Elasticity (physics)

Elasticity studies reversible deformation and the stresses produced by it.

The boundary of the loading range within which a material recovers its original configuration on unloading. Elasticity need not be linear; beyond the elastic range, permanent plastic deformation or damage may remain.

Failure by crack formation or propagation with limited distributed permanent deformation. Shear and opening fractures have different strength conditions; a frictional failure envelope should not be extrapolated indefinitely into tension.

Loss of load-bearing strength accompanied by relative displacement tangential to a surface, rather than primarily opening normal to it. The Mohr-Coulomb failure criterion models frictional shear failure by $|\tau |=c+\mu {\sigma }_n^{\prime }$, using cohesion, a friction coefficient and compressive effective stress normal to the potential sliding plane.

In the simplest strength-cutoff model, opening begins when the least compressive effective stress reaches minus the tensile strength: ${\sigma }_3^{\prime }=-T_0$ with compression positive. In Mohr's circle, this is contact with the vertical line ${\sigma }_n^{\prime }=-T_0$ at $\tau =0$. The criterion represents an opening mechanism and differs from frictional shear failure.

A coefficient relating an appropriate stress and strain in the linear elastic regime. Young's modulus, shear modulus and bulk modulus describe different deformation modes.

Loss of stability of a loaded configuration toward a bent or folded form. The threshold and deformation mode depend on geometry, material properties and boundary conditions.

A flat, unsupported film's bending and tension energies give modal stiffness $\kappa q^4+\sigma q^2$. Negative tension represents compression, destabilizing sufficiently long waves. A finite sheet restricts allowed wave numbers, while a supporting foundation adds stiffness and can select a wrinkle wavelength. This is generic continuum mechanics, independent of biological membrane structure.

An eigenstrain is a local preferred deformation in the absence of stress. Constraints or incompatibility can make the actual strain differ from it, generating stress. Magnetostrictive strain is one example.

Magnetoelastic coupling links a material's magnetic state to its strain and stress. Magnetostriction describes a resulting deformation; mechanical loading can also change the magnetic state.

An elastic deformation is recoverable when the deforming loads are removed. In a local elastic constitutive model the stress depends on the present deformation. When this response also derives from a stored elastic energy, the material is a hyperelastic material. In the small-deformation approximation, the relevant kinematic variable is the infinitesimal strain tensor.

A perfectly bonded elastic interface has continuous displacement field and continuous traction, with the same chosen normal on both sides. For piecewise affine displacement fields, the difference of their gradients annihilates every tangential vector, and hence has the form $a\otimes n$. Their infinitesimal strain tensors differ by $\mathrm{sym}(a\otimes n)$.

A circular elastic layer of thickness $h\ll a$ bonded between rigid faces develops shear strain $r\varphi /h$ under a relative twist $\varphi$. Integrating the shear stress times its moment arm gives $K_V=G{J}_p/h$. A small relative tilt produces longitudinal strain $y\theta /h$; when lateral deformation is constrained, the corresponding normal traction gives $K_H=MI/h$. The polar second moment of area is $J_p=\pi a^4/2$ and the diametric second moment of area is $I=\pi a^4/4$. Edge relaxation limits this thin-layer approximation, especially in a nearly incompressible material.

A symmetric stress tensor is in static elastic equilibrium when its divergence balances the prescribed body force. Its boundary traction is $t_i={\sigma }_{ij}{n}_j$. With zero body force, integration by parts gives ${\int }_{\Omega }\sigma :\mathrm{sym}\nabla v={\int }_{\partial \Omega }t\cdot v$. This is the weak principle of virtual work for a static elastic body.

An elastic beam is a slender body that supports bending moments and transverse loads. In the small-slope Euler-Bernoulli approximation its bending moment is $M=EI{y}_{xx}$, and transverse momentum balance gives $\rho A{y}_{tt}+EI{y}_{xxxx}=q$ for a uniform beam under distributed force per length $q$. Here $I$ is the second moment of area; boundary conditions specify the end constraints and applied moments and forces.

The component of the internal resultant force perpendicular to an elastic beam's tangent. For a planar elastic beam, one consistent sign convention has bending moment $M=-B{w}^{\prime \prime }$ and $M^{\prime }=N$, so $N=-B{w}^{\prime \prime \prime }$. Reversing the section or bending moment convention reverses the corresponding signs.

An Euler strut is an initially straight slender elastic column loaded in axial compression. Its lateral stability depends on bending stiffness, length and endpoint constraints. Euler buckling predicts a loss of straight-configuration stability before material failure only in the appropriate slender elastic regime; it is distinct from crushing or plastic yielding of a short column.

For the angular endpoint conditions $\theta (0)=\theta (1)=0$, take $U=(EI/L){\int }_0^1[{\theta }^{\prime 2}/2-{\pi }^2\lambda (1-\cos \theta )]d\xi$. The Poincaré inequality and $1-\cos t\le t^2/2$ give $U\ge 0$ for $0\le \lambda \le 1$. At $\lambda =1$ a nonzero continuous angle still has strictly positive energy because $1-\cos t<t^2/2$ for $t\ne 0$. Thus the straight state remains the unique global minimum at the critical parameter even though the quadratic variation vanishes on the first sine mode.

Along that mode the energy is ${\pi }^{2}EI{A}^4/(64L)+O(A^6)$ at criticality. For $\lambda >1$ a sufficiently small first-mode variation lowers energy, so the straight state loses local minimality. Near the first threshold, expansion of the Euler strut equation gives $A_0(A_0^2-8{\lambda }_1)=0$ when $\lambda =1+ϵ{\lambda }_1$ and $\theta \sim \sqrt{ϵ}{A}_0\sin \pi \xi$. The resulting supercritical pitchfork bifurcation is stabilized by the same positive quartic term. A zero second variation is therefore an inconclusive quadratic test, not proof of failure of minimization.

The planar equilibrium of an inextensible slender beam can have large tangent angle while retaining small elastic strain across its thickness. For constant horizontal compressive force $F$, moment balance and curvature ${\theta }^{\prime }=d\theta /ds$ give the displayed nonlinear equation. Multiplication by ${\theta }^{\prime }$ gives the first integral $EI{\theta }^{\prime 2}/2-F\cos \theta =C$. A tensile load has $F<0$ in this convention.

Specifying endpoint tangent angles of an inextensible rod does not by itself specify its endpoint positions. A planar Euler elastica satisfies $x^{\prime }=\cos \theta$, $y^{\prime }=\sin \theta$, so equal endpoint heights require the displayed integral to vanish. For constant compressive resultant $(-P,0)$, the angular equation is $EI{\theta }^{\prime \prime }+P\sin \theta =0$, giving $P\Delta y=-EI[{\theta }^{\prime }(L)-{\theta }^{\prime }(0)]$. Thus a nontrivial branch containing an odd number of pendulum half-periods cannot have equal endpoint heights and the same zero endpoint angles with zero transverse resultant. At small amplitude the tangent-only eigenfunctions are $\sin (n\pi s/L)$, with $P_n=EI(n\pi /L{)}^2$; equal endpoint heights retain even $n$ and give the first fixed-clamp load $4{\pi }^{2}EI/L^2$. Other clamp modes can require a nonzero transverse reaction. Axial shortening is also needed when the reference endpoint separation equals the rod length.

An ideal zero-thickness planar loop under tension $T$ is a separatrix of the elastica equation. Its tangent turns from zero to $2\pi$, the same physical orientation at both ends. With $\alpha =\sqrt{T/(EI)}$, integration gives $x=s-2\tanh (\alpha s)/\alpha$ and $y=2\mathrm{sech}(\alpha s)/\alpha$. Its height is $2\sqrt{EI/T}$. Self-contact and finite beam thickness are absent from this ideal crossing solution.

A beam fixed at one end and free at the other. A transverse end force $F$ gives bending moment $M(x)=F(L-x)$ and, for a rectangular section, maximum bending stress $6FL/(w{h}^2)$ at the fixed end. Longitudinal tension and compression are greatest on opposite outer fibres.

A uniform square section has second moment of area $a^4/12$ and gravitational load per length $\rho g{a}^2$. Euler-Bernoulli bending therefore gives $y^{\prime \prime \prime \prime }=12\rho g/(E{a}^2)$, with zero displacement and slope at the clamp and zero moment and shear at the free end. Integrating gives the displayed tip deflection and $y(x)=\rho g{x}^2(x^2-4lx+6l^2)/(2E{a}^2)$. Small slope and strain are required.

In an elastic boundary value problem, a dead load is a prescribed force independent of the incremental displacement. A dead reference traction has $\delta t^0=0$, even when the current face changes orientation. This differs from a follower load, whose direction follows the moving body, and from the structural-engineering use of the same phrase for permanent self-weight. Under pure dead tractions, a constant incremental translation remains a zero-frequency mode unless a displacement condition fixes it.

For $\Phi =\mu [{\sum }_i{v}_i^2/2+{\sum }_i{v}_i^{-1}]$ and positive stretch product one, the energy simplifies to $\mu S^2/2$, where $S={\sum }_i{v}_i$. Equal nominal tensile loading adds potential $-\tau S$. The homogeneous diagonal equilibrium equations reduce to $(\mu S-\tau )(v_i-v_j)=0$. Thus the unit triple exists for every load, while the other equilibria form the displayed sum-product family. Since $S\ge 3$, only the unit triple exists for $\tau \le 3\mu$. Above this load, a continuous family is obtained by choosing $v_1=t$ near one and taking $v_2,v_3$ as the two positive roots with sum $\tau /\mu -t$ and product $1/t$.

A dead-load traction is prescribed force per unit reference surface area, with its vector held fixed as the body deforms. Its boundary condition is $T_{R}N=t_R$. A follower load instead depends on the current deformation, for example through a current normal or current area.

Magnetostriction is a change of a material's shape or dimensions associated with its magnetization. A magnetic field can therefore produce strain without a contact force. The resulting magnetoelastic coupling lets a ferromagnet convert changes of magnetic state into mechanical motion; a constraint can instead turn the preferred deformation into stress. James Prescott Joule published his study of iron and steel bars in 1847.^[ref]

The surprising point is that reversing the magnetization need not reverse the deformation. A material can prefer a longer axis along either $+\mathbf{m}$ or $-\mathbf{m}$: an axis has no arrow. Isotropic linear magnetostriction vanishes excludes one particular linear constitutive term, while saturated isotropic magnetostrictive strain supplies a nonzero response with a quadratic dependence on the magnetic direction. The pulsed-field optical magnetostriction experiment shows how an actual length change can be resolved, and quadratic magnetostriction and frequency doubling separates an illustrative response law from a measured material curve.

Coupling in which a magnetic material's preferred strain transfers mechanical stress to a piezoelectric constituent, producing an electrical response. It depends on interfaces, elastic load sharing and electrical boundary conditions; a composite's measured voltage is not by itself an intrinsic single-phase magnetoelectric coefficient.

A magnetostrictive actuator uses magnetostriction to produce displacement or force. Its free motion and its blocked stress are different boundary conditions, as shown in free strain and blocked magnetostrictive stress.

Magnetostrictive strain is the preferred strain associated with a specified magnetic state relative to a stated reference. The measured strain also depends on mechanical constraints.

Volume magnetostriction is the relative volume change associated with a magnetic-state change. To first order it is the trace of the magnetostrictive strain. Zero trace is an additional restriction, not a consequence of isotropy alone.

In the saturated isotropic magnetostrictive strain convention, saturation magnetostriction is the longitudinal relative length change from a random-domain reference to alignment. A parallel-to-perpendicular length difference instead equals $3{\lambda }_s/2$ in that model.

Under proper rotations a third-rank isotropic tensor is proportional to the Levi-Civita symbol. Its contraction with a field vector is antisymmetric in the two stress indices. A physical symmetric stress therefore vanishes in this linear constitutive model. Under the full orthogonal group an ordinary isotropic third-rank tensor is already zero; the orientation-sensitive Levi-Civita tensor requires the usual pseudotensor distinction.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

A full $180^{\circ }$ reversal leaves saturated isotropic magnetostrictive strain unchanged. A $90^{\circ }$ rotation generally changes it. In the simplest unconstrained orientation average over magnetic domains, replace $m_i{m}_j$ by $⟨m_i{m}_j⟩$, not by $⟨m_i⟩⟨m_j⟩$. Equal populations pointing along $+z$ and $-z$ have zero net magnetization but the same preferred strain as a single domain along $z$. An isotropic orientation distribution instead has $⟨m_i{m}_j⟩={\delta }_{ij}/3$ and zero average preferred strain. Elastic compatibility between real magnetic domains can modify the macroscopic response.

Reversing an applied magnetic field does not guarantee that all microscopic orientations retrace the same path. Hysteresis, magnetic domain-wall pinning, and changes in domain populations can alter the measured curve. The reversal identity concerns the same pair of magnetic states with opposite local magnetization; it is not a universal prediction that an arbitrary experimental sweep is a single-valued even function of applied magnetic field.

A $3\times 1.5\times 1.5\mathrm{m}{\mathrm{m}}^3$ single crystal of gadolinium antimonide carried a $2\mathrm{mm}$ fibre Bragg grating cyanoacrylate-bonded along its $a$ axis. The pulsed magnetic field paralleled that axis. Broadband light, a circulator and a spectrometer resolved reflected-wavelength shifts at $5.4\mathrm{K}$. Pre-pulse readings measured noise; low fibre thermal expansion reduced thermal background. A residual post-pulse shift indicated possible heating. Spectral distortion checked bond uniformity, while the fibre's smaller cross-section limited loading; bonding was not assumed force-free. Longitudinal strain showed a kink near $32.6\mathrm{T}$.^[ref]

The distinction between readout and inference is essential. The fibre Bragg grating senses a change of reflected wavelength; interpreting this as the specimen's strain requires good strain transfer and a controlled temperature. A useful independent control would compare a bonded sensor with a nearby unloaded sensor through the same pulse. Another would vary bond stiffness or specimen cross-section to test loading. These are proposed checks, not additional procedures claimed for the experiment above. The pulsed magnet's winding details were not specified in the inspected account.

This experiment measures one longitudinal component of strain. Establishing the first-order volume change would require three orthogonal normal components under the same conditions. GdSb also has its own antiferromagnetism and field-dependent magnetic structure: this measured curve does not establish the simple saturated isotropic law. Its role here is to make the measurement chain and its limitations concrete.

As an original stipulated low-field model, take the axial magnetic field intensity $H$ in $A/m$ and prescribeAll strain components are dimensionless, and their trace vanishes. This is an even, quasistatic, unsaturated toy law with a fixed measuring axis. It is distinct from the fixed-magnitude directional model in saturated isotropic magnetostrictive strain. In particular, one must not replace local unit magnetization by normalized net magnetization in a many-domain specimen and assume this law follows.

For $H(t)=H_0\sin (2\pi ft)$, $H_0=20kA/m$, and $f=50\mathrm{Hz}$, a trigonometric identity givesThe axial strain ranges from zero to $20\times 10^{-6}$, with mean $10\times 10^{-6}$ and a $100\mathrm{Hz}$ component of amplitude $10\times 10^{-6}$. The cosine coefficient is negative with the chosen time origin. For $L=100\mathrm{mm}$ the length ranges from zero to $2\mu \mathrm{m}$ relative to $H=0$. There is no $50\mathrm{Hz}$ term in this unbiased model.

A static bias $H_b$ changes the conclusion. Expanding $\kappa [H_b+H_0\sin (2\pi ft){]}^2$ produces a fundamental-frequency term $2\kappa H_b{H}_0\sin (2\pi ft)$ as well as the doubled-frequency term. This is an incremental linear response about a state with a chosen magnetic axis; it does not contradict isotropic linear magnetostriction vanishes at an unbiased isotropic reference. Real hysteresis, thermal expansion from Joule heating, and mechanical resonance can add further components or change their phases. Frequency alone cannot identify the cause of motion: Joule heating also contains an even-in-electric current contribution.

An eigenstrain is the deformation a material would prefer locally without applied stress. In a one-dimensional linear elasticity model, let $E$ be Young's modulus, take tension as positive stress, and write $\sigma =E(\epsilon -{\epsilon }^{\mathrm{mag}})$. A free rod has $\sigma =0$ and follows $\epsilon ={\epsilon }^{\mathrm{mag}}$. A perfectly blocked rod has $\epsilon =0$ and $\sigma =-E{\epsilon }^{\mathrm{mag}}$. Thus positive preferred elongation produces compressive stress when blocked.

For stipulated $E=100\mathrm{GPa}$, cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$, and ${\epsilon }^{\mathrm{mag}}=20\times 10^{-6}$, blocking gives $\sigma =-2\mathrm{MPa}$ and axial force $F=\sigma A=-2\mathrm{N}$. This idealization ignores changes of magnetization caused by the load itself. A real magnetostrictive actuator therefore needs both magnetic and mechanical constitutive information: a free-length measurement alone does not determine its loaded motion.

For two Fourier elastic wave fields with the same real elastic stiffness tensor, its major symmetry cancels the cross-strain terms in the divergence displayed above. The force-balance equations supply the remaining volume terms, and the divergence theorem gives their boundary integral of reciprocal tractions. Taking one field to be a complex conjugate gives energy-work identities and weighted modal orthogonality. The bilinear identity uses no complex conjugation until that field is chosen explicitly.

A spherical radial deformation is $y(x)=r(R)x/R$, with $R=|x|$ and $\mathbf{\theta }=x/R$. Differentiation gives the displayed symmetric gradient. Its radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $r/R$; when these are positive, they are the principal stretches. The determinant is $r^{\prime }(r/R{)}^2$. Positivity distinguishes an orientation-preserving radial elastic deformation from a map that collapses an interior sphere.

For a radial deformation $y=r(R)x/R$ with positive radial derivative, put $q=r/R$ and $\delta =r^{\prime }{q}^2=\det Dy$. The principal nominal stresses are ${\Phi }_1=\mu +h^{\prime }(\delta )q^2$ and ${\Phi }_2=\mu +h^{\prime }(\delta )r^{\prime }q$. Radial equilibrium reads ${\Phi }_1^{\prime }=2({\Phi }_2-{\Phi }_1)/R$. Since $q^{\prime }=(r^{\prime }-q)/R$, substitution cancels the terms proportional to $h^{\prime }$, leaving $h^{\prime \prime }(\delta )q^2{\delta }^{\prime }=0$. Strict convexity of $h$ and positivity of $q$ force a constant positive determinant $D$.

Integration of $r^2{r}^{\prime }=D{R}^2$ gives $r^3=D{R}^3+C$. Two spherical boundary radii fix $D,C$ uniquely. A positive increasing cube-radius interpolation verifies positivity of $r$ and $r^{\prime }$, so the algebraic solution is an admissible smooth radial equilibrium. The result is restricted to this energy family and radial class; strict convexity of its determinant term alone does not imply uniqueness among arbitrary three-dimensional deformations.

Consider an orientation-preserving classical radial deformation of an elastic solid on a connected spherical annulus. If its nonnegative energy is zero everywhere and the material is isotropic with strict Baker-Ericksen inequalities, every gradient is a natural state, so $r^{\prime }=r/R$. Thus $(r/R{)}^{\prime }=0$ and $y=vx$.

A separate rigidity argument applies when the zero wells are $SO(3)\cup {\bigcup }_{i=1}^{N}SO(3)U_i$, with finitely many positive definite symmetric $U_i$. The radial gradient $F$ is positive symmetric; if $F=Q{U}_i$, then $F^2=U_i^2$, so uniqueness of the positive square root gives $F=U_i$. The rotation well similarly gives $F=I$. Hence the continuous ratio $r(R)/R$ takes values in a finite set of eigenvalues and must be constant on the connected radial interval. Differentiating $r=vR$ finishes the proof. This version needs neither material isotropy nor Baker-Ericksen inequalities and does not require an assumed continuity of $r^{\prime }$ beyond the radial-map regularity used for the ratio.

Let ${\mathbf{x}}_0(\mathbf{X};\tau )$ be a family of static elastic equilibrium solutions for a homogeneous material with incompressibility under a fixed spatial body force. Replacing $\tau$ by time gives a dynamic solution of the elastic field equations exactly when the resulting material acceleration has a global acceleration potential $\xi$. At the same deformation gradient, the Cauchy stress tensor can change only by the scalar pressure reaction. Subtracting static and dynamic balance of linear momentum gives $\nabla q=\rho \mathbf{a}$, and constant mass density then gives $q=-\rho \xi +C(t)$. Surface tractions must match the adjusted Cauchy stress tensor.

The opening of an internal cavity under tensile loading in a solid. For spherically symmetric deformation of an isotropic hyperelastic material with incompressibility, and an empty cavity without surface tension, radial static elastic equilibrium gives the displayed critical signed external pressure, where $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$ and positive $P$ denotes compression. Indeed, the principal stress formula for incompressible hyperelasticity gives $d{t}_r/d\lambda =h^{\prime }(\lambda )/(1-{\lambda }^3)$. Integration between the free inner surface and the outer surface, followed by the vanishing-initial-hole and vanishing-current-cavity limits, gives the formula. For a neo-Hookean solid with shear modulus $\mu >0$, the tensile magnitude is $-P_{\mathrm{crit}}=5\mu /2$.

For the unit ball and energy $W(F)=|F{|}^2+(\det F-1{)}^2$, boundary dilation $y=tx$ admits a singular incompressible radial competitor $r(R)=(R^3+t^3-1{)}^{1/3}$. Its radial stretch is $R^2/r^2$ and its double tangential stretch is $r/R$. Although the latter diverges at zero, the three-dimensional volume element makes its squared gradient integrable. Its energy is $4\pi {\int }_0^1[R^6/r^4+2r^2],dR\sim 8\pi t^2$.

The smooth homogeneous equilibrium $y=tx$ has constant stress and energy $(4\pi /3)[3t^2+(t^3-1{)}^2]$, of sixth-order growth. It is therefore beaten for large $t$ in the finite-energy Sobolev space class admitting the cavity. This conclusion depends on that class: the singular map is not continuous at the center and cannot simply be substituted as a smooth cavity-free competitor.

For strain energy density $W=\mu [({\lambda }_r^2+{\lambda }_{\theta }^2+{\lambda }_{\varphi }^2{)}^{\alpha }-3^{\alpha }]/2$ with $\mu >0$ and $\alpha >0$, the critical loading for elastic cavitation is finite exactly in the displayed range. To prove this, put $S={\lambda }^{-4}+2{\lambda }^2$ in the reduced strain energy density. The critical-pressure integrand is $2\mu \alpha S^{\alpha -1}({\lambda }^{-2}+{\lambda }^{-5})$. It is finite at $\lambda =1$ and asymptotic to $\mu \alpha 2^{\alpha }{\lambda }^{2\alpha -4}$ at infinity. The comparison test for the resulting improper integral gives $2\alpha -4<-1$.

For reference radii $a_0,b_0$, current inner radius $a$ and uniform axial stretch $z>0$, incompressibility integrates to $r^2=a^2+(R^2-a_0^2)/z$. The principal stretches are $r_{,R}=1/(z\lambda )$, $\lambda =r/R$ and $z$. Write $\hat{W}(\lambda ,z)=W((z\lambda {)}^{-1},\lambda ,z)$ and $\mathcal{E}=2\pi {\int }_{a_0}^{b_0}\hat{W}RdR$ per reference length. Virtual work gives $p={\mathcal{E}}_{,a}/(2\pi az)$ and the total axial wall load $N_{\mathrm{wall}}={\mathcal{E}}_{,z}$, both derivatives holding the other displayed parameter fixed. For pressurized closed ends the independently applied axial load is $N_{\mathrm{ext}}=N_{\mathrm{wall}}-\pi a^{2}p$. The distinction follows from the axial pressure power $\pi a^{2}p\dot{z}$.

For an equilibrium with $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$ and dead loads, the constrained second variation on ${\varphi }_{,F}:\delta F=0$ isIndeed, a constraint-preserving perturbation path has ${\varphi }_{,F}:{\delta }^{2}F=-{\varphi }_{,FF}[\delta F,\delta F]$, and stationarity of the total potential supplies the multiplier term. Linearized kinetic energy plus $\mathcal{Q}/2$ is conserved. Positivity gives the usual energy-norm linear stability; control in a specified displacement norm additionally requires the corresponding coercivity and treatment of rigid-motion modes.

For axial stretch $\lambda$ and twist per reference length $\alpha$, the deformation in local cylindrical orthonormal bases hasThe principal stretches follow from the eigenvalues of $F^{T}F$. For a cylinder of reference radius $A$, stored energy per reference length is $\mathcal{E}=2\pi {\int }_0^{A}W(F)RdR$. Reversible power $M\dot{\alpha }+N\dot{\lambda }=\dot{\mathcal{E}}$ gives torque $M={\mathcal{E}}_{,\alpha }$ and axial force $N={\mathcal{E}}_{,\lambda }$. The relevant derivative holds the reference twist fixed, not the twist per current length.

A deformation map twisting every cross-section of a cylinder through an angle proportional to its reference axial position, without radial or axial extension. In the local current and reference cylindrical coordinate systems, the deformation gradient is $F=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& \tau r\\ 0& 0& 1\end{array}\right)$. Its determinant is one. The twist per unit reference length is $\tau$; an end separation $L$ gives end rotation $\phi =\tau L$. The physical torque is $N=2\pi {\int }_0^A{T}_{\theta z}{r}^{2}dr$.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

In cylindrical orthonormal bases the deformation gradient is $F=\left(\begin{array}{ccc}\rho /r& 0& 0\\ 0& r/\rho & \alpha r\\ 0& 0& 1\end{array}\right)$ and has unit determinant. One principal stretch is $\rho /r$; the squares of the other two are the roots of $z^2-[(r/\rho {)}^2+{\alpha }^2{r}^2+1]z+(r/\rho {)}^2=0$. The reference-volume energy per height is $E=2\pi \int \rho Wd\rho$. Virtual work gives end torque $M={\partial }_{\alpha }E$ and inner pressure $p=(2\pi a{)}^{-1}{\partial }_{a}E$ at fixed reference radii and height.