A thin cylindrical shell (mild steel) of diameter 100 cm is subjected to an internal pressure of 10 kg/cm^2. Using the maximum principal stress (thin cylinder) theory with yield stress 2000 kg/cm^2 and factor of safety 4, what plate thickness is required?

Introduction / Context:
Design of thin cylindrical shells under internal pressure uses hoop (circumferential) and longitudinal stresses. For thin shells, the hoop stress is usually critical. Applying an allowable stress (yield divided by factor of safety) provides the required thickness.

Given Data / Assumptions:

• Diameter d = 100 cm; internal pressure p = 10 kg/cm^2.
• Yield stress σ_y = 2000 kg/cm^2; factor of safety = 4 → allowable σ_allow = 500 kg/cm^2.
• Thin cylinder assumption; hoop stress governs.

Concept / Approach:
For a thin cylinder, hoop stress σ_h = p * d / (2 * t). Set σ_h ≤ σ_allow and solve for thickness t.

Step-by-Step Solution:
σ_allow = σ_y / FOS = 2000 / 4 = 500 kg/cm^2.Use σ_h = p * d / (2t) ⇒ t = p * d / (2 * σ_allow).Compute t = (10 * 100) / (2 * 500) = 1000 / 1000 = 1.0 cm = 10 mm.Therefore, select a 10 mm plate.

Verification / Alternative check:
Longitudinal stress σ_L = p * d / (4t) would be 250 kg/cm^2 at t = 1 cm, below σ_allow, confirming hoop stress is critical.

Why Other Options Are Wrong:

• 5 mm is unsafe (σ_h would exceed allowable).
• 15 mm and 20 mm are conservative beyond what the given allowable demands.
• 8 mm still exceeds allowable hoop stress.

Common Pitfalls:
Using radius instead of diameter; mixing units (mm vs cm); forgetting factor of safety.

Final Answer:
10 mm