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.

The square example in chapter 4 states its two-dimensional law directly through $\mu$ and $\lambda$ and computes only the two in-plane displacement components. To interpret it as a slice of a three-dimensional body, you would choose one of the reductions above, which supplies the coefficients and the extra 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(...)
Traction on the boundary where the relation applies	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.

Previous: Loads and energy · Path · Next: Virtual work and finite elements