How do we find the body force potential? Before we proceed let us examine what conservative vector fields are.
Conservative vector fields
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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 (28) {\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
∇ × f = 0 (29) {\displaystyle {\boldsymbol {\nabla }}\times {\mathbf {f} }=0\qquad {\text{(29)}}}
The field has to be irrotational .
Determining the body force potential
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Suppose a body is rotating with an angular velocity θ ˙ {\displaystyle {\dot {\theta }}} and an angular acceleration of θ ¨ {\displaystyle {\ddot {\theta }}} . Then,
(30) a r = − θ ˙ 2 r e ^ r ; a θ = − θ ¨ r e ^ θ {\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 ( r , θ ) {\displaystyle (r,\theta )} coordinate system is oriented at an angle θ {\displaystyle \theta } to the ( x 1 , x 2 ) {\displaystyle (x_{1},x_{2})} system. Then,
(31) [ a 1 a 2 ] = [ cos θ − sin θ sin θ cos θ ] [ a r a θ ] {\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,
(32) [ a 1 a 2 ] = [ cos θ − sin θ sin θ cos θ ] [ − θ ˙ 2 r − θ ¨ r ] {\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,
(33) a 1 = − θ ˙ 2 r cos θ + θ ¨ r sin θ (34) a 2 = − θ ˙ 2 r sin θ − θ ¨ r cos θ {\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,
(35) a 1 = − θ ˙ 2 x 1 + θ ¨ x 2 (36) a 2 = − θ ˙ 2 x 2 − θ ¨ x 1 {\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 a 0 {\displaystyle \mathbf {a} _{0}} (for example, due to gravity), we have,
(37) a 1 = a 01 − θ ˙ 2 x 1 + θ ¨ x 2 (38) a 2 = a 02 − θ ˙ 2 x 2 − θ ¨ x 1 {\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
(39) f 1 = − ρ ( a 01 − θ ˙ 2 x 1 + θ ¨ x 2 ) (40) f 2 = − ρ ( a 02 − θ ˙ 2 x 2 − θ ¨ x 1 ) {\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,
f 1 , 2 − f 2 , 1 = 0 ⇒ 2 θ ¨ = 0 {\displaystyle f_{1,2}-f_{2,1}=0\Rightarrow 2{\ddot {\theta }}=0} Hence, the field f {\displaystyle \mathbf {f} } is conservative only if the rotational acceleration is zero, i.e. = the rotational velocity is constant.=
(41) f 1 = − ρ ( a 01 − θ ˙ 2 x 1 ) (42) f 2 = − ρ ( a 02 − θ ˙ 2 x 2 ) {\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,
f 1 = − V , 1 ; f 2 = − V , 2 {\displaystyle f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}} Hence,
(43) V , 1 = ρ ( a 01 − θ ˙ 2 x 1 ) (44) V , 2 = ρ ( a 02 − θ ˙ 2 x 2 ) {\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),
(45) V = ρ ( a 01 x 1 − θ ˙ 2 x 1 2 2 ) + h ( x 2 ) {\displaystyle {\text{(45)}}\qquad V=\rho \left(a_{01}x_{1}-{\dot {\theta }}^{2}{\cfrac {x_{1}^{2}}{2}}\right)+h(x_{2})} Hence,
( 46) V , 2 = h ′ ( x 2 ) = ρ ( a 02 − θ ˙ 2 x 2 ) {\displaystyle ({\text{46)}}\qquad V_{,2}=h^{'}(x_{2})=\rho \left(a_{02}-{\dot {\theta }}^{2}x_{2}\right)} Integrating,
(47) h ( x 2 ) = ρ ( a 02 x 2 − θ ˙ 2 x 2 2 2 ) + C {\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 C = 0 {\displaystyle C=0} . Then,
(48) V = ρ ( a 01 x 1 − θ ˙ 2 x 1 2 2 ) + ρ ( a 02 x 2 − θ ˙ 2 x 2 2 2 ) {\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,
(49) V = ρ [ a 01 x 1 + a 02 x 2 − θ ˙ 2 2 ( x 1 2 + x 2 2 ) ] {\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 a 01 = 0 {\displaystyle a_{01}=0} , a 02 = − g {\displaystyle a_{02}=-g} and θ ˙ = 0 {\displaystyle {\dot {\theta }}=0} , to get
(50) V = − ρ g x 2 {\displaystyle {\text{(50)}}\qquad V=-\rho gx_{2}\,} For a body loaded by rotational inertia only, we can set a 01 = 0 {\displaystyle a_{01}=0} , and a 02 = 0 {\displaystyle a_{02}=0} , and get
(51) V = − ρ θ ˙ 2 2 ( x 1 2 + x 2 2 ) {\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.