# Introduction to Elasticity/Equilibrium example 1

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## Example 1

Given:

Euler's second law for the conservation of angular momentum

${\displaystyle {\text{(1)}}\qquad \int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)}$

The divergence theorem

${\displaystyle {\text{(2)}}\qquad \int _{\partial B}n_{i}~\sigma _{ij}~dS=\int _{B}{\frac {\partial \sigma _{ij}}{\partial x_{i}}}~dV}$

The equilibrium equation (Cauchy's first law)

${\displaystyle {\text{(3)}}\qquad {\frac {\partial \sigma _{ij}}{\partial x_{i}}}+\rho ~b_{j}={\frac {d}{dt}}\left(\rho ~v_{j}\right)}$

Show:

${\displaystyle {\text{(4)}}\qquad \sigma _{ij}=\sigma _{ji}}$

### Solution

Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,

{\displaystyle {\begin{aligned}\int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS&=\int _{\partial B}n_{l}~(e_{ijk}~x_{j}~\sigma _{lk})~dS\\&=\int _{B}{\frac {\partial {(e_{ijk}~x_{j}~\sigma _{lk})}}{\partial x_{l}}}~dV\\&=\int _{B}\left(e_{ijk}~{\frac {\partial x_{j}}{\partial x_{l}}}~\sigma _{lk}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ijk}~\delta _{jl}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\end{aligned}}}

Plugging this back into equation~(1) gives

${\displaystyle \int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)}$

Therefore, bringing all terms to the left hand side,

${\displaystyle {\text{(5)}}\qquad \int _{B}\left[e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}\left({\frac {\partial \sigma _{lk}}{\partial x_{l}}}+\rho ~b_{k}-{\frac {d}{dt}}\left(\rho ~v_{k}\right)\right)\right]~dV=0}$

Using the equilibrium equations~(3), equation~(5) reduces to

${\displaystyle {\text{(6)}}\qquad \int _{B}e_{ilk}~\sigma _{lk}~dV=0}$

Since this holds for any ${\displaystyle B}$, we have

${\displaystyle {\text{(7)}}\qquad e_{ilk}~\sigma _{lk}=0}$

If you work this expression out, you will see that ${\displaystyle \sigma _{ij}=\sigma _{ji}}$. Hence, the stress tensor is symmetric.