Continuum Mechanics

The force per area on a surface is called the traction \( \mathbf{t} \). The Cauchy stress tensor \( \boldsymbol{\sigma} \) is defined to satisfy \[ \mathbf{t} = \boldsymbol{\sigma}\hat{\mathbf{n}} \] where \( \hat{\textbf{n}} \) is the unit normal vector of the surface.

The stress is being applied to a control volume which is a infinitesimal inertial volume. In particular, for a cube of side length \( \ell \), the torque is \( \ell^3(\sigma_{ij} - \sigma_{ji}) \) and the moment of inertia is proportional to \( \ell^5 \). That means the angular acceleration is infinite, if \( \sigma_{ij} \neq \sigma_{ji} \). Similarly the force is proportional to \( \ell^2 \) while the mass is proportional to \( \ell^3 \). This also suggest that the acceleration would be infinite. Therefore, the control volume is at a quasi-equalibrium, that is, the changes in tensile and shear stresses across the volume are infinitesimal and the forces and torques should sum to zero. This justifies the Cauchy stress tensor being symmetric.

In a coordinate system, \( \sigma_{ij} \) is the force in \( i \)th direction, when the surface is facing \( j \)th direction.

• Mean Normal Stress Tensor

The isotropic stress can change the volumn of the stressed body. \[ \pi \mathbf{I} := \frac{1}{3}\operatorname{tr} (\boldsymbol{\sigma}) \mathbf{I}. \] The three accounts for the three direction that the stress is applied.

In its full generality, \[ p := \zeta \nabla\cdot \mathbf{u} - \pi \] where \( \zeta \) is teh volume viscosity, \( \mathbf{u} \) is the flow velocity.

The isotropic stress from volume strain is already included in the total isotropic stress: \[ \pi = -p + \zeta \nabla\cdot \mathbf{u}. \]

• \( \boldsymbol{\tau} \), \( s_{ij} \)
• Shear Stress Tensor, Stress Deviator Tensor

Deviation of the Cauchy stress tensor from the isotropy. \[ \boldsymbol{\tau} := \boldsymbol{\sigma} - \pi \mathbf{I}. \]

\[ \boldsymbol{\varepsilon} := \frac{1}{2}(\nabla \mathbf{u} + (\nabla \mathbf{u})^{\rm T}), \qquad \varepsilon_{ij} = \frac{1}{2}(\partial_iu_j + \partial_ju_i). \] This is the symmetric part of the gradient of the displacement field \( \mathbf{u} \).

\( \mathbf{u} \) can be the flow velocity, in which case this is strain-rate tensor. The strain rate tensor generate stress the same way the strain would.

\[ \varepsilon_v := \varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz}. \] It tells the (infinitesimal) total volume change.

\( \nabla \vb{u} \) can be divided into four parts.

• Volumetric part: Uniformly stretched in all direction.
• Axial Deviatoric: Streched along each axis, with no overall volume change.
• Symmetric Shear: Non-rotational shear.
• Skew-symmetric Shear: rotation.

The last three collectively make up the deviatoric part.

By defining the infinitesimal strain in this way, the skew-symmetric shear is removed.

The ratio of pressure to the volumetric strain: \[ K := -V\dv{p}{V} \] where \( p \) is pressure, and \( V \) is the initial volume of system.

• Modulus of Rigidity

The ratio of shear stress to the shear strain: \[ G := \frac{\tau}{\gamma}. \] The shear stress \( \tau \) shears the material by the angle of \( \gamma \) which is the shear strain.

• \( \mathbf{C}, \mathbf{Y} \)
• Elastic Modulus Tensor, Stiffness Tensor

Rank-4 tensor \( C^{ijkl} \) that relates the linear stress-strain relation: \[ \sigma^{ij} = C^{ijkl}\varepsilon_{jk}. \]

• \( \mathbf{S}, \mathbf{K} \)

The "inverse" of elasticity tensor satisfying: \[ S_{ijpq}C^{pqkl} = \frac{1}{2} \left( \delta_i^k\delta_j^l + \delta_i^l\delta_j^k \right). \]

It describe the inverse stress-strain relation: \[ \varepsilon_{ij} = S_{ijkl}\sigma^{jk}. \]

Assume that there is stress only along a single axis. The Poisson's ratio here is the ratio of transverse strain to the axial strain, with negative sign: \[ \nu := -\frac{\varepsilon_{\rm trans}}{\varepsilon_{\rm axial}}. \]

The strain in one direction can be created by stress in multiple axes: \[ \varepsilon_{ii} = \frac{1}{E} \sigma_{ii} - \sum_{j\neq i}\frac{\nu}{E} \sigma_{jj} \] where \( E \) is the Young's modulus. The material is stretched by the axial stress, but squished by transverse stress because it is being stretched in other direction.

Using Voigt notation, we can write the stress-strain relation in two ways:

\begin{align} \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \end{bmatrix} &= \frac{1}{E} \begin{bmatrix} 1 & -\nu & -\nu \\ -\nu & 1 & -\nu \\ -\nu & -\nu & 1 \end{bmatrix} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \end{bmatrix}, \\ \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \end{bmatrix} &= \begin{bmatrix} K + \frac{4}{3}G & K - \frac{2}{3}G & K - \frac{2}{3}G \\ K - \frac{2}{3}G & K + \frac{4}{3}G & K - \frac{2}{3}G \\ K - \frac{2}{3}G & K - \frac{2}{3}G & K + \frac{4}{3}G \end{bmatrix} \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \end{bmatrix}. \end{align}

Notice that we are assuming linear isotropic material here.

These matrices are inverses of each other. In order to relate \( \nu, E \) to \( K, G \), let us the inverse of the second matrix. The determinant can be factorized with the formula \( a^3 -3ab^2 + 2b^3 = (a-b)(a^2 + ab - 2b^2) \) in mind, (The determinant is \( 2G ( 6 KG ) \).) and we obtain: \[ \frac{1}{E} = \frac{2K + \frac{2}{3}G}{6KG},\qquad \frac{\nu}{E} = \frac{K - \frac{2}{3}G}{6KG}. \]

Now any one of them can be expressed in terms of other two:

\begin{align*} \nu &= \frac{K - \frac{2}{3}G}{2K + \frac{2}{3}G} = \frac{E}{2G} - 1 = \frac{1}{2}\left( 1- \frac{E}{3K} \right),\\ E &= \frac{6KG}{2K + \frac{2}{3}G} = 2G(1+\nu) = 3K(1-2\nu), \\ G &= \frac{E}{2(1+\nu)}, \\ K &= \frac{E}{3(1-2\nu)}. \end{align*}

The volumetric strain is also related to the Poisson's ratio: \[ \varepsilon_v := \sum_i\varepsilon_{ii} = \frac{1-2\nu}{E}(\sigma_{xx} + \sigma_{yy} + \sigma_{zz}). \] Notice that it is zero when \( \nu = 0.5 \), which indicate that there is no volume change when stretched. Rubber is a good example that has similar Poisson's ratio. On the other hand cork has Poisson's ratio close to zero; they almost don't expand laterally.

\[ \vb*{\sigma} = 3K \left(\frac{1}{3}\tr(\vb*{\varepsilon}) \mathbf{I}\right) + 2G \left( \vb*{\varepsilon} - \frac{1}{3}\tr(\vb*{\varepsilon}) \mathbf{I} \right), \] where \( K \) is the bulk modulus, \( G \) is the shear modulus.

This provide a realistic description, since real material change their shape more easily than their volume.

The volumetric strain is three times the isotropic strain in each axis. The isotropic part of the constitutive equation looks like: \[ p = \frac{\sigma_{xx} + \sigma_{yy} +\sigma_{zz} }{3} = 3K \frac{\varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz}}{3} = K\varepsilon_v. \]

The shear strain is double the deviatoric strain by definition. A shear strain in one direction consists of rotation and deviatoric strain, each taking half of it.

We assume:

• The Cauchy stress tensor is Galilean invariant
• \( p \) is independent of the strain \( \boldsymbol{\varepsilon} \)
• The fluid is isotropic.

The relation (constitutive equation) between the strain and stress is given as: \[ \boldsymbol{\sigma} = -p\mathbf{I} + \lambda \operatorname{tr}(\boldsymbol{\varepsilon}) \mathbf{I} + 2\mu \boldsymbol{\varepsilon} \] where \( \lambda \) is the second viscosity.

The equation can also be written as: \[ \boldsymbol{\sigma} = -(p - \zeta \nabla\cdot \mathbf{u})\mathbf{I} + \mu \left( \nabla \mathbf{u} + (\nabla \mathbf{u})^{\mathsf{T}} - \frac{2}{3}(\nabla \cdot \mathbf{u})\mathbf{I} \right) \] using the bulk viscosity.

The linear relation between deviatoric stress and deviatoric strain: \[ \boldsymbol{\tau} = \mu \left[\nabla \mathbf{u} + (\nabla \mathbf{u})^{\mathsf{T}} - \frac{2}{3}(\nabla\cdot \vb{u})\vb{I}\right] \] where \(\boldsymbol{\tau}\) is the deviatoric stress, \( \mu \) is the dynamic viscosity, and \(\mathbf{u}\) is the flow velocity.

Coordinate representation would be \[ \tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right). \]

The fluid that satisfies the Newton's law of viscosity is called the Newtonian fluid.

• \(\mu\), \( \eta \)
• Shear Viscosity

The relational constant between stress and momentum change. It has the unit of \(\rm Pa\cdot s\) in SI, poise in cgs.

• Momentum Diffusivity

\[ \nu = \frac{\mu}{\rho}. \]

• \( \lambda \)

The ratio of additional axial stress due to the volumetric strain, not considering the axial stress due to deviatoric axial strain.

The stress occurs due to irreversible resistance, over the reversible resistance by ientropic bulk modulus.

• Volume Viscosity

\[ \zeta := \lambda + \frac{2}{3}\mu \] where \( \lambda \) is the second viscosity, and \( \mu \) is the dynamic viscosity.

It is the ratio of non-stress pressure to the volumetric strain.

For an object moving in a fluid with viscosity \( \mu \), if the size of the object is smaller than the mean free path of the fluid particles, the effective viscosity \( \mu_{\mathrm{eff}} \) deviates from the macroscopic viscosity \( \mu \). \[ \eta_{\mathrm{eff}} = \eta \frac{1}{1+\frac{b}{pr}} \] where \( p \) is the pressure, \( b = 8.22\times 10^{-3}\ \mathrm{Pa\cdot m}\) is a constant, and \( r \) is the radius of the particle.

This quantity is used when establishing the mechanical equalibrium using the Stokes' law: \[ 6\pi r \eta_{\mathrm{eff}} v = F. \] From the equalibrium, the radius \( r \) can be obtained. \[ r = \sqrt{\frac{9\eta_{\mathrm{eff}} v_t}{2g\rho}}. \] where \( v_t \) is the terminal velocity of free falling particle, \( g \) is the gravitational constant, \( \rho \) is the density of the fluid.

The viscosity for anisotropic fluid is given by a tensor \( \mu_{ij} \).

\[ \mathbf{J} = -D\nabla \varphi \] where \(\mathbf{J}\) is the diffusion flux, amount of substance per area per time, and \(D\) is the diffusion coefficient of diffusivity, and \(\nabla\varphi\) is the concentration gradient.

• \(D\) has the unit of area per time.
• \(\varphi\) has the unit of concentration, the amount of substance per volume.

• \[ \frac{\partial \varphi}{\partial t} = D\nabla^{\cdot 2}\varphi. \]
• This is the same form as the heat equation.

• \[ \frac{\partial u}{\partial t} = \alpha \nabla^{\cdot 2} u \]
  • where \(\alpha\) is the thermal diffusivity.
• \[ \alpha = \frac{k}{\rho c} \]
  • where \(k\) is the thermal conductivity, \(\rho\) is the density of the material, \(c\) is the specific heat capacity of the material.

• Fundamental solution to the heat equation.

• Material Coordinates

Lagrangian specification of the flow field represents individual fluid parcels. \[ \mathbf{X}(\mathbf{x}_0, t) \]

• Specification at fixed location.
• \[ \mathbf{u}(\mathbf{x}, t) \]

It is related to Mach number and Reynolds number by \[ \mathrm{Kn} = \frac{\mathrm{Ma}}{\mathrm{Re}} \sqrt{\frac{\gamma \pi}{2}}. \]

\[ \frac{D}{Dt} := \frac{\partial }{\partial t} + \mathbf{u}\cdot\nabla. \]

The first term describes the change due to time at a fixed position, and the second term describes the change due to movement in space at a fixed time, which combine to describe the rate of change while moving along with the flow.

For a system in which the velocity of the particles are not equilibrated, not even locally, the probability density function \( f(\mathbf{r}, \mathbf{p}, t) \) is a good description of the system.

It is the number density in the phase space, satisfying \[ dN = f(\mathbf{r}, \mathbf{p}, t)\, \mathrm{d}^3 \mathbf{r}\, \mathrm{d}^3 \mathbf{p} \] where \( \mathrm{d}N \) is the number of particles in a small chunk of the phase space \( \mathrm{d}^3\mathbf{r}\,\mathrm{d}^3\mathbf{p} \).

According to the Liouville's theorem, the density of state in the phase space does not change in a conservative system. When one is interested in the microscopic behavior, the force once called friction disappears and just molecular interactions remains, constituting a conservative system.

There's still collisions that can disturb the probability density function. So, \[ \frac{\mathrm{d}f}{\mathrm{d}t} = \left( \frac{\partial f}{\partial t} \right)_{\mathrm{coll}}. \] The change of \( f \) due to collision is swept under the simple looking collision term, and it is not an easy task to find the exact form of this term. The total derivative is taken along the trajectory of physical parcel in the phase space, therefore when expanded

\begin{align*} \frac{\partial f}{\partial t} &+ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d} t} \cdot \frac{\partial f}{\partial \mathbf{r}} + \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}\cdot \frac{\partial f}{\partial \mathbf{p}} = \left( \frac{\partial f}{\partial t} \right)_{\mathrm{coll}} \\ \implies \frac{\partial f}{\partial t} &+ \frac{\mathbf{p}}{m}\cdot \frac{\partial f}{\partial \mathbf{r}} + \mathbf{F}\cdot \frac{\partial f}{\partial \mathbf{p}} = \left( \frac{\partial f}{\partial t} \right)_{\mathrm{coll}}. \end{align*}

• Velocity \( \mathbf{v} := \mathbf{p}/m \)
• Flow Velocity \[ \mathbf{u} := \langle \mathbf{v} \rangle = \int \mathbf{v} f(\mathbf{r}, \mathbf{v}, t)\, \mathrm{d}^3 \mathbf{v}. \]
• Peculiar Velocity \[ \mathbf{w} := \mathbf{v} - \mathbf{u}. \]
• Pressure \[ p := \frac{1}{3} \rho \langle w^2 \rangle. \]
• Pressure Tensor \[ \mathbf{P} := -\boldsymbol{\sigma} = \rho \langle \mathbf{w} \mathbf{w}^{\mathsf{T}}\rangle \] where \( \boldsymbol{\sigma} \) is the stress tensor.
• Heat Flux \[ \mathbf{h} := \frac{1}{2} \rho \langle w^2 \mathbf{w} \rangle. \]

\[ \frac{\partial \rho}{\partial t} = -\nabla\cdot(\rho\mathbf{u}). \] It arises from the global mass conservation.

\[ \rho \frac{\mathrm{D}\mathbf{u}}{\mathrm{D}t} = \vec{f}_m(\rho, \mathbf{u}, p, T) \] where \(D/Dt\) is the material derivative, the rate of change while moving along with the flow.

• The Eulerian specification is used, since the Lagrangian specification is unstable.
• One form of this is Navier-Stokes equation.

\[ \rho\frac{\mathrm{D}}{\mathrm{D}t}f(\mathbf{u}, T) = f_e(\rho, \mathbf{u}, p, T) \] where the \(f\) is the local energy function.

• \[ f_{s,1}(p, \rho, T) = 0 \]
• \[ f_{s,2}(\mu, T) = 0 \]

Assuming the fluid is isothermal, \[ \rho\frac{\mathrm{D}\mathbf{u}}{\mathrm{D}t}=\nabla\cdot\boldsymbol{\sigma}+\rho\mathbf{f} \] where \( \rho \) is the mass density, \( \mathbf{u} \) is the flow velocity, \( \mathrm{D}/\mathrm{D} t \) is the material derivative, \(\boldsymbol{\sigma}\) is Cauchy stress tensor, and \( \mathbf{f} \) is the body force.

The divergence is applied to the index that represent the normal direction. The exact definition differs by author.

We assume the . Substituting, we get: \[ \rho \frac{\mathrm{D} \mathbf{u}}{\mathrm{D} t} = - \nabla p + \nabla \cdot \left[ \mu \left( \nabla \mathbf{u} + (\nabla \mathbf{u})^{\mathsf{T}} - \frac{2}{3}(\nabla \cdot \mathbf{u})\mathbf{I} \right) \right] + \nabla (\zeta \nabla \cdot \mathbf{u}) + \rho \mathbf{f}. \]

For a incompressible fluid \( \nabla\cdot \mathbf{u} = 0 \), simplifying the equation: \[ \rho\frac{\mathrm{D}\mathbf{u}}{\mathrm{D}t}=-\nabla p+\mu\nabla^2\mathbf{u}+\rho\mathbf{f}. \]

Note that \(\nabla\cdot \boldsymbol{\tau} = \mu\nabla^{\cdot 2}\mathbf{u}\) here, and this fluid can be called Newtonian.

The existence of smooth solution is proven for 2D, with the help of critical quantity that is invariant under scaling.

\[ \boldsymbol{\omega} = \nabla\times \mathbf{u}. \]

• The cross product of vorticity vector \(\boldsymbol{\omega}\) and velocity vector \(\mathbf{u}\) of the flow field: \[ \mathbf{l} = \mathbf{u}\times \boldsymbol{\omega}. \]
• This appears in the convective acceleration term of the material derivative in the Navier-Stokes equation.

• Flow in which the vorticity vector and the velocity vector are parallel.
• Flow in which the Lamb vector is zero.