Contents KBC 2009 Example 001

User Manual: KBC 2009 Example 001

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Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
KBC 2009 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The design flexural strengths are checked for the beam shown below. The beam
is loaded with a uniform load of 6.5 kN/m (D) and 11 kN/m (L). The flexural
moment capacity is checked for three unsupported lengths in the weak direction,
Lb = 1.75 m, 4 m and 12 m.
GEOMETRY, PROPERTIES AND LOADING
TECHNICAL FEATURES TESTED
Section Compactness Check (Bending)
Member Bending Capacities
Unsupported length factors
Member Properties
W460x74
E = 205,000 MPa
Fy = 345 MPa
Loading
w = 6.5 kN/m (D)
w = 11.0 kN/m (L)
Geometry
Span, L = 12m
KBC 2009 Example 001 - 1
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
RESULTS COMPARISON
Independent results are comparing with the results of ETABS.
Output Parameter
ETABS Independent Percent
Difference
Compactness Compact Compact 0.00%
Cb (
b
L
=1.75m) 1.004 1.002 0.20%
bn
M
φ
(
=1.75m) (kN-m) 515.43 515.43 0.00%
Cb (
b
L
=4m) 1.015 1.014 0.10%
bn
M
φ
(
=4m) (kN-m) 394.8 394.2 0.15%
Cb (
b
L
=12m) 1.136 1.136 0.00%
bn
M
φ
(
=12m) (kN-m) 113.48 113.45 0.03%
COMPUTER FILE: KBC 2009 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
KBC 2009 Example 001 - 2
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
HAND CALCULATION
Properties:
Material:
E = 205,000 MPa, Fy = 345 MPa
Section: W460x74
bf = 191 mm, tf = 14.5 mm, d = 457 mm, tw = 9 mm
2 457 2 14.5 428mm
f
hd t= − = −• =
0
457 14.5 442.5 mm
f
h dt
=−= − =
S33 = 1457.3 cm3, Z33 = 1660 cm3
Iy =1670 cm4, ry = 42 mm, Cw = 824296.4 cm6, J = 51.6 cm4
33
1670 824296.4 50.45mm
1457.3
yw
ts
IC
rS
•
= = =
0.1=
m
R
for doubly-symmetric sections
Other:
c = 1.0
L = 12 m
Loadings:
wu = (1.2wd + 1.6wl) = 1.2(6.5) + 1.6(11) = 25.4 kN/m
2
8
u
u
wL
M=
= 25.4∙122/8 = 457.2 kN-m
Section Compactness:
Localized Buckling for Flange:
191 6.586
2 2 14.5
f
f
b
t
λ
= = =
•
KBC 2009 Example 001 - 3
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
205,000
0.38 0.38 9.263
345
p
y
E
F
λ
= = =
p
λλ
<
, No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
428 47.56
9
w
h
t
λ
= = =
205,000
3.76 3.76 91.654
345
p
y
E
F
λ
= = =
p
λλ
<
, No localized web buckling
Web is Compact.
Section is Compact.
Section Bending Capacity:
33
345 1660 572.7kN-m
py
M FZ= =•=
Lateral-Torsional Buckling Parameters:
Critical Lengths:
205,000
1.76 1.76 42 1801.9 mm 1.8m
345
py
y
E
Lr
F
==•==
2
33
33
0.7
1.95 1 1 6.76
0.7
yo
r ts
yo
FSh
E Jc
Lr
F S h E Jc

= ++ 

2
205,000 51.6 1 0.7 345 1457.3 44.8
1.95 50.45 1 1 6.76
0.7 345 1457.3 44.25 205,000 51.6 1
r
L• ••

= • ++ 
•• •

KBC 2009 Example 001 - 4
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
5.25m
r
L=
Non-Uniform Moment Magnification Factor:
For the lateral-torsional buckling limit state, the non-uniform moment magnification factor is
calculated using the following equation:
0.3
3
435.2
5.12
max
max
≤
+++
=
m
CBA
b
R
M
MMM
M
C
Eqn. 1
Where MA = first quarter-span moment, MB = mid-span moment, MC = second quarter-span
moment.
The required moments for Eqn. 1 can be calculated as a percentage of the maximum mid-span
moment. Since the loading is uniform and the resulting moment is symmetric:
2
1
14
b
AC
L
MM L

= = − 

Member Bending Capacity for Lb = 1.75 m:
max 1.00
B
MM= =
22
1 1 1.75
1 1 0.995
4 4 12
b
AC
L
MM L
 
==−=− =

 

()
( ) ( ) ( ) ( )
12.5 1.00
2.5 1.00 3 0.995 4 1.00 3 0.995
b
C=+ ++
1.002
b
C=
pb
LL <
, Lateral-Torsional buckling capacity is as follows:
572.7kN-m
np
MM= =
0.9 572.7
bn
M
φ
= •
515.43 kN-m
bn
M
φ
=
KBC 2009 Example 001 - 5
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
Member Bending Capacity for Lb = 4 m:
max 1.00
B
MM= =
22
1 14
1 1 0.972
4 4 12
b
AC
L
MM L
 
==−=−=

 

( )
( ) ( ) ( ) ( )
12.5 1.00
2.5 1.00 3 0.972 4 1.00 3 0.972
b
C=+ ++
1.014
b
C=
rb
p
LL
L<
<
, Lateral-Torsional buckling capacity is as follows:
( )
p
pr
pb
yppbn M
LL
LL
SFMMCM ≤
−
−
−−= 33
7.0
( )
4.00 1.80
1.014 572.7 572.7 0.7 0.345 1457.3 437.97kN-m
5.25 1.80
n
M
−

= − −• • =


−


0.9 437.97
bn
M
φ
= •
394.2 kN-m
bn
M
φ
=
Member Bending Capacity for Lb = 12 m:
max 1.00
B
MM= =
22
1 1 12
1 1 0.750
4 4 12
b
AC
L
MM L
 
==−=−=

 

.
( )
() ( ) ( ) ( ) ( )
12.5 1.00 1.00
2.5 1.00 3 0.750 4 1.00 3 0.750
b
C=+ ++
1.136
b
C=
rb LL >
, Lateral-Torsional buckling capacity is as follows:
KBC 2009 Example 001 - 6
Software Verification
PROGRAM NAME:
ETABS
REVISION NO.:
0
2
2
2
1.136 205,000 51.6 1 12000
1 0.078 86.5MPa
1457.3 44.25 50.45
12000
50.45
cr
F
π
•• • 
=+=

•



pcrnMSFM ≤= 33
86.5 1457.3 126.056kN-m
n
M=•=
0.9 126.056
bn
M
φ
= •
113.45 kN-m
bn
M
φ
=
2
2
233
1 0.078
bb
cr
o ts
b
ts
CE L
Jc
FSh r
L
r
π

= + 
 


KBC 2009 Example 001 - 7

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