8.4 (A). A hollow shaft has to transmit 6 MW at 150 rev/min. The maximum allowable stress is not to exceed60 MN/m² nor the angle of twist 0.3° per metre length of shafting. If the outside diameter of the shaft is 300 mm findthe minimum thickness of the hollow shaft to satisfy the above conditions. G = 80 GN/m² [61.5 mm.]

According to the given data P=60MW N=150rpm τ=60MN/m2 θ=0.3°=0.3xπ180=0.005235rad L=1m D=300mm…

Answered: 8.4 (A). A hollow shaft has to transmit…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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For the circular shaft (G=12×106 psi) shown below: a. Compute the reaction torque and draw the torque (T) diagram. b. Find the max shear stress (Tmax) along the shaft. c. Compute the angle of twist AE at E. 1000 lb in 500 lb in 800 lb. in A B 2 in 2 in + 3 in D 5 in to E AC CD 2 DE
8.9 (A/B). A solid steel shaft of 200 mm diameter transmits 5 MW at 500 rev/min. It is proposed to alter the horsepower to 7 MW and the speed to 440 rev/min and to replace the solid shaft by a hollow shaft made of the same type of steel but having only 80% of the weight of the solid shaft. The length of both shafts is the same and the hollow shaft is to have the same maximum shear stress as the solid shaft. Find: (a) the ratio between the torque per unit angle of twist per metre for the two shafts; (b) the external and internal diameters for the hollow shaft. [L.Mech.E.] [2.085; 261, 190 mm.]
Q2(c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c). If the allowable shear stress of the shaft is 80kPa, i)Determine the maximum torque T that can be transmitted by the shaft. ii)What would be the maximum torque T' if a 25 mm diameter hole is bored through the shaft? Figure Q2(c.) |Q2 c) (i) T c The torsion formula :7 = J Т (0.019) T (0.019)^4 ; Therefore, the max torque T = x10 Nm 80 2 Q2 c) (ii) For point lying on the outside suurface of the shaft: x10 ="(0.019) * (0.019*-0.0125 ) ; Therefore the max torque T' Nm 80 2 Q2 c) (iii) The shear stress at point located at inner surface of shaft, T'c T = -= J ) (0.0125) = 7 (0.019*-0.01254) T'c kPa 4. 2 2.

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