Question

A transfer shaft with two spur gears shown below: Input Gear 1 Output Gear Torque Bearing 1 Gear 1 Gear 2 Bearing 2 B OTO 4" 4" 4" 4 . . Weight Ibf Teeth number Pressure angle Po (diametric pitch) teeth/in Gear A 50 20° 6 Gear C 2 25 20° 6 Deep groove ball bearings at B and D. Pitch diameter of the gear is equal to the teeth number divided by diametric pitch. Shaft speed = 3000 rpm. Torque=2500 lbf-in through Gear C to shaft portion of BC and then to Gear A. Shaft material = AISI 4130 Q&T CD, E= 29x 10psi, Sut=118x 103psi, Sy=102x 103psi, assume Poisson's ratio as 0.3. Reliability for fatigue life=99%, n=2. ng for other criteria=1. Each gear is attached to the shaft by a Woodruff key, assume r/d=0.02 Calculate the required diameter of the shaft for all the different considerations given on the provided answer sheet. Use these design limit values: 1. Use Table 7-2 on textbook for allowable slope at bearings and deflection at each gear. Use the minimum limit value for the design. . . . . v is the resulting Poisson's ratio, Etrans is transverse strain (negative for axial tension (stretching), positive for axial compression) Eaxial is axial strain (positive for axial tension, negative for axial compression).

Answer 1

The free body diagram for the system on the horizontal and vertical plane is prepared. Then, Shear Force and Bending Moment diagram is prepared for horizontal and vertical planes. The maximum combined stress due to torque and moment is computed at two points and then mean and amplitude stress are computed in terms of d. The Ka, Kb, Kc, Kd are computed using the tables and the data provided (Refer Shigley). Once Ka, Kb, Kc and Kd are known, the diameter is evaluated using the Goodman Criterion for fatigue failure with a factor of safety of 2 (as given in the problem). Using the given torsional wind up, the diameter is calculated by evaluating torsional rigidity as the Young's Modulus and poisson's ratio are given and the torque acting is known.

Note: The problem is lengthy and whatever was possible to solve in the given time is done. It's difficult to solve all the parts in two hours.

T= 2500 lbf-in T= 2500 llaf-in M = 1200+ 4 = 1204 lbf-in Sport (939) = (17954.500) M = 7200 + 16 = 7216 lbf-in 16 (M+ M² + 2) = 16 (M+/M²+72) nd a 27690.355 d? ud 2= 20263.95 = 75644.66 do 3 (Amplitude Streed i Mean Streas.) LE = 0.526 Ke (94d) = 0.02 K = 1.9 (Stress Couceuthation Focton) R = 99% Ke = 0.814 K = 0.85 -0.995 -0.995 +b Ka= a (Sut) Ka= (272) (Sut > K2=(272) (813.58) Sut = 813.58Mia Ka=0.3457. Using Se = Kakh Ke Kd Sei Sei = (0.5) Sut - (0 -357) (0.85)x( - 6 )x( 5 ) x ( 5 x 18 x 10" Se = 7423 pri (Endurance Strongth) Good Man Criterion for fatigue (+3) = 1 f = 2 (Facter of Safety) (47954.305/13) 118x103 26 (27690-355/2") --- 2 7423 d=2.0225" 0.4 (17/180) rad / inches 2500 T GI T GO l 12 32 19 - - = 0.4 degrees fft 1188103 ( +U) 2 (1 +0.3x2 = G = 45384.6 pri d=5.57" (from Torsional wind up)