Elasticity/Sample midterm3

Sample Homework Problem 3

For an isotropic material with $E=100$ GPa and $\nu =0.25$, find the stress tensor and strain energy density at a point in a body if the components of the strain tensor are given by

${\epsilon }_{ij}=\left[\begin{array}{ccc}200& 100& 0\\ 100& 200& 100\\ 0& 100& 0\end{array}\right]\times 10^{-6}.$

Solution

The shear modulus ($\mu$) is given by

$\mu =\frac{E}{2(1+\nu )}=\frac{100}{2.5}=40\text{GPa}=40\times 10^6\text{KPa}$

The Lamé modulus ($\lambda$) is given by

$\lambda =\frac{E\nu }{(1+\nu )(1-2\nu )}=\frac{25}{(1.25)(0.5)}=40\text{GPa}=40\times 10^6\text{KPa}$

The stress-strain relation for isotropic materials is

$\begin{array}{cc}{\sigma }_{ij}& =2\mu {\epsilon }_{ij}+\lambda {\epsilon }_{kk}{\delta }_{ij}\\ & =40(2{\epsilon }_{ij}+{\epsilon }_{kk}{\delta }_{ij})\end{array}$

Therefore, (after converting $\mu$ and $\lambda$ into KPa so that the $10^{-6}$ term in the strain cancels out),

$\begin{array}{cc}{\sigma }_{11}& =40(2{\epsilon }_{11}+{\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33})\\ & =40[(3)(200)+200+0]=(40)(800)=32000\\ {\sigma }_{22}& =40(2{\epsilon }_{22}+{\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33})\\ & =40[(3)(200)+200+0]=(40)(800)=32000\\ {\sigma }_{33}& =40(2{\epsilon }_{33}+{\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33})\\ & =40[(3)(0)+200+200]=(40)(400)=16000\\ {\sigma }_{23}& =40(2{\epsilon }_{23})=(40)(200)=8000\\ {\sigma }_{31}& =40(2{\epsilon }_{31})=(40)(0)=0\\ {\sigma }_{12}& =40(2{\epsilon }_{12})=(40)(200)=8000\end{array}$

In 3$\times$3 matrix form (after converting into MPa from KPa)

${\sigma }_{ij}=\left[\begin{array}{ccc}32& 8& 0\\ 8& 32& 8\\ 0& 8& 16\end{array}\right]\text{MPa}$

The strain energy density is given by

$U(\mathit{\epsilon })=\frac{1}{2}{\sigma }_{ij}{\epsilon }_{ij}$

Therefore,

$\begin{array}{cc}U(\mathit{\epsilon })& =\frac{1}{2}[{\sigma }_{11}{\epsilon }_{11}+{\sigma }_{22}{\epsilon }_{22}+{\sigma }_{33}{\epsilon }_{33}+2{\sigma }_{23}{\epsilon }_{23}+2{\sigma }_{31}{\epsilon }_{31}+2{\sigma }_{12}{\epsilon }_{12}]\\ & =\frac{1}{2}[(32)(200)+(32)(200)+(16)(0)+(2)(8)(100)+(2)(0)(0)+(2)(8)(100)]\text{Pa}\\ & =\frac{1}{2}[6400+6400+1600+1600]\text{Pa}\\ & =8000\text{Pa}=8\text{KPa}\end{array}$

The strain energy density is

$U=8\text{KPa}$