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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.

The first Piola-Kirchhoff stress tensor $T_R$ maps a reference normal to force per reference area. The Cauchy stress tensor $T$ maps a current normal to force per current area. Nanson's formula gives $\mathbf{n}da=J{F}^{-T}\mathbf{N}dA$. Equating the actual surface force in both descriptions gives $T_R=JT{F}^{-T}$, and thereforeFor incompressibility $\boxed{J=1}$. The strain-energy derivative only fixes the constitutive stress up to a constraint reaction in hyperelastic stress: $T_R=DW-p\mathrm{cof}F$, hence $T=DW{F}^T-pI$. The scalar $p$ is a Lagrange multiplier determined by equilibrium and boundary tractions, rather than by the isochoric energy alone.

In principal stretch axes, the extra principal stress is $v_i\partial W/\partial v_i=2\alpha v_i^2-2\beta v_i^{-2}$, up to that arbitrary spherical reaction. With $B=F{F}^T$, its eigenvalues are $v_i^2$, so isotropy yieldsFor the prescribed simple shear,The $x_3$ faces retain normal $\pm {\mathbf{e}}_3$, and zero traction there gives $p=2(\alpha -\beta )$. ThusThis constant stress satisfies the interior equilibrium equation, and the remaining face forces maintain the deformation.

To compute the angle, the material face $x_1=1$ becomes the spatial plane $y_1-\gamma y_2=1$. Its outward unit normal and an in-plane unit tangent areResolve the current traction $\mathbf{t}=T\mathbf{n}$ into these directions:Since $N<0$ for nonzero $\gamma$, the angle lies above $\pi /2$, andUsing $\arctan z\sim z$ at zero givesUsing the undeformed normal ${\mathbf{e}}_1$ would give the wrong coefficient. The transformed unit normal is essential to this simple-shear traction angle for a Mooney-Rivlin solid.

A homogeneous hyperelastic material has a strain energy density per unit reference volume $W(F)$ depending on the deformation gradient, with no explicit dependence on the reference configuration. For orientation-preserving gradients $F\in G{L}^{+}(3)$, its first Piola-Kirchhoff stress tensor is the derivative:Equivalently the infinitesimal stored-energy change is $DW(F)[H]=T:H$. Reference and spatial indices are distinguished here by upper- and lowercase index letters.

The material symmetry group may be written $G_W=\{H\in G{L}^{+}(3):W(FH)=W(F)\text{ for every }F\in G{L}^{+}(3)\}$. A volume-preserving relabelling convention restricts the candidate $H$ to $SL(3)$; the material isotropy criterion is the same. The material has material isotropy when $SO(3)\subseteq G_W$, meaning $W(FQ)=W(F)$ for every proper rotation $Q$. If only rotational symmetries are included in the definition of the group, material isotropy means that group equals $SO(3)$. This right-rotation invariance is distinct from material frame indifference, the left-rotation invariance $W(QF)=W(F)$.

The polar decomposition in continuum mechanics gives the unique factorizationswhere $U,V$ are positive-definite symmetric matrices. Their common positive eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ are the principal stretches, equivalently the singular values of $F$. They describe the local stretch magnitudes after separating out a rigid rotation.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

The radial unit vector in spherical coordinates is ${\mathbf{e}}_R=\mathbf{X}/R$, so $\mathbf{X}\otimes \mathbf{X}/R^2={\mathbf{e}}_R\otimes {\mathbf{e}}_R$. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient givesThe reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore ${\lambda }_r=f^{\prime }(R)$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=f(R)/R$. Incompressibility imposesIntegrating and using $f(A)=a$ givesIn particular $b^3-a^3=B^3-A^3$, expressing conservation of shell volume. Writing $\lambda =r/R$ also gives ${\lambda }_r={\lambda }^{-2}$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=\lambda$.

Let $W_r,W_{\theta },W_{\varphi }$ denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity isIt follows by applying $\mathbf{T}=\mathbf{F}({\partial }_{\mathbf{F}}W{)}^T-p1$ in principal axes. Equal tangential principal stretches and material isotropy give $W_{\theta }=W_{\varphi }$ at $({\lambda }^{-2},\lambda ,\lambda )$. Differentiating the reduced strain energy density by the chain rule givesThe unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, givesConsequently the desired ordinary differential equation isFor completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2={\lambda }^2$, and differentiating $\lambda =r/R$ gives $(1-{\lambda }^3)/R$. At the undeformed value $\lambda =1$ the division must be interpreted by a limit; the spherically symmetric deformation with $a=A$ has $\lambda \equiv 1$ and should be handled directly by the radial equilibrium equation.

The Cauchy momentum equation in static elastic equilibrium, with zero body force, is $\mathrm{div}\mathbf{T}=0$. The Cauchy stress tensor is symmetric. Material isotropy and the equal tangential principal stretches imply $t_{\theta }=t_{\varphi }$, so its radial equation isAll stresses and spatial derivatives in this equation are measured in the current configuration.

The neo-Hookean solid corresponds to $\alpha =1$, with positive shear modulus $\mu$. Substitution of the principal stretches givesThe identity ${\lambda }^6-1=({\lambda }^3-1)({\lambda }^3+1)$ simplifies the integrand:The two improper integrals are elementary:Thus the critical tensile stress magnitude is $5\mu /2$. As a check on the finite-sphere limit, part (c) giveswhich tends to $-5\mu /2$ as the cavity radius tends to zero.

The principal stretches are the eigenvalues of a stretch tensor, or equivalently the singular values of the deformation gradient. They measure local length ratios along the principal reference directions.

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.

The right Cauchy-Green tensor measures stretch on the reference configuration. Its eigenvalues are the same squared principal stretches as those of the left Cauchy-Green deformation tensor, but its eigenvectors are reference directions.

A radial deformation map preserving each particle's direction from a common centre. In an orthonormal basis of spherical coordinates, its deformation gradient has principal stretches $f^{\prime }(R),f(R)/R,f(R)/R$. Thus incompressibility gives $f^{\prime }(R)f(R{)}^2=R^2$, and integration gives $f(R{)}^3=R^3+C$.

The ratio of current to reference lengths of an infinitesimal material line element. Along an elastic rod, it is $\alpha =ds/dS=|d\mathbf{r}/dS|>0$. A principal stretch is a stretch ratio along a principal direction of the stretch tensor.

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.

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.