Neil McCartney

Properties for Design of Composite Structures

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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

      StartLayout 1st Row sigma Subscript r r Superscript f Baseline equals sigma Subscript theta theta Superscript f Baseline equals sigma Subscript z z Superscript f Baseline equals sigma Subscript r theta Superscript f Baseline identical-to 0 comma 2nd Row sigma Subscript r r Superscript m Baseline equals sigma Subscript theta theta Superscript m Baseline equals sigma Subscript z z Superscript m Baseline equals sigma Subscript r theta Superscript m Baseline identical-to 0 period EndLayout(4.75)

      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

      u Subscript z Superscript m Baseline left-parenthesis a comma z right-parenthesis equals u Subscript z Superscript f Baseline left-parenthesis a comma z right-parenthesis period(4.78)

      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

      upper A equals StartFraction mu Subscript m Baseline Over mu Subscript upper A Superscript f Baseline plus mu Subscript m Baseline EndFraction tau comma StartFraction beta Over a squared EndFraction zero width space equals one-half StartFraction mu Subscript m Baseline minus mu Subscript upper A Superscript f Baseline Over mu Subscript upper A Superscript f Baseline plus mu Subscript m Baseline EndFraction tau period(4.84)

      Substitution into (4.72) leads to the following expression for the displacement component uz: