---
title: "Constitutive relation for 2D isotropic material"
subject: "Computational Techniques in Civil Engineering"
chapter: 3
key: "semester_8/computational_techniques_in_civil_engineering:constitutive-2d-isotropic"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/constitutive-2d-isotropic
site: BCT Engineering
---

# Constitutive relation for 2D isotropic material

Computational Techniques in Civil Engineering · Chapter 3: Elasticity in solids

- Asked in: 2078 Kartik, 6 marks
- Video: [08.4 Generalized Hooke's Law](https://www.youtube.com/watch?v=UDMNrUt-0cE) — Introductory Engineering Mechanics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/constitutive-2d-isotropic
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-3

## Question

Derive the constitutive relation for a two-dimensional problem of isotropic material.

## Short answer

For plane stress: {σ} = (E/(1-ν^2))[[1,ν,0],[ν,1,0],[0,0,(1-ν)/2]]{ε}. For plane strain: {σ} = (E/((1+ν)(1-2ν)))[[1-ν,ν,0],[ν,1-ν,0],[0,0,(1-2ν)/2]]{ε}. Both derived from 3D Hooke's law by setting σ_z=0 or ε_z=0.

## Answer

For a two-dimensional problem the constitutive law depends on whether it is plane stress (thin body) or plane strain (long body). Start from the 3D Hooke's law of an isotropic material:
$$
\varepsilon_x=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)],\quad \varepsilon_y=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)],\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}
$$

#### Plane stress ($\sigma_z=0$)
For plane stress, put $\sigma_z=0$:
$$
\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E},\quad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E},\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}
$$
and $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$. In matrix form $\{\varepsilon\}=[C]\{\sigma\}$:
$$
\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}=\frac{1}{E}\begin{bmatrix}1&-\nu&0\\-\nu&1&0\\0&0&2(1+\nu)\end{bmatrix}\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}
$$
Invert the first two equations: from $E\varepsilon_x=\sigma_x-\nu\sigma_y$ and $E\varepsilon_y=\sigma_y-\nu\sigma_x$, we get $\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)$ and $\sigma_y=\dfrac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x)$. Also $\tau_{xy}=G\gamma_{xy}=\dfrac{E}{2(1+\nu)}\gamma_{xy}$. Hence
$$
\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\\nu&1&0\\0&0&\dfrac{1-\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}
$$
i.e. $\{\sigma\}=[D]\{\varepsilon\}$ with the plane stress matrix $[D]$ above.

#### Plane strain ($\varepsilon_z=0$)
From $\varepsilon_z=\frac{1}{E}[\sigma_z-\nu(\sigma_x+\sigma_y)]=0$ we get $\sigma_z=\nu(\sigma_x+\sigma_y)$. Substituting this into the first two equations,
$$
\varepsilon_x=\frac{1+\nu}{E}[(1-\nu)\sigma_x-\nu\sigma_y],\qquad \varepsilon_y=\frac{1+\nu}{E}[(1-\nu)\sigma_y-\nu\sigma_x]
$$
Solving these two equations for the stresses, and using $\tau_{xy}=G\gamma_{xy}$, gives
$$
\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&0\\\nu&1-\nu&0\\0&0&\dfrac{1-2\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}
$$
The same relation follows from the plane stress one by replacing $E$ with $E/(1-\nu^2)$ and $\nu$ with $\nu/(1-\nu)$.

## Check yourself

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

1. What decides which 2D constitutive law is used?
   A. Whether the material is isotropic
   B. Whether the body is thin (plane stress) or long (plane strain)
   C. The value of nu only
   D. The number of elements
   Answer: B. A thin body gives plane stress, a long body plane strain; both start from the same 3D Hooke's law. (Key point: Choice of 2D case)

2. In plane stress, what is epsilon_z?
   A. Zero
   B. nu(sigma_x + sigma_y)
   C. sigma_z/E
   D. -nu(sigma_x + sigma_y)/E
   Answer: D. With sigma_z = 0, epsilon_z = -nu(sigma_x + sigma_y)/E, non-zero but not needed in the analysis. (Key point: Plane stress compliance)

3. How are the plane strain strains in x and y found?
   A. Put sigma_z = 0
   B. Put sigma_z = nu(sigma_x + sigma_y) into Hooke's law, giving eps_x = ((1 + nu)/E)[(1 - nu) sigma_x - nu sigma_y]
   C. Put epsilon_z = 1
   D. Ignore sigma_z and use plane stress
   Answer: B. epsilon_z = 0 gives sigma_z = nu(sigma_x + sigma_y); substituting yields the stated form. (Key point: Plane strain substitution)

4. What is the plane strain shear diagonal term in [D]?
   A. (1 - nu)/2 times E/(1 - nu^2)
   B. (1 - 2 nu)/2 times E/((1 + nu)(1 - 2 nu))
   C. (1 - nu) times E/((1 + nu)(1 - 2 nu))
   D. 1 times E
   Answer: B. The plane strain matrix is E/((1 + nu)(1 - 2 nu)) with diagonal 1 - nu, 1 - nu and (1 - 2 nu)/2. (Key point: Plane strain D)

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