Airy stress function with body force potential
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If a body force exists, the Airy stress function (
φ
{\displaystyle \varphi }
V
{\displaystyle V}
σ
11
=
φ
,
22
+
V
;
σ
22
=
φ
,
11
+
V
;
σ
12
=
−
φ
,
12
(3)
{\displaystyle \sigma _{11}=\varphi _{,22}+V~;~~\sigma _{22}=\varphi _{,11}+V~;~~\sigma _{12}=-\varphi _{,12}\qquad {\text{(3)}}}
or,
σ
11
=
∂
2
φ
∂
x
2
2
+
V
;
σ
22
=
∂
2
φ
∂
x
1
2
+
V
;
σ
12
=
−
∂
2
φ
∂
x
1
∂
x
2
(4)
{\displaystyle \sigma _{11}={\cfrac {\partial ^{2}\varphi }{\partial x_{2}^{2}}}+V~;~~\sigma _{22}={\cfrac {\partial ^{2}\varphi }{\partial x_{1}^{2}}}+V~;~~\sigma _{12}=-{\cfrac {\partial ^{2}\varphi }{\partial x_{1}\partial x_{2}}}\qquad {\text{(4)}}}
Is equilibrium still satisfied ?
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Recall the equilibrium equation in two dimensions:
∇
∙
σ
+
f
=
0
;
σ
β
α
,
β
+
f
α
=
0
(5)
{\displaystyle {\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}+\mathbf {f} =0~;~~\sigma _{\beta \alpha ,\beta }+f_{\alpha }=0\qquad {\text{(5)}}}
or,
σ
11
,
1
+
σ
21
,
2
+
f
1
=
0
(6)
σ
12
,
1
+
σ
22
,
2
+
f
2
=
0
(7)
{\displaystyle {\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+f_{1}&=0\qquad {\text{(6)}}\\\sigma _{12,1}+\sigma _{22,2}+f_{2}&=0\qquad {\text{(7)}}\end{aligned}}}
In terms of
φ
{\displaystyle \varphi }
φ
,
122
+
V
,
1
−
φ
,
212
+
f
1
=
0
(8)
−
φ
,
112
+
φ
,
211
+
V
,
2
+
f
2
=
0
(9)
{\displaystyle {\begin{aligned}\varphi _{,122}+V_{,1}-\varphi _{,212}+f_{1}&=0\qquad {\text{(8)}}\\-\varphi _{,112}+\varphi _{,211}+V_{,2}+f_{2}&=0\qquad {\text{(9)}}\end{aligned}}}
or,
V
,
1
+
f
1
=
0
(10)
V
,
2
+
f
2
=
0
(11)
{\displaystyle {\begin{aligned}V_{,1}+f_{1}&=0\qquad {\text{(10)}}\\V_{,2}+f_{2}&=0\qquad {\text{(11)}}\end{aligned}}}
Therefore,
f
1
=
−
V
,
1
;
f
2
=
−
V
,
2
(12)
{\displaystyle f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}\qquad {\text{(12)}}}
or,
f
1
=
−
∂
V
∂
x
1
;
f
2
=
−
∂
V
∂
x
2
(13)
{\displaystyle f_{1}=-{\cfrac {\partial V}{\partial x_{1}}}~;~~f_{2}=-{\cfrac {\partial V}{\partial x_{2}}}\qquad {\text{(13)}}}
Hence,
f
=
−
∇
V
(14)
{\displaystyle \mathbf {f} =-{\boldsymbol {\nabla }}{V}\qquad {\text{(14)}}}
Equilibrium is satisfied only if the body force field can be expressed as the gradient of a scalar potential.
A force field that can be expressed as the gradient of a scalar potential is called conservative . What condition is needed to satisfy compatibility ?
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Recall that the compatibility condition in terms of the stresses can be written as
∇
2
σ
γ
γ
=
−
1
α
f
γ
,
γ
(15)
{\displaystyle \nabla ^{2}{\sigma _{\gamma \gamma }}=-{\cfrac {1}{\alpha }}f_{\gamma ,\gamma }\qquad {\text{(15)}}}
where,
α
=
{
1
−
ν
f
o
r
p
l
a
n
e
s
t
r
a
i
n
1
1
+
ν
f
o
r
p
l
a
n
e
s
t
r
e
s
s
(16)
{\displaystyle \alpha ={\begin{cases}1-\nu &{\rm {for~plane~strain}}\\{\cfrac {1}{1+\nu }}&{\rm {for~plane~stress}}\end{cases}}\qquad {\text{(16)}}}
or,
∇
2
(
σ
11
+
σ
22
)
+
1
α
(
f
1
,
1
+
f
2
,
2
)
=
0
(17)
{\displaystyle \nabla ^{2}{(\sigma _{11}+\sigma _{22})}+{\cfrac {1}{\alpha }}(f_{1,1}+f_{2,2})=0\qquad {\text{(17)}}}
or,
σ
11
,
11
+
σ
11
,
22
+
σ
22
,
11
+
σ
22
,
22
+
1
α
(
f
1
,
1
+
f
2
,
2
)
=
0
(18)
{\displaystyle \sigma _{11,11}+\sigma _{11,22}+\sigma _{22,11}+\sigma _{22,22}+{\cfrac {1}{\alpha }}(f_{1,1}+f_{2,2})=0\qquad {\text{(18)}}}
which is the same as,
∂
2
σ
11
∂
x
1
2
+
∂
2
σ
11
∂
x
2
2
+
∂
2
σ
22
∂
x
1
2
+
∂
2
σ
22
∂
x
2
2
+
1
α
(
∂
f
1
∂
x
1
+
∂
f
2
∂
x
2
)
=
0
(19)
{\displaystyle {\cfrac {\partial ^{2}\sigma _{11}}{\partial x_{1}^{2}}}+{\cfrac {\partial ^{2}\sigma _{11}}{\partial x_{2}^{2}}}+{\cfrac {\partial ^{2}\sigma _{22}}{\partial x_{1}^{2}}}+{\cfrac {\partial ^{2}\sigma _{22}}{\partial x_{2}^{2}}}+{\cfrac {1}{\alpha }}({\cfrac {\partial f_{1}}{\partial x_{1}}}+{\cfrac {\partial f_{2}}{\partial x_{2}}})=0\qquad {\text{(19)}}}
Plug in the stress potential and the body force potential in equation (18) to get
φ
,
2211
+
V
,
11
+
φ
,
2222
+
V
,
22
+
φ
,
1111
+
V
,
11
+
φ
,
1122
+
V
,
22
+
1
α
(
−
V
,
11
−
V
,
22
)
=
0
(20)
{\displaystyle \varphi _{,2211}+V_{,11}+\varphi _{,2222}+V_{,22}+\varphi _{,1111}+V_{,11}+\varphi _{,1122}+V_{,22}+{\cfrac {1}{\alpha }}(-V_{,11}-V_{,22})=0\qquad {\text{(20)}}}
or,
2
φ
,
2211
+
2
V
,
11
+
φ
,
2222
+
2
V
,
22
+
φ
,
1111
+
1
α
(
−
V
,
11
−
V
,
22
)
=
0
(21)
{\displaystyle 2\varphi _{,2211}+2V_{,11}+\varphi _{,2222}+2V_{,22}+\varphi _{,1111}+{\cfrac {1}{\alpha }}(-V_{,11}-V_{,22})=0\qquad {\text{(21)}}}
Rearrange to get
φ
,
1111
+
2
φ
,
2211
+
φ
,
2222
+
(
2
−
1
α
)
(
V
,
11
+
V
,
22
)
=
0
(22)
{\displaystyle \varphi _{,1111}+2\varphi _{,2211}+\varphi _{,2222}+\left(2-{\cfrac {1}{\alpha }}\right)(V_{,11}+V_{,22})=0\qquad {\text{(22)}}}
which is the same as,
∂
4
φ
∂
x
1
4
+
2
∂
4
φ
∂
x
1
2
∂
x
2
2
+
∂
4
φ
∂
x
2
4
+
(
2
−
1
α
)
(
∂
2
V
∂
x
1
2
+
∂
2
V
∂
x
2
2
)
=
0
(23)
{\displaystyle {\cfrac {\partial ^{4}\varphi }{\partial x_{1}^{4}}}+2{\cfrac {\partial ^{4}\varphi }{\partial x_{1}^{2}\partial x_{2}^{2}}}+{\cfrac {\partial ^{4}\varphi }{\partial x_{2}^{4}}}+\left(2-{\cfrac {1}{\alpha }}\right)\left({\cfrac {\partial ^{2}V}{\partial x_{1}^{2}}}+{\cfrac {\partial ^{2}V}{\partial x_{2}^{2}}}\right)=0\qquad {\text{(23)}}}
Therefore,
∇
4
φ
+
(
2
−
1
α
)
∇
2
V
=
0
(24)
{\displaystyle \nabla ^{4}{\varphi }+\left(2-{\cfrac {1}{\alpha }}\right)\nabla ^{2}{V}=0\qquad {\text{(24)}}}
Compatibility is satisfied only if equation (24) is satisfied.
Equations for Airy stress function with body force potential
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The relation between the Cauchy stress and the Airy stress function is (in direct tensor notation)
σ
=
∇
×
∇
×
φ
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\nabla }}\times {{\boldsymbol {\nabla }}\times {\varphi }}}
The relation between the body force and the body force potential is
f
=
−
∇
V
{\displaystyle \mathbf {f} =-{\boldsymbol {\nabla }}{V}}
We also have to satisfy the compatibility condition for the Airy stress function to be a true stress potential, i.e.,
∇
4
φ
+
(
2
−
1
α
)
∇
2
V
=
0
{\displaystyle \nabla ^{4}{\varphi }+\left(2-{\cfrac {1}{\alpha }}\right)\nabla ^{2}{V}=0}
In rectangular Cartesian coordinates
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In rectangular Cartesian coordinates, the relation between the Cauchy stress components and the Airy stress function + body force potential can be written as
σ
11
=
φ
,
22
+
V
;
σ
22
=
φ
,
11
+
V
;
σ
12
=
−
φ
,
12
{\displaystyle \sigma _{11}=\varphi _{,22}+V~;~~\sigma _{22}=\varphi _{,11}+V~;~~\sigma _{12}=-\varphi _{,12}}
or,
σ
x
x
=
∂
2
φ
∂
y
2
+
V
;
σ
y
y
=
∂
2
φ
∂
x
2
+
V
;
σ
x
y
=
−
∂
2
φ
∂
x
∂
y
{\displaystyle \sigma _{xx}={\cfrac {\partial ^{2}\varphi }{\partial y^{2}}}+V~;~~\sigma _{yy}={\cfrac {\partial ^{2}\varphi }{\partial x^{2}}}+V~;~~\sigma _{xy}=-{\cfrac {\partial ^{2}\varphi }{\partial x\partial y}}}
The relation between the body force components and the body force potential are:
f
1
=
−
V
,
1
;
f
2
=
−
V
,
2
{\displaystyle f_{1}=-V_{,1}~;~~f_{2}=-V_{,2}}
or,
f
x
=
−
∂
V
∂
x
;
f
y
=
−
∂
V
∂
y
{\displaystyle f_{x}=-{\cfrac {\partial V}{\partial x}}~;~~f_{y}=-{\cfrac {\partial V}{\partial y}}}
The compatibility condition is written as
φ
,
1111
+
2
φ
,
2211
+
φ
,
2222
+
(
2
−
1
α
)
(
V
,
11
+
V
,
22
)
=
0
{\displaystyle \varphi _{,1111}+2\varphi _{,2211}+\varphi _{,2222}+\left(2-{\cfrac {1}{\alpha }}\right)(V_{,11}+V_{,22})=0}
or,
∂
4
φ
∂
x
4
+
2
∂
4
φ
∂
x
2
∂
y
2
+
∂
4
φ
∂
y
4
+
(
2
−
1
α
)
(
∂
2
V
∂
x
2
+
∂
2
V
∂
y
2
)
=
0
{\displaystyle {\cfrac {\partial ^{4}\varphi }{\partial x^{4}}}+2{\cfrac {\partial ^{4}\varphi }{\partial x^{2}\partial y^{2}}}+{\cfrac {\partial ^{4}\varphi }{\partial y^{4}}}+\left(2-{\cfrac {1}{\alpha }}\right)\left({\cfrac {\partial ^{2}V}{\partial x^{2}}}+{\cfrac {\partial ^{2}V}{\partial y^{2}}}\right)=0}
In cylindrical coordinates
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In cylindrical coordinates, the relation between the Cauchy stresses and the Airy stress function + body force potential can be written as
σ
r
r
=
1
r
∂
φ
∂
r
+
1
r
2
∂
2
φ
∂
θ
2
+
V
;
σ
θ
θ
=
∂
2
φ
∂
r
2
+
V
;
σ
r
θ
=
−
∂
∂
r
(
1
r
∂
φ
∂
θ
)
(25)
{\displaystyle \sigma _{rr}={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}+V~;~~\sigma _{\theta \theta }={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}+V~;~~\sigma _{r\theta }=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)\qquad {\text{(25)}}}
The components of the body force are related to the body force potential via
f
r
=
−
∂
V
∂
r
;
f
θ
=
−
1
r
∂
V
∂
θ
(26)
{\displaystyle f_{r}=-{\cfrac {\partial V}{\partial r}}~;~~f_{\theta }=-{\cfrac {1}{r}}{\cfrac {\partial V}{\partial \theta }}\qquad {\text{(26)}}}
The compatibility condition can be expressed as
(
∂
2
∂
r
2
+
1
r
∂
∂
r
+
1
r
2
∂
2
∂
θ
2
)
(
∂
2
φ
∂
r
2
+
1
r
∂
φ
∂
r
+
1
r
2
∂
2
φ
∂
θ
2
)
+
(
2
−
1
α
)
(
∂
2
V
∂
r
2
+
1
r
∂
V
∂
r
+
1
r
2
∂
2
V
∂
θ
2
)
=
0
(27)
{\displaystyle {\begin{aligned}\left({\cfrac {\partial ^{2}}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}}{\partial \theta ^{2}}}\right)&\left({\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}\right)+\\&\left(2-{\cfrac {1}{\alpha }}\right)\left({\cfrac {\partial ^{2}V}{\partial r^{2}}}+{\cfrac {1}{r}}{\cfrac {\partial V}{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}V}{\partial \theta ^{2}}}\right)=0\qquad {\text{(27)}}\end{aligned}}}
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