# Introduction to 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
${\displaystyle f_{2,1}-f_{1,2}=0~;~~f_{3,2}-f_{2,3}=0~;~~f_{1,3}-f_{3,1}=0\qquad {\text{(28)}}}$

or

${\displaystyle {\boldsymbol {\nabla }}\times {\mathbf {f} }=0\qquad {\text{(29)}}}$

The field has to be irrotational.

### Determining the body force potential

Suppose a body is rotating with an angular velocity ${\displaystyle {\dot {\theta }}}$ and an angular acceleration of ${\displaystyle {\ddot {\theta }}}$. Then,

${\displaystyle {\text{(30)}}\qquad \mathbf {a} _{r}=-{\dot {\theta }}^{2}r{\widehat {\mathbf {e} }}_{r}~;~~\mathbf {a} _{\theta }=-{\ddot {\theta }}r{\widehat {\mathbf {e} }}_{\theta }}$

Let us assume that the ${\displaystyle (r,\theta )}$ coordinate system is oriented at an angle ${\displaystyle \theta }$ to the ${\displaystyle (x_{1},x_{2})}$ system. Then,

${\displaystyle {\text{(31)}}\qquad {\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}a_{r}\\a_{\theta }\end{bmatrix}}}$

or,

${\displaystyle {\text{(32)}}\qquad {\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}-{\dot {\theta }}^{2}r\\-{\ddot {\theta }}r\end{bmatrix}}}$

or,

{\displaystyle {\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,

{\displaystyle {\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 ${\displaystyle \mathbf {a} _{0}}$ (for example, due to gravity), we have,

{\displaystyle {\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

{\displaystyle {\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,

${\displaystyle f_{1,2}-f_{2,1}=0\Rightarrow 2{\ddot {\theta }}=0}$

Hence, the field ${\displaystyle \mathbf {f} }$ is conservative only if the rotational acceleration is zero, i.e. = the rotational velocity is constant.=

{\displaystyle {\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,

${\displaystyle f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}}$

Hence,

{\displaystyle {\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),

${\displaystyle {\text{(45)}}\qquad V=\rho \left(a_{01}x_{1}-{\dot {\theta }}^{2}{\cfrac {x_{1}^{2}}{2}}\right)+h(x_{2})}$

Hence,

${\displaystyle ({\text{46)}}\qquad V_{,2}=h^{'}(x_{2})=\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)}$

Integrating,

${\displaystyle {\text{(47)}}\qquad h(x_{2})=\rho \left(a_{02}x_{2}-{\dot {\theta }}^{2}{\cfrac {x_{2}^{2}}{2}}\right)+C}$

Without loss of generality, we can set ${\displaystyle C=0}$. Then,

${\displaystyle {\text{(48)}}\qquad V=\rho \left(a_{01}x_{1}-{\dot {\theta }}^{2}{\cfrac {x_{1}^{2}}{2}}\right)+\rho \left(a_{02}x_{2}-{\dot {\theta }}^{2}{\cfrac {x_{2}^{2}}{2}}\right)}$

or,

${\displaystyle {\text{(49)}}\qquad V=\rho \left[a_{01}x_{1}+a_{02}x_{2}-{\cfrac {{\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\right]}$

For a body loaded by gravity only, we can set ${\displaystyle a_{01}=0}$, ${\displaystyle a_{02}=-g}$ and ${\displaystyle {\dot {\theta }}=0}$, to get

${\displaystyle {\text{(50)}}\qquad V=-\rho gx_{2}\,}$

For a body loaded by rotational inertia only, we can set ${\displaystyle a_{01}=0}$, and ${\displaystyle a_{02}=0}$, and get

${\displaystyle {\text{(51)}}\qquad V=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)}$

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.