identical-to one-half left-parenthesis StartFraction partial-differential u Subscript theta Superscript m Baseline Over partial-differential z EndFraction plus StartFraction 1 Over r EndFraction StartFraction partial-differential u Subscript z Superscript m Baseline Over partial-differential theta EndFraction right-parenthesis equals minus one-half left-parenthesis alpha plus StartFraction beta Over r squared EndFraction right-parenthesis sine theta comma 4th Row epsilon Subscript r theta Superscript m Baseline identical-to one-half left-parenthesis StartFraction 1 Over r EndFraction StartFraction partial-differential u Subscript r Superscript m Baseline Over partial-differential theta EndFraction plus StartFraction partial-differential u Subscript theta Superscript m Baseline Over partial-differential r EndFraction minus StartFraction u Subscript theta Superscript m Baseline Over r EndFraction right-parenthesis equals 0 period EndLayout"/>(4.74)
The substitution of (4.80) into (4.79) then leads to
(4.81)
4.4.2 Solution in the Absence of Fibre
For loading conditions characterised by the shear stress τ applied to an infinite sample of matrix in the absence of fibre (filling the entire region of space), the solution is given by
(4.82)
(4.83)
A comparison of (4.72) and (4.82) with (4.77) and (4.83) indicates that the identification α=τ/2 can be made. It then follows from (4.81) that
(4.84)
Substitution into (4.72) leads to the following expression for the displacement component uz: