Question

Question 1. (a) A steel tube having inner and outer dimeters of 1.5 in. and 2 in., respectively, is turning at 45 rev/min. If the allowable shear stress (Tallow) of the tube material is 10 ksi, determine the horse power (hp) transmitted by the tube. (20 points)

Answer 1

*. Given Data o - 38.1 mm steel tube . inner dia di = los in • Outer dia = do = 2 in. Revoluticy = N = 45 rew/ min = 45/60 rer/s lo Ksi. so.8 mm. = allow but 1 ksi= 6.895 MPa ii 10 ksi 68.95 mpce soluting & Torsionul Formule T J Z R where, I= Torque Transmitted Shaft. allowable shear stren ? R = Outer rudius. Tp = polur moment of inertia

4 For hollow tube. I (dt - de In 32. * Now Put value in above Torsional Formulu. ri T ZX Jp R T zx T (dot - dit ) dole (50.84 38.14 68.95 x 7/8 50.8% T= 1213258.607 No mm T 1213.26 Nom

» Now, Power Equition for Torque.. power 2ANI 60 2 x 7 x 45 x 1213.26 60 all =5717.35 watt but I Watt H. p I 746 A 1 Hop = 746 Wutt . ,. power Hip 5717,35 746 + power power = 7.664 Horse power (H.p) * Answer