Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 70 / 1 / c

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.

For $0<\theta <\pi /2$ and $\lambda >\sin 2\theta$, the discriminant is strictly positive. The two roots have positive sum $\lambda /c^2$ and positive product $s^2/c^2$, so both roots are positive and distinct. Each gives exactly one positive $\alpha$ and hence a different right stretch tensor. If $\lambda \ne 1$, neither tensor is the undeformed identity because its third diagonal entry is $\lambda$. Thus two distinct deformed configurations are possible. Excluding $\lambda =1$ removes the case in which one of the two configurations is the undeformed body; it is not needed just for positivity of the roots.

The smallest admissible axial stretch is ${\lambda }_{\min }=\sin 2\theta$, where the two branches merge and ${\alpha }^2=\tan \theta =s/c$. The current unit fibre vectors are $m_{\pm }=U{l}_{\pm }$. Their dot product isAt the minimum, both terms in parentheses equal $cs$, soHence the two families are orthogonal at the maximum possible contraction. There is also a geometric proof of the bound: their reference parallelogram area is $\sin 2\theta$, and its current area is $\sin 2\theta /\lambda$, because the in-plane area stretch is $1/\lambda$. Two unit vectors span area at most one, with equality precisely when perpendicular. This gives $\lambda \ge \sin 2\theta$ and the same equality condition without solving the quadratic. At $\theta =\pi /4$ the minimum is one, so no axial contraction is possible and the fibres are already orthogonal.