In the case of pure fluid statics, the fluid is at rest and thus the velocity and acceleration of the fluid are both zero (note: actually, we only need a=0 for statics by Newton’s second law). Show that in this case, the only possible stress in the fluid is a hydrostatic pressure. Moreover,

Question

In the case of pure fluid statics, the fluid is at rest and thus the velocity and acceleration of the fluid are both zero (note: actually, we only need a=0 for statics by Newton’s second law). Show that in this case, the only possible stress in the fluid is a hydrostatic pressure. Moreover, show that if a cube of a static fluid is oriented with respect to an (o; x, y, z) coordinate system, then the only stresses with respect to an [latex](o; x^{\prime}, y^{\prime}, z^{\prime})[/latex] coordinate system are still hydrostatic.

Accepted Answer

If [latex] u =0,[/latex] then [latex] u_{x}= u_ {y}= u_{z}=0[/latex] and all components of [D] are zero. From Eq. (7.66), therefore,

[latex]\sigma _{xx}=-p+2\mu \frac{\partial u _{x}}{\partial x}, \sigma _{xy}=\mu \left(\frac{\partial u _{y}}{\partial x} +\frac{\partial u _{x}}{\partial y} ight)=\sigma _{yx},[/latex]

[latex]\sigma _{yy}=-p+2\mu \frac{\partial u _{y}}{\partial y}, \sigma _{xz}=\mu \left(\frac{\partial u _{x}}{\partial z} +\frac{\partial u _{z}}{\partial x} ight)=\sigma _{zx},[/latex]

[latex]\sigma _{zz}=-p+2\mu \frac{\partial u _{z}}{\partial z}, \sigma _{yz}=\mu \left(\frac{\partial u _{y}}{\partial z} +\frac{\partial u _{z}}{\partial y} ight)=\sigma _{zy},[/latex] (7.66)

[latex]\sigma _{xx}=-p, \sigma _{yy}=-p, \sigma _{zz}=-p,[/latex]

[latex]\sigma _{xy}=0=\sigma _{yx}, \sigma _{xz}=0=\sigma _{zx}, \sigma _{yz}=0=\sigma _{zy}[/latex]

and the stress is hydrostatic (i.e., the normal stresses at a point equal the negative of the pressure at that point). Moreover, if we consider a rotation about the z axis (i.e., in the x–y plane), then

[latex] \sigma ^{\prime }_{xx}=\sigma _{xx}\cos ^{2}\alpha +2\sigma _{xy}\cos \alpha \sin \alpha +\sigma _{yy}\sin ^{2}\alpha ,[/latex]

[latex] \sigma ^{\prime }_{xy}=(\sigma _{yy}-\sigma _{xx})\cos \alpha \sin \alpha +\sigma _{xy}(\cos ^{2}\alpha -\sin ^{2}\alpha ),[/latex]

[latex] \sigma ^{\prime }_{yy}=\sigma _{xx}\sin ^{2}\alpha +2\sigma _{xy}\cos \alpha \sin \alpha +\sigma _{yy}\cos ^{2}\alpha[/latex]

from Eqs. (2.13), (2.17), and (2.21). Hence, in our case,

[latex]\sigma ^{\prime }_{xx}=\sigma _{xx}\cos ^{2}\alpha +2\sigma _{xy}\sin \alpha \cos \alpha +\sigma _{yy}\sin ^{2}\alpha ,[/latex] (2.13)

[latex]\sigma ^{\prime }_{xy}=2\sin \alpha \cos \alpha \left(\frac{\sigma _{yy}-\sigma _{xx}}{2} ight)+(\cos ^{2}\alpha -\sin ^{2}\alpha )\sigma _{xy}.[/latex] (2.17)

[latex]\sigma ^{\prime }_{yy}=\sigma _{xx}\sin ^{2}\alpha -2\sigma _{xy}\sin \alpha \cos \alpha +_{yy}\cos ^{2}\alpha .[/latex] (2.21)

[latex]\sigma ^{\prime }_{xx}=-p\cos ^{2}\alpha +0-p\sin ^{2}\alpha =-p,[/latex]

[latex]\sigma ^{\prime }_{xy}=(-p+p)\cos \alpha \sin \alpha +0=0,[/latex]

[latex]\sigma ^{\prime }_{yy}=-p\sin ^{2}\alpha -0-p\cos ^{2}\alpha =-p[/latex]

for all α, and the stress is indeed hydrostatic in pure fluid statics regardless of the
coordinate system (Fig. 7.10). Although discussed in detail in Chap. 8, note that if μ is negligible, then

[latex]\sigma _{xx}=-p, \sigma _{yy}=-p, \sigma _{zz}=-p[/latex] (7.67)

and all shear stresses are zero. A fluid that can only support a pressure, not a shear stress, is said to be inviscid. Whereas no fluid has zero viscosity, the assumption of negligible viscosity (like that of a rigid member in a truss in statics even though no material is truly rigid) has proven useful in many areas of fluid mechanics, particularly in aerospace applications.

Question 7.6: In the case of pure fluid statics, the fluid is at rest and ...

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