# Elasticity/Body force potential
## Body force potential

How do we find the body force potential? Before we proceed let us examine what conservative vector fields are.

### Conservative vector fields

* Work done in moving a particle from point A to point B in the field should be path independent.
* The local potential at point P in the field is defined as the work done to move a particle from infinity to P.
* For a vector field to be conservative
$f_{2,1}-f_{1,2}=0~;~~f_{3,2}-f_{2,3}=0~;~~f_{1,3}-f_{3,1}=0\qquad {\text{(28)}}$
or
${\boldsymbol {\nabla }}\times {\mathbf {f} }=0\qquad {\text{(29)}}$
The field has to be irrotational.

### Determining the body force potential

Suppose a body is rotating with an angular velocity ${\dot {\theta }}$ and an angular acceleration of ${\ddot {\theta }}$. Then,
    ${\text{(30)}}\qquad \mathbf {a} _{r}=-{\dot {\theta }}^{2}r{\widehat {\mathbf {e} }}_{r}~;~~\mathbf {a} _{\theta }=-{\ddot {\theta }}r{\widehat {\mathbf {e} }}_{\theta }$
Let us assume that the $(r,\theta )$ coordinate system is oriented at an angle $\theta$ to the $(x_{1},x_{2})$ system. Then,
    ${\text{(31)}}\qquad {\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}a_{r}\\a_{\theta }\end{bmatrix}}$
or,
    ${\text{(32)}}\qquad {\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}-{\dot {\theta }}^{2}r\\-{\ddot {\theta }}r\end{bmatrix}}$
or,
    ${\begin{aligned}{\text{(33)}}\qquad a_{1}&=-{\dot {\theta }}^{2}r\cos \theta +{\ddot {\theta }}r\sin \theta \\{\text{(34)}}\qquad a_{2}&=-{\dot {\theta }}^{2}r\sin \theta -{\ddot {\theta }}r\cos \theta \end{aligned}}$
or,
    ${\begin{aligned}{\text{(35)}}\qquad a_{1}&=-{\dot {\theta }}^{2}x_{1}+{\ddot {\theta }}x_{2}\\{\text{(36)}}\qquad a_{2}&=-{\dot {\theta }}^{2}x_{2}-{\ddot {\theta }}x_{1}\end{aligned}}$
If the origin is accelerating with an acceleration $\mathbf {a} _{0}$ (for example, due to gravity), we have,
    ${\begin{aligned}{\text{(37)}}\qquad a_{1}&=a_{01}-{\dot {\theta }}^{2}x_{1}+{\ddot {\theta }}x_{2}\\{\text{(38)}}\qquad a_{2}&=a_{02}-{\dot {\theta }}^{2}x_{2}-{\ddot {\theta }}x_{1}\end{aligned}}$:
The body force field is given by
    ${\begin{aligned}{\text{(39)}}\qquad f_{1}&=-\rho \left(a_{01}-{\dot {\theta }}^{2}x_{1}+{\ddot {\theta }}x_{2}\right)\\{\text{(40)}}\qquad f_{2}&=-\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}-{\ddot {\theta }}x_{1}\right)\end{aligned}}$
For this vector body force field to be conservative, we require that,
    $f_{1,2}-f_{2,1}=0\Rightarrow 2{\ddot {\theta }}=0$
Hence, the field $\mathbf {f}$ is conservative only if the rotational acceleration is zero, i.e. = the rotational velocity is constant.=
    ${\begin{aligned}{\text{(41)}}\qquad f_{1}&=-\rho \left(a_{01}-{\dot {\theta }}^{2}x_{1}\right)\\{\text{(42)}}\qquad f_{2}&=-\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)\end{aligned}}$
Now,
    $f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}$
Hence,
    ${\begin{aligned}{\text{(43)}}\qquad V_{,1}&=\rho \left(a_{01}-{\dot {\theta }}^{2}x_{1}\right)\\{\text{(44)}}\qquad V_{,2}&=\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)\end{aligned}}$
Integrating equation (43),
    ${\text{(45)}}\qquad V=\rho \left(a_{01}x_{1}-{\dot {\theta }}^{2}{\cfrac {x_{1}^{2}}{2}}\right)+h(x_{2})$
Hence,
    $({\text{46)}}\qquad V_{,2}=h^{'}(x_{2})=\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)$
Integrating,
    ${\text{(47)}}\qquad h(x_{2})=\rho \left(a_{02}x_{2}-{\dot {\theta }}^{2}{\cfrac {x_{2}^{2}}{2}}\right)+C$
Without loss of generality, we can set $C=0$. Then,
    ${\text{(48)}}\qquad V=\rho \left(a_{01}x_{1}-{\dot {\theta }}^{2}{\cfrac {x_{1}^{2}}{2}}\right)+\rho \left(a_{02}x_{2}-{\dot {\theta }}^{2}{\cfrac {x_{2}^{2}}{2}}\right)$
or,
    ${\text{(49)}}\qquad V=\rho \left[a_{01}x_{1}+a_{02}x_{2}-{\cfrac {{\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\right]$
For a body loaded by gravity only, we can set $a_{01}=0$, $a_{02}=-g$ and ${\dot {\theta }}=0$, to get
    ${\text{(50)}}\qquad V=-\rho gx_{2}\,$
For a body loaded by rotational inertia only, we can set $a_{01}=0$, and $a_{02}=0$, and get
    ${\text{(51)}}\qquad V=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)$
We can see that an Airy stress function + a body force potential of the form shown in equation (49) can be used to solve two-dimensional elasticity problems of plane stress/plane strain.