 Myself, Mr. Akshay Kumar Suvade, Assistant Professor, Department of Mechanical Engineering. Today, we are going to study strain energy and impact loading. During outcome, at the end of the session, students will be able to estimate strain energy stored due to impact load. So, figure shows one vertical bar which is fixed at the upper end and there is a collar at the lower end of the bar. Let P is being the load dropped over the collar of a vertical bar from a height of h as shown in the figure. Now, at this moment, we will take a pause and think about what is the impact load and what are the various examples of impact load. So, when a particular load falls from a certain height on a particular object, that is impact load. For example, a cricket ball striking with a bat, that is an example of impact load or a certain amount of load falls through certain height on a particular object is also an impact load. For example, hammer blow or an accident take place in which the impact load is there in case of accident. So, consider a body subjected with an impact load and due to that, there will be extension in the bar by the amount x due to the impact load P as shown in the figure. Sigma stress developed in the body due to impact load. E, Young's modulus of elasticity of the material of the body. A, cross sectional area of the body. P, impact load. X, deformation of extension of the body. L, length of the body that is vertical bar. V, volume of the body that is vertical bar which is equal to A into L. U, strain energy stored in the body that is in the vertical bar. So, within the elastic limit strain energy stored in the vertical bar is equal to work done by the load in deforming the vertical bar. So, in this case, the load P is falling through height h on the collar and due to that impact, the bar will get deformed by the amount x which is shown in the figure. And therefore, strain energy stored in the vertical bar is nothing but load multiply by displacement. And therefore, load is P and displacement that load falls through height h and due to that extension of the bar take place by amount x and therefore, displacement will be h plus x. And hence, strain energy stored in the vertical bar U is equal to P multiply by inside the bracket h plus x. As we know that strain energy stored in the body U will be provided by the following expression as mentioned here. So, U is equal to sigma square upon 2 E into V, where volume is equal to A into L and strain energy stored in the vertical bar that is nothing but P into h plus x. So, putting the value of strain energy in the above equation, you will get P multiply by inside the bracket h plus x is equal to sigma square upon 2 E multiply by A into L. So, as per the Hooke's law, stress is equal to Young's modulus into strain, where capital letter E is Young's modulus of elasticity of the material and hence, sigma is equal to Young's modulus into strain. So, in this case, strain is nothing but ratio of extension due to the impact load which is given by x divided by original length L. And hence, x is equal to sigma multiplied by L upon E. Now, let us use this value of extension or deformation x in the above equation and we will have P multiplied by inside the bracket h plus sigma into L divided by E is equal to one half sigma square upon E multiplied by A into L. So, left side of the equation multiplied by P to the bracket, you will get P into h plus P into sigma into L divided by E which is equal to sigma square upon 2 E into A L. Divide this equation by A L, you will get sigma square upon 2 E minus P into sigma divided by A E is equal to P h divided by A L. And now, multiply this equation by 2 E. So, multiplying by 2 E, you will get sigma square minus 2 into P into sigma divided by A is equal to 2 into P into h multiplied by E divided by A L. And now, adding P square upon A square on both side of the equation, you will get sigma square minus 2 times P by A into sigma plus P square upon A square is equal to 2 multiplied by P into h into E divided by A L plus P square upon A square. Now, if you see the left side of the equation, this is the equation in terms of sigma and P upon A. So, it is nothing but A minus B square. So, this left side of the equation can be reduced to sigma minus P by A bracket square is equal to P square upon A square plus 2 P h E divided by A into L. So, taking the square root, you will get P sigma minus P by A is equal to square root P square upon A square plus 2 P h E divided by A L. And therefore, sigma is equal to P upon A plus under root P square upon A square plus 2 P h E divided by A L. And taking the term P by A outside the square root and rearranging the equation, you will get sigma is equal to P upon A bracket 1 plus under root 1 plus 2 A E h divided by P into L, where P is load dropper or impact. So, when the extension x is very small as compared to height h, then work done will be P into h. So, as we know work done is equal to strain energy stored. So, sigma square upon 2 E into A L is equal to P into h. And therefore, sigma square is equal to 2 times E into P into h divided by A into L. And therefore, sigma is equal to under root 2 into E into P into h divided by A L. And hence the strain energy stored is equal to sigma square upon 2 E into A L. So, by determining the stress induced due to the impact load, you can calculate the strain energy stored in the body due to the impact load. The material is referred from the book of Strength of Materials by Dr. Avkay Bansal and S S Bhavikatti. Thank you.