 Hello students. I am Dr. Bhagyaj Jaishmukh from Mechanical Engineering Department, Valchin Institute of Technology, Solapur. This session is on design of spur gear. We are going to solve a problem on gear design. At the end of this session, the learner will be able to design a gear for a typical application. The problem given is, it is required to design a spur gear speed reducer for a compressor. An application is a compressor. Running at 250 rpm driven by a 7.5 kW 1000 rpm electric motor. That means the driver element is motor driven is a compressor. Power required is 7.5 kW which is to be transmitted by the gear train. The center distance between the axis of the gear shaft should be exactly 250 mm. Center distance radius of the pinion plus radius of the gear that is exactly 250 mm. The starting torque of the motor can be assumed to be 150% of the rated torque. In this statement, CS is given. The gears are made of carbon steel 50C4 with SUT equals 700 Newton per mm square. The pressure angle is 20 degree. The factor of safety given is 2 for preliminary design based on the use of velocity factor. What is asked? Design the gears and specify their dimensions. Assume that the gears are manufactured to meet requirements of grade 6. Calculate the dynamic load by using Buckingham's equation. Calculate the effective load. What is the actual factor of safety against the bending failure? Using the same factor of safety against pitting failure, specify the suitable surface hardness for the gears. That is the BHN value you need to estimate. What is dynamic load? If you recall the concept, the methods to account for the dynamic load. First method was the approximate estimation by the velocity factor in the preliminary stages of gear design. Where you do not know the dimensions of the gear. Second method is precise calculation by Buckingham's equation in the final stages of gear design. Can we calculate the precise value of dynamic load in the given problem? Think upon it. Look towards the given data. The power to be transmitted is 7.5 kW. Speed 1000 rpm of pinion. Gear 250 rpm. Center distance is exactly 250 mm. Starting torque equals 150% of the rated torque. CSE is 1.5. SUT equals 700 Newton per mm square and factor of safety is 2. Grade of machining is 6. Then let us begin with the problem. First we need to estimate the module based on the beam strength. For this, we need to start with the first part. Center distance A equals half dp dash plus dg dash. We know that center distance is 250. Hence 250 equals half of dp dash plus dg dash. Therefore we can say that dp dash plus dg dash equals 500 mm. This is equation A. We also know that dg dash plus dp dash equals np upon ng that equals 1000 upon 250 that equals 4. This is equation B. With the help of equation A and B, we can calculate that dp dash equals 100 mm and dg dash equals 400 mm. That means we know now the diameter of the pinion and diameter of the gear. For pitch line velocity, V equals pi d dash np upon 60 into 10 to the power 3. V equals 5.236. Why we have calculated the velocity, the pitch line velocity? We need to calculate the velocity factor, Cv. Cv equals 3 upon 3 plus v by the Barth equation that equals 0.3643. We have not estimated the module right now. We are in the process. Face width B is assumed to be 10 times of module as it is not given. Levis form factor is unknown at this stage. We cannot get the Levis form factor. Then what to do? Why varies from 0.32 for 20 teeth to 0.358 for 30 teeth? Assuming an intermediate value, Y equals 0.34. See this is the beginning. We need to assume some value, then we need to verify whether the factor of safety is adequate, the beam strength. Sb equals mb sigma by m is not known. b is in the terms of module. Sut is known. Sut upon 3 is sigma b and y which is assumed value. We can get beam strength in the terms of module. The torque with the help of this equation mt equals 16 to 10 to the power 6 kA upon 2 pi np. As all the terms are available, we can calculate torque equals 71619.93 Newton mm. Remember this equation gives you torque directly in Newton mm. The tangential force we need to find out pt equals 2 mt upon dp dash. With the help of this equation, we can calculate pt equals 1432.39 Newton. Our interest is to calculate the effective load. p effective equals cspt upon cv. With the help of this equation, p effective equals 5897.85. p effective and sb. See the first equation, sb is in terms of module and we have obtained p effective 5897. Now, can we calculate the module with the help of this equation c and d? Let us see. The tangential force pt equals 2 mt upon dp dash. We have the tangential force. Effective load is given as cspt upon cv. Both the things are available. From equation c and d, p effective into fs equals sb or we can get the module by this equation m equals 3.86. Next step is getting the gear dimensions with the help of this module value. The choice, the preferred preference value of the module is 4 mm. We can get b equals 10 times module 40 mm. Zp, the number of teeth of pinion are known. Module is known. Then we can calculate Zp. Zg we can calculate diameter of the gear upon module. Number of teeth are known. Levi's form factor for 25 teeth. See what we are doing. Now we know what are the number of teeth on pinion and hence we can calculate y for that. Exact beam strength equals 1269 3.33 When we put this module value fs width, sigma b and the correct value of y the dynamic load using Buckingham's equation for grade 6 e equals 8 plus 0.635 for pinion we can get 5 equals m plus 0.25 square root of dp dash that equals 4 plus 0.25 square root of 100 e p equals 8 plus 0.635 we can calculate error for pinion 12.095 micron for gear we need to use the same method and calculate error in gear that equals 13.67 micron the total error is 25.765 micrometer or micron we need to convert it into millimeter by multiplying using a factor 10 to the power minus 3 deformation factor c equals 11400 Newton per mm square knowing these values we can calculate dynamic load using Buckingham's equation as 6448.30 the effective load is given as p effective equals cspt plus pd if we use this equation p effective equals 8596.89 the actual factor of safety against bending failure now we know the exact value of p effective we need to use this value exact p effective and the beam strength we can get factor of safety equals 1.48 we can conclude that the design is satisfactory for module equals 4 mm the surface hardness of gears ratio factor is obtained as 2 zg upon zg plus zp it is 1.6 effective load is given as p effective into fs that equals sw we are considering now wear strength as we need to find out surface hardness or p effective into fs equals bq dp dash k if we rearrange the terms and put the value of sw we can get bhn equals 352.49 or it is around 360 surface hardness of gear we can say that it is 360 bhn the last term is dimension of gears if we know the module all other terms can be calculated we have calculated number of teeth on pinion number of teeth on gear module face width pcd's addendum equals modum that equals 4 mm dredendum equals 1.25 times module 5 mm clearance to thickness and fillet radius can be obtained by such proportions thank you