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The torque which may be applied to a solid shaft of 90 mm outer diameter, without exceeding an allowable shearing stress of 75 MPa, is _____ kN-m. (Answer up to one decimal place)
Correct answer is '10.8'. Can you explain this answer?
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The torque which may be applied to a solid shaft of 90 mm outer diame...
Thus, Tmax = T X C / J
So that, T = Tmax j / C
= (75 x 106) (143.14 x 10-6) = 10.74 x 103 N - m
= 10.74 kN - m ≌ 10.8 kN - m
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The torque which may be applied to a solid shaft of 90 mm outer diame...
Torque and Shearing Stress in a Solid Shaft
To determine the torque that can be applied to a solid shaft without exceeding the allowable shearing stress, the following steps need to be followed:
1. Understand the problem: We are given the outer diameter of a solid shaft, the allowable shearing stress, and we need to calculate the torque that can be applied without exceeding this stress.
2. Identify the relevant formula: The formula for calculating the torque in a solid shaft is T = (π/16) * τ * d^3, where T is the torque, τ is the shearing stress, and d is the diameter of the shaft.
3. Convert the diameter to meters: The given diameter is in millimeters, so we need to convert it to meters. The outer diameter of 90 mm is equal to 0.09 meters.
4. Convert the shearing stress: The given shearing stress is in MPa, which stands for mega pascals. To convert it to pascals, we multiply it by 10^6. Therefore, the shearing stress is 75 MPa * 10^6 = 75,000,000 Pa.
5. Calculate the torque: Using the formula T = (π/16) * τ * d^3, we can substitute the values to calculate the torque.
T = (π/16) * 75,000,000 Pa * (0.09 m)^3
T ≈ 10.8 kN-m
Final Answer: The torque that can be applied to the solid shaft without exceeding the allowable shearing stress is approximately 10.8 kN-m.
To determine the torque that can be applied to a solid shaft without exceeding the allowable shearing stress, the following steps need to be followed:
1. Understand the problem: We are given the outer diameter of a solid shaft, the allowable shearing stress, and we need to calculate the torque that can be applied without exceeding this stress.
2. Identify the relevant formula: The formula for calculating the torque in a solid shaft is T = (π/16) * τ * d^3, where T is the torque, τ is the shearing stress, and d is the diameter of the shaft.
3. Convert the diameter to meters: The given diameter is in millimeters, so we need to convert it to meters. The outer diameter of 90 mm is equal to 0.09 meters.
4. Convert the shearing stress: The given shearing stress is in MPa, which stands for mega pascals. To convert it to pascals, we multiply it by 10^6. Therefore, the shearing stress is 75 MPa * 10^6 = 75,000,000 Pa.
5. Calculate the torque: Using the formula T = (π/16) * τ * d^3, we can substitute the values to calculate the torque.
T = (π/16) * 75,000,000 Pa * (0.09 m)^3
T ≈ 10.8 kN-m
Final Answer: The torque that can be applied to the solid shaft without exceeding the allowable shearing stress is approximately 10.8 kN-m.
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The torque which may be applied to a solid shaft of 90 mm outer diameter, without exceeding an allowable shearing stress of 75 MPa, is _____ kN-m. (Answer up to one decimal place)Correct answer is '10.8'. Can you explain this answer?
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