## Sunday, May 19, 2019

Let's think of an elastic member whose length is l, and cross-sectional area A, which subjected to an external axial load W. If the applied load is increased gradually from zero to the value W, the member also gradually increase be Î´.

So the work is done by the load is equal to the product of the average load and the displacement Î´.
External work is done = We =(1/2)⨯WÎ´ = 0.5⨯WÎ´
Let, the energy stored by the member be Wi. Since the work done by the external force on the member equals the energy stored by it, we have, We=Wi

Let the tension in the member be S.
For the equilibrium of the member, S=W.

Intensity of the tensile stress = f =S/A.
Now, tensile strain = e = f/E = f/AE.
Where E is the Young's Modulus of the material of the member.
So, change in length of the member =Î´= strain⨯ stress.
Or, Î´ = el = Sl/AE
Now, Strain energy stored = Work done = 0.5⨯WÎ´
After putting the value of W and Î´ in the above equation, we get
Strain energy stored = 0.5⨯ S(Sl/AE) = 0.5⨯(S2l/AE) =S2l/2AE.

In this case, the strain energy stored is due to axial loading on the member.
Strain energy stored per unit volume of the member = (S2l/2AE)/Al = S2l/2A2E = f2/2E.