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 (0) of the shaft. The shear modulus of elasticity (G) of the shaft material is 75 GPa. (10 points)

Answer 1

Given data- Inner diumeter (di)=1.5 in. outer diameter (Do) = 2 in. shear stress (2) - 10ksi cloxLo psi. hp? We we know that, - Torque (1b-in) x RPM power ( HP) 63,025 RPM =45 Now, calculating torque (for hollow shaft). x Zallow T= T (Dot-01 The 12.15] x luxor T 10737.86 lbiin from equation power 10737.86 x 45 63025 power = 7.6.6 hP

b) shart. Angle of twist For solid La 3.5 TE 524 Nm. ds GOMO) 0-0600 ra radius 0-08 m. G= 75 GPa = 75X100 N/m2 Now, we know the torosion equation. TL IG : © J (Polar moment of inertia for solid shaft) TE T (54) 2 JE IT Co-0334. Ja 1.273 X10 m4 from equation ③ e 524 (N-m) x (3:5)m 1273 X 10604)* 75X109 EN m2 tegres radlian. 10 = 0.0192 0 = 0.0192(10) 0 21.10° (degree) Angle of twist.