 Hello students, I am Bhagesh Deshmukh from Mechanical Engineering Department, Valchin Institute of Technology, Swalapur. This is the session on failure of machine elements, part 5. In this series, this is the last part. In this part, we will be going to study various equations for simple machine elements. Here, we will be going to write the design equations for typical machine elements under different type of loading. The equations developed over here will be useful in developing the design equations for combination of machine elements or simple assembly of machine elements. A typical case of tensile failure, let us consider that it is a threaded component. I will show threads over here. We usually do not show threads. We use conventions, but for drawing this component and finding out its design equation, it is essential for me to draw these threads. This is a screw thread. This is the shank diameter and this is the core diameter. I can say that the core diameter is over here and the component is loaded in tension, it is force P. Again, I need to find out that which is the area, the critical area. The component is not going to fail at the area indicated as by this line, but rather it is going to fail at the area I am showing right now. This length is going to limit not this, because this diameter is larger than that of this core diameter. Therefore, core diameter is going to govern the tensile failure of this rod. Hence, I can write the equation. Our equation is P equals pi by 4 dc square into sigma t. The terms, first term is the area resisting the failure and second is the corresponding stress. Here I am taking tensile failure. Again, our thumb rule can be used. If I extend this, it is perpendicular, nearby this is approximation. It is perpendicular to the line of action of the force and I can say that this failure is a tensile failure. Let us now check with some other component. I am saying that it is crushing. These are the threads. Let us start with it. If I apply a pull to this component P, there will be a resisting force at the other end. I am going to ignore it, but let us try to understand. Here I can have a contact, because at this location I am showing clearance, contact will happen at this location. Threads on other side will be clearance, contact, clearance, contact, clearance, then contact. Because this part is being pulled on the left, I am assuming that this is a stationary component. It is a nut. It is a screw. Core diameter is over dc, force P is applied to this component. The other part of the nut one can draw over here, but let us for this instance ignore it. Just consider that here is a contact, connection, connection, connection over here. Metal to metal contact happens at this location. I can draw the nut on the lower side also, but it is not going to make much difference. We need to design this component. For this, let us consider only one thread. This force is acting on the thread. This is the portion of the nut, as it was indicated by green color in the previous case. Let us shade it with green, so that it will be easy for us to understand. If I push it, earlier it was pulled on one end. Let us now consider it is pulled on the right end. This component will have a contact at this location. This is the outside diameter and this represents the core diameter. The contact is up to the core diameter. Therefore, I can find out the area which is available for taking this load. No doubt it is land, however, which we have assumed is straight. Outer diameter is small d, outer circle diameter is small d or this larger dimension. This dimension, if I take it is dc, the core diameter. Therefore, this much area is available for compression. Metal to metal contact is happening over there or material to metal contact is happening there. I need to find out this area. The area is indicated by area equals pi by 4 d square minus dc square. This is the net area. Then, if I want to design this rod or this screw thread under crushing, I need to design it as force p equals pi by 4 d square minus dc square is the net area available. Multiply by the stress which is sigma c. This failure is governed by crushing stress. If we make this safe under crushing, we can identify what kind of other failure will be there if the component is safe in crushing. Let us take another case. Now, here I have a component, a rod. This rod is pulled on the left and the diameter of this rod is capital D. I am going to drill this rod and insert another rod into it. These are the curves or interpenetration. Let us ignore for this instance. I want to apply a pull to this rod. This force is P. That means the rod is pulled on the left and this rod is going to be pulled on the right. The assembly we can see like this. Component is pulled on the left. This is pulled on the right. Now, here the assumption is this rod is safe. It is not going to fail under tension. We have already made the component safe. This component is not going to bend. It is pulled on the right. There will be a contact at this location. This will be the curve where the contact is happening. Metal of this rod and material metal of this rod is going to have a contact. There will be a compression at this location. Here I need to find out a typical area called as the projected area. This is the projected area of the component. This curvature, the projected area is indicated by this height and this width. This is going to govern the failure of this rod under compression. Thank you.