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.
Amol said:
7 years ago
Anyone Can explain it properly?
Usibha said:
6 years ago
Very well explained, Thanks @Asif.
Umesh said:
6 years ago
5wl^4/584. = 5wl^3/48.
Prachi said:
6 years ago
@Roy.
How Ra+Rb=2x?
Please explain it.
How Ra+Rb=2x?
Please explain it.
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?
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)
Samir said:
5 years ago
Thanks for explaining the answer @Asif.
Shweta said:
5 years ago
@Asif
By deflection of beams first find out value of middle prop reaction as:
(5wl^4)/384EI = (pl^3)/48EI.
p = 5wl/8.
After this B.M at the center is;
for UDL Wl^2/8 n for point load Pl/4.
therefore we have UDL minus point load.
Wl^2/8 - Pl/4.
Wl^2/8 - ((5/8) wl)/4 substitute value of P.
then we get WL/32.
By deflection of beams first find out value of middle prop reaction as:
(5wl^4)/384EI = (pl^3)/48EI.
p = 5wl/8.
After this B.M at the center is;
for UDL Wl^2/8 n for point load Pl/4.
therefore we have UDL minus point load.
Wl^2/8 - Pl/4.
Wl^2/8 - ((5/8) wl)/4 substitute value of P.
then we get WL/32.
(18)
Joti said:
4 years ago
Thanks for the explanation. @Asif & @Shweta.
MOHAN said:
4 years ago
Thanks for explaining @Asif.
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