CSA S16-14 Example 001 PDF
CSA S16-14 Example 001 PDF
CSA S16-14 Example 001 PDF
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-
14 steel frame design code. The moment and shear strengths are compared with
Handbook of Steel construction (9th Edition) results.
L = 8.0 m
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
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.
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 •11.2 381.6mm
Ag = 5890 mm2
I22 = 5,140,000 mm4
Z33 = 885,000 mm3
J = 192,000 mm4
Cw 1.976 •1011 mm6
Section: W410x60
bf = 178 mm, tf = 12.8 mm, d = 408 mm, tw = 7.7 mm
h d 2t f 408 2 •12.8 382.4mm
Ag = 7580 mm2
I22 = 12,000,000 mm4
Z33= 1,190,000 mm3
J = 328,000 mm4
Cw 4.698 •1011 mm6
Member:
L=8m
Ф = 0.9
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
Cl .1 7.75
Fy 350
W410x46
bf 140
6.25
2t f 2 • 11.2
1100 Cf 1100 0
Cl .1 1 0.39 1 0.39 58.8
Fy Cy 350 5890 • 350
W410x46
h 381.6
54.51
tw 7
Section is Class 1
W410x60
h 382.4
49.66
tw 7.7
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
2
M max 2 4M a 2 7 M b 2 4M c 2
where: Mmax = maximum moment
Ma = moment at ¼ unrestrained span
Mb = moment at ½ unrestrained span
Mc = moment at ¾ unrestrained span
E
2
Mu 2 EI 22GJ I 22Cw as L 0
L L
M p 33
M r 33 1.15M p 33 1 0.28 M p 33
Mu
M p 33
0.28 0 as M u
Mu
leading to M r 33 1.15 M p33 M p33
So
M r 33 M p33 278.775kN-m
4 • 250
2 1.008
250 4 • 246.0942 7 • 2502 4 • 246.094 2
2
ω2 = 1.008
2 E
2
Mu EI 22GJ I 22Cw
L L
1.008 •
Mu 2 • 10
2
5
2000 5.14 • 10 197.6 • 10
6 9
2000
M u 537.82 •106 N-mm = 537.82 kN-m
M p 33
M r 33 1.15M p 33 1 0.28 M 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 33 268.89kN-m
ω2 = 1.032
E
2
M u 2 EI 22GJ I 22Cw
Ly L
1.032 •
Mu 2 • 10
2
5
4000 12 • 10 469.8 • 10
6 9
4000
M u 362.06 •106 N-mm = 362.06 kN-m
M p 33
M r 33 1.15M p 33 1 0.28 M 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