 Hello students, I am Dr. Bhargesh Deshmukh from Walchand Institute of Technology, Swalapur Mechanical Engineering Department. This session is on design of spur gear. At the end of this session, the learner will be able to solve a problem on calculating rated power that a gear pair can transmit. Let us see the problem. It is asked to design a spur gear pair with 20 degree full depth Involved tooth consists of a 20-teeth pinion meshing with 41-teeth gear. The reduction is almost half. The module is 3 mm, m equals 3, while the phase width is 40 mm, 4-0. The material for pinion as well as gear is steel with an ultimate tensile strength of 600 Newton per mm square. The gears are heat treated to a surface hardness of 400 BHN. The pinion rotates at 1450 rpm and the service factor for the application is 1.75. Assume that velocity factor accounts for the dynamic load and the factor of safety is 1.5. Determine the rated power that the gear can transmit. See it is a very typical problem and it is asked to calculate the power, the rated power that the gear can transmit. You can recall what are the forces acting on the gear tooth. For a typical driven gear, Pn is the resultant force which is acting at an angle alpha, the pressure angle, at the pitch point. It has two components. One is Pt and another is Pr. The tangential force is given as Pt equals to mt upon d dash. The radial force is given as Pr equals Pt tan alpha, this radial force. But we know that radial force do not take part in the power transmission. Only the tangential force is contributing for power transmission. The tangential force is obtained from the equation of torque where torque mt equals 60 into 10 to the power 6 kW upon 2 pi n. The term kw is the power and n is the speed. If we know the speed and the power input, we can calculate what is the torque. The value of tangential component therefore depends upon the rated power and rated speed. Now in the question, it is asked to calculate the rated power. What do we need to do? Think upon it. The given data in the problem. The input speed is 1450 rpm. Zp number of teeth on pinion are equal to 20. Zg number of teeth on gear are 41. Module m equals 3 millimeter. Face width b equals 40 millimeter. Service factor cs equals 1.75. Factor of safety is given as 1.5. Surface hardness of the gear bhn equals 400. Sut of the material equals 600 Newton per mm square. With this available data, we need to calculate what is the rated power that the gear pair can transmit. Now to obtain the rated power, we need to obtain the rated torque. And to obtain the rated torque, we need to obtain the rated force. If I talk more precisely, to obtain the rated torque, we need to obtain the rated tangential force. That means our job is to find out pt somehow. Tangential force is given by beam strength or wear strength. Because we know that sb equals mb sigma by. But sb is nothing but maximum value of pt. Or sw equals bqdp dash k. But there also we know that sw equals pt max. Let us start solving the problem. Step one is beam strength. Can I calculate the beam strength? See the available data. Before beginning, I need to check this. Which element is weak? Either pinion or the gear. If the material for pinion and gear is same, then the pinion is weak element. Or it is weaker than the gear. The levies form factor is 0.32 for 20 teeth. You can refer it from the design data book. Permissible bending stress is one-third of sut. Use this formula. One-third of sut equals 200. And hence permissible bending stress is 200 newton per mm square. Beam strength is hence given as mb sigma by. y is 0.32. sut is 600. From that you have calculated sigma by as 200 is 240. 3 is the face width and 3 is the module value m. And hence we can get sb equals 7680 newton. Hence the beam strength is evaluated as 7680 newton. Next part is to calculate the wear strength. For the wear strength we need the ratio factor q, then k and the diameter. Diameter is the pitch circle diameter. q equals ratio factor that equals 2 zg upon zg plus zp. That equals 2 into 41 upon 41 plus 20. We can calculate q as 1.344. k 0.16 bhn upon 100 bracket square. That equals 0.16 400 by 100 bracket square. Then k equals to 0.56. The diameter is given as dp dash equals mzp. 3 into 20, 60. Hence we can calculate sw equals b q dp dash k. b is 40, q is 1.344, dp dash equals 60 and k equals to 0.56. sw equals 8257.54. Therefore the wear strength is around 8257.54 newton. Then the effective load. Effective load is governed by cs, cv and pt. v equals pi d dash np upon 60 into 10 to the power 3. We are calculating the pitch line velocity. Pitch line velocity is 4.5553 meter per second. Using this value we can calculate cv using Barth equation. cv equals 3 upon 3 plus v. That equals 3 upon 3 plus 4.5553. cv comes out to be 0.397. When we calculate cv, the p effective is given as cspt upon cv. cs is known, cv is known, pt we don't know. pt in the form of p effective we can get. We can find p effective and calculate tangential force to evaluate the rated power. This equation p effective is not known, pt is not known. cspt by some other equation we can calculate p effective. It is possible for us to calculate the tangential force pt to calculate the rated power. Beam strength equals 7680 newton which is less than the wear strength that equals 8257.54 newton. Hence beam strength is the criteria of design in the current problem. Remember students, you need to take the lower value of beam strength or wear strength. Compare these two, the lower value you need to select for the design. sb is also equal to p effective into fs, we know. Or sb is 7680, the value of p effective we have calculated as 4.41 times pt, fs is 1.5. From this equation we can calculate the tangential force. pt equals 1161. From that how to calculate the rated power? Rated power is given as first we need to calculate the torque mt equals pt tangential force multiplied by the radius dp dash by 2. We can get the torque. From the torque we can get the rated power. Rated power is 5.29. You need to write the answer as rated power that the gear can transmit is 5.29 kW. See how we have solved the problem. For the rated power you need to calculate the rated torque. And for the rated torque you have calculated the tangential force. This is what is the theme of the problem. Thank you.