Ex Welding.pdf →The Figure shows a 0.75-in-thick AISI 1020 HRsteel bar subjected to a load of (F = 1.4 kip) andwelded to a vertical support using a 7010electrode. Estimate the shear stress for the weld.For the AISI 1020 HR steel bar, S, = 30 kpsi, Su= 55 kpsi (from the table)From the pattern of weld shown below, b= d = 2 in (welding table)bd2* = b/2ý = d/2A = 1.414hdThen,A = 1.414hd = 1.414(5/16)(2) = 0.884 in?I= bd/2 = 2(4)/2 = 4 in1= 0.707hl, = 0.707(5/16)(4) = 0.884 inPrimary shear:Ft%3DA1.4= 1.584 kpsi0.884Secondary shear:Mr" =1.4(7)(1)= 11.086 kpsi0.884Then, the shear stress is1=Vr2 + t"2 = V2.509 + 122.9 = 11.2 kpsi . AnsII<>

Given: σy=30kpsi b=d=2in h=5/16in Load (F)=1.4kip Solution: Due to the symmetry of the weld joints,…

Answered: The Figure shows a 0.75-in-thick AISI…

The Figure shows a 0.75-in-thick AISI 1020 HR steel bar subjected to a load of (F = 1.4 kip) and welded to a vertical support using a 7010 electrode. Estimate the shear stress for the weld. 2in 2 in

Elements Of Electromagnetics

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Author:Sadiku, Matthew N. O.

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Question

Ex Welding.pdf →
The Figure shows a 0.75-in-thick AISI 1020 HR
steel bar subjected to a load of (F = 1.4 kip) and
welded to a vertical support using a 7010
electrode. Estimate the shear stress for the weld.
For the AISI 1020 HR steel bar, S, = 30 kpsi, Su= 55 kpsi (from the table)
From the pattern of weld shown below, b= d = 2 in (welding table)
bd2
* = b/2
ý = d/2
A = 1.414hd
Then,
A = 1.414hd = 1.414(5/16)(2) = 0.884 in?
I= bd/2 = 2(4)/2 = 4 in
1= 0.707hl, = 0.707(5/16)(4) = 0.884 in
Primary shear:
F
t%3D
A
1.4
= 1.584 kpsi
0.884
Secondary shear:
Mr
" =
1.4(7)(1)
= 11.086 kpsi
0.884
Then, the shear stress is
1=Vr2 + t"2 = V2.509 + 122.9 = 11.2 kpsi . Ans
II
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