 Mr. Akshay Kumar Sovade, Assistant Professor, Department of Mechanical Engineering. Today, we are going to study Strain Energy and Impact Load. Learning outcome, at the end of the session, student will be able to derive and determine the strain energy stored in a body due to shear stress. So, strain energy stored in a body due to shear stress. Consider a rectangular block A, B, C, D having length L, breadth B and height H. And this rectangular block is subjected to a tangential force acting along the top surface C, D. And the magnitude of the tangential force is P. The bottom phase of this rectangular block is kept fixed having length L, height H and width B. So, various terms involved in the expression of strain energy stored in the body due to shear stress are as follows. P is nothing but it is the shear force acting along phase C, D. L length of the block B, breadth H, height of the block U, strain energy stored in the block due to shear force P. Then C, C1 or D, D1 is nothing but shear deformation due to shear force P acting along phase C, D. Tau is the shear stress acting along C, D. V is the volume of the rectangular block which is equal to product of length into breadth into height. And A is the cross sectional area of the block which is equal to L into B. This is the area of the block. Now, as the shear force P is acting across the phase C, D, this particular block A, B, C, D will get distorted to a position A, B, C1, D1 where C, C1 or D, D1 is nothing but it is the lateral displacement of shear deformation which takes place due to this tangential force P. And shear stress Tau is nothing but it is the shear stress due to the shear force P acting along the phase C, D. And this force P which is acting along the phase C, D, this force P is applied gradually. And hence the strain energy stored in the body is nothing but it is the work done by the applied load in order to stretch the body. So, as this load P is applied gradually, the average load is to be taken, so average load into distance. This is nothing but strain energy stored in the body because strain energy is nothing but it is the work done by the applied load in stretching or deforming the body. And therefore, average load that is P upon 2 and the distortion which takes place due to the shear force is represented by C, C1 or D, D1. So, P divided by 2 into C, C1. Now, in order to determine the strain energy stored in the body, we have to determine the value of shear force and the value of C, C1 that is this lateral or shear deformation. So, as we know the shear force is equal to shear force is given by shear stress multiplied by area. So, in our case the shear force is P and due to the shear force P, the phase Cd is subjected to shear stress tau. So, P is equal to tau into area of this rectangular block. So, area of the rectangular block is nothing but L into B. So, it is L into B and therefore, the value of shear force is nothing but tau into L into B, this is my equation number 1. And now, in order to find out this lateral displacement of shear deformation from this particular diagram, as the block A, B, Cd is distorted to a position A, B, C1, D1 due to the tangential force P acting along the phase Cd. And from this particular triangle B, C, C1, the shear stress tan phi, shear strain tan phi is equal to C, C1 divided by Cb. As this angle phi is very small therefore, tan phi is equal to phi. And hence, we can write phi is equal to C, C1 divided by BC. And therefore, the value of C, C1 is phi into BC, where this BC is nothing but it is the height of the block. And therefore, C, C1 is equal to phi into H. So, this is my equation number 2. So, the equation number 1 and equation number 2 gives the value of shear force and the shear deformation. Now, as we know the modulus of rigidity tau, modulus of rigidity is represented by C and it is defined as it is the ratio of shear stress to shear strain. So, shear stress is tau and shear strain is phi. And therefore, you can write phi is equal to tau upon C. Now, what is the shear strain that is represented by the ratio C, C1 divided by BC? So, C, C1 divided by BC is equal to tau upon C. And therefore, what is the value of C, C1 that is tau upon C into BC that is H. So, this is the shear deformation C, C1. And therefore, strain energy stored in the block due to the shear stress that is equal to P divided by 2 into C, C1. So, what is the value of P that we have calculated that is tau into L into B that is shear stress multiplied by area into C, C1. So, value of C, C1 is tau upon C into H. So, here divided by 2. So, you can rewrite tau into L into B into H. So, tau multiplied by tau it is tau square and divided by 2 times C. Now, product of length, breadth and height that is nothing but it is the volume of this rectangular block. And therefore, strain energy stored in the block due to the shear stress that will be equal to tau square upon 2 C into V, where V is equal to volume of the block which is given by the product of length into breadth into height. So, by using this particular equation we can determine the strain energy stored in a rectangular block due to the shear stress. So, the material is referred from the book of Strength of Materials by Dr. R. K. Bansal and SS Bhavikatti. Thank you.