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A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%?
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A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear st...
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A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear st...
Introduction:
To determine the suitable diameter for the shaft, we need to consider the power transmitted, rotational speed, and shear stress. Additionally, we also need to account for the maximum torque transmitted at each revolution exceeding the mean by 20%.
Given:
- Power to be transmitted (P) = 112.5 kW
- Rotational speed (N) = 230 rpm
- Shear stress (τ) = 212 N/mm^2
Calculating the torque:
To calculate the torque, we can use the formula:
Torque (T) = (P × 60) / (2πN)
Substituting the given values:
T = (112.5 × 60) / (2π × 230)
T ≈ 68.62 kNm
Calculating the mean torque:
To calculate the mean torque, we can use the formula:
Mean torque (T_mean) = T / 2
T_mean ≈ 68.62 / 2
T_mean ≈ 34.31 kNm
Calculating the maximum torque:
To calculate the maximum torque, we can use the given information that it exceeds the mean torque by 20%. Therefore:
Maximum torque (T_max) = T_mean + (0.2 × T_mean)
T_max = 34.31 + (0.2 × 34.31)
T_max ≈ 41.17 kNm
Calculating the diameter:
To calculate the diameter of the shaft, we can use the formula:
T_max = (π / 16) × (d^3) × τ
Rearranging the formula to solve for diameter (d):
d^3 = (T_max × 16) / (π × τ)
d ≈ (∛((41.17 × 16) / (π × 212))) mm
Calculating the value:
d ≈ (∛0.392) mm
d ≈ 0.746 mm
Therefore, the suitable diameter for the shaft is approximately 0.746 mm.
Conclusion:
Based on the given power, rotational speed, and shear stress, the suitable diameter for the shaft is approximately 0.746 mm. This diameter ensures that the shaft can transmit 112.5 kW at 230 rpm while considering the maximum torque exceeding the mean by 20%.
To determine the suitable diameter for the shaft, we need to consider the power transmitted, rotational speed, and shear stress. Additionally, we also need to account for the maximum torque transmitted at each revolution exceeding the mean by 20%.
Given:
- Power to be transmitted (P) = 112.5 kW
- Rotational speed (N) = 230 rpm
- Shear stress (τ) = 212 N/mm^2
Calculating the torque:
To calculate the torque, we can use the formula:
Torque (T) = (P × 60) / (2πN)
Substituting the given values:
T = (112.5 × 60) / (2π × 230)
T ≈ 68.62 kNm
Calculating the mean torque:
To calculate the mean torque, we can use the formula:
Mean torque (T_mean) = T / 2
T_mean ≈ 68.62 / 2
T_mean ≈ 34.31 kNm
Calculating the maximum torque:
To calculate the maximum torque, we can use the given information that it exceeds the mean torque by 20%. Therefore:
Maximum torque (T_max) = T_mean + (0.2 × T_mean)
T_max = 34.31 + (0.2 × 34.31)
T_max ≈ 41.17 kNm
Calculating the diameter:
To calculate the diameter of the shaft, we can use the formula:
T_max = (π / 16) × (d^3) × τ
Rearranging the formula to solve for diameter (d):
d^3 = (T_max × 16) / (π × τ)
d ≈ (∛((41.17 × 16) / (π × 212))) mm
Calculating the value:
d ≈ (∛0.392) mm
d ≈ 0.746 mm
Therefore, the suitable diameter for the shaft is approximately 0.746 mm.
Conclusion:
Based on the given power, rotational speed, and shear stress, the suitable diameter for the shaft is approximately 0.746 mm. This diameter ensures that the shaft can transmit 112.5 kW at 230 rpm while considering the maximum torque exceeding the mean by 20%.
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A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%?
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A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%?.
A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid shift to transmit 112.5 kw at 230 rpm taking allowing shear stress as 212 N /mm2 , find suitable diameter for the shaft. If the maximum torque transmitted at each revolution exceeds the mean by 20%?.
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