Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 26)
26.
For a simply supported beam carrying uniformly distributed load W on it entire length L, the maximum bending moment is
Discussion:
9 comments Page 1 of 1.
Abhi said:
8 years ago
It i s WL^2/8.
Kallan said:
8 years ago
wl= W,
So WL/8.
So WL/8.
Sheela KR said:
7 years ago
I agree @Kallan.
Suraj Khatiwada said:
7 years ago
No, it should be WL^2/8.
Naveen said:
5 years ago
Wl^2/8 is the answer.
Enali said:
5 years ago
I think WL^2/8.
The reaction at support = WL/2.
Shear force D is a rectangular shape.
Bending moment = area of Rectangle.
BMD = 1/2 L/2 WL/2 = WL^2/4.
The reaction at support = WL/2.
Shear force D is a rectangular shape.
Bending moment = area of Rectangle.
BMD = 1/2 L/2 WL/2 = WL^2/4.
Milton shaikh said:
4 years ago
Wl^2/8 is the correct answer.
Balu said:
4 years ago
1/2*Wl/2*l/2.
= Wl^2/8.
= Wl^2/8.
Ravi Sundar said:
1 year ago
Load of the beam uniformly distributed throughout its length at intensity w per unit length then the maximum deflection of the beam will be (wL^2 /8).
Now total weight (W) = w.L.
Hence put (w = W/L) in the maximum bending moment formula you will get (WL/8).
Now total weight (W) = w.L.
Hence put (w = W/L) in the maximum bending moment formula you will get (WL/8).
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