What is value of L or length to get the results of 8.68kn.m and 0.753 kn.m in the green box   Topic: BIAXIAL BENDING- STEEL DESIGN

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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What is value of L or length to get the results of 8.68kn.m and 0.753 kn.m in the green box

 

Topic: BIAXIAL BENDING- STEEL DESIGN

 

Section
W6x8.5
W6x12
W4x13
For ASD.
Max
M₁ =
ax
M.
ay
+
Μ/ΩΜ/Ω
=
Weight
Area
(N/m) (mm²)
124.3
175.5
wL²
8
W, [²
32
nx
Ω
190.1
M M
ax
M
Ma
Depth
(mm)
1619.35 148.082 100.076
2283.97 152.400
2464.54 105.664
-= 8.68kN-m
M
ay
+
= 0.753kN-m
ny
≤1.0
Ω
bf (mm)
≤1.0
101.600
103.124
tf (mm)
4.928
7.087
8.763
tw (mm) Sx (mm³)
4.318
5.482
+
7.112
83246
8.68 0.753
30.277
6.35
1.67
1.67
118806
89309
Sy (mm³)
16551
24417
30316
Mpy = Fy(Zy) = 248×25.6(10³) = 6.35kN • m
Zx (mm³)
No LTB:M₁ = M₂
Mpx = Fy(Zx)=248×93.9(10³) = 23.29kN.m
For Cb: Cb = 1.30
M=1.30(23.29) = 30.277kN-m
= 0.677 <1.0
93900
136000
103000
Zy (mm³)
25600
38000
479000
W = 3282.28 N/m
X
=
W, 1138.87 N/m
=
check if section is compact
b₁ 100.026
2t, 2(4.928)
-=10.15
E
Fy
.. section is compact
.. section is adequate
Use W6x8.5
<0.38
Check the upper limit:
Z₁
= 1.55<1.6 :. M = M = 6.35kN m
ny
Py
S
= 10.79
Transcribed Image Text:Section W6x8.5 W6x12 W4x13 For ASD. Max M₁ = ax M. ay + Μ/ΩΜ/Ω = Weight Area (N/m) (mm²) 124.3 175.5 wL² 8 W, [² 32 nx Ω 190.1 M M ax M Ma Depth (mm) 1619.35 148.082 100.076 2283.97 152.400 2464.54 105.664 -= 8.68kN-m M ay + = 0.753kN-m ny ≤1.0 Ω bf (mm) ≤1.0 101.600 103.124 tf (mm) 4.928 7.087 8.763 tw (mm) Sx (mm³) 4.318 5.482 + 7.112 83246 8.68 0.753 30.277 6.35 1.67 1.67 118806 89309 Sy (mm³) 16551 24417 30316 Mpy = Fy(Zy) = 248×25.6(10³) = 6.35kN • m Zx (mm³) No LTB:M₁ = M₂ Mpx = Fy(Zx)=248×93.9(10³) = 23.29kN.m For Cb: Cb = 1.30 M=1.30(23.29) = 30.277kN-m = 0.677 <1.0 93900 136000 103000 Zy (mm³) 25600 38000 479000 W = 3282.28 N/m X = W, 1138.87 N/m = check if section is compact b₁ 100.026 2t, 2(4.928) -=10.15 E Fy .. section is compact .. section is adequate Use W6x8.5 <0.38 Check the upper limit: Z₁ = 1.55<1.6 :. M = M = 6.35kN m ny Py S = 10.79
Select a W section to serve as purlins between roof trusses spaced at 4.6 m on centers. Assume all loads passes through the
Centroid of the purlins. Spacing of purlins is 1.4 m on centers. Assume that the purlins are fully laterally supported and
subject to the following:
Loads
Slope of the roof truss is 1 vertical to 2 horizontal.
Sag rods are provided at the midpoints of the truss. A36 steel, Fy=248 MPa
Select from any W section listed in the table below. Use ASD
Section
W6x8.5
W6x12
Snow = 1440 N/m2 of roof surface
Roofing = 288 N/m2 of roof surface
Wind surface = 720 N/m2 perpendicular to roof surface
W4x13
Weight
(N/m)
124.3
175.5
190.1
Area
(mm²)
1619.35
2283.97
2464.54
Depth
(mm)
148.082
152.400
105.664
bf (mm)
100.076
101.600
103.124
tf (mm) tw (mm)
4.928
7.087
8.763
4.318
5.482
7.112
Sx (mm³)
83246
118806
89309
Sy (mm³) Zx (mm³)
16551
24417
30316
93900
136000
103000
Zy (mm³)
25600
38000
479000
Transcribed Image Text:Select a W section to serve as purlins between roof trusses spaced at 4.6 m on centers. Assume all loads passes through the Centroid of the purlins. Spacing of purlins is 1.4 m on centers. Assume that the purlins are fully laterally supported and subject to the following: Loads Slope of the roof truss is 1 vertical to 2 horizontal. Sag rods are provided at the midpoints of the truss. A36 steel, Fy=248 MPa Select from any W section listed in the table below. Use ASD Section W6x8.5 W6x12 Snow = 1440 N/m2 of roof surface Roofing = 288 N/m2 of roof surface Wind surface = 720 N/m2 perpendicular to roof surface W4x13 Weight (N/m) 124.3 175.5 190.1 Area (mm²) 1619.35 2283.97 2464.54 Depth (mm) 148.082 152.400 105.664 bf (mm) 100.076 101.600 103.124 tf (mm) tw (mm) 4.928 7.087 8.763 4.318 5.482 7.112 Sx (mm³) 83246 118806 89309 Sy (mm³) Zx (mm³) 16551 24417 30316 93900 136000 103000 Zy (mm³) 25600 38000 479000
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