Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 76 (Nonlinear Continuum Mechanics)

Let $F$ be an invertible, orientation-preserving deformation gradient, with $J=\det F>0$. To prove Nanson's formula, take two reference surface tangent vectors $a,b$. Their current images are $Fa,Fb$, and their oriented area vectors are proportional to $a\times b$ and $(Fa)\times (Fb)$. For every vector $c$, the scalar triple product givesThus $F^T[(Fa)\times (Fb)]=J(a\times b)$, andThis proof also identifies the area transformation as the cofactor of $F$, rather than a transformation by $F$ itself.

Use the paper's material-index-first nominal stress tensor $P_{Ii}$. Equality of force on the same material surface in its two configurations gives $P_{Ii}d{S}_{0I}={\sigma }_{ji}d{S}_j$. Substitute the area transformation to obtainHere $\tau$ is the Kirchhoff stress tensor. With symmetric Cauchy stress tensor, the matrix forms are $P=J{F}^{-1}\sigma$ and $\tau =FP$. This $P$ is the transpose of the first Piola-Kirchhoff stress tensor; keeping the index convention prevents a spurious transpose later.

For the velocity gradient $L=\dot{F}{F}^{-1}$, let $D=(L+L^T)/2$ be the rate-of-strain tensor. Then the reference-volume mechanical power isThe last step uses stress symmetry, as supplied by angular-momentum balance in an ordinary continuum without couple stresses: a symmetric tensor has zero contraction with the skew part of $L$.

By work-conjugate stress and strain, a pair $(T,E)$ must satisfy $T:\dot{E}=\tau :D$ for every deformation rate, with both sides measured per reference volume. For the Green-Lagrange strain tensor, set $C=F^{T}F$ and calculateCyclically rearranging the contraction shows $T^{(2)}:{\dot{E}}^{(2)}=(F{T}^{(2)}{F}^T):D$. Hence the symmetric conjugate isthe second Piola-Kirchhoff stress tensor.

The inverse right Cauchy-Green strain is $E^{(-2)}=(I-C^{-1})/2$, because $C^{-1}=F^{-1}{F}^{-T}$. Differentiating $C{C}^{-1}=I$ givesConsequently $T^{(-2)}:{\dot{E}}^{(-2)}=(F^{-T}{T}^{(-2)}{F}^{-1}):D$, andThe two stress measures have a geometric interpretation in the convected coordinate net. Its basis vectors are $g_I=F{e}_I$ and its reciprocal basis is $g^I=F^{-T}{e}_I$. The entries of $T^{(2)}$ are the contravariant components in $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$, whereas $T_{IJ}^{(-2)}=g_I\cdot \tau g_J$ are the covariant components, equivalently $\tau =T_{IJ}^{(-2)}{g}^I\otimes g^J$. They describe the same spatial Kirchhoff stress tensor in a deforming, generally nonorthonormal coordinate net.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

Use reference cylindrical coordinates $(R,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The prescribed map has $r={\lambda }^{-1/2}R$, $\theta =\Theta +\alpha Z$ and $z=\lambda Z$. In the corresponding orthonormal cylindrical bases, differentiate physical line elements to obtain the deformation gradientThe radial direction is an eigenvector of the right Cauchy-Green deformation tensor with eigenvalue ${\lambda }^{-1}$. Its other block isIts trace is $s={\lambda }^{-1}(1+{\alpha }^2{R}^2)+{\lambda }^2$ and its determinant is $\lambda$. Thus its eigenvalues solve $z^2-sz+\lambda =0$. The principal stretches are the positive square roots of these eigenvalues:Their product is one, independently checking incompressibility.

The strain energy density $W$ is per reference volume, so stored energy per initial cylinder height is $\mathcal{E}(\alpha ,\lambda )=2\pi {\int }_0^{A}W({\lambda }_1,{\lambda }_2,{\lambda }_3)RdR$. The end rotation is $\alpha H$ and the current height is $\lambda H$, giving applied power per initial height $M\dot{\alpha }+N\dot{\lambda }$. For quasistatic reversible loading, equate this with $\dot{\mathcal{E}}$ for independent changes of $\alpha ,\lambda$. This proves the torque and axial force of a twisted incompressible cylinder formulasIn the second derivative hold $\alpha$, the twist per initial length, fixed. Holding twist per current length instead would mix the generalized forces.

For the Mooney-Rivlin solid, compute invariants directly from $F$ rather than differentiating individual eigenvalue square roots:The second identity follows from $I_2=\mathrm{tr}(F^{T}F{)}^{-1}$ when $J=1$. Retain the paper's minus sign on its ${\mu }_2$ term. Integrating $R$ and $R^3$ givesConsequentlyAs checks, untwisted unit stretch gives $M=N=0$, and setting ${\mu }_2=0$ recovers the neo-Hookean solid formulas. A convention with a positive second-invariant coefficient corresponds to replacing this ${\mu }_2$ by its negative.

For fixed nominal boundary tractions and dead body forces, use the total potentialThe dead-load terms are linear in the deformation map. Let $x(\epsilon )$ be a constraint-preserving perturbation path through equilibrium, with $H=\delta F$ and $K={\delta }^{2}F$. Differentiating $\varphi (F)=0$ twice gives ${\varphi }_{,F}:H=0$ and ${\varphi }_{,F}:K=-{\varphi }_{,FF}[H,H]$. Equilibrium has nominal stress tensor $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$. Its virtual-work relation, applied to ${\delta }^{2}x$, gives the otherwise remaining first-derivative contribution ${\Pi }^{\prime }[{\delta }^{2}x]=-\int q{\varphi }_{,F}:KdX$. Therefore the constrained second variation isFor a time-dependent linearized perturbation $y$, the incremental reaction stress contains $\delta q{\varphi }_{,F}$, but its power vanishes because ${\varphi }_{,F}:\nabla \dot{y}=0$. Multiply the incremental momentum equation by $\dot{y}$, integrate by parts, and use zero incremental dead-load traction or the prescribed displacement condition. One obtainsPositive $\mathcal{Q}$ is therefore the standard constrained second-variation stability in elasticity criterion: perturbations cannot grow in this quadratic energy norm. A statement in a specified displacement norm also uses the usual coercivity and boundary conditions removing uncontrolled rigid-motion modes; translations with $\delta F=0$ are not tested by the displayed condition. This is a small-perturbation energy argument, not a claim of global nonlinear stability.

For the cube use incompressibility in the form $\varphi =\det F-1$. A neo-Hookean solid has $W_{,F}=\mu F$, and hence the spatial-index-first transpose of nominal stress is $\mu F+q{F}^{-T}$. On the diagonal family, equality to $TI$ givesThe first relation immediately gives $\boxed{q=T\lambda -\mu {\lambda }^2}$. Eliminating $q$ and factoring yieldsThus $\boxed{\lambda =1\text{or}T/\mu =\lambda +{\lambda }^{-1/2}}$.

Let $t=T/\mu$ and $h(\lambda )=\lambda +{\lambda }^{-1/2}$. Its derivative $h^{\prime }=1-\frac{1}{2}{\lambda }^{-3/2}$ vanishes only at ${\lambda }_c=2^{-2/3}$, where $h^{\prime \prime }>0$ and $h_c=32^{-2/3}$. Also $h\to \infty$ at both ends of $(0,\infty )$. Therefore $h(\lambda )=t$ has two distinct positive roots for $t>h_c$, one repeated root at $h_c$, and none below. There is an exception to the printed assertion that both roots differ from one: at $t=2$, one root is exactly $\lambda =1$. Indeed, with $x=\sqrt{\lambda }$, the equation becomes $(x-1)(x^2+x-1)=0$, so the only other positive stretch is $\lambda =(3-\sqrt{5})/2$.

Evaluate the stability form directly on the prescribed uniform perturbationIt satisfies $\mathrm{tr}(F^{-1}H)=0$. The determinant Hessian identity isMeanwhile $W_{,FF}[H,H]=\mu (1+\frac{1}{2}{\lambda }^{-3})(\delta \lambda {)}^2$. For the unit reference cube,On the trivial branch, $q=T-\mu$ andThus the undeformed state is stable in this perturbation family for $t<2$, unstable for $t>2$, and quadratically neutral at $t=2$.

On a nontrivial branch, substitute $t=h(\lambda )$ and put $z={\lambda }^{-3/2}$:The smaller root always has $\lambda <{\lambda }_c$, hence $z>2$, and is stable. For $h_c<t<2$, the larger root lies between ${\lambda }_c$ and one, so $1<z<2$ and it is unstable; the trivial branch is the second stable equilibrium. For $t>2$, the larger root lies above one and has $0<z<1$, so it is stable; the trivial branch is now unstable. Whenever there are three distinct equilibria in this family, precisely two are stable and one unstable against the specified uniform perturbations. This does not establish stability to every nonuniform or symmetry-breaking perturbation.

The homogeneous tensile bifurcation of a neo-Hookean cube has a nontrivial-branch fold at $t=32^{-2/3}$ and a meeting with the undeformed branch at $\boxed{t=2}$. The latter is the bifurcation from the undeformed state. Stating both values distinguishes this branch intersection from the earlier creation of the two nontrivial solutions.

Differentiate the pulled-back Kirchhoff stress tensor $T=F^T\tau F$, using the velocity gradient convention $L=\dot{F}{F}^{-1}$:Thus the lower-convected stress pull-back givesFor an explicit objective time derivative check, superpose $x^{\ast }=Q(t)x+c(t)$, with $Q^{T}Q=I$ and $\Omega =\dot{Q}{Q}^T$ skew. Then ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\Omega$, whereasThe additional terms in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$ are $-\Omega {\tau }^{\ast }+{\tau }^{\ast }\Omega$. They cancel the extra material-derivative terms, proving $\boxed{{\mathcal{D}}_l{\tau }^{\ast }=Q({\mathcal{D}}_l\tau )Q^T}$. The translation has no effect on a stress or velocity gradient.

For the incompressible Oldroyd-A model, avoid using the same symbol for a stress tensor and its relaxation time by referring to $\tau >0$ below only as the scalar relaxation time. Define $S={\sigma }^d-2{\mu }_{r}D$. The constitutive equation becomesBy the pull-back identity, $A=F^{T}SF$ satisfiesAn integrating factor givesPush forward by $F(t{)}^{-T}$ and $F(t{)}^{-1}$. For a convergent, fading remote-past history with the transported homogeneous term tending to zero as $t_0\to -\infty$, this proves the fading-memory representation of the Oldroyd-A model:For arbitrary finite-time initial data retain $e^{-(t-t_0)/\tau }F(t{)}^{-T}A(t_0)F(t{)}^{-1}$. Thus the printed integral-only expression incorporates an initial-memory condition; it is not an unconditional replacement for all initial-value solutions.

For time-dependent simple shear,Put $\Delta \gamma =\gamma (s)-\gamma (t)$. The relative deformation is $F(s)F(t{)}^{-1}=\left(\begin{array}{cc}1& \Delta \gamma \\ 0& 1\end{array}\right)$, so the entire transported integrand simplifies toConsequently the explicit nonpressure stresses areSymmetry gives ${\sigma }_{21}^d={\sigma }_{12}^d$, and the unshown third-direction nonpressure components vanish. The superscript $d$ here labels the paper's nonpressure constitutive stress; it need not have zero trace. Subtracting $pI$ supplies the total Cauchy stress tensor.

For constant shear rate $s_0$, use $\gamma (s)-\gamma (t)=-s_0(t-s)$ and the integrals ${\int }_0^{\infty }{e}^{-a/\tau }da=\tau$ and ${\int }_0^{\infty }a{e}^{-a/\tau }da={\tau }^2$. This gives the steady simple shear of an Oldroyd-A fluid resultand thereforeThe incompressibility pressure cancels from this normal-stress difference. With the usual $\mu \ge {\mu }_r\ge 0$, it is nonnegative, while the second normal-stress difference equals its negative.

In infinitesimal elasticity and plasticity in continuum mechanics, decompose the symmetric strain as $e=e^e+e^p$. For incompressible isotropic elasticity, ${\sigma }^{\prime }=2G{e}^e$, with $G$ the shear modulus and both strains traceless. The associated flow rule for the Von Mises yield criterion has ${\dot{e}}^p=\Lambda {\sigma }^{\prime }$ after absorbing a positive factor into $\Lambda \ge 0$. Plane strain has $e_{33}=0$, soIf ${\sigma }_{33}^{\prime }=0$ initially, it remains zero by this homogeneous scalar differential equation. Then $e_{33}^e=0$ and ${\dot{e}}_{33}^p=0$, so $e_{33}^p=0$ is consistent with zero initial plastic component. This proves consistency, not a claim that arbitrary prestressed plane-strain histories must have zero separate components. The out-of-plane shear components can consistently be zero by the same argument.

Write $p=-\mathrm{tr}\sigma /3$, so ${\sigma }_{33}=-p$. The proposed in-plane parameterization hasIts squared contraction is $2k^2({\sin }^{2}2\varphi +{\cos }^{2}2\varphi )=2k^2$, proving the yield condition identically. Its principal deviatoric stresses are $k,-k,0$. Define the two slip line families by tangent unit vectorsThus $\alpha$ lines have tangent angle $\varphi$ and $\beta$ lines have tangent angle $\varphi +\pi /2$. Their normal extensions from the associated plastic strain rate vanish, while their shear stresses have maximum magnitude $k$; they are not the principal stress directions.

In the absence of body force, differentiate the stress parameterization in the two equilibrium equations. One obtainsThe symmetric matrix multiplying $2k\nabla \varphi$ has eigenvectors $a,b$ with eigenvalues $+1,-1$. Taking its contraction with these vectors provesHence the Hencky stress relations areConstants can differ between lines, and these signs depend on the specified negative shear-stress convention.

Place the notch tip at the origin, with the opening towards negative $x_1$ and tensile loading in the transverse direction. For a usual reentrant notch, $0\le \gamma \le \pi /2$. The upper face has tangent angle ${\vartheta }_f=\pi -\gamma$. Choose its tangent $t=(\cos {\vartheta }_f,\sin {\vartheta }_f)$ directed away from the tip and outward material normal $n=(-\sin {\vartheta }_f,\cos {\vartheta }_f)$ pointing into the notch. Direct rotation of the stress gives the normal and tangential traction componentsTake the notch faces traction-free, as intended by the notch problem. Then the cosine vanishes. Select the tensile branch $\sin (2\varphi -2{\vartheta }_f)=1$, for which the face-tangential normal stress is $2k$. Thus the upper face has $\boxed{p=-k,{\varphi }_f=\pi /4-\gamma (\mathrm{mod}\pi )}$. Choosing the opposite branch would describe compression rather than the tensile field.

Construct the centred slip-line fan at a traction-free V-notch to connect this surface state to the forward symmetry axis. Near the upper face there is a constant-stress region. It joins a fan bounded by raysInside the fan choose $\beta$ lines radial and $\alpha$ lines circular about the tip, so their angle parameter is $\varphi =\vartheta -\pi /2$. Along a circular $\alpha$ line the first invariant givesThe second invariant is automatically constant along each radial $\beta$ line because $p,\varphi$ are independent of radius. Thus this is an actual equilibrium fan, not just an assumed transport of a surface value. At ${\vartheta }_2$ it matches the free-face constant state. At ${\vartheta }_1$ it matches a forward constant-stress region withReflect the upper field across the symmetry axis to obtain the lower field. The forward region has zero shear, respects this symmetry, and contains the positive $x_1$ axis. There $\sin 2\varphi =-1$, and henceThis value holds in the fully plastic notch-tip region described by the field. It is not a uniform stress assertion for the entire axis of an arbitrary finite specimen, including any elastic or far-field region.