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Let $F$ be an invertible, orientation-preserving deformation gradient, with $J=\det F>0$. To prove Nanson's formula, take two reference surface tangent vectors $a,b$. Their current images are $Fa,Fb$, and their oriented area vectors are proportional to $a\times b$ and $(Fa)\times (Fb)$. For every vector $c$, the scalar triple product givesThus $F^T[(Fa)\times (Fb)]=J(a\times b)$, andThis proof also identifies the area transformation as the cofactor of $F$, rather than a transformation by $F$ itself.

Use the paper's material-index-first nominal stress tensor $P_{Ii}$. Equality of force on the same material surface in its two configurations gives $P_{Ii}d{S}_{0I}={\sigma }_{ji}d{S}_j$. Substitute the area transformation to obtainHere $\tau$ is the Kirchhoff stress tensor. With symmetric Cauchy stress tensor, the matrix forms are $P=J{F}^{-1}\sigma$ and $\tau =FP$. This $P$ is the transpose of the first Piola-Kirchhoff stress tensor; keeping the index convention prevents a spurious transpose later.

For the velocity gradient $L=\dot{F}{F}^{-1}$, let $D=(L+L^T)/2$ be the rate-of-strain tensor. Then the reference-volume mechanical power isThe last step uses stress symmetry, as supplied by angular-momentum balance in an ordinary continuum without couple stresses: a symmetric tensor has zero contraction with the skew part of $L$.

By work-conjugate stress and strain, a pair $(T,E)$ must satisfy $T:\dot{E}=\tau :D$ for every deformation rate, with both sides measured per reference volume. For the Green-Lagrange strain tensor, set $C=F^{T}F$ and calculateCyclically rearranging the contraction shows $T^{(2)}:{\dot{E}}^{(2)}=(F{T}^{(2)}{F}^T):D$. Hence the symmetric conjugate isthe second Piola-Kirchhoff stress tensor.

The inverse right Cauchy-Green strain is $E^{(-2)}=(I-C^{-1})/2$, because $C^{-1}=F^{-1}{F}^{-T}$. Differentiating $C{C}^{-1}=I$ givesConsequently $T^{(-2)}:{\dot{E}}^{(-2)}=(F^{-T}{T}^{(-2)}{F}^{-1}):D$, andThe two stress measures have a geometric interpretation in the convected coordinate net. Its basis vectors are $g_I=F{e}_I$ and its reciprocal basis is $g^I=F^{-T}{e}_I$. The entries of $T^{(2)}$ are the contravariant components in $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$, whereas $T_{IJ}^{(-2)}=g_I\cdot \tau g_J$ are the covariant components, equivalently $\tau =T_{IJ}^{(-2)}{g}^I\otimes g^J$. They describe the same spatial Kirchhoff stress tensor in a deforming, generally nonorthonormal coordinate net.

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.

The inextensible fibre constraint gives $|{\mathbf{x}}_{,1}|=1$. Write the fibre direction and its transverse unit direction asThen $x_{1,1}=\cos \theta$ and $x_{2,1}=\sin \theta$. In plane strain, incompressibility gives $\det F=1$. Decomposing the second column of the deformation gradient along $e,n$ therefore gives ${\mathbf{x}}_{,2}=\gamma e+n$, where $\gamma ={\mathbf{x}}_{,2}\cdot e$. HenceThis is a local rotation followed by simple shear.

The compatibility of the deformation map requires ${\mathbf{x}}_{,12}={\mathbf{x}}_{,21}$. Since $e_{,j}={\theta }_{,j}n$ and $n_{,j}=-{\theta }_{,j}e$, this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus $\theta$ is constant when $d{X}_2/d{X}_1=-1/\gamma$. The equivalent condition $d{X}_1+\gamma d{X}_2=0$ also covers $\gamma =0$. Its current tangent is proportional to $F(-\gamma ,1{)}^T=n$, so these transverse curves are straight wherever the deformation is regular.

To construct the stress with zero body force, specify the local shear stress $h$ as the material's function or functional of $\gamma$ under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has $\nabla \theta =\kappa e$ with $\kappa ={\theta }_{,1}$, so $\nabla \cdot e=0$, $\nabla \cdot n=-\kappa$, $(e\cdot \nabla )e=\kappa n$ and $(n\cdot \nabla )e=0$. Resolving $\nabla \cdot \sigma =0$ along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines $s$ by integration along fibres and then $t$ along transverse curves. Thus any regular compatible deformation has a local zero-body-force stress construction once the shear law and suitable boundary integration data are supplied. There is no additional constitutive normal-stress law to specify: the two normal components are fixed by force balance and boundary tractions. This does not assert that arbitrary separately prescribed boundary tractions can accompany an arbitrary deformation.

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.

Use the material-first nominal stress tensor convention $P_{Ii}$, with $F_{iI}$ the deformation gradient. For any fixed reference material subvolume $V_0$ with outward unit normal $N$, conservation of energy isHere $b$ is body force per mass, $r$ is heat supply per mass and $q$ is nominal heat flux. Using momentum conservation to remove the kinetic energy balance gives the integral Lagrangian internal energy balance:The integral Lagrangian entropy inequality iswhere $\theta >0$ is temperature and $q/\theta$ is the thermodynamic entropy flux.

Since the subvolume is arbitrary, the divergence theorem givesExpand $\dot{U}=U_{,F_{iI}}{\dot{F}}_{iI}+U_{,\eta }\dot{\eta }+U_{,{\xi }_r}{\dot{\xi }}_r$ and eliminate the heat supply between these relations. The Clausius-Duhem inequality becomesUnder the usual local constitutive independence assumptions, the reversible mechanical and thermal variations can have either sign. The Coleman-Noll procedure therefore imposesDefine the thermodynamic force conjugate to an internal variable by $f_r=-{\rho }_0{U}_{,{\xi }_r}$. The energy balance and the remaining entropy constraint now give, respectively,

For an orientation-preserving invertible deformation gradient, the polar decomposition of an invertible real matrix separates a rotation $R$ from the right stretch tensor $U$ or the left stretch tensor $V$. The rotation is shared by the two factorizations, with $V=RU{R}^T$.

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 renovating shear flow holds a planar shear flow fixed for an interval $T$, then chooses a fresh orientation independently of preceding intervals. Each interval acts on a material line by a rotated copy of the displayed deformation gradient. The orientation law must be specified: independent uniformly distributed orientations define the isotropic model, whereas a biased orientation law generally gives another Lyapunov exponent.

A simple fluid has local stress determined by the history of deformation of a material element, rather than by spatial derivatives of that history. An isotropic incompressible simple fluid has an arbitrary isotropic pressure and an objective constitutive functional of the relative deformation gradients. Internal-variable equations can represent such a functional when the initial material state is fixed by the past history.

Simple shear displaces one coordinate direction in proportion to a transverse coordinate while preserving volume. Its deformation gradient has a single off-diagonal component $\gamma$. Simple shear flow describes the corresponding velocity field rather than the finite deformation.

The derivative of internal energy at fixed deformation gradient and entropy, or of specific Helmholtz free energy at fixed temperature, defines this force. Its product with the internal-variable rate contributes mechanical dissipation to the Clausius-Duhem inequality.

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.