Given:
If a material is incompressible (
ν
{\displaystyle \nu }
σ
11
=
σ
22
=
σ
33
{\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}}
σ
i
j
=
2
μ
ε
i
j
−
p
δ
i
j
{\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}-p\delta _{ij}}
where
p
{\displaystyle p}
e
=
ε
k
k
=
0
.
{\displaystyle e=\varepsilon _{kk}=0~.}
Show:
Show that the stress components and the hydrostatic pressure
p
{\displaystyle p}
∇
2
p
=
∇
∙
b
;
σ
11
+
σ
22
=
−
2
p
{\displaystyle \nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }~;~~\sigma _{11}+\sigma _{22}=-2p}
where
b
{\displaystyle \mathbf {b} }
We have,
e
=
ε
k
k
=
ε
11
+
ε
22
+
ε
33
=
0
.
{\displaystyle e=\varepsilon _{kk}=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0~.\,}
σ
11
=
2
μ
ε
11
−
p
;
σ
22
=
2
μ
ε
22
−
p
;
σ
33
=
2
μ
ε
33
−
p
.
{\displaystyle \sigma _{11}=2\mu \varepsilon _{11}-p~;~~\sigma _{22}=2\mu \varepsilon _{22}-p~;~~\sigma _{33}=2\mu \varepsilon _{33}-p~.\,}
Therefore,
σ
11
+
σ
22
+
σ
33
=
2
μ
(
ε
11
+
ε
22
+
ε
33
)
−
3
p
=
−
3
p
{\displaystyle {\begin{aligned}\sigma _{11}+\sigma _{22}+\sigma _{33}&=2\mu \left(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\right)-3p\\&=-3p\end{aligned}}}
Since
σ
11
=
σ
22
=
σ
33
{\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}\,}
σ
11
=
σ
22
=
σ
33
=
−
p
{\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}=-p\,}
σ
11
+
σ
22
=
−
2
p
{\displaystyle \sigma _{11}+\sigma _{22}=-2p\,}
The strain-stress relations are
2
μ
ε
11
=
σ
11
+
p
;
2
μ
ε
22
=
σ
22
+
p
;
2
μ
ε
12
=
σ
12
.
{\displaystyle 2\mu \varepsilon _{11}=\sigma _{11}+p~;~~2\mu \varepsilon _{22}=\sigma _{22}+p~;~~2\mu \varepsilon _{12}=\sigma _{12}~.}
Differentiating the strains so that they correspond to the compatibilityrelation is two-dimensions, we have
ε
11
,
22
=
1
2
μ
(
σ
11
,
22
+
p
,
22
)
;
ε
22
,
11
=
1
2
μ
(
σ
22
,
11
+
p
,
11
)
;
ε
12
,
12
=
1
2
μ
(
σ
12
,
12
)
.
{\displaystyle \varepsilon _{11,22}={\frac {1}{2\mu }}\left(\sigma _{11,22}+p_{,22}\right)~;~~\varepsilon _{22,11}={\frac {1}{2\mu }}\left(\sigma _{22,11}+p_{,11}\right)~;~~\varepsilon _{12,12}={\frac {1}{2\mu }}\left(\sigma _{12,12}\right)~.}
In terms of the compatibility equation,
ε
11
,
22
+
ε
22
,
11
−
2
ε
12
,
12
=
1
2
μ
(
σ
11
,
22
+
σ
22
,
11
−
2
σ
12
,
12
+
p
,
11
+
p
,
22
)
or,
0
=
σ
11
,
22
+
σ
22
,
11
−
2
σ
12
,
12
+
∇
2
p
{\displaystyle {\begin{aligned}&\varepsilon _{11,22}+\varepsilon _{22,11}-2\varepsilon _{12,12}={\frac {1}{2\mu }}\left(\sigma _{11,22}+\sigma _{22,11}-2\sigma _{12,12}+p_{,11}+p_{,22}\right)\\{\text{or,}}~&0=\sigma _{11,22}+\sigma _{22,11}-2\sigma _{12,12}+\nabla ^{2}{p}\end{aligned}}}
From the two-dimensional equilibrium equations,
σ
11
,
1
+
σ
12
,
2
+
b
1
=
0
;
σ
12
,
1
+
σ
22
,
2
+
b
2
=
0
{\displaystyle \sigma _{11,1}+\sigma _{12,2}+b_{1}=0~;~~\sigma _{12,1}+\sigma _{22,2}+b_{2}=0}
Therefore, differentiating w.r.t
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
σ
11
,
11
+
σ
12
,
21
+
b
1
,
1
=
0
;
σ
12
,
12
+
σ
22
,
22
+
b
2
,
2
=
0
{\displaystyle \sigma _{11,11}+\sigma _{12,21}+b_{1,1}=0~;~~\sigma _{12,12}+\sigma _{22,22}+b_{2,2}=0}
Adding,
2
σ
12
,
12
+
σ
11
,
11
+
σ
22
,
22
+
b
1
,
1
+
b
2
,
2
=
0
{\displaystyle 2\sigma _{12,12}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=0}
Hence,
σ
11
,
11
+
σ
22
,
22
+
b
1
,
1
+
b
2
,
2
=
−
2
σ
12
,
12
{\displaystyle \sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=-2\sigma _{12,12}}
Substituting back into the compatibility equation,
σ
11
,
22
+
σ
22
,
11
+
σ
11
,
11
+
σ
22
,
22
+
b
1
,
1
+
b
2
,
2
+
∇
2
p
=
0
or,
∇
2
σ
11
+
∇
2
σ
22
+
∇
2
p
+
∇
∙
b
=
0
or,
∇
2
(
σ
11
+
σ
22
+
p
)
+
∇
∙
b
=
0
or,
∇
2
(
−
2
p
+
p
)
+
∇
∙
b
=
0
or,
−
∇
2
p
+
∇
∙
b
=
0
{\displaystyle {\begin{aligned}&\sigma _{11,22}+\sigma _{22,11}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}+\nabla ^{2}{p}=0\\{\text{or,}}~&\nabla ^{2}{\sigma _{11}}+\nabla ^{2}{\sigma _{22}}+\nabla ^{2}{p}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&\nabla ^{2}{(\sigma _{11}+\sigma _{22}+p)}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&\nabla ^{2}{(-2p+p)}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&-\nabla ^{2}{p}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\end{aligned}}}
Hence,
∇
2
p
=
∇
∙
b
{\displaystyle {\nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }}}
