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A hollow shaft is to transmit 300kW at 80rpm. If shear stress is not to exceed 60 N/mm2 and internal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times the
mean torque.
mean torque.
- a)169.7mm, 101.8mm
- b)300mm, 180mm
- c)269.7mm, 161.82mm
- d)200mm, 120mm
Correct answer is option 'A'. Can you explain this answer?
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to ...
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to ...
Given:
- Power transmitted = 300 kW
- Speed = 80 rpm
- Maximum shear stress = 60 N/mm²
- Internal diameter (di) = 0.6 * External diameter (Do)
- Maximum torque = 1.4 * Mean torque
To find:
- External diameter (Do)
- Internal diameter (di)
Solution:
Step 1: Calculating the torque
Power = Torque * Angular velocity
Angular velocity = (2 * π * N) / 60
where, N = Speed in rpm
Angular velocity = (2 * π * 80) / 60 = 16.75 rad/s
Torque = (Power * 10^3) / Angular velocity
Torque = (300 * 10^3) / 16.75
Torque = 17,910 Nm
Step 2: Calculating mean torque
Mean torque = Torque / 2
Mean torque = 17,910 / 2
Mean torque = 8,955 Nm
Step 3: Calculating maximum torque
Maximum torque = 1.4 * Mean torque
Maximum torque = 1.4 * 8,955
Maximum torque = 12,537 Nm
Step 4: Finding the external diameter (Do)
Using the torsion formula:
T = (π / 16) * τmax * (Do⁴ - di⁴)
Rearranging and substituting the given values, we get:
Do⁴ - di⁴ = (16 * T) / (π * τmax)
Do⁴ - (0.6 * Do)⁴ = (16 * 12,537) / (π * 60)
Solving for Do, we get:
Do = 169.7 mm
Step 5: Finding the internal diameter (di)
di = 0.6 * Do
di = 0.6 * 169.7
di = 101.8 mm
Therefore, the external diameter (Do) is 169.7 mm and the internal diameter (di) is 101.8 mm.
Hence, option A is the correct answer.
- Power transmitted = 300 kW
- Speed = 80 rpm
- Maximum shear stress = 60 N/mm²
- Internal diameter (di) = 0.6 * External diameter (Do)
- Maximum torque = 1.4 * Mean torque
To find:
- External diameter (Do)
- Internal diameter (di)
Solution:
Step 1: Calculating the torque
Power = Torque * Angular velocity
Angular velocity = (2 * π * N) / 60
where, N = Speed in rpm
Angular velocity = (2 * π * 80) / 60 = 16.75 rad/s
Torque = (Power * 10^3) / Angular velocity
Torque = (300 * 10^3) / 16.75
Torque = 17,910 Nm
Step 2: Calculating mean torque
Mean torque = Torque / 2
Mean torque = 17,910 / 2
Mean torque = 8,955 Nm
Step 3: Calculating maximum torque
Maximum torque = 1.4 * Mean torque
Maximum torque = 1.4 * 8,955
Maximum torque = 12,537 Nm
Step 4: Finding the external diameter (Do)
Using the torsion formula:
T = (π / 16) * τmax * (Do⁴ - di⁴)
Rearranging and substituting the given values, we get:
Do⁴ - di⁴ = (16 * T) / (π * τmax)
Do⁴ - (0.6 * Do)⁴ = (16 * 12,537) / (π * 60)
Solving for Do, we get:
Do = 169.7 mm
Step 5: Finding the internal diameter (di)
di = 0.6 * Do
di = 0.6 * 169.7
di = 101.8 mm
Therefore, the external diameter (Do) is 169.7 mm and the internal diameter (di) is 101.8 mm.
Hence, option A is the correct answer.
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to ...
A hollow shaft is to transmit 500KW at 70 rpm. If the shear stress is not to exceed 150MN/m'2 and internal diameter is 0.7 of external diameter.find the external and internal diameter assuming that the maximum torgue is 1.6 times the mean
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer?
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer?.
A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A hollow shaft is to transmit300kW at 80rpm. If shear stress isnot to exceed 60 N/mm2 andinternal diameter is 0.6 of the external diameter, find external and internal diameters assuming maximum torque is 1.4 times themean torque.a)169.7mm, 101.8mmb)300mm, 180mmc)269.7mm, 161.82mmd)200mm, 120mmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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