Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 11)
11.
A uniform girder simply supported at its ends is subjected to a uniformly distributed load over its entire length and is propped at the centre so as to neutralise the deflection. The net B.M. at the centre will be
WL
Discussion:
37 comments Page 3 of 4.
Umesh said:
6 years ago
5wl^4/584. = 5wl^3/48.
Raghavan said:
9 years ago
@Asif, I think your answer is correct.
But the answer should be wl^2/32,
Because for uniform loading, BMD will be parabolic.
But the answer should be wl^2/32,
Because for uniform loading, BMD will be parabolic.
Amol said:
7 years ago
Anyone Can explain it properly?
Anommiii said:
7 years ago
@Roy:
(3wl/16)*(l/2)-(w*l/2)*((l/4) = Wl/32(ans).
(3wl/16)*(l/2)-(w*l/2)*((l/4) = Wl/32(ans).
Rahul MEHAR said:
8 years ago
Thanks for better explanation @Roy.
Roy said:
8 years ago
Ra+RB+5wl/8=wl so,2x+5wl/8=wl or,x=3wl/16. So, Ra=RB=3wl/16 @Anoop.
Anoop said:
8 years ago
Can anyone tell how reactions are 3wl/16 on each end?
Roy said:
8 years ago
Thanks @Asif.
Bhupendra singh said:
8 years ago
It should be WL^2/32.
Babypriyadarshini said:
8 years ago
I am not understanding it. Please, anyone explain it.
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