Question

Prob #5: The turbine develops 150 kW of power, which is transmitted to the gears such that C and D each receive 50%. The steel shaft has a 100 mm diameter, the rotation is ω = 700 rev/min., and the shear modulus is Gst-75 GPa. The journal bearing at E allows the shaft to turn freely about its axis rn 17· The torque delivered by gear D is A) 93.8 B)512 C)895 D1025 E) 1195 F) 1790 The torque in section BC of the shaft is A)512 B)895 C) 1025 D) 1380 E) 1790 F) 2050 The maximum shear stress in the shaft is A)4.56 B)5.20 C)9.12 D) 10.4 E) 12.2 F)24.4 N-m. 18. N-m 19· MPa. 20. The angle-of-twist of shaft end E relative to B is radians A) 0.014 B)0.018 0.039 0.162 )142 291

Answer 1

Power Developed = 150 kW N = 700 rpm d = 100 mm G = 75 GPa P = 2 pi NT/60 => T = 60 times P/2 pi N T = 60 times 150 times 10^3/2 times pi times 700 = 2046.28 N-m T TildeEqual 2050 N-m (approximate) Given Gear 'C' and 'D' are taking same power Therefore T_1 = T_2 = T/2 Therefore T_1 = T_2 = 2050/2 = 1025 N-m Therefore Torque developed by gear 'D' = 1025 N-m Torque loading diagram of the shaft Therefore The torque in section 'BC' = 2050 N-m \r\n The max. shear stress is given by 16T/pi d^3. Therefore In section BC, Torque is maximum. Therefore tau _max = 16 times 2050/pi times (0.1)^3 = 10.44 times 10^6 Pa tau _max = 10.44 MPa Angular twist At end 'E' is sum of angular twist due to torque T_1 and T_2 theta = theta _1 + theta _2 theta _1 = T_1L_1/GI = 1025 times 3/75 times 10^9 times (pi/32) (0.1)^4 = 0.0042 rad