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) (b) The A-36 solid steel shaft is 3.5 m long and has a diameter of 60 mm. The shaft is used to transmit a torque of T = 524 N-m from the engine to the generator. Determine the angle of twist () of the shaft. The shear modulus of elasticity (G) of the shaft material is 75 GPa. (10 points)

Answer 1

1 Question Solution : a) Innen diameter (da) +.5 in Outen diameter dz= 2 in allowable shear stress T - lo ksi speed IN= 45rpm Formula for Porque Poansmitted Hollow Pipe T= I el d2"-de") 16 by व 0 To Te dz}(1- di? (1-4) (2)*(1-(49)) T= Ix 10x Х 16 T= 10.7378655 kilo-Pound Inch 1 kilo-Pound Inch = 112.985014 Nom 12/3-2/788 N-m 10.7378655 kilo-Pound Inch Torque (T)- 12/3.21788N-m formula for Power (D) : Pe 27 NT 60

Х I HP Pe 2XU x 451213-21788 60 P=5717.1545 Watt conversion of ceealt Into Hourse Power 1 watt 746 5719.1545 = 5717.1545 Hp Watt 746 5717.1545 - 7.6637 Hp Watt Power Toansmitted is [P=706637 Hp Ans Formula For angle of Twist dis TOL T= 524 Nm T =524X10% N-mm 9.3 6- 524x102x 3500 757103x1272345-025 L= 3.5m = 3500 mm J = I xd4 1$ = 0.0192 19 Radian (60)4 1. 10117° J= /272345.025 mm (1 Radian= 180 dequee d=bomml Ans 16- (9= 75eppa = 75x10²N mm3 32 서식 32 T