The inextensible fibre constraint gives . Write the fibre direction and its transverse unit direction as
Then and . In plane strain, incompressibility gives . Decomposing the second column of the deformation gradient along therefore gives , where . Hence
This is a local rotation followed by simple shear.
The compatibility of the deformation map requires . Since and , this is
Taking 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. Write
The compatible geometry has with , so , , and . Resolving along the two directions gives
The 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.