Although the p that appears in Eq. (7.66) is actually a Lagrange multiplier that enforces the incompressibility constraint, similar to its role in Eq. (6.76) for nonlinear solids, its value is equivalent to the hydrostatic pressure for an incompressible Newtonian fluid. Prove this. \sigma _{xx}=-p+2\mu \frac{\partial\nu _{x}}{\partial x},\sigma _{xy}=\mu \left(\frac{\partial\nu _{y}}{\partial x} +\frac{\partial\nu _{x}}{\partial y} \right)=\sigma _{yx}, \sigma _{yy}=-p+2\mu \frac{\partial\nu _{y}}{\partial y},\sigma _{xz}=\mu \left(\frac{\partial\nu _{x}}{\partial z} +\frac{\partial\nu _{z}}{\partial x} \right)=\sigma _{zx}, \sigma _{zz}=-p+2\mu \frac{\partial\nu _{z}}{\partial z},\sigma _{yz}=\mu \left(\frac{\partial\nu _{y}}{\partial z} +\frac{\partial\nu _{z}}{\partial y} \right)=\sigma _{zy}, (7.66) \sigma _{z\theta }=2\mu \left[\frac{1}{2}\left(\frac{1}{r}\frac{\partial\nu _{z}}{\partial \theta }+\frac{\partial\nu _{\theta }}{\partial z} \right) \right]. (6.76)
Question 7.5: Although the p that appears in Eq. (7.66) is actually a Lagr...
Recall from Eq. (7.3) that the hydrostatic pressure is defined as
p=-\frac{1}{3}(\sigma _{xx}+\sigma _{yy}+\sigma _{zz}). (7.3)
hence, for the incompressible Navier–Poisson relation, we have
-\frac{1}{3}(\sigma _{xx}+\sigma _{yy}+\sigma _{zz})=-\frac{1}{3}(-p+2\mu D_{xx}-p+2\mu D_{yy}-p+2\mu D_{zz})=p-\frac{2}{3}\mu (D_{xx}+D_{yy}+D_{zz})
=p-\frac{2}{3}\mu \left(\frac{\partial\nu _{x}}{\partial x}+\frac{\partial\nu _{y}}{\partial y}+\frac{\partial\nu _{z}}{\partial z} \right)
=p-\frac{2}{3}\mu (\triangledown \cdot \nu ),
where we said that \triangledown \cdot \nu will be shown to be zero for incompressible flows. In this case, therefore, the p in Eq. (7.66) is the hydrostatic pressure. This is not true, in general, for incompressible solids. Finally, note the term-\frac{2}{3}\mu that multiplies the divergence of the velocity. This term reveals where the afore-mentioned Stokes’ hypothesis originated.
\sigma _{xx}=-p+2\mu \frac{\partial\nu _{x}}{\partial x}, \sigma _{xy}=\mu \left(\frac{\partial\nu _{y}}{\partial x} +\frac{\partial\nu _{x}}{\partial y} \right)=\sigma _{yx},\sigma _{yy}=-p+2\mu \frac{\partial\nu _{y}}{\partial y}, \sigma _{xz}=\mu \left(\frac{\partial\nu _{x}}{\partial z} +\frac{\partial\nu _{z}}{\partial x} \right)=\sigma _{zx},
\sigma _{zz}=-p+2\mu \frac{\partial\nu _{z}}{\partial z}, \sigma _{yz}=\mu \left(\frac{\partial\nu _{y}}{\partial z} +\frac{\partial\nu _{z}}{\partial y} \right)=\sigma _{zy}, (7.66)