Q2)A hollow shaft is required to transmit 600 kW at 110 r.p.m., the maximum torque being 20%greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of 3meters not to exceed 1.4 degrees. Find the external and internal diameter of the shaft, if youknow that the internal diameter is 37.5% of the external diameter. Take modulus of elasticityas 200 GPa and Poisson's ratio 0.19.

Step 1: Step 2: Step 3: Step 4:

Answered: Q2) A hollow shaft is required to…

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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4. For the solid shaft, 1 inch in diameter, subjected to bending and torsion, find the principal stresses, maximum shear stress and factors of safety using the maximum shear stress (Tresca) and distortion energy (von Mises) criteria if the yield strength is also 50,000 psi: M = 300 in-lb T = 500 in-lb
A shaft is required to transmit 1 MW power at 240 r.p.m. The shaft must not twist more than 1 degree on a length of 15 diameters. If the modulus of rigidity for material of the shaft is 80 GPa, find the MPa] diameter of the shaft and shear stress induced.
A 2-inch line shaft is transmitting 20 hp rotating at 200 rpm, carrying two pulleys spaced 30 inches apart which can cause a torsional deflection on it. If the shear modulus of elasticity is 12,000,000 psi, find the angle of twist in degrees. a. 0.573⁰ C. 1⁰ b. 0.75⁰ D. 0.25⁰
Submit correct and complete solutions. give propriate Explanation. Provide step-by-step detailed explanations.
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.
A shaft is required to transmit power 1MW at 240r.p.m. The shaft must not twist more than 1 degree on length of 15 diameters. If the modulus of rigidity for material of the shaft is 80GPa, find the diameter of the shaft and shear stress induced.
A shaft running at 400 r.p.m. transmits 10 kW. Assuming allowable shear stress in shaft as 40 MPa, rA- nml find the diameter of the shaft.

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