How do we find the body force potential? Before we proceed let us examine what conservative vector fields are.
Conservative vector fields
edit
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
edit
Suppose a body is rotating with an angular velocity
θ
˙
{\displaystyle {\dot {\theta }}}
θ
¨
{\displaystyle {\ddot {\theta }}}
(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 )}
θ
{\displaystyle \theta }
(
x
1
,
x
2
)
{\displaystyle (x_{1},x_{2})}
(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}}
(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} }
(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}
(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}
θ
˙
=
0
{\displaystyle {\dot {\theta }}=0}
(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}
a
02
=
0
{\displaystyle a_{02}=0}
(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.
![{\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)
