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. \[ \boldsymbol{\tau} := \boldsymbol{\sigma} - \pi \mathbf{I}. \]

Given three principal stresses \(\sigma_1, \sigma_2, \sigma_3\), the shear stress and normal stress inbetween the \(i\)th and \(j\)th principal axis are:

\begin{align*} \sigma_{\rm normal} &= \vu{n}\vdot (\vb*{\sigma} \vu{n}) \\ &= \vu{n} \vdot (\sigma_i \cos \theta \vb{e}_i + \sigma_j \sin\theta \vb{e}_j)\\ &= \sigma_i\cos^2\theta + \sigma_j\sin^2\theta \\ &= \frac{\sigma_i + \sigma_j}{2} + \frac{\sigma_i - \sigma_j}{2} \cos 2\theta \end{align*} \begin{align*} \tau &= \vu{T} \vdot (\vb*{\sigma}\vu{n}) \\ &= (-\sin\theta \vb{e}_i + \cos\theta \vb{e}_j) \vdot (\sigma_i \cos \theta \vb{e}_i + \sigma_j \sin\theta \vb{e}_j)\\ &= -\sigma_i\sin\theta\cos\theta + \sigma_j\cos\theta \sin\theta \\ &= - \frac{\sigma_i - \sigma_j}{2}\sin 2\theta \end{align*}

They together forms a circle centered at the average and radius \(|\sigma_i - \sigma_j|/2\).

The three circles are drawn on top of one another in normal stress-shear stress plane, which give us the Mohr's circle.

For principal stresses \(\sigma_3 < \sigma_2 < \sigma_1\), the shaded crescent is the set of admissible (normal, shear) stress states on any plane through the point; the two small circles are tangent at \(\sigma_2\).

• Rate of Deformation Tensor (when \(\mathbf{u}\) is strain-rate), Infinitesimal Strain

Given a displacement field (or strain-rate) \( \mathbf{u} \), \( \nabla \vb{u} \) 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.
• Antisymmetric Shear: rotation.

The second and third collectively make up the deviatoric part.

In a non-compressible material: \[ \boldsymbol{\varepsilon} := \frac{1}{2}(\nabla \mathbf{u} + (\nabla \mathbf{u})^{\rm T}), \qquad \varepsilon_{ij} = \frac{1}{2}(\partial_iu_j + \partial_ju_i). \] This is the symmetric part of the gradient that contains the first three parts: volumetric and deviatoric part.

\( \mathbf{u} \) 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.

The volumetric part of strain tensor is given by: \[ \vb{\varepsilon}_{\rm vol} = \frac{1}{3} (\nabla \cdot \mathbf{u}) \mathbf{I}. \]

The volumetric strain and deviatoric strain can be separated:

\begin{align*} \varepsilon &= \varepsilon_{\rm vol} + \varepsilon_{\rm dev} \\ &= \frac{1}{3}(\div{\vb{u}}) \vb{I} + \left( \frac{1}{2}(\grad{\vb{u}} + (\grad{\vb{u}})^{\mathsf{T}}) - \frac{1}{3}(\div{\vb{u}}) \vb{I} \right) \end{align*}

There exists separate viscosities for volumetric part and deviatoric part: bulk viscosity \(\zeta\) and shear viscosity \(2\mu\).

\[ E := \frac{\sigma}{\varepsilon} \] where \(\sigma\) is the tensile stress, and \(\varepsilon\) is the extensional strain.

It is directly related to the stiffness (or spring constant, rigidity) \(k\): \[ k = \frac{EA}{L}. \] where \(A\) is the cross-sectional area, and \(L\) is the length of the element.

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.

• 항복강도

Lowest stress that that produce a permanent deformation. Usually defined as the stress required to cause 0.2% plastic strain (0.2% proff stain).

Limit state of compressive stress that leads to ductile failure or brittle failure (rupture by crack propagation, sliding along a weak plane).

• Ultimate tensile strenght, ultimate strength, true stress, engineering stress.

Limit state of tensile stress that leads to ductile failure or brittle failure.

Failure within service period. In the case of cyclic loading, it is expressed as amplitude and number of cycles to failure.

Capability to withstand a sudden load, expressed in terms of energy.

• Izod impact strenght test
• Charpy impact test

It is a design criteria. \[ \mathrm{FS} = \frac{\mathrm{UAS}}{F} \] where \(\mathrm{UAS}\) is the ultimate alloable stress, \(F\) is applied stress.

• Fracture Mechanics

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

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

• Volume Viscosity
• \(\zeta\)

The ratio of isotropic axial stress to the volumetric strain: \[ \vb{\sigma} = \zeta \vb{\varepsilon}_{\rm vol} + 2\mu \vb{\varepsilon}_{\rm dev} \]

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

• Second Coefficient of Viscosity
• \( \lambda \)

The ratio of additional axial stress due to the volumetric strain, compared to the naive approach \(2\mu\varepsilon\): \[ \vb{\sigma} = 2\mu \vb{\varepsilon} + \lambda \vb{\varepsilon}_{\rm vol}. \]

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

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 \( \mu_{ij} \).

\[ K(t, x, y) = \exp(t \nabla^{\cdot 2})(x, y) = \frac{1}{(4\pi t)^{d/2}}e^{-\|x-y\|^2/4t}. \] For every smooth function \(\phi\) of compact support: \[ \lim_{t\to 0}\int_{\mathbb{R}^d}K(t, x, y)\phi(y)\,dy = \phi(x). \]