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A thin cylinder of internal diameter 60 mm and wall thickness 2mm is subjected to an internal pressure of 1.25 N/mm2. The cylinder is also subjected to a torque of 60 Nm, the axis of the torque coinciding with that of the cylinder. The minor principal stress is MPa.
  • a)
    6.98326
  • b)
    7.98326
  • c)
    8.98326
  • d)
    9.98326
Correct answer is option 'A'. Can you explain this answer?
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To find the minor principal stress in the thin cylinder, we need to consider the combined effect of pressure and torque. The formula to calculate the minor principal stress in a thin cylinder is given by:

σθ = (P * ri^2) / (ro^2 - ri^2) + (T * ri) / (2 * (ro^2 - ri^2))

Where:
σθ is the minor principal stress
P is the internal pressure
ri is the internal radius of the cylinder
ro is the external radius of the cylinder
T is the torque applied

Let's calculate the values using the given data:

Given:
Internal diameter of the cylinder = 60 mm
Wall thickness = 2 mm
Internal pressure = 1.25 N/mm^2
Torque = 60 Nm

Calculations:
1. Calculate the internal and external radius of the cylinder:
Internal radius (ri) = (Internal diameter / 2) - Wall thickness = (60 / 2) - 2 = 28 mm
External radius (ro) = Internal radius + Wall thickness = 28 + 2 = 30 mm

2. Convert the internal pressure from N/mm^2 to N/m^2:
Internal pressure = 1.25 * 10^6 N/m^2

3. Convert the internal radius from mm to m:
ri = 28 / 1000 m

4. Calculate the minor principal stress using the formula:
σθ = (P * ri^2) / (ro^2 - ri^2) + (T * ri) / (2 * (ro^2 - ri^2))
= (1.25 * 10^6 * (28 / 1000)^2) / ((30 / 1000)^2 - (28 / 1000)^2) + (60 * 28 / 1000) / (2 * ((30 / 1000)^2 - (28 / 1000)^2))
= 6.98326 MPa

Therefore, the minor principal stress in the thin cylinder is 6.98326 MPa, which is option 'A'.
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