Page 185 - Mechanics Analysis Composite Materials
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I70 Mechanics and analysis of composite materials
where EXi=Ex3 =E1 KX2 =E2, Ezi =E2, GXzi= Gxz3= G131 Gxz2= G23, vx2i = vxz3
= VI31 v,2 = v23 and El, E2, G13, G231 v13, ~23are elastic constants of a unidirec-
tional ply. Substituting stresses, Eqs. (4.1 14), into the functional in Eq. (4.1 15),
integrating with respect to z, and using traditional procedure of variational calculus
providing SW, = 0 we arrive at the following equation for ~(x):
where
General solution for this equation is
02 = e-klx(clsin k2x +c2 cos k2x) + eklx(c3sin k2x + cq cos k2x) , (4.116)
where
Assume that the strip shown in Fig. 4.37 is infinitely long in the x-direction. Then,
we should have 01 4 0 and a2 --f 0 for x 4 03 in Eqs. (4.110). This means that we
should put c3 = c4 = 0 in Eq. (4.1 16). The other two constants, CIand c2, should be
determined from the conditions on the crack surface (see Fig. 4.37), i.e.,
ox2(x = 0) = 0, ZX22(X = 0) = 0 .
Satisfying these conditions we obtain the following expressions for stresses:
4
zXz2 = --(k: +ki)ze-k1xsin kzx, (4.117)
k2
As an example, consider a glass-epoxy sandwich layer with the following
parameters: hl = 0.365 mm, h2 = 0.735 mm, EI = 56 GPa, E2 = 17 GPa, GI, =
