Concept of Strain Energy – Strain Energy – Strength of Materials


Hello students today we are starting with a new chapter in in sum the name of the chapter is Strain energy so let us start with this chapter we are starting with this new topic now let us understand what is Strain energy Strain energy is the energy which is stored in any material when it is loaded it means when we are loading we are applying a load on a material that material will get deform and because of the deformation there is some energy stored in that material and the moment we remove the load that material would be releasing its internal energy it means here I will write the definition of strain energy strain energy is the internal energy which is stored in any material which is loaded within its elastic limit so here is the basic definition the most simple definition of strain energy that it is defined as the internal energy which is stored in any material which is loaded within its elastic limit it means the condition for strain energy is that the material should be loaded within elastic limit it should not go into permanent deformation that the material will be deformed only temporarily it will store the energy and the moment we are removing the load the material will it will release the energy so that is the concept of strain energy now strain energy is also called as the work done by a loaded material so here strain energy we can also say that it is the work done by a loaded material next it is also called as resilience the other name for strain energy is resilience and since we are talking about energy E the unit of strain energy is either it will be in terms of newton meter or newton centimeter Newton mm or kilo Newton meter kilo Newton centimeter or kilo Newton mm since it is energy so here we have load into displacement and next for strain energy to be stored material must deform within elastic limit that is for strain energy to take place material should change its shape and the change of shape should be within elastic limit that is it should not get permanently deformed next point strain energy depends upon applied load the amount of strain energy stored in any material depends upon the load which is applied this applied load it can be gradually applied load it can be suddenly applied load or it can be impact load that is a load which is falling from a height now as far as application is concerned strain energy principle is used in Springs as we all know Springs are elastic members when we are applying loads spring will get d-formed and when the load is removed the spring will regain its original shape so you can treat a spring as a material which has strain energy or you can say stored energy because the other Memphis strain energies it is a terminal energy or you can say the stored energy so as we know spring when we are applying load it will change its shape this change in shape is called a strain and because of that whatever work is stored inside the spring that would be called a strain energy next when the load is removed material will regain its original shape next the load applied on a material can be we have three different cases in which the load can be applied the first one is gradually applied load next suddenly applied load and at last we have impact loads impact loads are also called as load falling from a height so we see that there are three methods or three ways in which load will be applied on a machine member or you can say any material and the load is gradually applied gradually applied means the load will be in the form of steps suddenly applied load the total value of load would be acting together on a member in impact load the load will fall from a height so all these in all these three cases the strain energy value will be different because as we know that strain energy the stored energy which is there inside a material it depends on the type of loading next if we say that it is resilience means strain energy is the resilience then you can say that resilience is the energy which is stored within elastic limit it is given by little capital you means we will be denoting resilience Australian energy by letter U next I will give you the formula of strain energy strain energy it is given by the formula u is equal to Sigma square 0.2 e multiplied by volume so here I have written the formula for strain energy so here we have written the formula for strain energy where I can say that Sigma it indicates stress in the material unit will be Newton per mm square capital e is the Youngs modulus of material and V is the volume of the material and strain energy unit it will be Newton mm if I put these values no after this I’ll write how this formula has come I’ll say that strain energy in and actually loaded member so how the strain energy formula is developed here if I consider a bar or a rod which is loaded actually and I will consider the load as tensile so here I can see that the stress which is stored in the bar that is Sigma is equal to P upon a and this stress which is stored I will call this as from this stress because of this there will be strain energy which I will denote it by later capital u now when this member is loaded we will be getting the behavior of this rod and I will plot it on a graph on y-axis we would be having load that is P load on x-axis we would be having the deflection deflection is Delta L now the graph would be linear since I am considering the load to be gradual so here I can say that I am getting this area shaded area because when you go on increasing the load deflection will go on increasing this is gradual load now from this I can say that area under the curve will give me the energy stored so I will say that area is equal to half into P into Delta L that is the area of this right angle triangle now since we know that stress is equal to node upon area so therefore load is equal to stress into area so this formula will be half into stress into area into Delta L we know that Delta L deflection is equal to PL upon ei so P upon a is equal to stress so it is equal to stress into L upon e stress into L upon e so therefore here I have this as equal to Sigma square upon 2 e a into L will give me the volume so here I am getting two conditions from the first condition I am considering that a member is subjected to external load because of this there will be internal stress and because of that internal stress there will be strain energy stored i give that value as you next the same behavior if I plot it on a graph I am getting the area as Sigma square 1 2 e into V this area is nothing but strain energy which is stored so if I give you as equation 1 and Sigma square 2 e upon Sigma square 2 e into V as equation 2 so from 1 & 2 I will be getting the same formula that is Sigma square 2 e into V so this is how the formula of strain energy is developed now after this I write down what is proof li resilience proof resilience it is the maximum strain energy stored in a material so in other words proof resilience becomes maximum strain energy and we can say that maximum strain energy will be given by it is U max which will be equal to Sigma max square upon 2 e into volume so this is the formula for maximum strain energy after this I’ll write another definition that is called as modulus of resilience modulus of resilience it is defined as the strain energy per unit volume mathematically it will be denoted by modulus of resilience is equal to it is equal to strain energy per unit volume we know that strain energy is Sigma square upon 2 e into V and here we are dividing by volume so the formula which I have of modulus of resilience therefore I will say mod of resilience that will be equal to just Sigma square upon 2 e so here I have written the definition of profe resilience and modulus of resilience now since we know the strain energy formula let me write three different formula of stresses for three different cases it means the heading is stresses in members for different cases here I will first write down the first case is gradually applied load as we know that strain energy depends on the method by which we are applying the load so first we are seeing gradually applied load so for gradually applied load stress is given by Sigma is equal to P upon e in Newton per mm square next for suddenly applied load stress is given by Sigma is equal to 2 P upon e and the last case we have that is for impact loads that is load which is falling from a height so for that stress is given by Sigma is equal to P Upon A plus square root of P upon a whole square plus 2 e pH upon a L so here I have the formula for impact load for gradually applied load we have stress as load upon area we can illustrate this with a diagram so here is the example of gradually applied load where the moment load goes on increasing you can say deflection will be increasing here I have shown area actually this is deflection so the graph is between load versus deflection and the moment you go on increasing the load deflection will go on increasing that is for gradually applied load next for suddenly applied load here as the load increases from 0 to a value suddenly there is increase in deflection so here we have the diagram for suddenly applied load next for impact load I’ll draw the diagram on the next page I will say that for impact loads or impact loading for studying impact load here we have a diagram in which there is a rod and one end of the rod is fixed to the other end we are attaching a collar now over this rod there is a weight which is placed I will denote it by P this P is at a distance of H from the collar and the moment we are releasing this load this weight will try to strike the collar and it will try to pull it so because of this there is change in shape of the rod and strain energy will be stored in the rod so here I will write the formula of strain energy as I have already written previously also so strain energy for impact load u it will be equal to Sigma square upon 2 e multiplied by volume and here stress for impact load is given by P Upon A plus square root of P upon a whole square plus 2 e pH upon a L so here is the formula to calculate stress for impact loads once we know the value of load which is falling from a height we know the area of the rod Young’s modulus for odd material H is the height between the weight and the collar area of the rod and length of this rod then we can easily calculate the value of stress and once this value of stress is known we can put it in the formula of strain energy and get the answer of energy which is stored in this rod so like these concepts would be there in strain energy and once we finish off the concepts we can easily start with the problems

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