Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 72 (Theory of Elastic Solids)

Write $B_{i\alpha }={\dot{A}}_{i\alpha }$ and assume that the material constraint is regular, so $F_{,A}\ne 0$. Differentiating the constraint shows that admissible rates of the deformation gradient satisfy $F_{,A_{i\alpha }}{B}_{i\alpha }=0$. The given stress-power identity says that $W_{,A}-N^T$ annihilates this tangent hyperplane. Its orthogonal complement is the one-dimensional span of $F_{,A}$, so a Lagrange multiplier gives the constraint reaction in hyperelastic stress:The sign of $q$ is a convention; the virtual work identity determines no constitutive value for this reaction. Regularity matters: a singular equation describing the same constraint need not have a nonzero gradient.

For incompressibility, take $F=\det A-1$. Differentiating the expression using the Levi-Civita symbol givesThe middle expression is the corresponding entry of the cofactor matrix. Since $\det A=1$, the nominal stress is thereforeFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$, so in the material-first convention $N=\mu A^T+q{A}^{-1}$. The identity matrix satisfies the prescribed dead loading with $q=T-\mu$, proving that the undeformed cube is always an equilibrium.

To obtain the homogeneous tensile bifurcation of a neo-Hookean cube, first seek $A=\mathrm{diag}(\lambda ,\lambda ,{\lambda }^{-2})$, with $\lambda >0$. This has unit determinant. The two distinct diagonal stress equations are $\mu \lambda +q/\lambda =T$ and $\mu {\lambda }^{-2}+q{\lambda }^2=T$. Eliminating $q$ givesThus the nontrivial branch isThe function $h(\lambda )=\lambda +{\lambda }^{-2}$ decreases on $(0,2^{1/3})$ and increases on $(2^{1/3},\infty )$, because $h^{\prime }=1-2{\lambda }^{-3}$. Its minimum is $h(2^{1/3})=3/2^{2/3}$. Above this threshold there are two positive roots ${\lambda }_{-}<2^{1/3}<{\lambda }_{+}$. For $t\ne 2$, neither root is one, and the three choices of the exceptional coordinate for each root give six distinct diagonal deformation gradients, in addition to $I$.

At $t=2$, however, ${\lambda }_{-}=1$ and the three smaller-root matrices coincide with $I$: it would be incorrect to count those as three new deformations. The requested existence of seven distinct homogeneous deformations nevertheless holds even at this value. For any unit vector $n$, putIts positive eigenvalues are ${\lambda }_{+},{\lambda }_{+},{\lambda }_{+}^{-2}$, and its determinant is one. Since $A(n)$ is an orthogonal conjugate of the diagonal solution, the same $q=\mu /{\lambda }_{+}$ gives $\mu A(n{)}^T+qA(n{)}^{-1}=TI$. Choose the six directionsTheir outer products are distinct, and ${\lambda }_{+}>2^{1/3}>1$, so the six $A(n)$ are distinct from one another and from $I$. Constant nominal stress satisfies equilibrium and the prescribed surface tractions. Hence seven distinct homogeneous equilibria exist whenever $T/\mu >3/2^{2/3}$. This is an existence count, not an exhaustive classification: rotated principal axes actually give a continuous family. The conventional seven axis-aligned positive-stretch branches coalesce to four distinct matrices at $T/\mu =2$.

Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

Work in the fixed reference domain ${\Omega }_0$, and write $v_i=\delta x_i$. For the unconstrained hyperelastic material considered here, the material-first nominal stress is $N_{\alpha i}=W_{,A_{i\alpha }}$. The increment of the deformation gradient and the incremental elastic moduli areAll coefficients are evaluated at the base deformation. The Hessian symmetry is $c_{\alpha i\beta j}=c_{\beta j\alpha i}$; minor interchange symmetry is unnecessary. Linearizing the momentum equation givesThe reference mass density and reference outward unit normal vector are fixed, so the incremental boundary conditions areIn particular, no variation of a current surface normal belongs in this material-coordinate traction condition. If a material constraint were imposed, one would additionally linearize its constraint reaction in hyperelastic stress and its admissibility equation; the present moduli formula is the unconstrained one.

For a uniform material and an affine static base deformation, the deformation gradient, incremental elastic moduli and reference mass density are constant. With zero incremental body force, the equation reduces to $c_{\alpha i\beta j}{v}_{j,\beta \alpha }={\rho }_0{\ddot{v}}_i$. Let $\nu$ be a unit material propagation direction, $m$ a constant polarization and $s=t-{\nu }_{\alpha }{\xi }_{\alpha }/c$. Direct differentiation of the plane wave givesThus every twice differentiable profile $f$ satisfies the incremental equation ifThis is the acoustic tensor eigenvalue problem. The acoustic tensor is a real symmetric matrix: interchange the index pairs in the incremental elastic moduli and then interchange the dummy material indices to obtain $Q_{ji}=Q_{ij}$. Its eigenvectors give the displacement polarizations. A positive eigenvalue gives a real nonzero speed $c$; arbitrary-profile propagation imposes precisely this algebraic condition. Boundary conditions determine which combinations of these local plane waves are allowed in a finite body.

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.

The fixed-displacement boundary value problem now has $u=0$ on the whole boundary. The elastic normal-mode energy identity still holds, because its boundary term contains $\overline{u}$. DefineFor real constant $K$, $B$ has major symmetry: both Levi-Civita symbols change sign upon exchanging the two index pairs. Its quadratic density is a quadratic minor null Lagrangian. Indeed,The boundary integral vanishes by the Dirichlet boundary condition, and the volume integral vanishes because mixed derivatives are symmetric in $\alpha ,\beta$ whereas ${ϵ}_{\alpha \beta \gamma }$ is antisymmetric. Constancy of $K$ is essential: otherwise its derivatives would produce extra terms. This calculation initially applies to smooth fields and extends to fields with zero Sobolev boundary trace by density.

It follows that the Rayleigh quotient numerator can be replaced by the quadratic form with $H$:The assumed positivity of $H$ on all nonzero real matrices also gives positivity on nonzero complex matrices by splitting real and imaginary parts. A nonzero $u$ satisfying $u=0$ on the boundary cannot have identically zero gradient. Hence the right side is strictly positive, andThis works even when the original moduli quadratic form is not pointwise positive on all matrices: its integrated energy on fixed-boundary gradients agrees with a positive density after adding the quadratic minor null Lagrangian.

At a fixed material point, freeze the incremental elastic moduli at their local values and fix a unit propagation direction $\nu$. For an arbitrary nonzero polarization $m$, the rank-one matrix $a_{i\alpha }=m_i{\nu }_{\alpha }$ annihilates the added quadratic minor null Lagrangian, sinceThe assumed positivity therefore givesThus the acoustic tensor is positive-definite; this is strong ellipticity in elasticity. Major symmetry also makes it a real symmetric matrix. By the spectral theorem for real symmetric matrices, it has three real positive eigenvalues $q_1,q_2,q_3$ and an orthonormal basis of real polarizations $m^{(1)},m^{(2)},m^{(3)}$. ThereforeEach polarization gives a real plane wave profile through the equation of part (a), in either sign of the propagation speed. Repeated eigenvalues still admit three independent polarizations; the claim is not that the three speeds must be distinct. For an inhomogeneous base deformation these are local, frozen-coefficient wave modes, rather than exact plane waves throughout the entire body.

For antiplane shear, the infinitesimal displacement is $(0,0,w(x_1,x_2))$. The isotropic linear elasticity constitutive law gives ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with the symmetric entries ${\sigma }_{31},{\sigma }_{32}$. All other stress tensor components vanish. Static force balance without body force becomes $\mu (w_{,11}+w_{,22})=0$, so $w$ satisfies the Laplace equation.

Locally choose a harmonic conjugate $\varphi$ by ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$. Compatibility follows from the Laplace equation. These are the Cauchy-Riemann equations for $F=\varphi +iw$, and henceThus the complex stress potential for antiplane shear supplies both stresses and, by integration, the displacement. A global potential must have the displacement periods appropriate to the cracked body; these periods also help fix homogeneous solutions.

For the additional crack-induced field, set $G=\mu F^{\prime }$ and $p(t)={\sigma }_{23}^A(t,0)$. Only $p$ loads the prospective faces, since ${\sigma }_{13}^A$ is tangential to their normal. Denote upper and lower boundary values by $G^{+},G^{-}$. The imposed cancellation of face tractions is $\mathrm{Re}{G}^{\pm }=-p$. The traction data and geometry are invariant under the displacement reflection $w(x_1,x_2)\mapsto -w(x_1,-x_2)$. For the induced correction with no independently added loading, uniqueness up to a rigid translation allows the odd displacement choice. Then $w_{,2}$ is even, $w_{,1}$ is odd, and $G^{-}=\overline{G^{+}}$ on the real axis. Therefore the Hilbert problem for an antiplane crack isThe second condition expresses continuation through uncracked material. There $G$ is real and the additional ${\sigma }_{13}$ vanishes. This scalar Riemann-Hilbert problem is supplemented by at most inverse-square-root stress singularities at the crack tips and the usual correction with no additional remote resultant or imposed dislocation. Without a far-field or period normalization, the face conditions alone permit homogeneous terms. The formulas below select this normalized induced field and explicitly identify the excluded terms.

For the finite crack use the branchcut on $[-a,a]$. For $-a<t<a$, its limiting values are ${\chi }^{+}(t)=i\sqrt{a^2-t^2}$ and ${\chi }^{-}(t)=-i\sqrt{a^2-t^2}$. Consequently $H=\chi G$ converts the sum condition of the Hilbert problem for an antiplane crack into an additive jump:By the Sokhotski–Plemelj theorem, the Cauchy integral with this jump isThus the finite antiplane crack Cauchy solution is $G=H/\chi$. For regular integrable loading, $H=O(z^{-1})$ and $G=O(z^{-2})$. A homogeneous term $C/\chi$ would instead produce a logarithmic potential. In the reflected real-loading class $C$ is real, and its $1/z$ term gives a nonzero displacement period around the crack. It is excluded for the induced, dislocation-free correction. A nonvanishing constant stress term would change the remote applied field and is likewise excluded.

For $x=x_1>a$, $\chi (x)=\sqrt{x^2-a^2}>0$, soThis is the additional stress, not the total stress; the applied field must still be added. The integral is real at these intact-axis points, so ${\sigma }_{13}(x,0)=0$. On the crack itself, the upper Cauchy principal value formula gives $H^{+}={\pi }^{-1}\mathrm{PV}\int \sqrt{a^2-t^2}p(t)/(x-t)dt-i\sqrt{a^2-x^2}p(x)$. Division by $i\sqrt{a^2-x^2}$ verifies $\mathrm{Re}{G}^{+}=-p$, checking the sign of the face traction cancellation.

For example, constant applied shear $p_0$ gives $H(z)=p_0(z-\chi (z))$ and $G(z)=p_0(z/\chi (z)-1)$. The total shear ahead of the crack is then $p_{0}x/\sqrt{x^2-a^2}$, providing a direct check of the normalization and the crack-tip singularity.

For the two semi-infinite cracks, choose insteadwith cuts on the two crack segments. Approaching from above givesThese opposite signs on the two cracks are essential. Multiplying the Hilbert problem for an antiplane crack by $\chi$ givesThe Sokhotski–Plemelj theorem therefore yields the antiplane ligament Cauchy solutionFor $-a<x=x_1<a$, $\chi (x)=\sqrt{a^2-x^2}$ and the kernel has no pole on either crack. HenceAgain this is the additional stress, and ${\sigma }_{13}=0$ on the intact ligament. The integrals have the Cauchy principal value interpretation wherever required, and their convergence at infinity is assumed. For a point on the right crack, the upper value of $H$ has imaginary part $+\sqrt{t^2-a^2}p(t)$; division by $-i\sqrt{t^2-a^2}$ gives real part $-p(t)$. On the left crack both signs reverse and the same cancellation follows.

The homogeneous ambiguity here is physically different from the finite-crack displacement period. A real termhas zero real traction on the crack faces but adds ligament shear with resultantIts potential is $(C/\mu )\arcsin (z/a)$, producing logarithmic displacement growth at infinity. Thus it represents a separately imposed load transfer between the two sides, even though its stress tends to zero pointwise at infinity. The induced correction with no added remote resultant selects $C=0$, giving the boxed formula. Merely requiring pointwise decay of stress would not fix this constant; the far-field load normalization is needed.