Given:
If a material is incompressible (
= 0.5), a state of hydrostatic stress (
) produces no strain. The corresponding stress-strain relation can be written as
![{\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}-p\delta _{ij}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8691e191c82f180e1015b177dd3baf08c3c8568)
where
is an unknown hydrostatic pressure which will generally vary with position. Also, the condition of incompressibility requires that the dilatation
![{\displaystyle e=\varepsilon _{kk}=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba2824d7198789696f7cf87155ccbd6b06b41590)
Show:
Show that the stress components and the hydrostatic pressure
must satisfy the equations
![{\displaystyle \nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }~;~~\sigma _{11}+\sigma _{22}=-2p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f1c136c9398cca41dab6b946ef4dec3e3f67870)
where
is the body force.
We have,
Also,
![{\displaystyle \sigma _{11}=2\mu \varepsilon _{11}-p~;~~\sigma _{22}=2\mu \varepsilon _{22}-p~;~~\sigma _{33}=2\mu \varepsilon _{33}-p~.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf86dde56d61495519fa42b6c7ca1941abbb130d)
Therefore,
![{\displaystyle {\begin{aligned}\sigma _{11}+\sigma _{22}+\sigma _{33}&=2\mu \left(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\right)-3p\\&=-3p\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66d9beb6b2c6ec3925a3cf181f589393b6d8f472)
Since
, the above relation
gives
.
Therefore,
![{\displaystyle \sigma _{11}+\sigma _{22}=-2p\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a74dbdc8205e9d41cea20ff7ad496ce4cf51762b)
The strain-stress relations are
![{\displaystyle 2\mu \varepsilon _{11}=\sigma _{11}+p~;~~2\mu \varepsilon _{22}=\sigma _{22}+p~;~~2\mu \varepsilon _{12}=\sigma _{12}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da965d00cfd72cc6843a61b4dd9df14ab27e6e22)
Differentiating the strains so that they correspond to the compatibilityrelation is two-dimensions, we have
![{\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)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc439c566bfcb76b6dc5fe6ba16f7fd52bd90d34)
In terms of the compatibility equation,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1254d9838c1636763a58f632416f34ab3a71e4cc)
From the two-dimensional equilibrium equations,
![{\displaystyle \sigma _{11,1}+\sigma _{12,2}+b_{1}=0~;~~\sigma _{12,1}+\sigma _{22,2}+b_{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f73cf2b7d04a25f4bbff80397320fe32fc642e15)
Therefore, differentiating w.r.t
and
respectively,
![{\displaystyle \sigma _{11,11}+\sigma _{12,21}+b_{1,1}=0~;~~\sigma _{12,12}+\sigma _{22,22}+b_{2,2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42d349c459d45e513ccf06c93845017fdace8eb8)
Adding,
![{\displaystyle 2\sigma _{12,12}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47a7b7cb86885a0114d1165d6728300edc116eb0)
Hence,
![{\displaystyle \sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=-2\sigma _{12,12}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d35f317d321ca8df528283eaeff10506007a118)
Substituting back into the compatibility equation,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baac9626c6509c98dd4a535ba6e1166e61dc35d5)
Hence,
![{\displaystyle {\nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c90d3f60afefa8b614282a105d29ded8a8dc77c8)