Example 3
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Given:
If a material is incompressible (ν {\displaystyle \nu } = 0.5), a state of hydrostatic stress (σ 11 = σ 22 = σ 33 {\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}} ) produces no strain. The corresponding stress-strain relation can be written as
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} is an unknown hydrostatic pressure which will generally vary with position. Also, the condition of incompressibility requires that the dilatation
e = ε k k = 0 . {\displaystyle e=\varepsilon _{kk}=0~.} Show:
Show that the stress components and the hydrostatic pressure p {\displaystyle p} must satisfy the equations
∇ 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} } is the body force.
Solution
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We have, e = ε k k = ε 11 + ε 22 + ε 33 = 0 . {\displaystyle e=\varepsilon _{kk}=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0~.\,}
Also,
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}\,} , the above relation
gives σ 11 = σ 22 = σ 33 = − p {\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}=-p\,} .
Therefore,
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}} and x 2 {\displaystyle x_{2}} respectively,
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} }}}