Thin-walled cylindrical pressure vessel: a cold-drawn seamless steel tube has diameter 6 cm, wall thickness 0.2 cm, and ultimate strength 3600 kg/cm^2. What is the bursting pressure (kg/cm^2)?

Introduction / Context:
For thin cylinders, the critical stress for bursting is the circumferential (hoop) stress. Using the thin-cylinder formula allows estimation of the internal pressure that will drive hoop stress up to the material’s ultimate strength.

Given Data / Assumptions:

• Diameter d = 6 cm, wall thickness t = 0.2 cm.
• Ultimate tensile strength σ_u = 3600 kg/cm^2.
• Thin-wall assumption (t ≪ d) holds; use membrane theory.

Concept / Approach:
Hoop stress for a thin cylinder: σ_h = p d / (2 t). Bursting pressure p_burst occurs when σ_h = σ_u. Solve for p at failure.

Step-by-Step Solution:
Set σ_u = p_burst * d / (2 t).Rearrange: p_burst = 2 t σ_u / d.Substitute: p_burst = 2 * 0.2 * 3600 / 6 = 1440 / 6 = 240 kg/cm^2.

Verification / Alternative check:
Longitudinal stress σ_l = p d /(4 t) is half the hoop stress; at bursting, σ_l = 1200 kg/cm^2, still below σ_u, consistent with hoop being critical.

Why Other Options Are Wrong:

• 120 kg/cm^2: corresponds roughly to longitudinal stress at burst, not hoop.
• 480 and 960 kg/cm^2: exceed the thin-wall hoop failure prediction.

Common Pitfalls:

• Using radius instead of diameter in σ_h = p d/(2 t) changes the factor.
• Mixing unit systems; all inputs are in consistent cm–kgf units.

Final Answer:
240