 Hello, my self Sachin Rathod working as assistant professor in mechanical engineering department from Walsh Institute of Technology, Sollapur. Today we will discuss the bit that is the selection of the belt part 6 from the course machine design one, the learning outcome. The learning outcome of this session is the learner will able to understand the constructions of cross bell drive. So, already in the previous part we had seen the construction of the open bell drive. Now, we will see the construction of the cross drive. So, from these figures you are getting that the two pulleys are there on which the belt is going to mount in the cross manner. The cross bell drive is used when the sense of the rotation of both pulleys are in opposite directions. Means suppose your the input pulley is rotated in the clockwise direction, your the output pulley will get rotated in the opposite direction that is in the anticlockwise direction. So, if the rotations required in the transmission of the power in the opposite we can use the cross bell drive. The angle of the wrap is more in case of the cross bell drive, the wrap angle is more as compared to the open bell drive. So, because of the wrap angle is more, the power transmitting capacity is more. In case of the cross bell drive, the belts rub against itself while crossing. Therefore, the wear rate increases which reduces the life of the belt. So, if you observe at this section the two belts are rub or the wear will get occurs. So, the life of this belt is less as compared to the open bell drive. Sorry, now we will see the construction of cross bell drive. What are the various terms or the various parameter they are going to use in the cross bell drive that we will see in this slide. So, here through these figures we are getting that this is the alpha s is nothing but the wrap angle for the smaller pulley. Next alpha b is nothing but the wrap angle for the larger pulley. So, you can pause this video and think about this. What is been by a wrap angle or how it will get affect on the power transmission of the belt. Next d is nothing but diameter of the larger pulley, small d is nothing but the diameter of the smaller pulley. d is nothing but the centre distance between these two pulley that the dimensions all were taken in the millimetre. Now, we will solve this construction of the cross bell drive through finding the length of the belt. So, if you observe the triangle o g o 1, I will use the pointer for showing that. So, if you observe this triangle o g o 1, just I will draw simply that triangle. So, in this triangle o g o 1, we are getting the sin beta the value of this angle sin beta is equal to opposite side that is o g o 1 divided by hypotenuse o o 1. So, here I have written o 1 g divided by o o 1 and the o 1 g is nothing but this distance that is the g e plus o 1 e divided by o o 1 is equal to o 1 c plus g c is nothing but your the o f, this distance g c is nothing but your this distance o f, this is your the distance is o f both are parallel to each other. Therefore, the o 1 c is nothing but your the radius is nothing but capital D by 2 and o 1 f this is a radius is equal to small d by 2, just put this value and o o 1 is nothing but the center distance is nothing but your the c value, put this value you will get the value of sin beta. Therefore, sin beta is equal to capital D plus small d divided by twice c also the alpha s that is a wrap angle for the smaller pulley. The wrap angle for the smaller pulley is nothing but this angle 180 plus beta plus this beta. So, 180 plus twice beta you will get the wrap angle for the smaller pulley is equal to pi plus 2 beta and the wrap angle for the larger pulley is nothing but total 180 degree plus beta plus beta. Therefore, we are getting 180 plus twice beta plus therefore, we are getting 180 plus twice beta is equal to pi plus twice beta. Therefore, we can easily find out the length of the belt. So, from this figures you will get the length of the belt is nothing but the arc FAB plus arc length BC plus arc CDE plus length EF. So, like this way we are getting the length of the belt. So just put that value arc FAB is nothing but arc FAB is nothing but the this radius into the angle therefore, d by 2 into alpha s plus OG length plus your the length BC is nothing but OG this is a length BC is nothing but your the length OG plus arc CDE this is arc CDE is equal to alpha B into radius that is a capital D by 2 plus EF. EF is nothing but again we can write down it as a OG. Just I have rewritten that equations once again. So just put the value of alpha s and alpha B where the alpha s is equal to pi plus twice beta and alpha B is equal to pi plus twice beta. Therefore, put that value of the alpha s and alpha B in the above equations you will get this equation and also you will put the value of OG. The length OG is nothing but we are getting O O 1 cos beta you will get the value of the OG just put that OG value C cos beta here also C cos beta. Therefore, just multiply this equation inside you will get this equation for the length of the belt. Again we have to get the parameter that is a capital D and small d together by considering the pi by 2 against the beta terms together. So you will get this new equation for the length of the belt. Again we can call this equation number 1. Again we have to simplify by considering that we are knowing sin beta is equal to capital D by small d divided by twice C. For the small angle at the beta we can call sin beta is equal to beta is equal to capital D plus small d divided by twice C and we are knowing that cos beta is equal to 1 minus 2 times sin square beta by 2. Just simplify this equation by putting the value of the beta that the sin beta is equal to capital D by twice C that the sin square beta by 2 is nothing but the beta square by 4. Again we have to multiply by 2 that is why I have written beta square by 2. Put the value of the beta beta is equal to capital D plus small d divided by twice C in this equations you will get this equation. So put the value of the beta and the cos beta in the equation number 1. So if I put that value in the equation number 1, this is the equation number 1 put the value of the beta and cos beta. So we have to put the value of the beta and the cos beta in this equation. So you will get this equation. Just I have put the value of the beta and cos beta. Again since you have to just multiply this 2 bracket and you will get the simplified manner. So we are getting these equations after multiplying terms. So just again we are going to simplify this equation. So finally we are getting this equation for finding the length of the cross bell drive. So this is the derivations for getting the value of length of the cross bell drive. So for this slide I have taken a reference from the VB Bhandari book from the design of machine element. So thank you.