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
$$ f_{2,1} - f_{1,2} = 0 ~;           f_{3,2} - f_{2,3} = 0 ~;           f_{1,3} - f_{3,1} = 0  \qquad \text{(28)} $$       or        $$ \boldsymbol{\nabla}\times{\mathbf{f}} = 0 \qquad \text{(29)}      $$ The field has to be irrotational.
 * 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

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}^2r\\-\ddot{\theta}r\end{bmatrix} $$ or,
 * $$\begin{align}

\text{(33)} \qquad a_1 & = -\dot{\theta}^2r\cos\theta + \ddot{\theta}r\sin\theta\\ \text{(34)} \qquad a_2 & = -\dot{\theta}^2r\sin\theta - \ddot{\theta}r\cos\theta \end{align}$$ or,
 * $$\begin{align}

\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{align}$$ If the origin is accelerating with an acceleration $$\mathbf{a}_0$$ (for example, due to gravity), we have,
 * $$\begin{align}

\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{align}$$: The body force field is given by
 * $$\begin{align}

\text{(39)} \qquad f_1 & = -\rho\left(a_{01}-\dot{\theta}^2x_1+\ddot{\theta}x_2\right) \\  \text{(40)} \qquad f_2 & = -\rho\left(a_{02}-\dot{\theta}^2x_2-\ddot{\theta}x_1\right) \end{align}$$ 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{align}

\text{(41)} \qquad f_1 & = -\rho\left(a_{01}-\dot{\theta}^2x_1\right)\\ \text{(42)} \qquad f_2 & = -\rho\left(a_{02}-\dot{\theta}^2x_2\right) \end{align}$$ Now,

f_1 = - V_{,1} ~; f_2 = - V_{,2} $$ Hence,
 * $$\begin{align}

\text{(43)} \qquad V_{,1} & = \rho\left(a_{01}-\dot{\theta}^2x_1\right)\\ \text{(44)} \qquad V_{,2} & = \rho\left(a_{02}-\dot{\theta}^2x_2\right) \end{align}$$ Integrating equation (43),
 * $$\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}^2x_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 g x_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.