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A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?
[Take G = 85 GPa]
- a)128
- b)155
- c)182
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. Wh...
As we know that, θ = TL/GJ
Thus, outside diameter (D0) = 2 x r0 = 182 mm
∴ Inside diameter,Di = 128.56 mm
Most Upvoted Answer
A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. Wh...
To solve this problem, we can use the torsion formula:
T/J = τ/R
Where:
T = torque transmitted (80 kN-m = 80,000 N-m)
J = polar moment of inertia of the shaft
τ = shear stress
R = radius of the shaft
To find the inside diameter of the shaft, we need to find the polar moment of inertia (J) and the radius (R).
The polar moment of inertia for a hollow cylinder is given by the formula:
J = π/2 * (R^4 - r^4)
Where:
R = outer radius of the shaft
r = inner radius of the shaft
We know that the length of the shaft (L) is 1.5 m.
The angle of twist (θ) is given by the formula:
θ = TL / (GJ)
Where:
θ = angle of twist (1 degree = 0.0174533 radians)
T = torque transmitted (80,000 N-m)
L = length of the shaft (1.5 m)
G = shear modulus of the material (assume 80 GPa for steel)
We can rearrange the formula to solve for J:
J = TL / (Gθ)
Let's substitute the given values into the formula and solve for J:
J = (80,000 N-m * 1.5 m) / (80 GPa * 0.0174533 radians)
J ≈ 2.7227 * 10^(-5) m^4
Now we can substitute this value into the formula for J and solve for the inner radius (r):
2.7227 * 10^(-5) m^4 = π/2 * ((R^2)^2 - r^4)
We know that the outer radius (R) is equal to the inner radius (r) plus the inside diameter (d):
R = r + d/2
Substituting this expression for R in terms of r and d into the equation:
2.7227 * 10^(-5) m^4 = π/2 * ((r + d/2)^2)^2 - r^4)
Simplifying the equation:
2.7227 * 10^(-5) m^4 = π/2 * (r^4 + 2r^3d + r^2d^2 + (d^4)/16) - r^4
Multiplying both sides by 2/π and simplifying:
0.0000434 m^4 = r^4 + 2r^3d + r^2d^2 + (d^4)/16 - r^4
0.0000434 m^4 = 2r^3d + r^2d^2 + (d^4)/16
Rearranging the equation to solve for d:
(d^4)/16 + r^2d^2 + 2r^3d - 0.0000434 m^4 = 0
This equation is a quartic equation, which can be solved using numerical methods.
However, without additional information about the dimensions of the hollow steel shaft or the specific material properties, it is not possible to determine the inside diameter of the shaft.
T/J = τ/R
Where:
T = torque transmitted (80 kN-m = 80,000 N-m)
J = polar moment of inertia of the shaft
τ = shear stress
R = radius of the shaft
To find the inside diameter of the shaft, we need to find the polar moment of inertia (J) and the radius (R).
The polar moment of inertia for a hollow cylinder is given by the formula:
J = π/2 * (R^4 - r^4)
Where:
R = outer radius of the shaft
r = inner radius of the shaft
We know that the length of the shaft (L) is 1.5 m.
The angle of twist (θ) is given by the formula:
θ = TL / (GJ)
Where:
θ = angle of twist (1 degree = 0.0174533 radians)
T = torque transmitted (80,000 N-m)
L = length of the shaft (1.5 m)
G = shear modulus of the material (assume 80 GPa for steel)
We can rearrange the formula to solve for J:
J = TL / (Gθ)
Let's substitute the given values into the formula and solve for J:
J = (80,000 N-m * 1.5 m) / (80 GPa * 0.0174533 radians)
J ≈ 2.7227 * 10^(-5) m^4
Now we can substitute this value into the formula for J and solve for the inner radius (r):
2.7227 * 10^(-5) m^4 = π/2 * ((R^2)^2 - r^4)
We know that the outer radius (R) is equal to the inner radius (r) plus the inside diameter (d):
R = r + d/2
Substituting this expression for R in terms of r and d into the equation:
2.7227 * 10^(-5) m^4 = π/2 * ((r + d/2)^2)^2 - r^4)
Simplifying the equation:
2.7227 * 10^(-5) m^4 = π/2 * (r^4 + 2r^3d + r^2d^2 + (d^4)/16) - r^4
Multiplying both sides by 2/π and simplifying:
0.0000434 m^4 = r^4 + 2r^3d + r^2d^2 + (d^4)/16 - r^4
0.0000434 m^4 = 2r^3d + r^2d^2 + (d^4)/16
Rearranging the equation to solve for d:
(d^4)/16 + r^2d^2 + 2r^3d - 0.0000434 m^4 = 0
This equation is a quartic equation, which can be solved using numerical methods.
However, without additional information about the dimensions of the hollow steel shaft or the specific material properties, it is not possible to determine the inside diameter of the shaft.
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A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hollow steel shaft 1.5 m long is to transmit a torque of 80 kN-m. What will be the inside diameter (in mm) of the shaft, if the angle of twist should not exceed 1 ° and the allowable shear stress is 90 MPa?[Take G = 85 GPa]a)128b)155c)182d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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