Chapter 10, Problem 10.1RP

In the case of plane stress, where the in-plane principal strains are given by ε₁ and ε₂, show that the third principal strain can be obtained from

${\epsilon }_3=\frac{-v\left(\epsilon +{\epsilon }_2\right)}{\left(1-v\right)}$

where v is Poisson’s ratio for the material.

To determine

To show that: The third principal strain can be obtained from ${\epsilon }_3=\frac{-\nu \left({\epsilon }_1+{\epsilon }_2\right)}{\left(1-\nu \right)}$.

Answer to Problem 10.1RP

The third principal strain can be obtained from $\underset{\_}{{\epsilon }_3=\frac{-\nu \left({\epsilon }_1+{\epsilon }_2\right)}{\left(1-\nu \right)}}$ is proved.

Explanation of Solution

Given information:

The third principal strain ${\epsilon }_3=\frac{-\nu \left({\epsilon }_1+{\epsilon }_2\right)}{\left(1-\nu \right)}$

Explanation:

For the case of plane stress ${\sigma }_3=0$.

Apply the normal strain in x direction as shown below.

${\epsilon }_1=\frac{1}{E}\left({\sigma }_1-\nu \left({\sigma }_2+{\sigma }_3\right)\right)$

Here, E is the modulus of elasticity, ${\sigma }_1$ is the normal stress in x direction, $\nu$ is the Poisson’s ratio, ${\sigma }_2$ is the normal stress in y direction, and ${\sigma }_3$ is the normal stress in z direction.

Substitute 0 for ${\sigma }_3$.

$\begin{array}{c}{\epsilon }_1=\frac{1}{E}\left({\sigma }_1-\nu \left({\sigma }_2+0\right)\right)\\ =\frac{1}{E}\left({\sigma }_1-\nu {\sigma }_2\right)\\ E{\epsilon }_1={\sigma }_1-\nu {\sigma }_2\end{array}$

Multiply both sides of the Equation by $\nu$.

$\begin{array}{c}\nu E{\epsilon }_1=\left({\sigma }_1-\nu {\sigma }_2\right)\nu \\ =\nu {\sigma }_1-{\nu }^2{\sigma }_2\end{array}$ (1)

Apply the normal strain in y direction as shown below.

${\epsilon }_2=\frac{1}{E}\left({\sigma }_2-\nu \left({\sigma }_1+{\sigma }_3\right)\right)$

Substitute 0 for ${\sigma }_3$.

$\begin{array}{c}{\epsilon }_2=\frac{1}{E}\left({\sigma }_2-\nu \left({\sigma }_1+0\right)\right)\\ =\frac{1}{E}\left({\sigma }_2-\nu {\sigma }_1\right)\\ E{\epsilon }_2={\sigma }_2-\nu {\sigma }_1\end{array}$ (2)

Apply the normal strain in z direction as shown below.

${\epsilon }_3=\frac{1}{E}\left({\sigma }_3-\nu \left({\sigma }_1+{\sigma }_2\right)\right)$

Substitute 0 for ${\sigma }_3$.

$\begin{array}{c}{\epsilon }_3=\frac{1}{E}\left(0-\nu \left({\sigma }_1+{\sigma }_2\right)\right)\\ =\frac{1}{E}\left(-\nu \left({\sigma }_1+{\sigma }_2\right)\right)\end{array}$ (3)

Adding Equation (1) and (2).

$\begin{array}{c}\nu E{\epsilon }_1+E{\epsilon }_2=\left(\nu {\sigma }_1-{\nu }^2{\sigma }_2\right)+\left({\sigma }_2-\nu {\sigma }_1\right)\\ =\nu {\sigma }_1-{\nu }^2{\sigma }_2+{\sigma }_2-\nu {\sigma }_1\\ ={\sigma }_2\left(1-{\nu }^2\right)\\ E\left(\nu {\epsilon }_1+{\epsilon }_2\right)={\sigma }_2\left(1-{\nu }^2\right)\end{array}$

${\sigma }_2=\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)$

Substitute $\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)$ for ${\sigma }_2$ in Equation (2).

$\begin{array}{l}E{\epsilon }_2=\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)-\nu {\sigma }_1\\ \nu {\sigma }_1=\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)-E{\epsilon }_2\end{array}$

$\begin{array}{c}{\sigma }_1=\frac{1}{\nu }\left(\frac{E\left(\nu {\epsilon }_1+{\epsilon }_2\right)-\left(1-{\nu }^2\right)E{\epsilon }_2}{\left(1-{\nu }^2\right)}\right)\\ =\frac{E}{\nu \left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2-{\epsilon }_2+{\nu }^2{\epsilon }_2\right)\\ =\frac{E}{\nu \left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\nu }^2{\epsilon }_2\right)\\ =\frac{E}{\left(1-{\nu }^2\right)}\left({\epsilon }_1+\nu {\epsilon }_2\right)\end{array}$

Substitute $\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)$ for ${\sigma }_2$ and $\frac{E}{\left(1-{\nu }^2\right)}\left({\epsilon }_1+\nu {\epsilon }_2\right)$ for ${\sigma }_1$ in Equation (3).

$\begin{array}{c}{\epsilon }_3=\frac{1}{E}\left(-\nu \left(\frac{E}{\left(1-{\nu }^2\right)}\left({\epsilon }_1+\nu {\epsilon }_2\right)+\frac{E}{\left(1-{\nu }^2\right)}\left(\nu {\epsilon }_1+{\epsilon }_2\right)\right)\right)\\ =\frac{-\nu E}{E\left(1-{\nu }^2\right)}\left({\epsilon }_1+\nu {\epsilon }_2+\nu {\epsilon }_1+{\epsilon }_2\right)\\ =\frac{-\nu }{\left(1-\nu \right)\left(1+\nu \right)}\left(\left(1+\nu \right){\epsilon }_1+\left(1+\nu \right){\epsilon }_2\right)\\ =\frac{-\nu \left(1+\nu \right)}{\left(1-\nu \right)\left(1+\nu \right)}\left({\epsilon }_1+{\epsilon }_2\right)\end{array}$

${\epsilon }_3=\frac{-\nu }{\left(1-\nu \right)}\left({\epsilon }_1+{\epsilon }_2\right)$

Therefore, the third principal strain can be obtained from $\underset{\_}{{\epsilon }_3=\frac{-\nu \left({\epsilon }_1+{\epsilon }_2\right)}{\left(1-\nu \right)}}$ is proved.