Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 6)
The bending moment in the centre of a simply supported beam carrying a uniformly distributed load of w per unit length is
Answer: Option
No answer description is available. Let's discuss.
8 comments Page 1 of 1.

Akshay said:   1 decade ago
The load is UDL then its point load acts at centre,

Moment at centre is zero.

Akshay said:   10 years ago
How can moment be zero? you have to consider the support loads at both ends at the centre of the beam.

Yadvendra Kaushik said:   9 years ago
At both ends reactions will be wl/2 and if we take moment at the centre it comes wl2/8. So answer D is correct.

Vinod rampa said:   9 years ago
1. Bending moment due to reaction force at one end is = (wl/2)*(l/2).

2. Bending moment due to udl up to l/2 will act at l/4 at with a magnitude of -((w/2)*(l/2))*(l/4) (-ve since clockwise).

So, net B.M. will be the sum of them, which is wl^2/4 - wl^2/8 = wl^2/8.

DEEPAK said:   9 years ago
Firstly calculate shear force i.e. wl/2.

Now calculate the area of shear force diagram you will get bending moment as the shear force diagram is in the form of triangle the area =1/2*base*height i.e 1/2*L/2*WL/2=WL2/8.

Vivek said:   7 years ago
If the centroidal axis and neutral axis of the beam concides, the the stresses at the centre of the beam will be zero.

Sravan said:   5 years ago
Simply supported beam(point load)then B.M = Wl/4.
Simply supported beam (U.D.L)then B.M = Wl^2/8.
Cantilever beam (point load)then B.M = Wl.
Cantilever beam (U.D.L)then B.M = wl^2/2.

Ashwer said:   3 years ago
Can anyone tell me which configuration has Wl/8 bending moment?

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