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.
Harish said:
9 years ago
Because it is simply supported to obtain reaction at a total load multiplied with centroidal distance to the apex (in this case) whole divided by length is the answer.
Suresh said:
9 years ago
The given answer is correct (option B). Because it is a simply supported beam, so we have to take support reactions also into account while calculating SF.
Rakesh kumar said:
9 years ago
If the force applied over the equal udl the derivation is wl/2 is right and here the udl is triangle and the right derivation will become wl/3.
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.
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.
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).
Ananth Smvec said:
9 years ago
Shear force at a =w, and the length at which the shear force acts is l/3.
Now calculate, shear force = w * l/3.
Now calculate, shear force = w * l/3.
Akash said:
1 decade ago
No it is right. Shear force at A is wl/3. Because at A-force is (2/3L)WL/2/L. Calculate it.
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.
Somesh said:
9 years ago
Answer is right. You have to subtract the shear force by the load taken by the beam b.
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