Question

The 50-mm-diameter shafts AB & DE are made of A-36 steel (G = 75 GPa) and are supported on journal bearings that allow the shafts to rotate freely and the turbine at Eis fixed from turning. If the allowable angle of twist of C with respect to Eis 0.2" and shaft AB rotates at 30 Hz, determine the maximum power (hp) that can be delivered from the motor, 0.75 m Tunin

Answer 1

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Solution. In the given figure two shafts are shown viz AB (connected to motor) and DE (connected to turbine) Two shafts are connected by gears (assuming cut B e c. As shafts AB rotates at 30 Hz

so WB LIIN = 4130 = 60 17 xadiseo As_size_o7 Bis_somm_ond Ćis. 75 mm hence ولا ولا تبالي GOT 5o с 75 We = speed of shaft DE = 4011 rad/sec Now let us find maxi for que that transmitted can be We are given with allowable angle of twist at C = 0.ge with respect_to E

С= 0-3 ут 180 = 3.49 x 103 radians Now applying pure torsion equation for shaft DE r 41 I- G8 All_Symbols axe__with J L stdnotations Strength Rigidity_coiterion Griterion Here we use L = length only for port CE = 0.75 m. = Allowable angle of twist = 3.49 x103 radions o e

G = 75 Gpa = 75 X10 N m2 32 J = Polax_moment_of_ inextio of_shalt DE LA Ix0.054 32 From Rigidity criterion I G. ㄴ Putting-valuex J ㅜ G. J. o L max toyque that can be transmitted T 75X100 x T yo ost x 3.49 x103 x L 32 0.75 T = 214.143 Nm Hence maximum power that can be be transmitted from motor P = T. Wc 8 14.143 X 40 π = 26.910 kW Answer - Mox power that can be transmitted is 26.990 kW