3 Engineering Example 1

3.1 Stresses and strains on a section of material

Introduction

An important engineering problem is to determine the effect on materials of different types of loading. One way of measuring the effects is through the strain or fractional change in dimensions in the material which can be measured using a strain gauge.

Problem in words

In a homogeneous, isotropic and linearly elastic material, the strains (i.e. fractional displacements) on a section of the material, represented by ${\epsilon }_x,{\epsilon }_y,{\epsilon }_z$ for the $x$ -,  $y$ -,  $z$ -directions respectively, can be related to the stresses (i.e. force per unit area), ${\sigma }_x,{\sigma }_y,{\sigma }_z$ by the following system of equations.

where $E$ is the modulus of elasticity (also called Young’s modulus) and $v$ is Poisson’s ratio which relates the lateral strain to the axial strain.

Find expressions for the stresses ${\sigma }_x,{\sigma }_y,{\sigma }_z,$ in terms of the strains ${\epsilon }_x,{\epsilon }_y,$ and ${\epsilon }_z.$

Mathematical statement of problem

The given system of equations can be written as a matrix equation:

$\left(\begin{array}{c}\hfill {\epsilon }_x\hfill \\ \hfill {\epsilon }_y\hfill \\ \hfill {\epsilon }_z\hfill \end{array}\right)=\frac{1}{E}\left(\begin{array}{ccc}\hfill 1& \hfill -v& \hfill -v\\ \hfill -v& \hfill 1& \hfill -v\\ \hfill -v& \hfill -v& \hfill 1\end{array}\right)\left(\begin{array}{c}\hfill {\sigma }_x\hfill \\ \hfill {\sigma }_y\hfill \\ \hfill {\sigma }_z\hfill \end{array}\right)$

We can write this equation as

$\epsilon =\frac{1}{E}A\sigma$

where $\epsilon =\left(\begin{array}{c}\hfill {\epsilon }_x\hfill \\ \hfill {\epsilon }_y\hfill \\ \hfill {\epsilon }_z\hfill \end{array}\right),A=\left(\begin{array}{ccc}\hfill 1& \hfill -v& \hfill -v\\ \hfill -v& \hfill 1& \hfill -v\\ \hfill -v& \hfill -v& \hfill 1\end{array}\right)$ and $\sigma =\left(\begin{array}{c}\hfill {\sigma }_x\hfill \\ \hfill {\sigma }_y\hfill \\ \hfill {\sigma }_z\hfill \end{array}\right)$

This matrix equation must be solved to find the vector $\sigma$ in terms of the vector $\epsilon$ and the inverse of the matrix $A$ .

Mathematical analysis

$\epsilon =\frac{1}{E}A\sigma$

Multiplying both sides of the expression by $E$ we get

$E\epsilon =A\sigma$

Multiplying both sides by $A^{-1}$ we find that:

$A^{-1}E\epsilon =A^{-1}A\sigma$

But $A^{-1}A=I$ so this becomes

$\sigma =E{A}^{-1}\epsilon$

To find expressions for the stresses ${\sigma }_x,{\sigma }_y,{\sigma }_z$ , in terms of the strains ${\epsilon }_x,{\epsilon }_y$ and ${\epsilon }_z$ , we must find the inverse of the matrix $A$ .

To find the inverse of $\left(\begin{array}{ccc}\hfill 1& \hfill -v& \hfill -v\\ \hfill -v& \hfill 1& \hfill -v\\ \hfill -v& \hfill -v& \hfill 1\end{array}\right)$ we first find the matrix of minors which is:

$\left(\begin{array}{ccc}\hfill \left|\begin{array}{cc}\hfill 1& \hfill -v\\ \hfill -v& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill -v& \hfill -v\\ \hfill -v& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill -v& \hfill 1\\ \hfill -v& \hfill -v\end{array}\right|\hfill \\ \hfill \hfill \\ \hfill \left|\begin{array}{cc}\hfill -v& \hfill -v\\ \hfill -v& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 1& \hfill -v\\ \hfill -v& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 1& \hfill -v\\ \hfill -v& \hfill -v\end{array}\right|\hfill \\ \hfill \hfill \\ \hfill \left|\begin{array}{cc}\hfill -v& \hfill -v\\ \hfill 1& \hfill -v\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 1& \hfill -v\\ \hfill -v& \hfill -v\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 1& \hfill -v\\ \hfill -v& \hfill 1\end{array}\right|\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1-v^2\hfill & \hfill -v-v^2\hfill & \hfill v^2+v\hfill \\ \hfill -v-v^2\hfill & \hfill 1-v^2\hfill & \hfill -v-v^2\hfill \\ \hfill v^2+v\hfill & \hfill -v-v^2\hfill & \hfill 1-v^2\hfill \end{array}\right).$

We then apply the pattern of signs:

$\left(\begin{array}{ccc}\hfill +\hfill & \hfill -\hfill & \hfill +\hfill \\ \hfill -\hfill & \hfill +\hfill & \hfill -\hfill \\ \hfill +\hfill & \hfill -\hfill & \hfill +\hfill \end{array}\right)$

to obtain the matrix of cofactors

$\left(\begin{array}{ccc}\hfill 1-v^2\hfill & \hfill v+v^2\hfill & \hfill v^2+v\hfill \\ \hfill v+v^2\hfill & \hfill 1-v^2\hfill & \hfill v+v^2\hfill \\ \hfill v^2+v\hfill & \hfill v+v^2\hfill & \hfill 1-v^2\hfill \end{array}\right).$

To find the adjoint we take the transpose of the above, (which is the same as the original matrix since the matrix is symmetric)

$\left(\begin{array}{ccc}\hfill 1-v^2\hfill & \hfill v+v^2\hfill & \hfill v^2+v\hfill \\ \hfill v+v^2\hfill & \hfill 1-v^2\hfill & \hfill v+v^2\hfill \\ \hfill v^2+v\hfill & \hfill v+v^2\hfill & \hfill 1-v^2\hfill \end{array}\right).$

The determinant of the original matrix is

$1\times \left(1-v^2\right)-v\left(v+v^2\right)-v\left(v^2+v\right)=1-3v^2-2v^3.$ Finally we divide the adjoint by the determinant to find the inverse, giving

$\frac{1}{1-3v^2-2v^3}\left(\begin{array}{ccc}\hfill 1-v^2\hfill & \hfill v+v^2\hfill & \hfill v+v^2\hfill \\ \hfill v+v^2\hfill & \hfill 1-v^2\hfill & \hfill v+v^2\hfill \\ \hfill v+v^2\hfill & \hfill v+v^2\hfill & \hfill 1-v^2\hfill \end{array}\right).$

Now we found that $\sigma =E{A}^{-1}\epsilon$ so $\left(\begin{array}{c}\hfill {\sigma }_x\hfill \\ \hfill {\sigma }_y\hfill \\ \hfill {\sigma }_z\hfill \end{array}\right)=\frac{E}{1-3v^2-2v^3}\left(\begin{array}{ccc}\hfill 1-v^2\hfill & \hfill v+v^2\hfill & \hfill v+v^2\hfill \\ \hfill v+v^2\hfill & \hfill 1-v^2\hfill & \hfill v+v^2\hfill \\ \hfill v+v^2\hfill & \hfill v+v^2\hfill & \hfill 1-v^2\hfill \end{array}\right)\left(\begin{array}{c}\hfill {\epsilon }_x\hfill \\ \hfill {\epsilon }_y\hfill \\ \hfill {\epsilon }_z\hfill \end{array}\right)$

We can write this matrix equation as 3 equations relating the stresses ${\sigma }_x,{\sigma }_y,{\sigma }_z,$ in terms of the strains ${\epsilon }_x,{\epsilon }_y$ and ${\epsilon }_z$ , by multiplying out this matrix expression, giving:

Interpretation

Matrix manipulation has been used to transform three simultaneous equations relating strain to stress into simultaneous equations relating stress to strain in terms of the elastic constants. These would be useful for deducing the applied stress if the strains are known. The original equations enable calculation of strains if the applied stresses are known.

Exercises

1. Solve the following using Cramer’s rule:
   1. $$\begin{array}{ccccc}\hfill 2x& \hfill -& \hfill 3y& \hfill =& \hfill 1\\ \hfill 4x& \hfill +& \hfill 4y& \hfill =& \hfill 2\\ \hfill \end{array}$$
   2. $$\begin{array}{ccccc}\hfill 2x& \hfill -& \hfill 5y& \hfill =& \hfill 2\\ \hfill -4x& \hfill +& \hfill 10y& \hfill =& \hfill 1\\ \hfill \end{array}$$
   3. $$\begin{array}{ccccc}\hfill 6x& \hfill -& \hfill y& \hfill =& \hfill 0\\ \hfill 2x& \hfill -& \hfill 4y& \hfill =& \hfill 1\\ \hfill \end{array}$$
2. Using Cramer’s rule obtain the solutions to the following sets of equations:
   1. $$\begin{array}{ccccccc}\hfill 2x_1& \hfill +& \hfill x_2& \hfill -& \hfill x_3& \hfill =& \hfill 0\\ \hfill x_1& \hfill & \hfill & \hfill +& \hfill x_3& \hfill =& \hfill 4\\ \hfill x_1& \hfill +& \hfill x_2& \hfill +& \hfill x_3& \hfill =& \hfill 0\end{array}$$
   2. $$\begin{array}{ccccccc}\hfill x_1& \hfill -& \hfill x_2& \hfill +& \hfill x_3& \hfill =& \hfill 1\\ \hfill -x_1& \hfill & \hfill & \hfill +& \hfill x_3& \hfill =& \hfill 1\\ \hfill x_1& \hfill +& \hfill x_2& \hfill -& \hfill x_3& \hfill =& \hfill 0\end{array}$$

Answer