dimension
of each term is consistent with the example given above.
d
km
H
L
0.375 7.42 0.3281 1.25
LT
1
ld
lt
At
kb
Ad
0.25 0.625 0.1104 0.0775 2.905
SAE
n0
d
C
N grade S p
Fi
P
np
nL
0.375 0.281 13
1
33 1.918 0.881 1.18 2.58 3.03
0.375 0.281 13
2
55 3.197 0.881 1.24 4.30 5.05
0.375 0.281 13
4
65 3.778 0.881 1.25 5.08 5.97
0.375 0.281 13
5
85 4.941 0.881 1.27 6.65 7.81
Note that changing the bolt grade only affects S p , F i , n p , n L , and n 0 . Any one of the
solutions is acceptable, especially the lowest grade bolt.
________________________________________________________________________
8-45 (a) Fb RFb,max sin
Half of the external moment is contributed by the line load in the interval 0 ≤ ≤
Chap. 8 Solutions - Rev. A, Page 38/69
M
FbR 2 sin d
0
2
M
Fb,max R 2
2
2
from which Fb,max
Fmax
0
Fb,max R 2 sin 2 d
M
R2
2
1
FbR sin d
M 2
M
R sin d
(cos 1 - cos 2 )
2
R 1
R
Noting 1 = 75, 2 = 105,
Fmax
12 000
(cos 75 - cos105 ) 494 lbf
(8 / 2)
Fmax Fb,max R
(b)
Fmax
Ans.
M
2M
2
( R)
2
N
RN
R
2(12 000)
500 lbf
(8 / 2)(12)
Ans.
(c) F = F max sin
M = 2 F max R [(1) sin2 90 + 2 sin2 60 + 2 sin2 30 + (1) sin2 (0)] = 6F max R
from which,
M
12 000
500 lbf
6R
6(8 / 2)
The simple general equation resulted from part (b)
Fmax
Ans.
2M
RN
________________________________________________________________________
Fmax
8-46
(a) From Table 8-11, S p = 600 MPa. From Table 8-1, A t = 353 mm2.
Eq. (8-31):
Table 8-15:
Eq. (8-27):
Fi 0.9 At S p 0.9 353 600 10 3 190.6 kN
K = 0.18
T = K F i d = 0.18(190.6)(24) = 823 Nm
Ans.
Chap. 8 Solutions - Rev. A, Page 39/69
(b) Washers: t = 4.6 mm, d = 24 mm, D = 1.5(24) = 36 mm, E = 207 GPa.
Eq. (8-20),
k1
0.5774 207 24
1.155 4.6 36 24 36 24
ln
1.155 4.6 36 24 36 24
31 990 MN/m
Cast iron: t = 20 mm, d = 24 mm, D = 36 + 2(4.6) tan 30 = 41.31 mm, E = 135 GPa.
Eq. (8-20) k 2 = 10 785 MN/m
Steel joist: t = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa. Eq. (8-20) k 3 = 16
537 MN/m
Eq. (8-18):
k m = (2 / 31 990 + 1 / 10 785 +1 / 16 537)1 = 4 636 MN/m
Bolt: l = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > 49.2 + 21.5 = 70.7
mm. From Table A-17, use L = 80 mm. From Eq. (8-14)
L T = 2(24) + 6 = 54 mm, l d = 80 54 = 26 mm, l t = 49.2 26 = 23.2 mm
From Table (8-1), A t = 353 mm2, A d = (242) / 4 = 452.4 mm2
Eq. (8-17):
kb
452.4 353 207
Ad At E
1680 MN/m
Ad lt At ld 452.4 23.2 353 26
C = k b / (k b + k m ) = 1680 / (1680 + 4636) = 0.266, S p = 600 MPa, F i = 190.6 kN,
P = P total / N = 18/4 = 4.5 kN
Yield: From Eq. (8-28)
np
S p At
CP Fi
600 353 10 3
1.10 Ans.
0.266 4.5 190.6
Load factor: From Eq. (8-29)
nL
S p At Fi
CP
600 353103 190.6
17.7
0.266 4.5
Ans.
Separation: From Eq. (8-30)
Chap. 8 Solutions - Rev. A, Page 40/69
n0
Fi
190.6
57.7
P 1 C 4.5 1 0.266
Ans.
As was stated in the text, bolts are typically preloaded such that the yielding factor of
safety is not much greater than unity which is the case for this problem. However, the
other load factors indicate that the bolts are oversized for the external load.
______________________________________________________________________________
8-47 (a) ISO M 20 2.5 grade 8.8 coarse pitch bolts, lubricated.
Table 8-2,
A t = 245 mm2
Table 8-11,
S p = 600 MPa
F i = 0.90 A t S p = 0.90(245)600(103) = 132.3 kN
Table 8-15,
K = 0.18
T = KF i d = 0.18(132.3)20 = 476 N m
Eq. (8-27),
Ans.
(b) Table A-31, H = 18 mm, L ≥ L G + H = 48 + 18 = 66 mm. Round up to L = 80 mm per
Table A-17.
LT 2d 6 2(20) 6 46 mm
ld L - LT 80 46 34 mm
lt l - ld 48 34 14 mm
A d = (202) /4 = 314.2 mm2,
kb
Ad At E
314.2(245)(207)
1251.9 MN/m
314.2(14) 245(34)
Ad lt Alt d
Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d =
20mm
km
0.5774 207 20
0.5774 Ed
4236 MN/m
0.5774l 0.5d
0.5774 48 0.5 20
2 ln 5
2 ln 5
0.5774l 2.5d
0.5774 48 2.5 20
kb
1251.9
0.228
kb km 1251.9 4236
/ N = 40/2 = 20 kN,
C
P = P total
Yield: From Eq. (8-28)
S p At
600 245 103
1.07 Ans.
np
CP Fi 0.228 20 132.3
Chap. 8 Solutions - Rev. A, Page 41/69
Load factor: From Eq. (8-29)
nL
S p At Fi
CP
600 245 10 3 132.3
3.22
0.228 20
Ans.
Separation: From Eq. (8-30)
Fi
132.3
8.57
Ans.
P 1 C 20 1 0.228
______________________________________________________________________________
n0
8-48 From Prob. 8-29 solution, P max =13.33 kips, C = 0.2, F i = 12.77 kips, A t = 0.141 9 in2
F
12.77
i i
90.0 kpsi
At 0.141 9
CP 0.2 13.33
Eq. (8-39),
a
9.39 kpsi
2 At 2 0.141 9
Eq. (8-41),
m a i 9.39 90.0 99.39 kpsi
(a) Goodman Eq. (8-45) for grade 8 bolts, S e = 23.2 kpsi (Table 8-17), S ut = 150 kpsi
(Table 8-9)
S S i 23.2 150 90.0
n f e ut
0.856
Ans.
a Sut Se 9.39 150 23.2
(b) Gerber Eq. (8-46)
1
nf
Sut Sut2 4Se Se i Sut2 2 i Se
2 a Se
1
150 1502 4 23.2 23.2 90.0 1502 2 90.0 23.2 1.32
2 9.39 23.2
Ans.
(c) ASME-elliptic Eq. (8-47) with S p = 120 kpsi (Table 8-9)
Se
nf
S p S p2 Se2 i2 i Se
2
2
a S p Se
23.2
120 120 2 23.2 2 902 90 23.2 1.30
2
2
9.39 120 23.2
Ans.
______________________________________________________________________________
8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength,
S e , of a class 5.8 bolt. Table 8-17 does not provide this and the student will be required to
estimate it by other means [see the solution of part (d)].
Per bolt, P bmax = 60/8 = 7.5 kN, P bmin = 20/8 = 2.5 kN
Chap. 8 Solutions - Rev. A, Page 42/69
kb
1
0.278
kb k m 1 2.6
(a) Table 8-1, A t = 20.1 mm2; Table 8-11, S p = 380 MPa
Eqs. (8-31) and (8-32), F i = 0.75 A t S p = 0.75(20.1)380(103) = 5.73 kN
S p At
380 20.1103
Yield, Eq. (8-28),
0.98
np
Ans.
CP Fi 0.278 7.5 5.73
C
S p At Fi
(b) Overload, Eq. (8-29),
nL
(c) Separation, Eq. (8-30),
n0
(d) Goodman, Eq. (8-35),
a
CP
380 20.1103 5.73
0.915
0.278 7.5
Fi
5.73
1.06
P 1 C 7.5 1 0.278
C Pb max Pb min
2 At
Ans.
Ans.
0.278 7.5 2.5 103
2 20.1
C Pb max Pb min Fi 0.278 7.5 2.5 103 5.73 10
Eq. (8-36), m
2 At
2 20.1
20.1
At
3
34.6 MPa
354.2
MPa
Table 8-11, S ut = 520 MPa, i = F i /A t = 5.73(103)/20.1 = 285 MPa
We have a problem for S e . Table 8-17 does not list S e for class 5.8 bolts. Here, we will
estimate S e using the methods of Chapter 6. Estimate S e from the,
Eq. (6-8), p. 282,
Se 0.5Sut 0.5 520 260 MPa .
Table 6-2, p. 288,
Eq. (6-19), p. 287,
a = 4.51, b = 0.265
ka aSutb 4.51 520 0.265 0.860
Eq. (6-21), p. 288,
kb = 1
Eq. (6-26), p.290,
k c = 0.85
The fatigue stress-concentration factor, from Table 8-16, is K f = 2.2. For simple axial
loading and infinite-life it is acceptable to reduce the endurance limit by K f and use the
nominal stresses in the stress/strength/design factor equations. Thus,
Eq. (6-18), p. 287,
S e = k a k b k c S e / K f = 0.86(1)0.85(260) / 2.2 = 86.4 MPa
Eq. (8-38),
nf
Se Sut i
86.4 520 285
0.847
Sut a Se m i 520 34.6 86.4 354.2 285
Ans.
It is obvious from the various answers obtained, the bolted assembly is undersized. This
can be rectified by a one or more of the following: more bolts, larger bolts, higher class
bolts.
______________________________________________________________________________
8-50 Per bolt, P bmax = P max /N = 80/10 = 8 kips, P bmin = P min /N = 20/10 = 2 kips
C = k b / (k b + k m ) = 4/(4 + 12) = 0.25
(a) Table 8-2, A t = 0.141 9 in2, Table 8-9, S p = 120 kpsi and S ut = 150 kpsi
Chap. 8 Solutions - Rev. A, Page 43/69
Table 8-17, S e = 23.2 kpsi
Eqs. (8-31) and (8-32), F i = 0.75 A t S p i = F i /A t = 0.75 S p = 0.75(120) =90 kpsi
Eq. (8-35),
a
Eq. (8-36),
m
C Pb max Pb min 0.25 8 2
5.29 kpsi
2 At
2 0.141 9
C Pb max Pb min
0.25 8 2
i
90 98.81 kpsi
2 At
2 0.141 9
Eq. (8-38),
Se Sut i
23.2 150 90
1.39
Ans.
Sut a Se m i 150 5.29 23.2 98.81 90
______________________________________________________________________________
nf
8-51 From Prob. 8-33, C = 0.263, P max = 4.712 kN / bolt, F i = 41.1 kN, S p = 650 MPa, and
A t = 84.3 mm2
i = 0.75 S p = 0.75(650) = 487.5 MPa
3
CP 0.263 4.712 10
Eq. (8-39):
a
7.350 MPa
2 At
2 84.3
m
Eq. (8-40)
CP Fi
7.350 487.5 494.9 MPa
2 At At
(a) Goodman: From Table 8-11, S ut = 900 MPa, and from Table 8-17, S e = 140 MPa
S S i 140 900 487.5
Eq. (8-45):
n f e ut
7.55
Ans.
a Sut Se 7.350 900 140
(b) Gerber:
Eq. (8-46):
1
nf
Sut Sut2 4Se Se i Sut2 2 i Se
2 a Se
1
900 9002 4 140 140 487.5 9002 2 487.5 140
2 7.350 140
11.4
Ans.
(c) ASME-elliptic:
Eq. (8-47):
Chap. 8 Solutions - Rev. A, Page 44/69
nf
Se
S p S p2 Se2 i2 i Se
2
2
a S p Se
140
650 6502 140 2 487.52 487.5 140 9.73
2
2
7.350 650 140
Ans.
______________________________________________________________________________
8-52 From Prob. 8-34, C = 0.299, P max = 1.443 kips/bolt,F i = 9.05 kips, S p = 85 kpsi, and
A t = 0.141 9 in2
i 0.75S p 0.75 85 63.75 kpsi
Eq. (8-37):
a
CP 0.299 1.443
1.520 kpsi
2 At
2 0.141 9
Eq. (8-38)
m
CP
i 1.520 63.75 65.27 kpsi
2 At
(a) Goodman: From Table 8-9, S ut = 120 kpsi, and from Table 8-17, S e = 18.8 kpsi
S S i 18.8 120 63.75
Eq. (8-45):
n f e ut
5.01
Ans.
a Sut Se 1.520 120 18.8
(b) Gerber:
Eq. (8-46):
1
nf
Sut Sut2 4Se Se i Sut2 2 i Se
2 a Se
1
120 1202 4 18.6 18.6 63.75 1202 2 63.75 18.6
2 1.520 18.6
7.45
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf
Se
S p S p2 Se2 i2 i Se
2
2
a S p Se
18.6
85 852 18.6 2 63.752 63.75 18.6 6.22
1.520 852 18.62
Ans.
______________________________________________________________________________
Chap. 8 Solutions - Rev. A, Page 45/69
8-53
From Prob. 8-35, C = 0.228, P max = 7.679 kN/bolt, F i = 36.1 kN, S p = 830 MPa, and
A t = 58.0 mm2
i = 0.75 S p = 0.75(830) = 622.5 MPa
3
CP 0.228 7.679 10
Eq. (8-37):
a
15.09 MPa
2 At
2 58.0
CP
i 15.09 622.5 637.6 MPa
2 At
(a) Goodman: From Table 8-11, S ut = 1040 MPa, and from Table 8-17, S e = 162 MPa
S S i 162 1040 622.5
Eq. (8-45):
n f e ut
3.73
Ans.
a Sut Se 15.09 1040 162
(b) Gerber:
Eq. (8-46):
1
nf
Sut Sut2 4Se Se i Sut2 2 i Se
2 a Se
m
Eq. (8-38)
1
1040 10402 4 162 162 622.5 10402 2 622.5 162
2 15.09 162
5.74
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf
Se
S p S p2 S e2 i2 i S e
2
2
a S p Se
162
830 830 2 1622 622.52 622.5 162 5.62
2
2
15.09 830 162
Ans.
______________________________________________________________________________
8-54 From Prob. 8-36, C = 0.291, P max = 1.244 kips/bolt, F i = 9.57 kips, S p = 120 kpsi, and
A t = 0.106 3 in2
i 0.75S p 0.75 120 90 kpsi
Eq. (8-37):
a
CP 0.2911.244
1.703 kpsi
2 At
2 0.106 3
Chap. 8 Solutions - Rev. A, Page 46/69
m
Eq. (8-38)
CP
i 1.703 90 91.70 kpsi
2 At
(a) Goodman: From Table 8-9, S ut = 150 kpsi, and from Table 8-17, S e = 23.2 kpsi
S S i
23.2 150 90
Eq. (8-45):
n f e ut
4.72
Ans.
a Sut Se 1.703 150 23.2
(b) Gerber:
Eq. (8-46):
1
nf
Sut Sut2 4Se Se i Sut2 2 i Se
2 a Se
1
150 1502 4 23.2 23.2 90 1502 2 90 23.2
2 1.703 23.2
7.28
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf
Se
S p S p2 Se2 i2 i Se
2
2
a S p Se
23.2
120 120 2 23.2 2 902 90 23.2 7.24
2
2
1.703 120 18.6
Ans.
______________________________________________________________________________
8-55 From Prob. 8-51, C = 0.263, S e = 140 MPa, S ut = 900 MPa,
487.5 MPa, and P max = 4.712 kN.
A t = 84.4 mm2, i =
P min = P max / 2 = 4.712/2 = 2.356 kN
Eq. (8-35):
a
C Pmax Pmin
2 At
0.263 4.712 2.356 103
2 84.3
3.675 MPa
Eq. (8-36):
Chap. 8 Solutions - Rev. A, Page 47/69
m
C Pmax Pmin
i
2 At
0.263 4.712 2.356 103
487.5 498.5 MPa
2 84.3
Eq. (8-38):
Se Sut i
140 900 487.5
11.9
Ans.
Sut a Se m i 900 3.675 140 498.5 487.5
______________________________________________________________________________
nf
8-56 From Prob. 8-52, C = 0.299, S e = 18.8 kpsi, S ut = 120 kpsi, A t = 0.141 9 in2, i = 63.75
kpsi, and P max = 1.443 kips
P min = P max / 2 = 1.443/2 = 0.722 kips
a
Eq. (8-35):
Eq. (8-36):
m
C Pmax Pmin 0.299 1.443 0.722
0.760 kpsi
2 At
2 0.141 9
C Pmax Pmin
i
2 At
0.299 1.443 0.722
63.75 66.03 kpsi
2 0.141 9
Eq. (8-38):
Se Sut i
18.8 120 63.75
7.89
Ans.
Sut a Se m i 120 0.760 18.8 66.03 63.75
______________________________________________________________________________
nf
8-57 From Prob. 8-53, C = 0.228, S e = 162 MPa, S ut = 1040 MPa, A t = 58.0 mm2, i = 622.5
MPa, and P max = 7.679 kN.
P min = P max / 2 = 7.679/2 = 3.840 kN
Eq. (8-35):
a
C Pmax Pmin
2 At
0.228 7.679 3.840 103
2 58.0
7.546 MPa
Chap. 8 Solutions - Rev. A, Page 48/69
Eq. (8-36):
m
C Pmax Pmin
i
2 At
0.228 7.679 3.840 103
622.5 645.1 MPa
2 58.0
Eq. (8-38):
Se Sut i
162 1040 622.5
5.88
Ans.
Sut a Se m i 1040 7.546 162 645.1 622.5
______________________________________________________________________________
nf
8-58 From Prob. 8-54, C = 0.291, S e = 23.2 kpsi, S ut = 150 kpsi, A t = 0.106 3 in2, i = 90
kpsi, and P max = 1.244 kips
P min = P max / 2 = 1.244/2 = 0.622 kips
a
Eq. (8-35):
Eq. (8-36):
m
C Pmax Pmin 0.2911.244 0.622
0.851 kpsi
2 At
2 0.106 3
C Pmax Pmin
i
2 At
0.2911.244 0.622
90 92.55 kpsi
2 0.106 3
Eq. (8-38):
Se Sut i
23.2 150 90
7.45
Ans.
Sut a Se m i 150 0.851 23.2 92.55 90
______________________________________________________________________________
nf
8-59 Let the repeatedly-applied load be designated as P. From Table A-22, S ut = 93.7 kpsi.
Referring to the Figure of Prob. 3-122, the following notation will be used for the radii of
Section AA.
r i = 1.5 in, r o = 2.5 in, r c = 2.0 in
From Table 3-4, p. 121, with R = 0.5 in
Chap. 8 Solutions - Rev. A, Page 49/69
rn
e
co
ci
A
R2
2 rc
rc2 R 2
0.52
2 2
22 0.52
1.968 246 in
rc rn 2.0 1.968 246 0.031 754 in
ro - rn 2.5 1.968 246 0.531 754 in
rn - ri 1.968 246 1.5 0.468 246 in
(12 ) / 4 0.7854 in 2
If P is the maximum load
M Prc 2 P
P
rc
P
2(0.468)
i 1 c i
1
26.29P
A
eri 0.785 4
0.031 754(1.5)
26.294P
a m i
13.15P
2
2
(a) Eye: Section AA,
Table 6-2, p. 288, a = 14.4 kpsi, b = 0.718
Eq. (6-19), p. 287,
k a 14.4(93.7) 0.718 0.553
Eq. (6-23), p. 289,
d e = 0.370 d
Eq. (6-20), p. 288,
0.37
kb
0.30
0.107
0.978
Eq. (6-26), p. 290,
Eq. (6-8), p. 282,
k c = 0.85
Se 0.5Sut 0.5 93.7 46.85 kpsi
Eq. (6-18) p. 287,
S e = 0.553(0.978)0.85(46.85) = 21.5 kpsi
From Table 6-7, p. 307, for Gerber
2
2
2 m Se
1 Sut a
nf
1 1
2 m Se
Sut a
With m = a ,
2
2
2S e 1
1 Sut2
93.7 2
2(21.5) 1.557
1 1
1 1
nf
2 a Se
P
93.7
Sut 2 13.15P(21.5)
where P is in kips.
Chap. 8 Solutions - Rev. A, Page 50/69
Thread: Die cut. Table 8-17 gives S e = 18.6 kpsi for rolled threads. Use Table 8-16 to find
S e for die cut threads
S e = 18.6(3.0/3.8) = 14.7 kpsi
Table 8-2, A t = 0.663 in2, = P/A t = P /0.663 = 1.51 P, a = m = /2 = 0.755 P
From Table 6-7, Gerber
2
2
2Se 1
1 Sut2
93.7 2
2(14.7) 19.01
1 1
nf
1
1
2 a Se
S
2
0.755
P
(14.7)
93.7
P
ut
Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans.
(b) Strengthening steps can include heat treatment, cold forming, cross section change (a
round is a poor cross section for a curved bar in bending because the bulk of the material
is located where the stress is small). Ans.
(c) For n f = 2
P
1.557 103
779 lbf, max. load Ans.
2
______________________________________________________________________________
8-60 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in
km
0.5774 16 0.75
0.5774 Ed
13.32 Mlbf/in
0.5774l 0.5d
0.5774 1.5 0.5 0.75
2 ln 5
2 ln 5
0.5774l 2.5d
0.5774 1.5 2.5 0.75
Bolt, Eq. (8-13),
L T = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in
l = 1.5 in
l d = L L T = 2.5 1.75 = 0.75 in
l t = l l d = 1.5 0.75 = 0.75 in
Table 8-2,
Eq. (8-17),
A t = 0.373 in2
A d = (0.752)/4 = 0.442 in2
Chap. 8 Solutions - Rev. A, Page 51/69
kb
0.442 0.373 30
Ad At E
8.09 Mlbf/in
Ad lt At ld 0.442 0.75 0.373 0.75
C
kb
8.09
0.378
kb k m 8.09 13.32
Eq. (8-35),
a
C Pmax Pmin 0.378 6 4
1.013 kpsi
2 At
2 0.373
m
C Pmax Pmin Fi 0.378 6 4
25
72.09 kpsi
2 At
At
2 0.373
0.373
Eq.(8-36),
(a) From Table 8-9, S p = 85 kpsi, and Eq. (8-51), the yielding factor of safety is
np
Sp
m a
85
1.16
72.09 1.013
Ans.
(b) From Eq. (8-29), the overload factor of safety is
nL
S p At Fi
CPmax
85 0.373 25
2.96
0.378 6
Ans.
(c) From Eq. (8-30), the factor of safety based on joint separation is
n0
Fi
25
6.70
Pmax 1 C 6 1 0.378
Ans.
(d) From Table 8-17, S e = 18.6 kpsi; Table 8-9, S ut = 120 kps; the preload stress is
i = F i / A t = 25/0.373 = 67.0 kpsi; and from Eq. (8-38)
Se Sut i
18.6 120 67.0
4.56
Ans.
Sut a Se m i 120 1.013 18.6 72.09 67.0
______________________________________________________________________________
nf
8-61 (a) Table 8-2,
A t = 0.1419 in2
Table 8-9,
S p = 120 kpsi, S ut = 150 kpsi
Table 8-17,
S e = 23.2 kpsi
i = 0.75 S p = 0.75(120) = 90 kpsi
Eqs. (8-31) and (8-32),
Chap. 8 Solutions - Rev. A, Page 52/69
kb
4
0.2
kb km
4 16
CP
0.2 P
0.705P kpsi
a
2 At
2(0.141 9)
C
Eq. (8-45) for the Goodman criterion,
S S i
23.2(150 90)
11.4
n f e ut
2
a Sut Se 0.705P(150 23.2)
P
P 5.70 kips
Ans.
(b) F i = 0.75A t S p = 0.75(0.141 9)120 = 12.77 kips
Yield, Eq. (8-28),
S p At
120 0.141 9
np
1.22
Ans.
CP Fi 0.2 5.70 12.77
Load factor, Eq. (8-29),
S A - Fi 120(0.141 9) 12.77
nL p t
3.74 Ans.
CP
0.2(5.70)
Separation load factor, Eq. (8-30)
Fi
12.77
2.80 Ans.
P(1 - C ) 5.70(1 0.2)
______________________________________________________________________________
n0
8-62 Table 8-2, A t = 0.969 in2 (coarse), A t = 1.073 in2 (fine)
Table 8-9,
S p = 74 kpsi, S ut = 105 kpsi
Table 8-17,
S e = 16.3 kpsi
Coarse thread,
F i = 0.75 A t S p = 0.75(0.969)74 = 53.78 kips
i = 0.75 S p = 0.75(74) = 55.5 kpsi
CP
0.30 P
a
0.155P kpsi
2 At
2(0.969)
Gerber, Eq. (8-46),
nf
1
S S 2 4 S S S 2 2 S
ut
ut
e
e
i
ut
i e
2 a Se
1
105 1052 4 16.3 16.3 55.5 1052 2 55.5 16.3 64.28
2 0.155 P 16.3
P
With n f =2,
Chap. 8 Solutions - Rev. A, Page 53/69
P
64.28
32.14 kip
2
Ans.
Fine thread,
F i = 0.75 A t S p = 0.75(1.073)74 = 59.55kips
i = 0.75 S p = 0.75(74) = 55.5 kpsi
CP
0.32 P
a
0.149P kpsi
2 At
2(1.073)
The only thing that changes in Eq. (8-46) is a . Thus,
0.155 64.28 66.87
nf
2 P 33.43 kips
P
0.149 P
Ans.
Percent improvement,
33.43 32.14
(100) 4% Ans.
32.14
______________________________________________________________________________
8-63 For an M 30 × 3.5 ISO 8.8 bolt with P = 65 kN/bolt and C = 0.28
Table 8-1,
Table 8-11,
Table 8-17,
A t = 561 mm2
S p = 600 MPa, S ut = 830 MPa
S e = 129 MPa
Eq. (8-31),
F i = 0.75F p = 0.75 A t S p
= 0.75(5610600(103) = 252.45 kN
i = 0.75 S p = 0.75(600) = 450 MPa
Eq. (8-39),
a
CP 0.28 65 10
16.22 MPa
2 At
2 561
3
Gerber, Eq. (8-46),
nf
1
S S 2 4 S S S 2 2 S
ut
ut
e
e
i
ut
i e
2 a Se
1
830 830 2 4 129 129 450 830 2 2 450 129
2 16.22 129
4.75
Ans.
The yielding factor of safety, from Eq. (8-28) is
Chap. 8 Solutions - Rev. A, Page 54/69
np
S p At
CP Fi
600 561103
0.28 65 252.45
1.24
Ans.
From Eq. (8-29), the load factor is
nL
S p At Fi
CP
600 561103 252.45
4.62
0.28 65
Ans.
The separation factor, from Eq. (8-30) is
Fi
252.45
5.39
Ans.
P 1 C 65 1 0.28
______________________________________________________________________________
n0
A t = 0.077 5 in2
S p = 85 kpsi, S ut = 120 kpsi
S e = 18.6 kpsi
8-64 (a) Table 8-2,
Table 8-9,
Table 8-17,
Unthreaded grip,
kb
Am
Ad E
(0.375) 2 (30)
0.245 Mlbf/in per bolt
l
4(13.5)
[( D 2t )2 - D 2 ]
(4.752 - 42 ) 5.154 in 2
4
4
Am E
5.154(30) 1
km
2.148 Mlbf/in/bolt.
l
12
6
(b)
Ans.
Ans.
F i = 0.75 A t S p = 0.75(0.0775)(85) = 4.94 kip
i 0.75S p 0.75(85) 63.75 kpsi
2000
2
P pA
(4) 4189 lbf/bolt
6 4
kb
0.245
C
0.102
kb k m
0.245 2.148
CP
0.102(4.189)
a
2.77 kpsi
2 At
2(0.0775)
From Eq. (8-46) for Gerber fatigue criterion,
nf
1
S S 2 4 S S S 2 2 S
ut
ut
e
e
i
ut
i e
2 a Se
1
120 120 2 4 18.6 18.6 63.75 120 2 2 63.75 18.6 4.09
2 2.77 18.6
Ans.
Chap. 8 Solutions - Rev. A, Page 55/69
(c) Pressure causing joint separation from Eq. (8-30)
Fi
1
P(1 C )
Fi
4.94
P
5.50 kip
1 C 1 0.102
P
5.50
6 2.63 kpsi Ans.
p
A (42 ) / 4
______________________________________________________________________________
n0
8-65 From the solution of Prob. 8-64, A t = 0.077 5 in2, S ut = 120 kpsi, S e = 18.6 kpsi, C =
0.102, i = 63.75 kpsi
P max = p max A = 2 (42)/4 = 25.13 kpsi, P min = p min A = 1.2 (42)/4 = 15.08 kpsi,
Eq. (8-35),
a
Eq. (8-36),
m
C Pmax Pmin 0.102 25.13 15.08
6.61 kpsi
2 At
2 0.077 5
C Pmax Pmin
0.102 25.13 15.08
i
63.75 90.21 kpsi
2 At
2 0.077 5
Eq. (8-38),
Se Sut i
18.6 120 63.75
nf
0.814
Sut a Se m i 120 6.61 18.6 90.21 63.75
Ans.
This predicts a fatigue failure.
______________________________________________________________________________
8-66 Members: S y = 57 kpsi, S sy = 0.577(57) = 32.89 kpsi.
Bolts: SAE grade 5, S y = 92 kpsi, S sy = 0.577(92) = 53.08 kpsi
Shear in bolts,
(0.252 )
0.0982 in 2
As 2
4
AS
0.0982(53.08)
Fs s sy
2.61 kips
n
2
Bearing on bolts,
A b = 2(0.25)0.25 = 0.125 in2
AS
0.125(92)
Fb b yc
5.75 kips
2
n
Bearing on member,
Chap. 8 Solutions - Rev. A, Page 56/69
0.125(57)
3.56 kips
2
Fb
Tension of members,
A t = (1.25 0.25)(0.25) = 0.25 in2
0.25(57)
7.13 kip
2
F min(2.61, 5.75, 3.56, 7.13) 2.61 kip
Ft
Ans.
The shear in the bolts controls the design.
______________________________________________________________________________
8-67 Members, Table A-20, S y = 42 kpsi
Bolts, Table 8-9, S y = 130 kpsi, S sy = 0.577(130) = 75.01 kpsi
Shear of bolts,
5 /16 2
As 2
0.1534 in 2
4
n
Fs
5
32.6 kpsi
As 0.1534
S sy
75.01
2.30
32.6
Ans.
Bearing on bolts,
A b = 2(0.25)(5/16) = 0.1563 in2
5
b
32.0 kpsi
0.1563
S
130
n y
Ans.
4.06
b 32.0
Bearing on members,
n
Tension of members,
Sy
b
42
1.31
32
Ans.
A t = [2.375 2(5/16)](1/4) = 0.4375 in2
t
5
11.4 kpsi
0.4375
Chap. 8 Solutions - Rev. A, Page 57/69
Sy
42
3.68
Ans.
t 11.4
______________________________________________________________________________
n
8-68 Members: Table A-20, S y = 490 MPa, S sy = 0.577(490) = 282.7 MPa
Bolts: Table 8-11, ISO class 5.8, S y = 420 MPa, S sy = 0.577(420) = 242.3 MPa
Shear in bolts,
(202 )
2
As 2
628.3 mm
4
AS
628.3(242.3)103
60.9 kN
Fs s sy
2.5
n
Bearing on bolts,
A b = 2(20)20 = 800 mm2
AS
800(420)103
134 kN
Fb b yc
2.5
n
Bearing on member,
800(490)103
Fb
157 kN
2.5
Tension of members,
A t = (80 20)(20) = 1 200 mm2
1 200(490)103
Ft
235 kN
2.5
F min(60.9, 134, 157, 235) 60.9 kN
Ans.
The shear in the bolts controls the design.
______________________________________________________________________________
8-69 Members: Table A-20, S y = 320 MPa
Bolts: Table 8-11, ISO class 5.8, S y = 420 MPa, S sy = 0.577(420) = 242.3 MPa
Shear of bolts,
A s = (202)/4 = 314.2 mm2
90 103
s
95.48 MPa
3 314.2
S
242.3
2.54 Ans.
n sy
95.48
s
Bearing on bolt,
A b = 3(20)15 = 900 mm2
Chap. 8 Solutions - Rev. A, Page 58/69
b
90 103
100 MPa
900
S
420
4.2 Ans.
n y
b
100
Bearing on members,
S
320
3.2 Ans.
n y
b
100
Tension on members,
90 103
F
t
46.15 MPa
A 15[190 3 20 ]
S
320
n y
6.93 Ans.
46.15
t
______________________________________________________________________________
8-70 Members: S y = 57 kpsi
Bolts: S y = 100 kpsi, S sy = 0.577(100) = 57.7 kpsi
Shear of bolts,
1/ 4 2
A 3
0.1473 in 2
4
F
5
s
33.94 kpsi
As 0.1473
n
Bearing on bolts,
S sy
s
57.7
1.70
33.94
Ans.
A b = 3(1/4)(5/16) = 0.2344 in2
b
n
Bearing on members,
Sy
b
F
5
21.3 kpsi
Ab
0.2344
100
4.69
21.3
Ans.
A b = 0.2344 in2 (From bearing on bolts calculation)
b = 21.3 kpsi (From bearing on bolts calculation)
Chap. 8 Solutions - Rev. A, Page 59/69
n
Sy
b
57
2.68
21.3
Ans.
Tension in members, failure across two bolts,
At
5
2.375 2 1/ 4 0.5859 in 2
16
F
5
8.534 kpsi
At 0.5859
Sy
57
6.68
n
Ans.
t 8.534
t
______________________________________________________________________________
8-71 By symmetry, the reactions at each support is 1.6 kN. The free-body diagram for the left
member is
M
M
B
0
1.6(250) 50 RA 0
RA 8 kN
A
0
200(1.6) 50 RB 0
RB 6.4 kN
Members: Table A-20, S y = 370 MPa
Bolts: Table 8-11, S y = 420 MPa, S sy = 0.577(420) = 242.3 MPa
Bolt shear,
As (122 ) 113.1 mm 2
4
F
8(103 )
max
70.73 MPa
As
113.1
S
242.3
3.43
n sy
70.73
A b = td = 10(12) = 120 mm2
8(103 )
b
66.67 MPa
120
S
370
5.55
n y
b
66.67
Bearing on member,
Chap. 8 Solutions - Rev. A, Page 60/69
Strength of member. The bending moments at the hole locations are:
in the left member at A, M A = 1.6(200) = 320 N · m. In the right member at B, M B =
8(50) = 400 N · m. The bending moment is greater at B
1
I B [10(503 ) 10(123 )] 102.7(103 ) mm 4
12
M c
400(25)
B A
(103 ) 97.37 MPa
3
IA
102.7(10 )
S
370
3.80
n y
A 97.37
At the center, call it point C,
M C = 1.6(350) = 560 N · m
1
IC
(10)(503 ) 104.2(103 ) mm 4
12
M c
560(25)
C C
(103 ) 134.4 MPa
IC
104.2(103 )
S
370
2.75 3.80 more critical at C
n y
C 134.4
n min(3.04, 3.80, 2.75) 2.72 Ans.
______________________________________________________________________________
8-72 The free-body diagram of the bracket, assuming the upper bolt takes all the shear and
tensile load is
F s = 2500 lbf
P
2500 3
1071 lbf
7
Table A-31, H = 7/16 = 0.4375 in. Grip, l = 2(1/2) = 1 in. L ≥ l + H = 1.4375 in. Use 1.5
in bolts.
Eq. (8-13),
L T = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in
Table 8-7,
l d = L L T = 1.5 1.25 = 0.25 in
Chap. 8 Solutions - Rev. A, Page 61/69
Table 8-2,
Eq. (8-17),
Eq. (8-22),
km
l t = l l d = 1 0.25 = 0.75 in
A t = 0.141 9 in2
A d = (0.52) /4 = 0.196 3 in2
0.196 3 0.141 9 30
Ad At E
kb
4.574 Mlbf/in
Ad lt At ld 0.196 3 0.75 0.141 9 0.25
0.5774 30 0.5
0.5774 Ed
16.65 Mlbf/in
0.5774 l 0.5d
0.5774 1 0.5 0.5
2 ln 5
2 ln 5
0.5774 l 2.5d
0.5774 1 2.5 0.5
kb
4.574
0.216
kb km 4.574 16.65
Table 8-9,
S p = 65 kpsi
Eqs. (8-31) and (8-32),
F i = 0.75 A t S p = 0.75(0.141 9)65 = 6.918 kips
i = 0.75 S p = 0.75(65) = 48.75 kips
C
CP Fi 0.216 1.071 6.918
50.38 kpsi
At
0.141 9
F
3
s s
21.14 kpsi
At 0.141 9
Eq. (a), p. 440, b
Direct shear,
von Mises stress, Eq. (5-15), p. 223
b2 3 s2 50.382 3 21.142
1/2
1/2
62.3 kpsi
Stress margin, m = S p = 65 62.3 = 3.7 kpsi Ans.
______________________________________________________________________________
8-73
2 P(200) 14(50)
14(50)
P
1.75 kN per bolt
2(200)
Fs 7 kN/bolt
S p 380 MPa
At 245 mm 2 , Ad
(202 ) 314.2 mm 2
4
Fi 0.75(245)(380)(103 ) 69.83 kN
i 0.75 380 285 MPa
Chap. 8 Solutions - Rev. A, Page 62/69
CP Fi 0.25(1.75) 69.83 3
(10 ) 287 MPa
At
245
3
F
7(10 )
22.3 MPa
s
Ad
314.2
[287 2 3(22.32 )]1/ 2 290 MPa
m S p 380 290 90 MPa
b
Ans.
Stress margin, m = S p = 380 90 = 90 MPa
______________________________________________________________________________
8-74 Using the result of Prob. 5-67 for lubricated assembly (replace 0.2 with 0.18 per Table
8-15)
2 f T
Fx
0.18d
With a design factor of n d gives
T
0.18nd Fx d
0.18(3)(1000)d
716d
2 f
2 (0.12)
or T/d = 716. Also,
T
K (0.75S p At )
d
0.18(0.75)(85 000) At
11 475 At
Form a table
Size
1
4 - 28
5
16
3
8
- 24
24
T/d = 11 475A t
n
At
0.0364
417.70 1.75
0.058
0.0878
665.55 2.8
1007.50 4.23
where the factor of safety in the last column of the table comes from
n
Select a
3"
8
2 f (T / d )
2 (0.12)(T / d )
0.0042(T / d )
0.18Fx
0.18(1000)
- 24 UNF cap screw. The setting is given by
T = (11 475A t )d = 1007.5(0.375) = 378 lbf · in
Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in.
Check the factor of safety
Chap. 8 Solutions - Rev. A, Page 63/69
2 f T
2 (0.12)(400)
4.47
0.18Fx d
0.18(1000)(0.375)
______________________________________________________________________________
n
8-75
Bolts, from Table 8-11, S y = 420 MPa
Channel, From Table A-20, S y = 170 MPa. From Table A-7, t = 6.4 mm
Cantilever, from Table A-20, S y = 190 MPa
F A = F B = F C = F / 3
M = (50 + 26 + 125) F = 201 F
FA FC
201F
2.01 F
2 50
1
(1)
FC FC FC 2.01 F 2.343F
3
Shear on Bolts: The shoulder bolt shear area, A s = (102) / 4 = 78.54 mm2
Max. force,
S sy = 0.577(420) = 242.3 KPa
max
FC S sy
As
n
From Eq. (1), F C = 2.343 F. Thus
F
S sy As 242.3 78.54 3
10 4.06 kN
n 2.343
2.0 2.343
Bearing on bolt: The bearing area is A b = td = 6.4(10) = 64 mm2. Similar to shear
Chap. 8 Solutions - Rev. A, Page 64/69
F
S y Ab 420 64 3
10 5.74 kN
n 2.343 2.0 2.343
Bearing on channel: A b = 64 mm2, S y = 170 MPa.
S A 170 64 3
F y b
10 2.32 kN
n 2.343 2.0 2.343
Bearing on cantilever: A b = 12(10) = 120 mm2, S y = 190 MPa.
F
S y Ab 190 120 3
10 4.87 kN
n 2.343 2.0 2.343
Bending of cantilever: At C
I
1
12 503 103 1.24 105 mm 4
12
max
Sy
n
Mc 151Fc
I
I
F
Sy I
n 151c
5
190 1.24 10 3
10 3.12 kN
F
2.0 151 25
So F = 2.32 kN based on bearing on channel. Ans.
______________________________________________________________________________
8-76 Bolts, from Table 8-11, S y = 420 MPa
Bracket, from Table A-20, S y = 210 MPa
12
4 kN; M 12(200) 2400 N · m
3
2400
FA FB
37.5 kN
64
FA FB (4) 2 (37.5) 2 37.7 kN
FO 4 kN
F
Bolt shear:
The shoulder bolt shear area, A s = (122) / 4 = 113.1 mm2
S sy = 0.577(420) = 242.3 KPa
Chap. 8 Solutions - Rev. A, Page 65/69
37.7(10)3
333 MPa
113
S
242.3
0.728
n sy
333
Ans.
Bearing on bolts:
Ab 12(8) 96 mm 2
37.7(10)3
b
393 MPa
96
S
420
n yc
1.07
Ans.
b
393
Bearing on member:
b 393 MPa
S
210
n yc
0.534
b
393
Ans.
Bending stress in plate:
bd 3
bh3 bd 3
2
a 2bd
12
12
12
3
3
3
8(12)
8(136)
8(12)
2
(32) 2 (8)(12)
12
12
12
6
4
1.48(10) mm
Ans.
2400(68)
Mc
(10)3 110 MPa
1.48(10)6
I
S
210
1.91
n y
Ans.
110
I
Failure is predicted for bolt shear and bearing on member.
______________________________________________________________________________
Chap. 8 Solutions - Rev. A, Page 66/69
8-77
FA 1208 125 1083 lbf,
Bolt shear:
3625
FA FB
1208 lbf
3
FB 1208 125 1333 lbf
A s = ( / 4)(0.3752) = 0.1104 in2
max
Fmax
1333
12 070 psi
0.1104
As
From Table 8-10, S y = 100 kpsi, S sy = 0.577(100) = 57.7 kpsi
n
S sy
max
57.7
4.78
12.07
Ans.
Bearing on bolt: Bearing area is A b = td = 0.375 (0.375) = 0.1406 in2.
b
n
Sy
b
F
1333
9 481 psi
0.1406
Ab
100
10.55
9.481
Ans.
Bearing on member: From Table A-20, S y = 54 kpsi. Bearing stress same as bolt
n
Sy
b
54
5.70
9.481
Ans.
Bending of member: At B, M = 250(13) = 3250 lbfin
Chap. 8 Solutions - Rev. A, Page 67/69
I
3
1 3 3 3
4
2
0.2484 in
12 8
8
Mc 3250 1
13 080 psi
I
0.2484
Sy
54
4.13
Ans.
13.08
______________________________________________________________________________
n
8-78 The direct shear load per bolt is F = 2000/6 = 333.3 lbf. The moment is taken only by the
four outside bolts. This moment is M = 2000(5) = 10 000 lbf · in.
10 000
1000 lbf and the resultant bolt load is
Thus F
2(5)
F
(333.3) 2 (1000) 2 1054 lbf
Bolt strength, Table 8-9, S y = 100 kpsi; Channel and Plate strength, S y = 42 kpsi
Shear of bolt: A s = (0.5)2/4 = 0.1963 in2
n
S sy
(0.577)(100)
10.7
1.054 / 0.1963
Ans.
Bearing on bolt: Channel thickness is t = 3/16 in, A b = 0.5(3/16) = 0.09375 in2
n
Bearing on channel:
Bearing on plate:
n
100
8.89
1.054 / 0.09375
42
3.74
1.054 / 0.09375
A b = 0.5(0.25) = 0.125 in2
n
42
4.98
1.054 / 0.125
Ans.
Ans.
Ans.
Strength of plate:
I
0.25(7.5)3 0.25(0.5)3
12
12
3
0.25(0.5)
2
0.25 0.5 (2.5) 2 7.219 in 4
12
Chap. 8 Solutions - Rev. A, Page 68/69
M 5000 lbf · in per plate
Mc 5000(3.75)
2597 psi
I
7.219
42
n
16.2 Ans.
2.597
______________________________________________________________________________
8-79 to 8-81 Specifying bolts, screws, dowels and rivets is the way a student learns about such
components. However, choosing an array a priori is based on experience. Here is a chance
for students to build some experience.
Chap. 8 Solutions - Rev. A, Page 69/69
Chapter 9
Figure for Probs.
9-1 to 9-4
9-1
Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hl allow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans.
______________________________________________________________________________
9-2
Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
F = 0.707 hl allow = 0.707(5/16)[2(2)](25) = 22.1 kip
Ans.
______________________________________________________________________________
9-3
Given, b = 50 mm, d = 30 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hl allow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans.
______________________________________________________________________________
9-4
Given, b = 4 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
F = 0.707 hl allow = 0.707(5/16)[2(4)](25) = 44.2 kip
Ans.
______________________________________________________________________________
9-5
Prob. 9-1 with E7010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in
= 2.923(4.45/25.4) = 0.512 kN/mm
F = f l = 0.512[2(50)] = 51.2 kN
Ans.
______________________________________________________________________________
9-6
Prob. 9-2 with E6010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in
Chapter 9, Page 1/36
F = f l = 4.64[2(2)] = 18.6 kip
Ans.
______________________________________________________________________________
9-7
Prob. 9-3 with E7010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in
= 2.923(4.45/25.4) = 0.512 kN/mm
F = f l = 0.512[2(50)] = 51.2 kN
Ans.
______________________________________________________________________________
9-8
Prob. 9-4 with E6010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in
F = f l = 4.64[2(4)] = 37.1 kip
Ans.
______________________________________________________________________________
9-9
Table A-20:
1018 CD: S ut = 440 MPa, S y = 370 MPa
1018 HR: S ut = 400 MPa, S y = 220 MPa
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
all min(0.30Sut , 0.40S y )
min[0.30(400), 0.40(220)]
min(120, 88) 88 MPa
for both materials.
Eq. (9-3):
F = 0.707hl all = 0.707(5)[2(50)](88)(103) = 31.1 kN Ans.
______________________________________________________________________________
9-10
Table A-20:
1020 CD: S ut = 68 kpsi, S y = 57 kpsi
1020 HR: S ut = 55 kpsi, S y = 30 kpsi
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
all min(0.30Sut , 0.40S y )
min[0.30(55), 0.40(30)]
min(16.5, 12.0) 12.0 kpsi
for both materials.
Eq. (9-3):
F = 0.707hl all = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans.
______________________________________________________________________________
Chapter 9, Page 2/36
9-11
Table A-20:
1035 HR: S ut = 500 MPa, S y = 270 MPa
1035 CD: S ut = 550 MPa, S y = 460 MPa
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
all min(0.30Sut , 0.40S y )
min[0.30(500), 0.40(270)]
min(150, 108) 108 MPa
for both materials.
Eq. (9-3):
F = 0.707hl all = 0.707(5)[2(50)](108)(103) = 38.2 kN Ans.
______________________________________________________________________________
9-12
Table A-20:
1035 HR: S ut = 72 kpsi, S y = 39.5 kpsi
1020 CD: S ut = 68 kpsi, S y = 57 kpsi, 1020 HR: S ut = 55 kpsi, S y = 30 kpsi
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
all min(0.30Sut , 0.40S y )
min[0.30(55), 0.40(30)]
min(16.5, 12.0) 12.0 kpsi
for both materials.
Eq. (9-3):
F = 0.707hl all = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans.
______________________________________________________________________________
9-13
2 100 103
2F
Ans.
141 MPa
hl
5 2 50 50
______________________________________________________________________________
Eq. (9-3):
Eq. (9-3):
9-14
2 40
2F
22.6 kpsi
hl
5 / 16 2 2 2
Ans.
______________________________________________________________________________
2 100 103
2F
Ans.
9-15
Eq. (9-3):
177 MPa
hl
5 2 50 30
______________________________________________________________________________
9-16
Eq. (9-3):
2 40
2F
15.1 kpsi
hl
5 / 16 2 4 2
Ans.
______________________________________________________________________________
Chapter 9, Page 3/36
9-17
b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 140 MPa.
(a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem
figure and table figure. Note, also, F in kN and in MPa):
F 103
V
2.829 F
y
A 1.414 5 50
Secondary shear, Table 9-1:
Ju
d 3b 2 d 2
6
50 3 502 502
6
83.33 103 mm3
J = 0.707 h J u = 0.707(5)(83.33)(103) = 294.6(103) mm4
x y
Mry
J
175 F 103 25
294.6 103
14.85 F
max x2 y y F 14.852 2.829 14.85 23.1F (1)
2
F
allow
23.1
2
140
6.06 kN Ans.
23.1
(b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa
1020 HR bar:
S ut = 380 MPa, S y = 210 MPa
1015 HR support: S ut = 340 MPa, S y = 190 MPa
Table 9-3, E7010 Electrode: S ut = 482 MPa, S y = 393 MPa
The support controls the design.
Table 9-4:
allow = min(0.30S ut , 0.40S y ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
allow
76
3.29 kN
Ans.
23.1 23.1
______________________________________________________________________________
F
9-18
b = d =2 in, c = 6 in, h = 5/16 in, and allow = 25 kpsi.
Chapter 9, Page 4/36
(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and in kpsi):
V
F
1.132 F
A 1.414 5 /16 2
Secondary shear, Table 9-1:
2
2
d 3b 2 d 2 2 3 2 2
5.333 in 3
Ju
6
6
y
J = 0.707 h J u = 0.707(5/16)(5.333) = 1.178 in4
x y
Mry
J
7 F 1
5.942 F
1.178
max x2 y y F 5.9422 1.132 5.942 9.24 F (1)
2
F
allow
9.24
2
25
2.71 kip Ans.
9.24
(b) For E7010 from Table 9-6, allow = 21 kpsi
1020 HR bar:
S ut = 55 kpsi, S y = 30 kpsi
1015 HR support: S ut = 50 kpsi, S y = 27.5 kpsi
Table 9-3, E7010 Electrode: S ut = 70 kpsi, S y = 57 kpsi
The support controls the design.
Table 9-4:
allow = min(0.30S ut , 0.40S y ) =min[0.30(50), 0.40(27.5) = min(15, 11)
= 11 kpsi
The allowable load, from Eq. (1) is
allow
11
1.19 kip
Ans.
9.24 9.24
______________________________________________________________________________
F
9-19
b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and allow = 140 MPa.
(a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem
figure and table figure. Note, also, F in kN and in MPa):
Chapter 9, Page 5/36
F 103
V
2.829 F
A 1.414 5 50
y
Secondary shear, Table 9-1:
Ju
d 3b 2 d 2
6
50 3 302 502
6
43.33 103 mm3
J = 0.707 h J u = 0.707(5)(43.33)(103) = 153.2(103) mm4
x
Mry
J
175F 103 15
153.2 103
17.13F
Mrx 175F 10 25
28.55 F
J
153.2 103
3
y
max x2 y y F 17.132 2.829 28.55 35.8F (1)
2
F
allow
35.8
2
140
3.91 kN Ans.
35.8
(b) For E7010 from Table 9-6, allow = 21 kpsi = 21(6.89) = 145 MPa
1020 HR bar:
S ut = 380 MPa, S y = 210 MPa
1015 HR support: S ut = 340 MPa, S y = 190 MPa
Table 9-3, E7010 Electrode: S ut = 482 MPa, S y = 393 MPa
The support controls the design.
Table 9-4:
allow = min(0.30S ut , 0.40S y ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
allow
76
2.12 kN
Ans.
35.8 35.8
______________________________________________________________________________
F
9-20
b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and allow = 25 kpsi.
Chapter 9, Page 6/36
(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and in kpsi):
V
F
0.5658F
A 1.414 5 /16 4
Secondary shear, Table 9-1:
y
Ju
d 3b 2 d 2
6
4 3 22 42
6
18.67 in 3
J = 0.707 h J u = 0.707(5/16)(18.67) = 4.125 in4
x
y
Mry
J
8 F 1
1.939 F
4.125
Mrx 8 F 2
3.879 F
J
4.125
max x2 y y F 1.9392 0.5658 3.879 4.85F (1)
2
F
allow
4.85
2
25
5.15 kip Ans.
4.85
(b) For E7010 from Table 9-6, allow = 21 kpsi
1020 HR bar:
S ut = 55 kpsi, S y = 30 kpsi
1015 HR support: S ut = 50 kpsi, S y = 27.5 kpsi
Table 9-3, E7010 Electrode: S ut = 70 kpsi, S y = 57 kpsi
The support controls the design.
Table 9-4:
allow = min(0.30S ut , 0.40S y ) =min[0.30(50), 0.40(27.5) = min(15, 11)
= 11 kpsi
The allowable load, from Eq. (1) is
allow
11
2.27 kip
Ans.
4.85 4.85
______________________________________________________________________________
F
Chapter 9, Page 7/36
9-21
Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
Primary shear (F in kN, in MPa, A in mm2):
F 103
V
y
1.414 F
A 1.414 5 50 50
Secondary shear:
Table 9-1:
Ju
b d
3
50 50
3
166.7 103 mm3
6
6
J = 0.707 h J u = 0.707(5)166.7(103) = 589.2(103) mm4
x y
Mry
J
175 F 103 (25)
589.2 103
7.425 F
Maximum shear:
max x2 y y F 7.4252 1.414 7.425 11.54 F
2
allow
2
140
12.1 kN Ans.
11.54 11.54
______________________________________________________________________________
F
9-22
Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
Primary shear:
y
V
F
0.5658 F
A 1.414 5 /16 2 2
Secondary shear:
Table 9-1:
Ju
b d
3
2 2
3
10.67 in 3
6
6
J = 0.707 h J u = 0.707(5/16)10.67 = 2.357 in4
x y
Mry
J
7 F (1)
2.970 F
2.357
Maximum shear:
max x2 y y F 2.9702 0.566 2.970 4.618 F
2
allow
2
25
5.41 kip Ans.
4.618 4.618
______________________________________________________________________________
F
9-23
Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa.
Chapter 9, Page 8/36
Primary shear (F in kN, in MPa, A in mm2):
F 103
V
y
1.768 F
A 1.414 5 50 30
Secondary shear:
Table 9-1:
Ju
b d
3
50 30
3
85.33 103 mm3
6
6
J = 0.707 h J u = 0.707(5)85.33(103) = 301.6(103) mm4
x
y
Mry
J
175 F 103 (15)
301.6 103
8.704 F
3
Mrx 175 F 10 (25)
14.51F
J
301.6 103
Maximum shear:
max x2 y y F 8.7042 1.768 14.51 18.46 F
2
allow
2
140
7.58 kN Ans.
18.46 18.46
______________________________________________________________________________
F
9-24
Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
Primary shear:
y
V
F
0.3772 F
A 1.414 5 /16 4 2
Secondary shear:
Table 9-1:
Ju
b d
3
4 2
3
36 in 3
6
6
J = 0.707 h J u = 0.707(5/16)36 = 7.954 in4
x
y
Mry
J
8 F (1)
1.006 F
7.954
Mrx 8 F (2)
2.012 F
J
7.954
Maximum shear:
max x2 y y F 1.0062 0.3772 2.012 2.592 F
2
2
Chapter 9, Page 9/36
allow
25
9.65 kip Ans.
2.592 2.592
______________________________________________________________________________
F
9-25
Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode.
A = 0.707(5)(50 +50 + 50) = 530.3 mm2
Member endurance limit: From Table A-20 for AISI 1010 HR, S ut = 320 MPa.
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 272(320)0.995 = 0.875
k b = 1 (uniform shear), k c = 0.59 (torsion, shear), k d = 1
Eqs. (6-8) and (6-18):
S e = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa
Electrode endurance: E6010, Table 9-3,
S ut = 427 MPa
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 272(427)0.995 = 0.657
As before, k b = 1 (direct shear), k c = 0.59 (torsion, shear), k d = 1
S e = 0.657(1)(0.59)(1)(0.5)(427) = 82.8 MPa
The members and electrode are basically of equal strength. We will use S e = 82.6 MPa.
For a factor of safety of 1, and with K fs = 2.7 (Table 9-5)
A 82.6 530.3
F allow
16.2 103 N 16.2 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-26
Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode.
A = 0.707(5/16)(2 +2 + 2) = 1.326 in2
Member endurance limit: From Table A-20 for AISI 1010 HR, S ut = 47 kpsi.
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 39.9(47)0.995 = 0.865
k b = 1 (uniform shear), k c = 0.59 (torsion, shear), k d = 1
Eqs. (6-8) and (6-18):
S e = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi
Electrode endurance: E6010, Table 9-3,
S ut = 62 kpsi
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 39.9(62)0.995 = 0.657
Chapter 9, Page 10/36
As before, k b = 1 (uniform shear), k c = 0.59 (torsion, shear), k d = 1
S e = 0.657(1)(0.59)(1)(0.5)(62) = 12.0 kpsi
Thus the members and electrode are of equal strength. For a factor of safety of 1, and
with K fs = 2.7 (Table 9-5)
A 12.0 1.326
F allow
5.89 kip
Ans.
K fs
2.7
______________________________________________________________________________
9-27
Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode.
A = 0.707(5)(50 +50 + 30) = 459.6 mm2
Member endurance limit: From Table A-20 for AISI 1010 HR, S ut = 320 MPa.
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 272(320)0.995 = 0.875
k b = 1 (direct shear), k c = 0.59 (torsion, shear), k d = 1
Eqs. (6-8) and (6-18):
S e = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa
Electrode endurance: E6010, Table 9-3,
S ut = 482 MPa
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 272(482)0.995 = 0.582
As before, k b = 1 (direct shear), k c = 0.59 (torsion, shear), k d = 1
S e = 0.582(1)(0.59)(1)(0.5)(482) = 82.7 MPa
The members and electrode are basically of equal strength. We will use S e =82.6 MPa.
For a factor of safety of 1, and with K fs = 2.7 (Table 9-5)
A 82.6 459.6
F allow
14.1103 N 14.1 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-28
Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode.
A = 0.707(5/16)(4 +4 + 2) = 2.209 in2
Member endurance limit: From Table A-20 for AISI 1010 HR, S ut = 47 kpsi.
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 39.9(47)0.995 = 0.865
k b = 1 (direct shear), k c = 0.59 (torsion, shear), k d = 1
Chapter 9, Page 11/36
S e = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi
Eqs. (6-8) and (6-18):
Electrode endurance: E7010, Table 9-3,
S ut = 70 kpsi
Eq. 6-19 and Table 6-2, pp. 287, 288:
k a = 39.9(70)0.995 = 0.582
As before, k b = 1 (direct shear), k c = 0.59 (torsion, shear), k d = 1
S e = 0.582(1)(0.59)(1)(0.5)(70) = 12.0 kpsi
Thus the members and electrode are of equal strength. For a factor of safety of 1, and
with K fs = 2.7 (Table 9-5)
A 12.0 2.209
F allow
9.82 kip
Ans.
K fs
2.7
______________________________________________________________________________
9-29
= 0 (why?)
Primary shear:
Secondary shear:
Table 9-1: J u = 2 r3 = 2 (1.5)3 = 21.21 in3
J = 0.707 h J u = 0.707(1/4)(21.21) = 3.749 in4
2 welds:
Mr 8 F 1.5
1.600 F
2 J 2 3.749
allow
1.600 F 20 F 12.5 kip
Ans.
______________________________________________________________________________
9-30
l = 2 + 4 + 4 = 10 in
2 1 4 0 4 2
1 in
10
2 4 4 2 4 0
y
1.6 in
10
x
M = FR = F(10 1) = 9 F
r1
1 1 4 1.6
r3
2 1
2
2
2
2.4 in,
r2 12 2 1.6 1.077 in
2
1.62 1.887 in
Chapter 9, Page 12/36
1
0.707 5 /16 23 0.1473 in 4
12
1
J G3 0.707 5 / 16 43 1.178 in 4
12
J G1
J G2
3
J J i Ai rG2i
i 1
0.1473 0.707 5 /16 2 2.42 1.178 0.707 5 /16 4 1.077 2
1.178 0.707 5 /16 4 1.887 2 9.220 in 4
1.6
o
28.07
4
1
tan 1
r 1.62 4 1 3.4 in
2
Primary shear ( in kpsi, F in kip) :
V
F
0.4526 F
A 0.707 5 /16 10
Secondary shear:
Mr 9 F 3.4
3.319 F
J
9.220
max
3.319 F sin 28.07 3.319 F cos 28.07
o 2
o
0.4526 F
2
3.724 F
max = allow
3.724 F = 25 F = 6.71 kip
Ans.
______________________________________________________________________________
Chapter 9, Page 13/36
9-31
l = 30 + 50 + 50 = 130 mm
30 15 50 0 50 25
13.08 mm
130
30 50 50 25 50 0
y
21.15 mm
130
x
M = FR = F(200 13.08)
= 186.92 F (M in Nm, F in kN)
r1
15 13.08 50 21.15
r3
25 13.08
2
2
28.92 mm, r2 13.082 25 21.15 13.63 mm
2
2
21.152 24.28 mm
1
0.707 5 303 7.954 103 mm 4
12
1
J G3 0.707 5 503 36.82 103 mm 4
12
J G1
J G2
3
J J i Ai rG2i
i 1
7.954 103 0.707 5 30 28.922 36.82 103 0.707 5 50 13.632
36.82 103 0.707 5 50 24.282 307.3 103 mm 4
21.15
o
29.81
50 13.08
tan 1
r 21.152 50 13.08 42.55 mm
2
Primary shear ( in MPa, F in kN) :
F 103
V
2.176 F
A 0.707 5 130
Secondary shear:
3
Mr 186.92 F 10 42.55
25.88 F
J
307.3 103
Chapter 9, Page 14/36
max
25.88F sin 29.81 25.88F cos 29.81
o 2
o
2.176 F
2
27.79 F
max = allow 27.79 F = 140 F = 5.04 kN
Ans.
______________________________________________________________________________
9-32
Weld
Pattern
1.
Figure of merit
a2
J
a 3 /12 a 2
fom u
0.0833
12h
lh
ah
h
a 3a 2 a 2
a2
a2
0.3333
3h
h
Rank______
5
2.
fom
3.
2a 6a 2 a 2 5a 2 0.2083 a 2
fom
12 a a 2ah 24h
h
4
4.
a2
a4
1 8a 3 6 a 3 a 3
fom
0.3056
3ah
12
2a a
h
2
5.
fom
6 2a h
1
4
2a
3
a2
1
8a 3
0.3333
6h 4a 24ah
h
1
2 a / 2
a2
a3
3
0.25
ah
4ah
h
______________________________________________________________________________
3
6.
fom
Chapter 9, Page 15/36
9-33
Weld
Pattern
1.
2.
3.
4.*
5. & 7.
6. & 8.
Figure of merit
3
a2
I u a / 12
fom
0.0833
lh
ah
h
3
a / 6 0.0833 a 2
fom
2ah
h
aa 2 / 2
a2
fom
0.25
2ah
h
a 2 / 12 6a a 7 a 2
a2
fom
0.1944
3ah
36h
h
a
a2
a
x , y
2
a 2a 3
2
2a 3
a3
a
2 a
Iu
2a a 2a
3
3
3
3
3
a2
I u a / 3 1 a 2
fom
0.1111
lh
3ah
9 h
h
2
a / 6 3a a 1 a 2 0.1667 a 2
fom
4ah
6 h
h
a / 2 a2
a2
fom
0.125
ah
8h
h
Rank______
6
6
1
2
5
3
3
9.
4
*Note. Because this section is not symmetric with the vertical axis, out-of-plane
deflection may occur unless special precautions are taken. See the topic of “shear center”
in books with more advanced treatments of mechanics of materials.
______________________________________________________________________________
9-34
Attachment and member (1018 HR), S y = 220 MPa and S ut = 400 MPa.
The member and attachment are weak compared to the properties of the lowest electrode.
Decision Specify the E6010 electrode
Controlling property, Table 9-4: all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa
For a static load, the parallel and transverse fillets are the same. Let the length of a bead
be l = 75 mm, and n be the number of beads.
Chapter 9, Page 16/36
F
all
n 0.707 hl
100 103
F
21.43
nh
0.707l all 0.707 75 88
where h is in millimeters. Make a table
Number of beads, n
1
2
3
4
Decision
Leg size, h (mm)
21.43
10.71
7.14
5.36 6 mm
Specify h = 6 mm on all four sides.
Weldment specification:
Pattern: All-around square, four beads each side, 75 mm long
Electrode: E6010
Leg size: h = 6 mm
______________________________________________________________________________
9-35
Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an
optimal pattern (see Prob. 9-32) and have thus reduced a synthesis problem to an analysis
problem:
Table 9-1, case 2, rotated 90: A = 1.414hd = 1.414(h)(75) = 106.05h mm2
Primary shear
12 103 113.2
V
y
A 106.05h
h
Secondary shear:
d (3b 2 d 2 )
6
75[3(752 ) 752 ]
281.3 103 mm3
6
J 0.707(h)(281.3) 103 198.8 103 h mm 4
Ju
With = 45,
Chapter 9, Page 17/36
x
3
Mry 12 10 (187.5)(37.5) 424.4
Mr cos 45o
y
J
J
h
198.8 103 h
max x 2 y y
2
1
684.9
424.42 (113.2 424.4) 2
h
h
Attachment and member (1018 HR): S y = 220 MPa, S ut = 400 MPa
Decision: Use E60XX electrode which is stronger
all min[0.3(400), 0.4(220)] 88 MPa
max all
h
684.9
88 MPa
h
684.9
7.78 mm
88
Decision: Specify 8 mm leg size
Weldment Specifications:
Pattern: Parallel horizontal fillet welds
Electrode: E6010
Type: Fillet
Length of each bead: 75 mm
Leg size: 8 mm
______________________________________________________________________________
9-36
Problem 9-35 solves the problem using parallel horizontal fillet welds, each 75 mm long
obtaining a leg size rounded up to 8 mm.
For this problem, since the width of the plate is fixed and the length has not been
determined, we will explore reducing the leg size by using two vertical beads 75 mm long
and two horizontal beads such that the beads have a leg size of 6 mm.
Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table
9-1 with b unknown and d = 75 mm).
Materials:
Attachment and member (1018 HR): S y = 220 MPa, S ut = 400 MPa
From Table 9-4, AISC welding code,
all = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa
Select a stronger electrode material from Table 9-3.
Decision: Specify E6010
Solving for b: In Prob. 9-35, every term was linear in the unknown h. This made solving
for h relatively easy. In this problem, the terms will not be linear in b, and so we will use
an iterative solution with a spreadsheet.
Throat area and other properties from Table 9-1:
A = 1.414(6)(b + 75) = 8.484(b + 75)
(1)
Chapter 9, Page 18/36
Ju
b 75
6
3
, J = 0.707 (6) J u = 0.707(b +75)3
(2)
Primary shear ( in MPa, h in mm):
y
12 103
V
A
A
(3)
Secondary shear (See Prob. 9-35 solution for the definition of ) :
Mr
J
3
Mry 12 10 150 b / 2 (37.5)
Mr
x cos
cos
3
J
J
0.707 b 75
3
Mr
Mrx 12 10 150 b / 2 (b / 2)
y sin
sin
3
J
J
0.707 b 75
max y 2 x y
2
(4)
(5)
(6)
Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of
the spreadsheet is shown below.
b (mm)
41
42
43
44
A (mm2)
984.144
992.628
1001.112
1009.596
J (mm4)
1103553.5
1132340.4
1161623.6
1191407.4
' y (Mpa) " y (Mpa) " x (Mpa)
12.19334
12.08912
11.98667
11.88594
69.5254
67.9566
66.43718
64.96518
38.00722
38.05569
38.09065
38.11291
max
(Mpa)
90.12492
88.63156
87.18485 < 88 Mpa
85.7828
We see that b 43 mm meets the strength goal.
Weldment Specifications:
Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mm
long
Electrode: E6010
Leg size: 6 mm
______________________________________________________________________________
9-37
Materials:
Member and attachment (1018 HR):
Table 9-4:
S y 32 kpsi,
Sut 58 kpsi
all min[0.3(58), 0.4(32)] 12.8 kpsi
Chapter 9, Page 19/36
Decision: Use E6010 electrode. From Table 9-3: S y 50 kpsi, Sut 62 kpsi,
all min[0.3(62), 0.4(50)] 20 kpsi
Decision: Since 1018 HR is weaker than the E6010 electrode, use all 12.8 kpsi
Decision: Use an all-around square weld bead track.
l 1 = 6 + a = 6 + 6.25 = 12.25 in
Throat area and other properties from Table 9-1:
A 1.414h(b d ) 1.414( h)(6 6) 16.97 h
Primary shear
3
1179
V F 20 10
psi
y
A A 16.97 h
h
Secondary shear
(b d )3 (6 6)3
288 in 3
6
6
J 0.707h(288) 203.6h in 4
Ju
x y
max
Mry
J
20 103 (6.25 3)(3)
2726
psi
h
203.6h
1
4762
x2 ( y y )2
27262 (1179 2726) 2
psi
h
h
Relate stress to strength
max all
4762
12.8 103
h
h
4762
0.372 in
12.8 103
Decision:
Specify 3 / 8 in leg size
Specifications:
Pattern: All-around square weld bead track
Electrode: E6010
Type of weld: Fillet
Weld bead length: 24 in
Leg size: 3 / 8 in
Attachment length: 12.25 in
______________________________________________________________________________
Chapter 9, Page 20/36
9-38
This is a good analysis task to test a student’s understanding.
(1) Solicit information related to a priori decisions.
(2) Solicit design variables b and d.
(3) Find h and round and output all parameters on a single screen. Allow return to Step 1
or Step 2.
(4) When the iteration is complete, the final display can be the bulk of your adequacy
assessment.
Such a program can teach too.
______________________________________________________________________________
9-39
The objective of this design task is to have the students teach themselves that the weld
patterns of Table 9-2 can be added or subtracted to obtain the properties of a
contemplated weld pattern. The instructor can control the level of complication. We have
left the presentation of the drawing to you. Here is one possibility. Study the problem’s
opportunities, and then present this (or your sketch) with the problem assignment.
Use b1 as the design variable. Express properties as a function of b1. From Table 9-3,
case 3:
A 1.414h(b b1 )
bd 2 b1d 2 (b b1 )d 2
Iu
2
2
2
I 0.707 hI u
V
F
A 1.414h(b b1 )
Mc Fa (d / 2)
I
0.707 hI u
Parametric study
Let a 10 in, b 8 in, d 8 in, b1 2 in, all 12.8 kpsi, l 2(8 2) 12 in
Chapter 9, Page 21/36
A 1.414h(8 2) 8.48h in 2
I u (8 2)(82 / 2) 192 in 3
I 0.707(h)(192) 135.7h in 4
10 000 1179
psi
8.48h
h
10 000(10)(8 / 2) 2948
psi
135.7h
h
1
3175
max
12 800 psi
11792 29482
h
h
from which h 0.248 in. Do not round off the leg size – something to learn.
Iu
192
64.5 in
hl 0.248(12)
A 8.48(0.248) 2.10 in 2
I 135.7(0.248) 33.65 in 4
fom '
h2
0.2482
l
12 0.369 in 3
2
2
I
33.65
eff
91.2 in
vol 0.369
1179
4754 psi
0.248
2948
11 887 psi
0.248
3175
max
12 800 psi
0.248
vol
Now consider the case of uninterrupted welds,
b1 0
A 1.414(h)(8 0) 11.31h
I u (8 0)(82 / 2) 256 in 3
I 0.707(256)h 181h in 4
10 000 884
11.31h
h
10 000(10)(8 / 2) 2210
181h
h
1
2380
max
all
8842 22102
h
h
2380
h max
0.186 in
all 12 800
Do not round off h.
Chapter 9, Page 22/36
A 11.31(0.186) 2.10 in 2
I 181(0.186) 33.67 in 4
884
0.1862
4753 psi, vol
16 0.277 in 3
0.186
2
2210
11882 psi
0.186
I
256
fom ' u
86.0 in
hl 0.186(16)
I
33.67
eff 2
121.7 in
(h / 2)l (0.1862 / 2)16
Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ.
To meet the shortened bead length, h is increased proportionately. However, volume of
bead laid down increases as h2. The uninterrupted bead is superior. In this example, we
did not round h and as a result we learned something. Our measures of merit are also
sensitive to rounding. When the design decision is made, rounding to the next larger
standard weld fillet size will decrease the merit.
Had the weld bead gone around the corners, the situation would change. Here is a follow
up task analyzing an alternative weld pattern.
______________________________________________________________________________
9-40
From Table 9-2
For the box
A 1.414h(b d )
Subtracting b1 from b and d1 from d
A 1.414h b b1 d d1
Iu
d3 b d2 1
d2
1
(3b d ) 1 1 b b1 d 2 d 3 d13
6
6
2
2
6
l 2(b b1 d d1 )
fom I u / hl
______________________________________________________________________________
Length of bead
Chapter 9, Page 23/36
9-41 Computer programs will vary.
______________________________________________________________________________
9-42
Note to the Instructor. In the first printing of the ninth edition, the loading was stated
incorrectly. In the fourth line, “bending moment of 100 kip ⋅ in in” should read, “10 kip
bending load 10 in from”. This will be corrected in the printings that follow. We
apologize if this has caused any inconvenience.
all = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ
beads and then
beads.
Horizontal parallel weld bead pattern
b = 3 in, d = 6 in
Table 9-2:
A 1.414hb 1.414( h)(3) 4.24h in 2
bd 2 3(6) 2
Iu
54 in 3
2
2
I 0.707 hI u 0.707( h)(54) 38.2h in 4
10
2.358
kpsi
4.24h
h
Mc 10(10)(6 / 2) 7.853
kpsi
38.2h
I
h
max 2 2
1
8.199
2.3582 7.8532
kpsi
h
h
Equate the maximum and allowable shear stresses.
max all
8.199
12
h
from which h 0.683 in. It follows that
I 38.2(0.683) 26.1 in 4
The volume of the weld metal is
h 2l (0.683) 2 (3 3)
1.40 in 3
2
2
The effectiveness, (eff) H , is
vol
Chapter 9, Page 24/36
I
26.1
18.6 in
vol 1.4
I
54
(fom ') H u
13.2 in
hl 0.683(3 3)
(eff) H
Vertical parallel weld beads
b 3 in
d 6 in
From Table 9-2, case 2
A 1.414hd 1.414(h)(6) 8.48h in 2
d 3 63
72 in 3
6
6
I 0.707 hI u 0.707(h)(72) 50.9h
10
1.179
psi
8.48h
h
Mc 10(10)(6 / 2) 5.894
psi
50.9h
I
h
1
6.011
1.1792 5.8942
kpsi
max 2 2
h
h
Iu
Equating max to all gives h 0.501 in. It follows that
I 50.9(0.501) 25.5 in 4
h 2l 0.5012
(6 6) 1.51 in 3
2
2
I
25.5
16.7 in
(eff ) V
vol 1.51
I
72
(fom ') V u
12.0 in
hl 0.501(6 6)
vol
The ratio of (eff ) V / (eff ) H is 16.7 /18.6 0.898. The ratio (fom ') V / (fom ') H is
12.0 /13.2 0.909. This is not surprising since
0.707hI u
I
I
I
2
2
1.414 u 1.414fom '
vol (h / 2)l (h / 2)l
hl
The ratios (eff ) V / (eff ) H and (fom ')V / (fom ')H give the same information.
______________________________________________________________________________
eff
Chapter 9, Page 25/36
9-43
F = 0, T = 15 kipin.
Table 9-1:
J u = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4
Tr 15 1
13.5 kpsi Ans.
J 1.111
______________________________________________________________________________
max
9-44
F = 2 kip, T = 0.
Table 9-2:
A = 1.414 h r = 1.414 (1/4)(1) = 1.111 in2
I u = r 3 = (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
V
2
1.80 kpsi
A 1.111
Mr 2 6 1
21.6 kpsi
I
0.5553
max = ( 2 + 2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi
Ans.
______________________________________________________________________________
9-45
F = 2 kip, T = 15 kipin.
Bending:
Table 9-2: A = 1.414 h r = 1.414 (1/4)(1) = 1.111 in2
I u = r 3 = (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
V
2
1.80 kpsi
A 1.111
M
Mr 2 6 1
21.6 kpsi
I
0.5553
Torsion:
Table 9-1:
J u = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4
T
Tr 15 1
13.5 kpsi
J 1.111
Chapter 9, Page 26/36
max 2 M T 1.802 21.62 13.52 25.5 kpsi
2
2
Ans.
______________________________________________________________________________
9-46
F = 2 kip, T = 15 kipin.
Bending:
Table 9-2: A = 1.414 h r = 1.414 h (1) = 4.442h in2
I u = r 3 = (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4
V
2
0.4502
kpsi
A 4.442h
h
M
Mr 2 6 1 5.403
kpsi
I
2.221h
h
Torsion:
J u = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4
Table 9-1:
T
Tr 15 1 3.377
kpsi
J 4.442h
h
max M T
2
2
max all
2
2
6.387
20
h
Should specify a 83 in weld.
2
2
6.387
0.4502 5.403 3.377
kpsi
h
h h h
h 0.319 in
Ans.
Ans.
______________________________________________________________________________
9-47
h 9 mm,
d 200 mm,
b 25 mm
From Table 9-2, case 2:
A = 1.414(9)(200) = 2.545(103) mm2
Iu
d 3 2003
1.333 106 mm3
6
6
I = 0.707h I u = 0.707(9)(1.333)(106) = 8.484(106) mm4
Chapter 9, Page 27/36
25 103
F
9.82 MPa
A 2.545(103 )
M = 25(150) = 3750 Nm
Mc 3750(100)
103 44.20 MPa
I
8.484(106 )
max 2 2 9.822 44.20 2 45.3 MPa Ans.
______________________________________________________________________________
9-48
Note to the Instructor. In the first printing of the ninth edition, the vertical dimension of
5 in should be to the top of the top plate. This will be corrected in the printings that
follow. We apologize if this has caused any inconvenience.
h = 0.25 in, b = 2.5 in, d = 5 in.
Table 9-2, case 5:
A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2
d2
52
y
2 in
b 2d 2.5 2 5
2d 3
2d 2 y b 2d y 2
3
2 53
2 52 2 2.5 2 5 22 33.33 in 3
3
Iu
I = 0.707 h I u = 0.707(1/4)(33.33) = 5.891 in4
Primary shear:
F
2
0.905 kpsi
A 2.209
Secondary shear (the critical location is at the bottom of the bracket):
y = 5 2 = 3 in
My 2 5 3
5.093 kpsi
I
5.891
max 2 2 0.9052 5.0932 5.173 kpsi
all
18
Ans.
3.48
max 5.173
______________________________________________________________________________
n
Chapter 9, Page 28/36
9-49
The largest possible weld size is 1/16 in. This is a small weld and thus difficult to
accomplish. The bracket’s load-carrying capability is not known. There are geometry
problems associated with sheet metal folding, load-placement and location of the center
of twist. This is not available to us. We will identify the strongest possible weldment.
Use a rectangular, weld-all-around pattern – Table 9-2, case 6:
A 1.414 h(b d ) 1.414(1 / 16)(1 7.5)
0.7512 in 2
x b / 2 0.5 in
y d / 2 7.5 / 2 3.75 in
d2
7.52
Iu
(3b d )
[3(1) 7.5] 98.44 in 3
6
6
I 0.707hI u 0.707(1 / 16)(98.44) 4.350 in 4
M (3.75 0.5)W 4.25W
V
W
1.331W
A 0.7512
Mc
4.25W (7.5 / 2)
3.664W
4.350
I
max 2 2 W 1.3312 3.6642 3.90W
Material properties: The allowable stress given is low. Let’s demonstrate that.
For the 1020 CD bracket, use HR properties of S y = 30 kpsi and S ut = 55. The 1030 HR
support, S y = 37.5 kpsi and S ut = 68. The E6010 electrode has strengths of S y = 50 and
S ut = 62 kpsi.
Allowable stresses:
1020 HR:
all = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi
1020 HR:
all = min[0.3(68), 0.4(37.5)] = min(20.4, 15) = 15 kpsi
E6010:
all = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi
Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value.
Therefore, the allowable shear stress is
all = min(14.4, 12, 18.0) = 12 kpsi
However, the allowable stress in the problem statement is 1.5 kpsi which is low from the
weldment perspective. The load associated with this strength is
max all 3.90W 1500
W
1500
385 lbf
3.90
Chapter 9, Page 29/36
If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is
12 000 psi and the load associated with this strength is W = 12 000/3.90 = 3077 lbf. Can
the bracket carry such a load?
There are geometry problems associated with sheet metal folding. Load placement is
important and the center of twist has not been identified. Also, the load-carrying
capability of the top bend is unknown.
These uncertainties may require the use of a different weld pattern. Our solution provides
the best weldment and thus insight for comparing a welded joint to one which employs
screw fasteners.
______________________________________________________________________________
9-50
F
FB
FBx
FBy
100 lbf , all 3 kpsi
100(16 / 3) 533.3 lbf
533.3cos 60 266.7 lbf
533.3cos 30 462 lbf
It follows that RAy 562 lbf and RAx 266.7 lbf, R A = 622 lbf
M = 100(16) = 1600 lbf · in
The OD of the tubes is 1 in. From Table 9-1, case 6:
A 2 1.414( hr ) 2(1.414)( h)(1 / 2) 4.442h in 2
J u 2 r 3 2 (1 / 2)3 0.7854 in 3
J 2(0.707)hJ u 1.414(0.7854)h 1.111h in 4
Chapter 9, Page 30/36
V
622
140.0
A 4.442h
h
Tc
Mc 1600(0.5) 720.1
J
J
1.111h
h
The shear stresses, and , are additive algebraically
1
860
(140.0 720.1)
psi
h
h
860
max all
3000
h
860
h
0.287 5 / 16 in
3000
max
Decision: Use 5/16 in fillet welds Ans.
______________________________________________________________________________
9-51
For the pattern in bending shown, find the centroid G of the weld group.
x
75 6 150 325 9 150
6 150 9 150
I 6mm 2 I G Ax 2
225 mm
6mm
0.707 6 1503
2
2
0.707 6 150 225 75 31.02 106 mm 4
12
I 9mm
0.707 9 1503
2
2
0.707 9 150 175 75 22.67 106 mm 4
12
I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4
The critical location is at B. With in MPa, and F in kN
Chapter 9, Page 31/36
F 103
V
0.3143F
A 2 0.707 6 9 150
3
Mc 200 F 10 225
0.8381F
I
53.69 106
max 2 2 F 0.31432 0.83812 0.8951F
Materials:
1015 HR (Table A-20): S y = 190 MPa, E6010 Electrode(Table 9-3): S y = 345 MPa
all = 0.577(190) = 109.6 MPa
Eq. (5-21), p. 225
all / n
109.6 / 2
61.2 kN
Ans.
0.8951 0.8951
______________________________________________________________________________
F
9-52
In the textbook, Fig. Problem 9-52b is a free-body diagram of the bracket. Forces and
moments that act on the welds are equal, but of opposite sense.
(a)
M = 1200(0.366) = 439 lbf · in Ans.
(b)
F y = 1200 sin 30 = 600 lbf Ans.
(c)
F x = 1200 cos 30 = 1039 lbf Ans.
(d) From Table 9-2, case 6:
A 1.414(0.25)(0.25 2.5) 0.972 in 2
d2
2.52
Iu
(3b d )
[3(0.25) 2.5] 3.39 in 3
6
6
The second area moment about an axis through G and parallel to z is
I 0.707hI u 0.707(0.25)(3.39) 0.599 in 4 Ans.
(e) Refer to Fig. Problem 9-52b. The shear stress due to F y is
1
Fy
A
600
617 psi
0.972
The shear stress along the throat due to F x is
2
Fx
1039
1069 psi
0.972
A
The resultant of 1 and 2 is in the throat plane
Chapter 9, Page 32/36
12 22 617 2 10692 1234 psi
The bending of the throat gives
Mc
439(1.25)
916 psi
I
0.599
The maximum shear stress is
max 2 2 1234 2 916 2 1537 psi
(f) Materials:
1018 HR Member:
E6010 Electrode:
n
S sy
max
Ans.
S y = 32 kpsi, S ut = 58 kpsi (Table A-20)
S y = 50 kpsi (Table 9-3)
0.577S y
max
0.577(32)
12.0
1.537
Ans.
(g) Bending in the attachment near the base. The cross-sectional area is approximately
equal to bh.
A1 bh 0.25(2.5) 0.625 in 2
F
1039
xy x
1662 psi
A1
0.625
I
bd 2
0.25(2.5) 2
0.260 in 3
c
6
6
At location A,
F
M
y y
A1
I /c
600
439
y
2648 psi
0.625 0.260
The von Mises stress is
y2 3 xy2
26482 3(1662) 2 3912 psi
Thus, the factor of safety is,
S
32
n y
8.18
3.912
Ans.
The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills
the hole
Chapter 9, Page 33/36
F
1200
9600 psi
td
0.25(0.50)
Sy
32(103 )
3.33 Ans.
n
9600
Further investigation of this situation requires more detail than is included in the task
statement.
(h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with S ut = 58
kpsi, Eq. (6-8), p. 282, gives
Se 0.504Sut 0.504(58) 29.2 kpsi
Eq. (6-19), p. 287:
k a = 14.4(58)-0.718 = 0.780
For the size factor estimate, we first employ Eq. (6-25), p. 289, for the equivalent
diameter
d e 0.808 0.707hb 0.808 0.707(2.5)(0.25) 0.537 in
Eq. (6-20), p. 288, is used next to find k b
d
kb e
0.30
-0.107
0.537
0.30
-0.107
0.940
Eq.(6-26), p. 290: k c = 0.59
From Eq. (6-18), p. 287, the endurance strength in shear is
S se = 0.780(0.940)(0.59)(29.2) = 12.6 kpsi
From Table 9-5, the shear stress-concentration factor is K f s = 2.7. The loading is
repeatedly-applied
1.537
a m K f s max 2.7
2.07 kpsi
2
2
Table 6-7, p. 307: Gerber factor of safety n f , adjusted for shear, with S su = 0.67S ut
2
1 S su a
1
nf
2 m S se
2 S
1 m se
S su a
2
2
2(2.07)(12.6)
1 0.67(58) 2.07
1 1
5.55 Ans.
2 2.07 12.6
0.67(58)(2.07)
Attachment metal should be checked for bending fatigue.
______________________________________________________________________________
9-53 (a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is
Chapter 9, Page 34/36
F
0.2829F
1.414(5 / 8)(4)
The secondary shear calculations, for a moment arm of 14 in give
4[3(42 ) 42 ]
Ju
42.67 in 3
6
J 0.707hJ u 0.707(5 / 8)42.67 18.85 in 4
Mry 14 F (2)
1.485F
x y
J
18.85
Thus, the maximum shear and allowable load are:
max F 1.4852 (0.2829 1.485) 2 2.309 F
all
25
F
10.8 kip Ans.
2.309 2.309
The load for part (a) has increased by a factor of 10.8/2.71 = 3.99
Ans.
(b) From Prob. 9-18b, all = 11 kpsi
Fall
all
2.309
11
4.76 kip
2.309
The allowable load in part (b) has increased by a factor of 4.76/1.19 = 4 Ans.
______________________________________________________________________________
Purchase the hook having the design shown in Fig. Problem 9-54b. Referring to text Fig.
9-29a, this design reduces peel stresses.
______________________________________________________________________________
9-54
9-55
(a)
l/2
l/2
1 l / 2 P cosh( x)
A
dx A1 cosh( x) dx 1 sinh( x)
l
/
2
l
/
2
l
4b sinh(l / 2)
l / 2
A
A
1 [sinh(l / 2) sinh(l / 2)] 1 [sinh(l / 2) ( sinh(l / 2))]
(b)
2 A1 sinh(l / 2)
P
P
[2sinh(l / 2)]
4bl sinh(l / 2)
2bl
P cosh(l / 2)
P
(l / 2)
Ans.
4b sinh(l / 2)
4b tanh(l / 2)
Ans.
Chapter 9, Page 35/36
(c)
K
(l / 2)
P
l / 2
2bl
4b tanh(l / 2) P tanh(l / 2)
Ans.
For computer programming, it can be useful to express the hyperbolic tangent in
terms of exponentials:
l exp(l / 2) exp(l / 2)
K
Ans.
2 exp(l / 2) exp(l / 2)
______________________________________________________________________________
9-56
This is a computer programming exercise. All programs will vary.
Chapter 9, Page 36/36
Chapter 10
10-1
From Eqs. (10-4) and (10-5)
KW K B
4C 1 0.615 4C 2
4C 4
C
4C 3
Plot 100(K W K B )/ K W vs. C for 4 C 12 obtaining
We see the maximum and minimum occur at C = 4 and 12 respectively where
Maximum = 1.36 % Ans.,
and Minimum = 0.743 % Ans.
______________________________________________________________________________
10-2
A = Sdm
dim(A uscu ) = [dim (S) dim(d m)] uscu = kpsiinm
dim(A SI ) = [dim (S) dim(d m)] SI = MPammm
ASI
MPa mm m
m
m
m Auscu 6.894 757 25.4 Auscu 6.895 25.4 Auscu
kpsi in
Ans.
For music wire, from Table 10-4:
A uscu = 201 kpsiinm,
m = 0.145;
what is A SI ?
A SI = 6.895(25.4)0.145 (201) = 2215 MPammm
Ans.
______________________________________________________________________________
10-3
Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, N t = 14 coils.
Chapter 10 - Rev. A, Page 1/41
(a) Table 10-1:
N a = N t 1 = 14 1 = 13 coils
L s = d N t = 2.5(14) = 35 mm
A = 2211 MPammm
Table 10-4:
m = 0.145,
Eq. (10-14):
Sut
Table 10-6:
S sy = 0.45(1936) = 871.2 MPa
A
2211
1936 MPa
m
d
2.50.145
D = OD d = 31 2.5 = 28.5 mm
C = D/d = 28.5/2.5 = 11.4
Eq. (10-5):
KB
4C 2 4 11.4 2
1.117
4C 3 4 11.4 3
d 3 S sy
Eq. (10-7):
Fs
Table 10-5):
d = 2.5/25.4 = 0.098 in
Eq. (10-9):
k
8K B D
2.53 871.2
8 1.117 28.5
167.9 N
G = 81.0(103) MPa
2.54 81103
d 4G
1.314 N / mm
8 D 3 N a 8 28.53 13
L0
(b)
F s = 167.9 N Ans.
(c)
k = 1.314 N/mm
Fs
167.9
Ls
35 162.8 mm
1.314
k
Ans.
Ans.
2.63 28.5
149.9 mm . Spring needs to be supported. Ans.
0.5
______________________________________________________________________________
(d)
10-4
L0 cr
Given: Design load, F 1 = 130 N.
Referring to Prob. 10-3 solution, C = 11.4, N a = 13 coils, S sy = 871.2 MPa, F s = 167.9 N,
L 0 = 162.8 mm and (L 0 ) cr = 149.9 mm.
Eq. (10-18): 4 ≤ C ≤ 12
C = 11.4
O.K.
Eq. (10-19): 3 ≤ N a ≤ 15 N a = 13
O.K.
Chapter 10 - Rev. A, Page 2/41
Fs
167.9
1
1 0.29
F1
130
0.15, 0.29 O.K.
Eq. (10-20):
From Eq. (10-7) for static service
Eq. (10-17):
8F D
1 K B 1 3 1.117
d
n
Eq. (10-21):
S sy
1
8(130)(28.5)
674 MPa
(2.5)3
871.2
1.29
674
n s ≥ 1.2, n = 1.29 O.K.
167.9
167.9
674
870.5 MPa
130
130
S sy / s 871.2 / 870.5 1
s 1
S sy / s ≥ (n s ) d : Not solid-safe (but was the basis of the design). Not O.K.
L 0 ≤ (L 0 ) cr : 162.8 149.9 Not O.K.
Design is unsatisfactory. Operate over a rod? Ans.
______________________________________________________________________________
10-5
Given: Oil-tempered wire, d = 0.2 in, D = 2 in, N t = 12 coils, L 0 = 5 in, squared ends.
(a) Table 10-1:
L s = d (N t + 1) = 0.2(12 + 1) = 2.6 in Ans.
(b) Table 10-1:
Table 10-5:
N a = N t 2 = 12 2 = 10 coils
G = 11.2 Mpsi
Eq. (10-9):
0.24 11.2 106
d 4G
k
28 lbf/in
8D3 N
8 23 10
F s = k y s = k (L 0 L s ) = 28(5 2.6) = 67.2 lbf
(c) Eq. (10-1):
Ans.
C = D/d = 2/0.2 = 10
4C 2 4 10 2
1.135
4C 3 4 10 3
Eq. (10-5):
KB
Eq. (10-7):
s KB
8 67.2 2
8 FD
1.135
48.56 103 psi
3
3
d
0.2
Chapter 10 - Rev. A, Page 3/41
Table 10-4:
m = 0.187, A = 147 kpsiinm
Eq. (10-14):
Sut
Table 10-6:
S sy = 0.50 S ut = 0.50(198.6) = 99.3 kpsi
A
147
198.6 kpsi
m
d
0.20.187
S sy
99.3
2.04
Ans.
s 48.56
______________________________________________________________________________
ns
10-6
Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L 0 = 80 mm, and at F = 50 N,
y = 15 mm.
(a)
k = F/y = 50/15 = 3.333 N/mm
(b)
D = Cd = 10(4) = 40 mm
Ans.
OD = D + d = 40 + 4 = 44 mm
Ans.
(c) From Table 10-5, G = 77.2 GPa
4
3
d 4G 4 77.2 10
11.6 coils
8kD 3 8 3.333 403
Eq. (10-9):
Na
Table 10-1:
N t = N a = 11.6 coils
Ans.
(d) Table 10-1:
L s = d (N t + 1) = 4(11.6 + 1) = 50. 4 mm
(e) Table 10-4:
m = 0.187, A = 1855 MPammm
Ans.
A 1855
1431 MPa
d m 40.187
Eq. (10-14):
Sut
Table 10-6:
S sy = 0.50 S ut = 0.50(1431) = 715.5 MPa
y s = L 0 L s = 80 50.4 = 29.6 mm
F s = k y s = 3.333(29.6) = 98.66 N
Eq. (10-5):
KB
4C 2 4(10) 2
1.135
4C 3 4(10) 3
Chapter 10 - Rev. A, Page 4/41
s KB
Eq. (10-7):
S sy
8 98.66 40
8 Fs D
1.135
178.2 MPa
3
d
43
715.5
4.02
Ans.
s 178.2
______________________________________________________________________________
ns
10-7
Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, N t = 8 coils, plain
and ground ends.
Preliminaries
Table 10-5: A = 140 kpsi · inm, m = 0.190
A
140
Eq. (10-14): Sut m
226.2 kpsi
d
0.0800.190
Table 10-6: S sy = 0.45(226.2) = 101.8 kpsi
Then,
D = OD d = 0.880 0.080 = 0.8 in
Eq. (10-1):
C = D/d = 0.8/0.08 = 10
4C 2 4(10) 2
Eq. (10-5):
KB
1.135
4C 3
4(10) 3
Table 10-1: N a = N t 1 = 8 1 = 7 coils
L s = dN t = 0.08(8) = 0.64 in
Eq. (10-7) For solid-safe, n s = 1.2 :
3
3
d 3S sy / ns 0.08 101.8 10 / 1.2
18.78 lbf
Fs
8K B D
8(1.135)(0.8)
Eq. (10-9):
k
0.084 11.5 106
d 4G
16.43 lbf/in
8D 3 N a
8 0.83 7
ys
Fs 18.78
1.14 in
k 16.43
(a) L 0 = y s + L s = 1.14 + 0.64 = 1.78 in Ans.
L
1.78
(b) Table 10-1: p 0
0.223 in Ans.
Nt
8
(c) From above: F s = 18.78 lbf Ans.
(d) From above: k = 16.43 lbf/in Ans.
2.63D
2.63(0.8)
(e) Table 10-2 and Eq. (10-13):
( L0 ) c r
4.21 in
0.5
Since L 0 < (L 0 ) cr , buckling is unlikely Ans.
______________________________________________________________________________
10-8
Given: Design load, F 1 = 16.5 lbf.
Referring to Prob. 10-7 solution, C = 10, N a = 7 coils, S sy = 101.8 kpsi, F s = 18.78 lbf,
y s = 1.14 in, L 0 = 1.78 in, and (L 0 ) cr = 4.208 in.
Chapter 10 - Rev. A, Page 5/41
Eq. (10-18):
Eq. (10-19):
4 ≤ C ≤ 12
3 ≤ N a ≤ 15
C = 10
Na = 7
O.K.
O.K.
Fs
18.78
1
1 0.14
F1
16.5
Eq. (10-20): 0.15, 0.14 not O.K . , but probably acceptable.
From Eq. (10-7) for static service
Eq. (10-17):
8(16.5)(0.8)
8F1D
1.135
74.5 103 psi 74.5 kpsi
3
3
(0.080)
d
S
101.8
n sy
1.37
74.5
1
1 K B
Eq. (10-21):
n s ≥ 1.2, n = 1.37 O.K.
18.78
18.78
74.5
84.8 kpsi
16.5
16.5
ns Ssy / s 101.8 / 84.8 1.20
s 1
Eq. (10-21):
n s ≥ 1.2, n s = 1.2 It is solid-safe (basis of design). O.K.
Eq. (10-13) and Table 10-2: L 0 ≤ (L 0 ) cr
1.78 in 4.208 in O.K.
______________________________________________________________________________
10-9
Given: A228 music wire, sq. and grd. ends, d = 0.007 in, OD = 0.038 in, L 0 = 0.58 in,
N t = 38 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 0.038 0.007 = 0.031 in
C = D/d = 0.031/0.007 = 4.429
4C 2 4 4.429 2
KB
1.340
4C 3 4 4.429 3
Table (10-1): N a = N t 2 = 38 2 = 36 coils
(high)
Table 10-5: G = 12.0 Mpsi
0.007 4 12.0 106
d 4G
Eq. (10-9):
k
3.358 lbf/in
8D3 N a
8 0.0313 36
Table (10-1): L s = dN t = 0.007(38) = 0.266 in
y s = L 0 L s = 0.58 0.266 = 0.314 in
F s = ky s = 3.358(0.314) = 1.054 lbf
8 1.054 0.031
8F D
Eq. (10-7):
s K B s 3 1.340
325.1103 psi
3
d
0.007
Table 10-4:
(1)
A = 201 kpsiinm, m = 0.145
Chapter 10 - Rev. A, Page 6/41
Eq. (10-14):
Table 10-6:
A
201
412.7 kpsi
m
d
0.007 0.145
S sy = 0.45 S ut = 0.45(412.7) = 185.7 kpsi
Sut
s > S sy , that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with
F s = ky s and s = S sy /n s , and solve for y s , giving
Ssy / ns d 3 185.7 103 /1.2 0.0073
ys
8 K B kD
The free length should be wound to
8 1.340 3.358 0.031
L 0 = L s + y s = 0.266 + 0.149 = 0.415 in
0.149 in
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-10 Given: B159 phosphor-bronze, sq. and grd. ends, d = 0.014 in, OD = 0.128 in, L 0 = 0.50
in, N t = 16 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 0.128 0.014 = 0.114 in
C = D/d = 0.114/0.014 = 8.143
4C 2 4 8.143 2
KB
1.169
4C 3 4 8.143 3
Table (10-1): N a = N t 2 = 16 2 = 14 coils
Table 10-5: G = 6 Mpsi
0.0144 6 106
d 4G
Eq. (10-9):
k
1.389 lbf/in
8D3 N a
8 0.1143 14
Table (10-1): L s = dN t = 0.014(16) = 0.224 in
y s = L 0 L s = 0.50 0.224 = 0.276 in
F s = ky s = 1.389(0.276) = 0.3834 lbf
8 0.3834 0.114
8F D
Eq. (10-7):
s K B s 3 1.169
47.42 103 psi
3
d
0.014
Table 10-4:
Eq. (10-14):
Table 10-6:
(1)
A = 145 kpsiinm, m = 0
A
145
Sut m
145 kpsi
d
0.0140
S sy = 0.35 S ut = 0.35(135) = 47.25 kpsi
s > S sy , that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with
F s = ky s and s = S sy /n s , and solve for y s , giving
Ssy / ns d 3 47.25 103 /1.2 0.0143
ys
8K B kD
The free length should be wound to
8 1.169 1.389 0.114
0.229 in
Chapter 10 - Rev. A, Page 7/41
L 0 = L s + y s = 0.224 + 0.229 = 0.453 in
Ans.
______________________________________________________________________________
10-11 Given: A313 stainless steel, sq. and grd. ends, d = 0.050 in, OD = 0.250 in, L 0 = 0.68 in,
N t = 11.2 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 0.250 0.050 = 0.200 in
C = D/d = 0.200/0.050 = 4
4C 2 4 4 2
KB
1.385
4C 3 4 4 3
Table (10-1): N a = N t 2 = 11.2 2 = 9.2 coils
Table 10-5: G = 10 Mpsi
0.0504 10 106
d 4G
Eq. (10-9):
k
106.1 lbf/in
8D3 N a
8 0.23 9.2
Table (10-1): L s = dN t = 0.050(11.2) = 0.56 in
y s = L 0 L s = 0.68 0.56 = 0.12 in
F s = ky s = 106.1(0.12) = 12.73 lbf
8 12.73 0.2
8F D
Eq. (10-7):
s K B s 3 1.385
71.8 103 psi
3
d
0.050
Table 10-4:
Eq. (10-14):
Table 10-6:
A = 169 kpsiinm, m = 0.146
A
169
Sut m
261.7 kpsi
d
0.0500.146
S sy = 0.35 S ut = 0.35(261.7) = 91.6 kpsi
S sy
91.6
1.28 Spring is solid-safe (n s > 1.2)
Ans.
s 71.8
______________________________________________________________________________
ns
10-12 Given: A227 hard-drawn wire, sq. and grd. ends, d = 0.148 in, OD = 2.12 in, L 0 = 2.5 in,
N t = 5.75 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 2.12 0.148 = 1.972 in
C = D/d = 1.972/0.148 = 13.32
(high)
4C 2 4 13.32 2
KB
1.099
4C 3 4 13.32 3
Table (10-1): N a = N t 2 = 5.75 2 = 3.75 coils
Table 10-5: G = 11.4 Mpsi
0.1484 11.4 106
d 4G
Eq. (10-9):
k
23.77 lbf/in
8D3 N a
8 1.9723 3.75
Table (10-1): L s = dN t = 0.148(5.75) = 0.851 in
y s = L 0 L s = 2.5 0.851 = 1.649 in
Chapter 10 - Rev. A, Page 8/41
Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
F s = ky s = 23.77(1.649) = 39.20 lbf
8 39.20 1.972
8F D
s K B s 3 1.099
66.7 103 psi
3
d
0.148
A = 140 kpsiinm, m = 0.190
A
140
Sut m
201.3 kpsi
d
0.1480.190
S sy = 0.35 S ut = 0.45(201.3) = 90.6 kpsi
S sy
90.6
1.36 Spring is solid-safe (n s > 1.2)
Ans.
s 66.7
______________________________________________________________________________
ns
10-13 Given: A229 OQ&T steel, sq. and grd. ends, d = 0.138 in, OD = 0.92 in, L 0 = 2.86 in,
N t = 12 coils.
D = OD d = 0.92 0.138 = 0.782 in
Eq. (10-1):
Eq. (10-5):
C = D/d = 0.782/0.138 = 5.667
4C 2 4 5.667 2
KB
1.254
4C 3 4 5.667 3
Table (10-1): N a = N t 2 = 12 2 = 10 coils
A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels,
G = 11.5 Mpsi.
0.1384 11.5 106
d 4G
Eq. (10-9):
k
109.0 lbf/in
8D3 N a
8 0.7823 10
Table (10-1): L s = dN t = 0.138(12) = 1.656 in
y s = L 0 L s = 2.86 1.656 = 1.204 in
F s = ky s = 109.0(1.204) = 131.2 lbf
8 131.2 0.782
8F D
Eq. (10-7):
s K B s 3 1.254
124.7 103 psi
3
d
0.138
Table 10-4:
Eq. (10-14):
Table 10-6:
(1)
A = 147 kpsiinm, m = 0.187
A
147
Sut m
212.9 kpsi
d
0.1380.187
S sy = 0.50 S ut = 0.50(212.9) = 106.5 kpsi
s > S sy , that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with
F s = ky s and s = S sy /n s , and solve for y s , giving
Ssy / ns d 3 106.5 103 /1.2 0.1383
ys
8K B kD
The free length should be wound to
8 1.254 109.0 0.782
0.857 in
Chapter 10 - Rev. A, Page 9/41
L 0 = L s + y s = 1.656 + 0.857 = 2.51 in
Ans.
______________________________________________________________________________
10-14 Given: A232 chrome-vanadium steel, sq. and grd. ends, d = 0.185 in, OD = 2.75 in, L 0 =
7.5 in, N t = 8 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 2.75 0.185 = 2.565 in
C = D/d = 2.565/0.185 = 13.86
(high)
4
13.86
2
1.095
4C 2
KB
4C 3 4 13.86 3
Table (10-1): N a = N t 2 = 8 2 = 6 coils
Table 10-5:
Eq. (10-9):
G = 11.2 Mpsi.
0.1854 11.2 106
d 4G
k
16.20 lbf/in
8D3 N a
8 2.5653 6
Table (10-1): L s = dN t = 0.185(8) = 1.48 in
y s = L 0 L s = 7.5 1.48 = 6.02 in
F s = ky s = 16.20(6.02) = 97.5 lbf
8 97.5 2.565
8F D
Eq. (10-7):
s K B s 3 1.095
110.1103 psi
3
d
0.185
(1)
A = 169 kpsiinm, m = 0.168
A
169
Eq. (10-14): Sut m
224.4 kpsi
d
0.1850.168
Table 10-6: S sy = 0.50 S ut = 0.50(224.4) = 112.2 kpsi
S
112.2
ns sy
1.02 Spring is not solid-safe (n s < 1.2)
s 110.1
Return to Eq. (1) with F s = ky s and s = S sy /n s , and solve for y s , giving
3
3
S sy / ns d 3 112.2 10 /1.2 0.185
ys
5.109 in
8K B kD
8 1.095 16.20 2.565
The free length should be wound to
Table 10-4:
L 0 = L s + y s = 1.48 + 5.109 = 6.59 in
Ans.
______________________________________________________________________________
10-15 Given: A313 stainless steel, sq. and grd. ends, d = 0.25 mm, OD = 0.95 mm, L 0 = 12.1
mm, N t = 38 coils.
D = OD d = 0.95 0.25 = 0.7 mm
Eq. (10-1):
C = D/d = 0.7/0.25 = 2.8
(low)
4C 2 4 2.8 2
Eq. (10-5):
KB
1.610
4C 3 4 2.8 3
Chapter 10 - Rev. A, Page 10/41
Table (10-1): N a = N t 2 = 38 2 = 36 coils
Table 10-5:
Eq. (10-9):
(high)
G = 69.0(103) MPa.
0.254 69.0 103
d 4G
k
2.728 N/mm
8D3 N a
8 0.73 36
Table (10-1): L s = dN t = 0.25(38) = 9.5 mm
y s = L 0 L s = 12.1 9.5 = 2.6 mm
F s = ky s = 2.728(2.6) = 7.093 N
8 7.093 0.7
8F D
Eq. (10-7):
s K B s 3 1.610
1303 MPa
d
0.253
(1)
Table 10-4 (dia. less than table):
A = 1867 MPammm, m = 0.146
A
1867
Eq. (10-14): Sut m
2286 MPa
d
0.250.146
Table 10-6: S sy = 0.35 S ut = 0.35(2286) = 734 MPa
s > S sy , that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with
F s = ky s and s = S sy /n s , and solve for y s , giving
Ssy / ns d 3 734 /1.2 0.253
ys
1.22 mm
8K B kD
8 1.610 2.728 0.7
The free length should be wound to
L 0 = L s + y s = 9.5 + 1.22 = 10.72 mm
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-16 Given: A228 music wire, sq. and grd. ends, d = 1.2 mm, OD = 6.5 mm, L 0 = 15.7 mm,
N t = 10.2 coils.
D = OD d = 6.5 1.2 = 5.3 mm
Eq. (10-1):
C = D/d = 5.3/1.2 = 4.417
4C 2 4 4.417 2
Eq. (10-5):
KB
1.368
4C 3 4 4.417 3
Table (10-1): N a = N t 2 = 10.2 2 = 8.2 coils
Table 10-5 (d = 1.2/25.4 = 0.0472 in):
G = 81.7(103) MPa.
1.24 81.7 103
d 4G
Eq. (10-9):
k
17.35 N/mm
8D3 N a
8 5.33 8.2
Table (10-1): L s = dN t = 1.2(10.2) = 12.24 mm
y s = L 0 L s = 15.7 12.24 = 3.46 mm
F s = ky s = 17.35(3.46) = 60.03 N
Chapter 10 - Rev. A, Page 11/41
8 60.03 5.3
8 Fs D
1.368
641.4 MPa
3
d
1.23
Eq. (10-7):
s KB
Table 10-4:
A = 2211 MPammm, m = 0.145
A
2211
Sut m 0.145 2153 MPa
d
1.2
S sy = 0.45 S ut = 0.45(2153) = 969 MPa
Eq. (10-14):
Table 10-6:
S sy
(1)
969
1.51 Spring is solid-safe (n s > 1.2) Ans.
s 641.4
______________________________________________________________________________
ns
10-17 Given: A229 OQ&T steel, sq. and grd. ends, d = 3.5 mm, OD = 50.6 mm, L 0 = 75.5 mm,
N t = 5.5 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 50.6 3.5 = 47.1 mm
C = D/d = 47.1/3.5 = 13.46 (high)
4C 2 4 13.46 2
KB
1.098
4C 3 4 13.46 3
Table (10-1): N a = N t 2 = 5.5 2 = 3.5 coils
A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels,
G = 79.3(103) MPa.
3.54 79.3103
d 4G
Eq. (10-9):
k
4.067 N/mm
8D3 N a
8 47.13 3.5
Table (10-1): L s = dN t = 3.5(5.5) = 19.25 mm
y s = L 0 L s = 75.5 19.25 = 56.25 mm
F s = ky s = 4.067(56.25) = 228.8 N
8 228.8 47.1
8F D
Eq. (10-7):
s K B s 3 1.098
702.8 MPa
d
3.53
(1)
A = 1855 MPammm, m = 0.187
A
1855
Eq. (10-14): Sut m
1468 MPa
d
3.50.187
Table 10-6: S sy = 0.50 S ut = 0.50(1468) = 734 MPa
S
734
ns sy
1.04 Spring is not solid-safe (n s < 1.2)
s 702.8
Return to Eq. (1) with F s = ky s and s = S sy /n s , and solve for y s , giving
S sy / ns d 3
734 /1.2 3.53
ys
48.96 mm
8K B kD
8 1.098 4.067 47.1
The free length should be wound to
Table 10-4:
Chapter 10 - Rev. A, Page 12/41
L 0 = L s + y s = 19.25 + 48.96 = 68.2 mm
Ans.
______________________________________________________________________________
10-18 Given: B159 phosphor-bronze, sq. and grd. ends, d = 3.8 mm, OD = 31.4 mm, L 0 = 71.4
mm, N t = 12.8 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 31.4 3.8 = 27.6 mm
C = D/d = 27.6/3.8 = 7.263
4C 2 4 7.263 2
KB
1.192
4C 3 4 7.263 3
Table (10-1): N a = N t 2 = 12.8 2 = 10.8 coils
Table 10-5:
Eq. (10-9):
G = 41.4(103) MPa.
3.84 41.4 103
d 4G
k
4.752 N/mm
8 D3 N a 8 27.63 10.8
Table (10-1): L s = dN t = 3.8(12.8) = 48.64 mm
y s = L 0 L s = 71.4 48.64 = 22.76 mm
F s = ky s = 4.752(22.76) = 108.2 N
8 108.2 27.6
8F D
Eq. (10-7):
s K B s 3 1.192
165.2 MPa
d
3.83
(1)
Table 10-4 (d = 3.8/25.4 = 0.150 in):
A = 932 MPammm, m = 0.064
A
932
Eq. (10-14): Sut m
855.7 MPa
d
3.80.064
Table 10-6: S sy = 0.35 S ut = 0.35(855.7) = 299.5 MPa
S
299.5
ns sy
1.81 Spring is solid-safe (n s > 1.2) Ans.
s 165.2
______________________________________________________________________________
10-19 Given: A232 chrome-vanadium steel, sq. and grd. ends, d = 4.5 mm, OD = 69.2 mm,
L 0 = 215.6 mm, N t = 8.2 coils.
Eq. (10-1):
Eq. (10-5):
D = OD d = 69.2 4.5 = 64.7 mm
C = D/d = 64.7/4.5 = 14.38 (high)
4C 2 4 14.38 2
KB
1.092
4C 3 4 14.38 3
Table (10-1): N a = N t 2 = 8.2 2 = 6.2 coils
Table 10-5:
Eq. (10-9):
G = 77.2(103) MPa.
4.54 77.2 103
d 4G
k
2.357 N/mm
8D3 N a
8 64.73 6.2
Table (10-1): L s = dN t = 4.5(8.2) = 36.9 mm
Chapter 10 - Rev. A, Page 13/41
Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
y s = L 0 L s = 215.6 36.9 = 178.7 mm
F s = ky s = 2.357(178.7) = 421.2 N
8 421.2 64.7
8F D
s K B s 3 1.092
832 MPa
d
4.53
(1)
A = 2005 MPammm, m = 0.168
A
2005
Sut m
1557 MPa
d
4.50.168
S sy = 0.50 S ut = 0.50(1557) = 779 MPa
s > S sy , that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with
F s = ky s and s = S sy /n s , and solve for y s , giving
Ssy / ns d 3 779 /1.2 4.53
ys
139.5 mm
8K B kD
8 1.092 2.357 64.7
The free length should be wound to
L 0 = L s + y s = 36.9 + 139.5 = 176.4 mm
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-20 Given: A227 HD steel.
From the figure: L 0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus
D = OD d = 2 0.135 = 1.865 in
(a) By counting, N t = 12.5 coils. Since the ends are squared along 1/4 turn on each end,
N a 12.5 0.5 12 turns Ans.
p 4.75 / 12 0.396 in Ans.
The solid stack is 13 wire diameters
L s = 13(0.135) = 1.755 in
Ans.
(b) From Table 10-5, G = 11.4 Mpsi
0.1354 (11.4) 106
d 4G
k
6.08 lbf/in
8D 3 N a
8 1.8653 (12)
(c) F s = k(L 0 - L s ) = 6.08(4.75 1.755)(10-3) = 18.2 lbf
(d) C = D/d = 1.865/0.135 = 13.81
Ans.
Ans.
Chapter 10 - Rev. A, Page 14/41
4(13.81) 2
1.096
4(13.81) 3
8F D
8(18.2)(1.865)
s K B s 3 1.096
38.5 103 psi 38.5 kpsi
3
d
0.135
KB
Ans.
______________________________________________________________________________
10-21 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground
ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For N a ,
k = F max /y = 20/2 = 10 lbf/in. For s , F = F s = 20(1 + ) = 20(1 + 0.15) = 23 lbf.
Source
Eq. (10-1)
Eq. (10-9)
Table 10-1
Table 10-1
1.15y + L s
Eq. (10-13)
(a) Spring over a Rod
Parameter Values
d
0.075
0.080
ID
0.800
0.800
D
0.875
0.880
C
11.667
11.000
Na
6.937
8.828
Nt
8.937
10.828
Ls
0.670
0.866
L0
2.970
3.166
(L 0 ) cr
4.603
4.629
Table 10-4
Table 10-4
Eq. (10-14)
A
m
S ut
201.000
0.145
292.626
Table 10-6
Eq. (10-5)
Eq. (10-7)
Eq. (10-3)
Eq. (10-22)
S sy
KB
131.681
1.115
135.335
0.973
0.282
s
ns
fom
For n s
Source
0.085
0.800
0.885
10.412
11.061
13.061
1.110
3.410
4.655
(b) Spring in a Hole
Parameter Values
d
0.075
0.080
OD
0.950
0.950
D
0.875
0.870
C
11.667
10.875
Na
6.937
9.136
Nt
8.937
11.136
Ls
0.670
0.891
L0
2.970
3.191
(L 0 ) cr
4.603
4.576
0.085
0.950
0.865
10.176
11.846
13.846
1.177
3.477
4.550
201.000
0.145
289.900
201.000
0.145
287.363
130.455
1.123
111.787
1.167
0.398
129.313
1.133
93.434
1.384
0.555
Eq. (10-1)
Eq. (10-9)
Table 10-1
Table 10-1
1.15y + L s
Eq. (1013)
201.000 201.000 Table 10-4
A
201.000
0.145
0.145
Table 10-4
m
0.145
289.900 287.363 Eq. (10S ut
292.626
14)
130.455 129.313 Table 10-6
S sy
131.681
1.122
1.129
Eq. (10-5)
KB
1.115
112.948 95.293 Eq. (10-7)
135.335
s
1.155
1.357
Eq. (10-3)
ns
0.973
Eq.
(10fom
0.391
0.536
0.282
22)
≥ 1.2, the optimal size is d = 0.085 in for both cases.
______________________________________________________________________________
10-22 In Prob. 10-21, there is an advantage of first selecting d as one can select from the
available sizes (Table A-28). Selecting C first, requires a calculation of d where then a
size must be selected from Table A-28.
Consider part (a) of the problem. It is required that
ID = D d = 0.800 in.
(1)
From Eq. (10-1), D = Cd. Substituting this into the first equation yields
d
0.800
C 1
(2)
Chapter 10 - Rev. A, Page 15/41
Starting with C = 10, from Eq. (2) we find that d = 0.089 in. From Table A-28, the closest
diameter is d = 0.090 in. Substituting this back into Eq. (1) gives D = 0.890 in, with C =
0.890/0.090 = 9.889, which are acceptable. From this point the solution is the same as
Prob. 10-21. For part (b), use
OD = D + d = 0.950 in.
(3)
d
and,
0.800
C 1
(a) Spring over a rod
Parameter Values
C
10.000
10.5
Eq. (2)
d
0.089
0.084
Table A-28
d
0.090
0.085
Eq. (1)
D
0.890
0.885
Eq. (10-1)
C
9.889
10.412
Eq. (10-9)
Na
13.669
11.061
Table 10-1
Nt
15.669
13.061
Table 10-1
Ls
1.410
1.110
1.15y + L s
L0
3.710
3.410
Eq. (10-13) (L 0 ) cr
4.681
4.655
Table 10-4
A
201.000 201.000
Table 10-4
m
0.145
0.145
Eq. (10-14)
S ut
284.991 287.363
Table 10-6
S sy
128.246 129.313
Eq. (10-5)
KB
1.135
1.128
Eq. (10-7)
81.167
95.223
s
ns
1.580
1.358
n s = S sy / s
Eq. (10-22) fom
-0.725
-0.536
Source
(4)
(b) Spring in a Hole
Source Parameter Values
C
10.000
Eq. (4)
d
0.086
Table A-28
d
0.085
Eq. (3)
D
0.865
Eq. (10-1)
C
10.176
Eq. (10-9)
Na
11.846
Table 10-1
Nt
13.846
Table 10-1
Ls
1.177
1.15y + L s
L0
3.477
Eq. (10-13)
(L 0 ) cr
4.550
Table 10-4
A
201.000
Table 10-4
m
0.145
Eq. (10-14)
S ut
287.363
Table 10-6
S sy
129.313
Eq. (10-5)
KB
1.135
Eq. (10-7)
93.643
s
ns
1.381
n s = S sy / s
Eq. (10-22)
fom
-0.555
Again, for n s 1.2, the optimal size is = 0.085 in.
Although this approach used less iterations than in Prob. 10-21, this was due to the initial
values picked and not the approach.
______________________________________________________________________________
10-23 One approach is to select A227 HD steel for its low cost. Try L 0 = 48 mm, then for
y = 48 37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286
N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with
d = 2 mm,
D = ID + d = 11.25 + 2 = 13.25 mm
C = D/d = 13.25/2 = 6.625
(acceptable)
Table 10-5 (d = 2/25.4 = 0.0787 in): G = 79.3 GPa
Chapter 10 - Rev. A, Page 16/41
Eq. (10-9):
Na
d 4G
24 (79.3)103
15.9 coils
8kD3 8(4.286)13.253
Assume squared and closed.
Table 10-1: N t = N a + 2 = 15.9 + 2 = 17.9 coils
L s = dN t = 2(17.9) =35.8 mm
Eq. (10-5):
y s = L 0 L s = 48 35.8 = 12.2 mm
F s = ky s = 4.286(12.2) = 52.29 N
4C 2 4 6.625 2
KB
1.213
4C 3 4 6.625 3
8(52.29)13.25
8Fs D
1.213
267.5 MPa
d3
23
Eq. (10-7):
s KB
Table 10-4:
A = 1783 MPa · mmm, m = 0.190
A 1783
Sut m 0.190 1563 MPa
d
2
S sy = 0.45S ut = 0.45(1563) = 703.3 MPa
Eq. (10-14):
Table 10-6:
ns
S sy
s
703.3
2.63 1.2
267.5
O.K .
No other diameters in the given range work. So specify
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
and closed, N t = 17.9 coils, N a = 15.9 coils, k = 4.286 N/mm, L s = 35.8 mm, and L 0 = 48
mm.
Ans.
______________________________________________________________________________
10-24 Select A227 HD steel for its low cost. Try L 0 = 48 mm, then for y = 48 37.5 = 10.5 mm
when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm.
and,
D d = 11.25
(1)
D =Cd
(2)
Starting with C = 8, gives D = 8d. Substitute into Eq. (1) resulting in d = 1.607 mm.
Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the
calculations are shown in the third column of the spreadsheet output shown. We see that
for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides
acceptable results with the specifications
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
Chapter 10 - Rev. A, Page 17/41
and closed, N t = 17.9 coils, N a = 15.9 coils, k = 4.286 N/mm, L s = 35.8 mm, and L 0 = 48
mm.
Ans.
Chapter 10 - Rev. A, Page 18/41
Source
Eq. (2)
Table A-28
Eq. (1)
Eq. (10-1)
Eq. (10-9)
Table 10-1
Table 10-1
L 0 L s
F s = ky s
Table 10-4
Table 10-4
Eq. (10-14)
Table 10-6
Eq. (10-5)
Eq. (10-7)
n s = S sy / s
C
d
d
D
C
Na
Nt
Ls
ys
Fs
A
m
S ut
S sy
KB
s
ns
Parameter Values
8.000
7
6.500
1.607
1.875
2.045
1.600
1.800
2.000
12.850
13.050
13.250
8.031
7.250
6.625
7.206
10.924
15.908
9.206
12.924
17.908
14.730
23.264
35.815
33.270
24.736
12.185
142.594
106.020
52.224
1783.000 1783.000 1783.000
0.190
0.190
0.190
1630.679 1594.592 1562.988
733.806
717.566
703.345
1.172
1.200
1.217
1335.568 724.943
268.145
0.549
0.990
2.623
The only difference between selecting C first rather than d as was done in Prob. 10-23, is
that once d is calculated, the closest wire size must be selected. Iterating on d uses
available wire sizes from the beginning.
______________________________________________________________________________
10-25 A stock spring catalog may have over two hundred pages of compression springs with up
to 80 springs per page listed.
• Students should be made aware that such catalogs exist.
• Many springs are selected from catalogs rather than designed.
• The wire size you want may not be listed.
• Catalogs may also be available on disk or the web through search routines. For
example, disks are available from Century Spring at
1 - (800) - 237 - 5225
www.centuryspring.com
• It is better to familiarize yourself with vendor resources rather than invent them
yourself.
• Sample catalog pages can be given to students for study.
______________________________________________________________________________
10-26 Given: ID = 0.6 in, C = 10, L 0 = 5 in, L s = 5 3 = 2 in, sq. & grd ends, unpeened, HD
A227 wire.
(a) With ID = D d = 0.6 in and C = D/d = 10 10 d d = 0.6 d = 0.0667 in Ans.,
and D = 0.667 in.
(b) Table 10-1:
L s = dN t = 2 in N t = 2/0.0667 30 coils Ans.
Chapter 10 - Rev. A, Page 19/41
N a = N t 2 = 30 2 = 28 coils
G = 11.5 Mpsi
0.0667 4 11.5 106
d 4G
k
3.424 lbf/in
8D3 N a
8 0.6673 28
(c) Table 10-1:
Table 10-5:
Eq. (10-9):
A = 140 kpsiinm, m = 0.190
A
140
Sut m
234.2 kpsi
d
0.06670.190
(d) Table 10-4:
Eq. (10-14):
Table 10-6:
Eq. (10-5):
Ans.
S sy = 0.45 S ut = 0.45 (234.2) = 105.4 kpsi
F s = ky s = 3.424(3) = 10.27 lbf
4C 2 4 10 2
KB
1.135
4C 3 4 10 3
Eq. (10-7):
s KB
8 10.27 0.667
8 Fs D
1.135
d3
0.06673
66.72 103 psi 66.72 kpsi
S sy
105.4
1.58
Ans.
s 66.72
(e) a = m = 0.5 s = 0.5(66.72) = 33.36 kpsi, r = a / m = 1. Using the Gerber fatigue
failure criterion with Zimmerli data,
ns
Eq. (10-30):
S su = 0.67 S ut = 0.67(234.2) = 156.9 kpsi
The Gerber ordinate intercept for the Zimmerli data is
S sa
35
Se
39.9 kpsi
2
2
1 S sm / S su 1 55 / 156.9
Table 6-7, p. 307,
2
2S se
r 2 S su2
S sa
1 1
2S se
rSsu
2
12 156.92
2 39.9
1 1
37.61 kpsi
2 39.9
1
156.9
S
37.61
n f sa
1.13 Ans.
a 33.36
______________________________________________________________________________
10-27 Given: OD 0.9 in, C = 8, L 0 = 3 in, L s = 1 in, y s = 3 1 = 2 in, sq. ends, unpeened,
music wire.
(a) Try OD = D + d = 0.9 in, C = D/d = 8 D = 8d 9d = 0.9 d = 0.1 Ans.
Chapter 10 - Rev. A, Page 20/41
D = 8(0.1) = 0.8 in
(b) Table 10-1:
L s = d (N t + 1) N t = L s / d 1 = 1/0.1 1 = 9 coils
Table 10-1:
(c) Table 10-5:
Eq. (10-9):
(d)
Ans.
N a = N t 2 = 9 2 = 7 coils
G = 11.75 Mpsi
0.14 11.75 106
d 4G
k
40.98 lbf/in
8D3 N a
8 0.83 7
Ans.
F s = ky s = 40.98(2) = 81.96 lbf
4C 2 4 8 2
1.172
4C 3 4 8 3
Eq. (10-5):
KB
Eq. (10-7):
s KB
Table 10-4:
A = 201 kpsiinm, m = 0.145
Eq. (10-14):
Sut
Table 10-6:
S sy = 0.45 S ut = 0.45(280.7) = 126.3 kpsi
ns
8 81.96 0.8
8Fs D
1.172
195.7 103 psi 195.7 kpsi
3
3
d
0.1
A
201
0.145 280.7 kpsi
m
0.1
d
S sy
s
126.3
0.645
195.7
Ans.
(e) a = m = s /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with
Zimmerli data,
Eq. (10-30):
S su = 0.67 S ut = 0.67(280.7) = 188.1 kpsi
The Gerber ordinate intercept for the Zimmerli data is
S sa
35
Se
36.83 kpsi
2
2
1 S sm / S su 1 55 / 188.1
Table 6-7, p. 307,
2
2S se
r 2 S su2
S sa
1 1
2S se
rSsu
2
12 188.12
2 38.3
1 1
36.83 kpsi
2 38.3
1
188.1
Chapter 10 - Rev. A, Page 21/41
nf
S sa
a
36.83
0.376
97.85
Ans.
Obviously, the spring is severely under designed and will fail statically and in fatigue.
Increasing C would improve matters. Try C = 12. This yields n s = 1.83 and n f = 1.00.
______________________________________________________________________________
10-28 Note to the Instructor: In the first printing of the text, the wire material was incorrectly
identified as music wire instead of oil-tempered wire. This will be corrected in
subsequent printings. We are sorry for any inconvenience.
Given: F max = 300 lbf, F min = 150 lbf, y = 1 in, OD = 2.1 0.2 = 1.9 in, C = 7,
unpeened, sq. & grd., oil-tempered wire.
(a)
D = OD d = 1.9 d
(1)
C = D/d = 7 D = 7d
(2)
Substitute Eq. (2) into (1)
7d = 1.9 d d = 1.9/8 = 0.2375 in Ans.
(b) From Eq. (2):
D = 7d = 7(0.2375) = 1.663 in
(c)
k
(d) Table 10-5:
G = 11.6 Mpsi
Eq. (10-9):
Table 10-1:
Ans.
F 300 150
150 lbf/in
y
1
Ans.
4
6
d 4G 0.2375 11.6 10
Na
6.69 coils
8D3k
8 1.6633 150
N t = N a + 2 = 8.69 coils
Ans.
Table 10-6:
A = 147 kpsiinm, m = 0.187
A
147
Sut m
192.3 kpsi
0.23750.187
d
S sy = 0.5 S ut = 0.5(192.3) = 96.15 kpsi
Eq. (10-5):
KB
(e) Table 10-4:
Eq. (10-14):
4C 2 4 7 2
1.2
4C 3 4 7 3
Chapter 10 - Rev. A, Page 22/41
s KB
Eq. (10-7):
Fs
8Fs D
S sy
d3
d 3 S sy
8K B D
0.23753 96.15 103
8 1.2 1.663
253.5 lbf
y s = F s / k = 253.5/150 = 1.69 in
Table 10-1:
L s = N t d = 8.46(0.2375) = 2.01 in
L 0 = L s + y s = 2.01 + 1.69 = 3.70 in
Ans.
______________________________________________________________________________
10-29 For a coil radius given by:
R R1
R2 - R1
2 N
The torsion of a section is T = PR where dL = R d
U
1
T
1 2 N 3
T
dL
PR d
P
GJ
P
GJ 0
3
P 2 N
R2 R1
R
1
d
GJ 0
2 N
P
P
GJ
4
R2 R1
1 2 N
R
1
2 N
4 R2 R1
PN
2GJ ( R2
R
R)
4
2
1
R14
PN
2GJ
2 N
0
( R1 R2 ) R12 R22
16 PN
( R1 R2 ) R12 R22
4
32
Gd
4
P
d G
k
Ans.
P 16 N ( R1 R2 ) R12 R22
J
d4
p
______________________________________________________________________________
10-30 Given: F min = 4 lbf, F max = 18 lbf, k = 9.5 lbf/in, OD 2.5 in, n f = 1.5.
For a food service machinery application select A313 Stainless wire.
Table 10-5:
G = 10(106) psi
Note that for
0.013 ≤ d ≤ 0.10 in A = 169,
m = 0.146
0.10 < d ≤ 0.20 in
A = 128,
m = 0.263
18 4
18 4
Fa
7 lbf , Fm
11 lbf , r 7 / 11
2
2
Chapter 10 - Rev. A, Page 23/41
Try,
169
244.4 kpsi
(0.08)0.146
S su = 0.67S ut = 163.7 kpsi, S sy = 0.35S ut = 85.5 kpsi
d 0.080 in, Sut
Try unpeened using Zimmerli’s endurance data: S sa = 35 kpsi, S sm = 55 kpsi
S sa
35
Gerber:
S se
39.5 kpsi
2
1 (S sm / S su )
1 (55 / 163.7) 2
2
(7 / 11) 2 (163.7) 2
2(39.5)
S sa
1 1
35.0 kpsi
2(39.5)
(7
/
11)(163.7)
S sa / n f 35.0 / 1.5 23.3 kpsi
8(7) 3
8Fa
(103 )
(10 ) 2.785 kpsi
2
2
d
(0.08 )
2
C
D
KB
a
nf
Na
Nt
ymax
ys
L0
( L0 )cr
s
ns
2(23.3) 2.785
2(23.3) 2.785
3(23.3)
6.97
4(2.785)
4(2.785)
4(2.785)
Cd 6.97(0.08) 0.558 in
4C 2 4(6.97) 2
1.201
4C 3
4(6.97) 3
8(7)(0.558) 3
8F D
(10 ) 23.3 kpsi
K B a 3 1.201
3
d
(0.08 )
35 / 23.3 1.50 checks
10(106 )(0.08) 4
Gd 4
31.02 coils
8kD 3
8(9.5)(0.558)3
31.02 2 33 coils, Ls dN t 0.08(33) 2.64 in
Fmax / k 18 / 9.5 1.895 in
(1 ) ymax (1 0.15)(1.895) 2.179 in
2.64 2.179 4.819 in
D
2.63(0.558)
2.63
2.935 in
0.5
1.15(18 / 7) a 1.15(18 / 7)(23.3) 68.9 kpsi
S sy / s 85.5 / 68.9 1.24
f
kg
d DN a
2
2
9.5(386)
109 Hz
(0.08 )(0.558)(31.02)(0.283)
2
2
These steps are easily implemented on a spreadsheet, as shown below, for different
diameters.
Chapter 10 - Rev. A, Page 24/41
d1
0.080
d
0.146
m
169.000
A
244.363
S ut
163.723
S su
85.527
S sy
39.452
S se
35.000
S sa
23.333
2.785
6.977
C
0.558
D
1.201
KB
23.333
a
1.500
nf
30.993
Na
32.993
Nt
2.639
LS
2.179
ys
L0
4.818
(L 0 ) cr
2.936
69.000
s
1.240
ns
f,(Hz) 108.895
d2
0.0915
0.146
169.000
239.618
160.544
83.866
39.654
35.000
23.333
2.129
9.603
0.879
1.141
23.333
1.500
13.594
15.594
1.427
2.179
3.606
4.622
69.000
1.215
114.578
d3
0.1055
0.263
128
231.257
154.942
80.940
40.046
35.000
23.333
1.602
13.244
1.397
1.100
23.333
1.500
5.975
7.975
0.841
2.179
3.020
7.350
69.000
1.173
118.863
d4
0.1205
0.263
128
223.311
149.618
78.159
40.469
35.000
23.333
1.228
17.702
2.133
1.074
23.333
1.500
2.858
4.858
0.585
2.179
2.764
11.220
69.000
1.133
121.775
The shaded areas depict conditions outside the recommended design conditions. Thus,
one spring is satisfactory. The specifications are: A313 stainless wire, unpeened, squared
and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L 0 = 3.606 in, and N t = 15.59
turns
Ans.
______________________________________________________________________________
10-31 The steps are the same as in Prob. 10-23 except that the Gerber-Zimmerli criterion is
replaced with Goodman-Zimmerli:
S se
S sa
1 S sm S su
Chapter 10 - Rev. A, Page 25/41
The problem then proceeds as in Prob. 10-23. The results for the wire sizes are shown
below (see solution to Prob. 10-23 for additional details).
d
m
A
S ut
S su
S sy
S se
S sa
C
D
Iteration of d for the first trial
d1
d2
d3
d4
d1
d2
d3
d4
0.080
0.0915 0.1055 0.1205 d
0.080 0.0915 0.1055 0.1205
0.146
0.146
0.263
0.263
1.151
1.108
1.078
1.058
KB
169.000 169.000 128.000 128.000 a
29.008 29.040 29.090 29.127
244.363 239.618 231.257 223.311 n f
1.500
1.500
1.500
1.500
163.723 160.544 154.942 149.618 N a
14.191
6.456
2.899
1.404
85.527 83.866 80.940 78.159 N t
16.191
8.456
4.899
3.404
52.706 53.239 54.261 55.345 L s
1.295
0.774
0.517
0.410
43.513 43.560 43.634 43.691 y max
2.875
2.875
2.875
2.875
29.008 29.040 29.090 29.127 L 0
4.170
3.649
3.392
3.285
2.785
2.129
1.602
1.228
(L 0 ) cr
3.809
5.924
9.354 14.219
9.052
12.309 16.856 22.433 s
85.782 85.876 86.022 86.133
0.724
1.126
1.778
2.703
0.997
0.977
0.941
0.907
ns
f (Hz) 140.040 145.559 149.938 152.966
Without checking all of the design conditions, it is obvious that none of the wire sizes
satisfy n s ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting
n f = 1.5 for Goodman makes it impossible to reach the yield line (n s < 1) . The table
below uses n f = 2.
d
m
A
S ut
S su
S sy
S se
S sa
C
D
Iteration of d for the second trial
d1
d2
d3
d4
d1
d2
d3
d4
0.080 0.0915 0.1055 0.1205 d
0.080 0.0915 0.1055 0.1205
0.146
0.146
0.263
0.263 K B
1.221
1.154
1.108
1.079
169.000 169.000 128.000 128.000 a
21.756 21.780 21.817 21.845
244.363 239.618 231.257 223.311 n f
2.000
2.000
2.000
2.000
163.723 160.544 154.942 149.618 N a
40.243 17.286
7.475
3.539
85.527 83.866 80.940 78.159 N t
42.243 19.286
9.475
5.539
52.706 53.239 54.261 55.345 L s
3.379
1.765
1.000
0.667
43.513 43.560 43.634 43.691 y max
2.875
2.875
2.875
2.875
21.756 21.780 21.817 21.845 L 0
6.254
4.640
3.875
3.542
2.785
2.129
1.602
1.228 (L 0 ) cr
2.691
4.266
6.821 10.449
6.395
8.864 12.292 16.485 s
64.336 64.407 64.517 64.600
0.512
0.811
1.297
1.986 n s
1.329
1.302
1.255
1.210
f (Hz) 98.936 104.827 109.340 112.409
The satisfactory spring has design specifications of: A313 stainless wire, unpeened,
squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, L 0 = 4.266 in, and
.N t = 19.6 turns. Ans.
______________________________________________________________________________
10-32 This is the same as Prob. 10-30 since S sa = 35 kpsi. Therefore, the specifications are:
Chapter 10 - Rev. A, Page 26/41
A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 =
0.971 in, L 0 = 3.606 in, and N t = 15.84 turns
Ans.
______________________________________________________________________________
10-33 For the Gerber fatigue-failure criterion, S su = 0.67S ut ,
S sa
S se
,
1 (S sm / S su ) 2
S sa
2
2S se
r 2 S su2
1 1
2S se
rS su
The equation for S sa is the basic difference. The last 2 columns of diameters of Ex. 10-5
are presented below with additional calculations.
0.105
0.112
0.105
0.112
d
278.691 276.096 N a
8.915
6.190
186.723 184.984 L s
1.146
0.917
38.325 38.394 L 0
3.446
3.217
125.411 124.243 (L 0 ) cr
6.630
8.160
34.658 34.652 K B
1.111
1.095
23.105 23.101 a
23.105 23.101
1.732
1.523 n f
1.500
1.500
12.004 13.851 s
70.855 70.844
C
1.260
1.551 n s
1.770
1.754
D
ID
1.155
1.439 f n
105.433 106.922
OD
1.365
1.663 fom
0.973 1.022
d
S ut
S su
S se
S sy
S sa
There are only slight changes in the results.
______________________________________________________________________________
10-34 As in Prob. 10-35, the basic change is S sa .
S sa
For Goodman,
S se
1 - ( S sm / S su )
Recalculate S sa with
rS se S su
S sa
rS su S se
Calculations for the last 2 diameters of Ex. 10-5 are given below.
Chapter 10 - Rev. A, Page 27/41
0.105
278.691
186.723
49.614
125.411
34.386
22.924
1.732
11.899
C
1.249
D
ID
1.144
OD
1.354
d
S ut
S su
S se
S sy
S sa
0.112
276.096
184.984
49.810
124.243
34.380
22.920
1.523
13.732
1.538
1.426
1.650
0.105
0.112
d
9.153
6.353
Na
1.171
0.936
Ls
L0
3.471
3.236
(L 0 ) cr
6.572
8.090
1.112
1.096
KB
22.924 22.920
a
1.500
1.500
nf
70.301 70.289
s
1.784
1.768
ns
104.509 106.000
fn
fom
0.986 1.034
There are only slight differences in the results.
______________________________________________________________________________
10-35 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1.
140
Try
d 0.067 in, Sut
234.0 kpsi
(0.067)0.190
Table 10-6: S sy = 0.45S ut = 105.3 kpsi
Table 10-7: S y = 0.75S ut = 175.5 kpsi
Eq. (10-34) with D/d = C and C 1 = C
S
F
A max2 [( K ) A (16C ) 4] y
d
ny
d 2S y
4C 2 C 1
(16C ) 4
n y Fmax
4C (C 1)
d 2S y
1
4C 2 C 1 (C 1)
4n y Fmax
2
2
d Sy
1
1 d Sy
C 2 1
1 C
2 0
4
4n y Fmax
4 4n y Fmax
2
d 2S y
1 d Sy
C
16n y Fmax
2 16n y Fmax
1 (0.067 2 )(175.5)(103 )
2
16(1.5)(18)
2
d 2S y
2 take positive root
4
n
F
y max
2
(0.067) 2 (175.5)(103 )
(0.067) 2 (175.5)(103 )
2 4.590
16(1.5)(18)
4(1.5)(18)
Chapter 10 - Rev. A, Page 28/41
D Cd 4.59 0.067 0.3075 in
Fi
d 3 i
8D
d3
33 500
C 3
1000 4
8D exp(0.105C )
6.5
Use the lowest F i in the preferred range. This results in the best fom.
Fi
(0.067)3
33 500
4.590 3
1000 4
6.505 lbf
8(0.3075) exp[0.105(4.590)]
6.5
For simplicity, we will round up to the next integer or half integer. Therefore, use F i = 7
lbf
18 7
k
22 lbf/in
0.5
d 4G
(0.067) 4 (11.5)(106 )
Na
45.28 turns
8kD 3
8(22)(0.3075)3
G
11.5
Nb N a
45.28
44.88 turns
E
28.6
L0 (2C 1 N b )d [2(4.590) 1 44.88](0.067) 3.555 in
L18 lbf 3.555 0.5 4.055 in
Body: K B
max
(n y ) body
r2
(K ) B
B
(n y ) B
fom
4C 2 4(4.590) 2
1.326
4C 3
4(4.590) 3
8K B Fmax D 8(1.326)(18)(0.3075) 3
(10 ) 62.1 kpsi
d3
(0.067)3
S
105.3
sy
1.70
62.1
max
2r
2(0.134)
2d 2(0.067) 0.134 in, C2 2
4
d
0.067
4C2 1
4(4) 1
1.25
4C2 4 4(4) 4
8(18)(0.3075) 3
8F D
( K ) B max3 1.25
(10 ) 58.58 kpsi
3
d
(0.067)
S
105.3
1.80
sy
58.58
B
2d 2 ( Nb 2) D
2 (0.067)2 (44.88 2)(0.3075)
(1)
0.160
4
4
Several diameters, evaluated using a spreadsheet, are shown below.
Chapter 10 - Rev. A, Page 29/41
0.067
0.072
0.076
0.081
0.085
0.09
0.095
0.104
d
233.977 230.799 228.441 225.692 223.634 221.219 218.958 215.224
S ut
105.290 103.860 102.798 101.561 100.635 99.548 98.531 96.851
S sy
175.483 173.100 171.331 169.269 167.726 165.914 164.218 161.418
Sy
4.589
5.412
6.099
6.993
7.738
8.708
9.721 11.650
C
0.307
0.390
0.463
0.566
0.658
0.784
0.923
1.212
D
F i (calc)
6.505
5.773
5.257
4.675
4.251
3.764
3.320
2.621
F i (rd)
7.0
6.0
5.5
5.0
4.5
4.0
3.5
3.0
22.000 24.000 25.000 26.000 27.000 28.000 29.000 30.000
k
45.29
27.20
19.27
13.10
9.77
7.00
5.13
3.15
Na
44.89
26.80
18.86
12.69
9.36
6.59
4.72
2.75
Nb
L0
3.556
2.637
2.285
2.080
2.026
2.071
2.201
2.605
L 18 lbf
4.056
3.137
2.785
2.580
2.526
2.571
2.701
3.105
1.326
1.268
1.234
1.200
1.179
1.157
1.139
1.115
KB
62.118 60.686 59.707 58.636 57.875 57.019 56.249 55.031
max
(n y ) body
1.695
1.711
1.722
1.732
1.739
1.746
1.752
1.760
58.576 59.820 60.495 61.067 61.367 61.598 61.712 61.712
B
(n y ) B
1.797
1.736
1.699
1.663
1.640
1.616
1.597
1.569
(n y ) A
1.500
1.500
1.500
1.500
1.500
1.500
1.500
1.500
fom
-0.138 0.135 0.133 0.135 0.138 0.154
0.160 0.144
Except for the 0.067 in wire, all springs satisfy the requirements of length and number of
coils. The 0.085 in wire has the highest fom.
______________________________________________________________________________
10-36 Given: N b = 84 coils, F i = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in.
D = OD d = 1.5 0.162 = 1.338 in
(a) Eq. (10-39):
L 0 = 2(D d) + (N b + 1)d
= 2(1.338 0.162) + (84 + 1)(0.162) = 16.12 in
2d + L 0 = 2(0.162) + 16.12 = 16.45 in overall
D 1.338
(b)
C
8.26
d
0.162
4C 2 4(8.26) 2
KB
1.166
4C 3
4(8.26) 3
8(16)(1.338)
8F D
i K B i 3 1.166
14 950 psi
(0.162)3
d
(c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi
Ans.
or
Ans.
Chapter 10 - Rev. A, Page 30/41
G
11.4
84
84.4 turns
E
28.5
d 4G
(0.162) 4 (11.4)(106 )
k
4.855 lbf/in
8D 3 N a
8(1.338)3 (84.4)
(d) Table 10-4:
A = 147 psi · inm ,
m = 0.187
147
Sut
207.1 kpsi
(0.162)0.187
S y 0.75(207.1) 155.3 kpsi
S sy 0.50(207.1) 103.5 kpsi
N a Nb
Ans.
Body
F
d 3S sy
KBD
(0.162)3 (103.5)(103 )
8(1.166)(1.338)
110.8 lbf
Torsional stress on hook point B
2r2
2(0.25 0.162 / 2)
4.086
d
0.162
4C2 1
4(4.086) 1
(K )B
1.243
4(4.086) 4
4C2 4
(0.162)3 (103.5)(103 )
103.9 lbf
F
8(1.243)(1.338)
C2
Normal stress on hook point A
2r1 1.338
8.26
d
0.162
4C12 C1 1 4(8.26) 2 8.26 1
1.099
(K ) A
4C1(C1 1)
4(8.26)(8.26 1)
4
16( K ) A D
S yt F
3
d 2
d
155.3(103 )
F
85.8 lbf
16(1.099)(1.338) / (0.162)3 4 / (0.162)2
C1
min(110.8, 103.9, 85.8) 85.8 lbf
Ans.
(e) Eq. (10-48):
F Fi
85.8 16
14.4 in Ans.
k
4.855
______________________________________________________________________________
y
Chapter 10 - Rev. A, Page 31/41
10-37 F min = 9 lbf,
F max = 18 lbf
18 9
18 9
Fa
4.5 lbf , Fm
13.5 lbf
2
2
A313 stainless:
0.013 ≤ d ≤ 0.1
A = 169 kpsi · inm , m = 0.146
0.1 ≤ d ≤ 0.2 A = 128 kpsi · inm , m = 0.263
E = 28 Mpsi, G = 10 Gpsi
Try d = 0.081 in and refer to the discussion following Ex. 10-7
169
Sut
243.9 kpsi
(0.081)0.146
S su 0.67 Sut 163.4 kpsi
S sy 0.35Sut 85.4 kpsi
S y 0.55Sut 134.2 kpsi
Table 10-8:
S r = 0.45S ut = 109.8 kpsi
Sr / 2
109.8 / 2
57.8 kpsi
Se
2
1 [Sr / (2Sut )]
1 [(109.8 / 2) / 243.9]2
r Fa / Fm 4.5 / 13.5 0.333
r 2 Sut2
Table 7-10:
Sa
1
2S e
(0.333)2 (243.92 )
Sa
1 1
2(57.8)
2
2Se
1
rSut
2(57.8)
0.333(243.9)
2
42.2 kpsi
Hook bending
16C
4
( a ) A Fa ( K ) A
2
d
d 2
4.5 (4C 2 - C - 1)16C
d 2
4C (C - 1)
Sa
S
a
(n f ) A
2
S
4 a
2
This equation reduces to a quadratic in C (see Prob. 10-35). The useable root for C is
Chapter 10 - Rev. A, Page 32/41
2
2
2
d
S
d
S
a
a
C 0.5
144
144
(0.081) 2 (42.2)(103 )
0.5
144
4.91
d 2Sa
36
2
2
(0.081) 2 (42.2)(103 )
(0.081) 2 (42.2)(103 )
2
144
36
D Cd 0.398 in
C 3
d 3 i d 3 33 500
Fi
1000 4
8D
8D exp(0.105C )
6.5
Use the lowest F i in the preferred range.
(0.081)3
33 500
4.91 3
1000 4
8(0.398) exp[0.105(4.91)]
6.5
8.55 lbf
Fi
For simplicity we will round up to next 1/4 integer.
Fi 8.75 lbf
18 9
k
36 lbf/in
0.25
d 4G
(0.081) 4 (10)(106 )
Na
23.7 turns
8kD 3
8(36)(0.398)3
G
10
Nb N a
23.7
23.3 turns
E
28
L0 (2C 1 N b )d [2(4.91) 1 23.3](0.081) 2.602 in
Lmax L0 ( Fmax Fi ) / k 2.602 (18 8.75) / 36 2.859 in
4.5(4) 4C 2 C 1
( a ) A
1
2
d C 1
18(10-3 ) 4(4.912 ) 4.91 1
1 21.1 kpsi
2
(0.081 )
4.91 1
Sa
42.2
(n f ) A
2 checks
( a ) A
21.1
Body:
KB
4C 2 4(4.91) 2
1.300
4C 3
4(4.91) 3
Chapter 10 - Rev. A, Page 33/41
8(1.300)(4.5)(0.398) 3
(10 ) 11.16 kpsi
(0.081)3
F
13.5
m m a
(11.16) 33.47 kpsi
Fa
4.5
a
The repeating allowable stress from Table 7-8 is
S sr = 0.30S ut = 0.30(243.9) = 73.17 kpsi
The Gerber intercept is
S se
73.17 / 2
38.5 kpsi
1 [(73.17 / 2) / 163.4]2
From Table 6-7,
2
2
2(33.47)(38.5)
1 163.4 11.16
(n f ) body
2.53
1 1
2 33.47 38.5
163.4(11.16)
Let r 2 = 2d = 2(0.081) = 0.162
2r
4(4) 1
C2 2 4, ( K ) B
1.25
d
4(4) 4
(K )B
1.25
( a ) B
a
(11.16) 10.73 kpsi
KB
1.30
(K )B
1.25
( m ) B
m
(33.47) 32.18 kpsi
KB
1.30
Table 10-8: (S sr ) B = 0.28S ut = 0.28(243.9) = 68.3 kpsi
68.3 / 2
( S se ) B
35.7 kpsi
1 [(68.3 / 2) / 163.4]2
2
2
2(32.18)(35.7)
1 163.4 10.73
(n f ) B
2.51
1 1
2 32.18 35.7
163.4(10.73)
Yield
Bending:
4 Fmax (4C 2 C 1)
( A ) max
1
2
d
C 1
4(18) 4(4.91) 2 4.91 1
1 (10-3 ) 84.4 kpsi
2
(0.081 )
4.91 1
134.2
(n y ) A
1.59
84.4
Body:
Chapter 10 - Rev. A, Page 34/41
i ( Fi / Fa ) a (8.75 / 4.5)(11.16) 21.7 kpsi
r a /( m i ) 11.16 / (33.47 21.7) 0.948
r
0.948
(S sy i )
(85.4 21.7) 31.0 kpsi
r 1
0.948 1
(S )
31.0
sa y
2.78
a
11.16
(S sa ) y
(n y ) body
Hook shear:
S sy 0.3Sut 0.3(243.9) 73.2 kpsi
max ( a ) B ( m ) B 10.73 32.18 42.9 kpsi
73.2
1.71
(n y ) B
42.9
7.6 2d 2 ( N b 2) D
7.6 2 (0.081) 2 (23.3 2)(0.398)
fom
1.239
4
4
A tabulation of several wire sizes follow
0.081
0.085
0.092
0.098
0.105
0.12
d
243.920 242.210 239.427 237.229 234.851 230.317
S ut
163.427 162.281 160.416 158.943 157.350 154.312
S su
109.764 108.994 107.742 106.753 105.683 103.643
Sr
57.809 57.403 56.744 56.223 55.659 54.585
Se
42.136 41.841 41.360 40.980 40.570 39.786
Sa
4.903
5.484
6.547
7.510
8.693 11.451
C
0.397
0.466
0.602
0.736
0.913
1.374
D
1.018
1.494
OD
0.478
0.551
0.694
0.834
8.572
7.874
6.798
5.987
5.141
3.637
F i (calc)
8.75
9.75
10.75
11.75
12.75
13.75
F i (rd)
36.000 36.000 36.000 36.000 36.000 36.000
k
23.86
17.90
11.38
8.03
5.55
2.77
Na
23.50
17.54
11.02
7.68
5.19
2.42
Nb
2.617
2.338
2.127
2.126
2.266
2.918
L0
2.874
2.567
2.328
2.300
2.412
3.036
L 18 lbf
21.068 20.920 20.680 20.490 20.285 19.893
( a ) A
2.000
2.000
2.000
2.000
2.000
2.000
(n f ) A
1.301
1.264
1.216
1.185
1.157
1.117
KB
11.141 10.994 10.775 10.617 10.457 10.177
( a ) body
33.424 32.982 32.326 31.852 31.372 30.532
( m ) body
73.176 72.663 71.828 71.169 70.455 69.095
S sr
38.519 38.249 37.809 37.462 37.087 36.371
S se
2.531
2.547
2.569
2.583
2.596
2.616
(n f ) body
1.250
1.250
1.250
1.250
1.250
1.250
(K) B
10.705 10.872 11.080 11.200 11.294 11.391
( a ) B
32.114 32.615 33.240 33.601 33.883 34.173
( m ) B
68.298 67.819 67.040 66.424 65.758 64.489
(S sr ) B
35.708 35.458 35.050 34.728 34.380 33.717
(S se ) B
Chapter 10 - Rev. A, Page 35/41
(n f ) B
Sy
( A ) max
(n y ) A
i
r
(S sy ) body
(S sa ) y
(n y ) body
(S sy ) B
( B ) max
(n y ) B
fom
2.519
2.463
2.388
2.341
2.298
2.235
134.156 133.215 131.685 130.476 129.168 126.674
84.273 83.682 82.720 81.961 81.139 79.573
1.592
1.592
1.592
1.592
1.592
1.592
21.663 23.820 25.741 27.723 29.629 31.097
0.945
1.157
1.444
1.942
2.906
4.703
85.372 84.773 83.800 83.030 82.198 80.611
30.958 32.688 34.302 36.507 39.109 40.832
2.779
2.973
3.183
3.438
3.740
4.012
73.176 72.663 71.828 71.169 70.455 69.095
42.819 43.486 44.321 44.801 45.177 45.564
1.709
1.671
1.621
1.589
1.560
1.516
1.246 1.234 1.245 1.283 1.357 1.639
optimal fom
The shaded areas show the conditions not satisfied.
______________________________________________________________________________
10-38 For the hook,
M = FR sin, ∂M/∂F = R sin
F
1
EI
/2
0
F R sin R d
2
FR3
2 EI
The total deflection of the body and the two hooks
FR 3 8FD 3 N b
8FD 3 N b
F ( D / 2)3
2
d 4G
d 4G
E ( / 64)(d 4 )
2 EI
8FD 3
G 8FD 3 N a
4 Nb
d G
E
d 4G
G
Q.E.D.
N a Nb
E
______________________________________________________________________________
10-39 Table 10-5 (d = 4 mm = 0.1575 in): E = 196.5 GPa
Table 10-4 for A227:
Eq. (10-14):
Eq. (10-57):
A = 1783 MPa · mmm,
m = 0.190
A
1783
Sut m 0.190 1370 MPa
d
4
S y = all = 0.78 S ut = 0.78(1370) = 1069 MPa
Chapter 10 - Rev. A, Page 36/41
D = OD d = 32 4 = 28 mm
C = D/d = 28/4 = 7
2
4C 2 C 1 4 7 7 1
1.119
Ki
4C (C 1)
4(7)(7 1)
Eq. (10-43):
32 Fr
d3
At yield, Fr = M y , = S y . Thus,
Ki
Eq. (10-44):
My
d 3S y
32Ki
43 1069 103
32(1.119)
6.00 N · m
Count the turns when M = 0
N 2.5
where from Eq. (10-51):
My
k
d 4E
k
10.8DN
Thus,
N 2.5
My
d 4 E / (10.8DN )
Solving for N gives
N
2.5
1 [10.8DM y / (d 4 E )]
2.5
1 10.8(28)(6.00) / 44 (196.5)
This means (2.5 - 2.413)(360) or 31.3 from closed.
2.413 turns
Ans.
Treating the hand force as in the middle of the grip,
87.5
68.75 mm
r 112.5 87.5
2
6.00 103
M
Fmax y
87.3 N Ans.
r
68.75
______________________________________________________________________________
10-40 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500,
(a) D = 0.500 0.081 = 0.419 in
Using E = 28.6 Mpsi for an estimate
Chapter 10 - Rev. A, Page 37/41
k
d 4E
(0.081) 4 (28.6)(106 )
24.7 lbf · in/turn
10.8DN
10.8(0.419)(11)
for each spring. The moment corresponding to a force of 8 lbf
Fr = (8/2)(3.3125) = 13.25 lbf · in/spring
The fraction windup turn is
n
Fr 13.25
0.536 turns
k
24.7
The arm swings through an arc of slightly less than 180, say 165. This uses up
165/360 or 0.458 turns. So n = 0.536 0.458 = 0.078 turns are left (or
0.078(360) = 28.1 ). The original configuration of the spring was
Ans.
(b)
D 0.419
5.17
d
0.081
4C 2 C 1 4(5.17) 2 5.17 1
1.168
Ki
4C C 1
4(5.17)(5.17 1)
C
Ki
32(13.25)
32M
1.168
297 103 psi 297 kpsi
3
3
d
(0.081)
Ans.
To achieve this stress level, the spring had to have set removed.
______________________________________________________________________________
10-41 (a) Consider half and double results
Straight section:
M = 3FR,
M
3R
P
Chapter 10 - Rev. A, Page 38/41
Upper 180 section:
M F [ R R(1 cos )]
M
FR(2 cos ),
R(2 cos )
F
M
R sin
F
M = FR sin ,
Lower section:
Considering bending only:
U
2 l/2
2
9
FR
dx
FR 2 (2 cos ) 2 R d
0
0
EI
F
2F 9 2
R l R 3 4 4sin 0 R 3
2
EI 2
4
2
2
2FR 19
9 FR
R l
(19 R 18l )
EI 4
2 2EI
/2
0
F ( R sin ) 2 R d
The spring rate is
k
F
2 EI
R (19 R 18 l )
2
Ans.
(b) Given: A227 HD wire, d = 2 mm, R = 6 mm, and l = 25 mm.
Table 10-5 (d = 2 mm = 0.0787 in):
k
2 197.2 109 0.0024 / 64
0.006 19 0.006 18 0.025
2
E = 197.2 MPa
10.65 103 N/m 10.65 N/mm
Ans.
(c) The maximum stress will occur at the bottom of the top hook where the bendingmoment is 3FR and the axial fore is F. Using curved beam theory for bending,
Eq. (3-65), p. 119:
Axial:
a
i
Mci
3FRci
Aeri d 2 / 4 e R d / 2
F
F
A d2 / 4
Chapter 10 - Rev. A, Page 39/41
Combining,
max i a
F
4F
d2
3Rci
1 S y
e R d / 2
d 2Sy
3Rci
1
4
e R d / 2
Ans.
(1)
For the clip in part (b),
S ut = A/dm = 1783/20.190 = 1563 MPa
Eq. (10-14) and Table 10-4:
Eq. (10-57):
S y = 0.78 S ut = 0.78(1563) = 1219 MPa
Table 3-4, p. 121:
rn
12
2 6 62 12
5.95804 mm
e = r c r n = 6 5.95804 = 0.04196 mm
c i = r n (R d /2) = 5.95804 (6 2/2) = 0.95804 mm
Eq. (1):
F
0.0022 1219 106
Ans.
46.0 N
3 6 0.95804
4
1
0.04196 6 1
______________________________________________________________________________
10-42 (a)
Chapter 10 - Rev. A, Page 40/41
M
x
F
M Fx,
0 xl
M Fl FR 1 cos ,
M
l R 1 cos
F
0 l
/2
2
1 l
(
)
1
cos
Fx
x
dx
F
l
R
Rd
0
EI 0
F
4l 3 3R 2 l 2 4 2 l R 3 8 R 2
12 EI
F
The spring rate is
k
F
F
12 EI
4l 3R 2 l 4 2 l R 3 8 R 2
3
Ans.
2
(b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in.
Table 10-5:
E = 28 Mpsi
I
k
64
d4
64
0.063 7.733 10 in
4
7
4
12 28 106 7.733107
4 0.53 3 0.625 2 0.52 4 2 0.5 0.625 3 8 0.6252
36.3 lbf/in
(c) Table 10-4:
Ans.
A = 169 kpsiinm, m = 0.146
Eq. (10-14):
S ut = A/ d m = 169/0.0630.146 = 253.0 kpsi
Eq. (10-57):
S y = 0.61 S ut = 0.61(253.0) = 154.4 kpsi
One can use curved beam theory as in the solution for Prob. 10-41. However, the
equations developed in Sec. 10-12 are equally valid.
C = D/d = 2(0.625 + 0.063/2)/0.063 = 20.8
Eq. (10-43):
Ki
2
4C 2 C 1 4 20.8 20.8 1
1.037
4C C 1
4 20.8 20.8 1
Eq. (10-44), setting = S y :
Chapter 10 - Rev. A, Page 41/41
Ki
32 Fr
Sy
d3
Solving for F yields
F = 3.25 lbf
1.037
32 F 0.5 0.625
0.063
3
154.4 103
Ans.
Try solving part (c) of this problem using curved beam theory. You should obtain the
same answer.
______________________________________________________________________________
10-43 (a)
M = Fx
M
Fx
Fx
2
I / c I / c bh / 6
Constant stress,
bh 2 Fx
6
6 Fx
b
h
(1)
Ans.
At x = l,
6 Fl
b
ho
(b)
h ho x / l
Ans.
M = Fx, M / F = x
l
y
0
M M / F
EI
dx
l
l
Fx x
1
12 Fl 3/2 1/ 2
dx
x dx
E 0 121 bho3 x / l 3/2
bho3 E 0
2 12 Fl 3/2 3/ 2 8 Fl 3
l 3
3 bho3 E
bho E
F bho3 E
Ans.
y
8l 3
______________________________________________________________________________
k
10-44 Computer programs will vary.
______________________________________________________________________________
10-45 Computer programs will vary.
Chapter 10 - Rev. A, Page 42/41
Chapter 11
11-1
For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples
of rating life, is
L
60D nD 60 25000 350
xD D
525 Ans.
LR
L10
106
The design radial load is
FD 1.2 2.5 3.0 kN
1/3
Eq. (11-6):
525
C10 3.0
1/1.483
0.02 4.459 0.02 ln 1/ 0.9
C 10 = 24.3 kN
Table 11-2:
Ans.
Choose an 02-35 mm bearing with C 10 = 25.5 kN.
Ans.
525 3 / 25.5 3 0.02 1.483
Eq. (11-18): R exp
Ans.
0.920
4.459 0.02
______________________________________________________________________________
11-2
For the angular-contact 02-series ball bearing as described, the rating life multiple is
L
60D nD 60 40 000 520
xD D
1248
LR
L10
106
The design radial load is
FD 1.4 725 1015 lbf 4.52 kN
Eq. (11-6):
1/3
1248
C10 1015
1/1.483
0.02 4.459 0.02 ln 1/ 0.9
10 930 lbf 48.6 kN
Select an 02-60 mm bearing with C 10 = 55.9 kN. Ans.
1.483
3
1248 4.52 / 55.9 0.02
Eq. (11-18): R exp
Ans.
0.945
4.439
______________________________________________________________________________
Table 11-2:
Chapter 11, Page 1/28
11-3
For the straight-roller 03-series bearing selection, x D = 1248 rating lives from Prob. 11-2
solution.
FD 1.4 2235 3129 lbf 13.92 kN
1248
C10 13.92
1
Table 11-3:
3/10
118 kN
Select an 03-60 mm bearing with C 10 = 123 kN.
Ans.
1.483
10/3
1248 13.92 /123 0.02
Eq. (11-18): R exp
0.917 Ans.
4.459 0.02
______________________________________________________________________________
11-4
The combined reliability of the two bearings selected in Probs. 11-2 and 11-3 is
R 0.945 0.917 0.867
Ans.
We can choose a reliability goal of 0.90 0.95 for each bearing. We make the
selections, find the existing reliabilities, multiply them together, and observe that the
reliability goal is exceeded due to the roundup of capacity upon table entry.
Another possibility is to use the reliability of one bearing, say R 1 . Then set the reliability
goal of the second as
R2
0.90
R1
or vice versa. This gives three pairs of selections to compare in terms of cost, geometry
implications, etc.
______________________________________________________________________________
11-5
Establish a reliability goal of
contact ball bearing,
0.90 0.95 for each bearing. For an 02-series angular
1/3
1248
C10 1015
1/1.483
0.02 4.439 ln 1/ 0.95
12822 lbf 57.1 kN
Select an 02-65 mm angular-contact bearing with C 10 = 63.7 kN.
1.483
3
1248 4.52 / 63.7 0.02
RA exp
0.962
4.439
Chapter 11, Page 2/28
For an 03-series straight roller bearing,
3/10
1248
C10 13.92
1/1.483
0.02 4.439 ln 1/ 0.95
136.5 kN
Select an 03-65 mm straight-roller bearing with C 10 = 138 kN.
1248 13.92 /138 10/3 0.02 1.483
RB exp
0.953
4.439
The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal.
______________________________________________________________________________
11-6
For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and
F R = 20 kN.
L
60D nD 60 8000 950
xD D
456
LR
L10
106
3/10
456
C10 20
145 kN
Ans.
1/1.483
0.02 4.439 ln 1/ 0.95
______________________________________________________________________________
11-7
Both bearings need to be rated in terms of the same catalog rating system in order to
compare them. Using a rating life of one million revolutions, both bearings can be rated
in terms of a Basic Load Rating.
1/ a
Eq. (11-3):
L
C A FA A
LR
8.96 kN
1/ a
n 60
FA A A
LR
3000 500 60
2.0
106
1/3
Bearing B already is rated at one million revolutions, so C B = 7.0 kN. Since C A > C B ,
bearing A can carry the larger load.
Ans.
______________________________________________________________________________
11-8
F D = 2 kN, L D = 109 rev, R = 0.90
1/ a
1/3
LD
109
Eq. (11-3):
C10 FD
2 6 20 kN Ans.
10
LR
______________________________________________________________________________
Chapter 11, Page 3/28
11-9
F D = 800 lbf, D = 12 000 hours, n D = 350 rev/min, R = 0.90
12 000 350 60
n 60
Eq. (11-3):
C10 FD D D 800
5050 lbf Ans
106
LR
______________________________________________________________________________
1/ a
1/3
11-10 F D = 4 kN, D = 8 000 hours, n D = 500 rev/min, R = 0.90
8 000 500 60
n 60
Eq. (11-3):
C10 FD D D 4
24.9 kN Ans
106
LR
______________________________________________________________________________
1/ a
1/3
11-11 F D = 650 lbf, n D = 400 rev/min, R = 0.95
D = (5 years)(40 h/week)(52 week/year) = 10 400 hours
Assume an application factor of one. The multiple of rating life is
xD
LD 10 400 400 60
249.6
LR
106
1/3
249.6
C10 1 650
Eq. (11-6):
1/1.483
0.02 4.439 ln 1/ 0.95
4800 lbf Ans.
______________________________________________________________________________
11-12 F D = 9 kN, L D = 108 rev, R = 0.99
Assume an application factor of one. The multiple of rating life is
xD
LD 108
100
LR 106
1/3
100
C10 1 9
Eq. (11-6):
1/1.483
0.02 4.439 ln 1/ 0.99
69.2 kN Ans.
______________________________________________________________________________
11-13 F D = 11 kips, D = 20 000 hours, n D = 200 rev/min, R = 0.99
Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 on
p. 608.
Chapter 11, Page 4/28
The multiple of rating life is
xD
LD 20 000 200 60
240
LR
106
1/3
240
C10 111
Eq. (11-6):
1/1.483
0.02 4.439 ln 1/ 0.99
113 kips Ans.
______________________________________________________________________________
11-14 From the solution to Prob. 3-68, the ground reaction force carried by the bearing at C is
R C = F D = 178 lbf. Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 15000 1200 60
1080
LR
106
1/ a
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/3
1080
C10 1.2 178
1/1.483
0.02 4.459 0.02 1 0.95
2590 lbf Ans.
______________________________________________________________________________
11-15 From the solution to Prob. 3-69, the ground reaction force carried by the bearing at C is
R C = F D = 1.794 kN. Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 15000 1200 60
1080
LR
106
1/ a
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/3
1080
C10 1.2 1.794
1/1.483
0.02 4.459 0.02 1 0.95
26.1 kN Ans.
______________________________________________________________________________
11-16 From the solution to Prob. 3-70, R Cz = –327.99 lbf, R Cy = –127.27 lbf
1/ 2
2
2
RC FD 327.99 127.27 351.8 lbf
Use the Weibull parameters for Manufacturer 2 on p. 608.
Chapter 11, Page 5/28
xD
LD 15000 1200 60
1080
LR
106
1/ a
Eq. (11-7):
xD
C10 a f FD
1/ b
xo xo 1 RD
1/3
1080
C10 1.2 351.8
1/1.483
0.02 4.459 0.02 1 0.95
5110 lbf Ans.
______________________________________________________________________________
11-17 From the solution to Prob. 3-71, R Cz = –150.7 N, R Cy = –86.10 N
1/2
2
2
RC FD 150.7 86.10 173.6 N
Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 15000 1200 60
1080
LR
106
1/ a
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/3
1080
C10 1.2 173.6
1/1.483
0.02 4.459 0.02 1 0.95
2520 N Ans.
______________________________________________________________________________
11-18 From the solution to Prob. 3-77, R Az = 444 N, R Ay = 2384 N
RA FD 4442 23842
1/2
2425 N 2.425 kN
Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to
the speed of shaft AD,
d
125
nD F ni
191 95.5 rev/min
dC
250
xD
LD 12 000 95.5 60
68.76
LR
106
1/ a
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
Chapter 11, Page 6/28
1/3
68.76
C10 1 2.425
1/1.483
0.02 4.459 0.02 1 0.95
11.7 kN Ans.
______________________________________________________________________________
11-19 From the solution to Prob. 3-79, R Az = 54.0 lbf, R Ay = 140 lbf
RA FD 54.02 1402
1/2
150.1 lbf
Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to
the speed of shaft AD,
d
10
nD F ni 280 560 rev/min
dC
5
xD
Eq. (11-7):
LD 14 000 560 60
470.4
LR
106
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/ a
3/10
470.4
C10 1150.1
1/1.483
0.02 4.459 0.02 1 0.98
1320 lbf Ans.
______________________________________________________________________________
11-20 (a) Fa 3 kN, Fr 7 kN, nD 500 rev/min, V 1.2
From Table 11-2, with a 65 mm bore, C 0 = 34.0 kN.
F a / C 0 = 3 / 34 = 0.088
From Table 11-1, 0.28 e 3.0.
Fa
3
0.357
VFr 1.2 7
Since this is greater than e, interpolating Table 11-1 with F a / C 0 = 0.088, we obtain
X 2 = 0.56 and Y 2 = 1.53.
Eq. (11-9):
Fe X iVFr Yi Fa 0.56 1.2 7 1.53 3 9.29 kN
F e > F r so use F e .
Ans.
(b) Use Eq. (11-7) to determine the necessary rated load the bearing should have to carry
the equivalent radial load for the desired life and reliability. Use the Weibull
parameters for Manufacturer 2 on p. 608.
Chapter 11, Page 7/28
xD
LD 10 000 500 60
300
LR
106
1/ a
xD
Eq. (11-7): C10 a f FD
1/ b
x0 x0 1 RD
300
C10 1 9.29
1/1.483
0.02 4.459 0.02 1 0.95
73.4 kN
1/3
From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the
necessary rating to meet the specifications. This bearing should not be expected to meet
the load, life, and reliability goals.
Ans.
______________________________________________________________________________
11-21 (a) Fa 2 kN, Fr 5 kN, nD 400 rev/min, V 1
From Table 11-2, 30 mm bore, C 10 = 19.5 kN, C 0 = 10.0 kN
F a / C 0 = 2 / 10 = 0.2
From Table 11-1, 0.34 e 0.38.
Fa
2
0.4
VFr 1 5
Since this is greater than e, interpolating Table 11-1, with F a / C 0 = 0.2, we obtain X 2 =
0.56 and Y 2 = 1.27.
Eq. (11-9): Fe X iVFr Yi Fa 0.56 1 5 1.27 2 5.34 kN
Ans.
F e > F r so use F e .
(b) Solve Eq. (11-7) for x D .
C
xD 10
a f FD
a
1/ b
x0 x0 1 RD
3
19.5
0.02 4.459 0.02 1 0.99 1/1.483
xD
1 5.34
xD 10.66
xD
LD D nD 60
LR
106
Chapter 11, Page 8/28
D
10.66 10 444 h
xD 106
6
Ans.
nD 60 400 60
______________________________________________________________________________
11-22 Fr 8 kN, R 0.9, LD 109 rev
1/ a
Eq. (11-3):
L
C10 FD D
LR
1/3
109
8 6
10
80 kN
From Table 11-2, select the 85 mm bore. Ans.
______________________________________________________________________________
11-23 Fr 8 kN, Fa 2 kN, V 1, R 0.99
Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 10 000 400 60
240
LR
106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-9):
Fe 0.56 1 8 1.63 2 7.74 kN
F e < F r , so just use F r as the design load.
Eq. (11-7):
xD
C10 a f FD
1/ b
xo xo 1 RD
1/ a
1/3
240
C10 1 8
82.5 kN
1/1.483
0.02 4.459 0.02 1 0.99
From Table 11-2, try 85 mm bore with C 10 = 83.2 kN, C 0 = 53.0 kN
Iterate the previous process:
F a / C 0 = 2 / 53 = 0.038
0.22 e 0.24
Fa
2
0.25 e
VFr 1 8
Interpolate Table 11-1 with F a / C 0 = 0.038 to obtain X 2 = 0.56 and Y 2 = 1.89.
Table 11-1:
Eq. (11-9):
Fe 0.56(1)8 1.89(2) 8.26 > Fr
Eq. (11-7):
240
C10 1 8.26
1/1.483
0.02 4.459 0.02 1 0.99
1/3
85.2 kN
Chapter 11, Page 9/28
Table 11-2: Move up to the 90 mm bore with C 10 = 95.6 kN, C 0 = 62.0 kN.
Iterate again:
F a / C 0 = 2 / 62 = 0.032
Again, 0.22 e 0.24
Fa
2
0.25 e
VFr 1 8
Interpolate Table 11-1 with F a / C 0 = 0.032 to obtain X 2 = 0.56 and Y 2 = 1.95.
Table 11-1:
Eq. (11-9):
Fe 0.56(1)8 1.95(2) 8.38 > Fr
1/3
240
Eq. (11-7):
C10 1 8.38
86.4 kN
1/1.483
0.02 4.459 0.02 1 0.99
The 90 mm bore is acceptable. Ans.
______________________________________________________________________________
11-24 Fr 8 kN, Fa 3 kN, V 1.2, R 0.9, LD 108 rev
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-9):
Fe 0.56 1.2 8 1.63 3 10.3 kN
Fe Fr
1/ a
Eq. (11-3):
L
C10 Fe D
LR
1/3
108
10.3 6
10
47.8 kN
From Table 11-2, try 60 mm with C 10 = 47.5 kN, C 0 = 28.0 kN
Iterate the previous process:
F a / C 0 = 3 / 28 = 0.107
0.28 e 0.30
Fa
3
0.313 e
VFr 1.2 8
Interpolate Table 11-1 with F a / C 0 = 0.107 to obtain X 2 = 0.56 and Y 2 = 1.46
Table 11-1:
Eq. (11-9):
Fe 0.56 1.2 8 1.46 3 9.76 kN > Fr
1/3
108
Eq. (11-3):
C10 9.76 6 45.3 kN
10
From Table 11-2, we have converged on the 60 mm bearing. Ans.
______________________________________________________________________________
Chapter 11, Page 10/28
11-25 Fr 10 kN, Fa 5 kN, V 1, R 0.95
Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 12 000 300 60
216
LR
106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-9):
Fe 0.56 110 1.63 5 13.75 kN
F e > F r , so use F e as the design load.
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/ a
1/3
216
C10 113.75
1/1.483
0.02 4.459 0.02 1 0.95
97.4 kN
From Table 11-2, try 95 mm bore with C 10 = 108 kN, C 0 = 69.5 kN
Iterate the previous process:
F a / C 0 = 5 / 69.5 = 0.072
Table 11-1:
0.27 e 0.28
Fa
5
0.5 e
VFr 110
Interpolate Table 11-1 with F a / C 0 = 0.072 to obtain X 2 = 0.56 and Y 2 = 1.62 1.63
Since this is where we started, we will converge back to the same bearing. The 95 mm
bore meets the requirements. Ans.
______________________________________________________________________________
11-26 Note to the Instructor. In the first printing of the 9th edition, the design life was
incorrectly given to be 109 rev and will be corrected to 108 rev in subsequent printings.
We apologize for the inconvenience.
Fr 9 kN, Fa 3 kN, V 1.2, R 0.99
Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 108
100
LR 106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Chapter 11, Page 11/28
Eq. (11-9):
Fe 0.56 1.2 9 1.63 3 10.9 kN
F e > F r , so use F e as the design load.
Eq. (11-7):
xD
C10 a f FD
1/ b
x0 x0 1 RD
1/ a
1/3
100
C10 110.9
1/1.483
0.02 4.459 0.02 1 0.99
83.9 kN
From Table 11-2, try 90 mm bore with C 10 = 95.6 kN, C 0 = 62.0 kN. Try this bearing.
Iterate the previous process:
F a / C 0 = 3 / 62 = 0.048
0.24 e 0.26
Fa
3
0.278 e
VFr 1.2 9
Interpolate Table 11-1 with F a / C 0 = 0.048 to obtain X 2 = 0.56 and Y 2 = 1.79
Table 11-1:
Eq. (11-9):
Fe 0.56 1.2 9 1.79 3 11.4 kN Fr
11.4
83.9 87.7 kN
10.9
From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the
requirements. Ans.
______________________________________________________________________________
C10
11-27 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2
From Prob. 3-72, R Cy = 183.1 lbf, R Cz = –861.5 lbf.
1/2
2
RC FD 183.12 861.5 881 lbf
15000 1200 60
L
xD D
1080
LR
106
1/3
1080
Eq. (11-7): C10 1.2 881
1/1.483
0.02 4.439 1 0.95
12800 lbf 12.8 kips Ans.
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
Chapter 11, Page 12/28
11-28 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2
From Prob. 3-72, R Oy = –208.5 lbf, R Oz = 259.3 lbf.
1/ 2
2
RC FD 259.32 208.5 333 lbf
15000 1200 60
L
xD D
1080
LR
106
1/3
1080
Eq. (11-7): C10 1.2 333
1/1.483
0.02 4.439 1 0.95
4837 lbf 4.84 kips Ans.
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-29 (a) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2
From Prob. 3-73, R Cy = 8.319 kN, R Cz = –10.830 kN.
2 1/ 2
RC FD 8.319 2 10.830 13.7 kN
12 000 900 60
L
xD D
648
LR
106
1/3
648
Eq. (11-7): C10 1.2 13.7
204 kN Ans.
1/1.483
0.02 4.439 1 0.98
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-30 (a) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2
From Prob. 3-73, R Oy = 5083 N, R Oz = 494 N.
RC FD 50832 4942
xD
1/2
5106 N 5.1 kN
LD 12 000 900 60
648
LR
106
1/3
648
Eq. (11-7): C10 1.2 5.1
76.1 kN Ans.
1/1.483
0.02 4.439 1 0.98
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
Chapter 11, Page 13/28
11-31 Assume concentrated forces as shown.
Pz 8 28 224 lbf
Py 8 35 280 lbf
T 224 2 448 lbf in
T x 448 1.5 F cos 20 0
448
F
318 lbf
1.5 0.940
M Oz 5.75Py 11.5RAy 14.25F sin 20 0
5.75 280 11.5RAy 14.25 318 0.342 0
RAy 5.24 lbf
M Oy 5.75 Pz 11.5 RAz 14.25 F cos 20 0
5.75 224 11.5RAz 14.25 318 0.940 0
1/2
2
2
RA 482 5.24
z
z
z
F RO Pz RA F cos 20 0
RAz 482 lbf;
482 lbf
ROz 224 482 318 0.940 0
ROz 40.9 lbf
F y ROy Py RAy F sin 20 0
ROy 280 5.24 318 0.342 0
ROy 166 lbf
1/2
2
2
RO 40.9 166
171 lbf
So the reaction at A governs.
Reliability Goal: 0.92 0.96
FD 1.2 482 578 lbf
xD 35 000 350 60 / 106 735
1/3
735
C10 578
1/1.483
0.02 4.459 0.02 ln 1/ 0.96
6431 lbf 28.6 kN
From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of
Chapter 11, Page 14/28
31.9 kN.
Ans.
______________________________________________________________________________
11-32 For a combined reliability goal of 0.95, use 0.95 0.975 for the individual bearings.
xD
40 000 420 60
106
1008
The resultant of the given forces are
R O = [(–387)2 + 4672]1/2 = 607 lbf
R B = [3162 + (–1615)2]1/2 = 1646 lbf
At O:
1/3
1008
C10 1.2 607
Eq. (11-6):
1/1.483
0.02 4.459 0.02 ln 1/ 0.975
9978 lbf 44.4 kN
From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load
rating of 46.2 kN. Ans.
At B:
3/10
Eq. (11-6):
1008
C10 1.2 1646
1/1.483
0.02 4.459 0.02 ln 1/ 0.975
20827 lbf 92.7 kN
From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller. Ans.
______________________________________________________________________________
11-33 The reliability of the individual bearings is R 0.98 0.9899
Chapter 11, Page 15/28
From statics,
T = (270 50) = (P 1 P 2 )125
= (P 1 0.15 P 1 )125
P 1 = 310.6 N,
P 2 = 0.15 (310.6) = 46.6 N
P 1 + P 2 = 357.2 N
FAy 357.2 sin 45 252.6 N FAz
M
F
M
F
850REy 300(252.6) 0 REy 89.2 N
z
O
y
252.6 89.2 ROy 0 ROy 163.4 N
850REz 700(320) 300(252.6) 0 REz 174.4 N
y
O
z
174.4 320 252.6 ROz 0 ROz 107 N
RO
163.4
2
RE
89.2
174.4 196 N
2
107 2 195 N
2
The radial loads are nearly the same at O and E. We can use the same bearing at both
locations.
xD
60 000 1500 60
106
5400
1/3
Eq. (11-6):
5400
C10 1 0.196
1/1.483
0.02 4.439 ln 1/ 0.9899
5.7 kN
From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating
of 6.89 kN. Ans.
______________________________________________________________________________
11-34
R 0.96 0.980
T 12(240 cos 20 ) 2706 lbf in
F
2706
498 lbf
6 cos 25
In xy-plane:
M Oz 16(82.1) 30(210) 42 RCy 0
Chapter 11, Page 16/28
RCy 181 lbf
ROy 82.1 210 181 111.1 lbf
In xz-plane:
M Oy 16(226) 30(451) 42 RCz 0
RCz 236 lbf
ROz 226 451 236 11 lbf
181
RO 111.12 112
RC
xD
2
236 2
1/ 2
112 lbf
Ans.
1/ 2
297 lbf
Ans.
50000 300 60
106
900
1/3
C10 O
900
1.2 112
1/1.483
0.02 4.439 ln 1/ 0.980
1860 lbf 8.28 kN
1/3
C10 C
900
1.2 297
1/1.483
0.02 4.439 ln 1/ 0.980
4932 lbf 21.9 kN
Bearing at O: Choose a deep-groove 02-17 mm. Ans.
Bearing at C: Choose a deep-groove 02-35 mm. Ans.
______________________________________________________________________________
11-35 Shafts subjected to thrust can be constrained by bearings, one of which supports the
thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust
force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is
heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may
not contribute measurably to the chance of failure. If this is the case, we may be able to
obtain the desired combined reliability with bearing A having a reliability near 0.99 and
bearing B having a reliability near 1. This would allow for bearing A to have a lower
capacity than if it needed to achieve a reliability of 0.99 . To determine if this is the
case, we will start with bearing B.
Bearing B (straight roller bearing)
30000 500 60
xD
900
106
Fr 36 2 67 2
1/ 2
76.1 lbf 0.339 kN
Try a reliability of 1 to see if it is readily obtainable with the available bearings.
Chapter 11, Page 17/28
3/10
Eq. (11-6):
900
C10 1.2 0.339
1/1.483
0.02 4.439 ln 1/1.0
10.1 kN
The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN.
Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans.
Bearing at A (angular-contact ball)
With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99
if bearing A has a reliability of 0.99.
Fr 36 2 2122
1/ 2
215 lbf 0.957 kN
Fa 555 lbf 2.47 kN
Trial #1:
Tentatively select an 02-85 mm angular-contact with C 10 = 90.4 kN and C 0 = 63.0 kN.
Fa 2.47
0.0392
C0 63.0
30000 500 60
xD
900
106
Table 11-1:
Interpolating, X 2 = 0.56, Y 2 = 1.88
Eq. (11-9):
Fe 0.56 0.957 1.88 2.47 5.18 kN
1/3
Eq. (11-6):
900
C10 1.2 5.18
1/1.483
0.02 4.439 ln 1/ 0.99
99.54 kN 90.4 kN
Trial #2:
Tentatively select a 02-90 mm angular-contact ball with C 10 = 106 kN and C 0 = 73.5 kN.
Fa 2.47
0.0336
C0 73.5
Table 11-1:
Interpolating, X 2 = 0.56, Y 2 = 1.93
Fe 0.56 0.957 1.93 2.47 5.30 kN
1/3
900
C10 1.2 5.30
1/1.483
0.02 4.439 ln 1/ 0.99
102 kN < 106 kN O.K.
Chapter 11, Page 18/28
Select an 02-90 mm angular-contact ball bearing. Ans.
______________________________________________________________________________
11-36 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use
line AB. In this case, B is to the right of A.
x 1
For F = 18 kN,
115 2000 60
106
13.8
This establishes point 1 on the R = 0.90 line.
The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter
Weibull distribution, x 0 = 0 and points A and B are related by [see Eq. (20-25)]:
x A ln 1/ 0.90
1/ b
(1)
xB ln 1/ 0.20
1/ b
and x B /x A is in the same ratio as 600/115. Eliminating θ,
b
ln ln 1/ 0.20 / ln 1/ 0.90
ln 600 /115
1.65
Ans.
Solving for θ in Eq. (1),
xA
ln 1/ RA
1/1.65
1
ln 1/ 0.90
1/1.65
3.91
Ans.
Chapter 11, Page 19/28
Therefore, for the data at hand,
x 1.65
R exp
3.91
Check R at point B: x B = (600/115) = 5.217
5.217 1.65
R exp
0.20
3.91
Note also, for point 2 on the R = 0.20 line,
log 5.217 log 1 log xm 2 log 13.8
xm 2 72
______________________________________________________________________________
11-37 This problem is rich in useful variations. Here is one.
Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of
(0.99)1/6 = 0.9983.
Shaft a
502
FAr 2392 1112
FBr
2
1/ 2
264 lbf 1.175 kN
1/ 2
10752
1186 lbf 5.28 kN
Thus the bearing at B controls.
xD
10 000 1200 60
106
720
0.02 4.439 ln 1/ 0.9983
1/1.483
720
C10 1.2 5.28
0.080 26
0.080 26
0.3
97.2 kN
Select either an 02-80 mm with C 10 = 106 kN or an 03-55 mm with C 10 = 102 kN.
Shaft b
393
FCr 874 2 2274 2
FDr
2
657 2
1/ 2
1/ 2
2436 lbf
766 lbf
or
or
Ans.
10.84 kN
3.41 kN
The bearing at C controls.
Chapter 11, Page 20/28
xD
10 000 240 60
106
144
144
C10 1.2 10.84
0.080 26
0.3
123 kN
Select either an 02-90 mm with C 10 = 142 kN or an 03-60 mm with C 10 = 123 kN.
Shaft c
417
FEr 11132 23852
FFr
8952
2
1/ 2
1/ 2
2632 lbf
987 lbf
or
or
Ans.
11.71 kN
4.39 kN
The bearing at E controls.
xD
10 000 80 60
106
48
48
C10 1.2 11.71
0.080 26
0.3
95.7 kN
Select an 02-80 mm with C 10 = 106 kN or an 03-60 mm with C 10 = 123 kN. Ans.
______________________________________________________________________________
11-38 Express Eq. (11-1) as
F1a L1 C10a L10 K
For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C 10 = 20.3 kN.
K 20.3 106 8.365 109
3
At a load of 18 kN, life L 1 is given by:
9
K 8.365 10
L1 a
1.434 106 rev
3
18
F1
For a load of 30 kN, life L 2 is:
L2
0.310 10 rev
30
8.365 109
6
3
In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be
expressed as
Chapter 11, Page 21/28
l1 l2
1
L1 L2
Substituting,
l2
200 000
1
6
1.434 10
0.310 106
l2
0.267 10 rev Ans.
6
______________________________________________________________________________
11-39 Total life in revolutions
Let:
l = total turns
f 1 = fraction of turns at F 1
f 2 = fraction of turns at F 2
From the solution of Prob. 11-38, L 1 = 1.434(106) rev and L 2 = 0.310(106) rev.
Palmgren-Miner rule:
l1 l2
fl f l
1 2 1
L1 L2 L1 L2
from which
l
1
f1 / L1 f 2 / L2
l
0.40 / 1.434 10 0.60 / 0.310 10
451 585 rev
1
6
6
Ans.
Total life in loading cycles
4 min at 2000 rev/min = 8000 rev/cycle
6 min at 2000 rev/min = 12 000 rev/cycle
Total rev/cycle = 8000 + 12 000 = 20 000
451 585 rev
22.58 cycles
20 000 rev/cycle
Ans.
Chapter 11, Page 22/28
Total life in hours
min 22.58 cycles
10
3.76 h Ans.
cycle 60 min/h
______________________________________________________________________________
FrA 560 lbf
FrB 1095 lbf
Fae 200 lbf
11-40
xD
LD 40 000 400 60
10.67
LR
90 106
R 0.90 0.949
Eq. (11-15):
Eq. (11-15):
0.47 FrA 0.47 560
175.5 lbf
KA
1.5
0.47 FrB 0.47 1095
FiB
343.1 lbf
KB
1.5
FiA
FiA ? FiB Fae
175.5 lbf 343.1 200 543.1 lbf, so Eq. (11-16) applies.
We will size bearing B first since its induced load will affect bearing A, but is not itself
affected by the induced load from bearing A [see Eq. (11-16)].
From Eq. (11-16b), F eB = F rB = 1095 lbf.
Eq. (11-7):
10.67
FRB 1.4 1095
4.48 1 0.949 1/1.5
3/10
3607 lbf
Ans.
Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95.
Ans.
With bearing B selected, we re-evaluate the induced load from bearing B using the actual
value for K.
0.47 FrB 0.47 1095
Eq. (11-15): FiB
263.9 lbf
KB
1.95
Find the equivalent radial load for bearing A from Eq. (11-16), which still applies.
Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
FeA 0.4 560 1.5 263.9 200 920 lbf
Chapter 11, Page 23/28
FeA FrA
Eq. (11-7):
10.67
FRA 1.4 920
4.48 1 0.949 1/1.5
3/10
3030 lbf
Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with
K = 1.07. Iterating with the new value for K, we get F eA = 702 lbf and F rA = 2312 lbf.
Ans.
By using a bearing with a lower K, the rated load decreased significantly, providing a
higher than requested reliability. Further examination with different combinations of
bearing choices could yield additional acceptable solutions.
______________________________________________________________________________
11-41 The thrust load on shaft CD is from the axial component of the force transmitted through
the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct
mounted bearings would allow bearing C to carry the thrust load. Ans.
From the solution to Prob. 3-74, the axial thrust load is F ae = 362.8 lbf, and the bearing
radial forces are F Cx = 287.2 lbf, F Cz = 500.9 lbf, F Dx = 194.4 lbf, and F Dz = 307.1 lbf.
Thus, the radial forces are
FrC 287.22 500.9 2 577 lbf
FrD 194.4 2 307.12 363 lbf
The induced loads are
0.47 FrC 0.47 577
Eq. (11-15): FiC
181 lbf
KC
1.5
0.47 FrD 0.47 363
Eq. (11-15): FiD
114 lbf
KD
1.5
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17), where bearings C
and D are substituted, respectively, for labels A and B in the equations.
FiC ? FiD Fae
181 lbf 114 362.8 476.8 lbf, so Eq.(11-16) applies
Eq. (11-16a): FeC 0.4 FrC KC FiD Fae
0.4 577 1.5 114 362.8 946 lbf FrC , so use FeC
Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608
are applicable.
Chapter 11, Page 24/28
xD
LD
108
1.11
LR 90 106
R 0.90 0.949
1.11
Eq. (11-7):
FRC 1 946
1/1.5
4.48 1 0.949
Eq. (11-16b): FeD FrD 363 lbf
3/10
1130 lbf
Ans.
3/10
1.11
433 lbf Ans.
Eq. (11-7):
FRD 1 363
1/1.5
4.48 1 0.949
______________________________________________________________________________
11-42 The thrust load on shaft AB is from the axial component of the force transmitted through
the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect
mounted bearings would allow bearing A to carry the thrust load. Ans.
From the solution to Prob. 3-76, the axial thrust load is F ae = 92.8 lbf, and the bearing
radial forces are F Ay = 639.4 lbf, F Az = 1513.7 lbf, F By = 276.6 lbf, and F Bz = 705.7 lbf.
Thus, the radial forces are
FrA 639.4 2 1513.7 2 1643 lbf
FrB 276.62 705.7 2 758 lbf
The induced loads are
0.47 FrA 0.47 1643
Eq. (11-15): FiA
515 lbf
KA
1.5
0.47 FrB 0.47 758
Eq. (11-15): FiB
238 lbf
KB
1.5
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17).
FiA ? FiB Fae
515 lbf 238 92.8 330.8 lbf, so Eq.(11-17) applies
Notice that the induced load from bearing A is sufficiently large to cause a net axial force
to the left, which must be supported by bearing B.
Eq. (11-17a): FeB 0.4 FrB K B FiA Fae
0.4 758 1.5 515 92.8 937 lbf FrB , so use FeB
Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608
are applicable.
Chapter 11, Page 25/28
6
LD 500 10
xD
5.56
LR
90 106
R 0.90 0.949
5.56
Eq. (11-7):
FRB 1 937
4.48 1 0.949 1/1.5
Eq. (11-16b): FeA FrA 1643 lbf
3/10
1810 lbf
Ans.
3/10
5.56
3180 lbf Ans.
Eq. (11-7):
FRA 11643
4.48 1 0.949 1/1.5
______________________________________________________________________________
11-43 The lower bearing is compressed by the axial load, so it is designated as bearing A.
FrA 25 kN
FrB 12 kN
Fae 5 kN
0.47 FrA 0.47 25
7.83 kN
KA
1.5
0.47 FrB 0.47 12
Eq. (11-15): FiB
3.76 kN
KB
1.5
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17)
Eq. (11-15):
FiA
FiA ? FiB Fae
7.83 kN 3.76 5 8.76 kN, so Eq.(11-16) applies
Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
0.4 25 1.5 3.76 5 23.1 kN FrA, so use FrA
60 min 8 hr 5 day 52 weeks
LD 250 rev/min
5 yrs
yr
hr day week
156 106 rev
Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608
are applicable.
Eq. (11-3):
L
FRA a f FD D
LR
3/10
156 106
1.2 25
90 106
3/10
35.4 kN
Ans.
Eq. (11-16b): FeB FrB 12 kN
Chapter 11, Page 26/28
3/10
156
FRB 1.2 12
17.0 kN Ans.
90
______________________________________________________________________________
Eq. (11-3):
11-44 The left bearing is compressed by the axial load, so it is properly designated as bearing A.
FrA 875 lbf
FrB 625 lbf
Fae 250 lbf
Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads.
Eq. (11-15):
Eq. (11-15):
0.47 FrA 0.47 875
274 lbf
KA
1.5
0.47 FrB 0.47 625
FiB
196 lbf
KB
1.5
FiA
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17).
FiA ? FiB Fae
274 lbf 196 250 lbf, so Eq.(11-16) applies
We will size bearing B first since its induced load will affect bearing A, but it is not
affected by the induced load from bearing A [see Eq. (11-16)].
From Eq. (11-16b), F eB = F rB = 625 lbf.
Eq. (11-3):
L
FRB a f FD D
LR
3/10
90 000 150 60
1 625
90 106
3/10
FRB 1208 lbf
Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans.
With bearing B selected, we re-evaluate the induced load from bearing B using the actual
value for K.
0.47 FrB 0.47 625
Eq. (11-15): FiB
203 lbf
KB
1.45
Find the equivalent radial load for bearing A from Eq. (11-16), which still applies.
Chapter 11, Page 27/28
Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
0.4 875 1.5 203 250 1030 lbf
FeA FrA
Eq. (11-3):
L
FRA a f FD D
LR
3/10
90 000 150 60
11030
90 106
3/10
FRA 1990 lbf
Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup
15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get F eA =
1120 lbf and F rA = 2160 lbf. The selected bearing is still adequate. Ans.
______________________________________________________________________________
Chapter 11, Page 28/28
Chapter 12
12-1
Given: d max = 25 mm, b min = 25.03 mm, l/d = 1/2, W = 1.2 kN, = 55 mPas, and N =
1100 rev/min.
b d max 25.03 25
0.015 mm
cmin min
2
2
r 25/2 = 12.5 mm
r/c = 12.5/0.015 = 833.3
N = 1100/60 = 18.33 rev/s
P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm2 = 3.84 MPa
Fig. 12-16:
h 0 /c = 0.3
h 0 = 0.3(0.015) = 0.0045 mm
Fig. 12-18:
f r/c = 5.4
f = 5.4/833.3 = 0.006 48
D
ra
ft
Eq. (12-7):
2
55 103 18.33
r N
2
0.182
833.3
S
6
c P
3.84 10
Ans.
T =f Wr = 0.006 48(1200)12.5(103) = 0.0972 Nm
H loss = 2 TN = 2 (0.0972)18.33 = 11.2 W
Fig. 12-19:
Q/(rcNl) = 5.1
Ans.
Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/s
Fig. 12-20:
Q s /Q = 0.81
Q s = 0.81(219) = 177 mm3/s
Ans.
______________________________________________________________________________
12-2
Given: d max = 32 mm, b min = 32.05 mm, l = 64 mm, W = 1.75 kN, = 55 mPas, and N =
900 rev/min.
b d max 32.05 32
0.025 mm
cmin min
2
2
r 32/2 = 16 mm
r/c = 16/0.025 = 640
N = 900/60 = 15 rev/s
Chapter 12, Page 1/26
P = W/ (ld) = 1750/ [32(64)] = 0.854 MPa
l/d = 64/32 = 2
Eq. (12-7):
2
55 103 15
r N
2
0.797
S
640
c P
0.854
Eq. (12-16), Figs. 12-16, 12-19, and 12-21
l/d
y
y1
y 1/2
y 1/4
y l/d
h 0 /c
2
0.98
0.83
0.61
0.36
0.92
P/p max
Q/rcNl
2
2
0.84
3.1
0.54
3.45
0.45
4.2
0.31
5.08
0.65
3.20
h 0 = 0.92 c = 0.92(0.025) = 0.023 mm
Ans.
Ans.
ft
p max = P / 0.065 = 0.854/0.65 = 1.31 MPa
12-3
D
ra
Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/s
Ans.
______________________________________________________________________________
Given: d max = 3.000 in, b min = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and
SAE 10 and SAE 40 at 150F.
b d max
3.005 3.000
cmin min
0.0025 in
2
2
r 3.000 / 2 1.500 in
l / d 1.5 / 3 0.5
r / c 1.5 / 0.0025 600
N 600 / 60 10 rev/s
W
800
P
177.78 psi
ld 1.5(3)
Fig. 12-12: SAE 10 at 150F, µ 1.75 µreyn
2
6
r N
2 1.75(10 )(10)
S
600
0.0354
c P
177.78
Figs. 12-16 and 12-21: h 0 /c = 0.11 and P/p max = 0.21
h0 0.11(0.0025) 0.000 275 in Ans.
pmax 177.78 / 0.21 847 psi Ans.
Fig. 12-12: SAE 40 at 150F, µ 4.5 µreyn
Chapter 12, Page 2/26
4.5
S 0.0354
0.0910
1.75
h0 / c 0.19, P / pmax 0.275
h0 0.19(0.0025) 0.000 475 in Ans.
pmax 177.78 / 0.275 646 psi Ans.
______________________________________________________________________________
2
ft
Given: d max = 3.250 in, b min = 3.256 in, l = 3.25 in, W = 800 lbf, and N = 1000 rev/min.
b d max
3.256 3.250
cmin min
0.003
2
2
r 3.250 / 2 1.625 in
l / d 3 / 3.250 0.923
r / c 1.625 / 0.003 542
N 1000 / 60 16.67 rev/s
W
800
P
82.05 psi
ld
3(3.25)
Fig. 12-14: SAE 20W at 150F, = 2.85 reyn
6
r N
2 2.85(10 )(16.67)
S
542
0.1701
82.05
c P
D
ra
12-4
From Eq. (12-16), and Figs. 12-16 and 12-21:
l/d
y
y1
y 1/2
y 1/4
y l/d
h o /c
0.923
0.85
0.48
0.28
0.15
0.46
P/p max
0.923
0.83
0.45
0.32
0.22
0.43
ho 0.46c 0.46(0.003) 0.001 38 in
P
82.05
pmax
191 psi Ans.
0.43
0.43
Ans.
Fig. 12-14: SAE 20W-40 at 150F, = 4.4 reyn
S 5422
4.4(10 6 )(16.67)
0.263
82.05
From Eq. (12-16), and Figs. 12-16 and 12-21:
l/d
y
y1
y 1/2
y 1/4
y l/d
h o /c
0.923
0.91
0.6
0.38
0.2
0.58
P/p max
0.923
0.83
0.48
0.35
0.24
0.46
Chapter 12, Page 3/26
h0 0.58c 0.58(0.003) 0.001 74 in Ans.
8205 82.05
pmax
178 psi Ans.
0.46
0.46
______________________________________________________________________________
Given: d max = 2.000 in, b min = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE
20 at 130F.
2.0024 2
b d max
0.0012 in
cmin min
2
2
d
2
r
1 in, l / d 1 / 2 0.50
2
2
r / c 1 / 0.0012 833
N 800 / 60 13.33 rev/s
W
600
P
300 psi
ld
2(1)
ft
12-5
Fig. 12-12: SAE 20 at 130F, µ 3.75 µreyn
2
D
ra
6
r N
2 3.75(10 )(13.3)
S
833
0.115
300
c P
From Figs. 12-16, 12-18 and 12-19:
h0 / c 0.23, r f / c 3.8, Q / (rcNl ) 5.3
h0 0.23(0.0012) 0.000 276 in Ans.
3.8
f
0.004 56
833
The power loss due to friction is
2 f WrN
2 (0.004 56)(600)(1)(13.33)
778(12)
778(12)
0.0245 Btu/s Ans.
Q 5.3rcNl
5.3(1)(0.0012)(13.33)(1)
0.0848 in 3 / s Ans.
______________________________________________________________________________
H
12-6
Given: d max = 25 mm, b min = 25.04 mm, l/d = 1, W = 1.25 kN, = 50 mPas, and N =
1200 rev/min.
Chapter 12, Page 4/26
bmin d max
25.04 25
0.02 mm
2
2
r d / 2 25 / 2 12.5 mm, l / d 1
r / c 12.5 / 0.02 625
N 1200 / 60 20 rev/s
W
1250
P
2 MPa
ld
252
cmin
2
50(10 3 )(20)
r N
S
6252
0.195
6
c P
2(10 )
From Figs. 12-16, 12-18 and 12-20:
For µ = 50 MPa · s,
ft
h0 / c 0.52, f r / c 4.5, Qs / Q 0.57
h0 0.52(0.02) 0.0104 mm Ans.
4.5
f
0.0072
625
T f Wr 0.0072(1.25)(12.5) 0.1125 N · m
The power loss due to friction is
D
ra
H = 2πT N = 2π (0.1125)(20) = 14.14 W
Ans.
Q s = 0.57Q The side flow is 57% of Q Ans.
______________________________________________________________________________
12-7
Given: d max = 1.25 in, b min = 1.252 in, l = 2 in, W = 620 lbf, = 8.5 reyn, and N =
1120 rev/min.
b d max 1.252 1.25
cmin min
0.001 in
2
2
r d / 2 1.25 / 2 0.625 in
r / c 0.625 / 0.001 625
N 1120 / 60 18.67 rev/s
W
620
P
248 psi
ld 1.25(2)
2
8.5(10 6 )(18.67)
r N
S
6252
0.250
248
c P
l / d 2 / 1.25 1.6
From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19
Chapter 12, Page 5/26
h 0 /c
fr/c
Q/rcNl
l/d
y
y1
y 1/2
y 1/4
y l/d
1.6
1.6
1.6
0.9
4.5
3
0.58
5.3
3.98
0.36
6.5
4.97
0.185
8
5.6
0.69
4.92
3.59
h 0 = 0.69 c = 0.69(0.001) =0.000 69 in
Ans.
f = 4.92/(r/c) = 4.92/625 = 0.007 87
Ans.
Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/s
Ans.
______________________________________________________________________________
Given: d max = 75.00 mm, b min = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min, and
SAE 20 and SAE 40 at 60C.
bmin d max
75.10 75
0.05 mm
2
2
l / d 36 / 75 0.48 0.5 (close enough)
r d / 2 75 / 2 37.5 mm
r / c 37.5 / 0.05 750
N 720 / 60 12 rev/s
W
2000
P
0.741 MPa
ld
75(36)
ft
cmin
D
ra
12-8
Fig. 12-13: SAE 20 at 60C, µ = 18.5 MPa · s
2
18.5(10 3 )(12)
r N
S
7502
0.169
6
c P
0.741(10 )
From Figures 12-16, 12-18 and 12-21:
h0 / c 0.29, f r / c 5.1, P / pmax 0.315
h0 0.29(0.05) 0.0145 mm Ans.
f 5.1 / 750 0.0068
T f Wr 0.0068(2)(37.5) 0.51 N · m
The heat loss rate equals the rate of work on the film
H loss = 2πT N = 2π(0.51)(12) = 38.5 W
p max = 0.741/0.315 = 2.35 MPa Ans.
Ans.
Fig. 12-13: SAE 40 at 60C, µ = 37 MPa · s
Chapter 12, Page 6/26
S = 0.169(37)/18.5 = 0.338
From Figures 12-16, 12-18 and 12-21:
h0 / c 0.42, f r / c 8.5, P / pmax 0.38
h0 0.42(0.05) 0.021 mm Ans.
f 8.5 / 750 0.0113
T f Wr 0.0113(2)(37.5) 0.85 N · m
H loss 2 TN 2 (0.85)(12) 64 W Ans.
pmax 0.741 / 0.38 1.95 MPa Ans.
_____________________________________________________________________________
Given: d max = 56.00 mm, b min = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and
SAE 40 at 65C.
b d max
56.05 56
0.025 mm
cmin min
2
2
r d / 2 56 / 2 28 mm
r / c 28 / 0.025 1120
l / d 28 / 56 0.5, N 900 / 60 15 rev/s
2400
1.53 MPa
P
28(56)
D
ra
ft
12-9
Fig. 12-13: SAE 40 at 65C, µ = 30 MPa · s
2
30(10 3 )(15)
r N
S
11202
0.369
6
c P
1.53(10 )
From Figures 12-16, 12-18, 12-19 and 12-20:
h0 / c 0.44, f r / c 8.5, Qs / Q 0.71, Q / (rcNl ) 4.85
h0 0.44(0.025) 0.011 mm Ans.
f 8.5 / 1000 0.007 59
T f Wr 0.007 59(2.4)(28) 0.51 N · m
H 2 TN 2 (0.51)(15) 48.1 W Ans.
Q 4.85 rcNl 4.85(28)(0.025)(15)(28) 1426 mm3 /s
Qs 0.71(1426) 1012 mm3 /s Ans.
_____________________________________________________________________________
12-10 Consider the bearings as specified by
minimum f :
d t0d , b0tb
maximum W:
d t0d , b0tb
Chapter 12, Page 7/26
and differing only in d and d .
Preliminaries:
l /d 1
P W / (ld ) 700 / (1.252 ) 448 psi
N 3600 / 60 60 rev/s
Fig. 12-16:
minimum f :
maximum W:
S 0.08
S 0.20
Fig. 12-12:
µ = 1.38(106) reyn
µN/P = 1.38(106)(60/448) = 0.185(106)
Eq. (12-7):
D
ra
For minimum f :
S
µN / P
ft
r
c
r
0.08
658
0.185(10 6 )
c
c 0.625 / 658 0.000 950 0.001 in
If this is c min ,
b d = 2(0.001) = 0.002 in
The median clearance is
c cmin
t d tb
t tb
0.001 d
2
2
and the clearance range for this bearing is
t tb
c d
2
which is a function only of the tolerances.
For maximum W:
r
0.2
1040
c
0.185(10 6 )
c 0.625 / 1040 0.000 600 0.0005 in
If this is c min
Chapter 12, Page 8/26
b d 2cmin 2(0.0005) 0.001 in
t tb
t tb
c cmin d
0.0005 d
2
2
t tb
c d
2
The difference (mean) in clearance between the two clearance ranges, c range , is
t d tb
t tb
0.0005 d
2
2
0.0005 in
crange 0.001
For the minimum f bearing
b d = 0.002 in
or
d = b 0.002 in
d = b 0.001 in
ft
For the maximum W bearing
For the same b, t b and t d , we need to change the journal diameter by 0.001 in.
D
ra
d d b 0.001 (b 0.002)
0.001 in
Increasing d of the minimum friction bearing by 0.001 in, defines d of the maximum
load bearing. Thus, the clearance range provides for bearing dimensions which are
attainable in manufacturing. Ans.
_____________________________________________________________________________
12-11 Given: SAE 40, N = 10 rev/s, T s = 140F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675
lbf.
3.003 3
b d max
cmin min
0.0015 in
2
2
r d / 2 3 / 2 1.5 in
r / c 1.5 / 0.0015 1000
675
W
P
75 psi
3(3)
ld
Trial #1: From Figure 12-12 for T = 160°F, µ = 3.5 µ reyn,
T 2(160 140) 40F
2
3.5(10 6 )(10)
r N
S
10002
0.4667
75
c P
From Fig. 12-24,
Chapter 12, Page 9/26
9.70T
0.349 109 6.009 40(0.4667) 0.047 467(0.4667) 2 3.16
P
P
75
T 3.16
3.16
24.4F
9.70
9.70
Discrepancy = 40 24.4 = 15.6°F
Trial #2: T = 150°F, µ = 4.5 µ reyn,
T 2(150 140) 20F
4.5 10 6 10
2
S 1000
0.6
75
From Fig. 12-24,
ft
9.70T
0.349 109 6.009 40(0.6) 0.047 467(0.6) 2 3.97
P
P
75
T 3.97
3.97
30.7F
9.70
9.70
D
ra
Discrepancy = 20 30.7 = 10.7°F
Trial #3: T = 154°F, µ = 4 µ reyn,
T 2(154 140) 28F
4 10 6 10
2
S 1000
0.533
75
From Fig. 12-24,
9.70T
0.349 109 6.009 40(0.533) 0.047 467(0.533)2 3.57
P
P
75
T 3.57
3.57
27.6F
9.70
9.70
Discrepancy = 28 27.6 = 0.4°F
O.K.
T av = 140 +28/2 = 154°F Ans.
T1 Tav T / 2 154 (28 / 2) 140F
T2 Tav T / 2 154 (28 / 2) 168F
S 0.4
From Figures 12-16, 12-18, to 12-20:
Chapter 12, Page 10/26
h0
fr
Q
Qs
0.75,
11,
3.6,
0.33
c
c
rcN l
Q
h0 0.75(0.0015) 0.001 13 in Ans.
11
f
0.011
1000
T f Wr 0.0075(3)(40) 0.9 N · m
2 0.011 675 1.5 10
2 f WrN
H loss
0.075 Btu/s
778 12
778 12
Ans.
Q 3.6rcN l 3.6(1.5)0.0015(10)3 0.243 in 3 /s
Ans.
3
Qs 0.33(0.243) 0.0802 in /s Ans.
_____________________________________________________________________________
12-12 Given: d = 2.5 in, b = 2.504 in, c min = 0.002 in, W = 1200 lbf, SAE = 20, T s = 110°F,
N = 1120 rev/min, and l = 2.5 in.
N = 1120/60 = 18.67 rev/s
ft
P = W/(ld) = 1200/(2.5)2 = 192 psi,
For a trial film temperature, let T f = 150°F
= 0.0136 exp[1271.6/(150 + 95)] = 2.441 reyn
D
ra
Table 12-1:
Eq. (12-7):
6
2
2
r N 2.5 / 2 2.44110 18.67
S
0.927
192
c P 0.002
Fig. 12-24:
192
0.349 109 6.009 40 0.0927 0.047 467 0.0927 2
9.70
17.9F
T
T
17.9
110
119.0F
2
2
T f Tav 150 119.0 31.0F
Tav Ts
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
(T f ) new
150 119.0
134.5F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
Chapter 12, Page 11/26
Trial
Tf
'
S
T
T av
T f T av
New
Tf
150.0
134.5
127.9
125.3
124.3
124.0
123.8
2.441
3.466
4.084
4.369
4.485
4.521
4.545
0.0927
0.1317
0.1551
0.1659
0.1704
0.1717
0.1726
17.9
22.6
25.4
26.7
27.2
27.4
27.5
119.0
121.3
122.7
123.3
123.6
123.7
123.7
31.0
13.2
5.2
2.0
0.7
0.3
0.1
134.5
127.9
125.3
124.3
124.0
123.8
123.8
Note that the convergence begins rapidly. There are ways to speed this, but at this point
they would only add complexity.
µ 4.545(106 ), S 0.1726
From Fig. 12-16:
ft
(a)
h0
0.482, h0 0.482(0.002) 0.000 964 in
c
D
ra
From Fig. 12-17: = 56°
Ans.
(b) e = c h 0 = 0.002 0.000 964 = 0.001 04 in Ans.
f r
(c) From Fig. 12-18:
4.10, f 4.10(0.002 /1.25) 0.006 56
c
Ans.
(d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in
H
(e) From Fig. 12-19:
2 T N
2 (9.84)(1120 / 60)
0.124 Btu/s
778(12)
778(12)
Ans.
Q
4.16
rcNl
1120
3
Q 4.16(1.25)(0.002)
(2.5) 0.485 in /s
60
Qs
From Fig. 12-20:
0.6, Qs 0.6(0.485) 0.291 in 3/s
Q
Ans.
Ans.
W / ld 1200 / 2.52
P
(f) From Fig. 12-21:
0.45, pmax
427 psi
pmax
0.45
0.45
From Fig. 12-22: pmax 16 Ans.
Ans.
Chapter 12, Page 12/26
(g) From Fig. 12-22: p0 82 Ans.
(h) From the trial table, T f = 123.8°F Ans.
(i) With T = 27.5°F from the trial table, T s + T = 110 + 27.5 = 137.5°F Ans.
_____________________________________________________________________________
12-13 Given: d = 1.250 in, t d = 0.001 in, b = 1.252 in, t b = 0.003 in, l = 1.25 in, W = 250 lbf,
N = 1750 rev/min, SAE 10 lubricant, sump temperature T s = 120°F.
P = W/(ld) = 250/1.252 = 160 psi,
N = 1750/60 = 29.17 rev/s
For the clearance, c = 0.002 0.001 in. Thus, c min = 0.001 in, c median = 0.002 in, and
c max = 0.003 in.
For c min = 0.001 in, start with a trial film temperature of T f = 135°F
= 0.0158 exp[1157.5/(135 + 95)] = 2.423 reyn
Eq. (12-7):
6
2
2
r N 1.25 / 2 2.423 10 29.17
S
0.1725
160
c P 0.001
ft
Table 12-1:
D
ra
Fig. 12-24:
160
0.349 109 6.009 40 0.1725 0.047 467 0.17252
9.70
22.9F
T
T
22.9
120
131.4F
2
2
T f Tav 135 131.4 3.6F
Tav Ts
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
135 131.4
133.2F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
(T f ) new
Trial
Tf
'
S
T
T av
T f T av
New
Tf
135.0
133.2
132.5
132.2
132.1
2.423
2.521
2.560
2.578
2.583
0.1725
0.1795
0.1823
0.1836
0.1840
22.9
23.6
23.9
24.0
24.0
131.4
131.8
131.9
132.0
132.0
3.6
1.4
0.6
0.2
0.1
133.2
132.5
132.2
132.1
132.1
Chapter 12, Page 13/26
With T f = 132.1°F, T = 24.0°F, = 2.583 reyn, S = 0.1840,
T max = T s + T = 120 + 24.0 = 144.0°F
Fig. 12-16:
h 0 /c = 0.50, h 0 = 0.50(0.001) = 0.000 50 in
= 1 h 0 /c = 1 0.50 = 0.05 in
Fig. 12-18:
r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8
Fig. 12-19:
Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/s
Fig. 12-20:
Q s /Q = 0.58, Q s = 0.58(0.0941) = 0.0546 in3/s
The above can be repeated for c median = 0.002 in, and c max = 0.003 in. The results are
shown below.
Tf
ft
c min 0.001 c median c max 0.003
in
0.002 in
in
132.1
2.583
0.184
24.0
125.6
3.002
0.0534
11.1
124.1
3.112
0.0246
8.2
h 0 /c
144.0
0.5
131.1
0.23
128.2
0.125
h0
0.00050
0.00069
0.00038
0.50
4.25
0.0068
4.13
0.0941
0.77
1.8
0.0058
4.55
0.207
0.88
1.22
0.0059
4.7
0.321
0.58
0.82
0.90
0.0546
0.170
0.289
D
ra
S
T max
fr/c
f
Q/(rcNl)
Q
Q s /Q
Qs
_____________________________________________________________________________
12-14 Computer programs will vary.
_____________________________________________________________________________
12-15 Note to the Instructor: In the first printing of the 9th edition, the l/d ratio and the
lubrication constant were omitted. The values to use are l/d = 1, and = 1. This will be
updated in the next printing. We apologize for any inconvenience this may have caused.
Chapter 12, Page 14/26
In a step-by-step fashion, we are building a skill for natural circulation bearings.
• Given the average film temperature, establish the bearing properties.
• Given a sump temperature, find the average film temperature, then establish the bearing
properties.
• Now we acknowledge the environmental temperature’s role in establishing the sump
temperature. Sec. 12-9 and Ex. 12-5 address this problem.
Given: d max = 2.500 in, b min = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W =
300 lbf, A = 60 in2, T = 70F, and = 1.
600 lbf load with minimal clearance: We will start by using W = 600 lbf (n d = 2). The
task is to iteratively find the average film temperature, T f , which makes H gen and
H loss equal.
b d max
2.504 2.500
cmin min
0.002 in
2
2
N = 1120/60 = 18.67 rev/s
W
600
96 psi
ld
2.52
ft
P
D
ra
6
2
2
r N 1.25 10 18.67
S
0.0760
96
c P
0.002
= 0.0136 exp[1271.6/(T f + 95)]
Table 12-1:
2545
fr
f r 2545
WNc
600 18.67 0.002
1050
c
c 1050
fr
54.3
c
H gen
H loss
2.7 60 / 144
CR A
T f T
T f 70
1
11
0.5625 T f 70
Start with trial values of T f of 220 and 240F.
Trial T f
220
240
0.770
0.605
S
0.059
0.046
f r/c
1.9
1.7
H gen
103.2
92.3
H loss
84.4
95.6
As a linear approximation, let H gen = mT f + b. Substituting the two sets of values of
T f and H gen we find that H gen = 0.545 T f +223.1. Setting this equal to H loss and
solving
for T f gives T f = 237F.
Chapter 12, Page 15/26
Trial T f
237
S
0.048
0.627
f r/c
1.73
H gen
93.9
H loss
94.0
which is satisfactory.
Table 12-16:
h 0 /c = 0.21,
h 0 = 0.21 (0.002) = 000 42 in
Fig. 12-24:
T
96
0.349 109 6.009 4 0.048 0.047 467 0.0482
9.7
6.31 F
T 1 = T s = T f T = 237 6.31/2 = 233.8F
T max = T 1 + T = 233.8 + 6.31 = 240.1F
Trumpler’s design criteria:
ft
0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h 0
O.K.
D
ra
T max = 240.1F < 250F O.K.
Wst
300
48 psi 300 psi
ld
2.52
O.K .
n d = 2 (assessed at W = 600 lbf)
O.K.
We see that the design passes Trumpler’s criteria and is deemed acceptable.
For an operating load of W = 300 lbf, it can be shown that T f = 219.3F, = 0.78, S =
0.118, f r/c = 3.09, H gen = H loss = 84 Btu/h, h 0 = , T = 10.5F, T 1 = 224.6F, and T max =
235.1F.
_____________________________________________________________________________
0.005
12-16 Given: d 3.5000.000
0.001 in, b 3.5050.000 in , SAE 30, T s = 120F, p s = 50 psi,
N = 2000/60 = 33.33 rev/s, W = 4600 lbf, bearing length = 2 in, groove width = 0.250 in,
and H loss 5000 Btu/hr.
b d max
3.505 3.500
cmin min
0.0025 in
2
2
r = d/ 2 = 3.500/2 = 1.750 in
r / c = 1.750/0.0025 = 700
l = (2 0.25)/2 = 0.875 in
Chapter 12, Page 16/26
l / d = 0.875/3.500 = 0.25
W
4600
P
751 psi
4rl 4 1.750 0.875
Trial #1: Choose (T f ) 1 = 150°F. From Table 12-1,
= 0.0141 exp[1360.0/(150 + 95)] = 3.63 µ reyn
2
6
r N
2 3.63(10 )(33.33)
S
700
0.0789
751
c P
From Figs. 12-16 and 12-18:
= 0.9, f r/ c = 3.6
From Eq. (12-24),
T
0.0123 3.6 0.0789 46002
1 1.5(0.9) 2 50 1.7504
71.2F
ft
0.0123( f r / c)SW 2
1 1.5 2 ps r 4
T av = T s + T / 2 = 120 + 71.2/2 = 155.6F
D
ra
Trial #2: Choose (T f ) 2 = 160°F. From Table 12-1
= 0.0141 exp[1360.0/(160 + 95)] = 2.92 µ reyn
2.92
S 0.0789
0.0635
3.63
From Figs. 12-16 and 12-18:
= 0.915, f r/ c =3
T
T av
0.0123 3 0.0635 46002
1 1.5 0.9152 50 1.7504
= 120 + 46.9/2 = 143.5F
46.9F
Chapter 12, Page 17/26
Trial #3: Thus, the plot gives (T f ) 3 = 152.5°F. From Table 12-1
= 0.0141 exp[1360.0/(152.5 + 95)] = 3.43 µ reyn
3.43
S 0.0789
0.0746
3.63
From Figs. 12-16 and 12-18:
= 0.905, f r/ c =3.4
T
T av
0.0123 3.4 0.0746 46002
1 1.5 0.9052 50 1.7504
= 120 + 63.2/2 = 151.6F
Result is close. Choose T f
Table 12-1:
63.2F
152.5 151.6
152.1F
2
Try 152F
= 0.0141 exp[1360.0/(152 + 95)] = 3.47 µ reyn
D
ra
ft
3.47
S 0.0789
0.0754
3.63
f r
h0
3.4, 0.902,
0.098
c
c
0.0123 3.4 0.0754 46002
T
64.1F
1 1.5 0.9022 50 1.7504
Tav 120 64.1 / 2 152.1F O.K.
h 0 = 0.098(0.0025) = 0.000 245 in
T max = T s + T = 120 + 64.1 = 184.1F
Eq. (12-22):
50 1.750 0.00253
ps rc3
2
1 1.5 0.9022
Qs
1 1.5
6
3 l
3 3.47 10 0.875
1.047 in 3 /s
H loss = C p Q s T = 0.0311(0.42)1.047(64.1) = 0.877 Btu/s
= 0.877(602) = 3160 Btu/h O.K.
Trumpler’s design criteria:
0.0002 + 0.000 04(3.5) = 0.000 34 in > 0.000 245
Not O.K.
T max = 184.1°F < 250°F O.K.
P st = 751 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
Chapter 12, Page 18/26
0.010
12-17 Given: d 50.000.00
0.05 mm, b 50.084 0.000 mm , SAE 30, T s = 55C, p s = 200 kPa,
N = 2880/60 = 48 rev/s, W = 10 kN, bearing length = 55 mm, groove width = 5 mm, and
H loss 300 W.
b d max
50.084 50
cmin min
0.042 mm
2
2
r = d/ 2 = 50/2 = 25 mm
r / c = 25/0.042 = 595
l = (55 5)/2 = 25 mm
l / d = 25/50 = 0.5
10 103
W
P
4 MPa
4rl 4 25 25
Trial #1: Choose (T f ) 1 = 79°C. From Fig. 12-13, µ = 13 MPa · s.
2
ft
13(103 )(48)
r N
S
5952
0.0552
6
c P
4(10 )
= 0.85, f r/ c = 2.3
D
ra
From Figs. 12-16 and 12-18:
From Eq. (12-25),
978(106 ) ( f r / c)SW 2
T
1 1.5 2
ps r 4
978(106 ) 2.3(0.0552)(102 )
76.3C
1 1.5(0.85) 2
200(25) 4
T av = T s + T / 2 = 55 + 76.3/2 = 93.2C
Trial #2: Choose (T f ) 2 = 100°C. From Fig. 12-13, µ = 7 MPa · s.
7
S 0.0552 0.0297
13
From Figs. 12-16 and 12-18:
= 0.90, f r/ c =1.6
T
978(106 ) 1.6(0.0297)(102 )
26.9C
200(25)4
1 1.5(0.9) 2
T av = 55 + 26.9/2 = 68.5C
Chapter 12, Page 19/26
Trial #3: Thus, the plot gives (T f ) 3 = 85.5°C. From Fig. 12-13, µ = 10.5 MPa · s.
10.5
S 0.0552
0.0446
13
From Figs. 12-16 and 12-18:
= 0.87, f r/ c =2.2
T
978(106 ) 2.2(0.0457)(102 )
58.9C
1 1.5(0.87 2 )
200(25) 4
ft
T av = 55 + 58.9/2 = 84.5C
85.5 84.5
85C
2
D
ra
Result is close. Choose T f
Fig. 12-13:
µ = 10.5 MPa · s
10.5
S 0.0552
0.0446
13
f r
h0
0.87,
2.2,
0.13
c
c
978(106 ) 2.2(0.0457)(102 )
T
58.9C
1 1.5(0.87 2 )
200(254 )
Tav 55 58.9 / 2 84.5C O.K.
or
138F
From Eq. (12-22)
h 0 = 0.13(0.042) = 0.005 46 mm or 0.000 215 in
T max = T s + T = 55 + 58.9 = 113.9C
or
237°F
200 25 0.0423
1 1.5 0.87 2
6
3µl
3 10.5 10 25
3156 mm3 /s 3156 25.4 3 0.193 in 3 /s
Qs (1 1.5 2 )
ps rc 3
H loss = C p Q s T = 0.0311(0.42)0.193(138) = 0.348 Btu/s
= 1.05(0.348) = 0.365 kW = 365 W not O.K.
Chapter 12, Page 20/26
Trumpler’s design criteria:
0.0002 + 0.000 04(50/25.4) = 0.000 279 in > h 0
Not O.K.
T max = 237°F O.K.
P st = 4000 kPa or 581 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
12-18 So far, we’ve performed elements of the design task. Now let’s do it more completely.
The values of the unilateral tolerances, t b and t d , reflect the routine capabilities of the
bushing vendor and the in-house capabilities. While the designer has to live with these,
his approach should not depend on them. They can be incorporated later.
First we shall find the minimum size of the journal which satisfies Trumpler’s constraint
of P st ≤ 300 psi.
W
300
2dl
ft
Pst
W
600(l / d )
D
ra
W
300 d
2d 2l / d
d min
900
1.73 in
2(300)(0.5)
In this problem we will take journal diameter as the nominal value and the bushing bore
as a variable. In the next problem, we will take the bushing bore as nominal and the
journal diameter as free.
To determine where the constraints are, we will set t b = t d = 0, and thereby shrink the
design window to a point.
We set
d = 2.000 in
b = d + 2c min = d + 2c
n d = 2 (This makes Trumpler’s n d ≤ 2 tight)
and construct a table.
Chapter 12, Page 21/26
c
b
d
Tf *
T max
h o P st T max n
fom
0.0010
0.0011
0.0012
0.0013
0.0014
0.0015
0.0016
0.0017
0.0018
0.0019
0.0020
2.0020
2.0022
2.0024
2.0026
2.0028
2.0030
2.0032
2.0034
2.0036
2.0038
2.0040
2
2
2
2
2
2
2
2
2
2
2
215.50
206.75
198.50
191.40
185.23
179.80
175.00
171.13
166.92
163.50
160.40
312.0
293.0
277.0
262.8
250.4
239.6
230.1
220.3
213.9
206.9
200.6
-5.74
-6.06
-6.37
-6.66
-6.94
-7.20
-7.45
-7.65
-7.91
-8.12
-8.32
*Sample calculation for the first entry of this column.
T f 215.5F
Iteration yields:
ft
With T f 215.5F , from Table 12-1
D
ra
µ 0.0136(10 6 ) exp[1271.6 / (215.5 95)] 0.817(10 6 ) reyn
900
N 3000 / 60 50 rev/s, P
225 psi
4
2
6
1 0.817(10 )(50)
S
0.182
225
0.001
From Figs. 12-16 and 12-18:
Eq. (12–24):
e = 0.7,
f r/c = 5.5
0.0123(5.5)(0.182)(9002 )
191.6F
[1 1.5(0.7 2 )](30)(14 )
191.6F
Tav 120F
215.8F 215.5F
2
TF
For the nominal 2-in bearing, the various clearances show that we have been in contact
with the recurving of (h o ) min . The figure of merit (the parasitic friction torque plus the
pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will
place the top of the design window at c min = 0.002 in, and b = d + 2(0.002) = 2.004 in. At
this point, add the b and d unilateral tolerances:
d 2.0000.000
b 2.0040.003
0.001 in,
0.000 in
Now we can check the performance at c min , c , and c max . Of immediate interest is the
fom of the median clearance assembly, 9.82, as compared to any other satisfactory
bearing ensemble.
Chapter 12, Page 22/26
If a nominal 1.875 in bearing is possible, construct another table with t b = 0 and t d = 0.
c
0.0020
0.0030
0.0035
0.0040
0.0050
0.0055
0.0060
b
1.879
1.881
1.882
1.883
1.885
1.886
1.887
d
1.875
1.875
1.875
1.875
1.875
1.875
1.875
Tf
157.2
138.6
133.5
130.0
125.7
124.4
123.4
T max h o P st T max n fom
7.36
194.30
8.64
157.10
9.05
147.10
9.32
140.10
9.59
131.45
9.63
128.80
9.64
126.80
The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our
design window.
b 1.8810.003
0.000 in
ft
d 1.8750.000
0.001 in,
D
ra
The ensemble median assembly has a fom = 9.31.
We just had room to fit in a design window based upon the (h 0 ) min constraint. Further
reduction in nominal diameter will preclude any smaller bearings. A table constructed for
a d = 1.750 in journal will prove this.
We choose the nominal 1.875-in bearing ensemble because it has the largest figure of
merit. Ans.
_____________________________________________________________________________
12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and
radial clearance c.
The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the
table for a nominal b = 1.875 in, note that at c = 0.003 in the constraints are “loose.” Set
b = 1.875 in
d = 1.875 2(0.003) = 1.869 in
For the ensemble
b 1.8750.003
0.001 in,
d 1.869 0.000
0.001 in
Analyze at c min = 0.003, c = 0.004 in and c max = 0.005 in
At cmin 0.003 in: T f 138.4, µ 3.160, S 0.0297, H loss 1035 Btu/h and the
Trumpler conditions are met.
At c 0.004 in: T f 130F, = 3.872, S = 0.0205, H loss = 1106 Btu/h, fom = 9.246
Chapter 12, Page 23/26
and the Trumpler conditions are O.K.
At cmax 0.005 in: T f 125.68F, = 4.325, S = 0.014 66, H loss = 1129 Btu/h and the
Trumpler conditions are O.K.
The ensemble figure of merit is slightly better; this bearing is slightly smaller. The
lubricant cooler has sufficient capacity.
_____________________________________________________________________________
12-20 Table 12-1:
( reyn) = 0 (106) exp [b / (T + 95)]
b and T in F
The conversion from reyn to mPas is given on p. 620. For a temperature of C degrees
Celsius, T = 1.8 C + 32. Substituting into the above equation gives
(mPas) = 6.89 0 (106) exp [b / (1.8 C + 32+ 95)]
= 6.89 0 (106) exp [b / (1.8 C + 127)]
Ans.
ft
For SAE 50 oil at 70C, from Table 12-1, 0 = 0.0170 (106) reyn, and b = 1509.6F.
From the equation,
D
ra
= 6.89(0.0170) 106(106) exp {1509.6/[1.8(70) + 127]}
= 45.7 mPas
Ans.
From Fig. 12-13, = 39 mPas
Ans.
The figure gives a value of about 15 % lower than the equation.
_____________________________________________________________________________
12-21 Originally
d 2.0000.000
b 2.0050.003
0.001 in,
0.000 in
Doubled,
d 4.000 0.000
b 4.010 0.006
0.002 in,
0.000 in
The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were
carried out. Some of the results are:
Part
(a)
(b)
c
0.007
0.0035
3.416
3.416
S
0.0310
0.0310
Tf
135.1
135.1
f r/c
0.1612
0.1612
Qs
6.56
0.870
h 0 /c
0.1032
0.1032
e
0.897
0.897
H loss
9898
1237
h0
0.000 722
0.000 361
Trumpler
h0
0.000 360
0.000 280
f
0.005 67
0.005 67
The side flow Q s differs because there is a c3 term and consequently an 8-fold increase.
H loss is related by a 9898/1237 or an 8-fold increase. The existing h 0 is related by a 2-fold
increase. Trumpler’s (h 0 ) min is related by a 1.286-fold increase.
Chapter 12, Page 24/26
_____________________________________________________________________________
12-22 Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T = 70F, F = 400
lbf, N = 250 rev/min, and w = 0.004 in.
Table 12-8:
K = 0.6(1010) in3min/(lbffth)
P = F/ (DL) = 400/ [0.75(0.75)] = 711 psi
V = DN/ 12 = (0.75)250/12 = 49.1 ft/min
Tables 12-10 and 12-11:
f 1 = 1.8, f 2 = 1.0
Table 12-12: PV max = 46 700 psift/min, P max = 3560 psi, V max = 100 ft/min
Pmax
4 F
4 400
905 psi 3560 psi O.K .
DL 0.752
Eq. (12-32) can be written as
O.K.
4 F
Vt
DL
D
ra
w f1 f 2 K
ft
PV = 711 (49.1) = 34 910 psift/min < 46 700 psift/min
Solving for t,
t
DLw
4 f1 f 2 KVF
0.75 0.75 0.004
4 1.8 1.0 0.6 1010 49.1 400
833.1 h 833.1 60 49 900 min
Cycles = Nt = 250 (49 900) = 12.5 (106) cycles
Ans.
_____________________________________________________________________________
12-23 Given: Oiles SP 500 alloy brass bushing, w max = 0.002 in for 1000 h, N = 400 rev/min, F
= 100 lbf, CR = 2.7 Btu/ (hft2F), T max = 300F, f s = 0.03, and n d = 2.
Estimate bushing length with f 1 = f 2 = 1, and K = 0.6(10-10) in3 · min/(lbf · ft · h)
Using Eq. (12-32) with n d F for F,
f1 f 2 Knd FNt 1(1)(0.6)(1010 )(2)(100)(400)(1000)
L
0.80 in
3w
3(0.002)
From Eq. (12-38), with f s = 0.03 from Table 12-9 applying n d = 2 to F
Chapter 12, Page 25/26
and CR 2.7 Btu/(h · ft 2 · °F)
L
720 f s nd FN
720(0.03)(2)(100)(400)
3.58 in
778(2.7)(300 70)
J CR T f T
0.80 L 3.58 in
Trial 1: Let L = 1 in, D = 1 in
4 nd F
4(2)(100)
255 psi 3560 psi O.K .
DL
(1)(1)
nF
2(100)
P d
200 psi
DL
1(1)
DN (1)(400)
V
104.7 ft/min 100 ft/min Not O.K .
12
12
Pmax
ft
Trial 2: Try D = 7/8 in = 0.875 in, L = 1 in
4(2)(100)
291 psi 3560 psi O.K .
(0.875)(1)
2(100)
P
229 psi
0.875(1)
(0.875)(400)
V
91.6 ft/min 100 ft/min
12
D
ra
Pmax
O.K .
PV = 229(91.6) = 20 976 psi · ft/min < 46 700 psi · ft/min
f1
V
33
1.3
91.6 f 1
100
1.8
O.K.
91.6 33
f1 1.3 (1.8 1.3)
1.74
100 33
Lnew f1Lold 1.74 0.80 1.39 in
Trial 3: Try D = 7/8 in = 0.875 in, L = 1.5 in
4(2)(100)
194 psi 3560 psi O.K .
(0.875)(1.5)
2(100)
P
152 psi, V 91.6 ft/min
0.875(1.5)
PV 152(91.6) 13 923 psi · ft/min 46 700 psi · ft/min
D 7 / 8 in, L 1.5 in is acceptable Ans.
Pmax
O.K .
Chapter 12, Page 26/26
D
ra
ft
Suggestion: Try smaller sizes.
Chapter 12, Page 27/26
Chapter 13
d P 17 / 8 2.125 in
N
1120
dG 2 d P
2.125 4.375 in
N3
544
13-1
NG PdG 8 4.375 35 teeth
Ans.
C 2.125 4.375 / 2 3.25 in
Ans.
______________________________________________________________________________
nG 1600 15 / 60 400 rev/min
p m 3 mm Ans.
13-2
Ans.
C 3 15 60 2 112.5 mm Ans.
______________________________________________________________________________
NG 16 4 64 teeth
13-3
Ans.
dG NG m 64 6 384 mm
Ans.
d P N P m 16 6 96 mm
Ans.
C 384 96 / 2 240 mm
Ans.
______________________________________________________________________________
13-4
Mesh:
a 1/ P 1/ 3 0.3333 in Ans.
b 1.25 / P 1.25 / 3 0.4167 in Ans.
c b a 0.0834 in Ans.
p / P / 3 1.047 in Ans.
t p / 2 1.047 / 2 0.523 in Ans.
Pinion Base-Circle:
d1 N1 / P 21/ 3 7 in
d1b 7 cos 20 6.578 in
Gear Base-Circle:
Ans.
d 2 N 2 / P 28 / 3 9.333 in
d 2 b 9.333cos 20 8.770 in
Base pitch:
pb pc cos / 3 cos 20 0.984 in
Ans.
Ans.
Contact Ratio:
mc Lab / pb 1.53 / 0.984 1.55 Ans.
See the following figure for a drawing of the gears and the arc lengths.
Chapter 13, Page 1/35
______________________________________________________________________________
13-5
1/2
(a)
14 / 6 2 32 / 6 2
A0
2 2
(b)
tan 1 14 / 32 23.63
Ans.
32 /14 66.37
Ans.
tan
1
(c)
d P 14 / 6 2.333 in
dG 32 / 6 5.333 in
2.910 in
Ans.
Ans.
Ans.
Chapter 13, Page 2/35
From Table 13-3, 0.3A 0 = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67
0.873 < 1.67 F 0.873 in Ans.
______________________________________________________________________________
(d)
13-6
(a)
pn / Pn / 4 0.7854 in
pt pn / cos 0.7854 / cos 30 0.9069 in
px pt / tan 0.9069 / tan 30 1.571 in
pnb pn cos n 0.7854 cos 25 0.7380 in
(b)
Eq. (13-7):
(c)
pt Pn cos 4 cos 30 3.464 teeth/in
Ans.
t tan 1 tan n / cos tan 1 (tan 25 / cos 30 ) 28.3 Ans.
Table 13-4:
a 1/ 4 0.250 in Ans.
b 1.25 / 4 0.3125 in Ans.
20
dP
5.774 in Ans.
4 cos 30
36
dG
10.39 in Ans.
4 cos 30
______________________________________________________________________________
(d)
13-7
N P 19 teeth, N G 57 teeth, n 20 , mn 2.5 mm
(a)
pn mn 2.5 7.854 mm
Ans.
pn
7.854
9.069 mm Ans.
cos cos 30
pt
9.069
px
15.71 mm Ans.
tan tan 30
mn
2.5
mt
2.887 mm Ans.
cos cos 30
pt
(b)
Chapter 13, Page 3/35
tan 20
22.80
cos
30
t tan 1
(c)
a mn 2.5 mm
Ans.
Ans.
b 1.25mn 1.25 2.5 3.125 mm
dP
Ans.
N
Nmt 19 2.887 =54.85 mm
Pt
Ans.
dG 57 2.887 164.6 mm Ans.
______________________________________________________________________________
13-8
(a)
Using Eq. (13-11) with k = 1, = 20º, and m = 2,
NP
2k
m m 2 1 2m sin 2
2
1 2m sin
2
1 2 2 sin 20
2 1
2
2
2
14.16 teeth
1 2 2 sin 2 20
Round up for the minimum integer number of teeth.
N P = 15 teeth
Ans.
(b)
(c)
(d)
Repeating (a) with m = 3, N P = 14.98 teeth. Rounding up, N P = 15 teeth. Ans.
Repeating (a) with m = 4, N P = 15.44 teeth. Rounding up, N P = 16 teeth. Ans.
Repeating (a) with m = 5, N P = 15.74 teeth. Rounding up, N P = 16 teeth. Ans.
Alternatively, a useful table can be generated to determine the largest gear that can mesh
with a specified pinion, and thus also the maximum gear ratio with a specified pinion.
The Max N G column was generated using Eq. (13-12) with k = 1, = 20º, and rounding
up to the next integer.
Min N P
13
14
15
16
17
18
Max N G
16
26
45
101
1309
unlimited
Max m = Max N G / Min N P
1.23
1.86
3.00
6.31
77.00
unlimited
With this table, we can readily see that gear ratios up to 3 can be obtained with a
minimum N P of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum N P
of 16 teeth. This is consistent with the results previously obtained.
______________________________________________________________________________
Chapter 13, Page 4/35
13-9
Repeating the process shown in the solution to Prob. 13-8, except with = 25º, we obtain
the following results.
(a)
For m = 2, N P = 9.43 teeth. Rounding up, N P = 10 teeth. Ans.
(b)
For m = 3, N P = 9.92 teeth. Rounding up, N P = 10 teeth. Ans.
(c)
For m = 4, N P = 10.20 teeth. Rounding up, N P = 11 teeth. Ans.
(d)
For m = 5, N P = 10.38 teeth. Rounding up, N P = 11 teeth. Ans.
For convenient reference, we will also generate the table from Eq. (13-12) for = 25º.
Min N P
Max N G
Max m = Max N G / Min N P
9
13
1.44
10
32
3.20
11
249
22.64
12
unlimited
unlimited
______________________________________________________________________________
13-10 (a)
The smallest pinion tooth count that will run with itself is found from Eq. (13-10).
NP
2k
1 1 3sin 2
3sin 2
2 1
1
3sin 20
2
12.32
(b)
1 3sin 2 20
13 teeth
Ans.
The smallest pinion that will mesh with a gear ratio of m G = 2.5, from Eq. (13-11)
is
NP
2k
m m 2 1 2m sin 2
2
1 2m sin
2 1
2.5 2.52 1 2 2.5 sin 2 20
1 2 2.5 sin 2 20
14.64
15 teeth Ans.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
NG
N P2 sin 2 4k 2
4k 2 N P sin 2
152 sin 2 20 4 1
2
4 1 2 15 sin 2 20
45.49 45 teeth
Ans.
Chapter 13, Page 5/35
(c)
The smallest pinion that will mesh with a rack, from Eq. (13-13),
2 1
2k
2
sin sin 2 20
17.097 18 teeth Ans.
______________________________________________________________________________
NP
13-11 n 20 , 30
From Eq. (13-19), t tan 1 tan 20 / cos 30 22.80
(a)
The smallest pinion tooth count that will run with itself, from Eq. (13-21) is
NP
2k cos
1 1 3sin 2 t
3sin 2 t
2 1 cos 30
1
3sin 22.80
2
1 3sin 2 22.80
8.48 9 teeth
(b)
Ans.
The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22)
is
2 1 cos 30
2.5 2.52 1 2 2.5 sin 2 22.80
NP
2
1 2 2.5 sin 22.80
9.95 10 teeth Ans.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is
NG
N P2 sin 2 t 4k 2 cos 2
4k cos 2 N P sin 2 t
102 sin 2 22.80 4 1 cos 2 30
4 1 cos 2 30 2 20 sin 2 22.80
26.08 26 teeth
(c)
Ans.
The smallest pinion that will mesh with a rack, from Eq. (13-24) is
2k cos 2 1 cos 30
NP
sin 2 t
sin 2 22.80
11.53 12 teeth Ans.
______________________________________________________________________________
Chapter 13, Page 6/35
tan n
1 tan 20
tan
22.796
cos
cos
30
Program Eq. (13-23) on a computer using a spreadsheet or code, and increment N P . The
first value of N P that can be doubled is N P = 10 teeth, where N G ≤ 26.01 teeth. So N G =
20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc.
13-12 From Eq. (13-19),
t tan 1
Use N P = 10 teeth, N G = 20 teeth
Ans.
Note that the given diametral pitch (tooth size) is not relevant to the interference problem.
______________________________________________________________________________
tan n
1 tan 20
13-13 From Eq. (13-19),
27.236
t tan
tan
cos
cos 45
Program Eq. (13-23) on a computer using a spreadsheet or code, and increment N P . The
first value of N P that can be doubled is N P = 6 teeth, where N G ≤ 17.6 teeth. So N G = 12
teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc.
1
Use N P = 6 teeth, N G = 12 teeth Ans.
______________________________________________________________________________
13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (1313).
2k
NP
sin 2
Setting k = 1 for full depth teeth, N P = 9 teeth, and solving for ,
2 1
2k
sin 1
28.126
Ans.
NP
9
______________________________________________________________________________
sin 1
13-15
(a)
(b)
Eq. (13-3):
pn mn 3 mm
Eq. (13-16):
pt pn / cos 3 / cos 25 10.40 mm
Eq. (13-17):
px pt / tan 10.40 / tan 25 22.30 mm
Eq. (13-3):
mt pt / 10.40 / 3.310 mm
Ans.
Ans.
Ans.
Ans.
Chapter 13, Page 7/35
Eq. (13-19):
t tan 1
tan n
tan 20
tan 1
21.88
cos
cos 25
Ans.
Eq. (13-2):
d p = m t N p = 3.310 (18) = 59.58 mm
Ans.
Eq. (13-2):
d G = m t N G = 3.310 (32) = 105.92 mm
Ans.
______________________________________________________________________________
(c)
13-16 (a)
Sketches of the figures are shown to
determine the axial forces by inspection.
The axial force of gear 2 on shaft a is in the
negative z-direction. The axial force of gear 3 on
shaft b is in the positive z-direction. Ans.
The axial force of gear 4 on shaft b is in the
positive z-direction. The axial force of gear 5 on
shaft c is in the negative z-direction. Ans.
12 16
700 77.78 rev/min ccw
48 36
(b)
nc n5
(c)
d P 2 12 / 12 cos 30 1.155 in
dG 3
Ans.
48 / 12 cos 30 4.619 in
1.155 4.619
2.887 in Ans.
2
16 / 8 cos 25 2.207 in
Cab
dP4
d G 5 36 / 8 cos 25 4.965 in
Cbc 3.586 in Ans.
______________________________________________________________________________
20 8 20 4
40 17 60 51
4
nd 00 47.06 rev/min cw Ans.
51
______________________________________________________________________________
13-17
e
6 18 20 3
3
10 38 48 36 304
3
n9
1200 11.84 rev/min cw Ans.
304
______________________________________________________________________________
13-18
e
Chapter 13, Page 8/35
13-19 (a)
(b)
12 1
540 162 rev/min cw about x. Ans.
40 1
d P 12 / 8 cos 23 1.630 in
nc
d G 40 / 8 cos 23 5.432 in
d P dG
3.531 in
2
Ans.
32
8 in at the large end of the teeth. Ans.
4
______________________________________________________________________________
(c)
d
13-20 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 45. For an exact ratio, we will choose to
factor the train value into integers, such that
N2 / N3 = 9
N4 / N5 = 5
(1)
(2)
Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the
in-line requirement of the compound reverted configuration is
N2 + N3 = N4 + N5
(3)
With three equations and four unknowns, one free choice is available. It is necessary that
all of the unknowns be integers. We will use a normalized approach to find the minimum
free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity,
thus N 3 = 1. From (1), N 2 = 9. From (3),
N 2 + N 3 = 9 + 1 = 10 = N 4 + N 5
Substituting N 4 = 5 N 5 from (2) gives
10 = 5 N 5 + N 5 = 6 N 5
N 5 = 10 / 6 = 5 / 3
To eliminate this fraction, we need to multiply the original free choice by a multiple of 3.
In addition, the smallest gear needs to have sufficient teeth to avoid interference. From
Eq. (13-11) with k = 1, = 20°, and m = 9, the minimum number of teeth on the pinion to
avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18.
Setting N 3 = 18 and repeating the solution of equations (1), (2), and (3) yields
N 2 = 162 teeth
N 3 = 18 teeth
N 4 = 150 teeth
N 5 = 30 teeth
Ans.
______________________________________________________________________________
Chapter 13, Page 9/35
13-21 The solution to Prob. 13-20 applies up to the point of determining the minimum number
of teeth to avoid interference. From Eq. (13-11), with k = 1, = 25°, and m = 9, the
minimum number of teeth on the pinion to avoid interference is 11. Therefore, the
smallest multiple of 3 greater than 11 is 12. Setting N 3 = 12 and repeating the solution of
equations (1), (2), and (3) yields
N 2 = 108 teeth
N 3 = 12 teeth
N 4 = 100 teeth
N 5 = 20 teeth
Ans.
______________________________________________________________________________
13-22 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 30. For an exact ratio, we will choose to
factor the train value into integers, such that
N2 / N3 = 6
N4 / N5 = 5
(1)
(2)
Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the
in-line requirement of the compound reverted configuration is
N2 + N3 = N4 + N5
(3)
With three equations and four unknowns, one free choice is available. It is necessary that
all of the unknowns be integers. We will use a normalized approach to find the minimum
free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity,
thus N 3 = 1. From (1), N 2 = 6. From (3),
N2 + N3 = 6 + 1 = 7 = N4 + N5
Substituting N 4 = 5 N 5 from (2) gives
7 = 5 N5 + N5 = 6 N5
N5 = 7 / 6
To eliminate this fraction, we need to multiply the original free choice by a multiple of 6.
In addition, the smallest gear needs to have sufficient teeth to avoid interference. From
Eq. (13-11) with k = 1, = 20°, and m = 6, the minimum number of teeth on the pinion to
avoid interference is 16. Therefore, the smallest multiple of 3 greater than 16 is 18.
Setting N 3 = 18 and repeating the solution of equations (1), (2), and (3) yields
N 2 = 108 teeth
N 3 = 18 teeth
N 4 = 105 teeth
N 5 = 21 teeth
Ans.
______________________________________________________________________________
Chapter 13, Page 10/35
13-23 Applying Eq. (13-30), e = (N 2 / N 3 ) (N 4 / N 5 ) = 45. For an approximate ratio, we will
choose to factor the train value into two equal stages, such that
N 2 / N3 N 4 / N5 45
If we choose identical pinions such that interference is avoided, both stages will be
identical and the in-line geometry condition will automatically be satisfied. From Eq.
(13-11) with k = 1, = 20°, and m 45 , the minimum number of teeth on the pinions
to avoid interference is 17. Setting N 3 = N 5 = 17, we get
N 2 N 4 17 45 114.04 teeth
Rounding to the nearest integer, we obtain
N 2 = N 4 = 114 teeth
N 3 = N 5 = 17 teeth
Ans.
Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97.
______________________________________________________________________________
13-24 H = 25 hp, i = 2500 rev/min
Let ω o = 300 rev/min for minimal gear ratio to minimize gear size.
o 300
1
i 2500 8.333
o
N N
1
2 4
i 8.333 N 3 N 5
Let
N2 N4
1
1
N3 N5
8.333 2.887
From Eq. (13-11) with k = 1, = 20°, and m = 2.887, the minimum number of teeth on
the pinions to avoid interference is 15.
Let
N 2 = N 4 = 15 teeth
N 3 = N 5 = 2.887(15) = 43.31 teeth
Try N 3 = N 5 = 43 teeth.
15 15
o 2500 304.2
43 43
Too big. Try N 3 = N 5 = 44.
Chapter 13, Page 11/35
15 15
o 2500 290.55 rev/min
44 44
N 2 = N 4 = 15 teeth, N 3 = N 5 = 44 teeth Ans.
______________________________________________________________________________
13-25 (a)
The planet gears act as keys and the wheel speeds are the same as that of the ring
gear. Thus,
nA n3 900 16 / 48 300 rev/min Ans.
(b)
nF n5 0, nL n6 , e 1
n 300
1 6
0 300
300 n6 300
n6 600 rev/min Ans.
The wheel spins freely on icy surfaces, leaving no traction for the other wheel.
The car is stalled. Ans.
______________________________________________________________________________
(c)
13-26 (a)
The motive power is divided equally among four wheels instead of two.
Locking the center differential causes 50 percent of the power to be applied to the
rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a
slippery surface such as ice, the other rear wheel has no traction. But the front
wheels still provide traction, and so you have two-wheel drive. However, if the
rear differential is locked, you have 3-wheel drive because the rear-wheel power
is now distributed 50-50.
______________________________________________________________________________
(b)
13-27 Let gear 2 be first, then n F = n 2 = 0. Let gear 6 be last, then n L = n 6 = –12 rev/min.
20 16 16 nL n A
30 34 51 nF nA
16
0 n A 12 n A
51
12
nA
17.49 rev/min (negative indicates cw) Ans.
35 / 51
______________________________________________________________________________
e
13-28 Let gear 2 be first, then n F = n 2 = 0 rev/min. Let gear 6 be last, then n L = n 6 = 85
rev/min.
Chapter 13, Page 12/35
20 16 16 nL n A
30 34 51 nF nA
16
0 nA 85 nA
51
16
nA nA 85
51
16
n A 1 85
51
85
nA
123.9 rev/min
16
1
51
e
The positive sign indicates the same direction as n 6 . nA 123.9 rev/min ccw Ans.
______________________________________________________________________________
13-29 The geometry condition is d5 / 2 d 2 / 2 d3 d 4 . Since all the gears are meshed, they
will all have the same diametral pitch. Applying d = N / P,
N5 / (2 P) N 2 / (2 P) N3 / P N 4 / P
N5 N 2 2 N3 2 N 4 12 2 16 2 12 68 teeth
Ans.
Let gear 2 be first, n F = n 2 = 320 rev/min. Let gear 5 be last, n L = n 5 = 0 rev/min.
e
12 16 12 3 nL nA
16 12 68 17 nF nA
320 nA
nA
17
0 nA
3
3
320 68.57 rev/min
14
The negative sign indicates opposite of n 2 . nA 68.57 rev/min cw Ans.
______________________________________________________________________________
13-30 Let n F = n 2 , then n L = n 7 = 0.
e
nL n5
20 16 36
0.5217
16 30 46
nF n5
0 n5
0.5217
10 n5
Chapter 13, Page 13/35
0.5217 10 n5 n5
5.217 0.5217n5 n5 0
n5 1 0.5217 5.217
5.217
1.5217
n5 nb 3.428 turns in same direction
______________________________________________________________________________
n5
13-31 (a)
2 n / 60
H T 2 Tn / 60
(T in N m, H in W)
60 H 103
So
T
So
F32t
2 n
9550 H / n (H in kW, n in rev/min)
9550 75
Ta
398 N m
1800
mN 2 5 17
r2
42.5 mm
2
2
Ta 398
9.36 kN
r2 42.5
F3b Fb3 2 9.36 18.73 kN in the positive x-direction.
(b)
Ans.
mN 4 5 51
127.5 mm
2
2
Tc 4 9.36 127.5 1193 N m ccw
r4
T4 c 1193 N m cw
Ans.
Note: The solution is independent of the pressure angle.
______________________________________________________________________________
Chapter 13, Page 14/35
N N
P 6
d 2 4 in, d 4 4 in, d5 6 in, d 6 24 in
d
13-32
24 24 36
e
1/ 6
24 36 144
nF n2 1000 rev/min
nL n6 0
n nA
0 nA
1
e L
nF nA 1000 nA 6
nA 200 rev/min
Noting that power equals torque times angular velocity, the input torque is
T2
550 lbf ft/s 60 s 1 rev 12 in
H
25 hp
1576 lbf in
n2 1000 rev/min
hp
min 2 rad ft
For 100 percent gear efficiency, the output power equals the input power, so
Tarm
H
25 hp 550 lbf ft/s 60 s 1 rev 12 in
7878 lbf in
n A 200 rev/min
hp
min 2 rad ft
Next, we’ll confirm the output torque as we work through the force analysis and
complete the free body diagrams.
Gear 2
1576
788 lbf
2
F32r 788 tan 20 287 lbf
Wt
Gear 4
FA4 2W t 2 788 1576 lbf
Chapter 13, Page 15/35
Gear 5
Arm
Tout 1576 9 1576 4 7880 lbf in
Ans.
______________________________________________________________________________
13-33 Given: m = 12 mm, n P = 1800 rev/min cw,
N 2 = 18T, N 3 = 32T, N 4 = 18T, N 5 = 48T
Pitch Diameters:
d 2 = 18(12) = 216 mm, d 3 = 32(12) = 384 mm,
d 4 = 18(12) = 216 mm, d 5 = 48(12) = 576 mm
Gear 2
From Eq. (13-36),
60 000 150
60 000 H
7.368 kN
Wt
dn
216 1800
d
216
Ta 2 Wt 2 7.368
795.7 N m
2
2
W r 7.368 tan 20 2.682 kN
Gears 3 and 4
384
216
Wt
7.368
2
2
t
W 13.10 kN
W r 13.10 tan 20 4.768 kN
Ans.
______________________________________________________________________________
Chapter 13, Page 16/35
13-34 Given: P = 5 teeth/in, N 2 = 18T, N 3 = 45T,
n 20 , H = 32 hp, n 2 = 1800 rev/min
Gear 2
Tin
63025 32
1800
1120 lbf in
18
3.600 in
5
45
dG
9.000 in
5
1120
W32t
622 lbf
3.6 / 2
W32r 622 tan 20 226 lbf
dP
Fat2 W32t 622 lbf,
Fa 2 6222 226 2
Far2 W32r 226 lbf
1/2
662 lbf
Each bearing on shaft a has the same radial load of R A = R B = 662/2 = 331 lbf.
Gear 3
W23t W32t 622 lbf
W23r W32r 226 lbf
Fb3 Fb 2 662 lbf
RC RD 662 / 2 331 lbf
Each bearing on shaft b has the same radial load which is equal to the radial load of
bearings A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans.
______________________________________________________________________________
13-35 Given: P = 4 teeth/in,
n 20 ,
N P = 20T,
n 2 = 900 rev/min
N P 20
5.000 in
P
4
63025 30 2
Tin
4202 lbf in
900
W32t Tin / d2 / 2 4202 / 5 / 2 1681 lbf
d2
W32r 1681 tan 20 612 lbf
Chapter 13, Page 17/35
The motor mount resists the equivalent forces and torque.
The radial force due to torque is
Fr
4202
150 lbf
14 2
Forces reverse with rotational sense as
torque reverses.
The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32t ,
the tensions in C and D are
M AB 0
1681 4.875 15.25 2 F 15.25 0
F 1109 lbf
Chapter 13, Page 18/35
If W32t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces
change direction. For A and B,
1681 2.875 2 F1 13.25 0 F1 182.4 lbf
For W32r ,
M 612 4.875 11.25 / 2 6426 lbf in
a
14 / 2 11.25 / 2
F2
6426
179 lbf
4 8.98
2
2
8.98 in
At C and D, the shear forces are:
FS 1 153 179 5.625 / 8.98 179 7 / 8.98
2
2
At A and B, the shear forces are:
FS 2 153 179 5.625 / 8.98 179 7 / 8.98
145 lbf
2
2
The shear forces are independent of the rotational sense.
The bolt tensions and the shear forces for cw rotation are,
Chapter 13, Page 19/35
For ccw rotation,
______________________________________________________________________________
13-36 (a)
N 2 = N 4 = 15 teeth,
P
N
d
d
N 3 = N 5 = 44 teeth
N
P
15
2.5 in Ans.
6
44
d3 d5
7.33 in Ans.
6
d2 d4
(b)
(c)
Vi V2 V3
d 2 n2
2.5 2500
1636 ft/min Ans.
12
d n 2.5 2500 15 / 44
Vo V4 V5 4 4
558 ft/min
12
12
Input gears:
Wti 33000
12
H 33000 25
504.3 lbf 504 lbf
Vi
1636
Wri Wti tan 504.3 tan 20 184 lbf
W
504.3
Wi ti
537 lbf
Ans.
cos cos 20
Output gears:
H 33000 25
Wto 33000
1478 lbf
Vo
558
Ans.
Ans.
Ans.
Ans.
Wro Wto tan 1478 tan 20 538 lbf Ans.
Wto
1478
Wo
1573 lbf Ans.
cos 20
cos 20
(d)
d
2.5
Ti Wti 2 504.3
630 lbf in
2
2
Ans.
Chapter 13, Page 20/35
2
2
44
44
To Ti 630 5420 lbf in Ans.
15
15
______________________________________________________________________________
(e)
13-37 H 35 hp, ni 1200 rev/min, =20
N 2 N 4 16 teeth, N3 N5 48 teeth, P 10 teeth/in
N
16
(a)
nintermediate n3 n4 2 ni 1200 400 rev/min
N3
48
N N
16 16
no 2 4 ni 1200 133.3 rev/min
N3 N5
48 48
(b)
P
N
d
d
Ans.
Ans.
N
P
16
1.6 in Ans.
10
48
d3 d5
4.8 in Ans.
10
d n 1.6 1200
Vi V2 V3 2 2
502.7 ft/min
12
12
d n 1.6 400
Vo V4 V5 4 4
167.6 ft/min
12
12
d2 d4
(c)
Wti 33000
H 33000 35
2298 lbf lbf
Vi
502.7
Wri Wti tan 2298 tan 20 836.4 lbf
W
2298
Wi ti
2445 lbf
Ans.
cos cos 20
Wto 33000
H 33000 35
6891 lbf
Vo
167.6
Ans.
Ans.
Ans.
Ans.
Ans.
Wro Wto tan 6891tan 20 2508 lbf Ans.
Wto
6891
Wo
7333 lbf Ans.
cos 20 cos 20
(d)
d
1.6
Ti Wti 2 2298
1838 lbf in
2
2
2
Ans.
2
48
48
(e)
To Ti 1838 16 540 lbf in Ans.
16
16
______________________________________________________________________________
Chapter 13, Page 21/35
13-38 (a)
For
o 2
, from Eq. (13-11), with m = 2, k = 1, 20
i 1
2 1
NP
2 22 1 2 2 sin 2 20 14.16
2
1 2 2 sin 20
So
N P min 15
(b)
P
Ans.
N 15
1.875 teeth/in
d
8
Ans.
(c)
To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
For A, with = 20°,
W tA = F A cos20° = 300 cos20° = 281.9 lbf
For A, with = 25°, same transmitted load,
F A = W tA /cos25° = 281.9/cos25° = 311.0 lbf
Ans.
Summing the torque about the shaft axis,
d
d
WtA A WtB B
2
2
d / 2 W d A 281.9 20 704.75 lbf
WtB WtA A
d B / 2 tA d B
8
WtB
704.75
FB
777.6 lbf Ans.
cos 25 cos 25
______________________________________________________________________________
13-39 (a)
For
o 5
, from Eq. (13-11), with m = 5, k = 1, 20
i 1
2 1
NP
5 52 1 2 5 sin 2 25 10.4
1 2 5 sin 2 25
So
N P min 11
(b)
m
Ans.
d 300
27.3 mm/tooth
N
11
Ans.
(d)
To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
Chapter 13, Page 22/35
For A, with = 20°,
W tA = F A cos20° = 11 cos20° = 10.33 kN
For A, with = 25°, same transmitted load,
F A = W tA /cos25° = 10.33 / cos 25° = 11.40 kN
Ans.
Summing the torque about the shaft axis,
d
d
WtA A WtB B
2
2
d / 2 W d A 11.40 600 22.80 kN
WtB WtA A
d B / 2 tA d B
300
WtB
22.80
FB
25.16 kN Ans.
cos 25 cos 25
______________________________________________________________________________
13-40 (a)
Using Eq. (13-11) with k = 1, = 20º, and m = 2,
NP
2k
m m 2 1 2m sin 2
2
1 2m sin
2
1 2 2 sin 20
2 1
2
2
2
1 2 2 sin 2 20 14.16 teeth
Round up for the minimum integer number of teeth.
N F = 15 teeth, N C = 30 teeth
Ans.
d 125
8.33 mm/tooth
N 15
(b)
m
(c)
T
(d)
From Eq. (13-36),
H
Wt
Ans.
2 kW
1000 W rev 60 s
100 N m
191 rev/min kW 2 rad min
60 000 2
60 000 H
1.60 kN 1600 N
dn
125 191
Ans.
Or, we could have obtained W t directly from the torque and radius,
Chapter 13, Page 23/35
Wt
T
100
1600 N
d / 2 0.125 / 2
Wr Wt tan 1600 tan 20 583 N Ans.
W
1600
W t
1700 N Ans.
cos cos 20
______________________________________________________________________________
13-41 (a)
Using Eq. (13-11) with k = 1, = 20º, and m = 2,
NP
2k
m m 2 1 2m sin 2
2
1 2m sin
2
1 2 2 sin 20
2 1
2
2
2
1 2 2 sin 2 20 14.16 teeth
Round up for the minimum integer number of teeth.
N C = 15 teeth, N F = 30 teeth
Ans.
(b)
(c)
(d)
N 30
3 teeth/in
d 10
P
Ans.
550 lbf ft/s 12 in rev 60 s
1 hp
70 rev/min
hp
ft 2 rad min
T 900 lbf in
Ans.
T
H
From Eqs. (13-34) and (13-35),
V
dn
10 70
183.3 ft/min
12
H 33000 1
Wt 33000
180 lbf
V
183.3
12
Ans.
Wr Wt tan 180 tan 20 65.5 lbf Ans.
W
180
W t
192 lbf Ans.
cos cos 20
______________________________________________________________________________
N
d
1.30
13-42 (a)
Eq. (13-14): tan 1 P tan 1 P tan 1
Ans.
18.5
3.88
N
d
G
G
(b)
Eq. (13-34):
V
dn
12
2 1.30 600
12
408.4 ft/min
Ans.
Chapter 13, Page 24/35
(c)
Eq. (13-38):
H
10
33 000
808 lbf
V
408.4
Wr Wt tan cos 808 tan 20 cos18.5 279 lbf
Eq. (13-38):
Wa Wt tan sin 808 tan 20 sin18.5 93.3 lbf
Eq. (13-35):
Wt 33 000
Ans.
Ans.
Ans.
The tangential and axial forces agree with Prob. 3-74, but the radial force given in Prob.
3-74 is shown here to be incorrect. Ans.
______________________________________________________________________________
Tin 63 025H / n 63025 2.5 / 240 656.5 lbf in
13-43
W t T / r 656.5 / 2 328.3 lbf
tan 1 2 / 4 26.565
tan 1 4 / 2 63.435
a 2 1.5cos 26.565 / 2 2.67 in
W r 328.3 tan 20 cos 26.565 106.9 lbf
W a 328.3 tan 20 sin 26.565 53.4 lbf
W = 106.9i – 53.4j + 328.3k lbf
R AG = –2i + 5.17j,
R AB = 2.5j
M 4 R AG W + R AB FB T 0
Solving gives
R AB FB 2.5 FBz i 2.5 FBx k
R AG W 1697i 656.6 j 445.9k
So
1697i 656.6 j 445.9k 2.5FBz i 2.5 FBxk Tj 0
FBz 1697 / 2.5 678.8 lbf
T 656.6 lbf in
FBx 445.9 / 2.5 178.4 lbf
So
1/ 2
2
2
FB 678.8 178.4
702 lbf
Ans.
F A = – (F B + W)
= – (–178.8i – 678.8k + 106.9i – 53.4j + 328.3k)
= 71.5i + 53.4j + 350.5k
Chapter 13, Page 25/35
FA (radial) 71.52 350.52
1/ 2
358 lbf
Ans.
FA (thrust) 53.4 lbf Ans.
______________________________________________________________________________
13-44
d 2 18 /10 1.8 in, d3 30 /10 3.0 in
d2 / 2
1 0.9
tan
30.96
d
/
2
1.5
3
tan 1
180 59.04
9
DE 0.5cos 59.04 0.8197 in
16
t
W 25 lbf
W r 25 tan 20 cos 59.04 4.681 lbf
W a 25 tan 20 sin 59.04 7.803 lbf
W = –4.681i – 7.803j +25k
R DG = 0.8197j + 1.25i
R DC = –0.625j
M D R DG W R DC FC T 0
R DG W 20.49i 31.25 j 5.917k
R DC FC 0.625 FCz i 0.625 FCx k
20.49i 31.25 j 5.917k 0.625 FCz i 0.625 FCx k Tj 0
T 31.25 lbf in
Ans.
FC 9.47i 32.8k lbf
FC 9.47 2 32.82
F 0
1/2
Ans.
34.1 lbf
Ans.
FD 4.79i 7.80 j 57.8k lbf
1/2
2
2
FD (radial) 4.79 57.8 58.0 lbf Ans.
a
FD (thrust) W 7.80 lbf Ans.
______________________________________________________________________________
Chapter 13, Page 26/35
13-45
Pt Pn cos 4 cos 30 3.464 teeth/in
t tan 1
dP
tan n
tan 20
tan 1
22.80
cos
cos 30
18
5.196 in
3.464
The forces on the shafts will be equal and opposite of the forces transmitted to the
gears through the meshing teeth.
Pinion (Gear 2)
W r W t tan t 800 tan 22.80 336 lbf
W a W t tan 800 tan 30 462 lbf
W 336i 462 j 800k lbf Ans.
1/2
W 336 462 8002
2
Gear 3
2
983 lbf
Ans.
W 336i 462 j 800k lbf Ans.
W 983 lbf Ans.
32
dG
9.238 in
3.464
TG W t r 800 9.238 7390 lbf in
______________________________________________________________________________
Chapter 13, Page 27/35
13-46 From Prob. 13-45 solution,
Notice that the idler shaft reaction contains a couple tending to turn the shaft end-overend. Also the idler teeth are bent both ways. Idlers are more severely loaded than other
gears, belying their name. Thus, be cautious.
______________________________________________________________________________
13-47 Gear 3:
Pt Pn cos 7 cos 30 6.062 teeth/in
tan 20
0.4203, t 22.8
cos 30
54
d3
8.908 in
6.062
W t 500 lbf
W a 500 tan 30 288.7 lbf
W r 500 tan 22.8 210.2 lbf
W3 210.2i 288.7 j 500k lbf Ans.
Gear 4:
14
d4
2.309 in
6.062
8.908
W t 500
1929 lbf
2.309
W a 1929 tan 30 1114 lbf
W r 1929 tan 22.8 811 lbf
W4 811i 1114 j 1929k lbf Ans.
______________________________________________________________________________
tan t
13-48
Pt 6 cos 30 5.196 teeth/in
42
d3
8.083 in
5.196
t 22.8
Chapter 13, Page 28/35
16
3.079 in
5.196
63025 25
T2
916 lbf in
1720
T
916
Wt
595 lbf
r 3.079 / 2
W a 595 tan 30 344 lbf
W r 595 tan 22.8 250 lbf
d2
W 344i 250 j 595k lbf
R DC 6i,
R DG 3i 4.04 j
M D R DC FC R DG W T 0
R DG W 2404i 1785 j 2140k
(1)
R DC FC 6 FCz j 6 FCy k
Substituting and solving Eq. (1) gives
T 2404i lbf in
FCz 297.5 lbf
FCy 365.7 lbf
F FD FC W 0
Substituting and solving gives
FCx 344 lbf
FDy 106.7 lbf
FDz 297.5 lbf
FC 344i 356.7 j 297.5k lbf Ans.
FD 106.7 j 297.5k lbf Ans.
______________________________________________________________________________
Chapter 13, Page 29/35
13-49
Since the transverse pressure angle is specified, we will assume the given module is also
in terms of the transverse orientation.
d 2 mN 2 4 16 64 mm
d3 mN3 4 36 144 mm
d 4 mN 4 4 28 112 mm
6 kW
1000 W rev 60 s
35.81 N m
1600 rev/min kW 2 rad min
T
35.81
Wt
1119 N
d 2 / 2 0.064 / 2
T
H
Chapter 13, Page 30/35
W r W t tan t 1119 tan 20 407.3 N
W a W t tan 1119 tan15 299.8 N
F2 a 1119i 407.3j 299.8k N Ans.
F3b 1119 407.3 i 1119 407.3 j
711.7i 711.7 j N Ans.
F4c 407.3i 1119 j 299.8k N Ans.
______________________________________________________________________________
13-50
N
14
36
2.021 in, d3
5.196 in
Pn cos 8cos 30
8cos 30
15
45
d4
3.106 in, d5
9.317 in
5cos15
5cos15
d2
For gears 2 and 3: t tan 1 tan n / cos tan 1 tan 20 / cos 30 22.8
For gears 4 and 5: t tan 1 tan 20 / cos15 20.6
F23t T2 / r2 1200 / 2.021/ 2 1188 lbf
5.196
1987 lbf
3.106
F23r F23t tan t 1188 tan 22.8 499 lbf
F54t 1188
F54r 1986 tan 20.6 746 lbf
F23a F23t tan 1188 tan 30 686 lbf
F54a 1986 tan15 532 lbf
Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as
shown. Position vectors are
Chapter 13, Page 31/35
R CG 1.553 j 3k
R CH 2.598 j 6.5k
R CD 8.5k
Force vectors are
F54 1986i 748 j 532k
F23 1188i 500 j 686k
FC FCx i FCy j
FD FDx i FDy j FDz k
Now, a summation of moments about bearing C gives
M C R CG F54 R CH F23 R CD FD 0
The terms for this equation are found to be
R CG F54 1412i 5961j 3086k
R CH F23 5026i 7722 j 3086k
R CD FD 8.5 FDy i 8.5 FDx j
When these terms are placed back into the moment equation, the k terms, representing
the shaft torque, cancel. The i and j terms give
3614
425 lbf Ans.
8.5
13683
FDx
1610 lbf Ans.
8.5
FDy
Next, we sum the forces to zero.
F FC F54 F23 FD 0
Substituting, gives
F i F j 1987i 746 j 532k 1188i 499 j 686k
1610i 425 j F k 0
x
C
y
C
z
D
Solving gives
FCx 1987 1188 1610 1565 lbf
F 746 499 425 672 lbf
y
C
z
D
Ans.
Ans.
F 532 686 154 lbf Ans.
______________________________________________________________________________
Chapter 13, Page 32/35
VW
13-51
WW t
dW nW
0.100 600
60
60
H 2000
637 N
VW
m/s
L px NW 25 1 25 mm
tan 1
tan 1
L
dW
25
4.550 lead angle
100
WW t
W
cos n sin f cos
V
VS W
3.152 m/s
cos cos 4.550
In ft/min: V S = 3.28(3.152) = 10.33 ft/s = 620 ft/min
Use f = 0.043 from curve A of Fig. 13-42. Then, from the first of Eq. (13-43)
W
637
5323 N
cos14.5 sin 4.55 0.043cos 4.55
W y W sin n 5323sin14.5 1333 N
W z 5323 cos14.5 cos 4.55 0.043 sin 4.55 5119 N
The force acting against the worm is
W 637i 1333 j 5119k N
Thus, A is the thrust bearing.
Ans.
R AG 0.05 j 0.10k , R AB 0.20k
M A R AG W R AB FB T 0
R AG W 122.6i 63.7 j 31.85k
R AB FB 0.2 FBy i 0.2 FBx j
Substituting and solving gives
T 31.85 N m Ans.
FBx 318.5 N, FBy 613 N
So
FB 318.5i 613 j N
Ans.
Chapter 13, Page 33/35
2
2
FB 613 318.5
F FA W R B 0
Or
1/2
691 N radial
FA W FB 637i 1333j 5119k 318.5i 613j
318.5i 1946 j 5119k
Ans.
FAr 318.5i 1946 j N
Radial
1/2
2
2
FAr 318.5 1946 1972 N
a
Thrust
FA 5119 N
______________________________________________________________________________
13-52 From Prob. 13-51,
WG 637i 1333j 5119k N
pt px
So
dG
N G px
48 25
382 mm
Bearing D takes the thrust load.
M D R DG WG R DC FC T 0
R DG 0.0725i 0.191j
R DC 0.1075i
The position vectors are in meters.
R DG WG 977.7i 371.1j 25.02k
R DC FC 0.1075 FCz j 0.1075 FCy k
Putting it together and solving,
T 977.7 N m Ans.
FC 233 j 3450k N,
F FC WG FD 0
FC 3460 N
Ans.
FD FC WG 637i 1566 j 1669k N
Ans.
Radial FDr 1566 j 1669k N
Or
FDr 15662 1669 2
FDt 637i N
1/ 2
2289 N (total radial)
(thrust)
Chapter 13, Page 34/35
______________________________________________________________________________
13-53
VW
1.5 600
235.7 ft/min
12
33000 0.75
W x WW t
105.0 lbf
235.7
pt px 0.3927 in
8
L 0.3927 2 0.7854 in
0.7854
9.46
1.5
105.0
W
515.3 lbf
cos 20 sin 9.46 0.05cos 9.46
W y 515.3sin 20 176.2 lbf
W z 515.3 cos 20 cos 9.46 0.05sin 9.46 473.4 lbf
tan 1
So
W 105i 176 j 473k lbf
Ans.
T 105 0.75 78.8 lbf in
Ans.
______________________________________________________________________________
13-54 Computer programs will vary.
Chapter 13, Page 35/35
Chapter 14
d
14-1
N
22
3.667 in
P
6
Table 14-2:
Y = 0.331
dn (3.667)(1200)
Eq. (13-34): V
1152 ft/min
12
12
1200 1152
Eq. (14-4b): K v
1.96
1200
H
15
Eq. (13-35) : W t 33 000
33 000
429.7 lbf
V
1152
K vW t P 1.96(429.7)(6)
7633 psi 7.63 kpsi Ans.
Eq. (14-7):
FY
2(0.331)
________________________________________________________________________
18
N
1.8 in
10
P
Table 14-2: Y = 0.309
dn (1.8)(600)
Eq. (13-34): V
282.7 ft/min
12
12
1200 282.7
Eq. (14-4b): K v
1.236
1200
H
2
Eq. (13-35) : W t 33 000
33 000
233.5 lbf
V
282.7
K W t P 1.236(233.5)(10)
Eq. (14-7):
v
9340 psi 9.34 kpsi Ans.
FY
1.0(0.309)
________________________________________________________________________
14-2
d
14-3
d mN 1.25(18) 22.5 mm
Y = 0.309
dn (22.5)(103 )(1800)
V
2.121 m/s
60
60
6.1 2.121
Kv
1.348
6.1
60 000H
60 000(0.5)
Wt
0.2358 kN 235.8 N
dn
(22.5)(1800)
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
K vW t
1.348(235.8)
68.6 MPa Ans.
FmY
12(1.25)(0.309)
________________________________________________________________________
Eq. (14-8):
Chapter 14, Page 1/39
14-4
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
d mN 8(16) 128 mm
Y = 0.296
dn (128)(103 )(150)
V
1.0053 m/s
60
60
6.1 1.0053
Kv
1.165
6.1
60 000H
60 000(6)
Wt
5.968 kN 5968 N
dn
(128)(150)
K vW t
1.165(5968)
32.6 MPa Ans.
FmY
90(8)(0.296)
________________________________________________________________________
Eq. (14-8):
14-5
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Eq. (14-8):
d mN 1(16) 16 mm
Y = 0.296
dn (16)(103 )(400)
V
0.335 m/s
60
60
6.1 0.335
Kv
1.055
6.1
60 000H
60 000(0.15)
Wt
0.4476 kN 447.6 N
dn
(16)(400)
K vW t
1.055(447.6)
F
10.6 mm
mY 150(1)(0.296)
From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans.
________________________________________________________________________
14-6
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Eq. (14-8):
d mN 2(20) 40 mm
Y = 0.322
dn (40)(103 )(200)
V
0.419 m/s
60
60
6.1 0.419
Kv
1.069
6.1
60 000H
60 000(0.5)
Wt
1.194 kN 1194 N
dn
(40)(200)
F
K vW t
1.069(1194)
26.4 mm
75(2.0)(0.322)
mY
From Table 13-2, use F = 28 mm. Ans.
________________________________________________________________________
Chapter 14, Page 2/39
14-7
24
N
4.8 in
5
P
Table 14-2: Y = 0.337
dn (4.8)(50)
Eq. (13-34): V
62.83 ft/min
12
12
1200 62.83
Eq. (14-4b): K v
1.052
1200
H
6
Eq. (13-35) : W t 33 000
33 000
3151 lbf
V
62.83
K vW t P 1.052(3151)(5)
F
Eq. (14-7):
2.46 in
20(103 )(0.337)
Y
d
Use F = 2.5 in Ans.
________________________________________________________________________
16
N
4.0 in
4
P
Table 14-2: Y = 0.296
dn (4.0)(400)
Eq. (13-34): V
418.9 ft/min
12
12
1200 418.9
Eq. (14-4b): K v
1.349
1200
H
20
Eq. (13-35) : W t 33 000
33 000
1575.6 lbf
V
418.9
K W t P 1.349(1575.6)(4)
F v
Eq. (14-7):
2.39 in
12(103 )(0.296)
Y
Use F = 2.5 in Ans.
________________________________________________________________________
d
14-8
14-9
Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.
Eq. (13-34):
Eq. (14-4b):
Eq. (13-35):
Eq. (14-7):
V
dn
12
(2.25)(600)
12
353.4 ft/min
1200 353.4
1.295
1200
H
2.5
W t 33 000
33 000
233.4 lbf
V
353.4
K W t P 1.295(233.4)(8)
F v
0.783 in
10(103 )(0.309)
Y
Kv
Using coarse integer pitches from Table 13-2, the following table is formed.
Chapter 14, Page 3/39
d
V
Wt
F
Kv
9.000 1413.717 2.178 58.356 0.082
6.000 942.478 1.785 87.535 0.152
4.500 706.858 1.589 116.713 0.240
3.000 471.239 1.393 175.069 0.473
2.250 353.429 1.295 233.426 0.782
1.800 282.743 1.236 291.782 1.167
1.500 235.619 1.196 350.139 1.627
1.125 176.715 1.147 466.852 2.773
P
2
3
4
6
8
10
12
16
Other considerations may dictate the selection. Good candidates are P = 8
(F = 7/8 in) and P =10 (F = 1.25 in). Ans.
________________________________________________________________________
14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309.
V
dn
(36)(103 )(900)
1.696 m/s
60
60
6.1 1.696
Eq. (14-6b): K v
1.278
6.1
60 000H
60 000(1.5)
Eq. (13-36): W t
0.884 kN 884 N
dn
(36)(900)
1.278(884)
Eq. (14-8):
F
24.4 mm
75(2)(0.309)
Using the preferred module sizes from Table 13-2:
m
1.00
1.25
1.50
2.00
3.00
4.00
5.00
6.00
8.00
10.00
12.00
16.00
20.00
25.00
32.00
40.00
50.00
d
18.0
22.5
27.0
36.0
54.0
72.0
90.0
108.0
144.0
180.0
216.0
288.0
360.0
450.0
576.0
720.0
900.0
V
0.848
1.060
1.272
1.696
2.545
3.393
4.241
5.089
6.786
8.482
10.179
13.572
16.965
21.206
27.143
33.929
42.412
Kv
Wt
F
1.139 1768.388 86.917
1.174 1414.711 57.324
1.209 1178.926 40.987
1.278 884.194 24.382
1.417 589.463 12.015
1.556 442.097 7.422
1.695 353.678 5.174
1.834 294.731 3.888
2.112 221.049 2.519
2.391 176.839 1.824
2.669 147.366 1.414
3.225 110.524 0.961
3.781
88.419 0.721
4.476
70.736 0.547
5.450
55.262 0.406
6.562
44.210 0.313
7.953
35.368 0.243
Chapter 14, Page 4/39
1.204(202.6) 1
1
C 2100
F cos 20 0.228 0.684
1/ 2
100(103 )
2
2100 1.204(202.6) 1
1
F
0.669 in
3
100(10 ) cos 20 0.228 0.684
Use F = 0.75 in Ans.
________________________________________________________________________
d p 5(24) 120 mm, dG 5(48) 240 mm
14-13
V
(120)(103 )(50)
0.3142 m/s
60
3.05 0.3142
1.103
3.05
60 000H
60(103 ) H
Wt
3.183H
dn
(120)(50)
where H is in kW and Wt is in kN
Eq. (14-6a):
Kv
Table 14-8: C p 163 MPa [Note: Using Eq. (14-13) can result in wide variation in
C p due to wide variation in cast iron properties].
120sin 20
240 sin 20
Eq. (14-12): r1
20.52 mm, r2
41.04 mm
2
2
Eq. (14-14):
1.103(3.183) 103 H
690 163
60 cos 20o
1/ 2
1
1
20.52 41.04
H 3.94 kW Ans.
________________________________________________________________________
14-14
Eq. (14-6a):
d P 4(20) 80 mm, dG 4(32) 128 mm
(80)(103 )(1000)
V
4.189 m/s
60
3.05 4.189
Kv
2.373
3.05
60(10)(103 )
Wt
2.387 kN 2387 N
(80)(1000)
Table 14-8: C p 163 MPa [Note: Using Eq. (14-13) can result in wide variation in
C p due to wide variation in cast iron properties.]
80 sin 20
128sin 20
Eq. (14-12): r1
13.68 mm, r2
21.89 mm
2
2
Chapter 14, Page 6/39
1/ 2
2.373(2387) 1
1
617 MPa Ans.
50 cos 20 13.68 21.89
________________________________________________________________________
Eq. (14-14):
C 163
14-15 The pinion controls the design.
Bending
Y P = 0.303,
Y G = 0.359
17
30
1.417 in, dG
2.500 in
12
12
d Pn (1.417)(525)
V
194.8 ft/min
12
12
1200 194.8
Eq. (14-4b): K v
1.162
1200
Eq. (6-8), p. 282:
Se 0.5(76) 38.0 kpsi
Eq. (6-19), p. 287:
k a = 2.70(76)–0.265 = 0.857
2.25 2.25
l
0.1875 in
12
Pd
3Y
3(0.303)
Eq. (14-3):
x P
0.0379 in
2P
2(12)
dP
Eq. (b), p. 737: t
Eq. (6-25), p. 289:
4lx
4(0.1875)(0.0379) 0.1686 in
de 0.808 hb 0.808 0.875(0.1686) 0.310 in
0.107
0.310
kb
Eq. (6-20), p. 288:
0.996
0.3
kc = kd = ke = 1
Account for one-way bending with k f = 1.66. (See Ex. 14-2.)
Eq. (6-18), p. 287:
S e = 0.857(0.996)(1)(1)(1)(1.66)(38.0) = 53.84 kpsi
For stress concentration, find the radius of the root fillet (See Ex. 14-2).
rf
0.300 0.300
0.025 in
P
12
From Fig. A-15-6,
r
r
0.025
f
0.148
d
t
0.1686
Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, K t = 1.68.
From Fig. 6-20, with S ut = 76 kpsi and r = 0.025 in, q = 0.62.
Eq. (6-32):
K f = 1 + 0.62 (1.68 – 1) = 1.42
Chapter 14, Page 7/39
Se
53.84
16.85 psi
1.42(2.25)
K f nd
0.875(0.303)(16 850)
FYP all
320.4 lbf
Wt
1.162(12)
K v Pd
320.4(194.8)
W tV
1.89 hp Ans.
H
33 000
33 000
all
Wear
1 = 2 = 0.292, E 1 = E 2 = 30(106) psi
1/ 2
Eq. (14-13):
1
Cp
2
1 0.292
2 30 106
Eq. (14-12):
r1
2285 psi
dP
1.417
sin
sin 20 0.242 in
2
2
d
2.500
r2 G sin
sin 20 0.428 in
2
2
1 1
1
1
6.469 in 1
r1 r2
0.242 0.428
(SC )108 0.4H B 10 kpsi [0.4(149) 10](103 ) 49 600 psi
From the discussion and equation developed on the bottom of p. 329,
S 8 49 600
C ,all C 10
33 067 psi
n
2.25
Eq. (6-68), p. 329:
2
33 067 0.875cos 20
Wt
22.6 lbf
2285 1.162(6.469)
W tV
22.6(194.8)
H
0.133 hp Ans.
33 000
33 000
Rating power (pinion controls):
Eq. (14-14):
H 1 = 1.89 hp
H 2 = 0.133 hp
H all = (min 1.89, 0.133) = 0.133 hp Ans.
________________________________________________________________________
14-16 See Prob. 14-15 solution for equation numbers.
Chapter 14, Page 8/39
Pinion controls:
Bending
Y P = 0.322,
Y G = 0.447
d P = 20/3 = 6.667 in, d G = 100/3 = 33.333 in
V d P n / 12 (6.667)(870) / 12 1519 ft/min
K v (1200 1519) / 1200 2.266
Se 0.5(113) 56.5 kpsi
ka 2.70(113) 0.265 0.771
l 2.25 / Pd 2.25 / 3 0.75 in
x 3(0.322) / [2(3)] 0.161 in
t 4(0.75)(0.161) 0.695 in
d e 0.808 2.5(0.695) 1.065 in
kb (1.065 / 0.30) 0.107 0.873
kc kd ke 1
k f = 1.66 (See Ex. 14-2.)
Se 0.771(0.873)(1)(1)(1)(1.66)(56.5) 63.1 kpsi
rf 0.300 / 3 0.100 in
r
r
0.100
f
0.144
d
t
0.695
K t = 1.75, q = 0.85, K f = 1.64
Se
63.1
all
25.7 kpsi
K f nd
1.64(1.5)
FYP all
2.5(0.322)(25 700)
Wt
3043 lbf
K v Pd
2.266(3)
H W tV / 33 000 3043(1519) / 33 000 140 hp
Ans.
Wear
1/ 2
Eq. (14-13):
Eq. (14-12):
1
Cp
2
1 0.292
2 30 106
2285 psi
r 1 = (6.667/2) sin 20° = 1.140 in
r 2 = (33.333/2) sin 20° = 5.700 in
Eq. (6-68), p. 329:
C ,all
S C = [0.4(262) – 10](103) = 94 800 psi
SC / nd 94 800 / 1.5 77 400 psi
Chapter 14, Page 9/39
2
F cos
1
W t C ,all
C p K v 1 / r1 1 / r2
2
1
77 400 2.5cos 20
2285 2.266 1 / 1.140 1 / 5.700
1130 lbf
W tV
1130(1519)
H
52.0 hp Ans.
33 000
33 000
For 108 cycles (revolutions of the pinion), the power based on wear is 52.0 hp.
Rating power (pinion controls):
H1 140 hp
H 2 52.0 hp
H rated min(140, 52.0) 52.0 hp Ans.
________________________________________________________________________
14-17 See Prob. 14-15 solution for equation numbers.
Given: = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, N P = 16 milled teeth,
N G = 30T, S ut = 900 MPa, H B = 260, n d = 3, Y P = 0.296, and Y G = 0.359.
Pinion bending
d P mN P 6(16) 96 mm
dG 6(30) 180 mm
d Pn (96)(103 )(1145)
5.76 m/s
V
60
(60)
6.1 5.76
Kv
1.944
6.1
Se 0.5(900) 450 MPa
ka 4.51(900)0.265 0.744
l 2.25m 2.25(6) 13.5 mm
x 3Ym / 2 3(0.296)6 / 2 2.664 mm
t 4lx 4(13.5)(2.664) 12.0 mm
de 0.808 75(12.0) 24.23 mm
0.107
24.23
0.884
kb
7.62
kc k d k e 1
k f = 1.66 (See Ex. 14-2)
S e 0.744(0.884)(1)(1)(1)(1.66)(450) 491.3 MPa
rf 0.300m 0.300(6) 1.8 mm
r/d = r f /t = 1.8/12 = 0.15, K t = 1.68, q = 0.86, K f = 1.58
Chapter 14, Page 10/39
all
Se
491.3
239.2 MPa
K f nd 1.58 1.3
Eq. (14-8):
Wt
FYm all
75(0.296)(6)(239.2)
16 390 N
Kv
1.944
Eq. (13-36):
H
W t dn 16.39 (96)(1145)
94.3 kW
60 000
60 000
Ans.
Wear: Pinion and gear
Eq. (14-12):
r 1 = (96/2) sin 20 = 16.42 mm
r 2 = (180/2) sin 20 = 30.78 mm
1/ 2
1
Eq. (14-13): C p
190 MPa
2
1
0.292
2 207 103
Eq. (6-68), p. 329:
S C = 6.89[0.4(260) – 10] = 647.7 MPa
647.7
C ,all SC / nd
568 MPa
1.3
Eq. (14-14):
W C ,all
Cp
2
t
F cos
1
K v 1 / r1 1 / r2
2
Eq. (13-36):
o
1
568 75cos 20
3469 N
190 1.944 1 / 16.42 1 / 30.78
W t dn 3.469 (96)(1145)
20.0 kW
H
60 000
60 000
Thus, wear controls the gearset power rating; H = 20.0 kW. Ans.
________________________________________________________________________
N P = 17 teeth, N G = 51 teeth
N 17
dP
2.833 in
P
6
51
dG
8.500 in
6
V d P n / 12 (2.833)(1120) / 12 830.7 ft/min
14-18
Eq. (14-4b):
K v = (1200 + 830.7)/1200 = 1.692
Chapter 14, Page 11/39
all
Sy
nd
90 000
45 000 psi
2
Table 14-2:
Y P = 0.303, Y G = 0.410
Eq. (14-7):
Wt
Eq. (13-35):
W tV
2686(830.7)
H
67.6 hp
33 000
33 000
FYP all
2(0.303)(45 000)
2686 lbf
KvP
1.692(6)
Based on yielding in bending, the power is 67.6 hp.
(a) Pinion fatigue
Bending
Eq. (2-121), p. 41: S ut = 0.5 H B = 0.5(232) = 116 kpsi
Eq. (6-8), p. 282: Se 0.5Sut 0.5(116) 58 kpsi
Eq. (6-19), p. 287: ka 2.70(116) 0.265 0.766
Table 13-1, p. 696: l
Eq. (14-3): x
1
1.25 2.25 2.25
0.375 in
6
Pd
Pd
Pd
3YP
3(0.303)
0.0758 in
2P
2(6)
Eq. (b), p. 737: t
4lx
4(0.375)(0.0758) 0.337 in
Eq. (6-25), p. 289: de 0.808 F t 0.808 2(0.337) 0.663 in
0.663
Eq. (6-20), p. 288: kb
0.30
kc = kd = ke = 1
0.107
0.919
Account for one-way bending with k f = 1.66. (See Ex. 14-2.)
Eq. (6-18): Se 0.766(0.919)(1)(1)(1)(1.66)(58) 67.8 kpsi
For stress concentration, find the radius of the root fillet (See Ex. 14-2).
0.300 0.300
rf
0.050 in
P
6
r
0.05
r
Fig. A-15-6:
f
0.148
0.338
d
t
Estimate D/d = ∞ by setting D/d = 3, K t = 1.68.
Chapter 14, Page 12/39
Fig. 6-20, p. 295: q = 0.86
Eq. (6-32), p. 295: K f 1 (0.86)(1.68 1) 1.58
all
Se
67.8
21.5 kpsi
K f nd
1.58(2)
Wt
FYP all
2(0.303)(21 500)
1283 lbf
K v Pd
1.692(6)
H
W tV
1283(830.7)
32.3 hp
33 000
33 000
Ans.
(b) Pinion fatigue
Wear
1/ 2
Eq. (14-13):
Eq. (14-12):
Eq. (6-68):
1
Cp
2
6
2 [(1 - 0.292 ) / 30(10 )]
2285 psi
dP
2.833
sin
sin 20o 0.485 in
2
2
dG
8.500
r2
sin
sin 20o 1.454 in
2
2
1 1
1
1
2.750 in
r1 r2 0.485 1.454
r1
(SC )108 0.4H B 10 kpsi
In terms of gear notation
C = [0.4(232) – 10]103 = 82 800 psi
We will introduce the design factor of n d = 2 and because it is a contact stress apply it
to the load Wt by dividing by 2 . (See p. 329.)
82 800
C,all c
58 548 psi
2
2
Solve Eq. (14-14) for Wt:
2
o
58 548 2 cos 20
W
265 lbf
2285 1.692(2.750)
W tV
265(830.7)
H all
6.67 hp Ans.
33 000
33 000
For 108 cycles (turns of pinion), the allowable power is 6.67 hp.
(c) Gear fatigue due to bending and wear
t
Chapter 14, Page 13/39
Bending
x
Eq. (14-3):
Eq. (b), p. 737: t
3YG
3(0.4103)
0.1026 in
2P
2(6)
4 lx
4(0.375)(0.1026) 0.392 in
Eq. (6-25): de 0.808 F t 0.808 2(0.392) 0.715 in
0.107
0.715
Eq. (6-20): kb
0.911
0.30
kc = kd = ke = 1
k f = 1.66. (See Ex. 14-2.)
Eq. (6-18): Se 0.766(0.911)(1)(1)(1)(1.66)(58) 67.2 kpsi
r
r
0.050
f
0.128
d
t
0.392
Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; K t = 1.80.
Fig. 6-20: q = 0.82
Eq. (6-32): K f 1 (0.82)(1.80 1) 1.66
all
Se
67.2
20.2 kpsi
K f nd
1.66(2)
Wt
FYP all
2(0.4103)(20 200)
1633 lbf
K v Pd
1.692(6)
W tV
1633(830.7)
41.1 hp
33 000
33 000
The gear is thus stronger than the pinion in bending.
H all
Ans.
Wear
Since the material of the pinion and the gear are the same, and the contact stresses are
the same, the allowable power transmission of both is the same. Thus, H all = 6.67 hp
for 108 revolutions of each. As yet, we have no way to establish S C for 108/3
revolutions.
(d)
Pinion bending:
H 1 = 32.3 hp
Pinion wear:
H 2 = 6.67 hp
Gear bending:
H 3 = 41.1 hp
Gear wear:
H 4 = 6.67 hp
Power rating of the gear set is thus
H rated = min(32.3, 6.67, 41.1, 6.67) = 6.67 hp Ans.
________________________________________________________________________
14-19
d P = 16/6 = 2.667 in, d G = 48/6 = 8 in
Chapter 14, Page 14/39
V
(2.667)(300)
209.4 ft/min
12
33 000(5)
Wt
787.8 lbf
209.4
Assuming uniform loading, K o = 1.
Eq. (14-28): Qv 6, B 0.25(12 6) 2 / 3 0.8255
A 50 56(1 0.8255) 59.77
0.8255
59.77 209.4
Eq. (14-27): K v
59.77
Table 14-2: YP 0.296, YG 0.4056
From Eq. (a), Sec. 14-10 with F = 2 in
2 0.296
( K s ) P 1.192
6
1.196
0.0535
1.088
0.0535
2 0.4056
( K s )G 1.192
1.097
6
From Eq. (14-30) with C mc = 1
2
0.0375 0.0125(2) 0.0625
Cp f
10(2.667)
C p m 1, Cma 0.093 (Fig. 14 - 11), Ce 1
K m 1 1[0.0625(1) 0.093(1)] 1.156
Assuming constant thickness of the gears → K B = 1
m G = N G /N P = 48/16 = 3
With N (pinion) = 108 cycles and N (gear) = 108/3, Fig. 14-14 provides the relations:
(YN ) P 1.3558(108 )0.0178 0.977
(YN )G 1.3558(108 / 3)0.0178 0.996
Fig. 14-6:
J P 0.27, J G 0.38
Table 14-10: K R = 0.85
KT = Cf = 1
Eq. (14-23):
Table 14-8:
I
cos 20o sin 20o 3
0.1205
2(1)
3 1
C p 2300 psi
Strength: Grade 1 steel with H BP = H BG = 200
Chapter 14, Page 15/39
Fig. 14-2:
(S t ) P = (S t ) G = 77.3(200) + 12 800 = 28 260 psi
Fig. 14-5:
(S c ) P = (S c ) G = 322(200) + 29 100 = 93 500 psi
Fig. 14-15:
(Z N ) P = 1.4488(108)–0.023 = 0.948
(Z N ) G = 1.4488(108/3)–0.023 = 0.973
Sec. 14-12:
H BP /H BG = 1 C H = 1
Pinion tooth bending
Eq. (14-15):
( ) P W t K o K v K s
Pd K m K B
F J
6 (1.156)(1)
787.8(1)(1.196)(1.088)
2 0.27
13 170 psi Ans.
Eq. (14-41):
S Y / ( KT K R )
(S F ) P t N
28 260(0.977) / [(1)(0.85)]
2.47
13 170
Ans.
Gear tooth bending
Eq. (14-15):
Eq. (14-41):
6 (1.156)(1)
( )G 787.8(1)(1.196)(1.097)
9433 psi
2 0.38
28 260(0.996) / [(1)(0.85)]
3.51 Ans.
( S F )G
9433
Ans.
Pinion tooth wear
1/ 2
Eq. (14-16):
K C
( c ) P C p W t K o K v K s m f
d P F I P
1/ 2
1.156 1
2300 787.8(1)(1.196)(1.088)
2.667(2) 0.1205
98 760 psi Ans.
Eq. (14-42):
S Z /( KT K R )
93 500(0.948) /[(1)(0.85)]
(S H ) P c N
1.06
98 760
c
P
Ans.
Gear tooth wear
Chapter 14, Page 16/39
1/ 2
1/ 2
(K )
1.097
( c )G s G ( c ) P
(98 760) 99 170 psi
1.088
(K s )P
93 500(0.973)(1) /[(1)(0.85)]
( S H )G
1.08 Ans.
99 170
Ans.
The hardness of the pinion and the gear should be increased.
________________________________________________________________________
14-20
d P = 2.5(20) = 50 mm, d G = 2.5(36) = 90 mm
d PnP (50)(103 )(100)
V
0.2618 m/s
60
60
60(120)
Wt
458.4 N
(50)(103 )(100)
With no specific information given to indicate otherwise, assume
KB = Ko = Y = ZR = 1
Eq. (14-28):
Eq. (14-27):
Table 14-2:
Q v = 6, B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
59.77 200(0.2618)
Kv
59.77
Y P = 0.322, Y G = 0.3775
0.8255
1.099
Similar to Eq. (a) of Sec. 14-10 but for SI units:
Ks
1
0.8433 mF Y
kb
0.0535
( K s ) P 0.8433 2.5(18) 0.322
0.0535
( K s )G 0.8433 2.5(18) 0.3775
Cmc Ce C pm 1
1.003
0.0535
=1.007
use 1
use 1
18
0.025 0.011
10(50)
0.247 0.0167(0.709) 0.765(104 )(0.7092 ) 0.259
K H 1 1[0.011(1) 0.259(1)] 1.27
F 18 / 25.4 0.709 in, C pf
Cma
Fig. 14-14:
(Y N ) P = 1.3558(108)–0.0178 = 0.977
(Y N ) G = 1.3558(108/1.8)–0.0178 = 0.987
Fig. 14-6:
Eq. (14-38):
(Y J ) P = 0.33, (Y J ) G = 0.38
Y Z = 0.658 – 0.0759 ln(1 – 0.95) = 0.885
Chapter 14, Page 17/39
cos 20o sin 20o 1.8
0.103
2(1)
1.8 1
Eq. (14-23):
ZI
Table 14-8:
Z E 191 MPa
Strength
Grade 1 steel, H BP = H BG = 200
Fig. 14-2:
Fig. 14-5:
Fig. 14-15:
(S t ) P = (S t ) G = 0.533(200) + 88.3 = 194.9 MPa
(S c ) P = (S c ) G = 2.22(200) + 200 = 644 MPa
(Z N ) P = 1.4488(108)–0.023 = 0.948
(Z N )G 1.4488(108 / 1.8)0.023 0.961
H BP / H BG 1 ZW CH 1
Fig. 14-12:
Pinion tooth bending
Eq. (14-15):
1 KH KB
( ) P W t K o K v K s
bmt YJ P
1 1.27(1)
458.4(1)(1.099)(1)
43.08 MPa
18(2.5) 0.33
S Y
194.9 0.977
Eq. (14-41) for SI: (S F ) P t N
4.99
43.08 1(0.885)
Y YZ P
Ans.
Ans.
Gear tooth bending
1 1.27(1)
( )G 458.4(1)(1.099)(1)
37.42 MPa
18(2.5) 0.38
194.9 0.987
( S F )G
5.81 Ans.
37.42 1(0.885)
Ans.
Pinion tooth wear
Eq. (14-16):
K Z
( c ) P Z E W t K o K v K s H R
d w1b Z I P
1.27 1
191 458.4(1)(1.099)(1)
501.8 MPa
50(18) 0.103
S Z Z
644 0.948(1)
Eq. (14-42) for SI: ( S H ) P c N W
1.37 Ans.
501.8 1(0.885)
c Y YZ P
Ans.
Gear tooth wear
(K )
( c )G s G
(K s )P
1/ 2
1/ 2
1
( c ) P
1
(501.8) 501.8 MPa
Ans.
Chapter 14, Page 18/39
644 0.961(1)
1.39 Ans.
501.8 1(0.885)
________________________________________________________________________
( S H )G
14-21
Pt Pn cos 6 cos 30 5.196 teeth/in
48
16
(3.079) 9.238 in
dP
3.079 in, dG
5.196
16
(3.079)(300)
241.8 ft/min
V
12
0.8255
59.77 241.8
33 000(5)
t
682.3 lbf , K v
1.210
W
241.8
59.77
From Prob. 14-19:
YP 0.296, YG 0.4056
( K s ) P 1.088, ( K s )G 1.097, K B 1
mG 3, (YN ) P 0.977, (YN )G 0.996, K R 0.85
( St ) P ( St )G 28 260 psi, CH 1, ( S c ) P ( S c )G 93 500 psi
( Z N ) P 0.948,
( Z N )G 0.973,
C p 2300 psi
The pressure angle is:
tan 20
t tan 1
22.80
cos 30
3.079
cos 22.8 1.419 in,
2
a 1 / Pn 1 / 6 0.167 in
(rb ) P
(rb )G 3(rb ) P 4.258 in
Eq. (14-25):
1/ 2
2
2
3.079
9.238
2
0.167 4.2582
Z
0.167 1.419
2
2
3.079 9.238
sin 22.8
2
2
0.9479 2.1852 2.3865 0.7466 Conditions O.K . for use
pN pn cos n
Eq. (14-21):
mN
6
1/ 2
cos 20 0.4920 in
pN
0.492
0.6937
0.95Z
0.95(0.7466)
Chapter 14, Page 19/39
Eq. (14-23):
sin 22.8 cos 22.8 3
I
3 1 0.193
2(0.6937)
Fig. 14-7:
J P 0.45,
J G 0.54
Fig. 14-8: Corrections are 0.94 and 0.98.
J P 0.45(0.94) 0.423, J G 0.54(0.98) 0.529
2
0.0375 0.0125(2) 0.0525
Cmc 1, C pf
10(3.079)
C pm 1, Cma 0.093, Ce 1
K m 1 (1)[0.0525(1) 0.093(1)] 1.146
Pinion tooth bending
5.196 1.146(1)
( ) P 682.3(1)(1.21)(1.088)
6323 psi
2 0.423
28 260(0.977) / [1(0.85)]
(S F ) P
5.14 Ans.
6323
Ans.
Gear tooth bending
5.196 1.146(1)
( )G 682.3(1)(1.21)(1.097)
5097 psi
2 0.529
28 260(0.996) / [1(0.85)]
( S F )G
6.50 Ans.
5097
Ans.
Pinion tooth wear
1/ 2
1.146 1
( c ) P 2300 682.3(1)(1.21)(1.088)
3.078(2) 0.193
93 500(0.948) / [(1)(0.85)]
(S H ) P
1.54 Ans.
67 700
67 700 psi
Ans.
Gear tooth wear
1/ 2
1.097
( c )G
(67 700) 67 980 psi Ans.
1.088
93 500(0.973) /[(1)(0.85)]
1.57 Ans.
( S H )G
67 980
________________________________________________________________________
14-22 Given: R = 0.99 at 108 cycles, H B = 232 through-hardening Grade 1, core and case, both
gears. N P = 17T, N G = 51T,
Table 14-2: Y P = 0.303, Y G = 0.4103
Chapter 14, Page 20/39
Fig. 14-6:
J P = 0.292, J G = 0.396
d P = N P / P = 17 / 6 = 2.833 in, d G = 51 / 6 = 8.500 in.
Pinion bending
From Fig. 14-2:
0.99
Fig. 14-14:
( St )107 77.3H B 12 800
77.3(232) 12 800 30 734 psi
Y N = 1.6831(108)–0.0323 = 0.928
V d P n / 12 (2.833)(1120 / 12) 830.7 ft/min
KT K R 1, S F 2, S H 2
30 734(0.928)
all
14 261 psi
2(1)(1)
Qv 5, B 0.25(12 5)2 / 3 0.9148
A 50 56(1 0.9148) 54.77
54.77 830.7
K v
54.77
0.9148
1.472
0.0535
2 0.303
K s 1.192
1.089 use 1
6
K m Cm f 1 Cmc (C p f C p m CmaCe )
Cmc 1
F
0.0375 0.0125F
10d
2
0.0375 0.0125(2) 0.0581
10(2.833)
1
C pf
C pm
Eq. (14-15):
Cma 0.127 0.0158(2) 0.093(104 )(22 ) 0.1586
Ce 1
K m 1 1[0.0581(1) 0.1586(1)] 1.217
KB 1
FJ P all
Wt
K o K v K s Pd K m K B
2(0.292)(14 261)
775 lbf
1(1.472)(1)(6)(1.217)(1)
775(830.7)
W tV
H
19.5 hp
33 000
33 000
Pinion wear
Chapter 14, Page 21/39
Fig. 14-15:
Eq. (14-23):
Fig. 14-5:
Z N = 2.466N–0.056 = 2.466(108)–0.056 = 0.879
mG 51 / 17 3
cos 20o sin 20o 3
I
1.205,
2
3 1
CH 1
(Sc )107 322H B 29 100
322(232) 29 100 103 804 psi
103 804(0.879)
c,all
64 519 psi
2(1)(1)
0.99
2
Eq. (14-16):
Fd P I
W c,all
C p K o K v K s K mC f
2
64 519 2(2.833)(0.1205)
2300 1(1.472)(1)(1.2167)(1)
300 lbf
W tV
300(830.7)
H
7.55 hp
33 000
33 000
t
The pinion controls, therefore H rated = 7.55 hp Ans.
________________________________________________________________________
14-23
l = 2.25/ P d ,
t
4 lx
x = 3Y / 2P d
2.25 3Y 3.674
4
Y
Pd
Pd 2Pd
3.674
F Y
d e 0.808 F t 0.808 F
Y 1.5487
Pd
Pd
0.107
0.0535
1.5487 F Y / P
F Y
d
kb
0.8389
Pd
0.30
0.0535
F Y
1
Ks
Ans.
1.192
kb
Pd
________________________________________________________________________
14-24 Y P = 0.331, Y G = 0.422, J P = 0.345, J G = 0.410, K o = 1.25. The service conditions are
adequately described by K o . Set S F = S H = 1.
d P = 22 / 4 = 5.500 in
d G = 60 / 4 = 15.000 in
Chapter 14, Page 22/39
V
Pinion bending
0.99
(5.5)(1145)
12
1649 ft/min
(St )107 77.3H B 12 800 77.3(250) 12 800 32 125 psi
YN 1.6831[3(109 )]0.0323 0.832
Eq. (14-17):
32 125(0.832)
26 728 psi
1(1)(1)
B 0.25(12 6)2 / 3 0.8255
A 50 56(1 0.8255) 59.77
all P
0.8255
59.77 1649
K v
1.534
59.77
K s 1, Cm 1
F
Cmc
0.0375 0.0125F
10d
3.25
0.0375 0.0125(3.25) 0.0622
10(5.5)
Cma 0.127 0.0158(3.25) 0.093(104 )(3.252 ) 0.178
Ce 1
K m Cm f 1 (1)[0.0622(1) 0.178(1)] 1.240
K B 1,
Eq. (14-15):
KT 1
26 728(3.25)(0.345)
3151 lbf
1.25(1.534)(1)(4)(1.240)
3151(1649)
H1
157.5 hp
33 000
W1t
Gear bending By similar reasoning, W2t 3861 lbf and H 2 192.9 hp
Pinion wear
mG 60 / 22 2.727
cos 20o sin 20o 2.727
0.1176
2
1 2.727
0.99 (S c )107 322(250) 29 100 109 600 psi
I
(Z N ) P 2.466[3(109 )]0.056 0.727
(Z N )G 2.466[3(109 ) / 2.727]0.056 0.769
109 600(0.727)
79 679 psi
( c,all ) P
1(1)(1)
Chapter 14, Page 23/39
2
Fd P I
W3t c,all
C KKKK C
p o v s m f
2
79 679 3.25(5.5)(0.1176)
1061 lbf
2300 1.25(1.534)(1)(1.24)(1)
1061(1649)
H3
53.0 hp
33 000
Gear wear
Similarly,
W4t 1182 lbf ,
H 4 59.0 hp
Rating
H rated min( H1, H 2 , H 3 , H 4 )
min(157.5, 192.9, 53, 59) 53 hp
Ans.
Note differing capacities. Can these be equalized?
________________________________________________________________________
14-25 From Prob. 14-24:
W1t 3151 lbf , W2t 3861 lbf ,
W3t 1061 lbf , W4t 1182 lbf
33 000K o H
33 000(1.25)(40)
1000 lbf
Wt
V
1649
Pinion bending: The factor of safety, based on load and stress, is
W1t
3151
(S F ) P
3.15
1000 1000
Gear bending based on load and stress
( S F )G
W2t
3861
3.86
1000 1000
Pinion wear
based on load:
based on stress:
W3t
1061
1.06
1000 1000
(S H ) P 1.06 1.03
n3
Gear wear
based on load:
n4
W4t
1182
1.18
1000 1000
Chapter 14, Page 24/39
(SH )G 1.18 1.09
based on stress:
Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06,
1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors
(S F ) P , (S F ) G , (S H ) P , (S H ) G
are
3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2
and the threat is again from pinion wear. Depending on the magnitude of the numbers,
using S F and S H as defined by AGMA, does not necessarily lead to the same conclusion
concerning threat. Therefore be cautious.
________________________________________________________________________
14-26 Solution summary from Prob. 14-24: n = 1145 rev/min, K o = 1.25, Grade 1 materials,
N P = 22T, N G = 60T, m G = 2.727, Y P = 0.331,Y G = 0.422, J P = 0.345, J G = 0.410, P d =
4T /in, F = 3.25 in, Q v = 6, (N c ) P = 3(109), R = 0.99, K m = 1.240, K T = 1, K B = 1,
d P = 5.500 in, d G = 15.000 in, V = 1649 ft/min, K v = 1.534, (K s ) P = (K s ) G = 1, (Y N ) P =
0.832, (Y N ) G = 0.859, K R = 1
Pinion H B : 250 core, 390 case
Gear H B : 250 core, 390 case
Bending
( all ) P
( all )G
W1t
W2t
26 728 psi
27 546 psi
3151 lbf ,
3861 lbf ,
(St ) P 32 125 psi
(St )G 32 125 psi
H1 157.5 hp
H 2 192.9 hp
Wear
20o ,
I 0.1176,
(Z N ) P 0.727
(Z N )G 0.769, CP 2300 psi
(Sc ) P Sc 322(390) 29 100 154 680 psi
154 680(0.727)
112 450 psi
( c,all ) P
1(1)(1)
154 680(0.769)
118 950 psi
( c,all )G
1(1)(1)
2
112 450
W3t
(1061) 2113 lbf ,
79 679
H3
2113(1649)
105.6 hp
33 000
2
118 950
W
(1182) 2354 lbf ,
109 600(0.769)
t
4
H4
2354(1649)
117.6 hp
33 000
Rated power
Chapter 14, Page 25/39
H rated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp
Ans.
Prob. 14-24:
H rated = min(157.5, 192.9, 53.0, 59.0) = 53 hp
The rated power approximately doubled.
________________________________________________________________________
14-27 The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell
285 core and Brinell 580–600 case.
Table 14-3:
0.99
(St )107 55 000 psi
Modification of S t by (Y N ) P = 0.832 produces
( all ) P 45 657 psi,
Similarly for (Y N ) G = 0.859
( all )G 47 161 psi,
W 4569 lbf ,
W 5668 lbf ,
t
1
t
2
and
H1 228 hp
H 2 283 hp
From Table 14-8, C p 2300 psi. Also, from Table 14-6:
0.99
(Sc )107 180 000 psi
Modification of S c by Y N produces
( c,all ) P 130 525 psi
( c,all )G 138 069 psi
and
W3t 2489 lbf ,
W4t 2767 lbf ,
H 3 124.3 hp
H 4 138.2 hp
Rating
H rated = min(228, 283, 124, 138) = 124 hp Ans.
________________________________________________________________________
14-28 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27.
Chapter 14, Page 26/39
Summary:
Table 14-3:
0.99
(St )107 65 000 psi
( all ) P 53 959 psi
( all )G 55 736 psi
and it follows that
W1t 5400 lbf ,
W2t 6699 lbf ,
H1 270 hp
H 2 335 hp
From Table 14-8, C p 2300 psi. Also, from Table 14-6:
Sc 225 000 psi
( c,all ) P 181 285 psi
( c,all )G 191 762 psi
Consequently,
W3t 4801 lbf ,
W4t 5337 lbf ,
H 3 240 hp
H 4 267 hp
Rating
H rated = min(270, 335, 240, 267) = 240 hp. Ans.
________________________________________________________________________
14-29 Given: n = 1145 rev/min, K o = 1.25, N P = 22T, N G = 60T, m G = 2.727, d P = 2.75 in, d G =
7.5 in, Y P = 0.331,Y G = 0.422, J P = 0.335, J G = 0.405, P = 8T /in, F = 1.625 in, H B = 250,
case and core, both gears. C m = 1, F/d P = 0.0591, C f = 0.0419, C pm = 1, C ma = 0.152,
C e = 1, K m = 1.1942, K T = 1, K B = 1, K s = 1,V = 824 ft/min, (Y N ) P = 0.8318, (Y N ) G =
0.859, K R = 1, I = 0.117 58
(St )107 32 125 psi
( all ) P 26 668 psi
( all )G 27 546 psi
0.99
and it follows that
W1t 879.3 lbf ,
W2t 1098 lbf ,
H1 21.97 hp
H 2 27.4 hp
For wear
Chapter 14, Page 27/39
W3t 304 lbf ,
W4t 340 lbf ,
H 3 7.59 hp
H 4 8.50 hp
Rating
H rated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp
In Prob. 14-24, H rated = 53 hp. Thus,
7.59
1
0.1432
,
53.0
6.98
not
1
8
Ans.
The transmitted load rating is
t
Wrated
min(879.3, 1098, 304, 340) 304 lbf
In Prob. 14-24
t
Wrated
1061 lbf
Thus
304
1
1
0.2865
Ans.
, not
1061
3.49
4
________________________________________________________________________
14-30 S P = S H = 1, P d = 4, J P = 0.345, J G = 0.410, K o = 1.25
Bending
Table 14-4:
0.99
(St )107 13 000 psi
13 000(1)
13 000 psi
1(1)(1)
all FJ P
13 000(3.25)(0.345)
1533 lbf
K o K v K s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1)
1533(1649)
76.6 hp
33 000
W1t J G / J P 1533(0.410) / 0.345 1822 lbf
H1J G / J P 76.6(0.410) / 0.345 91.0 hp
( all ) P ( all )G
W1t
H1
W2t
H2
Wear
Table 14-8:
Table 14-7:
C p 1960 psi
0.99
(Sc )107 75 000 psi ( c,all ) P ( c,all )G
Chapter 14, Page 28/39
2
( )
Fd p I
W3t c,all P
C p K o K v K s K mC f
2
75 000 3.25(5.5)(0.1176)
W
1295 lbf
1960 1.25(1.534)(1)(1.24)(1)
W4t W3t 1295 lbf
1295(1649)
H 4 H3
64.7 hp
33 000
t
3
Rating
H rated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp
Ans.
Notice that the balance between bending and wear power is improved due to CI’s more
favorable S c /S t ratio. Also note that the life is 107 pinion revolutions which is (1/300) of
3(109). Longer life goals require power de-rating.
________________________________________________________________________
14-31 From Table A-24a, E av = 11.8(106) Mpsi
For = 14.5 and H B = 156
SC
1.4(81)
51 693 psi
2sin14.5 / [11.8(106 )]
For = 20
1.4(112)
52 008 psi
2sin 20 / [11.8(106 )]
SC 0.32(156) 49.9 kpsi
The first two calculations were approximately 4 percent higher.
________________________________________________________________________
SC
14-32 Programs will vary.
________________________________________________________________________
14-33
(YN ) P 0.977, (YN )G 0.996
(St ) P ( St )G 82.3(250) 12 150 32 725 psi
32 725(0.977)
( all ) P
37 615 psi
1(0.85)
37 615(1.5)(0.423)
W1t
1558 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
1558(925)
H1
43.7 hp
33 000
Chapter 14, Page 29/39
32 725(0.996)
38 346 psi
1(0.85)
38 346(1.5)(0.5346)
W2t
2007 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
2007(925)
H2
56.3 hp
33 000
(Z N ) P 0.948, (Z N )G 0.973
( all )G
Table 14-6:
0.99
(Sc )107 150 000 psi
0.948(1)
( c,allow ) P 150 000
167 294 psi
1(0.85)
2
167 294 1.963(1.5)(0.195)
W3t
2074 lbf
2300 1(1.404)(1.043)
2074(925)
H3
58.1 hp
33 000
0.973
( c,allow )G
(167 294) 171 706 psi
0.948
2
171 706 1.963(1.5)(0.195)
t
W4
2167 lbf
2300 1(1.404)(1.052)
2167(925)
H4
60.7 hp
33 000
H rated min(43.7, 56.3, 58.1, 60.7) 43.7 hp Ans.
Pinion bending is controlling.
________________________________________________________________________
(YN ) P 1.6831(108 ) 0.0323 0.928
14-34
(YN )G 1.6831(108 / 3.059) 0.0323 0.962
Table 14-3:
S t = 55 000 psi
55 000(0.928)
60 047 psi
( all ) P
1(0.85)
60 047(1.5)(0.423)
W1t
2487 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
2487(925)
H1
69.7 hp
33 000
0.962
( all )G
(60 047) 62 247 psi
0.928
Chapter 14, Page 30/39
62 247 0.5346
(2487) 3258 lbf
60 047 0.423
3258
H2
(69.7) 91.3 hp
2487
W2t
Table 14-6:
S c = 180 000 psi
(Z N ) P 2.466(108 )0.056 0.8790
(Z N )G 2.466(108 / 3.059) 0.056 0.9358
180 000(0.8790)
186 141 psi
( c,all ) P
1(0.85)
2
186 141 1.963(1.5)(0.195)
W
2568 lbf
2300 1(1.404)(1.043)
2568(925)
H3
72.0 hp
33 000
0.9358
( c,all )G
(186 141) 198 169 psi
0.8790
t
3
2
198 169 1.043
W
(2568) 2886 lbf
186 141 1.052
2886(925)
H4
80.9 hp
33 000
H rated min(69.7, 91.3, 72, 80.9) 69.7 hp
t
4
Ans.
Pinion bending controlling
________________________________________________________________________
(Y N ) P = 0.928, (Y N ) G = 0.962 (See Prob. 14-34)
14-35
Table 14-3:
S t = 65 000 psi
65 000(0.928)
( all ) P
70 965 psi
1(0.85)
70 965(1.5)(0.423)
W1t
2939 lbf
1(1.404)(1.043)(8.66)(1.208)
2939(925)
H1
82.4 hp
33 000
65 000(0.962)
73 565 psi
( all )G
1(0.85)
73 565 0.5346
W2t
(2939) 3850 lbf
70 965 0.423
3850
H2
(82.4) 108 hp
2939
Chapter 14, Page 31/39
Table 14-6:
S c = 225 000 psi
( Z N ) P 0.8790, ( Z N )G 0.9358
225 000(0.879)
232 676 psi
( c,all ) P
1(0.85)
2
232 676 1.963(1.5)(0.195)
W3t
4013 lbf
2300 1(1.404)(1.043)
4013(925)
H3
112.5 hp
33 000
0.9358
( c,all )G
(232 676) 247 711 psi
0.8790
2
247 711 1.043
W
(4013) 4509 lbf
232 676 1.052
4509(925)
H4
126 hp
33 000
H rated min(82.4, 108, 112.5, 126) 82.4 hp
t
4
Ans.
The bending of the pinion is the controlling factor.
________________________________________________________________________
P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth
F 4 p 4 4 2
P
2
M x 0 10(300) cos 20 4FB cos20
14-36
F B = 750 lbf
W t FB cos 20 750 cos 20 705 lbf
n = 2400 / 2 = 1200 rev/min
dn (8)(1200)
2513 ft/min
V
12
12
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Fig. 14-2:
S t = 102(300) + 16 400 = 47 000 psi
Fig. 14-5:
Fig. 14-6:
S c = 349(300) + 34 300 = 139 000 psi
J = 0.27
cos 20o sin 20o 2
I
0.107
2(1)
2 1
C p 2300 psi
Eq. (14-23):
Table 14-8:
Assume a typical quality number of 6.
Eq. (14-28): B 0.25(12 Qv ) 2 / 3 0.25(12 6) 2 / 3 0.8255
Chapter 14, Page 32/39
A 50 56(1 B) 50 56(1 0.8255) 59.77
B
Eq. (14-27):
A V
59.77 2513
K v
A
59.77
0.8255
1.65
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296.
From Eq. (a), Sec. 14-10,
0.0535
0.0535
F Y
2 0.296
K s 1.192
1.192
1.23
2
P
The load distribution factor is applicable for straddle-mounted gears, which is not the
case here since the gear is mounted outboard of the bearings. Lacking anything better,
we will use the load distribution factor as a rough estimate.
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
C mc = 1 (uncrowned teeth)
2
Cp f
0.0375 0.0125(2 ) 0.1196
10(8)
C pm = 1.1
C ma = 0.23 (commercial enclosed gear unit)
Ce = 1
K m 1 1[0.1196(1.1) 0.23(1)] 1.36
For the stress-cycle factors, we need the desired number of load cycles.
Fig. 14-14:
Fig. 14-15:
Eq. 14-38:
N = 15 000 h (1200 rev/min)(60 min/h) = 1.1 (109) rev
Y N = 0.9
Z N = 0.8
K R 0.658 0.0759ln 1 R 0.658 0.0759ln 1 0.95 0.885
With no specific information given to indicate otherwise, assume K o = K B = K T = C f = 1
Tooth bending
Eq. (14-15):
Eq. (14-41):
Pd K m K B
F J
2 (1.36)(1)
705(1)(1.65)(1.23)
2294 psi
2 0.27
W t KoKvK s
S Y / ( KT K R )
SF t N
47 000(0.9) / [(1)(0.885)]
20.8
2294
Ans.
Tooth wear
Chapter 14, Page 33/39
1/ 2
Eq. (14-16):
K C
c C p W t Ko K v K s m f
dPF I
1/ 2
1.36 1
2300 705(1)(1.65)(1.23)
8(2 ) 0.107
43 750 psi
Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where
S c Z N / ( KT K R )
SH
c
139 000(0.8) / [(1)(0.885)]
2.9 Ans
43 750
________________________________________________________________________
m = 18.75 mm/tooth, d = 300 mm
N = d/m = 300 / 18.75 = 16 teeth
F b 4 p 4 m 4 18.75 236 mm
14-37
M
0 300(11) cos 20 150 FB cos25
F B = 22.81 kN
W t FB cos 25 22.81cos 25 20.67 kN
n = 1800 / 2 = 900 rev/min
dn (0.300)(900)
14.14 m/s
V
60
60
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Fig. 14-2:
Fig. 14-5:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
x
S t = 0.703(300) + 113 = 324 MPa
S c = 2.41(300) + 237 = 960 MPa
J = Y J = 0.27
cos 20o sin 20o 5
I ZI
0.134
2(1)
5 1
Z E 191 MPa
Assume a typical quality number of 6.
Eq. (14-28): B 0.25(12 Qv ) 2/ 3 0.25(12 6)2/ 3 0.8255
A 50 56(1 B) 50 56(1 0.8255) 59.77
Eq. (14-27):
A 200V
K v
A
B
59.77 200(14.14)
59.77
0.8255
1.69
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296.
Similar to Eq. (a) of Sec. 14-10 but for SI units:
Chapter 14, Page 34/39
Ks
1
0.8433 mF Y
kb
0.0535
K s 0.8433 18.75(236) 0.296
0.0535
1.28
Convert the diameter and facewidth to inches for use in the load-distribution factor
equations. d = 300/25.4 = 11.81 in, F = 236/25.4 = 9.29 in
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
C mc = 1 (uncrowned teeth)
9.29
C pf
0.0375 0.0125(9.29) 0.1573
10(11.81)
C pm = 1.1
C ma = 0.27 (commercial enclosed gear unit)
Ce = 1
K m K H 1 1[0.1573(1.1) 0.27(1)] 1.44
For the stress-cycle factors, we need the desired number of load cycles.
N = 12 000 h (900 rev/min)(60 min/h) = 6.48 (108) rev
Fig. 14-14:
Fig. 14-15:
Eq. 14-38:
Y N = 0.9
Z N = 0.85
K R 0.658 0.0759ln 1 R 0.658 0.0759ln 1 0.98 0.955
With no specific information given to indicate otherwise, assume K o = K B = K T = Z R = 1.
Tooth bending
Eq. (14-15):
W t KoKvKs
1 KH KB
bmt YJ
(1.44)(1)
1
20 670(1)(1.69)(1.28)
53.9 MPa
236(18.75) 0.27
Eq. (14-41):
S Y / ( KT K R )
SF t N
324(0.9) / [(1)(0.955)]
5.66
53.9
Ans.
Tooth wear
1/ 2
Eq. (14-16):
K Z
c Z E W t Ko K v K s H R
d w1b Z I
Chapter 14, Page 35/39
1/ 2
1.44 1
191 20 670(1)(1.69)(1.28)
300(236) 0.134
498 MPa
Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where
SH
S c Z N / ( KT K R )
c
960(0.85) / [(1)(0.955)]
1.72 Ans
498
________________________________________________________________________
14-38 From the solution to Prob. 13-40, n = 191 rev/min, Wt = 1600 N, d = 125 mm,
N = 15 teeth, m = 8.33 mm/tooth.
F b 4 p 4 m 4 8.33 105 mm
V
dn
60
(0.125)(191)
1.25 m/s
60
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Table 14-3:
Table 14-6:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
S t = 65 kpsi = 448 MPa
S c = 225 kpsi = 1550 MPa
J = Y J = 0.25
cos 20o sin 20o 2
I ZI
0.107
2(1)
2 1
Z E 191 MPa
Assume a typical quality number of 6.
Eq. (14-28):
Eq. (14-27):
B 0.25(12 Qv ) 2 / 3 0.25(12 6) 2 / 3 0.8255
A 50 56(1 B) 50 56(1 0.8255) 59.77
A 200V
K v
A
B
59.77 200(1.25)
59.77
0.8255
1.21
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290.
Similar to Eq. (a) of Sec. 14-10 but for SI units:
Ks
1
0.8433 mF Y
kb
0.0535
0.0535
K s 0.8433 8.33(105) 0.290
1.17
Convert the diameter and facewidth to inches for use in the load-distribution factor
Chapter 14, Page 36/39
equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in
Eq. (14-31): C mc = 1 (uncrowned teeth)
4.13
Eq. (14-32): C pf
0.0375 0.0125(4.13) 0.0981
10(4.92)
Eq. (14-33): C pm = 1
C ma = 0.32 (open gearing)
Fig. 14-11:
Eq. (14-35): C e = 1
Eq. (14-30): K m K H 1 1[0.0981(1) 0.32(1)] 1.42
For the stress-cycle factors, we need the desired number of load cycles.
N = 12 000 h (191 rev/min)(60 min/h) = 1.4 (108) rev
Fig. 14-14:
Y N = 0.95
Fig. 14-15:
Z N = 0.88
K R 0.658 0.0759ln 1 R 0.658 0.0759ln 1 0.95 0.885
Eq. 14-38:
With no specific information given to indicate otherwise, assume K o = K B = K T = Z R = 1.
Tooth bending
Eq. (14-15):
W t Ko K vKs
1 KH KB
bmt YJ
(1.42)(1)
1
1600(1)(1.21)(1.17)
14.7 MPa
105(8.33) 0.25
Since gear is a pinion, C H is not used in Eq. (14-42) (see p. 761), where
S Y / ( KT K R )
SF t N
448(0.95) / [(1)(0.885)]
32.7
14.7
Ans.
Tooth wear
1/ 2
Eq. (14-16):
K Z
c Z E W t Ko K v K s H R
d w1b Z I
1/ 2
1.42 1
191 1600(1)(1.21)(1.17)
125(105) 0.107
289 MPa
S Z / ( KT K R )
SH c N
c
1550(0.88) / [(1)(0.885)]
5.33 Ans
289
________________________________________________________________________
Eq. (14-42):
Chapter 14, Page 37/39
14-39 From the solution to Prob. 13-41, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in
N = 15 teeth, P = 3 teeth/in.
F 4 p 4 4 4.2 in
P
3
dn (5)(140)
V
183.3 ft/min
12
12
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Table 14-3:
Table 14-6:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
S t = 65 kpsi
S c = 225 kpsi
J = 0.25
cos 20o sin 20o 2
I
0.107
2(1)
2 1
C p 2300 psi
Assume a typical quality number of 6.
Eq. (14-28):
B 0.25(12 Qv ) 2 / 3 0.25(12 6) 2 / 3 0.8255
A 50 56(1 B) 50 56(1 0.8255) 59.77
B
Eq. (14-27):
A V
59.77 183.3
K v
A
59.77
0.8255
1.18
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290.
From Eq. (a), Sec. 14-10,
0.0535
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
F Y
4.2 0.290
1.192
K s 1.192
3
P
C mc = 1 (uncrowned teeth)
4.2
C pf
0.0375 0.0125(4.2) 0.099
10(5)
C pm = 1
C ma = 0.32 (Open gearing)
Ce = 1
K m 1 1[0.099(1) 0.32(1)] 1.42
0.0535
1.17
For the stress-cycle factors, we need the desired number of load cycles.
N = 14 000 h (140 rev/min)(60 min/h) = 1.2 (108) rev
Fig. 14-14:
Y N = 0.95
Fig. 14-15:
Z N = 0.88
K R 0.658 0.0759ln 1 R 0.658 0.0759ln 1 0.98 0.955
Eq. 14-38:
With no specific information given to indicate otherwise, assume K o = K B = K T = C f = 1.
Chapter 14, Page 38/39
Tooth bending
Eq. (14-15):
Eq. (14-41):
Pd K m K B
F J
3 (1.42)(1)
180(1)(1.18)(1.17)
1010 psi
4.2 0.25
W t KoKvK s
S Y / ( KT K R )
SF t N
65 000(0.95) / [(1)(0.955)]
64.0
1010
Ans.
Tooth wear
1/ 2
Eq. (14-16):
K C
c C p W t Ko K v K s m f
dPF I
1/ 2
1.42 1
2300 180(1)(1.18)(1.17)
5(4.2) 0.107
28 800 psi
Since gear B is a pinion, C H is not used in Eq. (14-42) (see p. 761), where
S Z / ( KT K R )
SH c N
c
225 000(0.88) / [(1)(0.955)]
7.28 Ans
28 800
________________________________________________________________________
Chapter 14, Page 39/39
Chapter 15
15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C =
109 rev of pinion at R = 0.999, N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6
teeth/in, shaft angle = 90°, n p = 900 rev/min, J P = 0.249 and J G = 0.216 (Fig.
15-7), F = 1.25 in, S F = S H = 1, K o = 1.
Mesh
d P = 20/6 = 3.333 in, d G = 60/6 = 10.000 in
Eq. (15-7):
v t = (3.333)(900/12) = 785.3 ft/min
Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
59.77 785.3
K v
59.77
Eq. (15-8):
v t,max = [59.77 + (6 – 3)]2 = 3940 ft/min
0.8255
1.374
Since 785.3 < 3904, K v = 1.374 is valid. The size factor for bending is:
Eq. (15-10):
K s = 0.4867 + 0.2132 / 6 = 0.5222
For one gear straddle-mounted, the load-distribution factor is:
Eq. (15-11):
K m = 1.10 + 0.0036 (1.25)2 = 1.106
Eq. (15-15):
(K L ) P = 1.6831(109)–0.0323 = 0.862
(K L ) G = 1.6831(109 / 3)–0.0323 = 0.893
Eq. (15-14):
(C L ) P = 3.4822(109)–0.0602 = 1
(C L ) G = 3.4822(109 / 3)–0.0602 = 1.069
Eq. (15-19):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3)
CR K R 1.25 1.118
Bending
St sat 44(300) 2100 15 300 psi
Fig. 15-13:
0.99
Eq. (15-4):
( all ) P sw t
sat K L
S F KT K R
15 300(0.862)
10 551 psi
1(1)(1.25)
Chapter 15, Page 1/20
Eq. (15-3):
Eq. (15-4):
( all ) P FK x J P
Pd K o K v K s K m
10 551(1.25)(1)(0.249)
690 lbf
6(1)(1.374)(0.5222)(1.106)
690(785.3)
H1
16.4 hp
33 000
WPt
15 300(0.893)
10 930 psi
1(1)(1.25)
10 930(1.25)(1)(0.216)
WGt
620 lbf
6(1)(1.374)(0.5222)(1.106)
620(785.3)
H2
14.8 hp Ans.
33 000
( all )G
The gear controls the bending rating.
________________________________________________________________________
15-2
Refer to Prob. 15-1 for the gearset specifications.
Wear
Fig. 15-12:
s ac = 341(300) + 23 620 = 125 920 psi
For the pinion, C H = 1. From Prob. 15-1, C R = 1.118. Thus, from Eq. (15-2):
sac (CL ) P CH
S H KT C R
125 920(1)(1)
( c,all ) P
112 630 psi
1(1)(1.118)
( c,all ) P
For the gear, from Eq. (15-16),
B1 0.008 98(300 / 300) 0.008 29 0.000 69
CH 1 0.000 69(3 1) 1.001 38
From Prob. 15-1, (C L ) G = 1.0685. Equation (15-2) thus gives
sac (CL )G CH
S H KT C R
125 920(1.0685)(1.001 38)
( c,all )G
120 511 psi
1(1)(1.118)
( c,all )G
For steel:
C p 2290 psi
Chapter 15, Page 2/20
Eq. (15-9):
Cs 0.125(1.25) 0.4375 0.593 75
Fig. 15-6:
I = 0.083
Eq. (15-12):
C xc = 2
Eq. (15-1):
( )
Fd P I
W c,all P
C p K o K v K mCsC xc
2
t
P
1.25(3.333)(0.083)
112 630
2290
1(1.374)(1.106)(0.5937)(2)
464 lbf
464(785.3)
11.0 hp
33 000
2
1.25(3.333)(0.083)
120 511
2290
1(1.374)(1.106)(0.593 75)(2)
531 lbf
531(785.3)
12.6 hp
33 000
2
H3
WGt
H4
The pinion controls wear:
H = 11.0 hp
Ans.
The power rating of the mesh, considering the power ratings found in Prob. 15-1,
is
H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans.
________________________________________________________________________
15-3
AGMA 2003-B97 does not fully address cast iron gears. However, approximate
comparisons can be useful. This problem is similar to Prob. 15-1, but not
identical. We will organize the method. A follow-up could consist of completing
Probs. 15-1 and 15-2 with identical pinions, and cast iron gears.
Given: Uncrowned, straight teeth, P d = 6 teeth/in, N P = 30 teeth, N G = 60 teeth,
ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, n P = 900
rev/min, n = 20, one gear straddle-mounted, K o = 1, J P = 0.268, J G = 0.228, S F
= 2, S H 2.
Mesh
d P = 30/6 = 5.000 in, d G = 60/6 = 10.000 in
v t = (5)(900 / 12) = 1178 ft/min
Set N L = 107 cycles for the pinion. For R = 0.99,
Table 15-7:
s at = 4500 psi
Chapter 15, Page 3/20
Table 15-5:
Eq. (15-4):
s ac = 50 000 psi
sat K L
4500(1)
swt
2250 psi
S F KT K R
2(1)(1)
The velocity factor K v represents stress augmentation due to mislocation of tooth
profiles along the pitch surface and the resulting “falling” of teeth into
engagement. Equation (5-67) shows that the induced bending moment in a
cantilever (tooth) varies directly with E of the tooth material. If only the
material varies (cast iron vs. steel) in the same geometry, I is the same. From the
Lewis equation of Section 14-1,
M
K W tP
v
I /c
FY
We expect the ratio CI / steel to be
CI
( K v )CI
ECI
steel ( K v )steel
Esteel
In the case of ASTM class 30, from Table A-24(a)
(E CI ) av = (13 + 16.2)/2 = 14.7 kpsi
Then,
( K v )CI
14.7
( K v )steel 0.7( K v )steel
30
Our modeling is rough, but it convinces us that (K v ) CI < (K v ) steel , but we are not
sure of the value of (K v ) CI . We will use K v for steel as a basis for a conservative
rating.
Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
59.77 1178
K v
59.77
Pinion bending
0.8255
1.454
( all ) P = s wt = 2250 psi
From Prob. 15-1, K x = 1, K m = 1.106, K s = 0.5222
Eq. (15-3):
( all ) P FK x J P
Pd K o K v K s K m
2250(1.25)(1)(0.268)
149.6 lbf
6(1)(1.454)(0.5222)(1.106)
WPt
Chapter 15, Page 4/20
H1
149.6(1178)
5.34 hp
33 000
Gear bending
JG
0.228
149.6
127.3 lbf
0.268
JP
127.3(1178)
H2
4.54 hp
33 000
WGt WPt
The gear controls in bending fatigue. H = 4.54 hp Ans.
________________________________________________________________________
15-4
Continuing Prob. 15-3,
Table 15-5:
s ac = 50 000 psi
50 000
sw t c,all
35 355 psi
2
2
Eq. (15-1):
Fd P I
W c,all
C p K o K v K mCsCxc
Fig. 15-6:
I = 0.86
t
From Probs. 15-1 and 15-2: C s = 0.593 75, K s = 0.5222, K m = 1.106, C xc = 2
From Table 14-8:
C p 1960 psi
1.25(5.000)(0.086)
35 355
91.6 lbf
Wt
1960 1(1.454)(1.106)(0.59375)(2)
91.6(1178)
H3 H 4
3.27 hp
33 000
2
Thus,
Rating
Based on results of Probs. 15-3 and 15-4,
H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp
Ans.
The mesh is weakest in wear fatigue.
________________________________________________________________________
15-5
Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of
pinion at R = 0.999, N P = z 1 = 22 teeth, N G = z 2 = 24 teeth, Q v = 5, m et = 4 mm,
shaft angle 90°, n 1 = 1800 rev/min, S F = 1, S H S F 1, J P = Y J1 = 0.23,
J G = Y J2 = 0.205, F = b = 25 mm, K o = K A = K T = K = 1 and C p 190 MPa .
Chapter 15, Page 5/20
Mesh
d P = d e1 = mz 1 = 4(22) = 88 mm,
d G = m et z 2 = 4(24) = 96 mm
Eq. (15-7):
v et = 5.236(10–5)(88)(1800) = 8.29 m/s
Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148
A = 50 + 56(1 – 0.9148) = 54.77
0.9148
Eq. (15-5):
Eq. (15-10):
54.77 200(8.29)
Kv
1.663
54.77
K s = Y x = 0.4867 + 0.008 339(4) = 0.520
Eq. (15-11): with K mb = 1 (both straddle-mounted),
K m = K H = 1 + 5.6(10–6)(252) = 1.0035
From Fig. 15-8,
(CL ) P (Z NT ) P 3.4822(109 ) 0.0602 1.00
(CL )G (Z NT )G 3.4822[109 (22 / 24)]0.0602 1.0054
Eq. (15-12):
C xc = Z xc = 2
(uncrowned)
Eq. (15-19):
K R = Y Z = 0.50 – 0.25 log (1 – 0.999) = 1.25
CR Z Z YZ 1.25 1.118
From Fig. 15-10, C H = Z w = 1
Eq. (15-9):
Z x = 0.004 92(25) + 0.4375 = 0.560
Wear of Pinion
Fig. 15-12:
H lim = 2.35H B + 162.89
= 2.35(180) + 162.89 = 585.9 MPa
Fig. 15-6:
I = Z I = 0.066
Eq. (15-2):
( H ) P
( H
) ( Z NT ) P ZW
S H K Z Z
585.9(1)(1)
524.1 MPa
1(1)(1.118)
lim P
2
bd e1Z I
Eq. (15-1):
W H
C p 1000 K A K v K H Z x Z xc
The constant 1000 expresses Wt in kN.
t
P
Chapter 15, Page 6/20
25(88)(0.066)
524.1
WPt
0.591 kN
190 1000(1)(1.663)(1.0035)(0.56)(2)
t
dnW
(88)(1800)(0.591)
1
H3
4.90 kW
60 000
60 000
2
Eq. (13-36):
Wear of Gear
H lim = 585.9 MPa
( H )G
585.9(1.0054)
526.9 MPa
1(1)(1.118)
( H )G
526.9
0.591
0.594 kN
( H ) P
524.1
(88)(1800)(0.594)
H4
4.93 kW
60 000
WGt WPt
Thus in wear, the pinion controls the power rating; H = 4.90 kW
Ans.
We will rate the gear set after solving Prob. 15-6.
________________________________________________________________________
15-6
Refer to Prob. 15-5 for terms not defined below.
Bending of Pinion
( K L ) P (YNT ) P 1.6831(109 ) 0.0323 0.862
( K L )G (YNT )G 1.6831[109 (22 / 24)]0.0323 0.864
Fig. 15-13:
F lim = 0.30H B + 14.48
= 0.30(180) + 14.48 = 68.5 MPa
Eq. (15-13):
Kx = Y = 1
From Prob. 15-5:
Y Z = 1.25, v et = 8.29 m/s,
K A 1, K v 1.663, K 1,
Yx 0.52, K H 1.0035, YJ 1 0.23
Eq. (5-4):
( F ) P
Eq. (5-3):
WPt
F limYNT
68.5(0.862)
47.2 MPa
1(1)(1.25)
S F K YZ
( F ) P bmetY YJ 1
1000K A K vYx K H
47.2(25)(4)(1)(0.23)
1.25 kN
1000(1)(1.663)(0.52)(1.0035)
Chapter 15, Page 7/20
H1
881800 1.25
60 000
10.37 kW
Bending of Gear
F lim 68.5 MPa
68.5(0.864)
47.3 MPa
1(1)(1.25)
47.3(25)(4)(1)(0.205)
WGt
1.12 kN
1000(1)(1.663)(0.52)(1.0035)
88 1800 1.12
H2
9.29 kW
60 000
Rating of mesh is
H rating = min(10.37, 9.29, 4.90, 4.93) = 4.90 kW Ans.
with pinion wear controlling.
________________________________________________________________________
( F )G
15-7
(a)
(S F ) P all (S F )G all
G
P
(sat K L / KT K R ) P
(sat K L / KT K R )G
t
t
(W Pd K o K v K s K m / FK x J ) P
(W Pd K o K v K s K m / FK x J )G
All terms cancel except for s at , K L , and J,
(s at ) P (K L ) P J P = (s at ) G (K L ) G J G
From which
(sat )G
(sat ) P ( K L ) P J P
J
( sat ) P P mG
( K L )G J G
JG
where = – 0.0178 or = – 0.0323 as appropriate. This equation is the same as
Eq. (14-44). Ans.
(b) In bending
sat K L
FK x J
FK x J
W t all
S F Pd K o K v K s K m 11 S F KT K R Pd K o K v K s K m 11
(1)
In wear
1/ 2
sacCLCU
W t K o K v K mCsCxc
C
p
Fd P I
S H KT CR 22
22
Chapter 15, Page 8/20
Squaring and solving for Wt gives
s 2 C 2C 2
Fd P I
W t 2 ac 2L 2H 2
S H KT CRCP 22 K o K v K mCsCxc 22
(2)
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and
recognizing that CR K R and P d d P = N P , we obtain
(sac )22
S H2 (sat )11( K L )11 K x J11KT CsCxc
SF
CH2 N P K s I
Cp
(CL )22
For equal Wt in bending and wear
S H2
SF
SF
2
SF
1
So we get
(sac )G
Cp
(CL )G CH
(sat ) P ( K L ) P J P K x KT CsCxc
N P IK s
Ans.
(c)
(S H ) P (S H )G c,all c,all
c P c G
Substituting in the right-hand equality gives
[ sacC L / (C R KT )]P
C W K K K C C / ( Fd I )
o v m s xc
P
p
P
t
[ sacCLC H / (C R KT )]G
C W t K K K C C / ( Fd I )
o v m s xc
P
p
G
Denominators cancel, leaving
(s ac ) P (C L ) P = (s ac ) G (C L ) G C H
Solving for (s ac ) P gives,
(sac ) P (sac )G
(CL )G
CH
(CL ) P
(1)
From Eq. (15-14), C L P 3.4822 N L0.0602 and C L G 3.4822 N L / mG
Thus,
0.0602
Ans.
C H sac G mG0.0602C H
sac P sac G 1 mG
0.0602
.
This equation is the transpose of Eq. (14-45).
Chapter 15, Page 9/20
________________________________________________________________________
Core
Case
Pinion (H B ) 11 (H B ) 12
Gear
(H B ) 21 (H B ) 22
15-8
Given (H B ) 11 = 300 Brinell
Eq. (15-23):
(s at ) P = 44(300) + 2100 = 15 300 psi
J P 0.0323
0.249 0.0323
mG
15 300
17 023 psi
3
JG
0.216
17 023 2100
( H B ) 21
339 Brinell Ans.
44
2290
15 300(0.862)(0.249)(1)(0.593 25)(2)
( sac )G
1.0685(1)
20(0.086)(0.5222)
141 160 psi
141 160 23 600
( H B ) 22
345 Brinell Ans.
341
(sac ) P (sac )G mG0.0602CH 141 160(30.0602 ) 1 150 811 psi
(sat )G (sat ) P
( H B )12
150 811 23 600
373 Brinell
341
Core Case
Pinion 300 373
Gear
339 345
Ans.
Ans.
________________________________________________________________________
15-9
Pinion core
(sat ) P 44(300) 2100 15 300 psi
15 300(0.862)
( all ) P
10 551 psi
1(1)(1.25)
10 551(1.25)(0.249)
Wt
689.7 lbf
6(1)(1.374)(0.5222)(1.106)
Gear core
(sat )G 44(352) 2100 17 588 psi
17 588(0.893)
( all )G
12 565 psi
1(1)(1.25)
12 565(1.25)(0.216)
Wt
712.5 lbf
6(1)(1.374)(0.5222)(1.106)
Chapter 15, Page 10/20
Pinion case
(sac ) P 341(372) 23 620 150 472 psi
150 472(1)
( c,all ) P
134 590 psi
1(1)(1.118)
1.25(3.333)(0.086)
134 590
Wt
685.8 lbf
2290 1(1.374)(1.106)(0.593 75)(2)
2
Gear case
(sac )G 341(344) 23 620 140 924 psi
140 924(1.0685)(1)
( c,all )G
134 685 psi
1(1)(1.118)
2
1.25(3.333)(0.086)
134 685
W
686.8 lbf
2290 1(1.374)(1.106)(0.593 75)(2)
t
The rating load would be
t
Wrated
min(689.7, 712.5, 685.8, 686.8) 685.8 lbf
which is slightly less than intended.
Pinion core
(sat ) P 15 300 psi
( all ) P 10 551 psi
(as before)
(as before)
W t 689.7 lbf
(as before)
Gear core
(sat )G 44(339) 2100 17 016 psi
17 016(0.893)
( all )G
12 156 psi
1(1)(1.25)
12 156(1.25)(0.216)
Wt
689.3 lbf
6(1)(1.374)(0.5222)(1.106)
Pinion case
(sac ) P 341(373) 23 620 150 813 psi
150 813(1)
( c,all ) P
134 895 psi
1(1)(1.118)
1.25(3.333)(0.086)
134 895
Wt
689.0 lbf
2290 1(1.374)(1.106)(0.593 75)(2)
2
Gear case
(sac )G 341(345) 23 620 141 265 psi
141 265(1.0685)(1)
( c,all )G
135 010 psi
1(1)(1.118)
Chapter 15, Page 11/20
1.25(3.333)(0.086)
135 010
Wt
690.1 lbf
2290 1(1.1374)(1.106)(0.593 75)(2)
2
The equations developed within Prob. 15-7 are effective.
________________________________________________________________________
15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also
given: N P = 20 teeth, N G = 40 teeth, n = 20, F = 0.71 in, J P = 0.241, J G = 0.201,
P d = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and
Q v = 5 uncrowned.
Mesh
d P 20 / 10 2.000 in, dG 40 / 10 4.000 in
d P nP (2)(1200)
vt
628.3 ft/min
12
12
K o 1, S F 1, S H 1
Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148
A = 50 + 56(1 – 0.9148) = 54.77
0.9148
Eq. (15-10):
54.77 628.3
K v
1.412
54.77
K s = 0.4867 + 0.2132/10 = 0.508
Eq. (15-11):
K m = 1.25 + 0.0036(0.71)2 = 1.252, where K mb = 1.25
Eq. (15-15):
(K L ) P = 1.6831(109)–0.0323 = 0.862
(K L ) G = 1.6831(109/2)–0.0323 = 0.881
Eq. (15-14):
(C L ) P = 3.4822(109)–0.0602 = 1.000
(C L ) G = 3.4822(109/2)–0.0602 = 1.043
Eq. (15-5):
Analyze for 109 pinion cycles at 0.999 reliability.
Eq. (15-19):
Bending
Pinion:
Eq. (15-23):
K R = 0.50 – 0.25 log(1 – 0.999) = 1.25
CR K R 1.25 1.118
(s at ) P = 44(300) + 2100 = 15 300 psi
Chapter 15, Page 12/20
15 300(0.862)
10 551 psi
1(1)(1.25)
(sw t ) P FK x J P
Eq. (15-4):
(swt ) P
Eq. (15-3):
Wt
Gear:
(s at ) G = 15 300 psi
15 300(0.881)
( sw t ) G
10 783 psi
1(1)(1.25)
10 783(0.71)(1)(0.201)
Wt
171.4 lbf
10(1)(1.412)(0.508)(1.252)
171.4(628.3)
H2
3.3 hp
33 000
Eq. (15-4):
Eq. (15-3):
Pd K o K v K s K m
10 551(0.71)(1)(0.241)
201 lbf
10(1)(1.412)(0.508)(1.252)
201(628.3)
H1
3.8 hp
33 000
Wear
Pinion:
(C H )G 1, I 0.078, C p 2290 psi,
Cs 0.125(0.71) 0.4375 0.526 25
Eq. (15-22):
C xc 2
(s ac ) P = 341(300) + 23 620 = 125 920 psi
125 920(1)(1)
( c,all ) P
112 630 psi
1(1)(1.118)
2
Eq. (15-1):
( )
Fd P I
W c,all P
C p K o K v K mCsC xc
t
0.71(2.000)(0.078)
112 630
2290 1(1.412)(1.252)(0.526 25)(2)
144.0 lbf
2
H3
144(628.3)
2.7 hp
33 000
Gear:
(sac )G 125 920 psi
125 920(1.043)(1)
( c,all )
117 473 psi
1(1)(1.118)
117 473
Wt
2290
2
0.71(2.000)(0.078)
1(1.412)(1.252)(0.526 25)(2) 156.6 lbf
Chapter 15, Page 13/20
H4
156.6(628.3)
3.0 hp
33 000
Rating:
H = min(3.8, 3.3, 2.7, 3.0) = 2.7 hp
Pinion wear controls the power rating. While the basis of the catalog rating is
unknown, it is overly optimistic (by a factor of 1.9).
________________________________________________________________________
15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So
(H B ) 11 and (H B ) 21 are 180 Brinell and the bending stress numbers are:
(sat ) P 44(180) 2100 10 020 psi
(sat )G 10 020 psi
The contact strength of the gear case, based upon the equation derived in Prob.
15-7, is
Cp
S H2 (sat ) P ( K L ) P K x J P KT CsCxc
(sac )G
(CL )G CH S F
N P IK s
Substituting (s at ) P from above and the values of the remaining terms from
Ex. 15-1,
2290 1.52 10 020(1)(1)(0.216)(1)(0.575)(2)
1.32(1) 1.5
25(0.065)(0.529)
114 331 psi
114 331 23 620
266 Brinell
(H B ) 22
341
(sac )G
The pinion contact strength is found using the relation from Prob. 15-7:
(sac ) P (sac )G mG0.0602CH 114 331(1)0.0602 (1) 114 331 psi
114 331 23 600
( H B )12
266 Brinell
341
Core Case
Pinion 180 266
Gear
180 266
Realization of hardnesses
The response of students to this part of the question would be a function of the
extent to which heat-treatment procedures were covered in their materials and
manufacturing prerequisites, and how quantitative it was. The most important
Chapter 15, Page 14/20
thing is to have the student think about it.
The instructor can comment in class when students’ curiosity is heightened.
Options that will surface may include:
(a) Select a through-hardening steel which will meet or exceed core hardness in
the hot-rolled condition, then heat-treating to gain the additional 86 points of
Brinell hardness by bath-quenching, then tempering, then generating the teeth in
the blank.
(b) Flame or induction hardening are possibilities.
(c) The hardness goal for the case is sufficiently modest that carburizing and case
hardening may be too costly. In this case the material selection will be different.
(d)The initial step in a nitriding process brings the core hardness to 33–38
Rockwell C-scale (about 300–350 Brinell), which is too much.
________________________________________________________________________
15-12 Computer programs will vary.
________________________________________________________________________
15-13 A design program would ask the user to make the a priori decisions, as indicated
in Sec. 15-5, p. 806, of the text. The decision set can be organized as follows:
A priori decisions:
• Function: H, K o , rpm, m G , temp., N L , R
• Design factor: n d (S F = n d , S H nd )
• Tooth system: Involute, Straight Teeth, Crowning, n
• Straddling: K mb
• Tooth count: N P (N G = m G N P )
Design decisions:
• Pitch and Face: P d , F
• Quality number: Q v
• Pinion hardness: (H B ) 1 , (H B ) 3
• Gear hardness: (H B ) 2 , (H B ) 4
First, gather all of the equations one needs, then arrange them before coding. Find
the required hardnesses, express the consequences of the chosen hardnesses, and
allow for revisions as appropriate.
Chapter 15, Page 15/20
Pinion Bending
Gear Bending
Pinion Wear
Gear Wear
1/ 2
Load-induced
stress (Allowable
stress)
Tabulated
strength
Associated
hardness
W t PK o K v K m K s
st
s11
FK x J P
( sat ) P
Factor of
safety
( sat )G
s21S F KT K R
( K L )G
W t K o K vCsC xc
Fd P I
c Cp
(sac ) P
s12
s 22 = s 12
s12 S H KT CR
(CL ) P (CH ) P
( sac )G
s22 S H KT CR
(CL )G (CH )G
(sat ) P 2100
44
Bhn
(sat ) P 5980
48
(sat )G 2100
44
Bhn
(sat )G 5980
48
(sac ) P 23 620
341
Bhn
(sac ) P 29 560
363.6
(sac ) P 23 620
341
Bhn
(sac ) P 29 560
363.6
(H B ) 11
(H B ) 21
(H B ) 12
(H B ) 22
44( H B )11 2100
(sat1) P
48( H B )11 5980
44( H B ) 21 2100
(sat1)G
48( H B ) 21 5980
341( H B )12 23 620
(sac1) P
363.6( H B )12 29 560
341( H B ) 22 23 620
(sac1)G
363.6( H B )22 29 560
(s ) ( K )
n11 all at1 P L P
s11KT K R
(s ) ( K )
n21 at1 G L G
s21KT K R
(s ) (C ) (C )
n12 ac1 P L P H P
s12 KT CR
Chosen
hardness
New tabulated
strength
s11S F KT K R
(K L )P
W t PK o K v K m K s
st
s21
FK x J G
Note: S F nd , S H
2
n22
(s ) (C ) (C )
ac1 G L G H G
s22 KT CR
2
SF
Chapter 15, Page 16/20
15-14 N W = 1, N G = 56, P t = 8 teeth/in, d = 1.5 in, H o = 1hp, n = 20, t a = 70F,
K a = 1.25, n d = 1, F e = 2 in, A = 850 in2
(a)
m G = N G /N W = 56, d G = N G /P t = 56/8 = 7.0 in
p x = / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in
Eq. (15-39):
a = p x / = 0.3927 / = 0.125 in
Eq. (15-40):
b = 0.3683 p x = 0.1446 in
Eq. (15-41):
h t = 0.6866 p x = 0.2696 in
Eq. (15-42):
d o = 1.5 + 2(0.125) = 1.75 in
Eq. (15-43):
d r = 3 – 2(0.1446) = 2.711 in
Eq. (15-44):
D t = 7 + 2(0.125) = 7.25 in
Eq. (15-45):
D r = 7 – 2(0.1446) = 6.711 in
Eq. (15-46):
c = 0.1446 – 0.125 = 0.0196 in
Eq. (15-47):
( FW ) max 2 2 7 0.125 2.646 in
Eq. (13-27):
VW (1.5)(1725 /12) 677.4 ft/min
(7)(1725 / 56)
VG
56.45 ft/min
12
L px NW 0.3927 in
Eq. (13-28):
0.3927
o
tan 1
4.764
(1.5)
Pn
pn
Eq. (15-62):
(b)
Eq. (15-38):
Pt
8
8.028
cos
cos 4.764
0.3913 in
Pn
(1.5)(1725)
Vs
679.8 ft/min
12 cos 4.764
f 0.103exp 0.110(679.8) 0.450 0.012 0.0250
Eq. (15-54):
cos n f tan
cos 20 0.0250 tan 4.764
e
0.7563
cos n f cot
cos 20 0.0250 cot 4.764
Ans.
Chapter 15, Page 17/20
Eq. (15-58):
Eq. (15-57):
33 000nd H o K a
33 000(1)(1)(1.25)
966 lbf
VG e
56.45(0.7563)
cos n sin f cos
WWt WGt
cos n cos f sin
WGt
Ans.
cos 20 sin 4.764 0.025cos 4.764
966
cos 20 cos 4.764 0.025sin 4.764
106.4 lbf Ans.
(c)
Eq. (15-33):
C s = 1190 – 477 log 7.0 = 787
Eq. (15-36):
Cm 0.0107 562 56(56) 5145 0.767
Eq. (15-37):
Cv 0.659 exp[0.0011(679.8)] 0.312
Eq. (15-38):
(Wt) all = 787(7)0.8(2)(0.767)(0.312) = 1787 lbf
Since WGt (W t )all , the mesh will survive at least 25 000 h.
Eq. (15-61):
Eq. (15-63):
0.025(966)
29.5 lbf
0.025sin 4.764 cos 20 cos 4.764
29.5(679.8)
Hf
0.608 hp
33 000
106.4(677.4)
HW
2.18 hp
33 000
966(56.45)
HG
1.65 hp
33 000
Wf
The mesh is sufficient
Ans.
Pn Pt / cos 8 / cos 4.764o 8.028
pn / 8.028 0.3913 in
966
G
39 500 psi
0.3913(0.5)(0.125)
The stress is high. At the rated horsepower,
G
1
39 500 23 940 psi
1.65
acceptable
(d)
Chapter 15, Page 18/20
Eq. (15-52):
A min = 43.2(8.5)1.7 = 1642 in2 < 1700 in2
Eq. (15-49):
H loss = 33 000(1 – 0.7563)(2.18) = 17 530 ft · lbf/min
Assuming a fan exists on the worm shaft,
1725
0.13 0.568 ft · lbf/(min · in 2 · oF)
3939
17 530
Eq. (15-51): ts 70
88.2o F Ans.
0.568(1700)
________________________________________________________________________
Eq. (15-50):
CR
Chapter 15, Page 19/20
15-15 Problem statement values of 25 hp, 1125 rev/min, m G = 10, K a = 1.25, n d = 1.1,
n = 20°, t a = 70°F are not referenced in the table. The first four parameters listed
in the table were selected as design decisions.
px
dW
FG
A
HW
HG
Hf
NW
NG
KW
Cs
Cm
Cv
VG
WGt
t
W
W
f
e
(P t ) G
Pn
C-to-C
ts
L
G
dG
15-15
1.75
3.60
2.40
2000
15-16
1.75
3.60
1.68
2000
15-17
1.75
3.60
1.43
2000
15-18
1.75
3.60
1.69
2000
15-19
1.75
3.60
2.40
2000
15-20
1.75
4.10
2.25
2000
38.0
36.1
1.85
3
30
50
15-21
1.75
3.60
2.4
2500
FAN
41.2
37.7
3.59
3
30
115
15-22
1.75
3.60
2.4
2600
FAN
41.2
37.7
3.59
3
30
185
38.2
36.2
1.87
3
30
38.2
36.2
1.47
3
30
38.2
36.2
1.97
3
30
38.2
36.2
1.97
3
30
125
38.2
36.2
1.97
3
30
80
607
0.759
0.236
492
854
0.759
0.236
492
1000
0.759
0.236
492
492
492
563
492
492
2430
2430
2430
2430
2430
2120
2524
2524
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
5103
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
7290
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
8565
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
7247
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
5103
16.71
1038
0.0183
0.951
1.571
1.732
11.6
171
6.0
24.98
4158
19.099
1284
0.034
0.913
1.795
1.979
10.156
179.6
5.25
24.9
5301
16.7
1284
0.034
0.913
1.795
1.979
10.156
179.6
5.25
24.9
5301
16.71
Chapter 15, Page 20/20
Chapter 16
16-1
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 0,
2 = 120, and a = 90.
From which, sin a = sin90 = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 120
0 sin (0.150 0.125 cos ) d
1
2.993 10 4 pa N · m
Mf
Eq. (16-3):
pa (0.040)(0.150)(0.125) 120 2
sin d 9.478 104 pa N · m
0
1
MN
c = 2(0.125 cos 30) = 0.2165 m
Eq. (16-4):
F
9.478 10 4 pa 2.993 10 4 pa
0.2165
2.995 10 3 pa
p a = F/ [2.995(103)] = 2200/ [2.995(103)]
= 734.5(103) Pa for cw rotation
Eq. (16-7):
2200
9.478 10 4 pa 2.993 10 4 pa
0.2165
p a = 381.9(103) Pa for ccw rotation
A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation.
(b) RH shoe:
Eq. (16-6):
0.28(734.5)103 (0.040)0.1502 (cos 0o cos120o )
TR
277.6 N · m
1
LH shoe:
381.9
TL 277.6
144.4 N · m Ans.
734.5
T total = 277.6 + 144.4 = 422 N · m Ans.
Ans.
Ans.
Chapter 16, Page 1/27
(c)
RH shoe:
F x = 2200 sin 30° = 1100 N, F y = 2200 cos 30° = 1905 N
Eqs. (16-8):
1
A sin 2
2
0o
Eqs. (16-9):
Rx
120o
Ry
0.375,
734.5 103 0.040(0.150)
1
3
734.5 10 0.04(0.150)
R [ 1007
LH shoe:
2 / 3 rad
1
B sin 2
2
4
0
2
1.264
[0.375 0.28(1.264)] 1100 1007 N
[1.264 0.28(0.375)] 1905 4128 N
1
41282 ]1/ 2 4249 N Ans.
F x = 1100 N, F y = 1905 N
Eqs. (16-10): Rx
381.9 103 0.040(0.150)
[0.375 0.28(1.264)] 1100 570 N
1
381.9 103 0.040(0.150)
[1.264 0.28(0.375)] 1905 751 N
Ry
1
1/ 2
R 597 2 7512 959 N Ans.
______________________________________________________________________________
16-2
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 15,
2 = 105, and a = 90.
From which, sin a = sin90 = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 105
4
Mf
15 sin (0.150 0.125 cos ) d 2.177 10 pa
1
Chapter 16, Page 2/27
MN
Eq. (16-3):
pa (0.040)(0.150)(0.125) 105 2
sin d 7.765 10 4 pa
15
1
c = 2(0.125) cos 30° = 0.2165 m
F
Eq. (16-4):
7.765 10 4 pa 2.177 10 4 pa
0.2165
2.58110 3 pa
p a = 2200/ [2.581(10 3)] = 852.4 (103) Pa
= 852.4 kPa on RH shoe for cw rotation Ans.
RH shoe:
TR
Eq. (16-6):
LH shoe:
2200
0.28(852.4)103 (0.040)(0.1502 )(cos15 cos105)
263 N · m
1
7.765 10 4 pa 2.177 10 4 pa
pa 479.110
3
0.2165
Pa 479.1 kPa on LH shoe for ccw rotation
Ans.
0.28(479.1)103(0.040)(0.150 2 )(cos15 cos105)
148 N · m
1
263 148 411 N · m Ans.
TL
Ttotal
Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by
using 25% less braking material.
______________________________________________________________________________
16-3
Given: 1 = 0°, 2 = 120°, a = 90°, sin a = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30,
F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation.
LH shoe:
Eq. (16-2), with 1 = 0:
f pabr 2
f pabr
a
sin r a cos d
Mf
r (1 cos 2 ) sin 2 2
sin a 1
sin a
2
0.30 pa (1.25)5.5
3.5 2
5.5(1 cos120o )
sin 120
1
2
14.31 pa lbf · in
Eq. (16-3), with 1 = 0:
pabra 2 2
p bra 2 1
MN
sin d a
sin 2 2
sin a 1
sin a 2
4
pa (1.25)5.5(3.5) 120 1
sin 2(120)
1
2 180 4
30.41 pa lbf · in
Chapter 16, Page 3/27
180o 2
o
c 2r cos
2(5.5) cos 30 9.526 in
2
30.41 pa 14.31 pa
1.690 pa
F 225
9.526
pa 225 / 1.690 133.1 psi
Eq. (16-6):
f pabr 2 (cos 1 cos 2 ) 0.30(133.1)1.25(5.52 )
[1 (0.5)]
sin a
1
2265 lbf · in 2.265 kip · in Ans.
TL
RH shoe:
30.41 pa 14.31 pa
4.694 pa
9.526
pa 225 / 4.694 47.93 psi
47.93
TR
2265 816 lbf ·in 0.816 kip·in
133.1
F 225
T total = 2.27 + 0.82 = 3.09 kip in
Ans.
______________________________________________________________________________
16-4
(a) Given: 1 = 10°, 2 = 75°, a = 75°, p a = 106 Pa, f = 0.24, b = 0.075 m (shoe width),
a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m.
Some of the terms needed are evaluated here:
2
2
2
2
1 2
A r sin d a sin cos d r cos a sin
1
1
1
2
1
75
1
200 cos 10 150 sin 2 77.5 mm
2
10
75 /180 rad
2
1
2
B sin d sin 2
0.528
1
2 4
10 /180 rad
75
C
2
sin cos d 0.4514
1
Now converting to Pascals and meters, we have from Eq. (16-2),
0.24 106 (0.075)(0.200)
f pabr
Mf
A
(0.0775) 289 N · m
sin a
sin 75
Chapter 16, Page 4/27
From Eq. (16-3),
MN
pabra
106 (0.075)(0.200)(0.150)
B
(0.528) 1230 N · m
sin a
sin 75
Finally, using Eq. (16-4), we have
F
MN M f
c
1230 289
5.70 kN
165
Ans.
(b) Use Eq. (16-6) for the primary shoe.
T
fpabr 2 (cos 1 cos 2 )
sin a
0.24 106 (0.075)(0.200)2 (cos 10 cos 75)
sin 75
541 N · m
For the secondary shoe, we must first find p a . Substituting
1230
289
pa and M f 6 pa into Eq. (16 - 7),
6
10
10
(1230 / 106 ) pa (289 / 106 ) pa
5.70
, solving gives pa 619 103 Pa
165
MN
Then
T
0.24 619 103 0.075 0.2002 cos 10 cos 75
sin 75
335 N · m
so the braking capacity is T total = 2(541) + 2(335) = 1750 N · m
Ans.
(c) Primary shoes:
pabr
C f B Fx
sin a
106 (0.075)0.200
[0.4514 0.24(0.528)](10 3 ) 5.70 0.658 kN
sin 75
pabr
( B f C ) Fy
Ry
sin a
106 (0.075)0.200
[0.528 0.24(0.4514)] 10 3 0 9.88 kN
sin 75
Rx
Chapter 16, Page 5/27
Secondary shoes:
Rx
pabr
(C f B) Fx
sin a
0.619 106 0.075(0.200)
sin 75
0.143 kN
p br
Ry a ( B f C ) Fy
sin a
0.619 106 0.075(0.200)
4.03 kN
sin 75
[0.4514 0.24(0.528)] 10 3 5.70
[0.528 0.24(0.4514)] 10 3 0
Note from figure that +y for secondary shoe is opposite to
+y for primary shoe.
Combining horizontal and vertical components,
RH 0.658 0.143 0.801 kN
RV 9.88 4.03 5.85 kN
R (0.801) 2 5.852
5.90 kN Ans.
______________________________________________________________________________
16-5
Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25.
Preliminaries: 1 = 45° tan1(6/8) = 8.13°, 2 = 98.13°, a = 90°,
a = (62 + 82)1/2 = 10 in
Eq. (16-2):
f pabr 2
0.25 pa (1.25)6
Mf
sin r a cos d
sin a 1
1
98.13
sin 6 10 cos d
8.13
3.728 pa lbf · in
Eq. (16-3):
MN
pabra 2 2
p (1.25)6(10)
sin d a
sin a 1
1
98.13
sin 2 d
8.13
69.405 pa lbf · in
Eq. (16-4): Using Fc = M N M f , we obtain
90(20) (69.405 3.728) pa
pa 27.4 psi
Ans.
Chapter 16, Page 6/27
Eq. (16-6):
fp br 2 cos 1 cos 2 0.25(27.4)1.25 6 cos8.13 cos 98.13
T a
sin a
1
348.7 lbf · in Ans.
______________________________________________________________________________
2
16-6
For 3ˆ f :
f f 3ˆ f 0.25 3(0.025) 0.325
From Prob. 16-5, with f = 0.25, M f = 3.728 p a . Thus, M f = (0.325/0.25) 3.728 p a =
4.846 p a . From Prob. 16-5, M N = 69.405 p a .
Eq. (16-4): Using Fc = M N M f , we obtain
90(20) (69.405 4.846) pa
pa 27.88 psi
Ans.
From Prob. 16-5, p a = 27.4 psi and T = 348.7 lbf⋅in. Thus,
0.325 27.88
T
348.7 461.3 lbf ·in
0.25 27.4
Ans.
Similarly, for 3ˆ f :
f f 3ˆ f 0.25 3(0.025) 0.175
M f (0.175 / 0.25) 3.728 pa 2.610 pa
90(20) = (69.405 2.610) p a
p a = 26.95 psi
0.175 26.95
T
348.7 240.1 lbf · in Ans.
0.25 27.4
______________________________________________________________________________
16-7
Preliminaries: 2 = 180° 30° tan1(3/12) = 136°, 1 = 20° tan1(3/12) = 6°,
a = 90, sin a = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, p a = 150 psi.
Eq. (16-2):
Mf
0.30(150)(2)(10) 136o
6 sin (10 12.37 cos ) d 12 800 lbf · in
sin 90
Eq. (16-3):
MN
150(2)(10)(12.37) 136 2
6 sin d 53 300 lbf · in
sin 90
LH shoe:
c L = 12 + 12 + 4 = 28 in
Chapter 16, Page 7/27
Now note that M f is cw and M N is ccw. Thus,
Eq. (16-6):
FL
53 300 12 800
1446 lbf
28
TL
0.30(150)(2)(10) 2 (cos 6 cos136)
15 420 lbf · in
sin 90
RH shoe:
M N 53 300
pa
355.3 pa ,
150
M f 12 800
pa
85.3 pa
150
On this shoe, both M N and M f are ccw. Also,
c R = (24 2 tan 14°) cos 14° = 22.8 in
Fact FL sin14 361 lbf Ans.
FR FL / cos14 1491 lbf
Thus,
1491
Then,
TR
355.3 85.3
pa pa 77.2 psi
22.8
0.30(77.2)(2)(10) 2 (cos 6 cos136)
7940 lbf · in
sin 90
T total = 15 420 + 7940 = 23 400 lbf · in Ans.
______________________________________________________________________________
16-8
2
M f 2 ( fdN )(a cos r )
0
where dN pbr d
2
2 fpbr (a cos r ) d 0
0
From which
2
2
0
0
a cos d r d
r 2
r (60)( / 180)
a
1.209r
sin 2
sin 60
Ans.
Chapter 16, Page 8/27
Eq. (16-15):
a
4r sin 60
1.170r
2(60)( / 180) sin[2(60)]
Ans.
a differs with a ¢ by 100(1.170 1.209)/1.209 = 3.23 %
Ans.
______________________________________________________________________________
16-9
(a) Counter-clockwise rotation, 2 = / 4 rad, r = 13.5/2 = 6.75 in
Eq. (16-15):
4r sin 2
4(6.75) sin( / 4)
7.426 in
a
2 2 sin 2 2
2 / 4 sin(2 / 4)
e 2a 2(7.426) 14.85 in
Ans.
(b)
= tan1(3/14.85) = 11.4°
M
F
x
R
0 3F x 6.375P
0 F x R x
F x 2.125P
R x F x 2.125P
F y F x tan11.4o 0.428P
Fy P F y R y
R y P 0.428P 1.428P
Left shoe lever.
M R 0 7.78S x 15.28F x
15.28
Sx
(2.125P) 4.174P
7.78
S y f S x 0.30(4.174 P) 1.252P
Fy 0 R y S y F y
R y F y S y 0.428P 1.252 P 1.68P
Fx 0 R x S x F x
R x S x F x 4.174P 2.125P 2.049P
Chapter 16, Page 9/27
(c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on
the brake shoe lever moment and hence, no effect on Sx or the brake torque.
The brake shoe levers carry identical bending moments but the left lever carries a
tension while the right carries compression (column loading). The right lever is
designed and used as a left lever, producing interchangeable levers (identical levers).
But do not infer from these identical loadings.
______________________________________________________________________________
16-10 r = 13.5/2 = 6.75 in,
b = 6 in,
2 = 45° = / 4 rad.
From Table 16-3 for a rigid, molded non-asbestos lining use a conservative estimate of
p a = 100 psi, f = 0.33.
Equation (16-16) gives the horizontal brake hinge pin reaction which corresponds to Sx in
Prob. 16-9. Thus,
p br
100(6)6.75
N S x a 2 2 sin 2 2
2 / 4 sin 2 45
2
2
5206 lbf
which, from Prob. 6-9 is 4.174 P. Therefore,
4.174 P = 5206
P = 1250 lbf = 1.25 kip
Ans.
Applying Eq. (16-18) for two shoes, where from Prob. 16-9, a = 7.426 in
T 2a f N 2(7.426)0.33(5206)
25 520 lbf · in 25.52 kip · in Ans.
______________________________________________________________________________
16-11 Given: D = 350 mm, b = 100 mm, p a = 620 kPa, f = 0.30, = 270.
Chapter 16, Page 10/27
Eq. (16-22):
pabD
620(0.100)0.350
10.85 kN
2
2
P1
Ans.
f 0.30(270)( / 180) 1.414
Eq. (16-19):
P 2 = P 1 exp( f ) = 10.85 exp( 1.414) = 2.64 kN
Ans.
T ( P1 P2 )( D / 2) (10.85 2.64)(0.350 / 2) 1.437 kN · m Ans.
______________________________________________________________________________
16-12 Given: D = 12 in,
Eq. (16-22):
f = 0.28, b = 3.25 in, = 270°, P 1 = 1800 lbf.
pa
2 P1
2(1800)
92.3 psi
bD 3.25(12)
Ans.
f 0.28(270o )( / 180o ) 1.319
P2 P1 exp( f ) 1800 exp(1.319) 481 lbf
T ( P1 P2 )( D / 2) (1800 481)(12 / 2)
7910 lbf · in 7.91 kip · in Ans.
______________________________________________________________________________
16-13
M O = 0 = 100 P 2 325 F P 2 = 325(300)/100 = 975 N Ans.
100
cos 1
51.32
160
270 51.32 218.7
f 0.30(218.7) / 180 1.145
P1 P2 exp( f ) 975exp(1.145) 3064 N
Ans.
T P1 P2 ( D / 2) (3064 975)(200 / 2)
209 103 N · mm 209 N · m
Ans.
______________________________________________________________________________
Chapter 16, Page 11/27
16-14 (a) D = 16 in, b = 3 in
n = 200 rev/min
f = 0.20, p a = 70 psi
Eq. (16-22):
P1
Eq. (16-14):
pabD
70(3)(16)
1680 lbf
2
2
f 0.20(3 / 2) 0.942
P2 P1 exp( f ) 1680 exp(0.942) 655 lbf
D
16
(1680 655)
2
2
8200 lbf · in Ans.
Tn
8200(200)
H
26.0 hp
63 025
63 025
3P
3(1680)
504 lbf Ans.
P 1
10
10
T ( P1 P2 )
Ans.
(b) Force of belt on the drum:
R = (16802 + 6552)1/2 = 1803 lbf
Force of shaft on the drum: 1680 and 655 lbf
TP1 1680(8) 13 440 lbf · in
TP2 655(8) 5240 lbf · in
Net torque on drum due to brake band:
T TP1 TP2
13 440 5240
8200 lbf · in
The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with
the drum at center span, the bearing radial load is 1803/2 = 901 lbf.
Chapter 16, Page 12/27
(c) Eq. (16-21):
2P
bD
2 P1
2(1680)
70 psi
3(16)
3(16)
p
p
0
Ans.
2P2
2(655)
27.3 psi Ans.
3(16)
3(16)
______________________________________________________________________________
p
270
16-15 Given: = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c 2 = 2.25 in (see
figure). Notice that the pivoting rocker is not located on the vertical centerline of the
drum.
(a) To have the band tighten for ccw rotation, it is necessary to have c 1 < c 2 . When
friction is fully developed,
P1 / P2 exp( f ) exp[0.2(3 / 2)] 2.566
If friction is not fully developed,
P 1 /P 2 ≤ exp( f )
To help visualize what is going on let’s add a force W parallel to P 1 , at a lever arm of
c 3 . Now sum moments about the rocker pivot.
M
0 c3W c1P1 c2 P2
From which
c2 P2 c1P1
c3
The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0.
It follows from the equation above
W
P1
c
2
P2
c1
When friction is fully developed
2.566 2.25 / c1
2.25
c1
0.877 in
2.566
When P 1 /P 2 is less than 2.566, friction is not fully developed. Suppose P 1 /P 2 = 2.25,
Chapter 16, Page 13/27
then
c1
2.25
1 in
2.25
We don’t want to be at the point of slip, and we need the band to tighten.
c2
c1 c2
P1 / P2
When the developed friction is very small, P 1 /P 2 → 1 and c 1 → c 2
Ans.
(b) Rocker has c 1 = 1 in
P1
c
2.25
2
2.25
P2
c1
1
ln( P1 / P2 ) ln 2.25
f
0.172
3 / 2
Friction is not fully developed, no slip.
T ( P1 P2 )
P
D
D
P2 1 1
2
P2
2
Solve for P 2
2T
2(150)(12)
349 lbf
[( P1 / P2 ) 1]D
(2.25 1)(8.25)
P1 2.25P2 2.25(349) 785 lbf
2P
2(785)
p 1
89.6 psi Ans.
2.125(8.25)
bD
P2
(c) The torque ratio is 150(12)/100 or 18-fold.
349
P2
19.4 lbf
18
P1 2.25P2 2.25(19.4) 43.6 lbf
89.6
p
4.98 psi Ans.
18
Comment:
As the torque opposed by the locked brake increases, P 2 and P 1 increase (although
ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be
provided by a shear key.
______________________________________________________________________________
16-16 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN.
Chapter 16, Page 14/27
(a) From Eq. (16-23),
2 4000
2F
pa
0.194 N/mm 2 194 kPa
Ans.
d ( D d ) (175)(250 175)
Eq. (16-25):
Ff
4000(0.30)
T
(D d )
(250 175)10 3 127.5 N · m Ans.
4
4
(b) From Eq. (16-26),
4F
4(4000)
0.159 N/mm 2 159 kPa
2
2
2
2
( D d ) (250 175 )
pa
Eq. (16-27):
T
f pa ( D3 d 3 )
(0.30)159 103 2503 1753 10 3
Ans.
3
12
12
128 N · m Ans.
______________________________________________________________________________
16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, p a = 120 psi.
(a) Eq. (16-23):
pa d
(120)(4)
F
(D d )
(6.5 4) 1885 lbf Ans.
2
2
Eq. (16-24) with N sliding planes:
fpa d 2
(0.24)(120)(4)
T
(D d 2 ) N
(6.52 42 )(6)
8
8
7125 lbf · in Ans.
(b)
T
(0.24)(120d )
8
(6.52 d 2 )(6)
d, in T, lbf · in
2
5191
3
6769
4
7125
Ans.
5
5853
6
2545
(c) The torque-diameter curve exhibits a stationary point maximum in the range of
diameter d. The clutch has nearly optimal proportions.
______________________________________________________________________________
16-18 (a) Eq. (16-24) with N sliding planes:
Chapter 16, Page 15/27
f pa d ( D 2 d 2 ) N
T
8
f pa N
8
D d d
2
3
Differentiating with respect to d and equating to zero gives
dT
f pa N 2
D 3d 2 0
8
dd
D
d*
Ans.
3
f pa N
3 f pa N
d 2T
6
d
d
2
8
4
dd
which is negative for all positive d. We have a stationary point maximum.
(b)
d*
6.5
3.75 in
3
Eq. (16-24):
T*
Ans.
(0.24)(120) 6.5 / 3 2
2
6.5 6.5 / 3 (6) 7173 lbf · in
8
(c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in
d
(d) Consider:
0.80
0.45
D
Multiply through by D,
0.45D d 0.80D
0.45(6.5) d 0.80(6.5)
2.925 d 5.2 in
*
1
d
0.577
d * /D
3
D
which lies within the common range of clutches.
Yes. Ans.
______________________________________________________________________________
16-19 Given: d = 11 in,
l = 2.25 in, T = 1800 lbf · in,
D = 12 in,
f = 0.28.
0.5
tan 1
12.53
2.25
Chapter 16, Page 16/27
Uniform wear
Eq. (16-45):
f pa d 2
D d2
8sin
(0.28) pa (11) 2
1800
12 112 128.2 pa
8sin12.53
1800
14.04 psi Ans.
pa
128.2
T
Eq. (16-44):
pa d
F
2
(D d )
(14.04)11
2
(12 11) 243 lbf
Ans.
Uniform pressure
Eq. (16-48):
f pa
D3 d 3
12 sin
(0.28) pa
1800
123 113 134.1 pa
12sin12.53
1800
pa
Ans.
13.42 psi
134.1
T
Eq. (16-47):
pa
(13.42)
122 112 242 lbf Ans.
4
4
______________________________________________________________________________
F
(D 2 d 2 )
16-20 Uniform wear
Eq. (16-34):
Eq. (16-33):
Thus,
1
( 2 1) f pa ri ro2 ri 2
2
F = ( 2 1 ) p a r i (r o r i )
T
(1 / 2)( 2 1) f pa ri ro2 ri 2
T
f ( 2 1) pa ri (ro ri )( D)
f FD
r ri
D / 2 d / 2 1
d
o
1 O.K .
D
2D
2D
4
Ans.
Uniform pressure
Eq. (16-38):
T
1
( 2 1) f pa ro3 ri3
3
Chapter 16, Page 17/27
F
Eq. (16-37):
1
( 2 1) pa ro2 ri 2
2
Thus,
(1 / 3)( 2 1) f pa ro3 ri3
T
2 ( D / 2)3 (d / 2)3
f FD (1 / 2) f ( 2 1) pa ro2 ri 2 D
3 ( D / 2) 2 (d / 2)2 D
2( D / 2)3 1 (d / D)3
1 1 (d / D) 3
O.K . Ans.
3( D / 2)2 1 (d / D) 2 D 3 1 (d / D) 2
______________________________________________________________________________
16-21
2 n / 60 2 500 / 60 52.4 rad/s
T
H
2(103 )
38.2 N· m
52.4
Key:
T
38.2
3.18 kN
r
12
Average shear stress in key is
3.18(103 )
13.2 MPa Ans.
6(40)
Average bearing stress is
F
3.18(103 )
b
26.5 MPa
Ab
3(40)
Let one jaw carry the entire load.
F
Ans.
1 26 45
17.75 mm
2 2
2
38.2
T
F
2.15 kN
rav 17.75
rav
The bearing and shear stress estimates are
b
2.15 103
22.6 MPa Ans.
10(22.5 13)
2.15(103 )
0.869 MPa Ans.
10 0.25 (17.75) 2
______________________________________________________________________________
Chapter 16, Page 18/27
16-22
1 2 n / 60 2 (1600) / 60 167.6 rad/s
2 0
From Eq. (16-51),
I1I 2
Tt1
2800(8)
133.7 lbf · in · s 2
I1 I 2 1 2 167.6 0
Eq. (16-52):
E
I1 I 2
133.7
2
(167.6 0) 2 1.877 106 lbf in
1 2
2 I1 I 2
2
H = E / 9336 = 1.877(106) / 9336 = 201 Btu
In Btu, Eq. (16-53):
Eq. (16-54):
H
201
41.9F Ans.
C pW
0.12(40)
______________________________________________________________________________
T
16-23
n1 n2
260 240
250 rev/min
2
2
C s = ( 2 1 ) / = (n 2 n 1 ) / n = (260 240) / 250 = 0.08
n
Eq. (16-62):
Ans.
= 2 (250) / 60 = 26.18 rad/s
From Eq. (16-64):
I
6.75 103
E2 E1
123.1 N · m · s 2
2
2
Cs
0.08(26.18)
I
m 2
do di2
8
m
8I
8(123.1)
233.9 kg
2
d di
1.52 1.42
Table A-5, cast iron unit weight = 70.6 kN/m3
Volume:
2
o
= 70.6(103) / 9.81 = 7197 kg / m3.
V = m / = 233.9 / 7197 = 0.0325 m3
V t d o2 d i2 / 4 t 1.52 1.4 2 / 4 0.2278t
Equating the expressions for volume and solving for t,
0.0325
t
0.143 m 143 mm
0.2278
Ans.
Chapter 16, Page 19/27
______________________________________________________________________________
16-24 (a) The useful work performed in one revolution of the crank shaft is
U = 320 (103) 200 (103) 0.15 = 9.6 (103) J
Accounting for friction, the total work done in one revolution is
U = 9.6(103) / (1 0.20) = 12.0(103) J
Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the
energy fluctuation is
E 2 E 1 = 9.6(103) 12.0(103)(0.075) = 8.70(103) J
(b) For the flywheel,
Since
Eq. (16-64):
Ans.
n 6(90) 540 rev/min
2 n
2 (540)
56.5 rad/s
60
60
C s = 0.10
E E
8.70(103 )
I 2 21
27.25 N · m · s 2
2
Cs
0.10(56.5)
Assuming all the mass is concentrated at the effective diameter, d,
md 2
I mr
4
4I
4(27.25)
75.7 kg Ans.
m 2
d
1.22
______________________________________________________________________________
2
16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine.
Cs 0.30
n 2400 rev/min or 251 rad/s
3(3368)
Tm
804 lbf · in
Ans.
4
E2 E1 3(3531) 10 590 in · lbf
E E
10 590
0.560 in · lbf · s 2 Ans.
I 2 21
2
0.30(251 )
Cs
______________________________________________________________________________
Chapter 16, Page 20/27
16-26 (a)
(1)
(T2 )1 F21rP
T2
T
rP 2
rG
n
Ans.
Equivalent energy
(2)
(1 / 2) I 222 (1 / 2)( I 2 )1 12
I
22
( I 2 )1 2 I 2 22
1
n
2
(3)
2
Ans.
2
I G rG mG rG rG
n4
I P rP mP rP rP
From (2)
( I 2 )1
(b) I e I M I P n 2 I P
IL
n2
IG
n4 I P
n2I P
n2
n2
Ans.
Ans.
______________________________________________________________________________
16-27 (a) Reflect I L , I G2 to the center shaft
Chapter 16, Page 21/27
Reflect the center shaft to the motor shaft
I e I M I P n2I P
I P m2
I
2 I P 2L 2
2
n
n
mn
(b) For R = constant = nm, I e I M
(c) For R = 10,
Ans.
IP
R2I P
I
I P n I P 2 4 L2
n
n
R
2
Ans.
I e
2(1) 4(102 )(1)
0 0 2n(1) 3
00
n
n
n5
n6 n2 200 = 0
From which
n* 2.430 Ans.
10
m*
4.115
2.430
Ans.
Notice that n*and m* are independent of I L .
______________________________________________________________________________
16-28 From Prob. 16-27,
IP
R2I P
I
L2
2
4
n
n
R
1
100(1)
100
10 1 n 2 (1) 2
2
n
n4
10
100
1
12 n 2 2 4
n
n
Ie I M I P n2I P
Chapter 16, Page 22/27
Optimizing the partitioning of a double reduction lowered the gear-train inertia to
20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two additional
gears.
______________________________________________________________________________
16-29 Figure 16-29 applies,
t2 10 s, t1 0.5 s
t2 - t1 10 0.5
19
t1
0.5
The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is
TL
1300(12)
1560 lbf · in
10
The rated motor torque T r is
Tr
63 025(3)
168.07 lbf · in
1125
For Eqs. (16-65):
2
(1125) 117.81 rad/s
60
2
(1200) 125.66 rad/s
s
60
168.07
Tr
a
21.41 lbf in s/rad
125.66 117.81
s r
168.07(125.66)
Trs
b
2690.4 lbf · in
s r 125.66 117.81
r
Chapter 16, Page 23/27
The linear portion of the squirrel-cage motor characteristic can now be expressed as
Eq. (16-68):
T M = 21.41 + 2690.4 lbf · in
19
1560 168.07
T2 168.07
1560 T2
One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a
successive substitution method
T2
0.00
19.30
24.40
26.00
26.50
New T 2
19.30
24.40
26.00
26.50
26.67
Continue until convergence to
T 2 = 26.771 lbf ⋅ in
Eq. (16-69):
a t2 t1
21.41(10 0.5)
110.72 lbf · in · s 2
ln T2 / Tr ln(26.771 / 168.07)
T b
a
T b 26.771 2690.4
max 2
124.41 rad/s Ans.
a
21.41
min 117.81 rad/s Ans.
124.41 117.81
121.11 rad/s
2
124.41 117.81
max min
Cs
0.0545 Ans.
(max min ) / 2 (124.41 117.81) / 2
1
1
E1 I r2 (110.72)(117.81) 2 768 352 in · lbf
2
2
1
1
E2 I 22 (110.72)(124.41) 2 856 854 in · lbf
2
2
E E2 E1 856 854 768 352 88 502 in · lbf
I
Eq. (16-64):
E Cs I 2 0.0545(110.72)(121.11) 2
88 508 in · lbf, close enough Ans.
Chapter 16, Page 24/27
During the punch
63 025H
n
TL (60 / 2 ) 1560(121.11)(60 / 2 )
H
28.6 hp
63 025
63 025
T
The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on
the motor shaft. From Table A-18,
m 2
W 2
d o di2
d o di2
8
8g
8gI
8(386)(110.72)
W 2
2
d o di
d o2 di2
I
If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in
di 30 (4 / 2) 28 in
d o 30 (4 / 2) 32 in
8(386)(110.72)
189.1 lbf
W
322 282
Rim volume V is given by
V
l
d
4
2
o
di2
l
4
(322 282 ) 188.5l
where l is the rim width as shown in Table A-18. The specific weight of cast iron is
= 0.260 lbf / in3, therefore the volume of cast iron is
V
W
189.1
727.3 in 3
0.260
Equating the volumes,
188.5 l 727.3
727.3
l
3.86 in wide
188.5
Proportions can be varied.
______________________________________________________________________________
16-30 Prob. 16-29 solution has I for the motor shaft flywheel as
Chapter 16, Page 25/27
I = 110.72 lbf · in · s2
A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2)
I = 102(110.72) = 11 072 lbf · in · s2
A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under
shock conditions.
Stating the problem is most of the solution. Satisfy yourself that on the crankshaft:
TL 1300(12) 15 600 lbf · in
Tr 10(168.07) 1680.7 lbf · in
r 117.81 / 10 11.781 rad/s
s 125.66 / 10 12.566 rad/s
a 21.41(100) 2141 lbf · in · s/rad
b 2690.35(10) 26903.5 lbf · in
TM 2141c 26 903.5 lbf · in
19
15 600 1680.5
T2 1680.6
15 600 T2
The root is 10(26.67) = 266.7 lbf · in
121.11 / 10 12.111 rad/s
Cs 0.0549 (same)
max 121.11 / 10 12.111 rad/s Ans.
min 117.81 / 10 11.781 rad/s Ans.
E 1 , E 2 , E and peak power are the same. From Table A-18
6
8gI
8(386)(11 072) 34.19 10
W 2
d o di2
d o2 di2
d o2 di2
Scaling will affect d o and d i , but the gear ratio changed I. Scale up the flywheel in the
Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in.
d 30(2.5) 75 in
do 75 (10 / 2) 80 in
di 75 (10 / 2) 70 in
Chapter 16, Page 26/27
W
34.19 106
3026 lbf
802 702
W
3026
V
11 638 in 3
0.260
V
l (802 702 ) 1178 l
4
11 638
l
9.88 in
1178
Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while
the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the
motor armature has its inertia magnified 100-fold, and during the punch there are
deceleration stresses in the train. With no motor armature information, we cannot
comment.
______________________________________________________________________________
16-31 This can be the basis for a class discussion.
Chapter 16, Page 27/27
Chapter 17
17-1
Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H nom = 2 hp, C = 9(12) =
108 in, velocity ratio = 0.5, K s = 1.25, n d = 1
V = d n / 12 = (2)(1750) / 12 = 916.3 ft/min
Eq. (17-1):
D = d / vel ratio = 2 / 0.5 = 4 in
42
Dd
d 2sin 1
2sin 1
3.123 rad
2C
2(108)
Table 17-2:
t = 0.05 in, d min = 1.0 in, F a = 35 lbf/in, = 0.035 lbf/in3, f = 0.5
w = 12 bt = 12(0.035)6(0.05) = 0.126 lbf/ft
2
2
0.126 916.3
w V
Fc
0.913 lbf
g 60
32.17 60
(a) Eq. (e), p. 885:
Ans.
63 025H nom K s nd
63 025(2)(1.25)(1)
90.0 lbf · in
n
1750
2T
2(90.0)
F F1 a F2
90.0 lbf
d
2
T
Table 17-4:
C p = 0.70
Eq. (17-12):
(F 1 ) a = bF a C p C v = 6(35)(0.70)(1) = 147 lbf
F 2 = (F 1 ) a [(F 1 ) a F 2 ] = 147 90 = 57 lbf
Ans.
Ans.
Do not use Eq. (17-9) because we do not yet know f
F1 a
F2
147 57
0.913 101.1 lbf
2
2
Using Eq. (17-7) solved for f ¢ (see step 8, p.888),
Fi
Eq. (i), p. 886:
f
Fc
Ans.
( F ) Fc
1
147 0.913
0.307
ln 1 a
ln
d F2 Fc 3.123 57 0.913
1
The friction is thus underdeveloped.
(b) The transmitted horsepower is, with F = (F 1 ) a F 2 = 90 lbf,
Chapter 17, Page 1/39
(F )V
90(916.3)
2.5 hp
33 000
33 000
H
2.5
1
H nom K s
2(1.25)
H
Eq. (j), p. 887:
nf s
D 2sin 1
Eq. (17-1):
Ans.
42
Dd
2sin 1
3.160 rad
2C
2(108)
L = [4C2 (D d)2]1/2 + (D D + d d )/2
Eq. (17-2):
= [4(108)2 (4 2)2]1/2 + [4(3.160) + 2(3.123)]/2 = 225.4 in
(c) Eq. (17-13):
3C 2 w 3(108 / 12)2 (0.126)
dip
0.151 in
2Fi
2(101.1)
Ans.
Ans.
Comment: The solution of the problem is finished; however, a note concerning the design
is presented here.
The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will
increase f . The limit of narrowing is b min = 4.680 in, whence
w 0.0983 lbf/ft
Fc 0.713 lbf
T 90 lbf · in (same)
F ( F1) a F2 90 lbf
Fi 68.9 lbf
( F1) a 114.7 lbf
F2 24.7 lbf
f f 0.50
dip 0.173 in
Longer life can be obtained with a 6-inch wide belt by reducing F i to attain f 0.50.
Prob. 17-8 develops an equation we can use here
(F Fc ) exp( f ) Fc
exp( f ) 1
F2 F1 F
F F2
Fi 1
Fc
2
1 F1 Fc
f
ln
d F2 Fc
F1
dip
3C 2 w
2Fi
which in this case, d = 3.123 rad, exp(f ) = exp[0.5(3.123)] = 4.766, w = 0.126 lbf/ft,
F = 90.0 lbf, F c = 0.913 lbf, and gives
Chapter 17, Page 2/39
F1
0.913 90 4.766 0.913
4.766 1
F 2 = 114.8 90 = 24.8 lbf
114.8 lbf
F i = (114.8 + 24.8)/ 2 0.913 = 68.9 lbf
f
1
114.8 0.913
ln
0.50
3.123 24.8 0.913
3 108 / 12 0.126
dip
0.222 in
2(68.9)
2
So, reducing F i from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50,
with a corresponding dip of 0.222 in. Having reduced F 1 and F 2 , the endurance of the
belt is improved. Power, service factor and design factor have remained intact.
______________________________________________________________________________
17-2
Double the dimensions of Prob. 17-1.
In Prob. 17-1, F-1 Polyamide was used with a thickness of 0.05 in. With what is available
in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let
b = 12 in, d = 4 in with n = 1750 rev/min, H nom = 2 hp, C = 18(12) = 216 in, velocity
ratio = 0.5, K s = 1.25, n d = 1.
V = d n / 12 = (4)(1750) / 12 = 1833 ft/min
Eq. (17-1):
D = d / vel ratio = 4 / 0.5 = 8 in
84
Dd
d 2sin 1
2sin 1
3.123 rad
2C
2(216)
Table 17-2:
t = 0.11 in, d min = 2.4 in, F a = 60 lbf/in, = 0.037 lbf/in3, f = 0.8
w = 12 bt = 12(0.037)12(0.11) = 0.586 lbf/ft
2
(a) Eq. (e), p. 885:
2
w V
0.586 1833
Fc
17.0 lbf
32.17 60
g 60
Ans.
63 025H nom K s nd
63 025(2)(1.25)(1)
90.0 lbf · in
n
1750
2T
2(90.0)
F F1 a F2
45.0 lbf
d
4
T
Table 17-4:
C p = 0.73
Chapter 17, Page 3/39
Eq. (17-12):
(F 1 ) a = bF a C p C v = 12(60)(0.73)(1) = 525.6 lbf
Ans.
F 2 = (F 1 ) a [(F 1 ) a F 2 ] = 525.6 45 = 480.6 lbf
Fi
Eq. (i), p. 886:
F1 a
F2
2
Fc
Ans.
525.6 480.6
17.0 486.1 lbf
2
Ans.
Eq. (17-9):
f
( F ) Fc
1
525.6 17.0
ln 1 a
ln
0.0297
d F2 Fc 3.123 480.6 17.0
1
The friction is thus underdeveloped.
(b) The transmitted horsepower is, with F = (F 1 ) a F 2 = 45 lbf,
(F )V
45(1833)
2.5 hp
33 000
33 000
H
2.5
1
H nom K s
2(1.25)
H
nf s
Eq. (17-1):
Eq. (17-2):
D 2sin 1
Ans.
84
Dd
2sin 1
3.160 rad
2C
2(216)
L = [4C2 (D d)2]1/2 + (D D + d d )/2
= [4(216)2 (8 4)2]1/2 + [8(3.160) + 4(3.123)]/2 = 450.9 in
Ans.
3C 2 w 3(216 / 12) 2 (0.586)
(c) Eq. (17-13):
dip
0.586 in Ans.
2 Fi
2(486.1)
______________________________________________________________________________
17-3
As a design task, the decision set on p. 893 is useful.
A priori decisions:
• Function: H nom = 60 hp, n = 380 rev/min, C = 192 in, K s = 1.1
• Design factor: n d = 1
• Initial tension: Catenary
• Belt material. Table 17-2: Polyamide A-3, F a = 100 lbf/in, = 0.042 lbf/in3, f = 0.8
• Drive geometry: d = D = 48 in
• Belt thickness: t = 0.13 in
Chapter 17, Page 4/39
Design variable: Belt width.
Use a method of trials. Initially, choose b = 6 in
dn (48)(380)
V
4775 ft/min
12
12
w 12 bt 12(0.042)(6)(0.13) 0.393 lbf/ft
wV 2
0.393(4775 / 60) 2
Fc
77.4 lbf
g
32.17
63 025H nom K s nd
63 025(60)(1.1)(1)
T
10 946 lbf · in
n
380
2T
2(10 946)
F
456.1 lbf
d
48
F1 ( F1) a bFaC pCv 6(100)(1)(1) 600 lbf
F2 F1 F 600 456.1 143.9 lbf
Transmitted power H
F (V )
H
33 000
F F2
Fi 1
2
1
F
f
ln 1
d F2
Eq. (17-2):
456.1(4775)
66 hp
33 000
600 143.9
Fc
77.4 294.6 lbf
2
Fc
1 600 77.4
ln
0.656
Fc
143.9 77.4
L = [4(192)2 (48 48)2]1/2 + [48() + 48()] / 2 = 534.8 in
Friction is not fully developed, so b min is just a little smaller than 6 in (5.7 in). Not having
a figure of merit, we choose the most narrow belt available (6 in). We can improve the
design by reducing the initial tension, which reduces F 1 and F 2 , thereby increasing belt
life (see the result of Prob. 17-8). This will bring f to 0.80
F1
F
Fc exp f Fc
exp f 1
exp f exp(0.80 ) 12.345
Therefore
(456.1 77.4)(12.345) 77.4
573.7 lbf
12.345 1
F2 F1 F 573.7 456.1 117.6 lbf
573.7 117.6
F F2
Fi 1
Fc
77.4 268.3 lbf
2
2
F1
These are small reductions since f is close to f , but improvements nevertheless.
Chapter 17, Page 5/39
f
1
d
ln
F1 Fc
1 573.7 77.4
ln
0.80
F2 Fc 117.6 77.4
3C 2 w 3(192 / 12) 2 (0.393)
0.562 in
2 Fi
2(268.3)
______________________________________________________________________________
dip
17-4
From the last equation given in the problem statement,
exp f
1
1 2T / [d (a0 a2 )b]
2T
1
exp f 1
d (a0 a2 )b
2T
exp f exp f 1
d (a0 a2 )b
b
1 2T exp f
a0 a2 d exp f 1
But 2T/d = 33 000H d /V. Thus,
exp f
Q.E.D.
exp f 1
______________________________________________________________________________
b
17-5
1 33 000 H d
a0 a2
V
Refer to Ex. 17-1 on p. 890 for the values used below.
(a) The maximum torque prior to slip is,
T
63 025H nom K s nd
63 025(15)(1.25)(1.1)
742.8 lbf · in
n
1750
Ans.
The corresponding initial tension, from Eq. (17-9), is,
Fi
T exp( f ) 1 742.8 11.17 1
148.1 lbf
d exp( f ) 1
6 11.17 1
Ans.
(b) See Prob. 17-4 statement. The final relation can be written
Chapter 17, Page 6/39
bmin
33 000H a exp f
1
2
FaC pCv (12 t / 32.174)(V / 60) V [exp f 1]
33 000(20.6)(11.17)
1
2
100(0.7)(1) [12(0.042)(0.13)] / 32.174 (2749 / 60) 2749(11.17 1)
4.13 in
Ans.
This is the minimum belt width since the belt is at the point of slip. The design must
round up to an available width.
Eq. (17-1):
18 6
D d
1
d 2sin 1
2sin
2C
2(96)
3.016 511 rad
D d
1 18 6
D 2sin 1
2sin
2C
2(96)
3.266 674 rad
Eq. (17-2):
L [4(96) 2 (18 6) 2 ]1/ 2
230.074 in
(c)
Ans.
1
[18(3.266 674) 6(3.016 511)]
2
2T
2(742.8)
247.6 lbf
d
6
( F1) a bFaC pCv F1 4.13(100)(0.70)(1) 289.1 lbf
F2 F1 F 289.1 247.6 41.5 lbf
w 12 bt 12(0.042)4.13(0.130) 0.271 lbf/ft
F
2
2
w V
0.271 2749
17.7 lbf
g 60
32.17 60
F F2
289.1 41.5
Fi 1
Fc
17.7 147.6 lbf
2
2
Fc
Transmitted belt power H
F (V )
247.6(2749)
20.6 hp
33 000
33 000
20.6
H
1.1
n fs
15(1.25)
H nom K s
H
Chapter 17, Page 7/39
Dip:
2
3C 2 w 3(96 / 12) 0.271
dip
0.176 in
2Fi
2(147.6)
(d) If you only change the belt width, the parameters in the following table change as
shown.
b
w
Fc
(F 1 ) a
F2
Fi
f
dip
Ex. 17-1 This Problem
6.00
4.13
0.393
0.271
25.6
17.7
420
289
172.4
41.5
270.6
147.6
0.33*
0.80**
0.139
0.176
*Friction underdeveloped
**Friction fully developed
______________________________________________________________________________
17-6
The transmitted power is the same.
Fc
Fi
(F 1 ) a
F2
Ha
n fs
f
dip
n-Fold
b = 6 in b = 12 in Change
25.65
51.3
2
270.35
664.9
2.46
420
840
2
172.4
592.4
3.44
20.62
20.62
1
1.1
1.1
1
0.139
0.125
0.90
0.328
0.114
0.34
If we relax F i to develop full friction (f = 0.80) and obtain longer life, then
n-Fold
b = 6 in b = 12 in Change
25.6
51.3
2
Fc
148.1
148.1
1
Fi
297.6
323.2
1.09
F1
50
75.6
1.51
F2
f
0.80
0.80
1
dip 0.255
0.503
2
______________________________________________________________________________
Chapter 17, Page 8/39
17-7
Find the resultant of F 1 and F 2 :
Dd
2C
Dd
sin
2C
2
1D d
cos 1
2 2C
sin 1
2
1 D d
R x F1 cos F2 cos ( F1 F2 ) 1
2 2C
Dd
R y F1 sin F2 sin ( F1 F2 )
Ans.
2C
Ans.
From Ex. 17-2, d = 16 in, D = 36 in, C = 16(12) = 192 in, F 1 = 940 lbf, F 2 = 276 lbf
36 16
o
sin 1
2.9855
2(192)
2
1 36 16
R (940 276) 1
1214.4 lbf
2 2(192)
x
36 16
R y (940 276)
34.6 lbf
2(192)
d
16
T ( F1 F2 ) (940 276) 5312 lbf · in
2
2
______________________________________________________________________________
17-8
Begin with Eq. (17-10),
F1 Fc Fi
2 exp( f )
exp( f ) 1
Introduce Eq. (17-9):
Chapter 17, Page 9/39
exp( f ) 1 2 exp( f )
2T exp( f )
Fc
F1 Fc d
d exp( f ) 1
exp( f ) 1 exp( f ) 1
exp( f )
F1 Fc F
exp( f ) 1
exp( f )
Now add and subtract Fc
exp( f ) 1
exp( f )
exp( f )
exp( f )
F1 Fc Fc
F
Fc
exp( f ) 1
exp( f ) 1
exp( f ) 1
exp( f )
exp( f )
( Fc F )
Fc Fc
exp( f ) 1
exp( f ) 1
exp( f )
Fc
( Fc F )
exp( f ) 1 exp( f ) 1
( F F ) exp( f ) Fc
c
Q.E.D.
exp( f ) 1
From Ex. 17-2: d = 3.037 rad, F = 664 lbf, exp( f ) = exp[0.80(3.037)] = 11.35, and
F c = 73.4 lbf.
(73.4 664)11.35 73.4
802 lbf
(11.35 1)
F2 F1 F 802 664 138 lbf
802 138
73.4 396.6 lbf
Fi
2
1 F1 Fc
1
802 73.4
ln
ln
0.80 Ans.
f
d F2 Fc 3.037 138 73.4
______________________________________________________________________________
F1
This is a good class project. Form four groups, each with a belt to design. Once each
group agrees internally, all four should report their designs including the forces and
torques on the line shaft. If you give them the pulley locations, they could design the line
shaft.
______________________________________________________________________________
17-9
17-10 If you have the students implement a computer program, the design problem selections
may differ, and the students will be able to explore them. For K s = 1.25, n d = 1.1, d = 14
in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do (b min = 6.58 in)
______________________________________________________________________________
17-11 An efficiency of less than unity lowers the output for a given input. Since the object of
Chapter 17, Page 10/39
the drive is the output, the efficiency must be incorporated such that the belt’s capacity is
increased. The design power would thus be expressed as
H nom K s nd
Ans.
eff
______________________________________________________________________________
Hd
17-12 Some perspective on the size of F c can be obtained from
2
w V
12 bt V
Fc
g 60
g 60
2
An approximate comparison of non-metal and metal belts is presented in the table below.
, lbf/in
3
b, in
t, in
Non-metal
0.04
5.00
0.20
Metal
0.280
1.000
0.005
The ratio w / w m is
w
12(0.04)(5)(0.2)
29
wm 12(0.28)(1)(0.005)
The second contribution to F c is the belt peripheral velocity which tends to be low in
metal belts used in instrument, printer, plotter and similar drives. The velocity ratio
squared influences any F c / (F c ) m ratio.
It is common for engineers to treat F c as negligible compared to other tensions in the
belting problem. However, when developing a computer code, one should include F c .
______________________________________________________________________________
17-13 Eq. (17-8):
exp( f ) 1
exp( f ) 1
F1
exp( f )
exp( f )
Assuming negligible centrifugal force and setting F 1 = ab from step 3, p. 897,
F F1 F2 ( F1 Fc )
bmin
Also,
F exp( f )
a exp( f ) 1
(1)
(F )V
33 000
33 000 H nom K s nd
F
V
H d H nom K s nd
Chapter 17, Page 11/39
1 33 000 H d exp( f )
Ans.
a
V
exp( f ) 1
______________________________________________________________________________
Substituting into Eq. (1),
bmin
17-14 The decision set for the friction metal flat-belt drive is:
A priori decisions
• Function: H nom = 1 hp, n = 1750 rev/min, VR = 2 , C 15 in, K s = 1.2 ,
N p = 106 belt passes.
• Design factor: n d = 1.05
• Belt material and properties: 301/302 stainless steel
Table 17-8:
S y = 175 kpsi, E = 28 Mpsi, = 0.285
• Drive geometry: d = 2 in, D = 4 in
• Belt thickness: t = 0.003 in
Design variables:
• Belt width, b
• Belt loop periphery
Preliminaries
H d H nom K s nd 1(1.2)(1.05) 1.26 hp
63 025(1.26)
T
45.38 lbf · in
1750
A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The
40 in loop available corresponds to a 15.254 in center distance.
42
d 2sin 1
3.010 rad
2(15.254)
42
D 2sin 1
3.273 rad
2(15.274)
For full friction development
exp( f d ) exp[0.35(3.010)] 2.868
dn (2)(1750)
V
916.3 ft/s
12
12
S y 175 kpsi
Eq. (17-15):
S y 14.17 106 N p0.407 14.17 106 106
0.407
51.212 103 psi
Chapter 17, Page 12/39
From selection step 3, p. 897,
Et
28(106 )(0.003)
3
a S f
t
51.212(10
)
(0.003)
(1 2 )d
(1 0.2852 )(2)
16.50 lbf/in of belt width
( F1) a ab 16.50b
For full friction development, from Prob. 17-13,
bmin
F exp( f d )
a exp( f d ) 1
F
2T
2(45.38)
45.38 lbf
d
2
bmin
45.38 2.868
4.23 in
16.50 2.868 1
So
Decision #1: b = 4.5 in
F1 ( F1)a ab 16.5(4.5) 74.25 lbf
F2 F1 F 74.25 45.38 28.87 lbf
F F2
74.25 28.87
Fi 1
51.56 lbf
2
2
Existing friction
F
1
74.25
ln 1
ln
0.314
d F2 3.010 28.87
(F )V
45.38(916.3)
1.26 hp
Ht
33 000
33 000
Ht
1.26
1.05
n fs
H nom K s 1(1.2)
f
1
This is a non-trivial point. The methodology preserved the factor of safety corresponding
to n d = 1.1 even as we rounded b min up to b.
Decision #2 was taken care of with the adjustment of the center-to-center distance to
accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this.
Remember to subsequently recalculate d and D .
______________________________________________________________________________
Chapter 17, Page 13/39
17-15 Decision set:
A priori decisions
• Function: H nom = 5 hp, N = 1125 rev/min, VR = 3, C 20 in, K s = 1.25,
N p = 106 belt passes
• Design factor: n d = 1.1
• Belt material: BeCu, S y = 170 kpsi, E = 17 Mpsi, = 0.220
• Belt geometry: d = 3 in, D = 9 in
• Belt thickness: t = 0.003 in
Design decisions
• Belt loop periphery
• Belt width b
Preliminaries:
H d H nom K s nd 5(1.25)(1.1) 6.875 hp
63 025(6.875)
T
385.2 lbf · in
1125
Decision #1: Choose a 60-in belt loop with a center-to-center distance of 20.3 in.
93
2.845 rad
2(20.3)
d 2sin 1
93
3.438 rad
2(20.3)
D 2sin 1
For full friction development:
exp( f d ) exp[0.32(2.845)] 2.485
dn (3)(1125)
883.6 ft/min
V
12
12
S f 56.67 kpsi
From selection step 3, p. 897,
Chapter 17, Page 14/39
Et
17(106 )(0.003)
3
a S f
t
56.67(10
)
(0.003) 116.4 lbf/in
(1 2 )d
(1 0.22 2 )(3)
2T
2(385.2)
F
256.8 lbf
d
3
F exp( f d ) 256.8 2.485
bmin
3.69 in
a exp( f d ) 1 116.4 2.485 1
Decision #2: b = 4 in
F1 ( F1) a ab 116.4(4) 465.6 lbf
F2 F1 F 465.6 256.8 208.8 lbf
F F2
465.6 208.8
Fi 1
337.3 lbf
2
2
Existing friction
F
1
465.6
ln 1
ln
0.282
d F2 2.845 208.8
(F )V
256.8(883.6)
6.88 hp
H
33 000
33 000
6.88
H
1.1
n fs
5(1.25) 5(1.25)
f
1
F i can be reduced only to the point at which f f 0.32. From Eq. (17-9)
Fi
T
d
exp( f d ) 1 385.2 2.485 1
301.3 lbf
3 2.485 1
exp( f d ) 1
Eq. (17-10):
2 exp( f d )
2(2.485)
F1 Fi
429.7 lbf
301.3
2.485 1
exp( f d ) 1
F2 F1 F 429.7 256.8 172.9 lbf
and
f f 0.32
______________________________________________________________________________
17-16 This solution is the result of a series of five design tasks involving different belt
thicknesses. The results are to be compared as a matter of perspective. These design tasks
are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions.
The details will not be presented here, but the table is provided as a means of learning.
Five groups of students could each be assigned a belt thickness. You can form a table
Chapter 17, Page 15/39
from their results or use the table given here.
Chapter 17, Page 16/39
0.002
4.000
20.300
109.700
3.000
9.000
310.600
439.000
182.200
1.100
60.000
0.309
301.200
429.600
172.800
0.320
b
CD
a
d
D
Fi
F1
F2
nf s
L
f
Fi
F1
F2
f
0.003
3.500
20.300
131.900
3.000
9.000
333.300
461.700
209.000
1.100
60.000
0.285
301.200
429.600
172.800
0.320
t, in
0.005
4.000
20.300
110.900
3.000
9.000
315.200
443.600
186.800
1.100
60.000
0.304
301.200
429.600
172.800
0.320
0.008
1.500
18.700
194.900
5.000
15.000
215.300
292.300
138.200
1.100
70.000
0.288
195.700
272.700
118.700
0.320
0.010
1.500
20.200
221.800
6.000
18.000
268.500
332.700
204.300
1.100
80.000
0.192
166.600
230.800
102.400
0.320
The first three thicknesses result in the same adjusted F i , F 1 and F 2 (why?). We have no
figure of merit, but the costs of the belt and pulleys are about the same for these three
thicknesses. Since the same power is transmitted and the belts are widening, belt forces
are lessening.
______________________________________________________________________________
17-17 This is a design task. The decision variables would be belt length and belt section, which
could be combined into one, such as B90. The number of belts is not an issue.
We have no figure of merit, which is not practical in a text for this application. It is
suggested that you gather sheave dimensions and costs and V-belt costs from a principal
vendor and construct a figure of merit based on the costs. Here is one trial.
Preliminaries: For a single V-belt drive with H nom = 3 hp, n = 3100 rev/min, D = 12 in,
and d = 6.2 in, choose a B90 belt, K s = 1.3 and n d = 1. From Table 17-10, select a
circumference of 90 in. From Table 17-11, add 1.8 in giving
L p = 90 + 1.8 = 91.8 in
Eq. (17-16b):
C 0.25 91.8 (12 6.2)
2
31.47 in
2
2
91.8
(12
6.2)
2(12
6.2)
2
Chapter 17, Page 17/39
12 6.2
d 2sin -1
2.9570 rad
2(31.47)
exp( f d ) exp[0.5123(2.9570)] 4.5489
dn (6.2)(3100)
V
12
12
5031.8 ft/min
Table 17-13:
Angle d
180
180
(2.957 rad)
169.42
The footnote regression equation of Table 17-13 gives K 1 without interpolation:
K 1 = 0.143 543 + 0.007 468(169.42°) 0.000 015 052(169.42°)2 = 0.9767
The design power is
H d = H nom K s n d = 3(1.3)(1) = 3.9 hp
From Table 17-14 for B90, K 2 = 1. From Table 17-12 take a marginal entry of H tab = 4,
although extrapolation would give a slightly lower H tab .
Eq. (17-17):
H a = K 1 K 2 H tab = 0.9767(1)(4) = 3.91 hp
The allowable F a is given by
Fa
63 025H a
63 025(3.91)
25.6 lbf
n(d / 2)
3100(6.2 / 2)
The allowable torque T a is
Ta
Fa d
25.6(6.2)
79.4 lbf · in
2
2
From Table 17-16, K c = 0.965. Thus, Eq. (17-21) gives,
2
2
V
5031.8
Fc K c
0.965
24.4 lbf
1000
1000
At incipient slip, Eq. (17-9) provides:
T exp( f ) 1 79.4 4.5489 1
Fi
20.0 lbf
d exp( f ) 1 6.2 4.5489 1
Eq. (17-10):
Chapter 17, Page 18/39
2 exp( f )
2(4.5489)
F1 Fc Fi
24.4 20
57.2 lbf
4.5489 1
exp( f ) 1
Thus, F 2 = F 1 F a = 57.2 25.6 = 31.6 lbf
Eq. (17-26):
n fs
H a Nb
(3.91)(1)
1.003
Hd
3.9
Ans.
If we had extrapolated for H tab , the factor of safety would have been slightly less than
one.
Life
Use Table 17-16 to find equivalent tensions T 1 and T 2 .
Kb
57.2
d
K
T2 F1 ( Fb ) 2 F1 b 57.2
D
T1 F1 ( Fb )1 F1
576
150.1 lbf
6.2
576
105.2 lbf
12
From Table 17-17, K = 1193, b = 10.926, and from Eq. (17-27), the number of belt passes
is:
1
K b K b
N P
T1
T2
1
1193 10.926 1193 10.926
9
6.72(10 ) passes
150.1
105.2
From Eq. (17-28) for N P > 109,
109 (91.8)
720V
720(5031.8)
t 25 340 h Ans.
t
N P Lp
Suppose n f s was too small. Compare these results with a 2-belt solution.
H tab 4 hp/belt, Ta 39.6 lbf · in/belt,
Fa 12.8 lbf/belt, H a 3.91 hp/belt
NH
Nb H a
2(3.91)
n fs b a
2.0
Hd
H nom K s
3(1.3)
Also,
F 1 = 40.8 lbf/belt,
F 2 = 28.0 lbf/belt
Chapter 17, Page 19/39
Fi 9.99 lbf/belt,
Fc 24.4 lbf/belt
( Fb )1 92.9 lbf/belt,
( Fb )2 48 lbf/belt
T1 133.7 lbf/belt,
T2 88.8 lbf/belt
10
N P 2.39(10 ) passes,
t 605 600 h
Initial tension of the drive:
(F i ) drive = N b F i = 2(9.99) = 20 lbf
______________________________________________________________________________
17-18 Given: two B85 V-belts with d = 5.4 in, D = 16 in, n = 1200 rev/min, and K s = 1.25
Table 17-11: L p = 85 + 1.8 = 86.8 in
Eq. (17-17b):
C 0.25 86.8 (16 5.4)
2
26.05 in Ans.
2
2
86.8 2 (16 5.4) 2(16 5.4)
Eq. (17-1):
16 5.4
d 180 2sin -1
156.5
2(26.05)
From table 17-13 footnote:
K 1 = 0.143 543 + 0.007 468(156.5°) 0.000 015 052(156.5°)2 = 0.944
Table 17-14:
K2 = 1
Belt speed:
V
(5.4)(1200)
12
1696 ft/min
Use Table 17-12 to interpolate for H tab .
2.62 1.59
H tab 1.59
(1696 1000) 2.31 hp/belt
2000 1000
Eq. (17-17) for two belts:
H a K1K 2 N b H tab 0.944(1)(2)(2.31) 4.36 hp
Assuming n d = 1,
H d = K s H nom n d = 1.25(1)H nom
For a factor of safety of one,
Chapter 17, Page 20/39
Ha Hd
4.36 1.25H nom
4.36
H nom
3.49 hp Ans.
1.25
______________________________________________________________________________
17-19 Given: H nom = 60 hp, n = 400 rev/min, K s = 1.4, d = D = 26 in on 12 ft centers.
Design task: specify V-belt and number of strands (belts). Tentative decision: Use D360
belts.
Table 17-11: L p = 360 + 3.3 = 363.3 in
Eq. (17-16b):
C 0.25 363.3 (26 26)
2
140.8 in (nearly 144 in)
2
2
363.3
(26
26)
2(26
26)
2
d , D , exp[0.5123 ] 5.0,
dn (26)(400)
V
12
12
2722.7 ft/min
Table 17-13: For = 180°, K 1 = 1
Table 17-14: For D360, K 2 = 1.10
Table 17-12: H tab = 16.94 hp by interpolation
Thus,
H a = K 1 K 2 H tab = 1(1.1)(16.94) = 18.63 hp / belt
Eq. (17-19):
H d = H nom K s n d = 60(1.4)(1) = 84 hp
Number of belts, N b
Nb
Hd
84
4.51
Ha
18.63
Round up to five belts. It is left to the reader to repeat the above for belts such as C360
and E360.
Chapter 17, Page 21/39
63 025H a
63 025(18.63)
225.8 lbf/belt
n(d / 2)
400(26 / 2)
(Fa )d
225.8(26)
Ta
2935 lbf · in/belt
2
2
Fa
Eq. (17-21):
2
2
V
2722.7
Fc 3.498
3.498
25.9 lbf/belt
1000
1000
At fully developed friction, Eq. (17-9) gives
Fi
T
d
exp( f ) 1 2935 5 1
exp( f ) 1 26 5 1 169.3 lbf/belt
2 exp( f )
2(5)
25.9 169.3
308.1 lbf/belt
F1 Fc Fi
5 1
exp( f ) 1
F2 F1 Fa 308.1 225.8 82.3 lbf/belt
18.63 5
H N
nf s a b
1.109 Ans.
84
Hd
Life From Table 17-16,
K
5 680
T1 T2 F1 b 308.1
526.6 lbf
d
26
Eq. (17-27):
Eq. (17-10):
K b K b
N P
T1
T2
Thus,
N P > 109 passes
Eq. (17-28):
1
5.28 109 passes
Ans.
109 (363.3)
t
720V
720(2722.7)
N P Lp
Thus,
t > 185 320 h Ans.
______________________________________________________________________________
17-20 Preliminaries: D 60 in, 14-in wide rim, H nom = 50 hp, n = 875 rev/min, K s = 1.2,
n d = 1.1, m G = 875/170 = 5.147, d 60 / 5.147 11.65 in
(a) From Table 17-9, an 11-in sheave exceeds C-section minimum diameter and
precludes D- and E-section V-belts.
Decision: Use d = 11 in, C270 belts
Chapter 17, Page 22/39
Table 17-11: L p = 270 + 2.9 = 272.9 in
Eq. (17-16b):
C 0.25 272.9 (60 11)
2
76.78 in
2
2
272.9
(60
11)
2(60
11)
2
This fits in the range
D C 3( D d ) 60 C 3(60 11) 60 in C 213 in
60 11
d 2sin 1
2.492 rad 142.8
2(76.78)
60 11
D 2sin 1
3.791 rad
2(76.78)
exp(f d ) = exp[0.5123(2.492)] = 3.5846
For the flat on flywheel, f = 0.13 (see p. 900), exp(f D ) = exp[0.13(3.791)] = 1.637.
The belt speed is
V
dn
12
(11)(875)
12
2520 ft/min
Table 17-13:
K 1 = 0.143 543 + 0.007 468(142.8°) 0.000 015 052(142.8°)2 = 0.903
Table 17-14: K 2 = 1.15
For interpolation of Table 17-12, let x be entry for d = 11.65 in and n = 2000 ft/min, and y
be entry for d = 11.65 in and n = 3000 ft/min. Then,
x 6.74
7.17 6.74
x 7.01 hp at 2000 ft/min
11.65 11
12 11
and
8.11 y
8.84 8.11
y 8.58 hp at 3000 ft/min
11.65 11
12 11
Interpolating these for 2520 ft/min gives
8.58 H tab
3000 2520
H tab 7.83 hp/belt
8.58 7.01 3000 2000
Eq. (17-17):
H a = K 1 K 2 H tab = 0.903(1.15)(7.83) = 8.13 hp
Chapter 17, Page 23/39
Eq. (17-19):
H d = H nom K s n d = 50(1.2)(1.1) = 66 hp
Eq. (17-20): N b
Hd
66
8.1 belts
Ha
8.13
Decision: Use 9 belts. On a per belt basis,
63 025H a
63 025(8.13)
106.5 lbf/belt
n(d / 2)
875(11 / 2)
Fa d 106.5(11)
Ta
586.8 lbf · in per belt
2
2
Table 17-16: K c = 1.716
Fa
2
Eq. (17-21):
2
V
2520
Fc 1.716
1.716
10.9 lbf/belt
1000
1000
At fully developed friction, Eq. (17-9) gives
Fi
T
d
exp( f d ) 1 586.9 3.5846 1
94.6 lbf/belt
11 3.5846 1
exp( f d ) 1
Eq. (17-10):
2 exp( f d )
2(3.5846)
158.8 lbf/belt
F1 Fc Fi
10.9 94.6
3.5846 1
exp( f d ) 1
F2 F1 Fa 158.8 106.7 52.1 lbf/belt
NH
9(8.13)
nf s b a
1.11 O.K . Ans.
Hd
66
Durability:
Fb 1 K b / d 1600 / 11 145.5 lbf/belt
Fb 2 K b / D 1600 / 60 26.7 lbf/belt
T1 F1 Fb 1 158.8 145.5 304.3 lbf/belt
T2 F1 Fb 2 158.8 26.7 185.5 lbf/belt
Eq. (17-27) with Table 17-17:
1
K b K b
2038 11.173 2038 11.173
N P
T
304.3
185.5
T1
2
9
9
1.68 10 passes 10 passes
Ans.
1
Since N P is greater than 109 passes and is out of the range of Table 17-17, life from Eq.
(17-27) is
Chapter 17, Page 24/39
t
N P Lp
720V
Remember:
109 (272.9)
150 103 h
720(2520)
(F i ) drive = 9(94.6) = 851.4 lbf
Table 17-9: C-section belts are 7/8 in wide. Check sheave groove spacing to see if 14 in
width is accommodating.
(b) The fully developed friction torque on the flywheel using the flats of the V-belts,
from Eq. (17-9), is
exp( f ) 1
1.637 1
Tflat Fi D
94.6(60)
1371 lbf · in per belt
1.637 1
exp( f ) 1
The flywheel torque should be
T fly = m G T a = 5.147(586.9) = 3021 lbf · in per belt
but it is not. There are applications, however, in which it will work. For example,
make the flywheel controlling. Yes. Ans.
______________________________________________________________________________
17-21
(a)
S is the spliced-in string segment length
D e is the equatorial diameter
D is the spliced string diameter
is the radial clearance
S + D e = D = (D e + 2) = D e + 2
From which
S
2
The radial clearance is thus independent of D e .
12(6)
11.5 in
2
Ans.
This is true whether the sphere is the earth, the moon or a marble. Thinking in terms of a
radial or diametral increment removes the basic size from the problem.
(b) and (c)
Chapter 17, Page 25/39
Table 17-9: For an E210 belt, the thickness is 1 in.
d P di
2
210 4.5
4.5
210
4.5
4.5
0.716 in
2
The pitch diameter of the flywheel is
DP 2 D DP D 2 60 2(0.716) 61.43 in
We could make a table:
Diametral
Growth
2
A
1.3
Section
B
C
D
1.8 2.9 3.3
E
4.5
The velocity ratio for the D-section belt of Prob. 17-20 is
mG
D 2
60 3.3 /
5.55
d
11
Ans.
for the V-flat drive as compared to m a = 60/11 = 5.455 for the VV drive.
The pitch diameter of the pulley is still d = 11 in, so the new angle of wrap, d , is
D 2 d
2C
1 D 2 d
D 2sin
2C
d 2sin 1
Ans.
Ans.
Chapter 17, Page 26/39
Equations (17-16a) and (17-16b) are modified as follows
(D d )2
2
4C
C p 0.25 L p ( D 2 d )
2
L p 2C
( D 2 d )
Ans.
2
L p ( D 2 d ) 2( D 2 d ) 2 Ans.
2
The changes are small, but if you are writing a computer code for a V-flat drive,
remember that d and D changes are exponential.
______________________________________________________________________________
17-22 This design task involves specifying a drive to couple an electric motor running at 1720
rev/min to a blower running at 240 rev/min, transmitting two horsepower with a center
distance of at least 22 inches. Instead of focusing on the steps, we will display two
different designs side-by-side for study. Parameters are in a “per belt” basis with per
drive quantities shown along side, where helpful.
Parameter
Four A-90 Belts Two A-120 Belts
7.33
7.142
mG
1.1
1.1
Ks
1.1
1.1
nd
0.877
0.869
K1
1.05
1.15
K2
d, in
3.0
4.2
D, in
22
30
2.333
2.287
d , rad
V, ft/min
1350.9
1891
3.304
3.2266
exp(f d )
91.3
101.3
L p , in
C, in
24.1
31
0.783
1.662
H tab , uncorr.
3.13
3.326
N b H tab , uncorr.
26.45(105.8)
60.87(121.7)
T a , lbf · in
17.6(70.4)
29.0(58)
F a , lbf
0.721(2.88)
1.667(3.33)
H a , hp
1.192
1.372
nf s
26.28(105.2)
44(88)
F 1 , lbf
8.67(34.7)
15(30)
F 2 , lbf
73.3(293.2)
52.4(109.8)
(F b ) 1 , lbf
10(40)
7.33(14.7)
(F b ) 2 , lbf
1.024
2.0
F c , lbf
16.45(65.8)
27.5(55)
F i , lbf
99.2
96.4
T 1 , lbf · in
Chapter 17, Page 27/39
T 2 , lbf · in
N , passes
t>h
36.3
1.61(109)
93 869
57.4
2.3(109)
89 080
Conclusions:
• Smaller sheaves lead to more belts.
• Larger sheaves lead to larger D and larger V.
• Larger sheaves lead to larger tabulated power.
• The discrete numbers of belts obscures some of the variation. The factors of safety
exceed the design factor by differing amounts.
______________________________________________________________________________
17-23 In Ex. 17-5 the selected chain was 140-3, making the pitch of this 140 chain14/8 = 1.75
in. Table 17-19 confirms.
______________________________________________________________________________
17-24 (a) Eq. (17-32): H1 0.004 N11.08n10.9 p (3 0.07 p )
1000K r N11.5 p 0.8
n11.5
Equating and solving for n 1 gives
Eq. (17-33):
H2
0.25(106 ) K r N10.42
n1
p (2.2 0.07 p )
1/ 2.4
Ans.
(b) For a No. 60 chain, p = 6/8 = 0.75 in, N 1 = 17, K r = 17
1/ 2.4
0.25(106 )(17)(17)0.42
n1
[2.2 0.07(0.75)]
0.75
1227 rev/min
Ans.
Table 17-20 confirms that this point occurs at 1200 ± 200 rev/min.
(c) Life predictions using Eq. (17-40) are possible at speeds greater than 1227 rev/min.
Ans.
______________________________________________________________________________
17-25 Given: a double strand No. 60 roller chain with p = 0.75 in, N 1 = 13 teeth at 300 rev/min,
N 2 = 52 teeth.
(a) Table 17-20:
Table 17-22:
Table 17-23:
Use
Eq. (17-37):
H tab = 6.20 hp
K 1 = 0.75
K 2 = 1.7
Ks = 1
H a = K 1 K 2 H tab = 0.75(1.7)(6.20) = 7.91 hp
Ans.
Chapter 17, Page 28/39
(b) Eqs. (17-35) and (17-36) with L/p = 82
13 52
82 49.5
2
2
p
52 13
2
49.5 49.5 8
C
23.95 p
4
2
C 23.95(0.75) 17.96 in, round up to 18 in Ans.
A
(c) For 30 percent less power transmission,
H 0.7(7.91) 5.54 hp
63 025(5.54)
T
1164 lbf · in
300
Ans.
Eq. (17-29):
0.75
3.13 in
sin(180o /13)
T
1164
F
744 lbf Ans.
3.13 / 2
r
______________________________________________________________________________
D
17-26 Given: No. 40-4 chain, N 1 = 21 teeth for n = 2000 rev/min, N 2 = 84 teeth, h = 20 000
hours.
(a) Chain pitch is p = 4/8 = 0.500 in and C 20 in.
Eq. (17-34):
L 2C N1 N 2 N1 N 2
p
p
2
4 2C / p
2(20) 21 84
(84 21) 2
135 pitches (or links)
0.5
2
4 2 (20 / 0.5)
L = 135(0.500) = 67.5 in Ans.
2
(b) Table 17-20:
H tab = 7.72 hp (post-extreme power)
Eq. (17-40): Since K 1 is required, the N13.75 term is omitted (see p. 914).
7.72 (15 000)
2.5
constant
18 399
135
1/ 2.5
18 399(135)
H tab
6.88 hp Ans.
20 000
Chapter 17, Page 29/39
(c) Table 17-22:
21
K1
17
Table 17-23:
1.5
1.37
K 2 = 3.3
H a K1K 2 H tab
1.37(3.3)(6.88) 31.1 hp
Ans.
N1 pn
21(0.5)(2000)
1750 ft/min
12
12
33 000(31.1)
F1
586 lbf Ans.
1750
______________________________________________________________________________
(d)
V
17-27 This is our first design/selection task for chain drives. A possible decision set:
A priori decisions
• Function: H nom , n 1 , space, life, K s
• Design factor: n d
• Sprockets: Tooth counts N 1 and N 2 , factors K 1 and K 2
Decision variables
• Chain number
• Strand count
• Lubrication type
• Chain length in pitches
Function: Motor with H nom = 25 hp at n = 700 rev/min; pump at n = 140 rev/min;
m G = 700/140 = 5
Design Factor: n d = 1.1
Sprockets: Tooth count N 2 = m G N 1 = 5(17) = 85 teeth–odd and unavailable. Choose
84 teeth. Decision: N 1 = 17, N 2 = 84
Evaluate K 1 and K 2
Eq. (17-38):
Eq. (17-37):
H d = H nom K s n d
H a = K 1 K 2 H tab
Equate H d to H a and solve for H tab :
KnH
H tab s d nom
K1K 2
Table 17-22:
K1 = 1
Table 17-23:
K 2 = 1, 1.7, 2.5, 3.3 for 1 through 4 strands
Chapter 17, Page 30/39
H tab
1.5(1.1)(25)
41.25
(1) K 2
K2
Prepare a table to help with the design decisions:
Lub.
Chain
No.
Strands K 2 H tab
H tab
n f s Type
1
1.0 41.3
100
59.4 1.58
B
2
1.7 24.3
80
31.0 1.40
B
3
2.5 16.5
80
31.0 2.07
B
4
3.3 12.5
60
13.3 1.17
B
Design Decisions
We need a figure of merit to help with the choice. If the best was 4 strands of No. 60
chain, then
Decision #1 and #2: Choose four strand No. 60 roller chain with n f s = 1.17.
n fs
K1K 2 H tab 1(3.3)(13.3)
1.17
K s H nom
1.5(25)
Decision #3: Choose Type B lubrication
Analysis:
Table 17-20:
H tab = 13.3 hp
Table 17-19:
p = 0.75 in
Try C = 30 in in Eq. (17-34):
2C N1 N 2 ( N 2 N1) 2
L
2
4 2C / p
p
p
17 84
(84 17) 2
2(30 / 0.75)
2
4 2 (30 / 0.75)
133.3
L = 0.75(133.3) = 100 in (no need to round)
Eq. (17-36) with p = 0.75 in: A
N1 N 2 L 17 84 100
82.83
2
p
2
0.75
Eq. (17-35):
p
N N1
A A2 8 2
4
2
2
0.75
2
84 17
82.83 82.83 8
30.0 in
4
2
2
C
Chapter 17, Page 31/39
Decision #4: Choose C = 30.0 in.
______________________________________________________________________________
17-28 Follow the decision set outlined in Prob. 17-27 solution. We will form two tables, the
first for a 15 000 h life goal, and a second for a 50 000 h life goal. The comparison is
useful.
Function: H nom = 50 hp at n = 1800 rev/min, n pump = 900 rev/min, m G = 1800/900 = 2,
K s = 1.2, life = 15 000 h, then repeat with life = 50 000 h
Design factor: n d = 1.1
Sprockets: N 1 = 19 teeth, N 2 = 38 teeth
Table 17-22 (post extreme):
1.5
1.5
N1
19
K1 1.18
17
17
Table 17-23:
K 2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0
Decision variables for 15 000 h life goal:
K s nd H nom 1.2(1.1)(50) 55.9
K1K 2
1.18K 2
K2
KK H
1.18K 2 H tab
1 2 tab
0.0197 K 2 H tab
K s H nom
1.2(50)
H tab
nf
s
(1)
Form a table for a 15 000 h life goal using these equations.
K2
H' tab
1
1.7
2.5
3.3
3.9
4.6
6
55.90
32.90
22.40
16.90
14.30
12.20
9.32
Chain #
120
120
120
120
80
60
60
H tab
nf s
Lub
21.6
21.6
21.6
21.6
15.6
12.4
12.4
0.423
0.923
1.064
1.404
1.106
1.126
1.416
C'
C'
C'
C'
C'
C'
C'
There are 4 possibilities where n f s ≥ 1.1
Decision variables for 50 000 h life goal
From Eq. (17-40), the power-life tradeoff is:
Chapter 17, Page 32/39
) 2.515 000 ( H tab
) 2.5 50 000
( H tab
1/ 2.5
15 000
) 2.5
H tab
( H tab
50 000
0.618 H tab
Substituting from (1),
55.9 34.5
H tab
0.618
K2
K 2
The H notation is only necessary because we constructed the first table, which we
normally would not do.
)
K K H
K K (0.618H tab
0.618[(0.0197) K 2 H tab ]
n f s 1 2 tab 1 2
K s H nom
K s H nom
0.0122 K 2 H tab
Form a table for a 50 000 h life goal.
K2
1
1.7
2.5
3.3
3.9
4.6
6
H'' tab
34.50
20.30
13.80
10.50
8.85
7.60
5.80
Chain #
120
120
120
120
120
120
80
H tab
21.6
21.6
21.6
21.6
21.6
21.6
15.6
nf s
0.264
0.448
0.656
0.870
1.028
1.210
1.140
Lub
C'
C'
C'
C'
C'
C'
C'
There are two possibilities in the second table with n f s ≥ 1.1. (The tables allow for the
identification of a longer life of the outcomes.) We need a figure of merit to help with
the choice; costs of sprockets and chains are thus needed, but is more information than
we have.
Decision #1: #80 Chain (smaller installation) Ans.
n f s = 0.0122K 2 H tab = 0.0122(8.0)(15.6) = 1.14 O.K.
Decision #2: 8-Strand, No. 80 Ans.
Decision #3: Type C Lubrication Ans.
Decision #4: p = 1.0 in, C is in midrange of 40 pitches
Chapter 17, Page 33/39
2C N1 N 2 ( N 2 N1) 2
L
2
4 2C / p
p
p
19 38 (38 19) 2
2(40)
2
4 2 (40)
108.7 110 even integer Ans.
Eq. (17-36):
A
Eq. (17-35):
N1 N 2 L 19 38 110
81.5
2
p
2
1
C
1
(81.5)
p
4
2
38 19
(81.5) 8
40.64
2
2
C = p(C/p) = 1.0(40.64/1.0) = 40.64 in (for reference) Ans.
______________________________________________________________________________
17-29 The objective of the problem is to explore factors of safety in wire rope. We will express
strengths as tensions.
(a) Monitor steel 2-in 6 19 rope, 480 ft long.
Table 17-2: Minimum diameter of a sheave is 30d = 30(2) = 60 in, preferably
45(2) = 90 in. The hoist abuses the wire when it is bent around a sheave. Table 17-24
gives the nominal tensile strength as 106 kpsi. The ultimate load is
Fu ( Su ) nom Anom
(2) 2
106
333 kip
4
Ans.
The tensile loading of the wire is given by Eq. (17-46)
a
W
w l 1
Ft
g
m
W 4(2) 8 kip, m 1
Table (17-24):
wl = 1.60d 2 l = 1.60(22)(480) = 3072 lbf = 3.072 kip
Therefore,
2
Ft (8 3.072) 1
11.76 kip
32.2
Eq. (17-48):
Ed A
Fb r w m
D
Ans.
Chapter 17, Page 34/39
and for the 72-in drum
12(106 )(2 / 13)(0.38)(22 )(103 )
39 kip
Fb
72
Ans.
For use in Eq. (17-44), from Fig. 17-21
(b) Factors of safety
Static, no bending:
( p / Su ) 0.0014
Su 240 kpsi, p. 920
0.0014(240)(2)(72)
Ff
24.2 kip
2
n
Static, with bending:
Eq. (17-49):
ns
Fu
333
28.3
Ft
11.76
Ans.
Fu Fb
333 39
25.0
Ft
11.76
Fatigue without bending:
F
24.2
nf f
2.06
Ft
11.76
Ans.
Ans.
Ans.
Fatigue, with bending: For a life of 0.1(106) cycles, from Fig. 17-21
( p / Su ) 4 / 1000 0.004
0.004(240)(2)(72)
Ff
69.1 kip
2
Eq. (17-50):
nf
69.1 39
2.56
11.76
Ans.
If we were to use the endurance strength at 106 cycles (F f = 24.2 kip) the factor of
safety would be less than 1 indicating 106 cycle life impossible.
Comments:
• There are a number of factors of safety used in wire rope analysis. They are different,
with different meanings. There is no substitute for knowing exactly which factor
of safety is written or spoken.
• Static performance of a rope in tension is impressive.
• In this problem, at the drum, we have a finite life.
• The remedy for fatigue is the use of smaller diameter ropes, with multiple ropes
Chapter 17, Page 35/39
supporting the load. See Ex. 17-6 for the effectiveness of this approach. It will also
be used in Prob. 17-30.
• Remind students that wire ropes do not fail suddenly due to fatigue. The outer
wires gradually show wear and breaks; such ropes should be retired. Periodic
inspections prevent fatigue failures by parting of the rope.
______________________________________________________________________________
17-30 Since this is a design task, a decision set is useful.
A priori decisions
• Function: load, height, acceleration, velocity, life goal
• Design Factor: n d
• Material: IPS, PS, MPS or other
• Rope: Lay, number of strands, number of wires per strand
Decision variables:
• Nominal wire size: d
• Number of load-supporting wires: m
From experience with Prob. 17-29, a 1-in diameter rope is not likely to have much of a
life, so approach the problem with the d and m decisions open.
Function: 5000 lbf load, 90 foot lift, acceleration = 4 ft/s2, velocity = 2 ft/s, life
goal = 105 cycles
Design Factor: n d = 2
Material: IPS
Rope: Regular lay, 1-in plow-steel 6 19 hoisting
Design variables
Choose 30-in D min . Table 17-27: w = 1.60d 2 lbf/ft
wl = 1.60d 2l = 1.60d 2(90) = 144d 2 lbf, each
Eq. (17-46):
a 5000
4
W
Ft w l 1
144d 2 1
g m
32.2
m
5620
162d 2 lbf, each wire
m
Eq. (17-47):
Ff
( p / Su ) Su Dd
2
From Fig. 17-21 for 105 cycles, p/S u = 0.004. From p. 920, S u = 240 kpsi, based on
metal area.
0.004(240 000)(30d )
Ff
14 400d lbf each wire
2
Eq. (17-48) and Table 17-27:
Chapter 17, Page 36/39
6
2
Ewd w Am 12 10 0.067d 0.4d
Fb
10 720d 3 lbf, each wire
D
30
Eq. (17-45):
nf
Ff Fb
Ft
14 400d 10 720d 3
(5620 / m) 162d 2
We could use a computer program to build a table similar to that of Ex. 17-6.
Alternatively, we could recognize that 162 d 2 is small compared to 5620 / m, and
therefore eliminate the 162d 2 term.
nf
14 400d 10 720d 3
m
(14 400d 10 720d 3 )
5620 / m
5620
Maximize n f ,
n f
d
0
m
[14 400 3(10 720)d 2 ]
5620
From which
d*
14 400
0.669 in
3(10 720)
Back-substituting
nf
m
[14 400(0.669) 10 720(0.6693 )] 1.14 m
5620
Thus n f = 1.14, 2.28, 3.42, 4.56 for m=1, 2, 3, 4 respectively. If we choose d = 0.50 in,
then m = 2.
14 400(0.5) 10 720(0.53 )
nf
2.06
(5620 / 2) 162(0.5) 2
This exceeds n d = 2
Decision #1: d = 1/2 in
Decision #2: m = 2 ropes supporting load. Rope should be inspected weekly for any
signs of fatigue (broken outer wires).
Comment: Table 17-25 gives n for freight elevators in terms of velocity.
d 2
83 252d 2 lbf, each wire
Fu (Su ) nom Anom 106 000
4
83 452(0.5) 2
Fu
n
7.32
Ft
(5620 / 2) 162(0.5) 2
Chapter 17, Page 37/39
By comparison, interpolation for 120 ft/min gives 7.08 - close. The category of
construction hoists is not addressed in Table 17-25. We should investigate this before
proceeding further.
______________________________________________________________________________
17-31 Given: 2000 ft lift, 72 in drum, 6 19 MS rope, cage and load 8000 lbf, accel. = 2 ft/s2.
(a) Table 17-24: (S u ) nom = 106 kpsi; S u = 240 kpsi (p. 920); Fig. 17-21: (p/S u )106 =
0.0014
Eq. (17-44):
Ff
p / Su Su dD
Table 17-24:
Eq. (17-46):
2
0.0014(240)d (72)
12.1 d kip
2
wl = 1.6d 2 2000(103) = 3.2d 2 kip
a
Ft (W wl ) 1
g
2
(8 3.2d 2 ) 1
32.2
8.5 3.4d 2 kip
Note that bending is not included.
F
12.1d
n f
Ft
8.5 3.4d 2
d, in
0.500
1.000
1.500
1.625
1.750
2.000
n
0.650
1.020
1.124
1.125 ← maximum n
1.120
1.095
Ans.
(b) Try m = 4 strands
Chapter 17, Page 38/39
2
8
Ft 3.2d 2 1
4
32.2
2.12 3.4d 2 kip
Ff 12.1d kip
12.1d
n
2.12 3.4d 2
d, in
0.5000
0.5625
0.6520
0.7500
0.8750
1.0000
n
2.037
2.130
2.193
2.250 ← maximum n
2.242
2.192
Ans.
Comparing tables, multiple ropes supporting the load increases the factor of safety,
and reduces the corresponding wire rope diameter, a useful perspective.
______________________________________________________________________________
Chapter 17, Page 39/39
17-32
ad
b / m cd 2
(b / m cd 2 )a ad (2cd )
dn
0
(b / m cd 2 )2
dd
n
From which
b
mc
d*
n*
Ans.
a b / (mc)
a m
(b / m) c [b / (mc)] 2 bc
Ans.
These results agree closely with the Prob. 17-31 solution. The small differences are due
to rounding in Prob. 17-31.
______________________________________________________________________________
17-33 From Prob. 17-32 solution:
n1
ad
b / m cd 2
Solve the above equation for m
b
(1)
ad / n1 cd 2
ad / n1 ad 2 (0) b a / n1 2cd
dm
0
2
dd
ad / n1 cd 2
m
From which
d*
a
2cn1
Ans.
Substituting this result for d into Eq. (1) gives
4bcn1
Ans.
a2
______________________________________________________________________________
m*
17-34 Note to the Instructor. In the first printing of the ninth edition, the wording of this
problem is incorrect. It should read “ For Prob. 17-29 estimate the elongation of the rope
if a 7000 lbf loaded mine cart is placed in the cage which weighs 1000 lbf. The results of
Prob. 4-7 may be useful”. This will be corrected in subsequent printings. We apologize
for any inconvenience encountered.
Chapter 17, Page 40/39
Table 17-27:
Am 0.40d 2 0.40(22 ) 1.6 in 2
Er 12 Mpsi, w 1.6d 2 1.6(2 2 ) 6.4 lbf/ft
w l 6.4(480) 3072 lbf
w l / Aml 3072 / 1.6(480)12 0.333 lbf/in 3
Treat the rest of the system as rigid, so that all of the stretch is due to the load of 7000 lbf,
the cage weighing 1000 lbf, and the wire’s weight. From the solution of Prob. 4-7,
l2
Wl
1
AE 2E
(1000 7000)(480)(12) 0.333(4802 )122
1.6(12)(106 )
2(12)(106 )
2.4 0.460 2.860 in
Ans.
______________________________________________________________________________
17-35 to 17-38 Computer programs will vary.
Chapter 17, Page 41/39
Chapter 20
20-1
(a)
(b) f / (Nx) = f / [69(10)] = f / 690
f x2
7200
4900
19200
40500
80000
145200
86400
169000
156800
112500
51200
86700
64800
36100
0
44100
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
f
2
1
3
5
8
12
6
10
8
5
2
3
2
1
0
1
fx
120
70
240
450
800
1320
720
1300
1120
750
320
510
360
130
0
210
69
8480 1 104 600
f / (Nx)
0.0029
0.0015
0.0043
0.0072
0.0116
0.0174
0.0087
0.0145
0.0116
0.0174
0.0029
0.0043
0.0029
0.0015
0
0.0015
Chapter 20, Page 1/29
x
Eq. (20-9):
8480
122.9 kcycles
69
12
1 104 600 8480 2 / 69
Eq. (20-10): sx
30.3 kcycles Ans.
69 1
______________________________________________________________________________
20-2
Data represents a 7-class histogram with N = 197.
x
174
182
190
198
206
214
220
x
6
9
44
67
53
12
6
fx
1044
1638
8360
13 266
10 918
2568
1320
f x2
181 656
298 116
1 588 400
2 626 688
2 249 108
549 552
290 400
197
39 114
7 789 900
f
39 114
198.55 kpsi Ans.
197
12
7 783 900 39 114 2 / 197
s
9.55 kpsi Ans.
197 1
______________________________________________________________________________
20-3
Form a Table:
x
64
68
72
76
80
84
88
92
f
2
6
6
9
19
10
4
2
58
fx
128
408
432
684
1520
840
352
184
4548
fx2
8192
27 744
31 104
51 984
121 600
70 560
30 976
16 928
359 088
Chapter 20, Page 2/29
x
4548
78.4 kpsi
58
Ans.
359 088 45482 / 58
sx
58 1
12
6.57 kpsi
Ans.
From Eq. 20-14
1 x 78.4
1
f x
exp
Ans.
6.57 2
2 6.57
______________________________________________________________________________
2
20-4
(a)
x
5.625
5.875
6.125
6.375
6.625
6.875
7.125
7.375
7.625
7.875
8.125
f
1
0
0
3
3
6
14
15
10
2
1
55
fy
5.625
0
0
19.125
19.875
41.25
99.75
110.625
76.25
15.75
8.125
396.375
f y2
31.64063
0
0
121.9219
131.6719
283.5938
710.7188
815.8594
581.4063
124.0313
66.015 63
2866.859
y
5.625
5.875
6.125
6.375
6.625
6.875
7.125
7.375
7.625
7.875
8.125
For a normal distribution,
y 396.375 / 55 7.207,
f / (Nw)
0.072 727
0
0
0.218 182
0.218 182
0.436 364
1.018 182
1.090 909
0.727 273
0.145 455
0.072 727
f (y)
0.001 262
0.008 586
0.042 038
0.148 106
0.375 493
0.685 057
0.899 389
0.849 697
0.577 665
0.282 608
0.099 492
2866.859 396.3752 / 55
sy
55 1
g(y)
0.000 295
0.004 088
0.031 194
0.140 262
0.393 667
0.725 002
0.915 128
0.822 462
0.544 251
0.273 138
0.106 720
12
0.4358
1 x 7.207 2
1
f y
exp
0.4358 2
2 0.4358
For a lognormal distribution,
x ln 7.206 818 ln 1 0.060 474 2 1.9732,
g y
1
x 0.0604
s x ln 1 0.060 4742 0.0604
1 ln x 1.9732 2
exp
2
2 0.0604
Chapter 20, Page 3/29
(b) Histogram
______________________________________________________________________________
20-5
Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000 in,
b = 0.5008 in.
a b 0.5000 0.5008
(a) Eq. (20-22)
0.5004
x
2
2
x
Eq. (20-23)
b a 0.5008 0.5000
0.000 231
2
2 3
(b) PDF, Eq. (20-20)
1250
f ( x)
0
0.5000 x 0.5008 in
otherwise
(c) CDF, Eq. (20-21)
0
F ( x ) ( x 0.5) / 0.0008
1
x 0.5000 in
0.5000 x 0.5008 in
x 0.5008 in
If all smaller diameters are removed by inspection, a = 0.5002 in, b = 0.5008 in,
0.5002 0.5008
0.5005 in
2
0.5008 0.5002
ˆ x
0.000 173 in
2 3
1666.7 0.5002 x 0.5008 in
f ( x)
otherwise
0
x
Chapter 20, Page 4/29
0
F ( x) 1666.7( x 0.5002)
1
x 0.5002 in
0.5002 x 0.5008 in
x 0.5008 in
______________________________________________________________________________
20-6
Dimensions produced are due to tool dulling and wear. When parts are mixed, the
distribution is uniform. From Eqs. (20-22) and (20-23),
a x 3s 0.6241 3 0.000 581 0.6231 in
b x 3s 0.6241 3 0.000 581 0.6251 in
0.623
in Ans.
0.625
______________________________________________________________________________
We suspect the dimension was
20-7
F(x) = 0.555x – 33 mm.
(a) Since F(x) is linear, distribution is uniform at x = a
F(a) = 0 = 0.555(a) – 33
a = 59.46 mm. Therefore at x = b
F(b) = 1= 0.555b – 33
b = 61.26 mm. Therefore,
0
F ( x) 0.555 x 33
1
x 59.46 mm
59.46 x 61.26 mm
x 61.26 mm
The PDF is dF/dx, thus the range numbers are:
0.555
f ( x)
0
59.46 x 61.26 mm
otherwise
Ans.
From the range numbers,
59.46 61.26
60.36 mm
2
61.26 59.46
ˆ x
0.520 mm
2 3
x
Ans.
Ans.
Chapter 20, Page 5/29
(b) is an uncorrelated quotient F 3600 lbf, A 0.112 in 2
CF 300 3600 0.083 33, C A 0.001 0.112 0.008 929
From Table 20-6, For
F 3600
32 143 psi
A 0.112
Ans.
0.083332 0.0089292
ˆ 32 143
1 0.0089292
C 2694 / 32 143 0.0838 Ans.
1/2
2694 psi
Ans.
Since F and A are lognormal, division is closed and is lognormal too.
= LN(32 143, 2694) psi
Ans.
______________________________________________________________________________
20-8
Cramer’s rule
a1
y
xy
x
x
x
x
x
x
2
2
a2
2
x
0
0
0
1
1
1
3
a1 =
y
0.01
0.15
0.25
0.25
0.17
0.01
0.82
x
x y x xy x
x
x x x
x
y
xy y xy y x
x x x x
x
2
3
3
2
3
2
2
2
Ans.
3
2
2
3
2
2
Ans.
3
x2
x3
xy
0
0.04
0.16
0.36
0.64
1.00
0
0.008
0.064
0.216
0.512
1.000
2.20
1.800
0
0.030
0.100
0.150
0.136
0.010
0.406
1.040 714
a 2 = 1.046 43
Ans.
Chapter 20, Page 6/29
Data
x
Regression
y
y
0.01
0
0.15 0.166 286
0.25 0.248 857
0.25 0.247 714
0.17 0.162 857
-0.01 -0.005 710
0
0.2
0.4
0.6
0.8
1.0
______________________________________________________________________________
20-9
Data
Su
0
60
64
65
82
101
119
120
130
134
145
180
S e
30
48
29.5
45
51
50
48
67
60
64
84
Regression
S e
Su2
Su S e
20.356 75
39.080 78
40.329 05
40.641 12
45.946 26
51.875 54
57.492 75
57.804 81
60.925 48
62.173 75
65.606 49
76.528 84
3 600
4 096
4 225
6 724
10 201
14 161
14 400
16 900
17 956
21 025
32 400
1 800
3 072
1 917.5
3 690
5 151
5 950
5 760
8 710
8 040
9 280
15 120
Chapter 20, Page 7/29
195
205
207
210
213
225
225
227
230
238
242
265
280
295
325
325
355
5462
78
96
87
87
75
99
87
116
105
109
106
105
96
99
114
117
122
2274.5
m = 0.312 067,
81.209 85
84.330 52
84.954 66
85.890 86
86.827 06
90.571 87
90.571 87
91.196 00
92.132 20
94.628 74
95.877 01
103.054 60
107.735 60
112.416 60
121.778 60
121.778 60
131.140 60
b = 20.356 75
38 025
42 025
42 849
44 100
45 369
50 625
50 625
51 529
52 900
56 644
58 564
70 225
78 400
87 025
105 625
105 625
126 025
1 251 868
15 210
19 680
18 009
18 270
15 975
22 275
19 575
26 332
24 150
25 942
25 652
27 825
26 880
29 205
37 050
38 025
43 310
501 855.5
Ans.
______________________________________________________________________________
20-10
y a0 a2 x 2
2
2 y a0 a2 x 2 0
ao
Chapter 20, Page 8/29
y na
0
a2 x 2 0
y na
0
a2 x 2
2 y a0 a2 x 2 2 x 0 xy a0 x a2 x3
a2
Cramer’s rule
y x2
xy x3 x 3 y x 2 xy
ao
n x3 x x 2
n
x2
x x
n
y
x xy
n
x
x x
Ans.
3
a1
2
n xy x y
n x3 x x 2
3
x
Data Regression
y
y
20
40
60
80
200
19
17
13
7
56
19.2
16.8
12.8
7.2
x2
400
1600
3600
6400
12 000
x3
8 000
64 000
216 000
512 000
800 000
xy
380
680
780
560
2400
800 000(56) 12 000(2400)
20
4(800 000) 200(12 000)
4(2400) 200(56)
0.002
a1
4(800 000) 200(12 000)
a0
______________________________________________________________________________
Chapter 20, Page 9/29
20-11
Data
x
0.2
0.4
0.6
0.8
1
2
5
Regression
y
7.1
10.3
12.1
13.8
16.2
25.2
84.7
y
7.931 803
9.884 918
11.838 032
13.791 147
15.744 262
25.509 836
mˆ k
x2
y2
0.04
50.41
0.16 106.09
0.36 146.41
0.64 190.44
1 262.44
4 635.04
6.2 1390.83
6 90.44 5 84.7
6 6.2 5
2
xy
1.42
4.12
7.26
11.04
16.2
50.4
90.44
xx
-0.633 333
-0.433 333
-0.233 333
-0.033 333
0.166 666
1.166 666
0
x x
2
0.401 111 111
0.187 777 778
0.054 444 444
0.001 111 111
0.027 777 778
1.361 111 111
2.033 333 333
9.7656
84.7 9.7656(5)
bˆ Fi
5.9787
6
(a)
5
x ;
6
Eq. (20-37):
y
84.7
14.117
6
1390.83 5.9787(84.7) 9.7656(90.44)
62
0.556
s yx
Eq. (20-36):
1 5 6
sbˆ 0.556
0.3964 lbf
6 2.0333
Fi 5.9787, 0.3964 lbf Ans.
2
Chapter 20, Page 10/29
(b) Eq. (20-35):
0.556
0.3899 lbf/in
2.0333
k (9.7656, 0.3899) lbf/in Ans.
______________________________________________________________________________
smˆ
20-12 The expression = / l is of the form x / y. Now = (0.0015, 0.000 092) in, unspecified
distribution; and l = (2, 000, 0.008 1) in, unspecified distribution;
C x = 0.000 092 / 0.0015 = 0.0613
C y = 0.0081 / 2.000 = 0.004 05
Table 20-6:
0.0015 / 2.000 0.000 75
0.06132 0.004 052
ˆ 0.000 75
2
1 0.004 05
1/2
4.607 105 0.000 046
We can predict and ˆ but not the distribution of .
______________________________________________________________________________
20-13 = E
= (0.0005, 0.000 034), distribution unspecified; E = (29.5, 0.885) Mpsi, distribution
unspecified;
C x = 0.000 034 / 0.0005 = 0.068
C y = 0.0885 / 29.5 = 0.03
is of the form xy
Table 20-6:
E 0.0005(29.5)106 14 750 psi
ˆ 14 750 0.0682 0.0302 0.0682 0.0302
1/ 2
1096.7 psi
C 1096.7 /14 750 0.074 35
______________________________________________________________________________
20-14
Fl
AE
where F = (14.7, 1.3) kip, A = (0.226, 0.003) in2, l = (1.5, 0.004) in, and
E = (29.5, 0.885) Mpsi, distributions unspecified.
Chapter 20, Page 11/29
C F = 1.3 / 14.7 = 0.0884; C A = 0.003 / 0.226 = 0.0133; C l = 0.004 / 1.5 = 0.00267;
C E =0.885 / 29.5 = 0.03
Table 20-6:
Fl
1 1
Fl
AE
A E
F l 1 A 1/ E F l 1 A 1 E
1
1
0.003 31 in.
14 700(1.5)
0.226 29.5 106
Ans.
For the standard deviation, using the first-order terms in Table 20-6,
12
12
Fl
ˆ
CF2 Cl2 C A2 CE2 CF2 Cl2 C A2 CE2
AE
ˆ 0.003 31 0.08442 0.002 67 2 0.01332 0.032
12
0.000 313 in
COV:
Ans.
C ˆ / 0.000 313 / 0.003 31 0.0945
Ans.
Force COV dominates. There is no distributional information on .
______________________________________________________________________________
20-15 M = (15 000, 1350) lbf ⋅ in, distribution unspecified; d = (2.00, 0.005) in, distribution
unspecified.
32M
d3
C M = 1350 / 15 000 = 0.09, C d = 0.005 / 2.00 = 0.0025
is of the form x/y3, Table 20-6.
Mean:
M 15 000 lbf in
1
1
1
2
2
3
3 3 1 6Cx 3 1 6 0.0025 0.125 in *
d
d
2
1
1
* Note: 3 3
d d
Chapter 20, Page 12/29
32M 32(15 000)
(0.125)
d 3
19 099 psi Ans.
Standard Deviation:
12
ˆ CM2 Cd2 / 1 Cd2
Table 20-6:
3
3
Cd 3 3Cd 3(0.0025) 0.0075
ˆ CM2 3Cd
2
1 3C
2
12
d
19 099 0.092 0.00752
1725 psi Ans.
1 0.0075
2
12
COV:
C
1725
0.0903
19 099
Ans.
Stress COV dominates. No information of distribution of .
______________________________________________________________________________
20-16
Fraction discarded is +. The area under the PDF was unity. Having discarded +
fraction, the ordinates to the truncated PDF are multiplied by a.
a
1
1
New PDF, g(x), is given by
f ( x) 1
g ( x)
0
x1 x x2
otherwise
A more formal proof: g(x) has the property
Chapter 20, Page 13/29
1 g x dx a
x2
x2
x1
x1
f x dx
x1
1 a f x dx f x dx f x dx
0
x2
1 a 1 F ( x1 ) 1 F ( x2 )
1
1
1
F ( x2 ) F ( x1 ) 1 1
______________________________________________________________________________
a
20-17 (a) d = U(0.748, 0.751)
0.751 0.748
0.7495 in
d
2
0.751 0.748
0.000 866 in
ˆ d
2 3
1
1
f x
333.3 in 1
b a 0.751 0.748
x 0.748
F ( x)
333.3( x 0.748)
0.751 0.748
(b)
F(x 1 ) = F(0.748) = 0
F(x 2 ) = (0.750 – 0.748) 333.3 = 0.6667
If g(x) is truncated, PDF becomes
f ( x)
333.3
500 in 1
F ( x2 ) F ( x1 ) 0.6667 0
a b 0.748 0.750
x
0.749 in
2
2
b a 0.750 0.748
ˆ x
0.000 577 in
2 3
2 3
______________________________________________________________________________
g ( x)
20-18 From Table A-10, 8.1% corresponds to z 1 = 1.4 and 5.5% corresponds to z 2 = +1.6.
k1 z1ˆ
k2 z2ˆ
From which
z2 k1 z1k2 1.6(9) (1.4)11
z2 z1
1.6 (1.4)
9.933
Chapter 20, Page 14/29
ˆ
k2 k1
11 9
0.6667
z2 z1 1.6 (1.4)
The original density function is
1 k 9.933 2
1
exp
f (k )
Ans.
0.6667 2
2 0.6667
______________________________________________________________________________
20-19 From Prob. 20-1, = 122.9 kcycles and ̂ = 30.3 kcycles.
x10 x10 122.9
ˆ
30.3
x10 122.9 30.3 z10
From Table A-10, for 10 percent failure, z 10 = 1.282
z10
x 10 = 122.9 + 30.3(1.282)
= 84.1 kcycles Ans.
______________________________________________________________________________
20-20
f (x)
x
f
fx
f x2
f / (Nw)
60 2 120
7200 0.002899 0.000399
70 1
70
4900 0.001449 0.001206
80 3 240 19200 0.004348 0.003009
90 5 450 40500 0.007246 0.006204
100 8 800 80000 0.011594 0.010567
110 12 1320 145200 0.017391 0.014871
120 6 720 86400 0.008696 0.017292
130 10 1300 169000 0.014493 0.016612
140 8 1120 156800 0.011594 0.013185
150 5 750 112500 0.007246 0.008647
160 2 320 51200 0.002899 0.004685
170 3 510 86700 0.004348 0.002097
180 2 360 64800 0.002899 0.000776
190 1 190 36100 0.001449 0.000237
200 0
0
0
0 5.98E-05
210 1 210 44100 0.001449 1.25E-05
69 8480
x = 122.8986
s x = 22.887 19
Chapter 20, Page 15/29
Eq. (20-14):
1 x 2
x
f ( x)
exp
ˆ x 2
2 ˆ x
1 x 122.8986 2
1
exp
22.88719 2
2 22.88719
1
x
55
55
65
65
75
75
85
85
95
95
105
105
115
115
125
125
135
135
f / (Nw)
0
0.002 899
0.002 899
0.001 449
0.001 449
0.004 348
0.004 348
0.007 246
0.007 246
0.011 594
0.011 594
0.017 391
0.017 391
0.008 696
0.008 696
0.014 493
0.014 493
0.011 594
f (x)
0.000 214
0.000 214
0.000 711
0.000 711
0.001 951
0.001 951
0.004 425
0.004 425
0.008 292
0.008 292
0.012 839
0.012 839
0.016 423
0.016 423
0.017 357
0.017 357
0.015 157
0.015 157
x
145
145
155
155
165
165
175
175
185
185
195
195
205
205
215
215
f / (Nw)
0.011 594
0.007 246
0.007 246
0.002 899
0.002 899
0.004 348
0.004 348
0.002 899
0.002 899
0.001 449
0.001 449
0
0
0.001 499
0.001 499
0
f (x)
0.010 935
0.010 935
0.006 518
0.006 518
0.002 21
0.003 21
0.001 306
0.001 306
0.000 439
0.000 439
0.000 122
0.000 122
2.8E-05
2.8E-05
5.31E-06
5.31E-06
______________________________________________________________________________
Chapter 20, Page 16/29
20-21
x
174
182
190
198
206
214
222
1386
f
6
9
44
67
53
12
6
197
x = 198.6091
x
170
170
178
178
186
186
194
194
202
202
210
210
218
218
226
226
f / (Nw)
0
0.003807
0.003807
0.005711
0.005711
0.027919
0.027919
0.042513
0.042513
0.033629
0.033629
0.007614
0.007614
0.003807
0.003807
0
fx
1044
1638
8360
13266
10918
2568
1332
39126
f x2
181656
298116
1588400
2626668
2249108
549552
295704
7789204
f / (Nw)
0.003807
0.005711
0.027919
0.042513
0.033629
0.007614
0.003807
f (x)
0.001642
0.009485
0.027742
0.041068
0.030773
0.011671
0.002241
s x = 9.695 071
f (x)
0.000529
0.000529
0.004297
0.004297
0.017663
0.017663
0.036752
0.036752
0.038708
0.038708
0.020635
0.020635
0.005568
0.005568
0.00076
0.00076
______________________________________________________________________________
20-22
x
64
68
72
76
80
84
88
92
624
f
2
6
6
9
19
10
4
2
58
fx
128
408
432
684
1520
840
352
184
4548
f x2
8192
27744
31104
51984
121600
70560
30976
16928
359088
f / (Nw)
0.008621
0.025862
0.025862
0.038793
0.081897
0.043103
0.017241
0.008621
f (x)
0.00548
0.017299
0.037705
0.056742
0.058959
0.042298
0.020952
0.007165
Chapter 20, Page 17/29
x = 78.041379
x
62
62
66
66
70
70
74
74
78
78
82
82
86
86
90
90
94
94
f / (Nw)
s x = 6.572 229
f (x)
0.002684
0.002684
0.010197
0.010197
0.026749
0.026749
0.048446
0.048446
0.060581
0.060581
0.052305
0.052305
0.03118
0.03118
0.012833
0.012833
0.003647
0.003647
0
0.008621
0.008621
0.025862
0.025862
0.025862
0.025862
0.038793
0.038793
0.0381897
0.081897
0.043103
0.043103
0.017241
0.017241
0.008621
0.008621
0
______________________________________________________________________________
20-23
4 P 4(40)
50.93 kpsi
d 2 12
ˆ
4ˆ P 4(8.5)
10.82 kpsi
d 2 12
ˆ S 5.9 kpsi
y
For no yield, m = S y 0
z
m m 0 m
m
ˆ m
ˆ m
ˆ m
m S y 78.4 50.93 27.47 kpsi
ˆ m ˆ2 S2
z
y
10.82
12
2
5.92 12.32 kpsi
2
m
27.47
2.230
ˆ m
12.32
Table A-10, p f = 0.0129
R = 1 – p f = 1 – 0.0129 = 0.987 Ans.
______________________________________________________________________________
Chapter 20, Page 18/29
20-24 For a lognormal distribution,
Eq. (20-18) y ln x ln 1 C x2
Eq. (20-19) ˆ y ln 1 C x2
From Prob. (20-23)
m S y x
y ln S y ln 1 CS2 ln ln 1 C2
y
S 1 C2
ln y
2
1 CS y
ˆ y ln 1 CS2 ln 1 C2
y
12
ln 1 CS2y 1 C2
S 1 C2
ln y
1 CS2y
z
ˆ
2
ln 1 CS y 1 C2
4 P 4(30)
38.197 kpsi
d 2 12
ˆ
4ˆ P 4(5.1)
6.494 kpsi
d 2 12
6.494
0.1700
38.197
3.81
CS y
0.076 81
49.6
49.6
1 0.1702
ln
2
38.197 1 0.07681
1.470
z
2
2
ln 1 0.076 81 1 0.170
C
Table A-10
p f = 0.0708
R = 1 – p f = 0.929 Ans.
______________________________________________________________________________
Chapter 20, Page 19/29
20-25
x
93
95
97
99
101
103
105
107
109
111
n
19
25
38
17
12
10
5
4
4
2
136
nx2
nx
1767
164 311
2375
225 625
3686
357 542
1683
166 617
1212
122 412
1030
106 090
525
55 125
428
45 796
436
47 524
222
24 642
13 364 1 315 704
x = 13 364/136 = 98.26 kpsi
12
1 315 704 13 364 2 / 136
sx
136 1
4.30 kpsi
Under normal hypothesis,
z0.01 x0.01 98.26 / 4.30
x0.01 98.26 4.30 z0.01
98.26 4.30 2.3267
88.26 88.3 kpsi
Ans.
______________________________________________________________________________
20-26 From Prob. 20.25, x = 98.26 kpsi, and ˆ x 4.30 kpsi.
C x ˆ x / x 4.30 / 98.26 0.043 76
From Eqs. (20-18) and (20-19),
y ln 98.26 0.043 762 / 2 4.587
ˆ y ln 1 0.043 762 0.043 74
For a yield strength exceeded by 99% of the population,
z0.01 ln x0.01 y / ˆ y ln x0.01 y ˆ y z0.01
From Table A-10, for 1% failure, z 0.01 = 2.326. Thus,
Chapter 20, Page 20/29
ln x0.01 4.587 0.043 74 2.326 4.485
x0.01 88.7 kpsi
Ans.
The normal PDF is given by Eq. (20-14) as
1 x 98.26 2
1
f x
exp
4.30 2
2 4.30
For the lognormal distribution, from Eq. (20-17), defining g(x),
1 ln x 4.587 2
1
g x
exp
x 0.043 74 2
2 0.043 74
x(kpsi)
92
92
94
94
96
96
98
98
100
100
102
f / (Nw)
0.000 00
0.069 85
0.069 85
0.091 91
0.091 91
0.139 71
0.139 71
0.062 50
0.062 50
0.044 12
0.044 12
f (x)
0.032 15
0.032 15
0.056 80
0.056 80
0.080 81
0.080 81
0.092 61
0.092 61
0.085 48
0.085 48
0.063 56
g(x)
0.032 63
0.032 63
0.058 90
0.058 90
0.083 08
0.083 08
0.092 97
0.092 97
0.083 67
0.083 67
0.061 34
x(kpsi)
102
104
104
106
106
108
108
110
110
112
112
f / (Nw)
0.036 76
0.036 76
0.018 38
0.018 38
0.014 71
0.014 71
0.014 71
0.014 71
0.007 35
0.007 35
0.000 00
f (x)
0.063 56
0.038 06
0.038 06
0.018 36
0.018 36
0.007 13
0.007 13
0.002 23
0.002 23
0.000 56
0.000 56
g(x)
0.061 34
0.037 08
0.037 08
0.018 69
0.018 69
0.007 93
0.007 93
0.002 86
0.002 86
0.000 89
0.000 89
Note: rows are repeated to draw histogram
The normal and lognormal are almost the same. However, the data is quite skewed and
perhaps a Weibull distribution should be explored. For a method of establishing the
Chapter 20, Page 21/29
Weibull parameters see Shigley, J. E., and C. R. Mishke, Mechanical Engineering
Design, McGraw-Hill, 5th ed., 1989, Sec. 4-12.
_____________________________________________________________________________
20-27 x Sf e
104
Eq. (20-28):
x 0 = 79 kpsi, = 86.2 kpsi, b = 2.6
x x0 x0 (1 1 b)
79 (86.2 79)(1 1 2.6)
79 7.2(1.38)
From Table A-34, Γ (1.38) = 0.888 54
x 79 7.2(0.888 54) 85.4 kpsi
Ans.
Eq. (20-29)
ˆ x x0 1 2 b 2 1 1 b
12
86.2 79 1 2 2.6 2 1 1 2.6
7.2 0.923 76 0.888 542
12
12
2.64 kpsi Ans.
ˆ
2.64
Cx x
0.031 Ans.
x 85.4
_____________________________________________________________________________
20-28 x = S ut
x 0 = 27.7 kpsi, = 46.2 kpsi, b = 4.38
x 27.7 46.2 27.7 1 1 4.38
27.7 18.5 (1.23)
27.7 18.5(0.910 75)
44.55 kpsi Ans.
ˆ x 46.2 27.7 1 2 4.38 2 1 1 4.38
12
12
18.5 (1.46) 2 (1.23)
12
18.5 0.8856 0.920 752
4.38 kpsi Ans.
4.38
Cx
0.098 Ans.
44.55
From the Weibull survival equation
Chapter 20, Page 22/29
x x b
0
R exp
1 p
x
0
x x b
R40 exp 40 0 1 p40
x0
40 27.7 4.38
exp
0.846
46.2 27.7
p40 1 R40 1 0.846 0.154 15.4% Ans.
_____________________________________________________________________________
20-29 x = S ut ,
x 0 = 151.9 kpsi, = 193.6 kpsi, b = 8
x 151.9 193.6 151.9 1 1 8
151.9 41.7 (1.125)
151.9 41.7(0.941 76)
191.2 kpsi Ans.
ˆ x 193.6 151.9 1 2 8 2 1 1 8
12
12
41.7 (1.25) 2 (1.125)
12
41.7 0.906 40 0.941 762
5.82 kpsi Ans.
5.82
Cx
0.030
191.2
_____________________________________________________________________________
20-30 x = S ut ,
x 0 = 47.6 kpsi, = 125.6 kpsi, b = 11.4
x 47.6 125.6 47.6 1 1 11.84
47.6 78 1.08
47.6 78 0.959 73 122.5 kpsi
ˆ x 125.6 47.6 1 2 11.84 2 1 1 11.84
12
12
78 (1.08) 2 (1.17)
12
78 0.959 73 0.936 702
22.4 kpsi
From Prob. 20-28,
x x b
100 47.6 11.84
0
1
exp
p 1 exp
0.0090
0
125.6 47.6
Ans.
Chapter 20, Page 23/29
y = Sy,
y 0 = 64.1 kpsi, = 81.0 kpsi, b = 3.77
y 64.1 (81.0 64.1) 1 1 3.77
64.1 16.9 (1.27)
64.1 16.9(0.902 50) 79.35 kpsi
y (81 64.1) (1 2 3.77) 1 1 3.77
12
16.9 0.887 57 0.902 502 4.57 kpsi
y y 3.77
0
p 1 exp
y0
70 64.1 3.77
1 exp
0.019 Ans.
81 64.1
_____________________________________________________________________________
12
20-31 x = S ut = W[122.3, 134.6, 3.64] kpsi, p(x > 120) = 1 = 100% since x 0 > 120 kpsi
133 122.3 3.64
p(x 133) exp
134.6 122.3
0.548 54.8% Ans.
_____________________________________________________________________________
20-32 Using Eqs. (20-28) and (20-29) and Table A-34,
n n0 n0 1 1 b 36.9 133.6 36.9 1 1 2.66
=122.85 kcycles
ˆ n n0 1 2 b 2 1 1 b 34.79 kcycles
For the Weibull density function, Eq. (20-27),
n 36.9 2.66
exp
133.6 36.9
For the lognormal distribution, Eqs. (20-18) and (20-19) give,
2.66
n 36.9
fW (n)
133.6 36.9 133.6 36.9
2.66 1
y ln 122.85 34.79 122.85 2 4.771
2
2
ˆ y 1 34.79 122.85 0.2778
From Eq. (20-17), the lognormal PDF is
Chapter 20, Page 24/29
1 ln n 4.771 2
1
exp
0.2778n 2
2 0.2778
We form a table of densities f W (n) and f LN (n) and plot.
f LN n
n(kcycles)
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
f W (n)
9.1E-05
0.000 991
0.002 498
0.004 380
0.006 401
0.008 301
0.009 822
0.010 750
0.010 965
0.010 459
0.009 346
0.007 827
0.006 139
0.004 507
0.003 092
0.001 979
0.001 180
0.000 654
0.000 336
f LN (n)
1.82E-05
0.000 241
0.001 233
0.003 501
0.006 739
0.009 913
0.012 022
0.012 644
0.011 947
0.010 399
0.008 492
0.006 597
0.004 926
0.003 564
0.002 515
0.001 739
0.001 184
0.000 795
0.000 529
The Weibull L10 life comes from Eq. (20-26) with reliability of R = 0.90. Thus,
n0.10 36.9 133.6 36.9 ln 1 0.90
1 2.66
78.4 kcycles
Ans.
The lognormal L10 life comes from the definition of the z variable. That is,
Chapter 20, Page 25/29
ln n0 y ˆ y z
or
n0 exp y ˆ y z
From Table A-10, for R = 0.90, z = 1.282. Thus,
n0 exp 4.771 0.2778 1.282 82.7 kcycles
Ans.
_____________________________________________________________________________
20-33 Form a table
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x
(105)L
3.05
3.55
4.05
4.55
5.05
5.55
6.05
6.55
7.05
7.55
8.05
8.55
9.05
9.55
10.05
fi
3
7
11
16
21
13
13
6
2
0
4
3
0
0
1
100
f i x⋅(105)
9.15
24.85
44.55
72.80
106.05
72.15
78.65
39.30
14.10
0
32.20
25.65
0
0
10.05
529.50
f i x2⋅(1010)
27.9075
88.2175
180.4275
331.24
535.5525
400.4325
475.8325
257.415
99.405
0
259.21
219.3075
0
0
101.0025
2975.95
g(x)⋅(105)
0.0557
0.1474
0.2514
0.3168
0.3216
0.2789
0.2151
0.1517
0.1000
0.0625
0.0375
0.0218
0.0124
0.0069
0.0038
Chapter 20, Page 26/29
x 529.5 105 100 5.295 105 cycles
Ans.
12
2975.95 1010 529.5 105 2 100
sx
100 1
1.319 105 cycles
Ans.
Cx s x 1.319 5.295 0.249
y ln 5.295 105 0.2492 2 13.149
ˆ y ln 1 0.2492 0.245
g ( x)
1
xˆ y
1 ln x
y
exp
2 ˆ y
2
2
1 ln x 13.149 2
1.628
exp
x
0.245
2
_____________________________________________________________________________
20-34 X = S u = W[70.3, 84.4, 2.01]
x 70.3 84.4 70.3 1 1 2.01
Eq. (2-28):
70.3 (84.4 70.3) 1.498
82.8 kpsi
Ans.
Chapter 20, Page 27/29
12
ˆ x (84.4 70.3) (1 2 2.01) 2 (1 1 2.01)
12
Eq. (2-29):
ˆ x 14.1 0.997 91 0.886 17 2
6.502 kpsi
6.502
0.079 Ans.
82.8
_____________________________________________________________________________
Cx
20-35 Take the Weibull equation for the standard deviation
ˆ x x0 (1 2 b) 2 (1 1 b)
12
and the mean equation solved for x x0
x x0 x0 1 1 b
and divide the first by the second,
1 2 b 2 1 1 b
1 1 b
x x0
ˆ x
12
1 2 b
4.2
1 R 0.2763
49 33.8
2 1 1 b
Make a table and solve for b iteratively
b
3
4
4.1
1 + 2/b
1.67
1.5
1.49
1 + 1/b 1 2 b
1.33
0.903 30
1.25
0.886 23
1.24
0.885 95
1 1 b
0.893 38
0.906 40
0.908 52
R
0.363
0.280
0.271
b 4.068 Using MathCad Ans.
x x0
49 33.8
33.8
x0
(1 1/ b)
1 1/ 4.068
49.8 kpsi Ans.
_____________________________________________________________________________
20-36 x = S y = W[34.7, 39, 2.93] kpsi
Chapter 20, Page 28/29
x 34.7 39 34.7 1 1 2.93 34.7 4.3 1.34
34.7 4.3 0.892 22 38.5 kpsi
ˆ x 39 34.7 1 2 2.93 2 1 1 2.93
12
4.3 1.68 2 1.34
12
12
4.3 0.905 00 0.892 222 1.42 kpsi Ans.
Cx 1.42 38.5 0.037 Ans.
_____________________________________________________________________________
20-37
x (Mrev)
1
2
3
4
5
6
7
8
9
10
11
12
Sum
78
f
fx
11
22
38
57
31
19
15
12
11
9
7
5
237
11
44
114
228
155
114
105
96
99
90
77
60
1193
f x2
11
88
342
912
775
684
735
768
891
900
847
720
7673
x 1193 106 / 237 5.034 106 cycles
2
7673 1012 1193 106 / 237
ˆ x
2.658 106 cycles
237 1
C x 2.658 / 5.034 0.528
From Eqs. (20-18) and (20-19),
y ln 5.034 106 0.5282 / 2 15.292
ˆ y ln 1 0.5282 0.496
From Eq. (20-17), defining g(x),
1 ln x 15.292 2
1
g ( x)
exp
0.496
x 0.496 2
2
x (Mrev) f / (Nw)
0.5
0.000 00
g(x)(106)
0.000 11
Chapter 20, Page 29/29
0.5
1.5
1.5
2.5
2.5
3.5
3.5
4.5
4.5
5.5
5.5
6.5
6.5
7.5
7.5
8.5
8.5
9.5
9.5
10.5
10.5
11.5
11.5
12.5
12.5
0.046414
0.046414
0.092827
0.092827
0.160338
0.160338
0.240506
0.240506
0.130802
0.130802
0.080 17
0.080 17
0.063 29
0.063 29
0.050 63
0.050 63
0.046 41
0.046 41
0.037 97
0.037 97
0.029 54
0.029 54
0.021 10
0.021 10
0.000 00
0.000 11
0.052 03
0.052 03
0.169 92
0.169 92
0.207 54
0.207 54
0.178 47
0.178 47
0.131 58
0.13158
0.090 11
0.090 11
0.059 53
0.059 53
0.038 69
0.038 69
0.025 01
0.025 01
0.016 18
0.016 18
0.010 51
0.010 51
0.006 87
0.006 87
z
ln x y
ln x y ˆ y z 15.292 0.496 z
ˆ y
L 10 life, where 10% of bearings fail, from Table A-10,
z = 1.282. Thus,
ln x = 15.292 + 0.496(1.282) = 14.66
x = 2.33 (106) rev Ans.
Chapter 20, Page 30/29
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