Solving for Stress and Strain in Three Dimensions

[KEØM]

Homework Statement

Knowing that the stress and strain for an isotropic media can be related with the following expressions:

$\sigma_{xx} = (\lambda + 2\mu)\varepsilon_{xx} + \lambda\varepsilon_{yy} + \lambda\varepsilon_{zz}$

$\sigma_{yy} = \lambda\varepsilon_{xx} + (\lambda + 2\mu)\varepsilon_{yy}) + \lambda\varepsilon_{zz}$

$\sigma_{zz} =\varepsilon_{xx}\lambda+ \lambda\varepsilon_{yy} + (\lambda + 2\mu)\varepsilon_{zz}$


Solve this system of equations for the three strain components to derive that:

$\varepsilon_{xx} = \frac{2(\lambda + \mu)\sigma_{xx} - \lambda(\sigma_{yy} + \sigma_{zz})}{2\mu(3\lambda + 2\mu)}$

$\varepsilon_{yy} = \frac{2(\lambda + \mu)\sigma_{yy} - \lambda(\sigma_{xx} + \sigma_{zz})}{2\mu(3\lambda + 2\mu)}$

$\varepsilon_{zz} = \frac{2(\lambda + \mu)\sigma_{xx} - \lambda(\sigma_{xx} + \sigma_{yy})}{2\mu(3\lambda + 2\mu)}$

Homework Equations

$\sigma_{xx} = (\lambda + 2\mu)(\varepsilon_{xx}) + \lambda\varepsilon_{yy} + \lambda\varepsilon_{zz}$

$\sigma_{yy} = \lambda\varepsilon_{xx} + (\lambda + 2\mu)(\varepsilon_{yy}) + \lambda\varepsilon_{zz}$

$\sigma_{zz} =(\varepsilon_{xx}) + \lambda\varepsilon_{yy} + (\lambda + 2\mu)\varepsilon_{zz}$

$\sigma_{ij} = \lambda\delta{ij}\varepsilon_{kk} + 2\mu\varepsilon{ij}$

Hint: $\sigma_{xx} = \sigma_{yy}$ due to symmetry.

The Attempt at a Solution

Well I put the three equations into matrix and tried to carry out row reduction and I came up with the following result:


$\left[ \begin{array}{c} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \end{array} \right] = \left[ \begin{array}{ccc} \lambda + 2\mu & \lambda & \lambda \\ \lambda & \lambda + 2\mu & \lambda \\ \lambda & \lambda & \lambda + 2\mu \\ \end{array} \right] \left[ \begin{array}{c} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \end{array} \right]$


I firstly attempted to remove the terms in the first column and the second and third rows, I did this by multiplying the top row by $\frac{\lambda}{\lambda + 2\mu}$ and subtracting that from the second and third rows which gave me the following matrix:
$= \left[ \begin{array}{ccc} \lambda + 2\mu & \lambda & \lambda \\ 0 & \lambda + 2\mu - \frac{\lambda^2}{\lambda + 2\mu} & \lambda - \frac{\lambda^2}{\lambda + 2\mu} \\ 0 & \lambda - \frac{\lambda^2}{\lambda + 2\mu} & \lambda + 2\mu - \frac{\lambda^2}{\lambda + 2\mu} \end{array} \right]$

In order to get rid of the term in the second column and the third row, I multiplied the second row by $\frac{\lambda(\lambda + 2\mu) - \lambda^2}{(\lambda + 2\mu)^2 - \lambda^2}$ and subtracted that row from the third row. This gave me the following matrix:

$= \left[ \begin{array}{ccc} \lambda + 2\mu & \lambda & \lambda \\ 0 & \lambda + 2\mu - \frac{\lambda^2}{\lambda + 2\mu} & \lambda - \frac{\lambda^2}{\lambda + 2\mu} \\ 0 & 0 & \frac{(\lambda(\lambda + 2\mu) - \lambda^2)^2}{(\lambda + 2\mu)((\lambda + 2\mu)^2 - \lambda^2)} - \left(\lambda + 2\mu - \frac{\lambda^2}{\lambda + 2\mu}\right) \end{array} \right]$


Therefore I can write my first solution out as:

$\sigma_{zz} = \left(\frac{(\lambda(\lambda + 2\mu) - \lambda^2)^2}{(\lambda + 2\mu)((\lambda + 2\mu)^2 - \lambda^2)} - \left(\lambda + 2\mu - \frac{\lambda^2}{\lambda + 2\mu}\right)\right)\varepsilon_{zz}$

which doesn't appear to be right. It would be of much help if someone could show me an easier method. Perhaps Kramer's Rule but I am unfamiliar with that method. Any help will be greatly appreciated.

Thank you,

KEØM

 

[SteamKing]

You haven't performed all the algebra you can on the coefficient of epsilon-zz. I would get a common denominator and add the two terms together to see if further simplification results.

Cramer's rule is fairly easy to remember. I am surprised you haven't com across it. It's the one where you compute all of the determinants and solve for the unknowns.

http://en.wikipedia.org/wiki/Cramer's_rule

 

[KEØM]

Thank you for your help steamengine, I was able to solve the problem by finding the inverse matrix using the method of determinants and cofactors. Thanks again for your reply.

 

[Chestermiller]

In your first attempt, you forgot to apply the operations to the left-hand sides of the equations. When you do something to the right hand side of an equation, you need to do the same thing to the left hand side.

For this particular set of 3 simultaneous linear equations, there is a simple trick for solving for the three strains. Just add the three equations together, to get:

$$(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})=(3\lambda+2\mu)(\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz})$$
or
$$(\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz})=\frac{(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})}{(3\lambda+2\mu)}$$

Then, rewrite the first equation as:

$$\sigma_{xx}=\lambda (\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz})+2\mu \epsilon_{xx}=\lambda \frac{(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})}{(3\lambda+2\mu)}+2\mu \epsilon_{xx}$$

 

[paranormal]

UN



Dear KEØMUN,

Thank you for sharing your solution attempt and for seeking help. Your approach using row reduction is a valid method for solving this system of equations. However, it seems that there may have been a mistake in your calculations. Here is a step-by-step solution using Kramer's Rule to solve for the three strain components:

1. First, we need to rewrite the system of equations in the form of a matrix equation:

\left[
\begin{array}{c}
\sigma_{xx} \\
\sigma_{yy} \\
\sigma_{zz}
\end{array}
\right]
=
\left[
\begin{array}{ccc}
\lambda + 2\mu & \lambda & \lambda \\
\lambda & \lambda + 2\mu & \lambda \\
\lambda & \lambda & \lambda + 2\mu \\
\end{array}
\right]
\left[
\begin{array}{c}
\varepsilon_{xx} \\
\varepsilon_{yy} \\
\varepsilon_{zz}
\end{array}
\right]

2. Now, we can use Kramer's Rule to solve for each strain component. For \varepsilon_{xx}, we replace the first column of the coefficient matrix with the vector of constants and solve for the determinant:

\varepsilon_{xx} = \frac{\left|
\begin{array}{ccc}
\sigma_{xx} & \lambda & \lambda \\
\sigma_{yy} & \lambda + 2\mu & \lambda \\
\sigma_{zz} & \lambda & \lambda + 2\mu \\
\end{array}
\right|}{\left|
\begin{array}{ccc}
\lambda + 2\mu & \lambda & \lambda \\
\lambda & \lambda + 2\mu & \lambda \\
\lambda & \lambda & \lambda + 2\mu \\
\end{array}
\right|}

3. Expanding the determinant, we get:

\varepsilon_{xx} = \frac{\sigma_{xx}[(\lambda + 2\mu)^2 - \lambda^2] - \lambda\sigma_{yy}(\lambda + 2\mu - \lambda) + \lambda\sigma_{zz}(\lambda - \lambda)}{(\lambda +