Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 29)
29.
For the beam shown in the below figure, the shear force at A is equal to
Discussion:
32 comments Page 2 of 4.
Dawood saleem said:
6 years ago
The force concentrated in the one-third near the base of the triangle.
King of Mechanical said:
8 years ago
wl/2 is correct. Refer strength of material by RS Khurmi page no:294.
Vin said:
7 years ago
Ra = WL/6 Rb = WL/3 so the shear force at A is WL/2(uvl)-WL/6 = WL/3.
Daka Hitesh said:
9 years ago
At B : {(WL/2) * (2L/3)}/L = WL/3.
At A : {(WL/2) * (L/3)}/L = WL/6.
At A : {(WL/2) * (L/3)}/L = WL/6.
Sonu said:
1 decade ago
Yes I think so wrong answer because shear force = (wL^2)/2L=wL/2.
Shweta said:
1 decade ago
No, the answer is right. WL/2 acts at 2L/3 not at point A.
Milindsardar said:
9 years ago
The given option B. wl/3 is the correct answer.
Vinod arya said:
1 decade ago
Options are wrong the shear force at A is Wl/2.
Gajendra said:
9 years ago
I think the given answer wl/3is correct.
Bharat khot said:
9 years ago
Wl/3 is the correct answer. I too agree.
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