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.
Ralph Puno said:
4 years ago
Thanks for the answer @Abidha.
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.
Nadeem said:
4 years ago
Thank you everyone.
MOHAN said:
4 years ago
Thanks for explaining @Asif.
Joti said:
4 years ago
Thanks for the explanation. @Asif & @Shweta.
Divya said:
1 decade ago
wl = W very simple.
Samir said:
5 years ago
Thanks for explaining the answer @Asif.
Sudip said:
9 years ago
@Subhasmita.
wXl = W, So wl^2 = Wl.
wXl = W, So wl^2 = Wl.
Ammu said:
6 years ago
@Asif.
Could you please explain how to equate deflection values at first step?
Could you please explain how to equate deflection values at first step?
Prachi said:
6 years ago
@Roy.
How Ra+Rb=2x?
Please explain it.
How Ra+Rb=2x?
Please explain it.
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