 Hello students, I am Bhagyash Deshpukh from Valchin Institute of Technology, Solapur Mechanical Engineering Department. The session is on failure of machine elements, Part 4. In this session, you will be able to write the design equations for typical machine elements under different type of loading. The simple design equations developed over here will be useful in developing the design equations for assembly of machine elements. That the component is loaded with a push force. It is a push force. In comparison with the tensile failure, the only change is instead of outweighing force, it is incoming or it is going to push. It is going to reduce the length and increase the lateral dimension. If I take some area over here, if I try to find out the area at this location, the component is having circular cross section. In the side view also, you can check the diameter is D. Therefore, I can write the equation force P is equal to 2 bracket. One is the area which is resisting the failure and second one is the corresponding stress. The equation is P equals pi by 4 D square multiplied by sigma C. The sigma C value comes because the component is under compression. The length is going to get reduced and the lateral dimension diameter is going to get increased. Therefore, the corresponding stress, this is stress. This is the area. Area is pi by 4 D square and the stress is sigma C. Here also, our thumb rule is that in compressive failure, the force is perpendicular to the area which is resisting the failure. However, if you compare with the tensile failure, the force is of pushed nature. In tensile, it was pulled in nature. That is only the difference. See, these are the thumb rules. The failure needs to be very critically analyzed in order to confirm that whether it is going to fail under tension, compression or shear. Let us move towards the next part. Now, let us see that there is a case. It is a pin which is inserted from this bracket. It is a bracket and from which the pin is inserted. What I am assuming that this link is fixed, this link is not going to move and this link is safe, I am going to apply a pull to this pin. Let us try with some other view. This is the fixed link I was talking about and this is the pin. The diameter of this pin is small d and its head is capital D. The bracket we are not going to consider for design, this is safe. I am going to assume this is my assumption that this bracket is safe and force P is applied over here as a pull force. You can see the other view how the construction is. If I take the cross section, it shows the pin and this is some part of the bracket. Now, what is going to happen if I apply a pull over here, the clearance over here, this is the clearance available, it nullifies and the component touches the bracket. This is the pin which is going to touch over here in the surface of the bracket. Let us see what is going to happen. When it touches the bracket, then the diameter of this pin we have taken as small d. Here the pin head diameter we have taken as capital D. The bracket we have considered over here as this. Let us show some clearance over here so that you will be able to imagine how the assembly is going to happen. Cross section I am not going to show it right now at the lower location. What will happen is there will be a compression, a compressive force between these two. It is due to the context. The component this part and this part, there is a compression at this location. Same thing is going to happen over here also. You can imagine that what will happen to this surface. I will request you to imagine that I am going to put some sugar cubes over here. If I pull it, the components can get crushed over here. In similar fashion, this compressive stress will cause crushing and slowly the particles will start getting crushed off this material. This is slow failure. This is going to happen from the top and here also it is going to happen same way. It is going to originate from here at one location or at one point of time. I can say that this part is crushed, this part is crushed. This is not expected and therefore I need to make this component safe for design because we have loaded it under P over here. Compression has happened over here and it caused crushing of this pin head. Therefore, now what I need to do? I need to find out what area is under compression. For that, I need to check the area which is under compression. It is this outer area of the pin and this is the head. I am not going to show over here, but this is the annular area which is under compression. Or crushing will happen at this location. This outer diameter is capital D. Pin head diameter over here, the pin is over here. This is the pin. This diameter is small d. This zone only is under compression. Therefore, if I want to find out the net area under compression, it is pi by 4 into d square minus d square. This will be the net area which is going to resist the failure. Then I need to write the equation for it. The equation for it will be force P equals. Area is pi by 4 d square minus d square is the annular area multiplied by sigma c is the compressive stress or the crushing over here. I can attribute it to crushing failure. This is a typical case of crushing failure. Let us move further. But before that, I want to tell you another thumb rule that we can use. See, the lateral dimension d, we have obtained considering the tensile failure. Considering this failure also, we are going to obtain this lateral dimension. What should be the limiting value of capital D? The pin head dimension, the failure is dominated this dimension capital D. This equation is governed by sigma c. That is what is the significance of this typical equation. Let us move further to the next problem. Thank you.