Properties for Design of Composite Structures. Neil McCartney
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)
It follows directly from (4.14)–(4.17), (4.73) and (4.74) that
In addition, it follows from (4.17), (4.73) and (4.74) that
It is easily shown that the stress field automatically satisfies the equilibrium equations (4.20)–(4.22) for any values of the parameters A, α and β.
The following continuity conditions must be satisfied at the interface r = a between fibre and matrix
It then follows from (4.76) that
and from (4.72) that
The substitution of (4.80) into (4.79) then leads to
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
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
Substitution into (4.72) leads to the following expression for the displacement component uz: