Continuum Mechanics §

Why two descriptions? In a solid, each atom stays near its equilibrium position, so it is natural to label particles by their reference location $\vec{X}$ and ask “where did this particle go?” In a fluid, particles flow past every point, so it is natural to fix a spatial location $\vec{x}$ and ask “what is the velocity here right now?”

Concrete example: a stretching rubber band. Consider a rubber band of rest length $L_0 = 1\;\text{m}$, fixed at the left end and pulled at the right end. At time $t$, the right end has moved to $x = L_0 + \delta(t)$.

• Lagrangian: a particle initially at $\vec{X} = (X, 0, 0)$ with $0 \le X \le L_0$ is now at $\vec{x} = (X(1 + \delta/L_0), 0, 0)$. The displacement is $\vec{u}(\vec{X}, t) = (X\,\delta/L_0, 0, 0)$.
• Eulerian: at a fixed spatial point $\vec{x} = (x, 0, 0)$ the velocity is $\vec{v}(\vec{x}, t) = (x\,\dot{\delta}/(L_0 + \delta), 0, 0)$.

The material derivative converts between them: for the velocity field, $Dv_x/Dt = \partial v_x/\partial t + v_x\,\partial v_x/\partial x$ gives the acceleration of the particle currently at $x$.

Example problem. A rod of length $L_0 = 2\;\text{m}$ undergoes the displacement field $u_x(X) = 0.01\,X^2/L_0$ (in metres). Find the Lagrangian strain $\varepsilon_{xx}$ at $X = L_0/2$.

Step 1: What is a displacement field? A displacement field assigns to every material point (labelled by $\vec{X}$) a vector $\vec{u}(\vec{X})$ telling how far and in what direction the point has moved. Physically, if you mark a grid on a rubber sheet and then stretch it, the displacement of each grid intersection is a point in the displacement field.

Step 2: Deriving strain from a line element. Consider two neighbouring material points $\vec{X}$ and $\vec{X} + d\vec{X}$. After deformation, the line element becomes:

The squared length changes from $|d\vec{X}|^2$ to $|d\vec{x}|^2 = d\vec{X}^T(\mathbf{I} + \nabla\vec{u})^T(\mathbf{I} + \nabla\vec{u})\,d\vec{X}$. Expanding and keeping only first-order terms (valid when $|\nabla\vec{u}| \ll 1$):

The symmetric part of $\nabla\vec{u}$ is the strain tensor $\boldsymbol{\varepsilon}$; the antisymmetric part is the infinitesimal rotation (which does not change lengths).

Step 3: 2-D example. A square element of side $\delta$ in the $x$-$y$ plane undergoes the displacement $u_x = \alpha\,x$, $u_y = \beta\,y$ (no shear). Then:

The square becomes a rectangle of sides $\delta(1+\alpha)$ and $\delta(1+\beta)$.

Step 4: Validity and finite deformation. The infinitesimal strain tensor $\boldsymbol{\varepsilon}$ is a linearization. It is valid when $|\nabla\vec{u}| \ll 1$ (typically strains below a few percent in engineering). For large deformations (rubber, biological tissue, metal forming), one must use:

• Deformation gradient: $\mathbf{F} = \partial \vec{x}/\partial \vec{X}$
• Green–Lagrange strain: $\mathbf{E} = \tfrac{1}{2}(\mathbf{F}^T\mathbf{F} - \mathbf{I})$
• Eulerian (Almansi) strain: $\mathbf{e} = \tfrac{1}{2}(\mathbf{I} - \mathbf{H}^T\mathbf{H})$ where $\mathbf{H} = \nabla\vec{v}$

The polar decomposition $\mathbf{F} = \mathbf{R}\mathbf{U}$ separates rotation $\mathbf{R}$ from stretch $\mathbf{U}$, which is essential for finite-strain constitutive models.

Step 5: Example problem. A 2-D plate undergoes the displacement field $u_x = 0.002\,x + 0.001\,y$, $u_y = 0.001\,x + 0.003\,y$ (all in metres, $x, y$ in metres). Compute the full strain tensor and check whether the small-strain assumption holds.

Step 1: Traction and stress. The traction vector $\vec{t}$ on a surface element $dA$ with normal $\hat{n}$ is the force per unit area exerted by the material on one side on the material on the other. Cauchy’s theorem states that $\vec{t}$ depends only on $\hat{n}$ (not on the shape of the surface) and is linear in $\hat{n}$: $\vec{t} = \boldsymbol{\sigma}\,\hat{n}$.

Step 2: Components. In 3-D, $\boldsymbol{\sigma}$ is a $3\times 3$ matrix:

Diagonal entries are normal stresses (tension positive); off-diagonal entries are shear stresses.

Step 3: Why symmetric? Consider an infinitesimal cube. The balance of angular momentum (no body couple) requires that the net torque about any axis vanishes. Taking moments about the $z$-axis: the shear stresses $\sigma_{xy}$ and $\sigma_{yx}$ produce torques that cancel only if $\sigma_{xy} = \sigma_{yx}$. Repeating for the other axes gives $\sigma_{ij} = \sigma_{ji}$ for all $i, j$. This reduces the 9 components to 6 independent ones.

Step 4: Relation to momentum balance. The stress tensor enters the local momentum balance (Cauchy’s equation of motion), presented in the Conservation laws section:

This is the continuum analogue of $F = ma$ from Dynamics, applied to an infinitesimal volume element.

Step 5: Example problem. A cube of side $a = 1\;\text{cm}$ is subjected to the stress state $\sigma_{xx} = 50\;\text{MPa}$, $\sigma_{yy} = 20\;\text{MPa}$, $\sigma_{xy} = 10\;\text{MPa}$ (all other components zero). Find the traction vector on a face with normal $\hat{n} = (1/\sqrt{2}, 1/\sqrt{2}, 0)$.

Step 1: From integral to local form. The integral balance laws (e.g., $\frac{d}{dt}\int_{V(t)} \rho\,dV = 0$ for mass) hold for any material volume $V(t)$. Applying the Reynolds transport theorem (converting the material derivative of an integral to an integral of the material derivative) and then the divergence theorem (converting the surface integral of flux to a volume integral of divergence), and requiring the result to hold for any sub-volume, yields the local (pointwise) form. The local form is what is used in PDE solvers.

Step 2: The material derivative. The operator $D/Dt = \partial/\partial t + \vec{v}\cdot\nabla$ is a pointwise operator that follows a material particle. In the Lagrangian description it reduces to $\partial/\partial t$ (since the label $\vec{X}$ is fixed); in the Eulerian description it includes the convective term $\vec{v}\cdot\nabla$.

Step 3: Boundary conditions. A well-posed continuum-mechanics problem requires boundary conditions on $\partial V$:

• Essential (Dirichlet): prescribe the displacement $\vec{u}$ (or velocity $\vec{v}$) on part of the boundary $\Gamma_u$.
• Natural (Neumann): prescribe the traction $\vec{t} = \boldsymbol{\sigma}\cdot\hat{n}$ on the remaining part $\Gamma_t$.

Together they close the system and are the natural input to the finite element method.

Step 4: Example problem. An incompressible fluid flows in a pipe of radius $R$ with velocity profile $v_z(r) = v_0(1 - r^2/R^2)$. Verify the continuity equation.

Step 1: Why a constitutive law is needed. The balance laws (mass, momentum, energy) are universal—they apply to any continuum. But they contain the stress tensor $\boldsymbol{\sigma}$ as an unknown. The constitutive law provides the material-specific relation between stress and deformation, closing the system.

Step 2: Hooke’s law in 1-D. For a uniaxial bar: $\sigma = E\,\varepsilon$. The single constant $E$ (Young’s modulus) replaces the two Lamé parameters because $\varepsilon_{kk} = \varepsilon$ and $\sigma_{ij} = 0$ for $i \neq j$ in 1-D.

Step 3: From Hooke’s law to Navier–Stokes. For a Newtonian fluid, substituting $\sigma_{ij} = -p\delta_{ij} + 2\mu D_{ij}$ into the momentum balance and using incompressibility ($\nabla\cdot\vec{v} = 0$) gives the Navier–Stokes equations:

Step 4: Elastodynamics. For elastic solids with small strain and no body force, the linearized equations reduce to the elastodynamic wave equation, connecting to Oscillations and Waves.

Step 5: Example problem. A copper rod ($E = 110\;\text{GPa}$, $\nu = 0.34$) is subjected to a uniaxial stress $\sigma_{xx} = 80\;\text{MPa}$. Find the Lamé parameters and the strain in the $x$-direction.

Step 1: From strong to weak form. Start with the strong form (local momentum balance): $\nabla\cdot\boldsymbol{\sigma} + \vec{f} = \rho\vec{a}$ in $\Omega$, with $\vec{u} = \bar{\vec{u}}$ on $\Gamma_u$ and $\boldsymbol{\sigma}\hat{n} = \bar{\vec{t}}$ on $\Gamma_t$. Multiply by a test function $\delta\vec{u}$ (zero on $\Gamma_u$) and integrate over $\Omega$:

Step 2: Integration by parts. Apply the divergence theorem to the stress term:

The boundary integral splits into $\Gamma_u$ (where $\delta\vec{u} = 0$, so it vanishes) and $\Gamma_t$ (where $\boldsymbol{\sigma}\hat{n} = \bar{\vec{t}}$). The result is the weak form stated above.

Step 3: Discretization (FEM). Approximate $\vec{u} = \sum_a u_a\,N_a(\vec{x})$ with shape functions $N_a$. Substituting into the weak form and choosing $\delta\vec{u} = N_b$ for each node $b$ gives:

where $K_{ba} = \int_\Omega \mathbf{B}_b^T \mathbf{D}\, \mathbf{B}_a\; dV$ is the element stiffness matrix ($\mathbf{D}$ is the constitutive matrix, $\mathbf{B}_a$ the strain-displacement matrix). Assembling over all elements gives the global system $\mathbf{K}\,\mathbf{u} = \mathbf{F}$.

Step 4: Connection to Lagrangian mechanics. The variational principle is the continuum analogue of the principle of virtual work in Dynamics and is closely related to Hamilton’s principle. In the static case, it is equivalent to minimizing the total potential energy $\Pi = U - W_{\text{ext}}$.

Step 5: Example problem. A 1-D bar of length $L$, cross-section $A$, modulus $E$, is fixed at $x=0$ and loaded by force $F$ at $x=L$. Using a single finite element (two nodes), find the nodal displacements.

A steel rod (cross-section $A = 10^{-4}\;\text{m}^2$, length $L = 2\;\text{m}$, $E = 200\;\text{GPa}$) is stretched by a force $F = 10\;\text{kN}$. Find the elongation and the maximum stress.