Thin-walled pressure conduit (cylinder) – hoop (circumferential) stress formula For a thin circular conduit carrying internal pressure, if p is the internal pressure, d is the internal diameter of the conduit, and t is the wall thickness, what is the hoop (circumferential) stress in the shell (assume thin-cylinder theory and uniform stress)?

Introduction / Context:
Design of pipes, penstocks, and pressure conduits often assumes a thin-walled cylinder to obtain quick yet safe estimates of wall stresses. The most critical membrane stress is the circumferential or hoop stress, which tends to split the cylinder longitudinally. Understanding the correct closed-form relation among internal pressure p, diameter d, and wall thickness t is essential to size the wall and select materials.

Given Data / Assumptions:

• Thin-walled circular cylinder (t << d), uniform thickness.
• Internal pressure = p (gauge).
• Internal diameter = d.
• Wall thickness = t.
• Linear elastic, small deformation; membrane theory holds.

Concept / Approach:
The hoop stress arises from equilibrium of half the cylinder split by a diametral longitudinal plane. Internal pressure produces a resultant force that is resisted by the tensile forces in the wall along the cut. For thin shells, stress is assumed uniform through thickness and along the width considered.

Step-by-Step Solution:
Resultant pressure on projected area = p * (d * L), where L is unit length along the axis.Resisting tensile force in two walls = 2 * (sigma_hoop * t * L).Equilibrium: p * d * L = 2 * sigma_hoop * t * L.Solve for sigma_hoop: sigma_hoop = p * d / (2 * t).

Verification / Alternative check:
Longitudinal (axial) stress for a closed-end thin cylinder is sigma_long = p * d / (4 * t), which is half the hoop stress. This consistent ratio verifies the derivation.

Why Other Options Are Wrong:

• p * d / t and 2 * p * d / t overestimate stress by ignoring the factor of 2 from two wall segments sharing load.
• p * d^2 / (4 * t) has the wrong dimension (involves area factor).
• p / (2 * t) omits diameter and is dimensionally inconsistent.

Common Pitfalls:
Using outside diameter instead of internal diameter; applying the thin-wall formula when t/d is not small (typically t/d < 1/10); forgetting that joints/corrosion allowances must be added in real design.

Final Answer:
sigma_hoop = p * d / (2 * t)