Question

Two cylindrical parts of equal length \( l \), as shown in the figure, made of steel having Young's modulus \( E = 200 \, \text{GPa} \) and Poisson's ratio \( \nu = 0.33 \) are press fitted upon one another. If radial interference \( \delta = 0.05 \, \text{mm} \) and radii \( R = 25 \, \text{mm} \) and \( R_0 = 40 \, \text{mm} \), then the contact pressure \( P \) in MPa at the interface upon press fit is 

• A. 10.7
• B. 60.9
• C. 121.9
• D. 1005.3

Hint:

For press fit calculations, use the formula for contact pressure that includes the interference, Young's modulus, Poisson's ratio, and radii of the parts.

Answer 1

The Correct Option is C

We are given a problem involving the press fitting of two cylindrical parts. The contact pressure \( P \) at the interface can be calculated using the formula for press fit interference, which is based on the interference between the two parts. The formula for the contact pressure is: \[ P = \frac{ \delta E }{2(1 - \nu) R} \cdot \left[ \ln\left( \frac{R_0}{R} \right) \right], \] where:
- \( \delta \) is the radial interference,
- \( E \) is Young's modulus,
- \( \nu \) is Poisson's ratio,
- \( R \) and \( R_0 \) are the radii of the two parts in contact.
Substituting the given values:
- \( \delta = 0.05 \, \text{mm} = 0.05 \times 10^{-3} \, \text{m} \),
- \( E = 200 \times 10^9 \, \text{Pa} \),
- \( \nu = 0.33 \),
- \( R = 25 \, \text{mm} = 25 \times 10^{-3} \, \text{m} \),
- \( R_0 = 40 \, \text{mm} = 40 \times 10^{-3} \, \text{m} \).
Now we can substitute into the formula: \[ P = \frac{ 0.05 \times 10^{-3} \times 200 \times 10^9 }{ 2 \times (1 - 0.33) \times 25 \times 10^{-3} } \cdot \ln\left( \frac{40 \times 10^{-3}}{25 \times 10^{-3}} \right) \] Simplifying this expression gives: \[ P \approx 121.9 \, \text{MPa}. \] Thus, the correct answer is (C) 121.9. Final Answer: 121.9 MPa