Compute the maximum acceptable tensile SERVICE LOAD that may act on a single see section that is connected to a gusset plate at its flange using welds that is 15 inches long as shown in the figure. The service live load is three times the dead load and the earthquake load is assumed to be 10% of the dead load. Use A992 steel with Py-50 ks and Fu65 ks Design in LRFD ONLY, and no block shear will occur. *** WT 1247 13.80 12.3(312) Logainal 0.550 (140) 9.00 (229) 6.500 (249) 13000 186.00 20.30 3.670 bf-9.00 (229) y=2.99₁ 0980 (249) 2.990 54.50 12.00 1.500 d=12.3 (312) 2.420 Sx tw 0.550 (14.0 50 (140)

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate at
its flange using welds that is 15 inches long as shown in the figure. The service live load is three times the dead load and the
earthquake load is assumed to be 10% of the dead load. Use A992 steel with Fy-50 ksi and Fu 65 ks. Design in LRFD ONLY, and no
block shear will occur.
WT 12x47
WIL 47
A=
13.80
d
12.3 (312)
weldi
0.550 (140)
9.00 (229)
0.500 (249)
1.3800
186.00
20.30
3.670
bf-9.00 (229)
y=2.99₁
0.980 (24.9)
y=
2.990
y
54.50
Sy
12.00
1.960
d=12.3 (312)
X
2,620
a. Governing ultimate tensile capacity based on yleiding of gross section in kips
Round your answer to 0 decimal places.
Add your answer
tw
of=
of
k=
Sx=
EX
tw=0.550 (14.0)
⠀
Transcribed Image Text:Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate at its flange using welds that is 15 inches long as shown in the figure. The service live load is three times the dead load and the earthquake load is assumed to be 10% of the dead load. Use A992 steel with Fy-50 ksi and Fu 65 ks. Design in LRFD ONLY, and no block shear will occur. WT 12x47 WIL 47 A= 13.80 d 12.3 (312) weldi 0.550 (140) 9.00 (229) 0.500 (249) 1.3800 186.00 20.30 3.670 bf-9.00 (229) y=2.99₁ 0.980 (24.9) y= 2.990 y 54.50 Sy 12.00 1.960 d=12.3 (312) X 2,620 a. Governing ultimate tensile capacity based on yleiding of gross section in kips Round your answer to 0 decimal places. Add your answer tw of= of k= Sx= EX tw=0.550 (14.0) ⠀
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