# Introduction to Elasticity/Sample midterm3

## Sample Homework Problem 3

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

${\displaystyle \varepsilon _{ij}={\begin{bmatrix}200&100&0\\100&200&100\\0&100&0\end{bmatrix}}\times 10^{-6}~~.}$

## Solution

The shear modulus (${\displaystyle \mu }$) is given by

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

The Lamé modulus (${\displaystyle \lambda }$) is given by

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

The stress-strain relation for isotropic materials is

{\displaystyle {\begin{aligned}\sigma _{ij}&=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}\\&=40(2\varepsilon _{ij}+\varepsilon _{kk}\delta _{ij})\end{aligned}}}

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

{\displaystyle {\begin{aligned}\sigma _{11}&=40(2\varepsilon _{11}+\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\\&=40\left[(3)(200)+200+0\right]=(40)(800)=32000\\\sigma _{22}&=40(2\varepsilon _{22}+\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\\&=40\left[(3)(200)+200+0\right]=(40)(800)=32000\\\sigma _{33}&=40(2\varepsilon _{33}+\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\\&=40\left[(3)(0)+200+200\right]=(40)(400)=16000\\\sigma _{23}&=40(2\varepsilon _{23})=(40)(200)=8000\\\sigma _{31}&=40(2\varepsilon _{31})=(40)(0)=0\\\sigma _{12}&=40(2\varepsilon _{12})=(40)(200)=8000\end{aligned}}}

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

${\displaystyle {\sigma _{ij}={\begin{bmatrix}32&8&0\\8&32&8\\0&8&16\end{bmatrix}}~{\text{MPa}}}}$

The strain energy density is given by

${\displaystyle U({\boldsymbol {\varepsilon }})={\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}}$

Therefore,

{\displaystyle {\begin{aligned}U({\boldsymbol {\varepsilon }})&={\frac {1}{2}}\left[\sigma _{11}\varepsilon _{11}+\sigma _{22}\varepsilon _{22}+\sigma _{33}\varepsilon _{33}+2\sigma _{23}\varepsilon _{23}+2\sigma _{31}\varepsilon _{31}+2\sigma _{12}\varepsilon _{12}\right]\\&={\frac {1}{2}}\left[(32)(200)+(32)(200)+(16)(0)+(2)(8)(100)+(2)(0)(0)+(2)(8)(100)\right]~{\text{Pa}}\\&={\frac {1}{2}}\left[6400+6400+1600+1600\right]~{\text{Pa}}\\&=8000~{\text{Pa}}=8~{\text{KPa}}\end{aligned}}}

The strain energy density is

${\displaystyle {U=8~{\text{KPa}}}}$