Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 76 2 Solution Created 2026-10-03 Updated 2026-10-14
The inextensible fibre constraint gives . Write the fibre direction and its transverse unit direction asThen and . In plane strain, incompressibility gives . Decomposing the second column of the deformation gradient along therefore gives , where . HenceThis is a local rotation followed by simple shear.
The compatibility of the deformation map requires . Since and , this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus is constant when . The equivalent condition also covers . Its current tangent is proportional to , so these transverse curves are straight wherever the deformation is regular.
To construct the stress with zero body force, specify the local shear stress as the material's function or functional of under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has with , so , , and . Resolving along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines by integration along fibres and then 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.