# Introduction to Elasticity/Sample midterm 4

## Sample Homework Problem 4

### Part (a)

A solid is subjected to stresses as shown by the arrows in the figure below. Indicate the indices for each of the stress components, and whether the stresses should be positive or negative.

### Part (b)

Suppose that the stress tensor field in a body is given by

$\sigma _{ij}={\begin{bmatrix}x_{1}^{2}+x_{2}&-2x_{3}^{2}&x_{1}^{2}+x_{2}\\-2x_{3}^{2}&x_{1}+x_{3}&-x_{3}^{2}+x_{2}\\x_{1}^{2}+x_{2}&-x_{3}^{2}+x_{2}&x_{1}^{2}+x_{2}\end{bmatrix}}({\text{MPa}})$ Find the body force distribution required to maintain equilibrium. ($x_{1}$ , $x_{2}$ , and $x_{3}$ are in meters). Show units.

#### Solution

The equation of equilibrium is

${\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}+\mathbf {b} =0~~{\text{or,}}~~\sigma _{ij,i}+b_{j}=0$ Therefore,

{\begin{aligned}{\frac {\partial (x_{1}^{2}+x_{2})}{\partial x_{1}}}+{\frac {\partial (-2x_{3}^{2})}{\partial x_{2}}}+{\frac {\partial (x_{1}^{2}+x_{2})}{\partial x_{3}}}+b_{1}&=0\\{\frac {\partial (-2x_{3}^{2})}{\partial x_{1}}}+{\frac {\partial (x_{1}+x_{3})}{\partial x_{2}}}+{\frac {\partial (-x_{3}^{2}+x_{2})}{\partial x_{3}}}+b_{2}&=0\\{\frac {\partial (x_{1}^{2}+x_{2})}{\partial x_{1}}}+{\frac {\partial (-x_{3}^{2}+x_{2})}{\partial x_{2}}}+{\frac {\partial (x_{1}^{2}+x_{2})}{\partial x_{3}}}+b_{3}&=0\end{aligned}} or,

{\begin{aligned}2~x_{1}+0+0+b_{1}&=0\\0+0-2~x_{3}+b_{2}&=0\\2~x_{1}+1+0+b_{3}&=0\end{aligned}} The required body forces are (in MN/m$^{3}$ )

${b_{1}=-2~x_{1}~;~~b_{2}=2~x_{3}~;~~b_{3}=-(1+2~x_{1})}$ ### Part (c)

Find the surface tractions at the internal point $\mathbf {r} =(1,-1,2)$ in the body on an internal surface with a surface normal ${\widehat {\mathbf {n} }}=(-1,-1,1)$ .

#### Solution

The surface traction is given by

$\mathbf {t} ={\widehat {\mathbf {n} }}\bullet {\boldsymbol {\sigma }}~~{\text{or,}}~~t_{j}=n_{i}\sigma _{ij}$ The stress at point $\mathbf {r}$ is

$\left[{\boldsymbol {\sigma }}\right]={\begin{bmatrix}0&-8&0\\-8&3&-5\\0&-5&0\end{bmatrix}}~({\text{MPa}})$ Therefore,

{\begin{aligned}t_{1}&=n_{1}\sigma _{11}+n_{2}\sigma _{21}+n_{3}\sigma _{31}\\&=(-1)(0)+(-1)(-8)+(1)(0)=8~{\text{MPa}}\\t_{2}&=n_{1}\sigma _{12}+n_{2}\sigma _{22}+n_{3}\sigma _{32}\\&=(-1)(-8)+(-1)(3)+(1)(-5)=0~{\text{MPa}}\\t_{3}&=n_{1}\sigma _{13}+n_{2}\sigma _{23}+n_{3}\sigma _{33}\\&=(-1)(0)+(-1)(-5)+(1)(0)=5~{\text{MPa}}\\\end{aligned}} The traction vector is (after converting ${\widehat {\mathbf {n} }}{}$ into a unit normal)

${\mathbf {t} =(1/{\sqrt {3}})(8,~0,~5)~~({\text{MPa}})}$ ### Part (d)

Find the hydrostatic and deviatoric stress at the point $\mathbf {r} =(1,-1,2)$ .

#### Solution

The hydrostatic stress is given by

${\boldsymbol {\sigma }}_{h}={\frac {{\text{tr}}~{\boldsymbol {\sigma }}}{3}}\mathbf {I}$ In this case,

${\text{tr}}{\boldsymbol {\sigma }}=\sigma _{11}+\sigma _{22}+\sigma _{33}=0+3+0=3$ Therefore,

${{\boldsymbol {\sigma }}_{h}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}~({\text{MPa}})}$ The deviatoric stress is given by

${\boldsymbol {\sigma }}_{d}={\boldsymbol {\sigma }}-{\boldsymbol {\sigma }}_{h}$ Therefore,

${{\boldsymbol {\sigma }}_{d}={\begin{bmatrix}-1&-8&0\\-8&2&-5\\0&-5&-1\end{bmatrix}}~({\text{MPa}})}$ ### Part (e)

Find the principal stresses at this point.

#### Solution

The principal stresses can be found using the equation

${\text{det}}({\boldsymbol {\sigma }}-\lambda \mathbf {I} )=0$ where $\lambda$ is a principal stress. In expanded form,

${\text{det}}{\begin{bmatrix}\sigma _{11}-\lambda &\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\lambda &\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\lambda \end{bmatrix}}=0$ Substituting the values of stress into the above equation,

${\text{det}}{\begin{bmatrix}-\lambda &-8&0\\-8&3-\lambda &-5\\0&-5&-\lambda \end{bmatrix}}=0$ Expanding out,

$-\lambda \left[(3-\lambda )(-\lambda )-(-5)(-5)\right]-(-8)\left[(-8)(-\lambda )-(-5)(0)\right]=0$ or,

$-\lambda (-3\lambda +\lambda ^{2}-25)+8(8\lambda )=0$ or,

$3\lambda ^{2}-\lambda ^{3}+25\lambda +64\lambda =0$ or,

$\lambda ^{3}-3\lambda ^{2}-89\lambda =0$ Thus, the first possible value of $\lambda =0$ MPa. Also,

$\lambda ^{2}-3\lambda -89=0$ Therefore,

$\lambda ={\frac {3\pm {\sqrt {9+356}}}{2}}=(11.05,-8.05){\text{MPa}}$ The principal stresses are (in MPa)

${\sigma _{1}=11.05~;~~\sigma _{2}=0~;~~\sigma _{3}=-8.05}$ ### Part (f)

Find the principal direction corresponding to the intermediate principal stress.

#### Solution

The directions of the principal stresses can be found using the equation

$({\boldsymbol {\sigma }}-\lambda \mathbf {I} ){\widehat {\mathbf {n} }}{}=0$ For the principal direction ${\widehat {\mathbf {n} }}{2}$ (corresponding to the principal stress $\sigma _{2}$ , we have,

${\begin{bmatrix}0&-8&0\\-8&3&-5\\0&-5&0\end{bmatrix}}{\begin{bmatrix}n^{1}\\n^{2}\\n^{3}\end{bmatrix}}=0$ Hence,

{\begin{aligned}(0)(n^{1})-(8)(n^{2})+(0)(n^{3})&=0\\(-8)(n^{1})+(3)(n^{2})+(-5)(n^{3})&=0\end{aligned}} gives us $n^{2}=0$ and

$8~n^{1}=-5~n^{3}$ Now, $n^{3}={\sqrt {1-(n^{1})^{2}-(n^{2})^{2}}}={\sqrt {1-(n^{1})^{2}}}$ . Therefore,

$8~n^{1}=-5~{\sqrt {1-(n^{1})^{2}}}$ Taking squares of both sides,

$64(n^{1})^{2}=25\left[1-(n^{1})^{2}\right]$ So we get,

$(n^{1})^{2}=25/89~~,{\text{or}}~~n^{1}=5/{\sqrt {89}}$ Therefore,

$n^{3}={\sqrt {1-(n^{1})^{2}}}=8/{\sqrt {89}}$ The direction corresponding to the intermediate principal stress is

${{\widehat {\mathbf {n} }}{2}=(5/{\sqrt {89}},~0,~8/{\sqrt {89}})}$ ### Part (g)

The symmetry of the stress tensor can be derived from a certain balance principle. Name the principle and write it down in index notation.

#### Solution

The balance principle is

${\text{Conservation of angular momentum}}$ In index notation

${\int _{\partial \Omega }e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~ds+\int _{\Omega }e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\int _{\Omega }\rho ~e_{ijk}~x_{j}~v_{k}~dV}$ 