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The two Lamé parameters $\lambda$ and $\mu$ determine the linear isotropic relation between stress and strain; $\mu$ is the shear modulus.

For inner and outer radii $a,b$, with internal and external pressures $p_1,p_0$, the boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. The radial equilibrium of a thick elastic cylinder gives $P=(p_1{a}^2-p_0{b}^2)/(b^2-a^2)$ and $Q=(p_1-p_0)a^2{b}^2/(b^2-a^2)$. The radial compressive pressure is $p(r)=-P+Q/r^2$. Since hoop stress is monotone in $r$ when $Q\ne 0$, its maximum absolute value is the larger of $|P+Q/a^2|$ and $|P+Q/b^2|$. With negligible external pressure, its maximum is the tensile inner-wall value $p_1(a^2+b^2)/(b^2-a^2)$. The stress distribution itself is independent of elastic moduli in this linear homogeneous axisymmetric model; its associated displacement depends on them.

Latent hardening is increased resistance on one crystal slip system caused by plastic slip on another slip system. The word latent emphasizes that the affected system need not have been active during the earlier loading. It is therefore a statement about interactions between deformation mechanisms, rather than simply about the total amount of deformation.

A simple crystal-plasticity model writes ${\dot{g}}_{\alpha }={\sum }_{\beta }{h}_{\alpha \beta }|{\dot{\gamma }}_{\beta }|$, where $g_{\alpha }$ is the critical resolved shear stress for system $\alpha$ and ${\gamma }_{\beta }$ is slip on system $\beta$. Off-diagonal coefficients represent cross-system hardening, while diagonal coefficients describe self-hardening.^[ref] Changing loading direction after predeformation can activate previously less-favoured slip systems and reveal this history dependence. Predicting the next system to move therefore requires both resolved stress geometry and the evolving resistances.

In a neutral turbulent constant-stress region, the stress velocity and wall distance give mean shear $U_z=u_{\ast }/(\kappa z)$. Its logarithmic integral uses the roughness length. The law applies outside the inner sublayer and below the outer boundary-layer scale, not all the way to a smooth wall.

Lead zirconate titanate is an oxide solid solution commonly written $\mathrm{Pb}(\mathrm{Z}{\mathrm{r}}_{\mathrm{x}}\mathrm{T}{\mathrm{i}}_{1-\mathrm{x}}){\mathrm{O}}_3$. Its perovskite structure can support switchable electric polarization and a strong piezoelectricity response.

Changing the zirconium-to-titanium ratio changes the competing crystal structures and the response to an applied field or stress. Composition, domain configuration, and processing therefore matter when using the material as a transducer.

A requirement that a crack penetrate a pressure boundary and produce a detectable leak before unstable fracture. A simple screening comparison uses critical crack size versus wall thickness, but the result depends on crack geometry and on whether stress or pressure is held fixed.

Linearizing the differential pom-pom model about its relaxed state leaves no first-order stretch perturbation in incompressible flow. Its anisotropic orientation obeys $\dot{a}+a/{\tau }_1=A_1$, producing the displayed relaxation modulus. In a slow-flow expansion, $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Trace normalization and the first stretch correction affect only isotropic stress at this order, giving the displayed second-order fluid coefficients. Linear small-amplitude response allows arbitrary frequency, whereas this local second-order expansion also requires slow time dependence.

Linear elastic fracture mechanics predicts the loading of a crack using linear elasticity outside a small region near its tip. The near-tip stress field has a characteristic $r^{-1/2}$ dependence; its amplitude is the stress intensity factor. For an opening crack under nominal tensile stress $\sigma$,where $a$ is a specified crack-length parameter and $Y$ a dimensionless geometry factor. A central crack in an effectively infinite plate has total length $2a$ and $Y=1$. Crack advance is assessed by comparing $K_I$ with the appropriate measured fracture toughness.^[ref]

For $a=1\mathrm{mm}$ and $\sigma =100\mathrm{MPa}$ in that ideal geometry, $K_I\simeq 5.60\mathrm{MPa}\sqrt{\mathrm{m}}$. Increasing the crack's half-length fourfold doubles $K_I$ without changing the applied stress. This explains why a structure can fail at a stress below the strength of an uncracked specimen.

A toughness measurement loads a precracked specimen of known geometry and records the force associated with crack extension. Thickness, remaining ligament and crack length must be large relative to the crack-tip plastic region before the plane-strain value $K_{Ic}$ is transferable in this model. Large-scale plastic deformation requires a different description; the elastic singularity is a useful outer approximation, not an assertion of physically infinite stress inside the material.^[ref]

Linear elasticity describes small deformations using a stress that depends linearly on the infinitesimal strain tensor. If $\mathbf{u}$ is the displacement field, thenFor an unstressed isotropic material, the constitutive equation iswhere $\lambda ,\mu$ are the Lamé parameters, ${\delta }_{ij}$ is the Kronecker delta and repeated indices are summed. In particular, $\mu$ is the shear modulus.

The small-displacement-gradient approximation permits superposition: for the same linear material and linear boundary conditions, the sum of two displacement solutions responds to the sum of their applied loads. This makes it possible to calculate bending, vibration and elastic waves from common material coefficients. Small strain alone is not sufficient when rotations are large; a slender elastic rod can turn through a large angle while stretching very little, requiring geometrically nonlinear kinematics.

A linear Maxwell fluid has constitutive equation $\mathbf{\sigma }+\lambda {\partial }_t\mathbf{\sigma }=\mu \dot{\mathbf{\gamma }}$, with $\dot{\mathbf{\gamma }}$ twice the rate-of-strain tensor. Here $\mu$ is the zero-frequency shear viscosity and $\lambda >0$ is the viscoelastic relaxation time. Stress at fixed strain decays as $e^{-t/\lambda }$; the corresponding series spring and dashpot have shear modulus $G=\mu /\lambda$. In small-amplitude simple shear from an isotropic relaxed state, the first-order forcing is off-diagonal, so both normal-stress differences vanish at linear order. This is not a statement about the nonlinear objective model at second order in shear rate.

Linearity and reversibility answer different questions. The proportional limit ends the interval in which axial stress is proportional to axial strain. Elastic deformation means that deformation is recovered after unloading; a nonlinear response can still be elastic. Conversely, a short straight portion of a loading curve does not by itself demonstrate recovery.

To check recovery, unload and examine the local extensometer strain relative to its initial value, allowing for instrument resolution, drift and any time dependence. A residual extension can indicate plasticity in continuum mechanics, but grip slip can produce a residual in crosshead displacement without the same local residual. Compare the channels before assigning the effect to the specimen. Yield point concerns the onset of irreversible deformation; a specified offset proof stress is an operational convention rather than Young's modulus. For example, a $0.2\%$ offset proof stress is the intersection of the loading curve with $\sigma =E(\epsilon -0.002)$, using the correctly referenced strain. Report the fitted stress/strain window and the unloading evidence alongside the modulus, so that apparatus seating, departure from linearity and onset of plasticity in continuum mechanics remain distinguishable.

A linear viscoelastic fluid has causal stress response linear in the deformation history, often expressed as ${\sigma }^d(t)=2{\int }_0^{\infty }G(s)E(t-s)ds$. The approximation requires small deformation and rotation over the relevant memory times. High frequency is allowed at sufficiently small amplitude. An integrable decaying relaxation modulus gives a finite zero-shear viscosity; an instantaneous solvent term may be written separately.

stress is a linear causal functional of deformation history: in simple shear, $\sigma (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. Superposition holds and complex viscoelastic modulus is amplitude-independent within the small-deformation regime.

A liquid flows under a sustained shear stress while usually resisting a change of volume much more strongly than a gas. It adopts the shape of a container but, when it coexists with a gas and gravity permits, can form a free surface rather than expanding to fill all available space. These are mechanical observations, not a requirement that every liquid be exactly incompressible.

At rest in a uniform gravitational field, a liquid develops the hydrostatic pressure variation needed to support its weight. During flow, viscosity relates rate of deformation to viscous stress for a Newtonian fluid. Releasing a small surface disturbance probes its dynamical response through gravity waves and surface tension. Thus its static pressure, resistance to rapid compression and resistance to shear rate are distinct measurable properties. Liquid and gas can also connect continuously around a liquid-gas critical point, so the broad labels do not imply a first-order boundary along every thermodynamic path.

The antiplane field $w=b\arg (x+iy)/(2\pi )$ has a smooth local gradient on the punctured plane and satisfies the differential strain compatibility condition there. Its circulation around the defect axis is nevertheless $b\ne 0$, excluding a globally single-valued displacement on that domain. A radial cut permits a branch with an axial jump $b$ between the two cut faces; joining faces with that mismatch is the Volterra construction of a screw dislocation with Burgers vector parallel to the axis. The stress has a singular core where linear elasticity requires microscopic or plastic regularization. This example separates a local differential integrability test from global closure around a hole.

A long-fibre composite contains reinforcing fibres whose lengths are large enough that end regions do not dominate their load-carrying response. In a well-bonded continuous aligned specimen pulled along the fibres, the fibres and composite matrix undergo approximately the same axial strain. Their different Young's moduli then produce different axial stresses, and adding their forces gives the longitudinal rule of mixtures.

Load enters a fibre through shear stress at its interface with the composite matrix. Near a free fibre end the axial stress must rise from zero, so a short fibre can spend much of its length below its useful load. The critical fibre length expresses the associated scale in a simple strength-transfer model. Whether a fibre is mechanically long therefore depends on its diameter, the interface and the required stress, rather than on an absolute length in millimetres. Changing fibre direction still changes the response: long reinforcement does not remove anisotropy.

The longitudinal rule of mixtures estimates the axial Young's modulus of continuous aligned fibres in a composite matrix:The model assumes perfect bonding, small elastic strain, no voids and loading parallel to the fibres. Equal extension over a common initial length gives equal axial strain $\epsilon$ in both phases. Adding their forces gives the average axial stress $\sigma =V_f{E}_f\epsilon +(1-V_f)E_m\epsilon$, which proves the formula. It is the axial Voigt elastic approximation.

For example, $V_f=0.6$, $E_f=70\mathrm{GPa}$ and $E_m=3\mathrm{GPa}$ give $E_L=43.2\mathrm{GPa}$. These are illustrative inputs, not universal constituent values. The formula makes reinforcement efficient because the stiffer constituent carries the larger stress at the common strain. Loading across the fibres changes the compatibility and traction conditions, so the same formula is not a general transverse-modulus law. Fibre waviness or poor fibre-matrix load transfer can also invalidate the ideal axial estimate.

The magnetic tension stress transports axial angular momentum along a poloidal magnetic field. In purely azimuthal ideal flow the poloidal energy flux is ${\mathbf{F}}_{E,p}=\Omega {\mathbf{F}}_{J,p}$. This proportionality is valid even at vanishing flux, where an ordinary ratio is undefined. The coupled fluxes explain how a torsional Alfvén wave redistributes rotation.

In vacuum or an approximately nonmagnetizable plasma, the field energy per volume is $B^2/(2{\mu }_0)$. magnetic pressure has the same scale, but a magnetic field also exerts directional tension, so its stress is not universally an isotropic pressure. Material magnetization can require a different constitutive treatment.

Magnetic pressure is the isotropic part of magnetic stress. It is $p_B=B^2/(2{\mu }_0)$ in SI units and $B^2/(8\pi )$ in Gaussian units.