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

\varepsilon_{ij} = \begin{bmatrix} 200 & 100 & 0 \\                      100 & 200 & 100 \\                       0 & 100 & 0                      \end{bmatrix} \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{align}

\sigma_{ij} & = 2\mu\varepsilon_{ij} + \lambda\varepsilon_{kk}\delta_{ij} \\ & = 40(2\varepsilon_{ij} + \varepsilon_{kk}\delta_{ij}) \end{align}$$ Therefore, (after converting $$\mu$$ and $$\lambda$$ into KPa so that the $$10^{-6}$$ term in the strain cancels out),
 * $$\begin{align}

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

{ \sigma_{ij} = \begin{bmatrix} 32 & 8 & 0 \\               8 & 32 & 8 \\                0 & 8 & 16                 \end{bmatrix} ~\text{MPa} } $$ The strain energy density is given by

U(\boldsymbol{\varepsilon}) = \frac{1}{2} \sigma_{ij}\varepsilon_{ij} $$ Therefore,
 * $$\begin{align}

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{align}$$ The strain energy density is

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