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

Template:Center top$f_{2,1}-f_{1,2}=0;f_{3,2}-f_{2,3}=0;f_{1,3}-f_{3,1}=0\text{(28)}$Template:Center bottom or Template:Center top$\mathrm{\nabla }\times 𝐟=0\text{(29)}$Template:Center bottom 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,

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

Let us assume that the $(r,\theta )$ coordinate system is oriented at an angle $\theta$ to the $(x_1,x_2)$ system. Then,

$\text{(31)}\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}{a}_r\\ a_{\theta }\end{array}\right]$

or,

$\text{(32)}\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}-{\dot{\theta }}^{2}r\\ -\ddot{\theta }r\end{array}\right]$

or,

$\begin{array}{cc}\text{(33)}{a}_1& =-{\dot{\theta }}^{2}r\cos \theta +\ddot{\theta }r\sin \theta \\ \text{(34)}{a}_2& =-{\dot{\theta }}^{2}r\sin \theta -\ddot{\theta }r\cos \theta \end{array}$

or,

$\begin{array}{cc}\text{(35)}{a}_1& =-{\dot{\theta }}^2{x}_1+\ddot{\theta }{x}_2\\ \text{(36)}{a}_2& =-{\dot{\theta }}^2{x}_2-\ddot{\theta }{x}_1\end{array}$

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

$\begin{array}{cc}\text{(37)}{a}_1& =a_{01}-{\dot{\theta }}^2{x}_1+\ddot{\theta }{x}_2\\ \text{(38)}{a}_2& =a_{02}-{\dot{\theta }}^2{x}_2-\ddot{\theta }{x}_1\end{array}$:

The body force field is given by

$\begin{array}{cc}\text{(39)}{f}_1& =-\rho (a_{01}-{\dot{\theta }}^2{x}_1+\ddot{\theta }{x}_2)\\ \text{(40)}{f}_2& =-\rho (a_{02}-{\dot{\theta }}^2{x}_2-\ddot{\theta }{x}_1)\end{array}$

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 $𝐟$ is conservative only if the rotational acceleration is zero, i.e. = the rotational velocity is constant.=

$\begin{array}{cc}\text{(41)}{f}_1& =-\rho (a_{01}-{\dot{\theta }}^2{x}_1)\\ \text{(42)}{f}_2& =-\rho (a_{02}-{\dot{\theta }}^2{x}_2)\end{array}$

Now,

$f_1=-V_{,1};f_2=-V_{,2}$

Hence,

$\begin{array}{cc}\text{(43)}{V}_{,1}& =\rho (a_{01}-{\dot{\theta }}^2{x}_1)\\ \text{(44)}{V}_{,2}& =\rho (a_{02}-{\dot{\theta }}^2{x}_2)\end{array}$

Integrating equation (43),

$\text{(45)}V=\rho (a_{01}{x}_1-{\dot{\theta }}^2\frac{x_1^2}{2})+h(x_2)$

Hence,

$(\text{46)}{V}_{,2}=h^{\text{'}}(x_2)=\rho (a_{02}-{\dot{\theta }}^2{x}_2)$

Integrating,

$\text{(47)}h(x_2)=\rho (a_{02}{x}_2-{\dot{\theta }}^2\frac{x_2^2}{2})+C$

Without loss of generality, we can set $C=0$. Then,

$\text{(48)}V=\rho (a_{01}{x}_1-{\dot{\theta }}^2\frac{x_1^2}{2})+\rho (a_{02}{x}_2-{\dot{\theta }}^2\frac{x_2^2}{2})$

or,

$\text{(49)}V=\rho [a_{01}{x}_1+a_{02}{x}_2-\frac{{\dot{\theta }}^2}{2}(x_1^2+x_2^2)]$

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

$\text{(50)}V=-\rho g{x}_2$

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

$\text{(51)}V=-\frac{\rho {\dot{\theta }}^2}{2}(x_1^2+x_2^2)$

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.