 Hello students, I am Bhagyaj Deshmukh, Professor, Department of Mechanical Engineering, Walshan Institute of Technology, Solapur. This session is on design of lever. It is from the course of machine design one. At the end of this session, we will be able to design a bell crank lever. You know that bell crank lever is used in railway signaling system, used for probably the levers of a vehicle's clutch lever, brake lever of a two wheeler. Those kind of levers you will be able to design at the end of this session. Let us see a problem, a right angle bell crank lever. That means the angle between the two arm is confined to 90 degree. It is to be designed to raise a load of 5 kilo Newton. At the short arm end, we know that there are two arm ends for the right angle bell crank lever. One is the short arm and another is a long arm. This load is attached to the short arm end. The lengths of short and long arms are 150mm respectively. The lever and the pins are made of 30C8 for which the SYT value is 400 Newton per mm square. And the factor of safety is 5. The next data is the permissible bearing pressure on the pin, which is given as 10 Newton per mm square. The lever has a rectangular cross section and the ratio of width to the thickness is 3 is to 1. The length to the diameter ratio for the fulcrum pin is given as 1.25 is to 1. What we are supposed to do, we are supposed to calculate the first one, the dimension of the fulcrum pin. Then after the shear stress in the pin, which is the induced shear stress, we need to prove that the pin is safe for shear. Then the boss dimensions at the fulcrum. We need to define the diameter of the pin at the fulcrum, we need to calculate it and then after what are the dimensions of the boss at the fulcrum. The cross section of the lever, what should be the width and height of the lever. Cross section. We need to assume that the arm of bending moment on the lever extends up to the axis of the fulcrum. Let us begin with the design of the lever. As per the given data, the syat value is 400 Newton per mm square. The factor of safety is given as 5 and the force or the load is 5 kilo Newton. This is the given data. Then for the lever, the short arm is given as 100 millimeter and long arm is 450 millimeter. The ratio d by b is given as 3. For the pin, the bearing pressure is 10 Newton per mm square and the length by diameter ratio for the pin is 1.25. Let us begin with the design of the lever. A bell crank lever. What is a bell crank lever? This point is the point where the lever is attached to the fixed link. Short arm. The long arm. As per the given data, the load 5 kilo Newton is attached at the end of the short arm. The short arm is 100 millimeter. The long arm is 450 millimeter and at the end of this 450 millimeter long arm, we need to attach the effort P. P, the effort and 5 kilo Newton is the load. Those will provide the resultant which is acting at the fulcrum point. Further width and thickness of the lever we need to define. Then to begin with, the first step is calculation of permissible stresses. Permissible stresses are obtained from the values of the SYT. We know that SYT divided by FS, we can get the working tensile stress. SYT value is 400 and factor of safety is 5. We can calculate sigma T equals 18 Newton per mm square. This is the permissible value. Similarly, applying the maximum shear stress theory, we know that SSY upon FS that will give us the tau, but how to get SSY? SSY equals 0.5 SYT by maximum shear stress theory divided by the factor of safety. We can get 0.5 into 400 divided by 5 that will give us the working shear stress value. These two values are taken as the permissible stresses for the design of the lever. Then the next step is to calculate the forces on the lever. It is a very important step because with the known forces we can then calculate what are the resultant forces acting at the fulcrum. Now the force P, the load is given to us, which is acting at a length of 100 mm from the fulcrum. The other arm is 450 mm long. We can take the moment at about point R, this point, fulcrum, force F and P we have. This force is F, this force is P, 5000 into 100, 5000 is the force 5 kilo Newton, which is 5000 Newton multiplied by 100, we can get the moment of the force that is equal to P which is unknown multiplied by the arm length 450. We can get P equals around 1111 Newton. Then we need to find out what is the resultant R which is acting at the fulcrum point. R is given as R equals square root of P square plus F square. This is R, this is P and this is F, P is 5000 and F is 1111 point 11. We can calculate the value of R with the help of these two. We can calculate R using P square plus F square square root of that, where R equals square root of 5000 square plus 1111 point 11 square, square root of that we can get the resultant equals 51 to 1.97, this is the critical step. We need to be very careful about calculating the value of resultant. The next step to calculate the pin dimensions, how to calculate these pin dimensions? Using the bearing pressure, we need to consider the bearing failure wherein the force acting at the pin is R which is equated to the pressure multiplied by the projected area of the pin. As this is a bearing failure, we need to consider the projected area of the pin. Therefore R equals P into D1 into L1, where D1 L1 represents the projected area of the pin. We can get R equals 51 to 1.97. We already have it. Therefore, if we equate this value to bearing pressure is 10 Newton per mm square multiplied by L1 D1, but we have the relation for L and D and hence we can put L equals 1.25 D1. We can get D1 equals 20.24 the unit is obviously millimeter L1 we can calculate L1 equals 1.25 times the diameter of the pin and hence it is 25.30. Next part is the pin dimensions. We can calculate the pin dimensions as per the bearing failure and need to check the pin dimensions for shear. The equation is shear stress tau is given as force divided by 2 times pi by 4 D1 square the diameter of the pin square. We know the value of R 2 times pi by 4 diameter square. We can get the induced shear stress as 7.96 Newton per mm square. This induced shear stress must be less than the permissible shear stress which is in this case it is less than the permissible stress. Hence we can conclude that the pin design is safe. We can think upon why here is 2 in the equation. Then the lever dimension, the dimension of the boss. Pin dimension is 20.24 let us take it as 21. Pin length is 25.30 let us take it as 26 round it off. Inner diameter of the boss equals diameter of the pin 21. Outer diameter of the boss equals 2 times the id it is 42. Length of the boss equals length of the pin which is equal to 26 mm. Now dimensions of the lever. The given data is lever has a rectangular cross section and the ratio of width to the thickness is 3 is to 1. Bending moment is given as bending moment equals 5000 into 100 that much Newton millimetre. We know that sigma b equals mb into y by i sigma b permissible is 80 moment is known width and depth are not known solving this equation we can get b equals 17 millimetre approximately it is 16.9 let us take it as 17 and the d value is 3 times b 51 millimetre thank you.