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Plane strain sets all strains involving $z$ to zero, including $e_{zz}=0$, while $e_{xx}=u_x$, $e_{yy}=v_y$ and $e_{xy}=(u_y+v_x)/2$. ThusIn particular, ${\tau }_{zz}$ generally does not vanish: the suppressed out-of-plane strain requires a reaction stress. Setting that entry to zero would instead be a plane-stress assumption.

The source uses the forward Fourier phase $e^{ikx}$, so $\widehat{{\partial }_{x}A}=-ik\hat{A}$. For a nonzero wavenumber put $q=|k|$. Biharmonicity becomes $({\partial }_y^2-q^2{)}^2\hat{A}=0$. The decaying solutions are $(C+Dy)e^{-qy}$. The normal-stress condition gives $-k^2\hat{A}(k,0)=-\hat{P}(k)$, hence $C=\hat{P}/k^2$. The shear condition gives ${\hat{A}}_y(k,0)=0$, so $D=qC$. ConsequentlyIts second derivatives are $\widehat{A_{xx}}=-\hat{P}(1+qy)e^{-qy}$ and $\widehat{A_{yy}}=\hat{P}(qy-1)e^{-qy}$. The given strain/stress inversion therefore yieldsIntegrating in $y$ with the stated decaying-mode condition givesThis is the Fourier solution of a loaded elastic half-space.

The zero wavenumber needs separate treatment: these displayed divisions by $k^2$ and $|k|$ are not formulas at $k=0$. In particular a nonzero resultant $\hat{P}(0)=\int Pdx$ is incompatible with an ordinary decaying zero-mode displacement of the stipulated kind; two-dimensional net loading produces a logarithmic displacement reference issue. Balanced loads or an appropriate distributional/reference-displacement formulation are needed for a global transform solution. The nonzero modes and the displacement-gradient kernel below remain the correct formal calculation; the singular zero mode must not be silently assigned a finite value.

The Airy stress function determines the in-plane stress through its second derivatives. If two functions give the same stress, their difference $B$ has $B_{xx}=B_{xy}=B_{yy}=0$. On a connected domain this implies that its gradient is constant. Thereforeis exactly the local affine gauge freedom. It also preserves biharmonicity. An imposed decay condition for $A$ can fix this freedom; adding an arbitrary biharmonic function would usually change the physical stress and is not a gauge transformation.

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.

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.

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.

The multiplier enforcing elastic incompressibility contributes an arbitrary hydrostatic stress. Denote it by $q(r)$ to distinguish it from the applied pressure $P$. For the neo-Hookean solid, the Cauchy stress tensor is $\tau =-qI+\mu F{F}^T$. Thus its principal components areThe axial component is $t_z=-q+\mu$, despite the unit axial stretch. The quantity $q$ is a Lagrange multiplier determined together with equilibrium and the traction conditions; it is not obtained by differentiating the reduced stretch energy while ignoring its constraint.

In particular the stress difference eliminates this multiplier:

The radial equilibrium equation gives $d{t}_r/dr=(t_{\theta }-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the neo-Hookean solid stress difference eliminates both $q$ and $t_{\theta }$:This is a first-order ordinary differential equation for the radial stress, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner pressure to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\ge A^2>0$.

Integrate the radial equilibrium equation across the thickness, retaining the integral rather than evaluating its antiderivative. The traction conditions give $t_r(b)-t_r(a)=P$. Hence $a$ must satisfyAll endpoints and coefficients here depend only on $a$ and the specified material and reference geometry. This is the required implicit pressure–radius relation for inflation and extension of an incompressible tube with unit axial stretch. In particular $a=A$ gives zero integrand and $P=0$. Once an admissible root is selected, the radial stress follows by integrating from the outer face, $t_r(r)=-{\int }_r^b(d{t}_r/ds)ds$, and the remaining stress components follow from the constitutive equation.

The material occupies $Y\ge 0$, so its outward unit normal at the surface is $n=-e_Y$. A free surface has zero traction, $\tau n=0$. ThereforeThe third condition is automatic for the two-dimensional displacement field with zero third component and no dependence on the third coordinate. The first two impose zero shear and normal stress at the free surface.

In the completely elastic exterior, radial elastic compatibility makes $q+2t$ constant. The far-field stress vanishes, so $q+2t=0$ and $t=-q/2$. Radial force balance then gives $q^{\prime }+3q/r=0$, whence $q=C/r^3$. The pressure boundary condition is $q(a)=-P$: positive applied pressure is compressive. ThereforeFor completeness, the corresponding radial displacement field is $u=P{a}^3/(4\mu r^2)$, which gives these components through Hooke's law and decays at infinity.

The Tresca yield criterion becomesFor $P\ge 0$ this is largest at the cavity wall and decreases strictly with radius when $P>0$. The first yield point is therefore at $r=a$, where $3P/4={\tau }_Y$. HenceThe normalization matters: ${\tau }_Y$ in this paper is a shear yield stress, rather than the uniaxial yield strength.

During increasing pressure, the elastic solution reaches the Tresca yield criterion with $t-q>0$. In the plastic shell of a material with perfect plasticity, this sign is retained andRadial force balance becomes $q^{\prime }-4{\tau }_Y/r=0$. Integrating from the cavity wall, where $q(a)=-P$, givesOutside the plastic shell, radial elastic compatibility and vanishing far-field stress again give $q=-C/r^3$ and $t=C/(2r^3)$. At the advancing plastic front $r=s$, the Tresca yield criterion gives $3C/(4s^3)={\tau }_Y$, so $C=P_c{s}^3$. Continuity of radial traction now givesThusThe complete stress field of this spherical elastoplastic cavity isAt $s$, the tangential stress also matches because $-P_c+2{\tau }_Y=P_c/2$. Outside $s$, the function in the Tresca yield criterion is ${\tau }_Y(s/r{)}^3<{\tau }_Y$, which verifies that the remaining material is elastic. As $P\downarrow P_c$, $s\downarrow a$ and the solution joins the entirely elastic regime. The radius $a$ is the fixed reference radius of the small-strain model used by the paper.

LetAt maximum pressure, the stress is the field from part (c) with $P=P_m$ and $s=s_m$. Under the assumed elastic unloading, previously accumulated plastic strain is frozen. Therefore it is the stress increment, rather than the total stress, that satisfies homogeneous linear elasticity and radial elastic compatibility. Applying the superposition principle, a pressure change $-\Delta$ produces the elastic incrementsThey have zero far-field stress and change the wall traction from $-P_m$ to $-P$. These increments are the same in the formerly plastic and formerly elastic regions because their elastic Lamé parameters agree. Adding them to the maximum-load field givesandThese formulae retain $s_m$: it records the plastic loading history and is not the instantaneous loading-front formula with $P$ substituted for $P_m$. If the entire return is elastic, setting $\Delta =P_m$ gives the residual stress after removal of the applied pressure.

To check admissibility, examine the signed difference of the principal stresses in the formerly plastic shell:It immediately moves into the interior of the Tresca yield criterion when unloading starts. It can next reach the opposite limit $D=+2{\tau }_Y$. For fixed $\Delta >0$, $D$ is largest at $r=a$ and stays above $-2{\tau }_Y$. At the wall,There is no earlier violation in the outer region. Indeed,so its magnitude is no larger than $|D(s_m)|$, which has already been controlled by the shell calculation. Hence the first reverse yielding occurs at the cavity wall, withIf $P_c<P_m<2P_c$, the stress stays within the yield surface throughout unloading to zero. If $P_m=2P_c$, it reaches reverse yield exactly at $P=0$; further unloading would produce reverse plastic deformation. If $P_m>2P_c$, reverse plastic deformation starts while the applied pressure is still positive. ThusIf “yield again while unloaded” includes first touching the yield surface at the final endpoint, the corresponding condition is $P_m\ge 8{\tau }_Y/3$. The displayed elastic unloading fields are physically valid only for $0\le \Delta \le \min (P_m,2P_c)$; beyond that interval a new reverse-plastic region must be solved for. Continuing the elastic formula to $P=0$ when $P_m>2P_c$ would violate the given Tresca yield criterion.

Let $H$ be the typical thickness, ${\ell }_r$ the radial variation length, $U$ the horizontal speed, and $W$ the vertical speed. For a thin smooth profile, $\epsilon =H/{\ell }_r\ll 1$. The axisymmetric continuity equation isAway from singular boundary regions, balancing its terms gives $U/{\ell }_r\sim W/H$, hence $\boxed{W/U\sim \epsilon \ll 1}$. Near $r=0$, regularity gives $v_r=O(r)$ and the same conclusion follows without a divergent $v_r/r$ term.

Vertical viscous derivatives dominate horizontal ones by ${\epsilon }^{-2}$. The ratio of convective acceleration to vertical viscous acceleration isFor thickness-evolution time $T$, the local inertial ratio is $H^2/(\nu T)$. When both ratios are small, lubrication theory reduces the Navier-Stokes equation toThinness must therefore accompany a slow enough motion; it is not a justification for dropping arbitrarily rapid imposed acceleration. The small vertical velocity also makes the leading vertical balance hydrostatic balance, $p_z=-\rho g$. The neglected normal viscous stress at the interface is smaller than the characteristic lubrication pressure by order ${\epsilon }^2$.

Use the source's curvature sign: $p_{\mathrm{air}}-p_{\mathrm{liquid}}=-\gamma \kappa$. The Young–Laplace equation and atmospheric pressure $p_0$ thus give $p(r,h,t)=p_0+\gamma \kappa$. Integrating the hydrostatic balance yieldsConsequently $p_r=\gamma {\kappa }_r+\rho g{h}_r$, independent of $z$ at this order. The radial Stokes equation and boundary conditions areThe first is the no-slip boundary condition; the second is the leading stress-free boundary condition, since the atmosphere has negligible dynamic viscosity and the surface tension is constant. Integrating twice givesThis is the horizontal velocity only. A smaller vertical velocity is generally required by the continuity equation.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

The materials interpretation is a cooling-rate gradient in a near-eutectoid steel. Before transformation, the parent thermodynamic phase is austenite. A small section can cool rapidly enough to suppress diffusional decomposition and form martensite, a supersaturated, highly strained transformation product. A much thicker interior cools more slowly and can transform diffusively to pearlite: alternating ferrite and cementite lamellae grouped into colonies. Bainite or mixed products are also possible for intermediate histories.

The likely contrast is therefore a martensite-rich rapidly cooled region and a more pearlitic slowly cooled interior. Brittle fracture is promoted by the highly strained martensitic state, residual stresses and local stress concentration. A larger intact interior need not share the same microstructure or local loading. The observations are consistent with this contrast, not a measurement proving pure thermodynamic phases everywhere.

The sketches below are generic steel microstructures, rather than a component design or fabrication prescription. No cooling curve, dimensions or transformation diagram is provided, so exact thermodynamic phase fractions, plate sizes and a universally pearlitic centre cannot be inferred.

For a slowly varying load with $\omega \tau \ll 1$, the dashpot stress is small relative to the spring stress. The strain approaches $\sigma /E$ almost in phase, so the low-frequency response is predominantly elastic. A constant load approaches the finite equilibrium strain ${\sigma }_0/E$; it does not sustain steady viscous flow in this parallel network.

On a timescale short compared with $\tau =\eta /E$, or for $\omega \tau \gg 1$, the dashpot strongly resists rapid deformation. The imposed oscillating stress produces a small strain, of amplitude approximately ${\sigma }_0/(\eta \omega )$, nearly90degrees behind the stress. The network appears stiff and predominantly viscous in this stress-controlled high-frequency limit. It is not an ideal perfectly rigid body at finite frequency.

The original diagram is a Kelvin-Voigt model: a spring and a dashpot act in parallel. The spring is an ideal elastic element, ${\sigma }_s=E{\epsilon }_s$; under its own sinusoidal stress its strain is in phase with stress and energy is stored reversibly. The dashpot is viscous, ${\sigma }_d=\eta {\dot{\epsilon }}_d$; its strain under an oscillating stress lags by90degrees and mechanical energy is dissipated.

In parallel the strain is common and the branch stresses add:For imposed harmonic stress, the strain amplitude and phase lag areThe short- and long-timescale limits follow under the printed roman labels below.

A sudden constant stress ${\sigma }_0$ gives delayed, bounded creep:The strain is initially zero because an instantaneous change would require an infinite dashpot stress. Initially the dashpot carries the load, with finite strain rate ${\sigma }_0/\eta$; the spring gradually carries more of it as the rate decays. On unloading, the strain recovers exponentially to zero. The network does not produce the instantaneous strain jump or unbounded constant-stress creep of a series Maxwell element.

A multi-anvil press transmits a large external load through several hard anvils onto a confined pressure medium surrounding the sample. Reducing the loaded area raises pressure, while the surrounding medium and gasket limit lateral escape. A resistive heater within the assembly can heat the sample, with insulation and temperature sensing arranged around it. It provides relatively large sample volumes and useful bulk, high-temperature experimental conditions, but access to the sample and optical paths is restricted, pressure gradients and calibration matter, and the maximum pressure is constrained by anvil strength and geometry.

A diamond anvil cell squeezes a very small sample and pressure medium in a gasket hole between opposed diamond culets. Diamond combines high strength with useful optical transparency, permitting spectroscopy and optical pressure calibration as well as suitable X-ray access. Heating can be external or laser-based. The very small contact area permits high pressure from a modest total force, but gives a tiny sample volume. Nonhydrostatic stress, pressure gradients, gasket deformation, heating gradients, chemical compatibility and diamond failure are important limitations.

Neither device generates pressure by force alone independent of area and confinement. Nor does a nominal force-to-area ratio guarantee a known hydrostatic sample pressure; experimental pressure and temperature must be calibrated. The two assemblies trade sample volume and access against pressure reach rather than one being universally preferable.