2. Displacement, strain and stress

Can a body move without straining?

Slide a square rigidly across a table. Every point moves, but no pair of points changes its separation. Displacement alone cannot measure deformation; we need differences in displacement. Our continuum model will turn those differences into strain, turn strain into stress, then balance the resulting forces.

Write the in-plane displacement as $\boldsymbol u=(u_x,u_y)$, in metres. Coordinates $x,y$ describe the reference, undeformed body. The gradient convention is $(\nabla\boldsymbol u)_{ij}=\partial u_i/\partial x_j$.

Strain removes infinitesimal rigid rotation

For a short material vector $\mathrm{d}\boldsymbol x$, the change in its squared length is, to first order in the displacement gradient,

$$|\mathrm{d}\boldsymbol x+\nabla\boldsymbol u\,\mathrm{d}\boldsymbol x|^2 -|\mathrm{d}\boldsymbol x|^2 \simeq 2\,\mathrm{d}\boldsymbol x^{\mathsf T} \boldsymbol\varepsilon\,\mathrm{d}\boldsymbol x, \qquad \boldsymbol\varepsilon=\frac12(\nabla\boldsymbol u+\nabla\boldsymbol u^{\mathsf T}).$$

Thus

$$\varepsilon_{xx}=\partial_xu_x,\qquad \varepsilon_{yy}=\partial_yu_y,\qquad \varepsilon_{xy}=\frac12(\partial_yu_x+\partial_xu_y).$$

Strain is dimensionless. The engineering shear strain is $\gamma_{xy}=2\varepsilon_{xy}$; confusing the two creates a factor-of-two error. These are the infinitesimal-strain relations described in Roylance’s The Kinematic Equations, pp. 1–3.

Try $\boldsymbol u=(-\theta y,\theta x)$. The gradient is skew-symmetric, so $\boldsymbol\varepsilon=0$: this is an infinitesimal rotation. The approximation discards terms quadratic in the gradient. For a finite rigid rotation, those terms matter, which is why a small-strain law is not a finite-rotation model.

Stress tells us the force on a cut

A stress tensor $\boldsymbol\sigma$ has units of Pa. Given a surface with outward unit normal $\boldsymbol n$, its traction is

$$\boldsymbol t=\boldsymbol\sigma\boldsymbol n.$$

Traction is force per area, a vector. Reversing the normal reverses the traction on the same cut. On a vertical left edge $\boldsymbol n=(-1,0)$, positive $\sigma_{xx}$ therefore gives a leftward traction. On the right edge the same stress gives a rightward traction. This orientation rule is Cauchy’s relation; see Roylance’s The Equilibrium Equations, pp. 1–3.

Balance an arbitrary piece of stationary material:

$$\int_{\partial\Omega}\boldsymbol\sigma\boldsymbol n\,\mathrm{d}S +\int_\Omega\boldsymbol b\,\mathrm{d}V=\boldsymbol0.$$

Here $\boldsymbol b$ is body force per volume, in N/m³. The divergence theorem gives $\nabla\cdot\boldsymbol\sigma+\boldsymbol b=\boldsymbol0$. This is force balance, independent of the material law. In a two-dimensional calculation, area and edge integrals instead yield forces per out-of-plane thickness, in N/m. Multiplying by a chosen physical thickness gives forces in N.

Isotropic elasticity closes the equations

Our in-plane law is

$$\boldsymbol\sigma=2\mu\boldsymbol\varepsilon +\lambda\,\operatorname{tr}(\boldsymbol\varepsilon)\boldsymbol I, \qquad \operatorname{tr}(\boldsymbol\varepsilon)=\nabla\cdot\boldsymbol u.$$

Both $\mu$ and $\lambda$ have units of Pa. The identity tensor $\boldsymbol I$ spreads the scalar trace over diagonal components. Stress is symmetric, so $\sigma_{xy}=\sigma_{yx}$. The law stores the energy density

$$W=\frac12\boldsymbol\sigma:\boldsymbol\varepsilon =\mu\boldsymbol\varepsilon:\boldsymbol\varepsilon +\frac\lambda2(\operatorname{tr}\boldsymbol\varepsilon)^2.$$

The colon sums products of matching tensor entries, including both off-diagonal entries. $W$ has units of J/m³, the same dimensions as Pa. The square problem uses $\mu>0$ and $\lambda=0$, making every nonzero symmetric strain cost energy. Roylance’s Constitutive Equations, pp. 1–4 develops isotropic elasticity and its shear/volume decomposition.

A two-dimensional drawing does not choose a reduction

Starting with a three-dimensional isotropic material, write $\mu=E/[2(1+\nu)]$ and $\lambda_3=E\nu/[(1+\nu)(1-2\nu)]$, where $\nu$ is dimensionless Poisson ratio. Two different physical assumptions lead to different in-plane coefficients:

Assumption	Out-of-plane condition	Coefficient in the in-plane law
Plane strain	$\varepsilon_{zz}=\varepsilon_{xz}=\varepsilon_{yz}=0$	$\lambda=\lambda_3$
Plane stress	$\sigma_{zz}=\sigma_{xz}=\sigma_{yz}=0$	$\lambda=E\nu/(1-\nu^2)$

Plane strain describes suppressed out-of-plane deformation; it need not mean zero out-of-plane stress. Plane stress can describe a thin sheet loaded in its plane with free broad faces; its thickness can change.

We can derive the second coefficient rather than memorize it. From $0=\sigma_{zz}=2\mu\varepsilon_{zz}+ \lambda_3(\varepsilon_{xx}+\varepsilon_{yy}+\varepsilon_{zz})$,

$$\varepsilon_{zz}=-\frac{\lambda_3}{2\mu+\lambda_3} (\varepsilon_{xx}+\varepsilon_{yy}).$$

Substitution into the two in-plane normal stresses changes the trace coefficient to $2\mu\lambda_3/(2\mu+\lambda_3)=E\nu/(1-\nu^2)$. For $\nu=0.3$, the plane-strain coefficient is about $0.577E$, while the plane-stress coefficient is about $0.330E$. Choosing the wrong reduction changes the model even when the mesh is identical.

Our forthcoming example states its two-dimensional law directly through $\mu$ and $\lambda$. It computes two displacement components; there is no out-of-plane field to inspect. The derivation above explains how a physical three-dimensional interpretation would supply coefficients and additional assumptions.

Read the operators as mechanics

The .eqi spelling maps directly to these definitions:

Mathematical meaning	Source expression
Small strain	symmetric_part(grad(displacement))
Trace of strain	div(displacement)
Scalar multiplied by identity	isotropic_lift(...)
Stress traction on the owning boundary	normal(...)

Open the direct elastic component. The declaration vector<m, 2> gives displacement both its unit and its two components. load_potential has units of Pa, so its gradient has the correct N/m³ body-force units.

For a Newtonian fluid the same traction rule uses a different stress law: $\boldsymbol\sigma=-p\boldsymbol I+2\mu_f\operatorname{sym}(\nabla\boldsymbol v)$. The fluid’s $\mu_f$ is a viscosity in Pa·s, while the solid’s $\mu$ is a shear modulus in Pa. The fluid path uses velocity $\boldsymbol v$, in m/s; substituting velocity into an elastic law would change the dimensions.

Exercises

1. Compute strain for $\boldsymbol u=(a x,b y)$ and for a constant translation.
2. For $\boldsymbol\sigma=\begin{pmatrix}4&1\\1&2\end{pmatrix}\,\mathrm{Pa}$, find traction on the left and upper edges. Check: $(-4,-1)\,\mathrm{Pa}$ and $(1,2)\,\mathrm{Pa}$.
3. For simple shear $\boldsymbol u=(c y,0)$, show that $\sigma_{xy}=\mu c$ and $W=\mu c^2/2$.
4. Why does setting both $\varepsilon_{zz}=0$ and $\sigma_{zz}=0$ generally impose more than either plane reduction alone?
5. In the square example $\lambda=0$. Show why a displacement depending only on $x$ can leave the upper and lower edges traction-free. What changes if $\lambda$ becomes positive?

Reading

• David Roylance, The Kinematic Equations, MIT, September 19, 2000: module and PDF.
• David Roylance, The Equilibrium Equations, MIT, September 26, 2000: module and PDF.
• David Roylance, Constitutive Equations, MIT, October 4, 2000: module and PDF.

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