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 1 of 4.
Pritisha said:
1 decade ago
How to solve?
Asif said:
1 decade ago
By deflection of beams first find out value of middle prop reaction as:
(5wl^4)/384EI = (pl^3)/48EI.
p = 5wl/8.
Then reaction at each end is 3wl/16.
Therefore BM at centre will be = (3wl/16)*(l/2)-(wl/2)*(l/4) is the answer wl/32.
(5wl^4)/384EI = (pl^3)/48EI.
p = 5wl/8.
Then reaction at each end is 3wl/16.
Therefore BM at centre will be = (3wl/16)*(l/2)-(wl/2)*(l/4) is the answer wl/32.
(3)
Subhasmita said:
1 decade ago
If we consider your answer to be the correct one, then the answer will be coming as -wl^2/32.
Divya said:
1 decade ago
wl = W very simple.
Sudip said:
9 years ago
@Subhasmita.
wXl = W, So wl^2 = Wl.
wXl = W, So wl^2 = Wl.
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.
Suhas said:
9 years ago
You are right @Raghavan.
Naren said:
9 years ago
Please give me the clear explanation.
Honey said:
9 years ago
its in Kn, not in kn/m2
So W=wl that's why answer is Wl/32.
So W=wl that's why answer is Wl/32.
RAM said:
9 years ago
It must be (WL^2) /32.
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