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.
Samir said:
5 years ago
Thanks for explaining the answer @Asif.
Abidha said:
6 years ago
BM= wl2/8 for simply supported beams.
When l = l/2.
Then BM = wl2/32.
= Wl/32( where W=wl).
When l = l/2.
Then BM = wl2/32.
= Wl/32( where W=wl).
(2)
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.
Umesh said:
6 years ago
5wl^4/584. = 5wl^3/48.
Usibha said:
6 years ago
Very well explained, Thanks @Asif.
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.
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