7. Constitutive laws

Chapter status

Illustrative · compile-only. The fluid and solid excerpts target the current 0.1.0a7 standard packages.

Conservation needs material behavior

Conservation constrains storage and transport, but usually does not say how a material carries flux or stress. A constitutive law supplies that closure. Its parameters and assumptions describe the modeled material, not a universal identity.

Learning outcomes

After this chapter, you should be able to:

• separate a balance law from the constitutive relation that closes it;
• read Newtonian viscosity and isotropic linear elasticity in residual form;
• check the dimensions of material parameters; and
• identify when a constitutive assumption is unsuitable even though its source is well typed.

Newtonian viscous stress

For the steady fluid component, the Cauchy stress is represented by

$$\boldsymbol{\sigma} = 2\mu\,\operatorname{sym}(\nabla\boldsymbol{u}) - p\boldsymbol{I}.$$

Velocity gradient has dimension $1/T$. Dynamic viscosity $\mu$ has dimension $M/(LT)$, so their product has pressure dimension $M/(LT^2)$. The isotropic pressure term has the same dimension and tensor shape.

The package places that expression inside momentum balance:

eqiora

2 * dynamic_viscosity * symmetric_part(grad(velocity))
  - isotropic_lift(pressure)

This is a Newtonian, isotropic relation. Selecting it is a modeling decision.

Isotropic linear elasticity

For small displacement $\boldsymbol{d}$, the solid package uses

$$\boldsymbol{\sigma} =2\mu_s\,\operatorname{sym}(\nabla\boldsymbol{d}) +\lambda\,(\nabla\!\cdot\boldsymbol{d})\boldsymbol{I}.$$

Both strain measures are dimensionless. The Lamé parameters $\mu_s$ and $\lambda$ therefore have stress dimension. Eqiora expresses the balance as:

eqiora

relation balance on body {
  -div(
    2 * shear_modulus * symmetric_part(grad(displacement))
    + first_lame_parameter * isotropic_lift(div(displacement))
  ) - grad(load_potential) = 0;
}

The plane-strain and plane-stress components derive different Lamé parameters from Young’s modulus and Poisson’s ratio. They share an equilibrium structure while making different kinematic assumptions.

Read the standard solid components.

Parameters do not validate the material model

Dimensional checking can reject a viscosity supplied in metres or a Young’s modulus supplied in seconds. It cannot establish that a fluid is Newtonian, that deformation is small, or that an isotropic material adequately describes the specimen. Those judgments come from the phenomenon, scale, and intended use of the model.

Deliberate failure: give viscosity the pressure dimension

If dynamic_viscosity is declared as kg / (m * s ^ 2), multiplying it by grad(velocity) produces pressure per time rather than pressure. It can no longer be subtracted from isotropic_lift(pressure). The dimensional error points to the parameter declaration; rearranging the momentum equation does not repair it.

Exercises

1. Verify the dimensions of both terms in the Newtonian stress.
2. Explain which part of the fluid example is conservation and which part is constitutive closure.
3. Show that displacement gradient is dimensionless.
4. Compare the plane-stress and plane-strain parameter expressions in the standard package. Which balance and interface structures stay unchanged?
5. Name a material behavior for which isotropic linear elasticity would be a poor assumption, even if every unit checks.

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