# 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/cm

^{2}. The intensity of the load per metre length, isDiscussion:

12 comments Page 1 of 2.
George Zac said:
1 decade 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.

(2)

Tanmoy Ghosh said:
3 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:
8 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:
6 years ago

Why you takes length in meters? All dimension take into cm than the answer is different.

Ranjit said:
7 years ago

But the type of beam is not given here then how can we use simply supported condition?

Shahsi Ranjan said:
6 years ago

Every time, assume SSB if there is not any information about support.

Sai madhav said:
7 years ago

In question there is no information about beam type.

John parker said:
9 years ago

Correct option is (D) , 270 Kg.

Bibek said:
6 years ago

D. 270 is the answer I think.

SHUBHAM said:
5 years ago

270 kg is the correct answer.

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