Properties for Design of Composite Structures. Neil McCartney

Properties for Design of Composite Structures

Properties for Design of Composite Structures - Neil McCartney

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overTilde Subscript normal upper T Baseline upper Delta upper T comma"/>(2.221)

      where

      It is noted that the in-plane stresses are only linear functions of x3. It then follows, on using (2.210), that all the equilibrium equations (2.120)–(2.122) are satisfied.

      It now only remains to determine the value of the displacement component u3 describing the deflection of the beam. From (2.216) and (2.220)

      On using (2.215) and (2.222) it can be shown that

      nu Subscript normal t Baseline plus nu Subscript normal upper A Baseline StartFraction upper E Subscript normal upper T Baseline Over upper E Subscript normal upper A Baseline EndFraction nu overTilde Subscript normal a Baseline equals upper E overTilde Subscript normal upper T Baseline left-parenthesis StartFraction nu Subscript normal t Baseline Over upper E Subscript normal upper T Baseline EndFraction plus StartFraction nu Subscript normal a Baseline nu Subscript normal upper A Baseline Over upper E Subscript normal upper A Baseline EndFraction right-parenthesis equals nu overTilde Subscript normal t Baseline comma(2.224)

      so that (2.223) may now be written as

      Integration with respect to x3 then leads to

      As the shear stresses σ13 and σ23 are everywhere zero, it follows from (2.143), (2.209) and (2.211) that

      StartFraction partial-differential u 3 Over partial-differential x 1 EndFraction equals minus StartFraction partial-differential u 1 Over partial-differential x 3 EndFraction equals minus ModifyingAbove epsilon With caret Subscript normal upper A Baseline x 1 comma StartFraction partial-differential u 3 Over partial-differential x 2 EndFraction equals minus StartFraction partial-differential u 2 Over partial-differential x 3 EndFraction equals minus ModifyingAbove epsilon With caret Subscript normal upper T Baseline x 2 period(2.227)

      Integration then yields

      u 3 equals minus one-half ModifyingAbove epsilon With caret Subscript normal upper A Baseline x 1 squared plus g 1 left-parenthesis x 2 comma x 3 right-parenthesis comma u 3 equals minus one-half ModifyingAbove epsilon With caret Subscript normal upper T Baseline x 2 squared plus g 2 left-parenthesis x 1 comma x 3 right-parenthesis period(2.228)

      These relations must be consistent with (2.226) so that