Inextensible fibre constraint ($|F{m}_0|=1$)

A unit reference fibre direction $m_0$ retains its length under this constraint on the deformation gradient. Its reaction contributes an undetermined tension along the current fibre direction, independently of the pressure reaction enforcing incompressibility.

For two distinct reference fibre directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0)$, with $0<\theta <\pi /2$, consider the volume-preserving deformation with right stretch tensor $U=\mathrm{diag}({\lambda }^{-1/2}\alpha ,{\lambda }^{-1/2}{\alpha }^{-1},\lambda )$, with positive $\lambda ,\alpha$. The inextensible fibre constraint gives ${\alpha }^2{\cos }^2\theta +{\alpha }^{-2}{\sin }^2\theta =\lambda$. Setting $b={\alpha }^2$ gives a quadratic with the displayed roots. There are two distinct positive branches for $\lambda >\sin 2\theta$. The smallest possible axial stretch is $\lambda =\sin 2\theta$, at which ${\alpha }^2=\tan \theta$ and $(U{l}_{+})\cdot (U{l}_{-})=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin 2\theta /\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.

Let $e=(\cos \theta ,\sin \theta )$ be the current fibre direction and $n=(-\sin \theta ,\cos \theta )$. Inextensibility fixes the first column of the deformation gradient to $e$, and incompressibility fixes the second to $\gamma e+n$. Commuting mixed derivatives gives ${\gamma }_{,1}={\theta }_{,1}$ and ${\theta }_{,2}=\gamma {\theta }_{,1}$. Thus $\gamma =\theta +f(X_2)$, and curves tangent to $(-\gamma ,1)$ map to straight transverse curves with constant $\theta$.

Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.