Past exam of the elasticity and plasticity course of the University of Oxford / 2010 (C6.2b, DMAC 2467, DMSU 2467)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

The displacement follows a material point from its reference position to its current position:The deformation gradient is the Jacobian matrix $F=\partial \mathbf{x}/\partial \mathbf{X}=I+{\nabla }_X\mathbf{u}$, and the volume Jacobian isFor an orientation-preserving physical deformation, $J>0$ measures the local volume expansion factor. Incompressibility requires $J=1$; conservation of mass gives $\rho (\mathbf{x},t)J={\rho }_0(\mathbf{X})$. For small displacement gradients, $J=1+{\nabla }_X\cdot \mathbf{u}+O(|{\nabla }_X\mathbf{u}{|}^2)$. If a spatial representation is desired, the same displacement is $\mathbf{u}(\mathbf{x},t)=\mathbf{x}-\mathbf{X}(\mathbf{x},t)$.

A Cartesian Cartesian second-rank tensor has components transforming under a constant orthogonal change of axes asHere ${\mathbf{X}}^{\prime }=P\mathbf{X}$ and ${\mathbf{u}}^{\prime }=P\mathbf{u}$. By the chain rule and orthogonality,Thus the displacement gradient is a Cartesian second-rank tensor. Its symmetric part transforms in the same way:Hence the infinitesimal strain tensor is also second-rank. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-deformation frame indifference of the linearized strain formula.

The traction on the oriented solid element isWith constant Lamé coefficients and derivatives in the reference variables,Substitution into balance of linear momentum givesTo check the vector identity componentwise, contract the two alternating tensors:Rearranging proves ${\nabla }^2\mathbf{u}=\nabla (\nabla \cdot \mathbf{u})-\nabla \times (\nabla \times \mathbf{u})$. Therefore

Assume $\mathbf{k}\ne 0$. Dotting the decomposition with $\mathbf{k}$ givesCrossing it on the left by $\mathbf{k}$ and using $\mathbf{B}\cdot \mathbf{k}=0$ gives $\mathbf{k}\times \mathbf{a}=|\mathbf{k}{|}^2\mathbf{B}$. ThusIndeed $(\mathbf{k}\times \mathbf{a})\times \mathbf{k}/|\mathbf{k}{|}^2=\mathbf{a}-A\mathbf{k}$, verifying the decomposition and the imposed transverse condition.

Apply the wave ansatz to the oscillatory displacement about a static body-force balance, or equivalently to the zero-body-force homogeneous equation. The parallel and perpendicular polarizations have speedsThe first is longitudinal and the second transverse. A wave family is dispersive when its phase speed $\omega /|\mathbf{k}|$ varies with wavenumber. In each of these homogeneous bulk families, $\omega =c|\mathbf{k}|$, so both are nondispersive and group and phase speeds coincide.

The positivity assumptions imply $\lambda >-2\mu /3$, hence $\lambda +\mu >\mu /3>0$. Therefore $c_p^2-c_s^2=(\lambda +\mu )/\rho >0$: longitudinal waves travel faster.

Substitution of $w_i=f_i(y)e^{i(kx-\omega t)}$ givesThe oscillatory seam solution requires $m^2={\omega }^2/c_{s1}^2-k^2$, givingFor an exponentially decaying exterior, $f_2^{\prime \prime }={\ell }^2{f}_2$ with $\ell >0$, so the same differential equation instead requiresThe printed exterior plus sign is inconsistent with its decaying ansatz and must be a sign error. The guided mode has $c_{s1}<c_L<c_{s2}$, so a slower shear-wave seam than its cladding is necessary.

For an even seam mode $f_1=A\cos (my)$, upper-interface displacement and traction continuity giveThe lower interface gives the same equations when $B_2=A_2$. Eliminating the amplitudes yieldsFor an odd seam mode $f_1=B\sin (my)$, the equations areWith $B_2=-A_2$ the lower condition agrees, and elimination givesThese interface conditions together with the two corrected dispersion identities determine the guided-wave branches.

In the formal stiff-cladding limit with a nonvanishing exterior decay rate, the interface displacement tends to zero. The even seam mode then has $\cos (mh)=0$, whereas the odd mode has $\sin (mh)=0$. The positive wavenumbers areThe zero odd index would give an identically zero mode and is excluded. The smallest even transverse wavenumber is $\pi /(2h)$, so its lowest branch satisfiesThis is the symmetric-mode cutoff: the actual limiting branches have the discrete listed values of $m$. If the exterior shear speed remains finite, the separate decay requirement ${\omega }^2<c_{s2}^2{k}^2$ still applies; the cutoff alone does not replace that guided-wave condition.

Differentiate the displacement components. The shear strains involving $z$ vanish because $u_z=-w_x$ and $v_z=-w_y$. The full infinitesimal strain tensor isIn particular $\mathrm{tr}\mathcal{E}=(2\nu -1)(a+c)z/(1-\nu )$. Substituting the Lamé coefficients into the constitutive law givesFor example ${\tau }_{zz}=\lambda \mathrm{tr}\mathcal{E}+2\mu e_{zz}=0$: the thickness-direction strain supplies exactly the plane-stress Poisson correction. Also ${\tau }_{xy}=-Ebz/(1+\nu )$, confirming the off-diagonal factor.

Contract the tensors, counting both off-diagonal shear terms:Since ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$, integration givesFor the general midplane deflection, expand the energy density as $w_{xx}^2+2\nu w_{xx}{w}_{yy}+w_{yy}^2+2(1-\nu )w_{xy}^2$. A variation $w+\epsilon \eta$ givesFor clamped variations, $\eta ={\partial }_n\eta =0$ on the boundary, so integrating twice transfers all derivatives to $w$ without a boundary contribution. The bulk coefficient isStationarity for all compactly supported variations and the fundamental lemma therefore give

Use the same first variation on the half-plane, with no restrictions on the edge traces of $\eta$. Write $M=w_{xx}+\nu w_{yy}$. Two $x$ integrations of the $M{\eta }_{xx}$ term produce the boundary term $-M{\eta }_x+M_x\eta$ at $x=0$. The mixed term, after one integration in $x$ and one along the edge in $y$, contributes $2(1-\nu )w_{xyy}\eta$. All other boundary terms vanish at infinity. HenceThe bulk equation makes its area term zero. Stationarity therefore gives exactlyThe two edge traces can be prescribed independently: use $\eta (x,y)=\chi (x)[f(y)+xg(y)]$ with $\chi =1$ near zero and compactly supported smooth $f,g$. The fundamental lemma on the edge then givesThey are zero bending moment and zero effective transverse shear at a free straight edge. The second includes the edge derivative of the twisting moment; omitting the mixed contribution would give the wrong coefficient.