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A circular shaft of 50 mm diameter is required to transmit a torque from one shaft to another. If shear stress is not to exceed 40 Mpa, the safe torque is:
  • a)
    0.882 kN-m
  • b)
    0.982 kN-m
  • c)
    1.982 kN-m
  • d)
    2.00 kN-m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A circular shaft of 50 mm diameter is required to transmit a torque fr...
Using the relation: 
Where,
T = Torque applied to the shaft
IP = Polar section modulus
τ = shear stress in the material
R = radius of the shaft

= 0.982 kN-m
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Most Upvoted Answer
A circular shaft of 50 mm diameter is required to transmit a torque fr...
Given: Diameter of circular shaft, D = 50 mm = 0.05 m
Shear stress, τ = 40 MPa = 40 N/mm²
We need to find the safe torque that can be transmitted by the shaft.

Formula used:
The formula for torsional or twisting moment (T) is:
T = π/16 * τ * D^3
where,
π = 3.14 (pi)
τ = Shear stress
D = Diameter of shaft

Calculation:
Substituting the given values in the above formula, we get
T = π/16 * τ * D^3
T = 3.14/16 * 40 * (0.05)^3
T = 0.9818 N-m ≈ 0.982 kN-m

Therefore, the safe torque that can be transmitted by the shaft is 0.982 kN-m.

Hence, option B is the correct answer.
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A circular shaft of 50 mm diameter is required to transmit a torque from one shaft to another. If shear stress is not to exceed 40 Mpa, the safe torque is:a)0.882 kN-mb)0.982 kN-mc)1.982 kN-md)2.00 kN-mCorrect answer is option 'B'. Can you explain this answer?
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