Chapter 6, Problem 6.5P

To determine

Compare the true and engineering strain energy density, the strain energy absorbed in natural rubber and energy stored in steel specimen.

Answer to Problem 6.5P

The true and engineering strain energy density is 1.6 times the engineering strain density for natural rubber and the strain energy absorbed in natural rubber and energy stored in steel specimen have a huge difference because rubber can strain more and can store more energy inside it due to application of force on it.

Explanation of Solution

Given:

Engineering strain is $6.0$ .

Write the expression for engineering secant elastic moduli of natural rubber.

  $E_{\sec }\left(\epsilon \right)=\frac{\sigma }{\epsilon }$ .......... (1)

Here, $E_{\sec }$ are the engineering secant elastic moduli, $\sigma$ is the engineering stress of natural rubber and $\epsilon$ is the engineering strain.

Write the expression for true secant elastic moduli of natural rubber.

  $E_{\sec }\left({\epsilon }_t\right)=\frac{{\sigma }_t}{{\epsilon }_t}$ .......... (2)

Here, $E_{\sec }$ are the true secant elastic moduli, ${\sigma }_t$ is the true stress of natural rubber and ${\epsilon }_t$ is the true strain.

Write the expression for engineering strain energy absorbed in natural rubber.

  $SE=\frac{{\sigma }^2}{2E_{\sec }\left(\epsilon \right)}$ .......... (3)

Here, $SE$ is the engineering strain energy absorbed by natural rubber.

Write the expression for true strain energy absorbed in natural rubber.

  ${\left(SE\right)}_t=\frac{{\sigma }_t^2}{2E_{\sec }\left({\epsilon }_t\right)}$ .......... (4)

Here, ${\left(SE\right)}_t$ is the true strain energy absorbed by natural rubber.

Refer to Example 6.2, the engineering stress for natural rubber is $60\times {10}^6\text{Pa}$ .

Substitute $60\times {10}^6\text{Pa}$ for $\sigma$ and $6.0$ for $\epsilon$ in equation (1).

  $\begin{array}{c}{E}_{\sec }\left(6.0\right)=\frac{60\times {10}^6\text{ Pa}}{6.0}\\ =10\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ \text{=10}\text{MPa}\end{array}$

Refer to Example 6.2, the true stress for natural rubber is $293\times {10}^6\text{ Pa}$ and true strain for natural rubber is $1.95$ .

Substitute $293\times {10}^6\text{Pa}$ for ${\sigma }_t$ and $1.95$ for ${\epsilon }_t$ in equation (1).

  $\begin{array}{c}{E}_{\sec }\left(1.95\right)=\frac{293\times {10}^6\text{ Pa}}{1.95}\\ =150\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ \text{=150}\text{MPa}\end{array}$

Substitute $60\times {10}^6\text{Pa}$ for $\sigma$ and $10\text{ MPa}$ for $E_{\sec }\left(\epsilon \right)$ in equation (3).

  $\begin{array}{c}SE=\frac{{\left(60\times {10}^6\right)}^2}{2\left(10\text{MPa}\right)\left(\frac{1{\text{0}}^6\text{ Pa}}{1\text{}\text{MPa}}\right)}\\ =\frac{3600\times {10}^{12}}{20\times {10}^6}\\ =180\times {10}^6{\text{J/m}}^3\end{array}$

Substitute $293\times {10}^6\text{Pa}$ for ${\sigma }_t$ and $150\text{ MPa}$ for $E_{\sec }\left({\epsilon }_t\right)$ in equation (4).

  $\begin{array}{c}{\left(SE\right)}_t=\frac{{\left(293\times {10}^6\right)}^2}{2\left(150\text{MPa}\right)\left(\frac{1{\text{0}}^6\text{ Pa}}{1\text{}\text{MPa}}\right)}\\ =\frac{85849\times {10}^{12}}{300\times {10}^6}\\ =286\times {10}^6{\text{J/m}}^3\end{array}$

The true strain energy absorbed by natural rubber is 1.6 times the engineering strain energy absorbed by natural rubber. The variation in values indicates the difference in measurement through theoretical and practical approach.

Refer to example 6.5, the value of strain energy for steel specimen is $0.245\times {10}^6{\text{J/m}}^3$ . The strain energy of rubber is greater than the strain energy for natural rubber because rubber can strain more than the steel specimen under the application of load.

Thus, the true and engineering strain energy density is 1.6 times the engineering strain density for natural rubber and the strain energy absorbed in natural rubber and energy stored in steel specimen have a huge difference because rubber can strain more and can store more energy inside it due to application of force on it.