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.

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 $\mathrm{div}\tau =0$.

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.

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.

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.

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

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.

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.

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.

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.

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.

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

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 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 $\mathbf{u}(\mathbf{x},t)=\mathbf{x}-\mathbf{X}(\mathbf{x},t)$.

Insert the two constitutive stresses into the radial component of the Cauchy momentum equation. After collecting terms, $(\lambda +2\mu )(u_r^{\prime \prime }+u_r^{\prime }/r-u_r/r^2)=0$, so $u_r=Ar+B/r$. Its stresses areThe far pressure gives $A=-p_{\infty }/[2(\lambda +\mu )]$. At the tunnel wall, compressive bracing imposes ${\tau }_{rr}(a)=-P$, so $B=(P-p_{\infty })a^2/(2\mu )$. HenceWith no shear and ${\tau }_{\theta \theta }=k{\tau }_{rr}$, divide the yield identity by the nonzero ${\tau }_{rr}^2$ to obtain $4k=(1+k{)}^2{\cos }^2\varphi$. Solving its quadratic givesTo locate first yield, the stress-circle center is the constant $-p_{\infty }$, while its radius is $|P-p_{\infty }|a^2/r^2$. The radius is maximal at $r=a$, so that is where it first meets the friction envelope as the bracing departs from the hydrostatic value. At the wall ${\tau }_{rr}=-P$ and ${\tau }_{\theta \theta }=P-2p_{\infty }$; hence $k=(2p_{\infty }-P)/P$ andFor lowering bracing from $P=p_{\infty }$, the hoop stress becomes more compressive, so $k=k_{+}>1$ and $\boxed{P_{\mathrm{crit}}=p_{\infty }(1-\sin \varphi )}$. Increasing bracing instead selects $k_{-}$ and reaches the other threshold $p_{\infty }(1+\sin \varphi )$. Stating the loading direction removes any ambiguity between the two reciprocal stress ratios. These conclusions describe first yield of the elastic field, not the later plastic-zone solution.

The Mohr circle has center $C=({\tau }_{xx}+{\tau }_{yy})/2$ on the normal-stress axis and radius $R=\sqrt{({\tau }_{xx}-{\tau }_{yy}{)}^2/4+{\tau }_{xy}^2}$. In the compressive regime $C<0$, yield occurs when the circle first touches either friction line $F=\pm (-N)\tan \varphi$. The perpendicular distance of its center from either line is $|C|\sin \varphi$. Tangency therefore gives $R^2=C^2{\sin }^2\varphi$, or $C^2-R^2=C^2{\cos }^2\varphi$. But $C^2-R^2={\tau }_{xx}{\tau }_{yy}-{\tau }_{xy}^2$. ThusThe compressive sign restriction remains part of this equivalence: squaring alone would also produce algebraic branches that do not represent cohesionless compressive friction.

For dry friction, the tangential contact force opposes sliding and has limiting magnitude ${\mu }_f{N}_c$, where $N_c\ge 0$ is the compressive normal force and ${\mu }_f=\tan \varphi$. Sticking requires $|F|\le {\mu }_f{N}_c$; impending sliding reaches equality.

The cohesionless Mohr-Coulomb failure criterion applies the same local condition on internal planes: yield begins when some plane has $|{\tau }_{\mathrm{shear}}|=N_c\tan \varphi$. In the tension-positive Cauchy convention used here, $N_c=-N$ for a compressive plane, soA cohesive material would add a positive intercept, but the later formulas in this paper are explicitly the zero-cohesion case. The model does not supply a tensile strength law.