 Hello students, I am Bhagyash Deshmukh from Valchain Institute of Technology, Swalapur Mechanical Engineering Department. This session is on estimation of endurance limit, how to estimate the endurance limit of a specimen and a component that we are going to study. At the end of this session you will be able to use the laboratory method to calculate the endurance limit. The fatigue or endurance limit of a material is defined as the maximum amplitude of completely reversed is that the standard specimen can sustain for an unlimited number of cycles without fatigue failure. What does it mean? Since the fatigue test cannot be conducted for unlimited or infinite number of cycles, 10 to the power 6 cycle is considered as a sufficient number of cycles to define the endurance limit. There is another term which is called as the fatigue life, which is frequently used with the endurance limit. Let us define the fatigue life. The fatigue life is defined as the number of stage cycles that a standard specimen can complete during the test before the appearance of the first fatigue crack. We need to see the rotating beam machine developed by R R Moore. R R Moore has developed a machine to estimate the endurance limit by laboratory method. A standard specimen is shown wherein the minimum diameter is 7.5 which is subjected to bending moment. The specimen is machine polished and it is placed on this machine by R R Moore. Here we can see that there exists a specimen which is under loaded by the weights which causes the bending moment acted upon by these weights on the specimen. Electric motor is used to rotate the specimen and revolution counter is used to count the revolutions that has been seen before the first evadence of the fatigue crack. What happens in this rotating beam? This beam is loaded under the bending. However, the component is continuously under rotation. This is the beam which is under rotation. We can see that this is rotation. It is subjected to counter clockwise direction. This component is subjected to the stress which is a reverse stress. Let us take an example of reverse stress. Here the magnitude is 0 and if I say that it is starting at this location 0, it is going to reach the maximum value. Again it is going to come at 0. Then the value is going to be negative max. Then the value of the stress is going to be equal to 0 and this cycle is going to get repeated. This is the sine wave. The plus end is maximum stress plus on plus end it is considered as tensile and on the negative end it is the compressive stress. That means in one rotation the component is subjected to plus and minus both type of stresses. Therefore, this is called as the reverse stress completely reverse. That means plus changes to minus. The magnitude is say if it is plus 100, it is minus 100. Therefore, it is the same type of stress. Now, what we need to do is let us consider that this component is rotating for one full revolution and this beam that was under bending. Let us assume that this is the beam which is under bending. I need to consider for this for one full rotation. The components side view is presented over here. Let us assume that there is one point at this location. After 90 degree the point has to travel on the top at this location. Then it is going to move again on the down end this location and after full 270 degree it is going to come over here and after 360 degree completion the point will be again this point A which changes to A1. The positions A1, A2, A3 when the component rotates to 90, 180, 270 and 360 degree at the end the component is again at point A. That means it is going to repeat the cycle. For this case I need to check that what is going to happen. At the top most point it is the full tensile stress. That means if I show the stress distribution it is the full tensile stress sigma t and it is full sigma c at this location plus on the top end minus on the lower end it is going to change its magnitude. Then we need to conduct a trial this kind of trials and count the rotations till which the component is going to sustain this load. Then how to draw this SN curve? For drawing the SN curve the specimen as shown in the figure I need to load it as shown right now. On the y-axis I need to count the number of revolutions and on the y-axis it is stress over here on the x-axis it is the number of cycles. I need to take log of it but however let us first see how to plot this SN curve. First I need to load the component up to approximately SUT but we cannot load up to SUT because the component is going to fail anyhow. I am going to load it at 0.9 SUT. I need to take the log of it. I will get the first point over here. At x-axis I need to take the point 3 and it is nothing but 10 to the power 3 cycles. If I take the log of it log of 10 to the power 3 cycle it is going to be equal to 3. Therefore this point I am going to say that it is 3 and log of 0.9 SUT. Similar to that I need to check for the next locations. I need to do I need to load the component less than this load. Then for the lower load I need to calculate the number of stress cycles. Here I am going to get number of cycles on the revolution counter. I need to draw a horizontal line and from the x-axis I need to draw a vertical line. I will get one more point. Then I need to go for the further. I need to lower the stress value. I need to draw a horizontal line for that. I need to check the number of stress cycles that the component has sustained. Draw a vertical line. I am going to get the second point. This is how I am going to get the first point, second point and over here at 0.9 SUT somewhere I am going to get one more point. Then if I go on reducing the load at some case condition starts that the points obtained are parallel to the x-axis. The line if I draw through this it is asymptotic. If I extend these join and extend these points it is going to cut over here. Here if you observe it is typically 6 or 10 to the power 6 cycles and for this case we can say that the stress corresponding to this level over here it is SE. Then what is SE? SE is the endurance limit. SE is our interest, SE is the endurance limit. Now what is the benefit to get this SE? Why to get this SE value? Because in case of the static loading we are obtained these values that means the dimensions of the components were obtained on the basis of SYT or SUT as in case of the requirement. However something has changed the changes in the type of loading. Therefore in fluctuating load we need to change the methodology. The methodology change to the fluctuating load. In fluctuating load then what happens is we need to lower these values SYT and SUT no longer remains the criteria for design. Then what remains the criteria for design is the endurance limit SE. Our interest is to get this SE value. Though in this laboratory method I had shown that the point is this point is around 0.9 SUT log of that. But actually what we need to do is we need to load the component just below the SUT value. If I load it just below the SUT value the component will not break by SUT but it is going to fail by pure fatigue loading. That means we need to attribute this failure to fluctuating load. Hence the endurance limit this becomes the criteria for design and hence our interest is to calculate this SE value. For each component you need to draw this SN curve get the value of SE and then SE upon the factor of safety as per the requirement can give you the working stress under the fluctuating load or the reverse loading in this case. This is what is the outcome of rotating beam machine the laboratory method to get SN curve the method is also called as the laboratory method by RR Moore. Here we get this value of SE for the standard specimen. But our interest is not the standard specimen our interest is to get the SE value for a component to be used in service. Therefore it becomes necessary to calculate the endurance limit of a component. Thank you. We will stop over here. Thank you very much.