 Hello students, I am Bagesh Deshmukh from Mechanical Engineering Department of Valtian Institute of Technology, Solaapur. I am taking this session on failure of machine elements. These are simple failures. This is part one of the series. At the end of this session, you will be able to write the design equation for typical machine elements. Where the typical elements are under tension, under compression, under bending, under shear. We will be discussing some cases related to the tensile failure in this part. Start with the rod under tension. This is the typical case when we consider a rod under tension. Let us start with the first case. Here I am taking the first rod which has a circular cross section and it is actually loaded. The force acting on this rod is P. If I draw its side view, it is a circle having the diameter equal to small d. If you look towards its area, if I draw a revolve section over here or remove section over here, I can see that this is a cross section which is circular along this plane. If I do it, section over here is circular. Now let us see how to begin with the design of a rod under tension. We will choose this format of equation. Force P equals two brackets, first and second. The first bracket will always correspond to the area which is resisting the failure and second is the corresponding stress if it is a case of typical static loading. If I start writing with this equation, then force P I need to equate. Force P is equal to two brackets, first bracket is over here. I will begin with the first bracket. It is pi by 4 into d square because this typical circle is having the area of cross section as pi by 4 d square. Using this as cross section, I need to multiply by using a corresponding stress. I am going to use sigma t as the stress. Now why to take sigma t? Because if I see the loading condition, the component is actually loaded. This is the axis of the component. The load is applied actually. It is bound to happen that the length of this bar is going to get increased and the diameter is going to get decreased. This is the typical case of tension. Therefore, I need to use stress as sigma t. Now here I want to establish a thumb rule which is useful for further calculations. With the loading axis, this is the load is perpendicular to the plane of area resisting the failure of this plane. It is this plane which is perpendicular to the axis of loading. It is a case of tension. Therefore, one has to check what is the loading direction and accordingly which is the plane which resist the failure, you will be able to identify the type of failure. Therefore, this stress is sigma t. Let us move for the next. I will change the case of loading and the component. Now the component over here, I am going to take it as a rod. You may check the first case, the rod. The only change is this rod is now made hollow. We can establish its side view. The component is loaded axially as like in the previous case and the force acting is p. The outer circle diameter is capital D and the inside circle diameter is small d. The area of cross section, if I take a plane over here at this plane, the area of this component is or the cross section of this component is shown as this is the area which is resisting the failure. I need to establish this area, outer circle diameter is capital D and the inner circle diameter is small d. Therefore, as per our rule, the equation can be obtained by force p is equal to two brackets. First bracket represents the area which is resisting the failure and second bracket represents the corresponding stress. Therefore, if I go on writing the equation, it is pi by four into outer circle diameter is capital D square minus inside circle because there is no material. I need to reduce the corresponding area which is pi by four small d square. Therefore, the equation is pi by four d square minus d square. This is the total area available multiplied by the corresponding stress. Here again our thumb rule is applicable. If I use extend this line, check the area, the area of cross section. If I say that this is the area of cross section, the force is perpendicular to that and therefore, I can say that this is the component no doubt and this was the cross section. The force is perpendicular to the area and therefore, this stress must be sigma t. Remember if the area is reduced, if there is no material over here, I need to reduce that from the outer area, outer area is pi by four capital D square and inside area is pi by four small d square. Let us check another case with some more complications. Again I have changed the type of bar. Now it is a square rod or a bar obviously under tension. Let us see this is the bar, square bar I am using. The square bar is loaded under tension. This is the pull force P applied to the bar. This cross section if I draw the side view, the component is rectangular. The dimensions are this is W and if you take this as H, this is the cross section. If I again go for its available cross section, if I take this plane here, if I take a plane and check what is the exact cross section on this component, the area which is resisting the failure is this much. The dimensions are available as W and H. As per our rule force P equals two brackets. I have used force P equals W into H which is the area of this rectangular cross section multiplied by sigma t which represents the component loaded under tension. Why it is loaded under tension? The reason is if you see the cross section again, if I extend this line you can see that it is perpendicular, the component will fail like this, the force, the cross section is this component has failed over here after loading. I need to avoid that failure, this is force P. I want to avoid the failure and hence I will consider that the failure domination is in tension and hence I am going to use sigma t at this location. Remember when you use the component under tension or compression, the orientation whether this width and this height, the plane has no effect on the resistance. That means if you change H and W, if the orientation is changed instead of this plane, if I orient the component like this where this is H and this is W, it is not going to have any impact on the design equation because here only cross section is important. We are going to see a case where orientation matters a lot. We will see that kind of failure also. Thank you.