Question

Q- 1. A flexible coupling is to be used to connect two shafts, carrying 24 kW power @ 1200 rpm. If the connecting strip is placed 2.5” away from the center of the shafts, how many folds of 0.015” by 0.15” steel made from SAE 1045, quenched in water and drawn at 600°F) are required? (SAE 1045, quenched in water and drawn at 600°F has S = 1 GPa and s = 775 MPa) NOTE: Solve this problem (in SI Units) by using a Factor of Safety of 2.5 on s and knowing that the Maximum Shear Stress is the half of the Ultimate Tensile Stress.

Answer 1

Answer? Given, power = 1200 pm 26 KN strip - 2,5" folds 0.015" by 0.15" quenched in water - boor Su = 1 Gla = 775 MPa Sy Mein Toogel Tmean = 60P 27 Qux 103 x60 21 X1200 600 Nm Tmean Service factor = 135 Tmax Tmax = 135 x 60orm 810 Nm Tmax 810x10 30mm Fox Txd 7 8 x 1 x 10*x d 8102 103 x 16 71x106 3 810X103 % 16 TX 106 8107103 d3 D2 20 = 816038 1D - 3.207410 221.50 = 1.5x1.6038 29.4057 mm 1 d OTT 1.60380 mm