Neil McCartney

Properties for Design of Composite Structures


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alt="sigma Subscript theta theta Superscript f Baseline equals sigma Subscript upper T Baseline minus phi period"/>(4.42)

      The substitution of (4.29)3, (4.41) and (4.42) into (4.16) leads to

      sigma Subscript z z Superscript f Baseline equals upper E Subscript upper A Superscript f Baseline left-parenthesis epsilon minus alpha Subscript upper A Superscript f Baseline upper Delta upper T right-parenthesis plus 2 nu Subscript upper A Superscript f Baseline left-parenthesis sigma Subscript upper T Baseline minus phi right-parenthesis period(4.43)

      It is clear from (4.41) that σzzf is a constant, automatically satisfying (4.37)2. The substitution of (4.31)3, (4.39) and (4.40) into (4.16), applied to the matrix, leads to

      thus automatically satisfying (4.38)2. The substitution of (4.29), (4.31), (4.39)–(4.44) into relation (4.14), applied to the fibre and matrix, leads to

      where kTf is the plane strain bulk modulus for the transverse isotropic fibre and kTm is the plane strain bulk modulus of the isotropic matrix, defined by (see (2.202))

      fraction numerator 1 over denominator k subscript text T end text end subscript superscript text m end text end superscript end fraction equals fraction numerator 2 left parenthesis 1 minus nu subscript text m end text end subscript right parenthesis over denominator E subscript text m end text end subscript end fraction minus fraction numerator 4 nu subscript text m end text end subscript superscript 2 over denominator E subscript text m end text end subscript end fraction.(4.48)

      It should be noted that

      where km is the bulk modulus of the matrix defined by (see (2.205))

      k Subscript m Baseline equals StartFraction upper E Subscript m Baseline Over 3 left-parenthesis 1 minus 2 nu Subscript m Baseline right-parenthesis EndFraction equals StartFraction 2 mu Subscript m Baseline left-parenthesis 1 plus nu Subscript m Baseline right-parenthesis Over 3 left-parenthesis 1 minus 2 nu Subscript m Baseline right-parenthesis EndFraction period(4.50)

      It now only remains to determine the constant ϕ, which can be specified on applying the remaining condition (4.28)5, because the conditions (4.28)2, (4.28)4 and (4.28)6 are automatically satisfied by (4.25) and (4.33). It follows from (4.26), (4.27), (4.28)5, (4.45) and (4.46) that

      phi equals upper Lamda left-bracket left-parenthesis nu Subscript m Baseline minus nu Subscript upper A Superscript f Baseline right-parenthesis epsilon plus left-parenthesis alpha Subscript upper T Superscript f Baseline plus nu Subscript upper A Superscript f Baseline alpha Subscript upper A Superscript f Baseline right-parenthesis upper Delta upper T minus left-parenthesis 1 zero width space zero width space plus nu Subscript m Baseline right-parenthesis alpha Subscript m Baseline upper Delta upper T zero width space plus one-half left-parenthesis StartFraction 1 Over k Subscript upper T Superscript f Baseline EndFraction minus StartFraction 1 Over k Subscript upper T Superscript m Baseline EndFraction right-parenthesis sigma Subscript upper T Baseline right-bracket comma(4.51)

      where

      StartFraction 1 Over upper Lamda EndFraction equals one-half left-parenthesis StartFraction 1 Over k Subscript upper T Superscript f Baseline EndFraction plus StartFraction 1 Over mu Subscript m Baseline EndFraction right-parenthesis period(4.52)

      The displacement distribution is specified by (4.25)–(