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A circular shaft of 60mm diameter is running at 150rpm. If the shear stress is not to exceed 50MPa, find the power which can be transmitted by the shaft.?
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A circular shaft of 60mm diameter is running at 150rpm. If the shear s...
Given Data
- Diameter of the shaft, d = 60 mm = 0.06 m
- Radius, r = d/2 = 0.03 m
- Speed, N = 150 rpm
- Maximum shear stress, τ = 50 MPa = 50 × 106 Pa

Calculate Polar Moment of Inertia
The polar moment of inertia (J) for a circular shaft is given by:
J = (π/32) × d4

Substituting the value of d:
J = (π/32) × (0.06)4 = 1.13097 × 10-9 m4


Calculate Maximum Torque (T)
The maximum torque (T) can be calculated using the formula:
T = τ × J / r

Substituting the known values:
T = (50 × 106) × (1.13097 × 10-9) / 0.03

Calculating T:
T = 188.49 Nm


Calculate Power Transmitted (P)
Power (P) transmitted by the shaft is given by:
P = (2πNT) / 60

Substituting N and T:
P = (2π × 150 × 188.49) / 60

Calculating P:
P ≈ 59.06 kW


Conclusion
The power which can be transmitted by the shaft, with a maximum shear stress of 50 MPa, is approximately **59.06 kW**.
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A circular shaft of 60mm diameter is running at 150rpm. If the shear stress is not to exceed 50MPa, find the power which can be transmitted by the shaft.?
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A circular shaft of 60mm diameter is running at 150rpm. If the shear stress is not to exceed 50MPa, find the power which can be transmitted by the shaft.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A circular shaft of 60mm diameter is running at 150rpm. If the shear stress is not to exceed 50MPa, find the power which can be transmitted by the shaft.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular shaft of 60mm diameter is running at 150rpm. If the shear stress is not to exceed 50MPa, find the power which can be transmitted by the shaft.?.
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