Thin cylinder hoop stress – water main thickness: A water main of internal diameter 80 cm carries a pressure of 10 kg/cm2. If the permissible circumferential (hoop) stress in the metal is limited to 200 kg/cm2, what is the minimum required thickness of the pipe wall (ignore corrosion allowance)?

Introduction / Context:
Water mains are commonly designed using thin-cylinder theory for hoop stress. The wall thickness must be sufficient so that the circumferential stress due to internal pressure does not exceed the allowable stress in the metal. This is a direct substitution problem widely used in exams.

Given Data / Assumptions:

• Internal pressure p = 10 kg/cm2.
• Internal diameter D = 80 cm.
• Allowable hoop stress σ_allow = 200 kg/cm2.
• Thin cylinder approximation; neglect joint efficiency and corrosion allowance for this calculation step.

Concept / Approach:
For a thin cylinder, hoop stress σ_h = p * D / (2 * t). Solving for thickness t gives t = p * D / (2 * σ_allow). This is adequate for preliminary sizing; refined formulas include a small addition of pressure in the denominator, but the thin-wall formula is standard in objective problems.

Step-by-Step Solution:
Write: t = p * D / (2 * σ_allow).Insert values: t = 10 * 80 / (2 * 200).Compute numerator: 10 * 80 = 800.Compute denominator: 2 * 200 = 400.Thus t = 800 / 400 = 2.0 cm.

Verification / Alternative check:
A refined expression t = p * D / (2 * σ_allow + p) gives 800 / (400 + 10) ≈ 1.95 cm. Rounding and provision for allowances would still make a 2.0 cm plate the practical choice among the options.

Why Other Options Are Wrong:

• 1.0 or 1.5 cm: would over-stress the metal above the 200 kg/cm2 limit.
• 2.5 or 3.0 cm: overly conservative for the stated limits; increase cost and weight without need.

Common Pitfalls:

• Using radius instead of diameter in the thin-cylinder formula.
• Mixing units (e.g., MPa versus kg/cm2).

Final Answer:
2.0 cm