Chapter 2, Problem 2.49P

To determine

To show that the total true strain is additive and the engineering strain is not additive.

Answer to Problem 2.49P

It is proved that the true strain is additive in nature and the engineering strain is not additive in nature.

Explanation of Solution

Given:

Original length of strip is $1.0\text{m}$ .

Length of strip after first step is $1.5\text{ m}$ .

Length of strip after second step is $2.5\text{ m}$ .

Length of strip after third step is $3.0\text{m}$ .

Concept used:

Write the expression for true strain in material.

  ${\epsilon }_f=\ln \left(\frac{l_f}{l_0}\right)$ ..............(1)

Here, ${\epsilon }_f$ is the true strain in specimen, $l_f$ is the final length of specimen and $l_0$ is the original length of specimen.

Write the expression for total true strain.

  ${\epsilon }_{total}={\epsilon }_1+{\epsilon }_2+{\epsilon }_3$ ..............(2)

Here, ${\epsilon }_{total}$ is the total strain, ${\epsilon }_1$ is strain after first step, ${\epsilon }_2$ is the true strain after second step and ${\epsilon }_3$ is the true strain after third step.

Write the expression for engineering strain of material.

  ${{\epsilon }^{\prime }}_e=\frac{l_f-l_0}{l_0}$ ..............(3)

Here, ${{\epsilon }^{\prime }}_e$ is the engineering strain.

Write the expression for total engineering strain.

  ${{\epsilon }^{\prime }}_{total}={{\epsilon }^{\prime }}_1+{{\epsilon }^{\prime }}_2+{{\epsilon }^{\prime }}_3$ ..............(4)

Here, ${{\epsilon }^{\prime }}_{total}$ is the total engineering strain.

Calculation:

Substitute $1.5\text{m}$ for $l_f$ , ${\epsilon }_1$ for ${\epsilon }_f$ and $1.0\text{}\text{m}$ for $l_0$ in equation (1).

  $\begin{array}{c}{\epsilon }_1=\ln \left(\frac{1.5}{1}\right)\\ =0.4055\end{array}$

Substitute $2.5\text{m}$ for $l_f$ , ${\epsilon }_2$ for ${\epsilon }_f$ and $1.5\text{}\text{m}$ for $l_0$ in equation (1).

  $\begin{array}{c}{\epsilon }_2=\ln \left(\frac{2.5}{1.5}\right)\\ =0.5108\end{array}$

Substitute $3.0\text{m}$ for $l_f$ , ${\epsilon }_3$ for ${\epsilon }_f$ and $2.5\text{}\text{m}$ for $l_0$ in equation (1).

  $\begin{array}{c}{\epsilon }_1=\ln \left(\frac{3.0}{2.5}\right)\\ =0.1823\end{array}$

Substitute $3.0\text{m}$ for $l_f$ , ${\epsilon }_{total}$ for ${\epsilon }_f$ and $1.0\text{}\text{m}$ for $l_0$ in equation (1).

  $\begin{array}{c}{\epsilon }_{total}=\ln \left(\frac{3.0}{1.0}\right)\\ =1.0986\end{array}$

Substitute $0.4055$ for ${\epsilon }_1$ , $0.5108$ for ${\epsilon }_2$ and $0.1823$ for ${\epsilon }_3$ in equation (2).

  $\begin{array}{c}{\epsilon }_{total}=0.4055+0.5108+0.1823\\ =1.0986\end{array}$

Thus, as the value of true strain is same in both conditions it can be concluded that the true strain is additive in nature.

Substitute $1.5\text{m}$ for $l_f$ , ${{\epsilon }^{\prime }}_1$ for ${{\epsilon }^{\prime }}_e$ and $1.0\text{}\text{m}$ for $l_0$ in equation (3).

  $\begin{array}{c}{{\epsilon }^{\prime }}_1=\frac{1.5-1.0}{1.0}\\ =0.5\end{array}$

Substitute $2.5\text{m}$ for $l_f$ , ${{\epsilon }^{\prime }}_2$ for ${{\epsilon }^{\prime }}_e$ and $1.5\text{}\text{m}$ for $l_0$ in equation (3).

  $\begin{array}{c}{{\epsilon }^{\prime }}_2=\frac{2.5-1.5}{1.5}\\ =0.67\end{array}$

Substitute $3.0\text{m}$ for $l_f$ , ${{\epsilon }^{\prime }}_3$ for ${{\epsilon }^{\prime }}_e$ and $2.5\text{}\text{m}$ for $l_0$ in equation (3).

  $\begin{array}{c}{{\epsilon }^{\prime }}_3=\frac{3.0-2.5}{2.5}\\ =0.2\end{array}$

Substitute $3.0\text{m}$ for $l_f$ , ${{\epsilon }^{\prime }}_{total}$ for ${{\epsilon }^{\prime }}_e$ and $1.0\text{}\text{m}$ for $l_0$ in equation (3).

  $\begin{array}{c}{{\epsilon }^{\prime }}_{total}=\frac{3.0-1.0}{1.0}\\ =2.0\end{array}$

Substitute $0.5$ for ${{\epsilon }^{\prime }}_1$ , $0.67$ for ${{\epsilon }^{\prime }}_2$ and $0.2$ for ${{\epsilon }^{\prime }}_3$ in equation (4).

  $\begin{array}{c}{{\epsilon }^{\prime }}_{total}=0.5+0.67+0.2\\ =1.37\end{array}$

Thus, as the value of engineering strain is not same in both conditions it can be concluded that the engineering strain is not additivein nature.

Conclusion:

Thus, it is proved that the true strain is additive in nature and the engineering strain is not additive in nature.