Question

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

[A]. WL

[B]. wl/8

[C]. wl/24

[D]. wl/32

[E]. wl/64​

give correct answer with explanation

Answer 1

Answer:

(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