Application of external loads to a member causes deformation of the member but the member has a natural tendency to oppose the deformation and in doing so it develops internal stresses. These internal stresses have the capacity to do work and as such, the member has energy stored in it. Thus work done upon a member in straining it is called as Strain Energy.

Volume of bar= A.L Strain Energy when loading is gradually applied load to the member.

A)Gradually applied loads:

Let us consider a vertical bar of X-sectional area 'A' be rigidly held at one end and carry a gradually applied load at the other end as shown in fig.

Let $\delta l$ be extension in length L

Strain energy stored in the member=world done by gradually applied load.

u=avg. load*distance moved by it=$\frac{0+P}{2}*\delta l$

u=$\frac{0+P}{2}*\delta l$

u=$\frac{P^2 L}{2AE}$

u=$\frac{p^2LA}{2AE.A}$

=$\frac{P^2}{A^2}.\frac{1}{2E}.*AL$

Strain energy, u=$\frac{p^2}{2E}$*volume of bar

Hence modulus of resilience=$\frac{p^2}{2E}$

Strain energy when loading is suddenly applied load to the member

Suddenly Applied Load: Let the load P be suddenly applied. Let the extension of member be $\delta l$. In this case, the magnitude of load is constant throughout the process of extension. Let 'p' be the maximum stress included. Equating strain energy stored by the member to the work done.

$\frac{p^2}{2E}*AL=\frac{(p+p) \delta l}{2}=p \delta l$

i.e $\frac{p^2}{2E}*AL=\frac{p.L.P}{E}$

i.e $p=\frac{1}{2} \frac{P}{A}$

Hence, the maximum stress intensity due to a suddenly applied load is twice the stress intensity produced by the load of the same magnitude applied gradually.

Strain Energy when loading is impacted load to the member.

c) Impact load:

In this case, the load P is dropped from a height h before it commences to stretch the bar. A vertical bar whose upper end is fixed at the top and collar is provided at the lower end.

Load P drops by a height h on the collar and this extend the member by $\delta l$

Let p=maximum stress intensity produced in the bar/ Hence extension at bar

$\delta l=\frac{p.l}{E}$

Equating the loadd of PE to strain energy storoed by the road.

$P(h+\delta l)=\frac{p^2}{@E}*AL$

$P(h+\frac{pl}{E})=\frac{p^2}{@E}*AL$

$\frac{p^2}{@E}*AL-\frac{p.Pl}{E}=Ph$ (divided by $\frac{AL}{2E}$ throughout)

Adding $\frac{p^2}{A^2}$ to both sides

$p+\frac{p^2}{A^2}-\frac{2.pP}{A}=\frac{2.PhE}{AL}+\frac{p^2}{A^2}$

$(p-\frac{p}{A})^2=[\frac{p^2}{A^2}+\frac{2PhE}{AL}]$ (taking square root)

i.e $p=\frac{P}{A}+\sqrt{\frac{p^2}{A^2}+\frac{2PhE}{AL}}$

Note: For all three types of load, $\delta l=\frac{p.l}{E}$ and $u=\frac{p^2}{2E}$*volume of bar

Application of external loads to a member causes deformation of the member but the member has a natural tendency to oppose the deformation and in doing so it develops internal stresses. These internal stresses have the capacity to do work and as such, the member has energy stored in it. Thus work done upon a member in straining it is called as Strain Energy.

Volume of bar= A.L Strain Energy when loading is gradually applied load to the member. Strain Energy when loading is suddenly applied load to the member. Strain Energy when loading is impacted load to the member.