Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 5 (Q.No. 42)
42.
The shear force in the centre of a simply supported beam carrying a uniformly distributed load of w per unit length, is
Discussion:
8 comments Page 1 of 1.
Lilly said:
4 years ago
If you consider till left side it will be 0 only.
SF@L = wl/2-wl/2 = 0.
SF@R = wl/2-wl = wl/2.
As wl/2 is not given in the option, the answer should be 0.
So, the given answer is correct.
SF@L = wl/2-wl/2 = 0.
SF@R = wl/2-wl = wl/2.
As wl/2 is not given in the option, the answer should be 0.
So, the given answer is correct.
(1)
L Rajendra said:
4 years ago
We will get an answer from the shear force diagram.
Hiren Bavaliya said:
4 years ago
It's WL/2.
How Can Say Zero is the right answer?
How Can Say Zero is the right answer?
VVG said:
4 years ago
The point of inflexion lies at the centre of the beam in this case. So, it is the right answer.
Neeraj said:
9 years ago
Answer is right it asks at centre of the beam that is l/2, not at x=0 at x=0 WL/2.
Nilesh said:
9 years ago
It has to be (Wl)/2.
Josh said:
9 years ago
Where shear force minimum it has Max bending momentum.
Vijay said:
9 years ago
It should be wl/2.
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