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A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to
  • a)
    41 MPa
  • b)
    82 MPa
  • c)
    164 MPa
  • d)
    204 MPa
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A solid circular shaft of diameter 100 mm is subjected to an axial str...
Shear Stress 
Maximum principal Stress 
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A solid circular shaft of diameter 100 mm is subjected to an axial str...
Given:
Diameter of the circular shaft, d = 100 mm
Axial stress, σa = 50 MPa
Torque, T = 10 kNm

To find: Maximum principal stress experienced on the shaft

Assumptions:
The shaft is made up of homogeneous and isotropic material.
The torque is applied at the end of the shaft.

Formula used:
Maximum shear stress theory or Tresca's theory states that the maximum shear stress in a shaft should be less than or equal to the yield strength of the material.

Maximum shear stress, τmax = (σa/2) + sqrt[(σa/2)^2 + (T/2*(π/4)*d^3)/(π/32*(d^4/4))]

Where,
σa = Axial stress
T = Torque
d = Diameter of the shaft

Calculation:
τmax = (50/2) + sqrt[(50/2)^2 + (10*10^3/2*(π/4)*(100^3))/(π/32*(100^4/4))]
τmax = 25 + sqrt[(25)^2 + (10*10^3*4)/(100^2*4)]
τmax = 25 + sqrt[625 + 25]
τmax = 25 + sqrt(650)
τmax = 41.68 MPa (approx)

The maximum principal stress can be calculated as σ1 = σa/2 + τmax and σ2 = σa/2 - τmax.

σ1 = 50/2 + 41.68 = 66.68 MPa
σ2 = 50/2 - 41.68 = 33.32 MPa

The maximum principal stress is the greater of σ1 and σ2 which is 66.68 MPa (approx).

Therefore, the closest answer is option 'B' which is 82 MPa.

Conclusion:
The maximum principal stress experienced on the shaft is 66.68 MPa (approx) which is closest to option 'B' which is 82 MPa.
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