Elasticity/Sample midterm 4

The equation of equilibrium is

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

Therefore,

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

or,

${\displaystyle {\begin{aligned}2~x_{1}+0+0+b_{1}&=0\\0+0-2~x_{3}+b_{2}&=0\\2~x_{1}+1+0+b_{3}&=0\end{aligned}}}$

The required body forces are (in MN/m${\displaystyle ^{3}}$)

${\displaystyle {b_{1}=-2~x_{1}~;~~b_{2}=2~x_{3}~;~~b_{3}=-(1+2~x_{1})}}$

The surface traction is given by

${\displaystyle \mathbf {t} ={\widehat {\mathbf {n} }}\bullet {\boldsymbol {\sigma }}~~{\text{or,}}~~t_{j}=n_{i}\sigma _{ij}}$

The stress at point ${\displaystyle \mathbf {r} }$ is

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

Therefore,

${\displaystyle {\begin{aligned}t_{1}&=n_{1}\sigma _{11}+n_{2}\sigma _{21}+n_{3}\sigma _{31}\\&=(-1)(0)+(-1)(-8)+(1)(0)=8~{\text{MPa}}\\t_{2}&=n_{1}\sigma _{12}+n_{2}\sigma _{22}+n_{3}\sigma _{32}\\&=(-1)(-8)+(-1)(3)+(1)(-5)=0~{\text{MPa}}\\t_{3}&=n_{1}\sigma _{13}+n_{2}\sigma _{23}+n_{3}\sigma _{33}\\&=(-1)(0)+(-1)(-5)+(1)(0)=5~{\text{MPa}}\\\end{aligned}}}$

The traction vector is (after converting ${\displaystyle {\widehat {\mathbf {n} }}{}}$ into a unit normal)

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

The hydrostatic stress is given by

${\displaystyle {\boldsymbol {\sigma }}_{h}={\frac {{\text{tr}}~{\boldsymbol {\sigma }}}{3}}\mathbf {I} }$

In this case,

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

Therefore,

${\displaystyle {{\boldsymbol {\sigma }}_{h}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}~({\text{MPa}})}}$

The deviatoric stress is given by

${\displaystyle {\boldsymbol {\sigma }}_{d}={\boldsymbol {\sigma }}-{\boldsymbol {\sigma }}_{h}}$

Therefore,

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

The principal stresses can be found using the equation

${\displaystyle {\text{det}}({\boldsymbol {\sigma }}-\lambda \mathbf {I} )=0}$

where ${\displaystyle \lambda }$ is a principal stress. In expanded form,

${\displaystyle {\text{det}}{\begin{bmatrix}\sigma _{11}-\lambda &\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\lambda &\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\lambda \end{bmatrix}}=0}$

Substituting the values of stress into the above equation,

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

Expanding out,

${\displaystyle -\lambda \left[(3-\lambda )(-\lambda )-(-5)(-5)\right]-(-8)\left[(-8)(-\lambda )-(-5)(0)\right]=0}$

or,

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

or,

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

or,

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

Thus, the first possible value of ${\displaystyle \lambda =0}$ MPa. Also,

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

Therefore,

${\displaystyle \lambda ={\frac {3\pm {\sqrt {9+356}}}{2}}=(11.05,-8.05){\text{MPa}}}$

The principal stresses are (in MPa)

${\displaystyle {\sigma _{1}=11.05~;~~\sigma _{2}=0~;~~\sigma _{3}=-8.05}}$

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

${\displaystyle ({\boldsymbol {\sigma }}-\lambda \mathbf {I} ){\widehat {\mathbf {n} }}{}=0}$

For the principal direction ${\displaystyle {\widehat {\mathbf {n} }}{2}}$ (corresponding to the principal stress ${\displaystyle \sigma _{2}}$, we have,

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

Hence,

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

gives us ${\displaystyle n^{2}=0}$ and

${\displaystyle 8~n^{1}=-5~n^{3}}$

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

${\displaystyle 8~n^{1}=-5~{\sqrt {1-(n^{1})^{2}}}}$

Taking squares of both sides,

${\displaystyle 64(n^{1})^{2}=25\left[1-(n^{1})^{2}\right]}$

So we get,

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

Therefore,

${\displaystyle n^{3}={\sqrt {1-(n^{1})^{2}}}=8/{\sqrt {89}}}$

The direction corresponding to the intermediate principal stress is

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

The balance principle is

${\displaystyle {\text{Conservation of angular momentum}}}$

In index notation

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