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 1 of 4.
Vinay BEL said:
5 years ago
wl/3 is right. I agree.
(1)
Manoj badu said:
5 years ago
F (a)+Fb=wl/2.
M (b)=0.
Then, Fa*l-wl/2*2l/3=0.
We get Fa=wl/3.
And Fb=wl/2-wl/3=wl/6.
M (b)=0.
Then, Fa*l-wl/2*2l/3=0.
We get Fa=wl/3.
And Fb=wl/2-wl/3=wl/6.
Dawood saleem said:
6 years ago
The force concentrated in the one-third near the base of the triangle.
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.
Manish said:
7 years ago
Here it is asking about the shear force which will be WL/2. To calculate BM. It will be - WLsq/3 and at A- WLsq/6.
Suraj said:
7 years ago
It should be WL/2.
(1)
V.S said:
8 years ago
If we are finding shear force at a point just before A then the answer is right and at Exactly point A it will be 0.
King of Mechanical said:
8 years ago
wl/2 is correct. Refer strength of material by RS Khurmi page no:294.
Jitendra kumar said:
8 years ago
The given answer is right because span load is uvl (uniformly varying load) not a udl (uniformly distributed load).
Bharat khot said:
9 years ago
Wl/3 is the correct answer. I too agree.
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