---
title: "Plane stress and plane strain differences"
subject: "Finite Element Method"
chapter: 6
key: "semester_8/finite_element_method:plane-stress-plane-strain-difference"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/plane-stress-plane-strain-difference
site: BCT Engineering
---

# Plane stress and plane strain differences

Finite Element Method · Chapter 6: Applications in Solid Mechanics

- Asked in: year not given, 5 marks
- Video: [(18B) Introduction to plane stress, plane strain, and uniaxial stress](https://www.youtube.com/watch?v=QHyUv49jRBI) — Sanjay Govindjee
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/plane-stress-plane-strain-difference
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6

## Question

Differentiate between plane stress and plane strain conditions with practical examples. Write the stress-strain relations for each.

## Short answer

Plane stress assumes sigma_z=0 in a thin body, giving epsilon_z=-nu/E(sigma_x+sigma_y); plane strain assumes epsilon_z=0 in a long body, giving sigma_z=nu(sigma_x+sigma_y). Their constitutive matrices differ: plane stress uses E/(1-nu^2) with 1, nu, 0; plane strain uses E/[(1+nu)(1-2nu)] with 1-nu, nu, 0. Examples: thin plate vs dam.

## Answer

Both are two-dimensional idealisations of a 3D elasticity problem, which reduce the unknowns from six stress components to three.

#### Comparison

| Point | Plane stress | Plane strain |
|---|---|---|
| Definition | Stress components in the z-direction are zero: $\sigma_z = \tau_{xz} = \tau_{yz} = 0$ | Strain components in the z-direction are zero: $\varepsilon_z = \gamma_{xz} = \gamma_{yz} = 0$ |
| Geometry | Thin body (thickness much smaller than other dimensions) | Very long body, with constant section and load along its length |
| Loading | In-plane loads only | Loads perpendicular to the length, not varying along it |
| Out-of-plane value | $\varepsilon_z = -\dfrac{\nu}{E}(\sigma_x + \sigma_y) \neq 0$ | $\sigma_z = \nu(\sigma_x + \sigma_y) \neq 0$ |
| Examples | Thin plate with a hole, gusset plate, deep beam web | Dam section, retaining wall, long cylinder or tunnel under pressure, rolling process |
| Unknowns | $\sigma_x, \sigma_y, \tau_{xy}$ | $\sigma_x, \sigma_y, \tau_{xy}$ |

#### Constitutive matrices

**Plane stress**
$$
\begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{1 - \nu^2}\begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1 - \nu}{2} \end{bmatrix}\begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix}
$$

**Plane strain**
$$
\begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{(1 + \nu)(1 - 2\nu)}\begin{bmatrix} 1 - \nu & \nu & 0 \\ \nu & 1 - \nu & 0 \\ 0 & 0 & \frac{1 - 2\nu}{2} \end{bmatrix}\begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix}
$$

The plane strain matrix is obtained from the plane stress matrix by replacing $E \to \dfrac{E}{1 - \nu^2}$ and $\nu \to \dfrac{\nu}{1 - \nu}$. Plane strain is stiffer for the same $E$ and $\nu$, because the z-direction is constrained.

## Check yourself

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

1. Which statement defines plane stress and plane strain correctly?
   A. Plane stress: sigma_z = tau_xz = tau_yz = 0; plane strain: eps_z = gamma_xz = gamma_yz = 0
   B. Both have eps_z = 0
   C. Both have sigma_z = 0
   D. Plane stress: eps_z = 0; plane strain: sigma_z = 0
   Answer: A. Plane stress sets out-of-plane stresses to zero; plane strain sets out-of-plane strains to zero. (Key point: Definitions)

2. Which pairing of body type and idealisation is correct?
   A. Thin plate with a hole: plane stress; long dam section: plane strain
   B. Thin plate with a hole: plane strain; long dam section: plane stress
   C. Retaining wall: plane stress; gusset plate: plane strain
   D. Both are plane stress
   Answer: A. Thin plates loaded in their plane are plane stress; long prismatic bodies like dams and retaining walls are plane strain. (Key point: Examples)

3. In plane strain, what is the non-zero out-of-plane quantity?
   A. sigma_z = nu(sigma_x + sigma_y)
   B. eps_z = -nu(sigma_x + sigma_y)/E
   C. tau_xz = nu sigma_x
   D. sigma_z = E eps_z
   Answer: A. Constraining eps_z = 0 produces sigma_z = nu(sigma_x + sigma_y); the eps_z formula belongs to plane stress. (Key point: Out-of-plane value)

4. How is the plane strain [D] obtained from plane stress [D], and which is stiffer?
   A. Replace only nu by 2 nu; equal stiffness
   B. Replace E by E(1 - nu^2) and nu by nu(1 - nu); plane stress is stiffer
   C. Multiply by (1 - 2nu); plane stress is stiffer
   D. Replace E by E/(1 - nu^2) and nu by nu/(1 - nu); plane strain is stiffer
   Answer: D. The substitution converts one matrix to the other, and plane strain is stiffer because the z-direction is constrained. (Key point: Relation between matrices)

## Related in Finite Element Method

- [Axisymmetric stress analysis formulation](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aaxisymmetric-analysis-strain-displacement) — Finite Element Method, chapter 6
- [Compute D matrix for plane problems](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aelasticity-matrix-plane-stress-strain) — Finite Element Method, chapter 6
- [Essential and natural boundary conditions](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aessential-natural-boundary-conditions) — Finite Element Method, chapter 2
- [Truss, beam and frame element differences](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Atruss-beam-frame-element-differences) — Finite Element Method, chapter 3
- [Derive CST element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Acst-stiffness-matrix-derivation) — Finite Element Method, chapter 6

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