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The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal to
- a)59 mm
- b)118 mm
- c)177 mm
- d)236 mm
Correct answer is option 'B'. Can you explain this answer?
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The minimum diameter of solid steel shaft that will not twist through...
Given, θ = 3° = 3 × 
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I = 6m; T = 14 kN − m; G = 83GN/m2
= 118mm
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The minimum diameter of solid steel shaft that will not twist through...
Calculation of Minimum Diameter of Solid Steel Shaft
Given parameters:
- Torque (T) = 14 kN-m
- Length (L) = 6 m
- Maximum angle of twist (θ) = 3°
- Modulus of rigidity (G) = 83 GN/m²
Formula:
The maximum angle of twist (θ) can be calculated using the following formula:
θ = TL/GJ
where J = (πd⁴)/32 is the polar moment of inertia of the solid shaft.
Rearranging the formula, we get:
d = [(32θGJ)/π]^(1/4)
Solution:
First, we need to calculate the polar moment of inertia (J) for the solid shaft:
J = (πd⁴)/32
= (π/32) x (0.01⁴) [taking diameter d = 10 mm]
= 8.18 x 10^-10 m^4
Now, substituting the given values in the formula:
d = [(32 x 3 x (π/180) x 83 x 10^9 x 8.18 x 10^-10)/π]^(1/4)
= 118 mm
Therefore, the minimum diameter of the solid steel shaft that will not twist through more than 3° in a 6 metre length, when subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m², is 118 mm.
Answer: Option B (118 mm)
Given parameters:
- Torque (T) = 14 kN-m
- Length (L) = 6 m
- Maximum angle of twist (θ) = 3°
- Modulus of rigidity (G) = 83 GN/m²
Formula:
The maximum angle of twist (θ) can be calculated using the following formula:
θ = TL/GJ
where J = (πd⁴)/32 is the polar moment of inertia of the solid shaft.
Rearranging the formula, we get:
d = [(32θGJ)/π]^(1/4)
Solution:
First, we need to calculate the polar moment of inertia (J) for the solid shaft:
J = (πd⁴)/32
= (π/32) x (0.01⁴) [taking diameter d = 10 mm]
= 8.18 x 10^-10 m^4
Now, substituting the given values in the formula:
d = [(32 x 3 x (π/180) x 83 x 10^9 x 8.18 x 10^-10)/π]^(1/4)
= 118 mm
Therefore, the minimum diameter of the solid steel shaft that will not twist through more than 3° in a 6 metre length, when subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m², is 118 mm.
Answer: Option B (118 mm)
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The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer?.
The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum diameter of solid steel shaft that will not twist through more than 3° (degrees) in a 6 metre length. When subjected to a torque of 14 kN-m with modulus of rigidity G = 83 GN/m2, is equal toa) 59 mmb) 118 mmc) 177 mmd) 236 mmCorrect answer is option 'B'. Can you explain this answer?.
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