Topic: Strength of Materials
A thin-walled cylinder with closed ends for which inner radius is 0.50 meter and outer radius is 0.52 meter, is subjected to internal pressure 2 MPa. The absolute maximum shearing stress on the inner surface of the cylinder shall be
- A. 37.5 MPa
- B. 75 MPa
- C. 26 MPa
- D. 24 MPa
Answer: C. 26 MPa
Explanation: The inner radius r = 0.50 m, so the inner diameter d = 1.0 m. The outer radius is R = 0.52 m, meaning the outer diameter D = 1.04 m. The thickness of the cylinder wall is t = R - r = 0.52 - 0.50 = 0.02 m. The absolute maximum shear stress for a thin cylinder is derived from the hoop stress and radial pressure, frequently approximated using the outer diameter for conservative safety in this context: tau_max = P * D / (4 * t). Substituting the values: tau_max = (2 * 1.04) / (4 * 0.02) = 26 MPa.