 Hello everyone, myself R.C. B. Radar working as an assistant professor in Walsh Indian Staff Technology, Sallapur. In this session, we will discuss about strain energy due to impact load. At the end of this session, students will be able to determine the strain energy stored in an object under impact load. Now, we can derive the formula for stress, so that can be used in a strain energy formula that is already known to us. So if you derive the formula for sigma, so we can find the energy equation. Now we can consider, let us consider a rod. So at the end of the rod, a collar is attached. If a load P acting in downward direction, it falls through a distance of H onto a collar. And due to the load impact on a collar, it is assumed that the rod will extend with a deformation called delta L. This is a deformation of a rod under the impact load and this is the length of a rod that we have. Now let us consider, so this rod, so it is acting at a load P and through height of H. So as already we know that the strain energy u equal to sigma square by 2e into volume and also we know that the strain energy is equal to the work done on a rod and the work done on a rod can be calculated, so if you know the area under the P delta diagram. That is if you plot a diagram for load versus deformation, so this I can get, so delta L on x axis and P on y axis. So this sudden load is applied on an impact rod with a distance H, so it impacts on a collar, so it increases the strain energy in a rod. So with this, so this area under this is called as a work done on a rod. So now I can equate strain energy is nothing but the work done and work done can be formulated as area under the P delta diagram. So now I can write that, so this can be equal to work done and this can be equal to area under the area under P delta diagram. Now already we know that strain energy sigma square by 2e into volume and this can be equal to area under this one and area for this is H plus delta L. So here if H is 0 the load becomes sudden type, if H will be having some value the load becomes impact load. Now in order to find the sigma value I need to replace this delta L. So now I can write this as pin bracket of H plus sigma L divided by E because E equal to sigma by epsilon and that implies delta L equal to sigma L divided by E. So if I replace this, so this I can write it as sigma square by 2e into area into volume that is area into length. So just rearrange the equation to get P sigma L divided by E equal to sigma square by 2e into AL. So take this term here, I will bring this RHS term to LHS side. This becomes AL by 2e minus sigma in bracket of PL by E minus pH equal to 0. So from this above equation I can say that this is a quadratic equation form. So here to make this term A equal to 1, so multiply both side by 2e by AL times. And I get that implies sigma square into 1 minus sigma in bracket of PL by E into 2e by PL minus pH 2e by AL equal to 0. So if I simplify this again, so I get sigma square minus of, so this will get cancelled 2p by AL will be remaining here minus 2p he by AL will be remaining, this will be equal to 0. This I will term it as equation 1. Now so if I see here, so solving the above equation, solving the above quadratic equation, to get the roots, so now roots, so that is in the form of, so in that above equation 1, so A equal to 1, B equal to minus 2p by A and C equal to what, minus 2p he by AL, these are the terms. Now if I have, so if I to get the roots, roots that is sigma equal to minus B, so as if I have 2p by A plus or minus square root of 2p by A bracket square minus 4 into A into C that is minus 2p he by AL because roots equal to minus B plus or minus square root of B square minus 4A C divided by 2A. So on simplification, 2p by A will be remaining here plus, so I can take 4 out sides, so the root is nothing but 2, I can take outside, so the remaining is p by a bracket square, so minus minus becomes plus 2p he divided by AL will be remaining here inside that and here divided by, here I need to have 2 into A, that is 2 will be remaining. So I can cancel 2, 2, so the remaining term is p by A plus or minus p by A bracket square plus 2p he by AL will be remaining. So again I can bring p by A term outside the square root, so if I multiplied by p by A and A by p, this becomes p by a bracket square, then I can take p by A outside the bracket and it becomes 1 plus under square root 1 plus 2 into hAe divided by p into AL will be remaining inside the bracket and now this is nothing but the sigma, so now this is a value we should use for to find the value of sigma in order to find the strain energy and already we know that strain energy is, strain energy is sigma square by 2e into Array into length, so this term sigma, so we should use from this formula. So when h is 0, so sigma is p by A 1 plus square root of 1, it becomes 2 times p by A and this is clearly the case of sudden type of load and when delta L is very small, is very small in comparison with, in comparison with h, so I can put that that, so I can use sigma square by 2e into Array into length equal to p into h plus 0, so it is equal to p h, so I can get sigma equal to 2 p h e divided by A into L under square root, so this is a value of sigma, so when delta L is small, this is a value of sigma when h is 0 and if we use this value of sigma in this equation 2, we can get the strain energy for impact load, so now you can think that, so what is the strain energy equation for sudden and gradual load case, thank you, thank you.