Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 2 (Q.No. 32)
32.
A bar l metre long and having its area of cross-section A, is subjected to a gradually applied tensile load W. The strain energy stored in the bar is
Discussion:
6 comments Page 1 of 1.
Shiv said:
1 year ago
@All.
Here, it will act like a self-weight of the bar so, the strain energy should be W^2L/6AE.
Here, it will act like a self-weight of the bar so, the strain energy should be W^2L/6AE.
Fahad said:
4 years ago
Thanks @Abhi @Lalit.
Lalit said:
5 years ago
Strain energy formula = {( sigma) ^2/(2E) }*Volume.
= {( sigma) ^2/(2E) }*A*L.
= {(W/A) ^2/(2E) }*A*E.
== {W^2L/(2AE) }.
= {( sigma) ^2/(2E) }*A*L.
= {(W/A) ^2/(2E) }*A*E.
== {W^2L/(2AE) }.
Amu bokato said:
6 years ago
Thanks @Abhi.
Abhi said:
7 years ago
Deformation = WL/AE.
Strain Energy = 1/2 x W x Deformation.
Therefore W2L/2AE.
Strain Energy = 1/2 x W x Deformation.
Therefore W2L/2AE.
CHHAYA said:
7 years ago
It is 1/2*W*WL/AE.
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