Past exam of the solid mechanics course of the University of Oxford / 2011 / 3

The displacement field is $u(x)=y(x)-x$, and its gradient is $Du=Dy-I$. Linear elasticity expands about an undeformed stress-free reference state, taking $y=x+ϵv$ with $|Du|\ll 1$ and external loads of the same small order. A rigid translation need not itself be small; it is the deformation gradient that controls the infinitesimal strain tensor.

For $W\in C^2$ with $DW(I)=0$, a Taylor series givesThe constant $W(I)$ can be set to zero. The tensor $\mathbb{A}$ is the elasticity tensor. Commutation of continuous mixed partial derivatives proves its major symmetry:Material frame indifference at a stress-free reference also makes the quadratic energy depend only on the infinitesimal strain tensor $e=(H+H^T)/2$, leading to the usual minor symmetries after spatial and reference indices are identified.

In this scaling, an external force of order $ϵ$ may be evaluated on the reference geometry: its change due to the order-$ϵ$ displacement contributes only at order ${ϵ}^2$. Thus the leading forces are fixed dead loads. Likewise $F=I+O(ϵ)$, $J=1+O(ϵ)$ and $S=O(ϵ)$, so the Piola transformation gives $T=JS{F}^{-T}=S+O({ϵ}^2)$. Hence Cauchy and first Piola stresses agree to linear order. This reasoning uses the usual small-load, stress-free linearization; an order-one prestress or leading follower-load effect requires a different expansion.

Let the boundary split into a displacement part ${\Gamma }_D$ and a traction part ${\Gamma }_N$. The linearized stored elastic energy functional isIts corresponding total potential subtracts the work of fixed body and surface loads:The problem in the calculus of variations is to minimize $\Pi$ among admissible displacements with $u=u_D$ on ${\Gamma }_D$. For a stable quadratic elastic energy this is the usual energy minimization formulation. Its stationarity condition suffices to derive equilibrium.

Vary by $u+sv$, with $v=0$ on ${\Gamma }_D$. Major symmetry combines the two differentiated quadratic terms, givingIntegration by parts, first with test functions compactly supported in the interior, gives $-\mathrm{Div}{S}^{\mathrm{lin}}=f$. Then arbitrary traces on ${\Gamma }_N$ give the natural traction condition. The complete boundary value problem isFor a homogeneous hyperelastic material the elasticity tensor is constant. In isotropic linear elasticity the constitutive equation is $S^{\mathrm{lin}}=\lambda \mathrm{div}uI+\mu (Du+D{u}^T)$, giving the familiar Navier-Cauchy equation $-\mu \Delta u-(\lambda +\mu )\nabla \mathrm{div}u=f$.

Taking traces of the constitutive equation givesThe assumed finite limits of stress and strain make the right side bounded. Therefore $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$, and the limiting strain satisfies $\mathrm{tr}e=0$. The permitted interchange of derivatives and limits gives $\mathrm{tr}e=\mathrm{div}u$. DefineThen the limit stress is $S=-pI+2\mu e$. Passing to the limit in force balance and using $2\mathrm{Div}e=\Delta u+\nabla \mathrm{div}u$ givesThe pressure acts as the Lagrange multiplier enforcing the volume constraint. The differential equations determine it up to an additive constant, while a specified absolute limit stress can fix that constant.

Nonlinear elastic incompressibility is $\det Dy=1$. Expanding the determinant for $Dy=I+Du$ gives $\det (I+Du)=1+\mathrm{div}u+O(|Du{|}^2)$, so its first-order form is precisely $\mathrm{div}u=0$. The volumetric energy term $\lambda (\mathrm{div}u{)}^2/2$ becomes increasingly expensive as $\lambda$ grows, while the fixed positive $\mu$ leaves shear strain possible. Thus the infinite-bulk-stiffness limit is the incompressible limit of isotropic linear elasticity, with the same equilibrium operator as the steady Stokes equations.