Past exam of the elasticity and plasticity course of the University of Oxford / 2011 / 1

The deformation gradient is the derivative of the deformation map with respect to the reference configuration coordinates, at fixed time:Its determinant $J=\det F$ is the local ratio of current to reference volume: $dv=JdV$ for an orientation-preserving deformation in continuum mechanics. Thus elastic incompressibility is the pointwise constraint $\boxed{J=1}$, rather than a requirement that each individual principal stretch be one.

Choose the outward unit normal $n$ to the boundary of the chosen current material subbody. The traction $t(x,n)$ is the contact force exerted on that subbody per unit current area at the boundary point. Reversing the subbody side reverses $n$ and $t$.

By Cauchy's stress theorem, the dependence on $n$ is linear. With the usual column-vector convention the Cauchy stress tensor satisfiesThe same point has different traction vectors on differently oriented planes, but a single Cauchy stress tensor encodes all of them. In the absence of body couples, conservation of angular momentum makes $\tau$ symmetric.

In the current configuration, balance of linear momentum is the Cauchy momentum equationHere $\tau$ is the Cauchy stress tensor, $\rho$ is current mass density, $b$ is body force per unit mass, and $v(x,t)$ is the velocity of the material particle currently at $x$. The expression in parentheses is the material acceleration. The divergence is taken in current spatial coordinates, with $(\mathrm{div}\tau {)}_i={\partial }_j{\tau }_{ij}$.

Equivalently the right side is $\rho {\partial }_t^{2}x(X,t)$ evaluated at the particle, because this is the same material acceleration in Lagrangian coordinates. For the subsequent static inflation with negligible body force, the equation reduces to $\mathrm{div}\tau =0$.

Write $B=A+H$ for the reference outer radius and $b=a+h$ for the current outer radius. The inner material normal is $-e_r$. A fluid pressure $P$ applies traction $-Pn=P{e}_r$ there, so the radial Cauchy stress tensor component must be negative:The second condition uses zero external gauge pressure. Both faces also have zero shear traction; the axisymmetric diagonal stress field satisfies those conditions automatically. The axial constraint is plane strain, so supports may supply an axial reaction; it does not imply zero axial stress.

For the axisymmetric deformation map, take $\theta =\Theta$, $z=Z$, and $r=r(R)$. A reference cylindrical shell of unit axial length has volume $2\pi RdR$, and the corresponding current shell has volume $2\pi rdr$. Elastic incompressibility therefore gives $rdr=RdR$.

Integrating from the inner surface, where $r(A)=a$, gives $r^2-a^2=R^2-A^2$. The positive radius is consequentlyThe outer surface is not an additional independent unknown: $\boxed{b=\sqrt{B^2+a^2-A^2}}$, and hence $h=b-a$.

Use orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

The multiplier enforcing elastic incompressibility contributes an arbitrary hydrostatic stress. Denote it by $q(r)$ to distinguish it from the applied pressure $P$. For the neo-Hookean solid, the Cauchy stress tensor is $\tau =-qI+\mu F{F}^T$. Thus its principal components areThe axial component is $t_z=-q+\mu$, despite the unit axial stretch. The quantity $q$ is a Lagrange multiplier determined together with equilibrium and the traction conditions; it is not obtained by differentiating the reduced stretch energy while ignoring its constraint.

In particular the stress difference eliminates this multiplier:

The radial equilibrium equation gives $d{t}_r/dr=(t_{\theta }-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the neo-Hookean solid stress difference eliminates both $q$ and $t_{\theta }$:This is a first-order ordinary differential equation for the radial stress, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner pressure to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\ge A^2>0$.

Integrate the radial equilibrium equation across the thickness, retaining the integral rather than evaluating its antiderivative. The traction conditions give $t_r(b)-t_r(a)=P$. Hence $a$ must satisfyAll endpoints and coefficients here depend only on $a$ and the specified material and reference geometry. This is the required implicit pressure–radius relation for inflation and extension of an incompressible tube with unit axial stretch. In particular $a=A$ gives zero integrand and $P=0$. Once an admissible root is selected, the radial stress follows by integrating from the outer face, $t_r(r)=-{\int }_r^b(d{t}_r/ds)ds$, and the remaining stress components follow from the constitutive equation.