Civil Engineering - Strength of Materials - Discussion

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

[A]. 240 kg
[B]. 250 kg
[C]. 260 kg
[D]. 270 kg
[E]. 280 kg.

Answer: Option D

Explanation:

No answer description available for this question.

George Zac said: (Feb 27, 2014)  
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.

John Parker said: (Sep 25, 2015)  
Correct option is (D) , 270 Kg.

Krishna said: (May 12, 2016)  
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.

Ranjit said: (Dec 17, 2016)  
But the type of beam is not given here then how can we use simply supported condition?

Sai Madhav said: (Sep 24, 2017)  
In question there is no information about beam type.

Shahsi Ranjan said: (Nov 15, 2017)  
Every time, assume SSB if there is not any information about support.

Bibek said: (Mar 10, 2018)  
D. 270 is the answer I think.

Satish said: (Mar 17, 2018)  
W=270 is right.

Deepak said: (Jul 5, 2018)  
Why you takes length in meters? All dimension take into cm than the answer is different.

Rahul said: (Aug 24, 2019)  
Thanks for the answer.

Shubham said: (Sep 14, 2019)  
270 kg is the correct answer.

Tanmoy Ghosh said: (Jan 3, 2021)  
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.

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