Cylindrical Pressure Vessel

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

13 formulas

10 exam-style examples

~60 min

All Mechanics of Materials lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Cylindrical Pressure Vessel within Mechanics of Materials. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what cylindrical pressure vessel describes physically and when it applies.
State every one of the 13 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Cylindrical Pressure Vessel is the part of Mechanics of Materials that lets you connect an axially loaded, bent or twisted member to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a stress or deformation at one point of one member. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. psi vs ksi and kip vs lb decide the answer choice. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: cylindrical pressure vessel.
Wikimedia Commons, public domain

Mechanics of Materials — Cylindrical Pressure Vessel: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an axially loaded, bent or twisted member. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 13 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

vt	Quantity produced by "vt = Pi o2 and vr = - Pi" — read its definition and unit from the handbook line directly above the equation.
σt	Quantity produced by "σt =−Po o2 i2 and σr =−Po" — read its definition and unit from the handbook line directly above the equation.
σr	Quantity produced by "σr = radial stress" — read its definition and unit from the handbook line directly above the equation.
Pi	Quantity produced by "Pi = internal pressure" — read its definition and unit from the handbook line directly above the equation.
Po	Quantity produced by "Po = external pressure" — read its definition and unit from the handbook line directly above the equation.
ri	Quantity produced by "ri = inside radius" — read its definition and unit from the handbook line directly above the equation.
ro	Quantity produced by "ro = outside radius" — read its definition and unit from the handbook line directly above the equation.
va	Quantity produced by "va = Pi" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = wall thickness" — read its definition and unit from the handbook line directly above the equation.
r	Quantity produced by "r = i2o" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

For internal pressure only, the stresses at the inside wall are:
r 2 + ri2
ro - ri2
For external pressure only, the stresses at the outside wall are:
r2+r2
r o − ri
where
For vessels with end caps, the axial stress is:
ro2 - ri2
When the thickness of the cylinder wall is about one-tenth or less of inside radius, the cylinder can be considered as thin-walled.
In which case, the internal pressure is resisted by the hoop stress and the axial stress.
Pr Pi r
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel

A thin-walled cylindrical pressure vessel of 72 in inside diameter and 0.75 in wall thickness carries an internal pressure of 518 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 518 psi
D = 72 in
t = 0.75 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 518 (72) / (2 \times 0.75) = 24, 864 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 518 (72) / (4 \times 0.75) = 12, 432 p s i$
Thin-wall check
$D / t = 96.0 > 20 ✓$

Answer: σ_hoop = 24,864 psi, σ_long = 12,432 psi

Why the other options are there

49,728 psi (factor of 2 omitted)
hoop = 12,432 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 2

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (2)

A thin-walled cylindrical pressure vessel of 56 in inside diameter and 0.65 in wall thickness carries an internal pressure of 317 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 317 psi
D = 56 in
t = 0.65 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 317 (56) / (2 \times 0.65) = 13, 655 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 317 (56) / (4 \times 0.65) = 6, 828 p s i$
Thin-wall check
$D / t = 86.2 > 20 ✓$

Answer: σ_hoop = 13,655 psi, σ_long = 6,828 psi

Why the other options are there

27,311 psi (factor of 2 omitted)
hoop = 6,828 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 3

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (3)

A thin-walled cylindrical pressure vessel of 90 in inside diameter and 0.25 in wall thickness carries an internal pressure of 150 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 150 psi
D = 90 in
t = 0.25 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 150 (90) / (2 \times 0.25) = 27, 000 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 150 (90) / (4 \times 0.25) = 13, 500 p s i$
Thin-wall check
$D / t = 360.0 > 20 ✓$

Answer: σ_hoop = 27,000 psi, σ_long = 13,500 psi

Why the other options are there

54,000 psi (factor of 2 omitted)
hoop = 13,500 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 4

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (4)

A thin-walled cylindrical pressure vessel of 64 in inside diameter and 0.50 in wall thickness carries an internal pressure of 481 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 481 psi
D = 64 in
t = 0.50 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 481 (64) / (2 \times 0.50) = 30, 784 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 481 (64) / (4 \times 0.50) = 15, 392 p s i$
Thin-wall check
$D / t = 128.0 > 20 ✓$

Answer: σ_hoop = 30,784 psi, σ_long = 15,392 psi

Why the other options are there

61,568 psi (factor of 2 omitted)
hoop = 15,392 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 5

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (5)

A thin-walled cylindrical pressure vessel of 30 in inside diameter and 0.75 in wall thickness carries an internal pressure of 146 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 146 psi
D = 30 in
t = 0.75 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 146 (30) / (2 \times 0.75) = 2, 920 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 146 (30) / (4 \times 0.75) = 1, 460 p s i$
Thin-wall check
$D / t = 40.0 > 20 ✓$

Answer: σ_hoop = 2,920 psi, σ_long = 1,460 psi

Why the other options are there

5,840 psi (factor of 2 omitted)
hoop = 1,460 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 6

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (6)

A thin-walled cylindrical pressure vessel of 68 in inside diameter and 0.85 in wall thickness carries an internal pressure of 159 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 159 psi
D = 68 in
t = 0.85 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 159 (68) / (2 \times 0.85) = 6, 360 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 159 (68) / (4 \times 0.85) = 3, 180 p s i$
Thin-wall check
$D / t = 80.0 > 20 ✓$

Answer: σ_hoop = 6,360 psi, σ_long = 3,180 psi

Why the other options are there

12,720 psi (factor of 2 omitted)
hoop = 3,180 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 7

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (7)

A thin-walled cylindrical pressure vessel of 88 in inside diameter and 0.20 in wall thickness carries an internal pressure of 293 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 293 psi
D = 88 in
t = 0.20 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 293 (88) / (2 \times 0.20) = 64, 460 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 293 (88) / (4 \times 0.20) = 32, 230 p s i$
Thin-wall check
$D / t = 440.0 > 20 ✓$

Answer: σ_hoop = 64,460 psi, σ_long = 32,230 psi

Why the other options are there

128,920 psi (factor of 2 omitted)
hoop = 32,230 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 8

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (8)

A thin-walled cylindrical pressure vessel of 96 in inside diameter and 0.25 in wall thickness carries an internal pressure of 317 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 317 psi
D = 96 in
t = 0.25 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 317 (96) / (2 \times 0.25) = 60, 864 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 317 (96) / (4 \times 0.25) = 30, 432 p s i$
Thin-wall check
$D / t = 384.0 > 20 ✓$

Answer: σ_hoop = 60,864 psi, σ_long = 30,432 psi

Why the other options are there

121,728 psi (factor of 2 omitted)
hoop = 30,432 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 9

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (9)

A thin-walled cylindrical pressure vessel of 60 in inside diameter and 0.90 in wall thickness carries an internal pressure of 396 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 396 psi
D = 60 in
t = 0.90 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 396 (60) / (2 \times 0.90) = 13, 200 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 396 (60) / (4 \times 0.90) = 6, 600 p s i$
Thin-wall check
$D / t = 66.7 > 20 ✓$

Answer: σ_hoop = 13,200 psi, σ_long = 6,600 psi

Why the other options are there

26,400 psi (factor of 2 omitted)
hoop = 6,600 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Example 10

Hoop and longitudinal stress in a cylindrical pressure vessel — Cylindrical Pressure Vessel (10)

A thin-walled cylindrical pressure vessel of 54 in inside diameter and 0.60 in wall thickness carries an internal pressure of 264 psi. Compute the hoop stress, the longitudinal stress, and verify the thin-wall assumption.

Given

p = 264 psi
D = 54 in
t = 0.60 in

Find

Hoop stress, longitudinal stress and the D/t ratio

Start with the thinking

In a cylindrical pressure vessel the hoop stress is exactly twice the longitudinal stress.
Thin-wall theory is valid when D/t exceeds about 20.

Step-by-step solution

Formula
$σ_{h} = p D / (2 t)$
Substituting
$σ_{h} = 264 (54) / (2 \times 0.60) = 11, 880 p s i$
Formula
$σ_{l} = p D / (4 t)$
Substituting
$σ_{l} = 264 (54) / (4 \times 0.60) = 5, 940 p s i$
Thin-wall check
$D / t = 90.0 > 20 ✓$

Answer: σ_hoop = 11,880 psi, σ_long = 5,940 psi

Why the other options are there

23,760 psi (factor of 2 omitted)
hoop = 5,940 psi (hoop and longitudinal swapped)

Reference: FE Reference Handbook — Mechanics of Materials → Cylindrical Pressure Vessel

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an axially loaded, bent or twisted member, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Cylindrical Pressure Vessel contains 13 relations; you must be able to find this page in under 15 seconds.
Exam style: a stress or deformation at one point of one member.
Unit rule: psi vs ksi and kip vs lb decide the answer choice.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

psi vs ksi and kip vs lb decide the answer choice
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.