# 0.4 Stress in fluids  (Page 2/4)

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$\underset{d\to 0}{lim}\frac{1}{{d}^{2}}\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}{\mathbf{t}}_{\left(n\right)}dS=0$

or, the stresses are locally in equilibrium .

## The stress tensor

To elucidate the nature of the stress system at a point $P$ we consider a small tetrahedron with three of its faces parallel to the coordinate planes through $P$ and the fourth with normal $n$ (see Fig. 5.1 of Aris). If $dA$ is the area of the slanted face, the areas of the faces perpendicular to the coordinate axis Pi is

$d{A}_{i}={n}_{i}dA.$

The outward normals to these faces are $-{\mathbf{e}}_{\left(i\right)}$ and we may denote the stress vector over these faces by $-{\mathbf{t}}_{\left(i\right)}$ . ( ${\mathbf{t}}_{\left(i\right)}$ denotes the stress vector when $+{\mathbf{e}}_{\left(i\right)}$ is the outward normal.) Then applying the principle of local equilibrium to the stress forces when the tetrahedron is very small we have

$\begin{array}{c}\hfill {\mathbf{t}}_{\left(n\right)}dA-{\mathbf{t}}_{\left(1\right)}d{A}_{1}-{\mathbf{t}}_{\left(2\right)}d{A}_{2}-{\mathbf{t}}_{\left(3\right)}d{A}_{3}\\ \hfill =\left({\mathbf{t}}_{\left(n\right)}-{\mathbf{t}}_{\left(1\right)}{n}_{1}-{\mathbf{t}}_{\left(2\right)}{n}_{2}{\mathbf{t}}_{\left(3\right)}{n}_{3}\right)dA=0\end{array}$

Now let ${T}_{ji}$ denote the ${i}^{th}$ component of ${\mathbf{t}}_{\left(j\right)}$ and ${t}_{\left(n\right)i}$ the ${i}^{th}$ component of ${\mathbf{t}}_{\left(n\right)}$ so that this equation can be written

${t}_{\left(n\right)i}={T}_{ji}{n}_{j}.$

However, ${\mathbf{t}}_{\left(n\right)}$ is a vector and $\mathbf{n}$ is a unit vector quite independent of the ${T}_{ji}$ so that by the quotient rule the ${T}_{ji}$ are components of a second order tensor $\mathbf{T}$ . In dyadic notation we might write

${\mathbf{t}}_{\left(n\right)}=\mathbf{T}•\mathbf{n}.$

This tells us that the system of stresses in a fluid is not so complicated as to demand a whole table of functions ${\mathbf{t}}_{\left(n\right)}\left(\mathbf{x},\mathbf{n}\right)$ at any given instant, but that it depends rather simply on $\mathbf{n}$ through the nine quantities ${T}_{ji}\left(\mathbf{x}\right)$ . Moreover, because these are components of a tensor, any equation we derive with them will be true under any rotation of the coordinate axis.

Inserting the tensor expression for the stress into the momentum balance and using the equation of continuity and Green's theorem we have

$\begin{array}{ccc}\hfill \frac{D}{Dt}\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{v}dV& =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{f}dV+\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}{\mathbf{t}}_{\left(n\right)}dS\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{f}dV+\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}\mathbf{n}•\mathbf{T}dS\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\left(\rho \mathbf{f}+\nabla •\mathbf{T}dV\hfill \\ \hfill \frac{D}{Dt}\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{v}dV& =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\left[\frac{D}{Dt},\left(\rho \mathbf{v}\right),+,\rho ,\mathbf{v},\left(\nabla •\mathbf{v}\right)\right]dV\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\left[\rho ,\frac{D\mathbf{v}}{Dt},+,\mathbf{v},\left(\frac{D\rho }{Dt},+,\rho ,\nabla ,•,\mathbf{v}\right)\right]dV\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \frac{D\mathbf{v}}{Dt}dV\hfill \end{array}$

Since all the integrals are now volume integrals, they can be combined as a single integrand.

$\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\left(\rho \frac{d\mathbf{v}}{Dt}-\rho \mathbf{f}-\nabla •\mathbf{T}\right)dV=0$

However, since $V$ is an arbitrary volume this equation is satisfied only if the integrand vanishes identically.

$\begin{array}{ccc}\hfill \rho \frac{D\mathbf{v}}{Dt}& =& \rho \mathbf{f}+\nabla •\mathbf{T}\hfill \\ & =& \rho \mathbf{a}\hfill \\ & \text{or}& \\ \hfill \rho \frac{D{v}_{i}}{Dt}& =& \rho {f}_{i}+{T}_{ji,j}\hfill \\ & =& \rho {\alpha }_{i}\hfill \end{array}$

This is Cauchy's equation of motion and $\mathbf{a}$ is the acceleration. It holds for any continuum no matter how the stress tensor $\mathbf{T}$ is connected with the rate of strain.

## The symmetry of the stress tensor

A polar fluid is one that is capable of transmitting stress couples and being subject to body torques, as in magnetic fluids. In case of a polar fluid we must introduce a body torque per unit mass in addition to the body force and a couple stress in addition to the normal stress ${\mathbf{t}}_{\left(n\right)}$ . The stress for polar fluids is discussed by Aris.

A fluid is nonpolar if the torques within it arise only as the moments of direct forces. For the nonpolar fluid we can make the assumption either that angular momentum is conserved or that the stress tensor is symmetric. We will make the first assumption and deduce the symmetry.

Return now to the integral linear momentum balance with the internal force expressed as a surface integral. If we assume that all torques arise from macroscopic forces, then not only linear momentum but also the angular momentum $\mathbf{x}×\left(\rho \mathbf{v}\right)$ are expressible in terms of $\mathbf{f}$ and ${\mathbf{t}}_{\left(n\right)}$ .

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