Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
CSA S16-19 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object moment and shear strength is tested in this example.
A simply supported beam is (a) laterally restrained along its full length, (b)
laterally restrained along its quarter points, at mid-span, and at the ends (c)
laterally restrained along mid-span, and is subjected to a uniform factored load of
DL = 7 kN/m and LL = 15 kN/m. This example was tested using the CSA S16-
19 steel frame design code. The moment and shear strengths are compared with
Handbook of Steel construction (9th Edition) results.
GEOMETRY, PROPERTIES AND LOADING
L = 8.0 m
Material Properties Loading Design Properties
E= 2x108 kN/m2 WD = 7 kN/m ASTM A992
Fy = 350 kN/m2 WL = 15 kN/m CSA G40.21 350W
W410X46
W410X60
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacity, Mr (fully restrained)
Member bending capacity, Mr (buckling)
Member bending capacity, Mr (LTB)
CSA S16-19 Example 001 - 1
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
RESULT COMPARISON
Independent results are taken from Examples 1, 2 and 3 on pages 5-84 and 5-85
of the Hand Book of Steel Construction to CSA S16-01 published by Canadian
Institute of Steel Construction.
Percent
Output Parameter SAP2000 Independent Difference
Compactness Class 1 Class 1 0.00%
Design Moment, Mf (kN-m) 250.0 250.0 0.00%
(a) Moment Capacity, Mr33 of 278.775 278.775 0.00%
W410X46 (kN-m) w/ lb = 0 m
(b) Moment Capacity, Mr33 of 268.97 268.89 0.03%
W410X46 (kN-m) w/ lb = 2 m
(c) Moment Capacity, Mr33 of 292.10 292.23 -0.04%
W410X60 (kN-m) w/ lb = 4 m
COMPUTER FILE: CSA S16-19 EX001
CONCLUSION
The results show an acceptable comparison with the independent results. The
minor differences are accounted for by the fact that the program uses a slightly
different value for Cw.
CSA S16-19 Example 001 - 2
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
HAND CALCULATION
Properties:
Material: CSA G40.21 Grade 350W
fy = 350 MPa
E = 200,000 MPa
G = 76923 MPa
Section: W410x46
bf = 140 mm, tf = 11.2 mm, d = 404 mm, tw = 7 mm
h = d − 2t f = 404 − 2 • 11.2 = 381.6 mm
Ag = 5890 mm2
I22 = 5,140,000 mm4
Z33 = 885,000 mm3
J = 192,000 mm4
Cw 1.976 • 1011 mm 6
=
Section: W410x60
bf = 178 mm, tf = 12.8 mm, d = 408 mm, tw = 7.7 mm
h = d − 2t f = 408 − 2 • 12.8 = 382.4 mm
Ag = 7580 mm2
I22 = 12,000,000 mm4
Z33= 1,190,000 mm3
J = 328,000 mm4
Cw 4.698 • 1011 mm 6
=
Member:
L=8m
Ф = 0.9
CSA S16-19 Example 001 - 3
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
Loadings:
wf = (1.25wd + 1.5wl) = 1.25(7) + 1.5(15) = 31.25 kN/m
w f L2 31.25 • 82
Mf
= =
8 8
M f = 250 kN-m
Section Compactness:
Localized Buckling for Flange:
145 145
λCl .1
= = = 7.75
Fy 350
W410x46
bf 140
λ
= = = 6.25
2t f 2 • 11.2
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
W410x60
bf 178
λ
= = = 6.95
2t f 2 • 12.8
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
Localized Buckling for Web:
1100 Cf 1100 0
λCl .1 = 1 − 0.39 = 1 − 0.39 =58.8
Fy Cy 350 5890 • 350
W410x46
h 381.6
λ
= = = 54.51
tw 7
CSA S16-19 Example 001 - 4
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
W410x60
h 382.4
λ
= = = 49.66
tw 7.7
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
Calculation of ω2:
ω2 is calculated from the moment profile so is independent of cross section and is
calculated as:
4 • M max
ω2 =
M max 2 + 4 M a 2 + 7 M b 2 + 4 M c 2
where: Mmax = maximum moment
Ma = moment at ¼ unrestrained span
Mb = moment at ½ unrestrained span
Mc = moment at ¾ unrestrained span
Section Bending Capacity for W410x46:
M p = Fy Z 33 = 350 • 885,000 / 10 6 = 309.75 kN-m
φM p = 0.9 • 309.75 = 278.775 kN-m
Member Bending Capacity for Lb = 0 mm (Fully Restrained):
Lb = 0, so Mmax = Ma = Mb = Mc = Mu = 250 kN-m and ω2 = 1.000
CSA S16-19 Example 001 - 5
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
2
ω2 π πE
=Mu EI 22GJ + I 22Cw → ∞ as L → 0
L L
M p 33
M r=
33 1.15φM p 33 1 − 0.28 ≤ φM p 33
Mu
M p 33
0.28 → 0 as M u → ∞
Mu
leading to M =
r 33 1.15 φ M p 33 > φ M p 33
So
M r 33 = 278.775 kN-m
φ M p 33 =
Member Bending Capacity for Lb = 2000 mm:
L − Lb Lb 8 − 2 2
M a @ x=
a + = + = 3.5 m
2 4 2 4
ω f Lxa ω f xa 2 31.25 • 8 • 3.5 31.25 • 3.52
Ma = − = − = 246.094 kN-m
2 2 2 2
Ma = Mc = 246.094 kN-m @ 3500 mm and 4500 mm
Mmax = Mb = 250 kN-m @ 4000 mm
4 • 250
ω2 = 1.008
250 + 4 • 246.0942 + 7 • 2502 + 4 • 246.0942
2
ω2 = 1.008
2
ω2 π πE
=Mu EI 22GJ + I 22Cw
L L
CSA S16-19 Example 001 - 6
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
( 2 •10 ) • ( 5.14 •10 ) • 76923 • (192 •10 )
5 6 3
1.008 • π
π ( 2 • 10 )
2
Mu = 5
2000 + ( 5.14 • 10 )(197.6 • 10 )
6 9
2000
M u 537.82 • 106 N-mm = 537.82 kN-m
=
0.67 M p = 0.67 • 309.75 = 208 < M u = 537.82 kN-m, so
M p 33
M r=
33 1.15φM p 33 1 − 0.28 ≤ φM p 33
Mu
309.75
M r 33 = 1.15 • 0.9 • 309.75 1 − 0.28 = 268.89 kN-m < 278.775 kN-m
537.82
M r 33 = 268.89 kN-m
Section Capacity for W410x60:
M p = Fy Z 33 = 350 • 1190,000 / 10 6 = 416.5 kN-m
φM p = 0.9 • 416.5 = 374.85 kN-m
Member Bending Capacity for Lb = 4000 m:
L − Lb Lb 8 − 4 4
M a @ x=
a + = + = 3m
2 4 2 4
ω f Lxa ω f xa 2 31.25 • 8 • 3 31.25 • 32
Ma = − = − = 234.375 kN-m
2 2 2 2
Ma = Mc = 234.375 kN-m @ 3000 mm and 5000 mm
Mmax = Mb = 250 kN-m @ 4000 mm
4 • 250
ω2 = 1.032
250 + 4 • 234.3752 + 7 • 2502 + 4 • 234.3752
2
ω2 = 1.032
CSA S16-19 Example 001 - 7
Software Verification
PROGRAM NAME: SAP2000
REVISION NO.: 0
2
ω2 π πE
=Mu EI 22GJ + I 22Cw
Ly L
( 2 •10 ) • (12 •10 ) • 76923 • ( 328 •10 )
5 6 3
1.032 • π
π ( 2 • 10 )
2
Mu = 5
4000 + (12 • 10 )( 469.8 • 10 )
6 9
4000
M u 362.06 • 106 N-mm = 362.06 kN-m
=
0.67 M p = 0.67 • 309.75 = 279 < M u = 362.06 kN-m, so
M p 33
M r=
33 1.15φM p 33 1 − 0.28 ≤ φM p 33
Mu
416.5
M r 33 = 1.15 • 0.9 • 416.5 1 − 0.28
362.06
M r 33 = 292.23kN-m
CSA S16-19 Example 001 - 8
