Past exam of the elasticity and plasticity course of the University of Oxford / 2023 / 1

For antiplane shear, the displacement field is $\mathbf{u}=(0,0,w(x,y,t))$, independent of $z$. Its divergence is zero. Consequently the only possibly nonzero components of the stress tensor are the shear stressesThe $x$ and $y$ components of the Navier-Cauchy equation vanish identically. Its $z$ component gives $\rho w_{tt}={\partial }_x(\mu w_x)+{\partial }_y(\mu w_y)$. Each material has constant shear modulus and mass density, so the required wave equation and wave speeds are

At a perfectly bonded elastic interface, both the displacement field and the traction are continuous. Using the same normal ${\mathbf{e}}_y$ on both sides, the only nonzero traction component is ${\tau }_{zy}$. Thus the interface boundary conditions areContinuity of “stress” here means continuity of traction; it does not require equality of the full stress tensors. The rigid plate supplies the remaining boundary condition $w_1(x,0,t)=0$.

Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when $c<c_2$. Withits wave amplitude is $f_2(y)=D{e}^{-\ell (y-h)}$. At $c=c_2$ the exterior wave amplitude is affine and cannot decay; at $c>c_2$ it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both $f_1(h)=0$ and $f_1^{\prime }(h)=0$, forcing the entire normal mode to vanish.

We also need $c>c_1$. Indeed, if $c<c_1$, the rigid boundary condition gives $f_1(y)=A\sinh (qy)$ with $q=k\sqrt{1-c^2/c_1^2}>0$. At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At $c=c_1$, the layer wave amplitude is $Ay$ and the same contradiction follows from $f_1^{\prime }(h)/f_1(h)=1/h>0$.

Therefore $c_1<c<c_2$. Define the positive transverse wavenumberThe rigid boundary condition now gives $f_1(y)=A\sin (my)$. The interface boundary conditions giveNeither $\sin (mh)$ nor $\cos (mh)$ can vanish for a nontrivial normal mode with $m,\ell >0$. Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive $m,\ell$ satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.

Eliminating the phase velocity from the first two relations yieldsWith $\theta =mh$, putThen $\ell =(c_1/(c_{2}h))\sqrt{a^2{k}^2-{\theta }^2}$. The dispersion relation becomesHere $a>0$ follows from $c_1<c_2$. The parameter $a$ has dimensions of length; $A$ and $ϵ$ are dimensionless.

For the strict mode count for a rigid-backed elastic layer, considerThe right side is negative, so intersections occur only on the negative branches of the tangent,On each such branch $\tan \theta -g(\theta )$ is strictly increasing. If $p_n<A$, its limit at $p_n^{+}$ is negative infinity. At the other end of the available interval, it is positive: either $\theta \to n{\pi }^{-}$ with $n\pi <A$, or $\theta \to A^{-}$ with $A\le n\pi$. The intermediate value theorem and strict monotonicity give exactly one intersection. No such interval exists when $p_n\ge A$.

Thus the exact answer, including all cutoff values, isAway from $A=(n-1/2)\pi$, this equals the printed floor function expression $\lfloor A/\pi +1/2\rfloor$. At $A=(n-1/2)\pi$, the printed expression is one too large: the $n$th interval is empty and the limiting state has $\ell =0$, contradicting exponential decay. This exception is present in the native paper, rather than introduced by transcription.

Now hold $A$ fixed and let $ϵ\to 0^{+}$. Neglecting the right side gives ${\theta }_n\simeq n\pi$ and hence the candidate phase velocitiesThe zero $\theta =0$ is excluded: it would require $c=c_1$, already ruled out. For every $n\pi <A$, the candidate is a valid leading asymptotic expansion. A first correction, obtained by expanding the tangent about $n\pi$, isThe big O notation here assumes a fixed positive gap between $n\pi$ and $A$. In particular, when $A>N\pi$, all $N$ leading approximations are valid. The phase velocity increases strictly with $\theta$, so only the last branch can cause an endpoint difficulty.

If $(N-1/2)\pi <A<N\pi$, the formal last equation root $N\pi$ lies outside the allowed interval. The actual last equation root approaches $A$, where the small factor $ϵ$ is balanced by the vanishing denominator. This is a singular perturbation. For fixed $A$ strictly inside that interval, write ${\theta }_N=A-\delta$. Since $\tan A<0$,The valid endpoint approximation for a rigid-backed shear mode consequently givesHere we used $c^2=c_1^2(1+{\theta }^2/(kh{)}^2)$ and $A^2/(kh{)}^2=c_2^2/c_1^2-1$. Thus this normal mode approaches the upper wave speed from below; it does not approach the inadmissible value obtained from $\theta =N\pi >A$.

There is a further exceptional fixed value $A=N\pi$. Put ${\theta }_N=A-\delta$ again. Now $\tan (A-\delta )\sim -\delta$, so the balance changes toIn the last formula the bracket specifies the first nonzero correction. The zeroth-order candidate $c_N=c_2$ is correct as a limit but misses this correction and the positive exterior exponential decay rate. More generally, close to $A=N\pi$, let $\Delta =A-N\pi$ and ${\theta }_N=N\pi -\delta$. For $|\Delta |,\delta \ll 1$, a transition approximation valid on both sides isIt recovers the ${ϵ}^{2/3}$ balance at $\Delta =0$ and the ordinary expansion above the transition. The fixed-endpoint formula involving $\tan A$ should be used when $A$ is below, and bounded away from, $N\pi$.