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

# Constitutive relations for 3D elastic solid

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

- Asked in: 2078 Chaitra, 5 marks; 2071 Bhadra, 6 marks (asked 2 times)
- 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-3d-solid
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-3

## Question

Derive the constitutive relations for the 3D state of a solid (i.e. $\{\sigma\} = [D]\{\varepsilon\}$ for an elastic isotropic material).

## Short answer

The 3D constitutive matrix for isotropic linear elasticity is [D] = (E/[(1+ν)(1-2ν)]) × a 6×6 matrix with diagonal entries (1-ν, 1-ν, 1-ν, (1-2ν)/2, (1-2ν)/2, (1-2ν)/2) and off-diagonal ν for normal stress components, relating {σ} = [D]{ε}.

## Answer

The constitutive relation links stress and strain. For a linear elastic isotropic material the stresses are related to strains by Hooke's law with two constants, $E$ and $\nu$.

#### Strains from stresses
By superposition of the effect of each normal stress (a stress $\sigma_x$ gives $\varepsilon_x=\sigma_x/E$ and lateral strains $-\nu\sigma_x/E$):
$$
\begin{aligned}
\varepsilon_x&=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]\\
\varepsilon_y&=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)]\\
\varepsilon_z&=\frac{1}{E}[\sigma_z-\nu(\sigma_x+\sigma_y)]
\end{aligned}
$$
and shear strains $\gamma_{xy}=\dfrac{\tau_{xy}}{G},\ \gamma_{yz}=\dfrac{\tau_{yz}}{G},\ \gamma_{zx}=\dfrac{\tau_{zx}}{G}$ with $G=\dfrac{E}{2(1+\nu)}$.

#### Stresses from strains
Add the three normal strain equations. With $e=\varepsilon_x+\varepsilon_y+\varepsilon_z$ and $\Theta=\sigma_x+\sigma_y+\sigma_z$:
$$
e=\frac{1-2\nu}{E}\Theta\ \Rightarrow\ \Theta=\frac{E}{1-2\nu}e
$$
Write the first equation as $E\varepsilon_x=(1+\nu)\sigma_x-\nu\Theta$, so
$$
\sigma_x=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu}{1+\nu}\Theta=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu E}{(1+\nu)(1-2\nu)}e
$$
Substituting $e$ and collecting terms gives $\sigma_x=\dfrac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\varepsilon_x+\nu\varepsilon_y+\nu\varepsilon_z]$, and similarly for $\sigma_y,\sigma_z$.

#### Matrix form $\{\sigma\}=[D]\{\varepsilon\}$
$$
\begin{Bmatrix}\sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{yz}\\\tau_{zx}\end{Bmatrix}=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&\nu&0&0&0\\\nu&1-\nu&\nu&0&0&0\\\nu&\nu&1-\nu&0&0&0\\0&0&0&\frac{1-2\nu}{2}&0&0\\0&0&0&0&\frac{1-2\nu}{2}&0\\0&0&0&0&0&\frac{1-2\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\\gamma_{yz}\\\gamma_{zx}\end{Bmatrix}
$$
The matrix $[D]$ is symmetric. In terms of Lame's constants, $\sigma_x=\lambda e+2\mu\varepsilon_x$ and $\tau_{xy}=\mu\gamma_{xy}$ with $\lambda=\dfrac{\nu E}{(1+\nu)(1-2\nu)}$, $\mu=G$.

## Check yourself

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

1. How is epsilon_x obtained in 3D Hooke's law?
   A. epsilon_x = sigma_x/E only
   B. epsilon_x = (1/E)[sigma_x + nu(sigma_y + sigma_z)]
   C. epsilon_x = (1/E)[sigma_x - nu(sigma_y + sigma_z)]
   D. epsilon_x = (nu/E)[sigma_x - sigma_y - sigma_z]
   Answer: C. sigma_x gives sigma_x/E, while sigma_y and sigma_z each give a lateral strain -nu sigma/E, which are superposed. (Key point: Strain from stress by superposition)

2. Adding the three normal strain equations gives which result?
   A. e = ((1 - 2 nu)/E) Theta, so Theta = E e/(1 - 2 nu)
   B. e = (E/(1 - 2 nu)) Theta
   C. e = ((1 + nu)/E) Theta
   D. e = Theta/E
   Answer: A. Summing gives e = (1 - 2 nu) Theta/E, where Theta is the sum of normal stresses; this is used to eliminate Theta. (Key point: Sum of normal strains)

3. How are shear strains related to shear stresses and what is G?
   A. gamma = tau/E, G = E
   B. gamma = tau/G, G = E/(2(1 + nu))
   C. gamma = G tau, G = E/(1 - 2 nu)
   D. gamma = tau/G, G = E/(1 - nu)
   Answer: B. gamma_xy = tau_xy/G with G = E/(2(1 + nu)) for an isotropic material. (Key point: Shear modulus)

4. In the 3D [D] matrix, what is the shear diagonal term and its factor?
   A. (1 - nu) times E/((1 + nu)(1 - 2 nu))
   B. nu times E/((1 + nu)(1 - 2 nu))
   C. (1 - 2 nu)/2 times E/((1 + nu)(1 - 2 nu))
   D. 1/2 times E
   Answer: C. The shear diagonal entries are (1 - 2 nu)/2 times the common factor, which equals G. The matrix is symmetric with (1 - nu) on the normal diagonal and nu off-diagonal. (Key point: D matrix shape)

## Related in Computational Techniques in Civil Engineering

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- [Isotropic vs anisotropic; Lame's constants](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aisotropic-anisotropic-lame) — Computational Techniques in Civil Engineering, chapter 3
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