 Hello students, I am Professor Bhagyash Deshmukh from Mechanical Engineering Department, Valchan Institute of Technology, Solapur. In this session, we will be studying the failure of simple machine elements. This is part 3 of the failure of machine elements. At the end of this session, you will be able to design the equations for typical machine elements under various conditions of loading. These failures and the equations developed will be useful for developing the design equation for typical assembly of machine elements. Let us move further. Let us make the earlier component more complicated now. We have seen a bar which is loaded under tension. We made that bar hollow. The side view of the bar was like this, inner circle and outer circle. The top view if I try to indicate is like this. I need to make this component hollow. Then further, due to additional need, we drill a hole in this component. In the front view, you can see it like this. There will be no hole over here. The reason is already the material is removed. In the side view also, I can show the component. This is the change in the component. If one try to identify the area, one need to think upon it. The force is P, the component is loaded as in tension. Outside diameter over here for this circle is D1. The inside circle diameter for this circle or this hollow portion is D2. If I extend this dimension is small d. You can see over here the circle is having the diameter equal to small d. If I take some area or take some section over here, some plane and try to look over here the area. The area is like this, inside circle, outside circle, but I need to reduce. This is the drilled hole. Therefore, if I take a cross section over here, this is going to become dark lines. Only the shaded area is available to resist the applied force. Here again the outside circle diameter is D1. This inside circle diameter is D2 and this dimension is small d. I need to find out that area. Let us try writing this equation P equals pi by 4 D1 square outer circle diameter minus inner circle diameter pi by 4 D1 square minus D2 square. Let us try establishing it. This is pi by 4 D1 square minus pi by 4 D2 square. Hence, here is a hole. Therefore, net area available is this much. I need to reduce a strip over here, this small strip and this small strip at the bottom. If I go to find out the area, it is D1 minus D2. Why? Because this dimension is D1 minus this dimension is D2, but it represents only the length. I need to add to its width and therefore, this dimension is small d. I need to multiply with small d. This is the area of solid bar. This represents the area of hollow bar or the part removed and the removed area is represented by pi by 4 D2 square. Therefore, this is a net area available minus this strip and this strip D1 minus D2 into small d. But as the component is loaded under tension, why? Because if I again extend it, the force line of action of the force acting on the component is perpendicular to the area resisting the failure. I can say that this failure is tensile failure. Therefore, I can use sigma t as the corresponding stress and mention that this component is under tension. Let us go for the next case. Another case of tensile failure is indicated for a bracket. Let us see a bracket. This is the bracket and at this location, I want to apply a pull. If I want to apply a pull, this portion is to be constrained. Here, the diameter of the outer circle, if I take the dimension, this diameter, it is capital D. The dimension over here is small d and the thickness of this plate is small d. Let us see this component, how it is loaded. The top view of the component, if I draw and the front view, then you will understand how the loading is happening. This is the component. I am going to load with a force equal to p. Therefore, this fixed portion will have a reaction equal to p. It is the resistance offered. As indicated over here, this dimension will show that this dimension is capital D and the inside circle diameter is small d. And the thickness of this plate is small d. Now, what will happen to this component? The component under tension I am thinking of. It is possible that I am going to fit a pin over here. The pin may be located like this. It will come out from the bottom and to this pin, I am going to apply a pull and that is going to cause tension in this bracket. This is obviously a solid pin and I am going to pull it. So, that the pin is going to impart load on this bracket. Here, the assumption is pin is safe, pin is not going to fail. Therefore, it is the bracket that is going to fail. Let us see how to proceed with this. The bracket will fail as shown here. I am showing the failure of the bracket and the remaining part, the failed portion. The pin is going to pull this half circle, a semicircular plate, the piece. It is pulled over here and the failure is going to happen at this location. And therefore, I need to write the equation for it. As the force is perpendicular to this area, I need to use sigma t as the stress p. capital D minus small d, y capital D minus small d. Let us see this dimension is capital D. This dimension is going to be equal to small d and the thickness of the plate is indicated as small t. Therefore, the net area of this zone is capital D minus small d multiplied by t. Therefore, it is capital D minus small d, linear dimension multiplied by the other dimension to get the rectangular area. D minus d into t multiplied by sigma t, that is going to give me the design equation for this component. And as the component is having the load, the failure zone will be having the line of action of force perpendicular to it. I can say that it is a case of tensile failure. Thank you.