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For a smooth objective isotropic hyperelastic material, the strict Baker-Ericksen inequalities require larger distinct principal stretches to have larger principal stresses. For $W=\Phi (v_1,v_2,v_3)$ and $J=v_1{v}_2{v}_3$, ${\sigma }_i=J^{-1}{v}_i{\Phi }_i$. The strict convention implies every stress-free state has equal stretches: if $v_i\ne v_j$ but both stresses are zero, the displayed product would vanish. Thus a natural state of a hyperelastic material has $F=vQ$, $v>0$, $Q\in SO(3)$. A weak non-strict inequality alone does not imply this rigidity.

A thin isotropic hyperelastic material subject to incompressibility, with equal in-plane principal stretches $\lambda$ has transverse stretch ${\lambda }^{-2}$. With unloaded transverse faces, $N_{33}=W_3+q{\lambda }^2=0$. Hence $N_{11}=N_{22}=W_1-{\lambda }^{-3}{W}_3$, using isotropy to set $W_1=W_2$. The derivative along the constrained path is $W_1+W_2-2{\lambda }^{-3}{W}_3$, proving the displayed formula. The two equal nominal tensions do work $2Nd\lambda$ per reference volume, exactly the change in strain energy density.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

The first principal invariant of the left Cauchy-Green deformation tensor is the sum of the squares of the principal stretches. Since $B=F{F}^T$, it also equals ${\sum }_{i,I}{F}_{iI}^2$. Its derivative with respect to the deformation gradient is $2F$. For a strain energy density $W=w(I_1)$ with incompressibility, the Cauchy stress tensor is $T=2w^{\prime }(I_1)B-pI$.

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 $\pi a^{2}p\dot{z}$.

The left Cauchy-Green tensor measures stretch on the current configuration. It is a positive-definite symmetric matrix for an invertible deformation gradient, with eigenvalues equal to the squares of the principal stretches.

For $\gamma >0$, direct multiplication gives $|FN{|}^2-|N{|}^2=\gamma N_2(2N_1+\gamma N_2)$. Thus length-preserving line directions form the union of two planes in direction space: $N_2=0$ and $2N_1+\gamma N_2=0$. Their intersection is the third coordinate axis. This concerns segment length rather than pointwise fixed lines; a segment can move and turn while keeping its length.

The two nontrivial principal stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, with $v_{+}{v}_{-}=1$. The maximizing reference direction is proportional to $(1,v_{+},0)$. Consequently it tends to $(1,1,0)$ as $\gamma$ tends to zero from above. At zero itself all stretches equal one, so every direction maximizes length; the limit selects a direction from a degenerate endpoint. This follows from the largest eigenvector of the right Cauchy-Green deformation tensor.

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 orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

Here a prime denotes $d/d\sigma$. It is useful to retain terms through $O({ϵ}^2)$ in the normal mechanical equilibrium equation, even though only first-order shape corrections are sought. The undeformed elastic membrane is tensionless, so a first-order change of curvature contributes to normal force only at second order.

The geometric equations written in reference arc length areTheir first-order Taylor expansions giveand henceThe principal stretches and the first-order constitutive equation areBecause $t_s=t_{\phi }=t$ exactly, the tangential mechanical equilibrium equation gives ${\partial }_{s}t=0$. Thus both $t_1$ and $t_2$ are spatial constants: $t_1^{\prime }=t_2^{\prime }=0$.

Expand the geometric factors in the normal balance:ConsequentlyMatching the first-order coefficient gives the requested relationSince the prescribed pressure has no second-order term, and ${\theta }_1^{\prime }=-r_1^{\prime \prime }$, its second-order coefficient givesA closed boundary value problem for the three requested unknowns is thereforeThe first equation has order two, the second order one, and the last order one, so the four boundary conditions match the total order. The unknown constant $t_2$ is selected by the axial endpoint constraint, rather than by an additional specified traction.

For $p\ne 0$, an equivalent formulation eliminates $z_1$ and displays that constraint as an integral:After solving this system, recover $z_1$ by integration from $z_1(0)=0$. No second derivatives of the unspecified function $f$ are required: they enter the next-order shape corrections through the second-order constitutive equation, whereas the displayed normal balance already determines the leading shape. The case $p=0$ has the undeformed solution $r_1=z_1=t_2=0$.

Take $s=\sigma$ up to an irrelevant additive constant, $r=a$ constant, $z=z_0-\sigma$, and $\theta =\pi /2$. These satisfy the geometric relations and give the principal stretchesThe constitutive equations therefore produce constant membrane tensionsThe tangential mechanical equilibrium equations hold automatically, while the normal equation reduces toEvery positive root of this algebraic equation provides a cylindrical solution with no axial stretch ratio change. There is no further constraint on $t_s$ in the interior equations when end boundary conditions are discarded. In particular, a closed-end pressure balance would impose additional axial loading data that are absent here. Existence of a positive root for an arbitrary specified $P$ depends on the given constitutive equation.

Let $B=A^{\mathsf{T}}A$. Since $\det A>0$, $B$ is a positive-definite matrix. The spectral theorem gives a unique principal square root of a positive semidefinite matrixThen $R^{\mathsf{T}}R=U^{-1}B{U}^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a rotation matrix. Define $V=RU{R}^{\mathsf{T}}$. It is a symmetric matrix, is positive definite, and satisfies $V^2=A{A}^{\mathsf{T}}$. Consequently the polar decomposition in continuum mechanics isHere $U$ and $V$ are the right stretch tensor and left stretch tensor.

For the Cauchy stress tensor, use both material isotropy and material frame indifference. Their respective transformation laws areThe first law and $A=VR$ give $\sigma (A)=\sigma (V)$. Combining the two laws gives $\sigma (QV{Q}^{\mathsf{T}})=Q\sigma (V)Q^{\mathsf{T}}$. In a principal stretch basis, $V=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$. Each half-turn about a coordinate axis leaves $V$ unchanged; its conjugation changes the signs of the corresponding off-diagonal stress tensor entries. Those entries must therefore vanish. If two principal stretches coincide, rotations within their eigenspace additionally force the stress tensor to be scalar on that eigenspace. Thus the Cauchy stress tensor and left stretch tensor have common principal axes, including at repeated principal stretches. This proves coaxiality of isotropic elastic stress.

The precise symmetry conclusion is equivariance of isotropic principal stresses. Every permutation of the principal stretches can be realized by a signed permutation rotation matrix, sowhen the indexing is permuted consistently. Equivalently, permuting the arguments permutes the labelled principal stresses in the same way. In particular, ${\sigma }_1$ is symmetric in ${\lambda }_2,{\lambda }_3$, and analogous statements hold for ${\sigma }_2,{\sigma }_3$.

For distinct principal stretches, interpolate the three values ${\sigma }_i$ at the three points ${\lambda }_i$:Uniqueness of this interpolation implies that the scalar coefficients ${\beta }_j$ are symmetric functions of the unordered principal stretches; they may be expressed through the elementary symmetric invariants. Smooth isotropic constitutive laws have the corresponding invariant representation across repeated stretches as well. The wording about symmetric functions must be understood in this collective sense: each labelled principal stress need not be symmetric in all three arguments. For example, the objective isotropic law $\sigma =V^2$ has ${\sigma }_i={\lambda }_i^2$, which disproves that stronger interpretation.

Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

Use $z={\lambda }_z>0$. In cylindrical orthonormal bases the principal stretches are $r_{,R}$, $r/R$ and $z$. Incompressibility givesIntegrate from the inner surface to obtain the inflation and extension of an incompressible tube kinematicsIn particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial principal stretch is $1/(z\lambda )$, so the reference-volume strain energy density is $\hat{W}(\lambda ,z)$ with the three arguments $(1/(z\lambda ),\lambda ,z)$.

For reference length $L_0$, the stored energy is $L_0\mathcal{E}$, whereLet $N_{\mathrm{wall}}$ denote the total axial resultant acting on the material annulus. The work of the inner pressure on the cylindrical wall is $2\pi az{L}_{0}p\dot{a}$, and axial work is $N_{\mathrm{wall}}{L}_0\dot{z}$. Reversible virtual work therefore saysThe independent derivatives of the circumferential stretch areEquating the coefficients of $\dot{a}$ and $\dot{z}$ provesand the corresponding axial resultant isHere ${\hat{W}}_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the pressure also does axial end-cap work $\pi a^{2}p{L}_0\dot{z}$. Equivalently its complete work is $pd(\pi a^{2}z{L}_0)/dt$. In that convention,If $N$ includes the end-pressure resultant, it is $N_{\mathrm{wall}}$ itself. Stating this distinction is necessary because the ends are not specified; the pressure formula is the same in both conventions.

For the specified Mooney-Rivlin solid convention, differentiating the constrained strain energy density givesThusPut $c=z{a}^2-a_0^2$. If $c\ne 0$, $z{\lambda }^2-1=c/R^2$ yields $dR/R=-z\lambda d\lambda /(z{\lambda }^2-1)$. The factor cancels becauseThe limits are ${\lambda }_a=a/a_0$ and ${\lambda }_b=b/b_0$, so direct integration givesFor $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.

Use reference cylindrical coordinates $(R,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The prescribed map has $r={\lambda }^{-1/2}R$, $\theta =\Theta +\alpha Z$ and $z=\lambda Z$. In the corresponding orthonormal cylindrical bases, differentiate physical line elements to obtain the deformation gradientThe radial direction is an eigenvector of the right Cauchy-Green deformation tensor with eigenvalue ${\lambda }^{-1}$. Its other block isIts trace is $s={\lambda }^{-1}(1+{\alpha }^2{R}^2)+{\lambda }^2$ and its determinant is $\lambda$. Thus its eigenvalues solve $z^2-sz+\lambda =0$. The principal stretches are the positive square roots of these eigenvalues:Their product is one, independently checking incompressibility.

The strain energy density $W$ is per reference volume, so stored energy per initial cylinder height is $\mathcal{E}(\alpha ,\lambda )=2\pi {\int }_0^{A}W({\lambda }_1,{\lambda }_2,{\lambda }_3)RdR$. The end rotation is $\alpha H$ and the current height is $\lambda H$, giving applied power per initial height $M\dot{\alpha }+N\dot{\lambda }$. For quasistatic reversible loading, equate this with $\dot{\mathcal{E}}$ for independent changes of $\alpha ,\lambda$. This proves the torque and axial force of a twisted incompressible cylinder formulasIn the second derivative hold $\alpha$, the twist per initial length, fixed. Holding twist per current length instead would mix the generalized forces.

For the Mooney-Rivlin solid, compute invariants directly from $F$ rather than differentiating individual eigenvalue square roots:The second identity follows from $I_2=\mathrm{tr}(F^{T}F{)}^{-1}$ when $J=1$. Retain the paper's minus sign on its ${\mu }_2$ term. Integrating $R$ and $R^3$ givesConsequentlyAs checks, untwisted unit stretch gives $M=N=0$, and setting ${\mu }_2=0$ recovers the neo-Hookean solid formulas. A convention with a positive second-invariant coefficient corresponds to replacing this ${\mu }_2$ by its negative.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

Put $c=a^2-a_0^2$ and use reference cylindrical coordinates $(\rho ,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The deformation map is $r=\sqrt{{\rho }^2+c}$, $\theta =\Theta +\alpha Z$, $z=Z$. In the associated orthonormal bases its deformation gradient isIts determinant is one, as required by incompressibility. The radial principal stretch is ${\lambda }_0=\rho /r$. The other two squared principal stretches are the eigenvalues of the tangential-axial block of $F{F}^T$:Writing $K=(r/\rho {)}^2+{\alpha }^2{r}^2+1$, they areThe positive square roots give ${\lambda }_{\pm }$, and ${\lambda }_0{\lambda }_{+}{\lambda }_{-}=1$.

The strain energy density is per reference volume, so the energy per height isThe work rates per height are $M\dot{\alpha }$ from the end torque and $2\pi ap\dot{a}$ from the inner pressure. The outer surface contributes no work because it is traction free, and the fixed height has zero axial work. Thus the virtual work identities for twist and inflation of an incompressible tube are

For a neo-Hookean solid, an irrelevant energy constant can be omitted:Differentiating under the finite integral givesAn antiderivative divided by $\mu$ is $\log \rho -\frac{1}{2}\log ({\rho }^2+c)-c/[2({\rho }^2+c)]+{\alpha }^2{\rho }^2/2$. With $b^2=b_0^2+c$, the neo-Hookean tube pressure under twist and inflation is thereforeIt vanishes for an untwisted undeformed tube, and the twist contributes the last term at fixed radii.