# Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 2 (Q.No. 34)

34.

A simply supported beam which carries a uniformly distributed load has two equal overhangs. To have maximum B.M. produced in the beam least possible, the ratio of the length of the overhang to the total length of the beam, is

Discussion:

7 comments Page 1 of 1.
Vinuthna said:
1 year ago

For overhanging beam on both sides;

If a = length of overhang.

L= length of the overall span.

l = length of span b/w supports = L-2a;

(+ve) Max BM = (-ve) Max BM.

Wl^2/8 - Wa^2/2 = Wa^2/2.

a = l/2*2=0.35 * l.

a = 0.35(L-2a).

a = 0.207L.

Ratio; a/L = 0.207.

If a = length of overhang.

L= length of the overall span.

l = length of span b/w supports = L-2a;

(+ve) Max BM = (-ve) Max BM.

Wl^2/8 - Wa^2/2 = Wa^2/2.

a = l/2*2=0.35 * l.

a = 0.35(L-2a).

a = 0.207L.

Ratio; a/L = 0.207.

Madhan said:
3 years ago

Answer is not clear. Please explain.

Shubham said:
3 years ago

Max. BM = Min. BM.

If l= length of overhang.

L= length of ssb span.

wl2/8 - wl2/2 = wl2/2,

Solving this;

l/(L+2l) = 0.207.

If l= length of overhang.

L= length of ssb span.

wl2/8 - wl2/2 = wl2/2,

Solving this;

l/(L+2l) = 0.207.

Akshi said:
4 years ago

Explain the answer in detail.

Lufura said:
4 years ago

Give Explanation.

Monika said:
5 years ago

Please, can anyone explain this?

Taha said:
6 years ago

Please give explanation.

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