Chapter 15, Problem 15D.3BE

Interpretation Introduction

Interpretation:

The change in volume of lead after applying a uniaxial stress has to be calculated.

Concept Introduction:

Poison ratio can be calculated as follows,

  $\text{poison}\text{}\text{ratio, σ =}\frac{\text{transverse}\text{strain}}{\text{uniaxial}\text{strain}}$

Strain can be calculated as follows,

  $strain=\frac{changein\dim ension}{original\dim ension}$

Contraction in both transverse direction occur during transverse strain.

Answer to Problem 15D.3BE

The change in volume is calculated as $3.3\times 10^{-3}d{m}^3.$

Explanation of Solution

Consider ${\left(\frac{\Delta L}{L}\right)}_z$ is the uniaxial strain, and  ${\left(\frac{\Delta L}{L}\right)}_x$, ${\left(\frac{\Delta L}{L}\right)}_y$ are the transverse strain.

${\left(\frac{\Delta L}{L}\right)}_x={\left(\frac{\Delta L}{L}\right)}_y.$

When the stress is applied the length in the transverse direction decreases, but length in axial direction increases.

Given information is shown below,

    $\begin{array}{l}{\left(\frac{\Delta L}{L}\right)}_z=0.02\\ \\ \sigma =0.41\\ \\ Initialvolumeofthecube,V=1d{m}^3\\ \\ Initialsidelengthofthecube,L_x=L_y=L_z=1dm\end{array}$

Therefore transverse strain can be calculated as follows,

  $\begin{array}{c}transversestrain=\sigma \times uniaxialstrain\\ \\ =0.41\times 0.02\\ \\ =8.2\times 10^{-3}\end{array}$

So, ${\left(\frac{\Delta L}{L}\right)}_x= {\left(\frac{\Delta L}{L}\right)}_y=8.2\times {10}^{-3}.$

Change in length in z direction can be calculated as follows,

  $\begin{array}{c}\Delta L_z=0.02\times L_z\\ \\ =0.02\times 1\text{}dm\\ \\ =0.02dm\end{array}$

Therefore the new length in the axial direction ($L_z^{\text{'}}$) can be calculated as follows,

  $\begin{array}{c}\Delta L_z=L_z^{\text{'}}-L_z\\ \\ L_z^{\text{'}}=\Delta L_z+L_z\\ \\ =0.02dm+1dm\\ \\ =1.02dm\end{array}$

Change in length in x and y direction can be calculated as follows,

  $\begin{array}{c}\Delta L_x=8.2\times {10}^{-3}\times L_x\\ \\ =8.2\times {10}^{-3}\times 1\text{}dm\\ \\ =8.2\times 10^{-3}dm\end{array}$

  $\begin{array}{c}\Delta L_y=8.2\times {10}^{-3}\times L_y\\ \\ =8.2\times {10}^{-3}\times 1\text{}dm\\ \\ =8.2\times 10^{-3}dm\end{array}$

Therefore the new lengths in the transverse directions ($L_x^{\text{'}}and{L}_y^{\text{'}}$) can be calculated as follows,

  $\begin{array}{c}\Delta L_x=L_x-L_x^{\text{'}}\\ \\ L_x^{\text{'}}=L_x-\Delta L_x\\ \\ =1dm-\left(8.2\times {10}^{-3}\right)dm\\ \\ =0.9918dm\end{array}$

  $\begin{array}{c}\Delta L_y=L_y-L_y^{\text{'}}\\ \\ L_y^{\text{'}}=L_y-\Delta L_y\\ \\ =1dm-\left(8.2\times {10}^{-3}\right)dm\\ \\ =0.9918dm\end{array}$

The new volume of the solid can be calculated as follows,

  $\begin{array}{c}V\text{'}=L_x^{\text{'}}\times L_y^{\text{'}}\times L_z^{\text{'}}\\ \\ =0.9918dm\times 0.9918dm\times 1.02dm\\ \\ =1.0033d{m}^3\end{array}$

Therefore the change in volume can be calculated as follows,

  $\begin{array}{c}changeinvolume,\Delta V=V\text{'}-V\\ =\text{1}\text{.0033}{\text{dm}}^{\text{3}}\text{-1}{\text{dm}}^{\text{3}}\\ =\text{0}\text{.0033}{\text{dm}}^{\text{3}}\\ =3.3\times 10^{-3}d{m}^3.\end{array}$

The change in volume is calculated as $3.3\times 10^{-3}d{m}^3.$