Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 3 (Q.No. 41)
41.
A steel bar 20 mm in diameter simply-supported at its ends over a total span of 40 cm carries a load at its centre. If the maximum stress induced in the bar is limited to N/mm2, the bending strain energy stored in the bar, is
Discussion:
12 comments Page 1 of 2.
Vishnoi said:
11 months ago
Stress/y = M/I
Stress = 480/Π.
y= 20/2.
I= π*20^4 /64.
Put all values in the above formula and get the value of M= 120000.
Maximum bending moment = PL/4 for the simply supported beam so by this get the value of P= 1200.
Now strain energy due to bending = ∫M^2dx/2EI.
M =(Px/2) put value in strain energy formula.
Limit L/2 to 0 and integrate.
And get the value of U =610.400N-mm.
Stress = 480/Π.
y= 20/2.
I= π*20^4 /64.
Put all values in the above formula and get the value of M= 120000.
Maximum bending moment = PL/4 for the simply supported beam so by this get the value of P= 1200.
Now strain energy due to bending = ∫M^2dx/2EI.
M =(Px/2) put value in strain energy formula.
Limit L/2 to 0 and integrate.
And get the value of U =610.400N-mm.
Sulav said:
1 year ago
Strain Energy = workdone = 1/2 * P * deflection.
For P;
(stress/y) = (M/I);
Maximum moment = (PL/4).
For deflection:
(PL^3)/(48EI).
Answer= 611.16 Nmm => option C.
For P;
(stress/y) = (M/I);
Maximum moment = (PL/4).
For deflection:
(PL^3)/(48EI).
Answer= 611.16 Nmm => option C.
Ankit Pandey said:
4 years ago
Does anyone know the Formula of bending strain energy? Please explain.
Abhay said:
6 years ago
It is impossible to solve, the question is wrong.
Kartik said:
6 years ago
I got an answer in 1833, now why it is divided by 3 to get 611.
U.S.lakshmi said:
6 years ago
Please check the question there is data deficiency that I think.
Dilip said:
6 years ago
Explain me the solution.
Sneha said:
7 years ago
It is limited to 480/π N/mm2.
(1)
Skm said:
8 years ago
Please give me solution in detail.
Hasan said:
8 years ago
Please explain how to solve this?
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