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A solid circular shaft needs to be designed to transmit a torque of 50 N.m. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is
- a)8
- b)16
- c)24
- d)32
Correct answer is option 'B'. Can you explain this answer?
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A solid circular shaft needs to be designed to transmit a torque of 50...
To design a solid circular shaft to transmit a torque of 50 N.m, we need to determine the minimum allowable design diameter considering the allowable shear stress of the material and a factor of safety of 2.
Given:
Torque (T) = 50 N.m
Allowable shear stress (τ) = 140 MPa
Factor of safety (FoS) = 2
The formula to calculate the torque transmitted by a solid circular shaft is:
T = (π/16) * τ * d^3
where,
T is the torque transmitted
τ is the shear stress
d is the diameter of the shaft
To find the minimum allowable design diameter, we rearrange the formula:
d = [(16 * T)/(π * τ)]^(1/3)
Substituting the given values:
d = [(16 * 50)/(π * 140)]^(1/3)
d ≈ 16.28 mm
Since the diameter should be a whole number, we round up to the nearest whole number, which is 17 mm.
Considering the factor of safety, the minimum allowable design diameter is:
Minimum allowable design diameter = d * FoS
Minimum allowable design diameter = 17 mm * 2
Minimum allowable design diameter = 34 mm
Therefore, the correct answer is option 'B' (16).
Given:
Torque (T) = 50 N.m
Allowable shear stress (τ) = 140 MPa
Factor of safety (FoS) = 2
The formula to calculate the torque transmitted by a solid circular shaft is:
T = (π/16) * τ * d^3
where,
T is the torque transmitted
τ is the shear stress
d is the diameter of the shaft
To find the minimum allowable design diameter, we rearrange the formula:
d = [(16 * T)/(π * τ)]^(1/3)
Substituting the given values:
d = [(16 * 50)/(π * 140)]^(1/3)
d ≈ 16.28 mm
Since the diameter should be a whole number, we round up to the nearest whole number, which is 17 mm.
Considering the factor of safety, the minimum allowable design diameter is:
Minimum allowable design diameter = d * FoS
Minimum allowable design diameter = 17 mm * 2
Minimum allowable design diameter = 34 mm
Therefore, the correct answer is option 'B' (16).
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A solid circular shaft needs to be designed to transmit a torque of 50 N.m. If the allowable shearstress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable designdiameter in mm isa)8b)16c)24d)32Correct answer is option 'B'. Can you explain this answer?
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