Contents AISC LRFD 93 Example 002

User Manual: AISC LRFD-93 Example 002

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AISC LRFD-93 Example 002
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BIAXIAL BENDING
EXAMPLE DESCRIPTION
A check of the column adequacy is checked for combined axial compression and
flexural loads. The column is 14 feet tall and loaded with an axial load,
Pu = 1400 kips and bending, M ux , M uy = 200k-ft and 70k-ft, respectively. It is
assumed that there is reverse-curvature bending with equal end moments about
both axes and no loads along the member. The column demand/capacity ratio is
checked against the results of Example 6.2 in the 3rd Edition, LRFD Manual of
Steel Construction, pages 6-6 to 6-8.
GEOMETRY, PROPERTIES AND LOADING

Member Properties
W14X176
E = 29000 ksi
Fy = 50 ksi

Loading
Pu = 1,400 kips
Mux = 200 kip-ft
Muy = 70 kip-ft

Geometry
H = 14.0 ft

TECHNICAL FEATURES TESTED
 Section compactness check (compression)
 Member compression capacity
 Member bending capacity
 Demand/capacity ratio, D/C
AISC LRFD-93 Example 002 - 1

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RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
Example 6.2 in the 3rd Edition, LRFD Manual of Steel Construction, pages 6-6 to
6-8.

SAP2000

Independent

Percent
Difference

Compactness

Compact

Compact

0.00%

φc Pn (kips)

1937.84

1937.84

0.00%

φb M nx (k-ft)

1200

1200

0.00%

φb M ny (k-ft)

600.478

600.478

0.00%

0.974

0.974

0.00%

Output Parameter

D/C
COMPUTER FILE: AISC LRFD-93 EX002

CONCLUSION
The results show an exact comparison with the independent results.

AISC LRFD-93 Example 002 - 2

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HAND CALCULATION
Properties:
Material: ASTM A992 Grade 50 Steel
Fy = 50 ksi, E = 29,000 ksi
Section: W14x176
A = 51.8 in2,
bf = 15.7 in, tf = 1.31 in, d = 15.2 in, tw = 0.83 in
hc = d − 2t f = 15.2 − 2 • 1.31 = 12.58 in
Ix = 2,140 in4, Iy = 838 in4, rx = 6.4275 in, ry = 4.0221 in
Sx = 281.579 in3, Sy = 106.7516 in3, Zx = 320.0 in3, Zy = 163.0 in3.
Member:
Kx = Ky = 1.0
L = Lb = 14 ft
Other
φc =0.85
φb =0.9
Loadings:
Pu = 1400 kips
Mux = 200 k-ft
Muy = 70 k-ft
Section Compactness:
Localized Buckling for Flange:
/ 2)
( b=
(15.7
f / 2)
=
λ
= 5.99
tf
1.31
λ=
p

65
=
Fy

65
= 9.19
50

λ < λ p , No localized flange buckling
Flange is Compact.

AISC LRFD-93 Example 002 - 3

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Localized Buckling for Web:
h 12.58
λ= c=
= 15.16
tw 0.83
φb Py =
φb Ag Fy =
0.9 • 51.8 • 50 =
2331 kips

Pu
1400
=
= 0.601
φb Py 2331
Pu
Since =
0.601 > 0.125
φb Py
=
λp

191
Fy


P
 2.33 − u
φb Py


 253
 ≥
Fy


191
253
=
( 2.33 − 0.601
) 46.714 ≥ = 35.780
50
50
λ < λ p , No localized web buckling
Web is Compact.

=
λp

Section is Compact.
Member Compression Capacity:
For braced frames, K = 1.0 and KxLx = KyLy = 14.0 ft, From AISC Table 4-2,
φc Pn =
1940 kips
Or by hand,
=
λc

K y L Fy 1.0 • 14 • 12
50
=
= 0.552
ry π E
4.022 • π 29000

Since λ c < 1.5,

(

Fcr =
Fy 0.658λc

2

50 • 0.658
)=

0.5522

=
44.012 ksi

φc Pn =
φc Fcr Ag =
0.85 • 44.012 • 51.8

φc Pn =
1937.84 kips

AISC LRFD-93 Example 002 - 4

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From LRFD Specification Section H1.2,
Pu
1400
=
= 0.722 > 0.2
φc Pn 1937.84

Therefore, LRFD Specification Equation H1-1a governs.
Section Bending Capacity
50 • 310
=
= 1333.333 k-ft
M
F=
px
yZx
12
M py = Fy Z y
Zy
163
However,=
= 1.527 > 1.5,
S y 106.7516
So
Zy =
1.5 S y =
1.5 • 106.7516 =
160.1274in 3
=
M py

50 • 160.1274
= 667.198 k-ft
12

Member Bending Capacity
From LRFD Specification Equation F1-4,
L p = 1.76ry

E
Fyf

1.76 • 4.02
L=
p

29000 1
• = 14.2 ft > L=
14 ft
b
12
50

φb M nx =
φb M px

φb M nx =0.9 • 1333.333
φb M nx =
1200 k-ft
φb M ny =
φb M py
φb M ny =
0.9 • 667.198

φb M ny =
600.478 k-ft

AISC LRFD-93 Example 002 - 5

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Interaction Capacity: Compression & Bending
From LRFD Specification section C1.2, for a braced frame, Mlt = 0.
M ux = B1x M ntx , where M ntx = 200 kip-ft; and

M uy = B1 y M nty , where M nty = 70 kip-ft
B1 =

Cm

P 
1 − u 
 Pe1 

≥1

For reverse curvature bending and equal end moments:
M1
= +1.0
M2

M 
C m = 0.6 − 0.4 1 
 M2 
C m = 0.6 − 0.4(1.0 ) = 0.2
pe1 =

=
pe1x

π2 EI

( KL )

2

π2 • 29000 • 2140
=
21, 702 kips
2
(14.0 • 12 )

π2 • 29000 • 838
=
8, 498
pe1 y =
2
(14.0 • 12 )
B1x =

C mx

≥1

Pu 
1 −

 Pe1x 
0.2
=
= 0.214 ≥ 1
B1x
1400 

1 −

 21702 

AISC LRFD-93 Example 002 - 6

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B1x = 1
C my

≥1


P
u

1 −
 P 
e1 y 

0.2
B1 y
=
= 0.239 ≥ 1
 1400 
1 −

 8498 
B1 y = 1
B1 y =

M ux = 1.0 • 200 = 200 kip-ft;
and
M uy = 1.0 • 70 = 70 kip-ft

From LRFD Specification Equation H1-1a,
1400 8  200
70 
+ 
+
0.974 < 1.0 , OK
=
1940 9  1200 600.478 
D
= 0.974
C

AISC LRFD-93 Example 002 - 7



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