Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 35)
35.
If a rectangular beam measuring 10 x 18 x 400 cm carries a unformly distributed load such that the bending stress developed is 100 kg/cm2. The intensity of the load per metre length, is
Discussion:
12 comments Page 1 of 2.
George Zac said:
9 years ago
The correct answer is 270 kg.
Bending Stress = f = M/Z (Moment / section Modulus).
100 kg/cm2 = (WL^2/8)/(1/6 bd^3) M = 100/6 x 10 x 18^2.
M = 54,000 Kg.cm - 540 Kg.m.
M = WL^2/8= 540.
W = 8 * M / L^2.
W = 8 * 540 / 4^2.
W = 270 Kg.
Bending Stress = f = M/Z (Moment / section Modulus).
100 kg/cm2 = (WL^2/8)/(1/6 bd^3) M = 100/6 x 10 x 18^2.
M = 54,000 Kg.cm - 540 Kg.m.
M = WL^2/8= 540.
W = 8 * M / L^2.
W = 8 * 540 / 4^2.
W = 270 Kg.
(1)
Tanmoy Ghosh said:
2 years ago
WL^2/8.
Z = I/Y.
Z called section modulus.
F = M/Z.
Z =(BD^2/6)=540.
M = WL^2/8,
(WL^2/8)/(10*18^2/6).
W = 540 * 8/4^2,
W = 270KG,
W = 540 * 8.
Z = I/Y.
Z called section modulus.
F = M/Z.
Z =(BD^2/6)=540.
M = WL^2/8,
(WL^2/8)/(10*18^2/6).
W = 540 * 8/4^2,
W = 270KG,
W = 540 * 8.
(1)
Krishna said:
7 years ago
F= My/I.
M = (WL^2) /8.
M = fI/y = wl^2/8.
I = bh^3/12.
Y = h/2.
F and l are known, then we get,
W= 270 kg.
M = (WL^2) /8.
M = fI/y = wl^2/8.
I = bh^3/12.
Y = h/2.
F and l are known, then we get,
W= 270 kg.
Deepak said:
5 years ago
Why you takes length in meters? All dimension take into cm than the answer is different.
Ranjit said:
6 years ago
But the type of beam is not given here then how can we use simply supported condition?
Shahsi Ranjan said:
5 years ago
Every time, assume SSB if there is not any information about support.
Sai madhav said:
6 years ago
In question there is no information about beam type.
John parker said:
8 years ago
Correct option is (D) , 270 Kg.
Bibek said:
5 years ago
D. 270 is the answer I think.
SHUBHAM said:
4 years ago
270 kg is the correct answer.
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