Question
Download Solution PDFA motor driving a solid circular shaft transmits 30 kW at 500 r.p.m. What is the torque activity on the shaft, if allowable shear stress is 42 MPa?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Power (P)
P = T × W
Torsion equation
\(\frac{T}{J} = \frac{{G\theta }}{L} = \frac{\tau }{r}\)
Calculation:
Given:
P = 30 kW, N = 500 rpm, D = 40 mm
T = ?
As, P = T × W
\(30 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} \)
T = 572.9 kN ≈ 573 N-m
Last updated on Jun 23, 2025
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