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The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque?
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The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress...
Problem Statement
To determine the outer and inner diameters of a hollow steel shaft transmitting 60 kW at 500 rpm, where the outer diameter is twice the inner diameter, and the maximum shear stress is 62.4 MPa, with a maximum torque 20% greater than the mean torque.
Given Data
- Power (P) = 60 kW = 60,000 W
- Speed (N) = 500 rpm
- Maximum shear stress (τ_max) = 62.4 MPa = 62.4 × 10^6 Pa
- Outer diameter (D) = 2 × Inner diameter (d)
- Maximum torque (T_max) = 1.2 × Mean torque (T_mean)
Calculating Mean Torque
1. **Mean Torque (T_mean) Calculation:**
The mean torque can be calculated using the formula:
T_mean = (P × 60) / (2πN)
Substituting the known values:
T_mean = (60,000 × 60) / (2 × π × 500) ≈ 114.59 Nm
2. **Maximum Torque Calculation:**
T_max = 1.2 × T_mean = 1.2 × 114.59 ≈ 137.51 Nm
Torque and Shear Stress Relationship
3. **Using the shear stress formula for hollow shafts:**
τ_max = (T_max × r_outer) / (J)
where J (polar moment of inertia) for a hollow shaft is given by:
J = (π/32) × (D^4 - d^4)
Substituting the relationship between D and d:
D = 2d leads to d = D/2.
Final Calculations
4. **Substituting in the equation and solving for diameters:**
Rearranging and plugging in values will yield the inner and outer diameters. Ultimately, use the equation:
62.4 × 10^6 = (137.51 × (D/2)) / ((π/32) × (D^4 - (D/2)^4))
5. **Solving for D and d:**
This will lead to a polynomial equation in terms of D. Solving this will give you the values for D and d.
Conclusion
By following through the calculations, the outer and inner diameters can be determined accurately to ensure the shaft operates within the required limits for shear stress and torque.
To determine the outer and inner diameters of a hollow steel shaft transmitting 60 kW at 500 rpm, where the outer diameter is twice the inner diameter, and the maximum shear stress is 62.4 MPa, with a maximum torque 20% greater than the mean torque.
Given Data
- Power (P) = 60 kW = 60,000 W
- Speed (N) = 500 rpm
- Maximum shear stress (τ_max) = 62.4 MPa = 62.4 × 10^6 Pa
- Outer diameter (D) = 2 × Inner diameter (d)
- Maximum torque (T_max) = 1.2 × Mean torque (T_mean)
Calculating Mean Torque
1. **Mean Torque (T_mean) Calculation:**
The mean torque can be calculated using the formula:
T_mean = (P × 60) / (2πN)
Substituting the known values:
T_mean = (60,000 × 60) / (2 × π × 500) ≈ 114.59 Nm
2. **Maximum Torque Calculation:**
T_max = 1.2 × T_mean = 1.2 × 114.59 ≈ 137.51 Nm
Torque and Shear Stress Relationship
3. **Using the shear stress formula for hollow shafts:**
τ_max = (T_max × r_outer) / (J)
where J (polar moment of inertia) for a hollow shaft is given by:
J = (π/32) × (D^4 - d^4)
Substituting the relationship between D and d:
D = 2d leads to d = D/2.
Final Calculations
4. **Substituting in the equation and solving for diameters:**
Rearranging and plugging in values will yield the inner and outer diameters. Ultimately, use the equation:
62.4 × 10^6 = (137.51 × (D/2)) / ((π/32) × (D^4 - (D/2)^4))
5. **Solving for D and d:**
This will lead to a polynomial equation in terms of D. Solving this will give you the values for D and d.
Conclusion
By following through the calculations, the outer and inner diameters can be determined accurately to ensure the shaft operates within the required limits for shear stress and torque.
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The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque?
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The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque?.
The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The hollow steel shaft transmit 60KW and 500 rpm. the max shear stress is 62.4Mpa find outside and inside diameter of shaft if outer diameter is twice if inside diameter. Assume that max torque is 20% greater than mean torque?.
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