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:
2 years 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:
4 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:
6 years ago
Please, can anyone explain this?
Taha said:
7 years ago
Please give explanation.
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