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A material fibre segment of reference length $L$ and unit direction $l$ has reference vector $Ll$. The uniform deformation gradient maps it to $LAl$, whose current length isTherefore its length is preserved if and only ifThis is the inextensible fibre constraint. The quadratic form uses the right Cauchy-Green deformation tensor; a subsequent rigid rotation does not change it, consistently with the polar factorization $A=RU$.

For two distinct symmetric fibre families use reference directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0{)}^T$, with $0<\theta <\pi /2$. The second sign is needed for the orthogonality conclusion: the PDF prints only one direction. If both families literally had that same direction, their images under an invertible deformation would remain parallel, so part (ii) would be false. The following derivation states the intended two-family geometry explicitly.

Write $c=\cos \theta$, $s=\sin \theta$ and take positive $\lambda ,\alpha$, as required for a positive stretch tensor. Applying the inextensible fibre constraint to either family givesWith $b={\alpha }^2$, multiplication by $b$ yields $c^2{b}^2-\lambda b+s^2=0$. Its discriminant is ${\lambda }^2-4c^2{s}^2={\lambda }^2-{\sin }^{2}2\theta$, so the two-family inextensible-fibre pure stretch hasThis also shows why the angle endpoints need separate treatment: at $\theta =0$ the admissible root is ${\alpha }^2=\lambda$, and at $\theta =\pi /2$ it is ${\alpha }^2=1/\lambda$. The nominal two families then represent the same unoriented fibre direction, and the two-branch/orthogonality assertions do not apply.

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.

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.