 Hello students. I am Bhargesh Deshmukh, Professor of Mechanical Engineering Department at Valshan Institute of Technology, Sola. This session is on design of lever. It is from the course of machine design. At the end of this session, you will be able to design a lever safety valve. Let us check a problem. A lever loaded safety valve is mounted on a boiler to blow off at a pressure of 1.5 megapascal gauge. It is expected that this valve should blow off at a gauge pressure of 1.5 megapascal. The effective diameter of the opening of the valve is 50 millimeter. That means valve diameter is 50. The distance between the fulcrum and the dead weights on the lever is 1000 millimeter. The distance between the fulcrum and the pin connecting valve spindle to the lever is 100 millimeter. The lever and the pin are made of plain carbon steel .38 that is SYT equals 400 Newton per mm square. And the factor of safety is 5. The permissible bearing pressure at the pins in the lever is 25 Newton per mm square. The lever has a rectangular cross section and the ratio of width to thickness is 3 is to 1. Design a suitable lever for the safety valve. Here the rectangular cross section is mentioned and the ratio of width to thickness is given and it is 3 is to 1. We need to design a suitable lever for the safety valve. Now let us see what are the given things. SYT value is 400 Newton per mm square and factor of safety is given as 5. For valve the diameter is 50 millimeter and the valve pressure is 1.5 megapascal. For the lever L1 is 1000, L2 length is 100 millimeter and the cross section D by B equals 3. This is the given value. And for the pin bearing pressure P is 25 Newton per mm square. Let us start with the problem. A lever loaded safety valve. This is the shell or it is the steam space of the boiler. The steam force is exerted upwards. This is a typical valve. The force is exerted upwards. At this fulcrum this lever is allowed to swivel. The effort with the dead weights is applied at the free end. The length of effort from the fulcrum is taken as L1 and the length L2 is the length of the wall spindle to the fulcrum and the cross section is B into B or B by B is the cross section, the rectangular cross section is B. Let us first find out the permissible stresses in the lever. Sigma T we need to calculate the bending stress. Sigma T equals SYT upon FS. SYT is given as 400 and factor of safety is 5. We can calculate the permissible stress 80 Newton per mm square. Then the tau value we know that tau is given by SSY upon FS or it is 0.5 SYT upon FS. As per the maximum shear stress theory we can calculate permissible shear stress equals 40 Newton per mm square. Then we need to calculate what are the forces acting on the lever. We must be very careful about calculating the forces. The wall blows off at a blow off pressure. The force is equal to area pi by 4 d square of the wall multiplied by the blow off pressure. The diameter of the wall is 50 and the blow off pressure is 1.5. The force is around 2945 Newton. The distance L1 and the dead weights P are kept such that when the steam pressure inside the boiler reaches the limiting value the moment F into L2 overcomes the moment P into L1. The limiting moment of forces P and F about the fulcrum F into L1 equals P into L1 or we can calculate the P putting the other values or P comes out to be around 294.52. You can see that this is the force and comparatively what is the effort value. It is around 10 times less than that of the force. This is due to the lever length, lever arm. Continuing with the forces on the lever, the forces acting on the lever we can calculate as the force exerted by the steam reaction will be at the fulcrum place and this is the effort. Distances are 100 and 1000. Considering the equilibrium of forces F equals R plus P or R equals F minus P, you can calculate that the other force R is 2650.72 Newton. The next part is the pin dimension. The forces acting on the lever we know that these are the three forces R, F and P. These are to be used to calculate the pin dimensions. Force F is 2945, force P is around 294 and force R is around 2650 Newton. It is clear that the pin at the point F is subjected to the maximum force. D1 and L1 are the diameter and the length of the pin at F. We need to assume that L1 equals D1. If we assume L1 equals D1, the pin dimensions are obtained as per the bearing concentration. Force F equals bearing pressure multiplied by D1 into L1, which is the projected area of the pin. We know that for bearing concentration we need to take the projected area, which is given by D1 into L1. Therefore, D1 is around 10.85. Let us select some standard value, 12 millimeter. Then checking this pin, as we have designed the pin, we need to check this pin for the shear. Shear stress is given by force upon. As it is a double shear, we need to take two times. The cross section is in the shear pi by 4 D1 square, where D1 is the diameter of the pin. We can solve and get that tau is 13.02 Newton per mm square. This is the induced shear stress, which is well below the permissible shear stress. Therefore, the pin design is safe in shear. We have designed the pin for bending and bearing concentration and we have proven that this pin is safe for shear. The forces on the fulcrum pin, the R, it is less than the force acting on the pin at spindle. Hence we can say that pin dimension at the fulcrum will be slightly less, but you know that. You can think upon this, what is the interchangeability and how interchangeability and parts, part count reduction are related. We have taken diameter of the pin at R equals diameter of the pin at f. Then the dimensions of the lever. A gunmetal bush of 2 millimeter thickness is press fitted at the both pin zone in order to reduce the friction. I D of the boss D1 plus 2 times the thickness. Therefore, it comes out to be 12 plus 4 16. O D of the boss, it is to be taken as 2 times the I D, which is equal to 32 millimeter. We need to calculate the bending moment diagram for the lever. Bending moment will be given by P multiplied by 1000 minus 100 that comes out to be 2650 68 Newton. This is at the point f, the fulcrum point. The given data is lever as a rectangular cross section and the ratio of width to thickness is 3 is to 1 or in other words D equals 3 times B. We can use that and calculate sigma B equals m by z. Use the equation and find out what is B and D. Now the pinhole at the wall spindle makes the lever B. We need to check the bending stress at the pinhole section. The dimensions of the lever are 15 and 45. These are the dimensions. We need to add the boss on both the ends so that there is sufficient strength at the pinhole. The hole for the pin, the cross section is like this 16, 32, 45, 15 and boss increases the overall thickness of the lever becomes 20 at the pin section. Dimensions of the lever and then we can check mB is known, y is half of 45. We can calculate I and then the induced bending stress. It is well below the permissible bending stress and hence again we can state that the design is safe.