 Hello everyone, myself Sachin Rathod working in Department of Mechanical Engineering from Walsh Institute of Technology, Sallapur. Today we are dealing with the chapter that is the selection of the belt under the course of machine design one. So, the learning outcome of this session is student will able to design a V belt from manufacturing catalog. The numerical is it is required to design a V belt drive to connect a 7.5 kilowatt 1,440 rpm induction motor to a fan running at approximately 480 rpm for a service of 24 hours per day. The space is available for a center distance of about 1 meter. So, in these they had given us the power, the speed of the faster pulley and the speed of the smaller pulley and service for the working hour of the service that is the 24 hours per day as well as they had given us the center distance that is the 1 meter. So, these are the given data. So, the step number one we have to determine the correction factor according to the service. So, from this tables we have to find out the correction factor that is F A. So, as they had given us 7.5 kilowatt the speed of the faster pulley is 1,440 and the fan service is 24 hours per day. So, we are getting the F A is equal to 1.3. So, as the we are getting this for the 24 hours we are getting the service factor A is equal to F A is equal to 1.3. Step number two we have to find out the design power. So, we are knowing the equation design power is equal to F A into transmitted power as already we had calculated the value of F A that is the 1.3 and the transmitted power they had given us 7.5. So, you have to calculate the design power. So, it is coming at 9.75 kilowatt. Step number three determine the cross section of the belt based on design power and high speed of the shaft in rpm from the following table. So, in this table we have to find out the section of the belt as we are knowing there are the 5 section of the belts are available that is A, B, C, D, E. So, from that section belts we have to find out the appropriate cross section of the belt for the for our application. So, as they had given us the design power is 9.75 kilowatt. So, this horizontal axis indicating the line for the design power and the vertical axis is indicating for the speed of the faster shaft. So, as they had given us the design power is 9.75 kilowatt and speed of the faster pulley is 1440 rpm. So, this vertical line indicating the design power is 9.75 kilowatt just we are projecting this line and the speed of 1440. So, we have to project this line from 1440. So, it is going to intersect at this junction. So, the cross section of the belt is B belt. So, for the B belt we have to find out the recommanded minimum pitch diameter of the pulley. So, the belt section for our application is B belt. So, for that B belt the pitch width Wp is 14 mm this is Wp. The nominal top width W is equal to 17 mm. The nominal height is 11 mm t and the recommanded minimum pitch diameter is 200 mm. So, the value of small d is 200 mm. Just now we have to check out whether the 200 mm for the B section belt is available or not. Yes, it is available. Now, the step number 4 calculate the diameter of the bigger pulley as we are knowing the equation n1d is equal to n2d just put down the value of n1n2 as we are knowing the input speed and the output space as well as we have calculated the value of small d you have to find out the value of capital D. So, by putting this value we are getting the capital D is equal to 600 mm. So, again we have to check out whether the capital D 600 mm is available or not from the manufacturing data book. So, for the B section belt we have to check out 600 mm is available so we are getting capital D is equal to 600 mm. Step number 5 calculate the pitch length of the belt so as the application is for the open belt drive we have to select the equation for the open belt drive that L is equal to 2c plus pi in bracket d plus d divided by 2 plus d minus d bracket square divided by 4c. As we are knowing the value of centre distance between a two shaft and the diameter of smaller and the bigger pulley put this value you will get the answer L is equal to 3296.64 mm. Then the step number 6 select the standard pitch length so as this pitch length L is equal to 3296.64 mm we are getting by the calculation then we have to check out whether this pitch length is available or not so that data we are getting from the manufacturing data book. So, from this tables we have to check out whether that pitch length is available or not. So, for the B section belt it is not available so we have to select the pitch length as 3200 so we have selected the pitch length for the V belt is 3200 mm so now we have to find out the correct centre distance by considering the standard pitch length. So, you can think about this can we adjust the centre distance the answer is yes we can adjust the centre distance by approximate smaller value. So, in this equation we have to put the value of the standard pitch length and we have to calculate the C value. So, we are getting the C value is equal to 950.64 mm the previously in the example they had given us the centre distance is 1 meter approximately 1 meter they had given us. So, by considering the standard pitch length we are getting the distance between input shaft and output shaft is 950.64 mm. Step number 8 determine the correction factor of the belt pitch length that is indicated by the letter fc. So, we have to select the manufacturing catalog so from this catalogs we have to find out the value of fc so for the B section belts we are getting the standard pitch length as 3200 so we have to select the value of fc so we are getting the value of fc is 1.08 for the 3200 mm standard pitch length. Step number 9 calculate the angle of contact for smaller pulley that is indicated by the letter alpha s so we are getting this value alpha s is equal to 155.71 degree then step number 10 find the arc of contact factor fd from the manufacturing data as we are getting the value of wrap angle alpha s you have to calculate the value of fd so the value of alpha s is 155.71 so 155.71 it comes in between 154 and 157 so you are getting the value in between 0.93 to 0.94 so by using the trial method or interpolation method you are getting the value of fd is 0.9357 then step number 11 find the power rating pr from the manufacturing data so this table is indicating for the B section v belt so as they had given us the speed of the smaller pulley or the faster pulley is 140 rpm and the diameter is 200 mm for the B section so we have to find out the value of pr so the value of the speed of the faster shaft is 140 for diameter of the 200 so the pr value is 5.90 as well as we have to find out the speed ratio that is the speed of the faster pulley divided by the speed of the smaller pulley so we are getting the speed ratio is 3 so in the tables the speed ratio is 2 and over as we are getting the speed ratio is 3 so for the 1440 you are getting value as 0.46 so total pr value is equal to 5.90 plus 0.46 so we are getting the value as 6.36 kilo watt then step number 12 determine the number of the belt so as we are knowing this equation for finding the number of the belt calculate the number of the belt we are getting 1.51 so we have to consider the number of the belt required for that much power transmission is 2 belts so in this figures we have to use the 2 belt as we are getting the value of the diameter of the smaller pulley diameter of the larger pulley pitch length we are knowing center distance also we had calculated and number of the belt is 2 so at last the design solution will be obtained as the diameter of the smaller pulley small d is equal to 200 mm diameter of larger pulley capital D is equal to 600 mm belt specification is B3200 LP V belt the number of the belt required is equal to 2 center distance is equal to 950.64 mm so the reference I have taken as a VB Bhandari book