---
title: "Volumetric strain of thin cylindrical vessel"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:volumetric-strain-thin-cylinder-derive"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/volumetric-strain-thin-cylinder-derive
site: BCT Engineering
---

# Volumetric strain of thin cylindrical vessel

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2078 Kartik, 6 marks; 2075 Asoj, 7 marks (asked 2 times)
- Video: [VOLUMETRIC STRAIN FOR PRESSURE VESSELS (THIN CYLINDER)](https://www.youtube.com/watch?v=jo6fPw1yEHs) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/volumetric-strain-thin-cylinder-derive
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

In a thin walled cylindrical vessel show that the volumetric strain is equal to two times circumferential strain plus longitudinal strain.

## Short answer

Volume V=(π/4)d²L. Differentiating, dV/V=2(dd/d)+dL/L=2ε_c+ε_l because ε_c=δd/d. Hence volumetric strain equals twice circumferential strain plus longitudinal strain.

## Answer

**To show:** $\varepsilon_v = 2\varepsilon_c + \varepsilon_l$, where $\varepsilon_c$ is the circumferential (hoop) strain and $\varepsilon_l$ the longitudinal strain.

Consider a thin cylinder of internal diameter $d$ and length $L$. Its volume is

$$
V = \frac{\pi}{4}d^2L
$$

#### Derivation
Under internal pressure the diameter changes to $d + \delta d$ and the length to $L + \delta L$. The strains are

$$
\varepsilon_c = \frac{\pi(d+\delta d) - \pi d}{\pi d} = \frac{\delta d}{d}, \qquad \varepsilon_l = \frac{\delta L}{L}
$$

(circumferential strain equals diametral strain, since circumference $= \pi d$). The new volume is

$$
V' = \frac{\pi}{4}(d+\delta d)^2(L+\delta L) = V\left(1+\varepsilon_c\right)^2\left(1+\varepsilon_l\right)
$$

Expand and neglect products of small strains:

$$
V' = V\left(1 + 2\varepsilon_c + \varepsilon_l + \varepsilon_c^2 + 2\varepsilon_c\varepsilon_l + \varepsilon_c^2\varepsilon_l\right) \approx V\left(1 + 2\varepsilon_c + \varepsilon_l\right)
$$

$$
\varepsilon_v = \frac{V'-V}{V} = 2\varepsilon_c + \varepsilon_l
$$

**Alternative (differentiation):** from $V = \frac{\pi}{4}d^2L$,

$$
\frac{dV}{V} = 2\frac{dd}{d} + \frac{dL}{L} = 2\varepsilon_c + \varepsilon_l
$$

#### In terms of pressure
With $\sigma_c = \dfrac{pd}{2t}$ and $\sigma_l = \dfrac{pd}{4t}$:

$$
\varepsilon_c = \frac{pd}{2tE}\left(1-\frac{\nu}{2}\right), \qquad \varepsilon_l = \frac{pd}{2tE}\left(\frac{1}{2}-\nu\right)
$$

$$
\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{2tE}\left(\frac{5}{2} - 2\nu\right) = \frac{pd}{4tE}\left(5-4\nu\right)
$$

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does circumferential strain equal the diametral strain delta d / d?
   A. Circumference = pi d, so its fractional change equals that of d
   B. Because length does not change
   C. Because hoop stress equals longitudinal stress
   D. Because Poisson's ratio is zero
   Answer: A. (pi(d+dd) - pi d)/(pi d) = dd/d. (Key point: Circumferential strain)

2. How is the new volume written?
   A. V' = V(1 + 2e_c)(1 + e_l)^2
   B. V' = V(1 + e_c)^2 (1 + e_l)
   C. V' = V(1 + e_c)(1 + e_l)
   D. V' = V(1 + e_c + e_l)
   Answer: B. Volume depends on d squared and on L, so the d-factor is squared. (Key point: Volume expression)

3. Why can e_c^2, e_c e_l and e_c^2 e_l be dropped?
   A. They equal zero exactly
   B. They cancel with the linear terms
   C. They are products of small strains and negligible
   D. They are only valid for spheres
   Answer: C. Strains are tiny, so their products are far smaller than the linear terms. (Key point: Neglecting small terms)

4. Which differentiation gives the same result?
   A. dV/V = 2(dd/d) + dL/L
   B. dV/V = dd/d + 2(dL/L)
   C. dV/V = dd/d + dL/L
   D. dV/V = 3(dd/d)
   Answer: A. Differentiating V = (pi/4)d^2 L gives 2 d dd L + d^2 dL, dividing by V. (Key point: Differentiation check)

## Related in Strength of Materials

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- [Max shear stress and volumetric strain in thin cylinder](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Amax-shear-half-longitudinal-volumetric-strain) — Strength of Materials, chapter 5
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