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 2 of 4.
Babypriyadarshini said:
8 years ago
I am not understanding it. Please, anyone explain it.
Anoop said:
8 years ago
Can anyone tell how reactions are 3wl/16 on each end?
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).
Palash said:
2 years ago
@Asif.
By solving the equation, we get wl^2/32.
By solving the equation, we get wl^2/32.
(2)
Joti said:
4 years ago
Thanks for the explanation. @Asif & @Shweta.
Prachi said:
6 years ago
@Roy.
How Ra+Rb=2x?
Please explain it.
How Ra+Rb=2x?
Please explain it.
Samir said:
5 years ago
Thanks for explaining the answer @Asif.
Rayudu said:
8 years ago
Please, give me the clear explanation.
Naren said:
9 years ago
Please give me the clear explanation.
Sudip said:
9 years ago
@Subhasmita.
wXl = W, So wl^2 = Wl.
wXl = W, So wl^2 = Wl.
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