How do we find the body force potential? Before we proceed let us examine what conservative vector fields are.
- 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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b6918dea8d46a6f1a751e3c41c22dc853ae5ca5)
or
![{\displaystyle {\boldsymbol {\nabla }}\times {\mathbf {f} }=0\qquad {\text{(29)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc0ee00ef58f9ed72636e372a3b3bf327f61d56e)
The field has to be irrotational.
Suppose a body is rotating with an angular velocity
and an angular acceleration of
. 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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30e73ec8403666b0e311aade26a8b4e36983d467)
Let us assume that the
coordinate system is oriented at an angle
to the
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdb66ffae5c3722cb6d7ee81425798aab685eb7e)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c45934d477d687fbae726cfedfb514c7bc65d7b)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faecf0296f0b21563c0d2219dd1af07fe651b3af)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15257b82bc5e70fb2d014200d2e28a26c649a2d7)
If the origin is accelerating with an acceleration
(for example, due to gravity), we have,
:
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71534405424596172a4d4674a438be2bf5d9e5cc)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b3d8682347671d225779a27aea9d54b2b24b0c7)
Hence, the field
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5deb5216d8a55ae70361b5dfee94cbbe5d26190e)
Now,
![{\displaystyle f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf61f1622f65dac1f4269a5601258981e3ff65f6)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f7785f75ccaa1846df15199622fa1ebb549a121)
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})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/740521064e20ac570a0c441f084dc3aa55330898)
Hence,
![{\displaystyle ({\text{46)}}\qquad V_{,2}=h^{'}(x_{2})=\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6335690bc70ee0297fa5c5605b729dce2847de4)
Integrating,
![{\displaystyle {\text{(47)}}\qquad h(x_{2})=\rho \left(a_{02}x_{2}-{\dot {\theta }}^{2}{\cfrac {x_{2}^{2}}{2}}\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb0f6de78b1c38ce059dd40b13e56132bd5276de)
Without loss of generality, we can set
. 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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00bce052b4366143ae949cb434ea3dfae9f1b6da)
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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e9eb2b055a506dc329164341c75c0167fd5c9c1)
For a body loaded by gravity only, we can set
,
and
, to get
![{\displaystyle {\text{(50)}}\qquad V=-\rho gx_{2}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbf453506616929874ae103bb5f61260e7041fe2)
For a body loaded by rotational inertia only, we can set
, and
, and get
![{\displaystyle {\text{(51)}}\qquad V=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/734f32eafc355376b33cc9f7f052d69ed9aed8a6)
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.