 Myself, Mr. Akshay Kumar Suvde, Assistant Professor, Department of Mechanical Engineering. Today, we will study Strain Energy and Impact Load. Learning outcomes. At the end of the session, student will be able to determine the strain energy stored in the body due to different types of loading. So, we will see certain numericals. The statement of the numerical is an axial pull of 50 kilo Newton is suddenly applied on a steel bar of 2 meter long and 1000 mm square in cross section. If the modulus of elasticity of a steel is 200 kilo Newton per mm square, find maximum instantaneous stress, maximum instantaneous extension, strain energy and modulus of resilience. So, from this particular statement, the data which is given to you, the load P is 50 kilo Newton and this load is applied suddenly. So, the condition of load which is applied that is the load is suddenly applied. The length of the bar is 2 meter. So, convert this into mm by multiplying 10 raise to 3 that is 2000 mm. Area of the bar is 1000 mm square and modulus of elasticity E is equal to 200 kilo Newton per mm square. This is to be converted into Newton per mm square. So, 1 kilo is equal to 1000 and therefore, this will be 200 into 10 raise to 3 Newton per mm square which is equal to 2 into 10 raise to 5 Newton per mm square. What is to be calculated? Stress sigma, maximum extension dL, strain energy stored in the body U and modulus of resilience. Now, as we have studied in case of suddenly applied load, the stress is equal to 2 times P by A and therefore, instantaneous stress sigma is equal to 2 times P upon A and therefore, putting the value of P 2 into 50 into 10 raise to 3 Newton divided by 1000. So, calculating this particular equation, you will get stress is equal to 108 Newton per mm square. Now, extension dL as per the Hooke's law, stress is directly proportional to strain and therefore, deformation dL or extension dL is equal to sigma upon Young's modulus multiplied by length of the body. So, already we have calculated the stress 100 Newton per mm square. So, 100 into its length 2000 divided by Young's modulus 2 into 10 raise to 5 and that is equal to by solving this particular equation, you will get the deformation dL is equal to 1 mm. And now, we have to calculate the strain energy stored in the bar, which is given by U is equal to sigma square upon 2 E multiplied by volume, where volume is equal to A into L and therefore, this equation you can write as sigma square upon 2 E into A into L. So, putting the value of stress, which we have already calculated divided by 2 times Young's modulus 2 into 10 raise to 5 multiplied by area 1000 and length 2000 mm. So, calculating this particular equation, you will get the strain energy stored in the body is equal to 50000 Newton mm. So, this is the strain energy stored in the body. Now, the next thing which we have to calculate that is modulus of resilience. So, modulus of resilience is equal to sigma square upon 2 times E, because modulus of resilience is nothing but strain energy stored per unit volume and therefore, putting the value of sigma 100 square divided by 2 times Young's modulus 2 into 10 raise to 5. Calculating this particular equation, you will get strain modulus of resilience that will be equal to 0.025. So, this is how we can calculate the various is an unknown weight falls by 30 mm on a collar which is rigidly fixed at the lower end of vertical bar having length 4 meter and 1000 mm square in area. If the instantaneous elongation is 3.66 mm, determine the stress produced and unknown weight, take E is equal to 2 into 10 raise to 5 Newton per mm square. So, in this case, the weight of load is falling through height as 30 mm, length of the bar is 4 meter, area of the bar is 1000 mm square, instantaneous elongation DL is equal to 3.66 mm, we have to calculate the instantaneous stress sigma and unknown weight P, where Young's modulus E is equal to 2 into 10 raise to 5 Newton per mm square. If I draw the diagram of a bar whose length is 4 meter and at its lower end, the collar is attached. The upper end is kept fixed. This length is L and now a load of P Newton is falling through a height of 30 mm and due to that the bar will get extended by 3.6 mm. X is equal to or DL is equal to 3.66 mm and therefore, DL is equal to sigma upon E into L and therefore, deformation 3.66 is equal to sigma upon 2 into 10 raise to 5 into length of the bar is 4 meter. So, convert it into mm and from this equation calculate the stress sigma and the stress is 183 Newton per mm square. So, this is the value of stress and now we have to calculate the unknown weight. So, unknown weight can be calculated from the equation strain energy stored in the body is equal to work done by the load. So, work done is P into H plus X or DL, but at this moment we do not know the strain energy. Hence, strain energy U is equal to sigma square upon 2 E into A into L. So, put the value 183 bracket square divided by 2 into 10 raise to 5 into area 1000 into length 4000. So, solving this particular equation you will get the strain energy is equal to 334890 Newton mm and now from this you can calculate what is the unknown load and therefore, putting the value of strain energy in the above equation and value of H 30 plus 3.66 is the extension. So, solve this equation and calculate the value of P and that is equal to 9449.19 Newton. So, you can calculate the value of the load. So, this material is referred from the book of Strength of Material by Dr. R. K. Bunsell and S. S. Bhavikatti. Thank you.