Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- A solid circular drive shaft of length = 0.5 m, is to be designed to transmit up to 45000 W at 1800 rev/min. If the maximum allowable shear stress is 60 MN/m? and the modulus of rigidity, G = 79 GPa, determine the following: Q4 b) The minimum diameter of the shaftarrow_forward3. A solid circular shaft is 1.8 m and is made of steel with G = 70 GPa. When rotating at w = 70 rad/s, it transmits 28 kW of power from engine A to generator B. As the design engineer, you are asked to: (a) Determine the shaft's smallest allowable radius if the allowable shear stress is T allow = 80 MPa. (b) Determine the shaft's smallest allowable radius If the allowable angle of twist is allow = 0.1 rad (in other words, the shaft is restricted not to twist more than 0.1 rad). (c) If both conditions (a) and (b) are required simultaneously, which radius will you select as the allowable radius. (In other words which condition controls the situation)? Briefly explain your answer. (1 kW = 1000 Watts) Barrow_forwardThe internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consist of an axial force of 12.5 kN, a bending moment of 1.2 kNm, and a torsional moment of 2.25 kN. m. ( Figure 1). Figure 1.2 kN-m 12.5 kN 2.25 kN-m 1 of 1 Part A Determine the principal stresses at point B. Express your answers using three significant figures separated by a comma. 01,02 = IVE ΑΣΦ vec Submit Previous Answers Request Answer ? MPaarrow_forward
- When a steel rod below is subjected to torque (T = 1 kN · m) and bending moment (M = 4 kN · m) simultaneously, calculate equivalent stresses using Tresca's theory and Von-Mises' theory 2. M M 50mmarrow_forwardQ2(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.arrow_forward
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