Page 182 - Mechanics Analysis Composite Materials

Chapter 4. Mechanics of a composire layer 167

where E1.2 = E1.2/(1- ~12~21).We assume that strains and E,, do not change
through the laminate thickness. Substituting Eqs. (4.107) into Eqs. (4.106) we can
find strains and then stresses using again Eqs. (4.107). The final result is

To simplify the analysis, neglect Poisson's effect taking v12 = v21 = 0. Then

(4.108)

Consider, for example, the case hl = h2 = 0.5 and find the ultimate stresses
corresponding to the failure of longitudinal plies or to the failure of the transverse
ply. Putting oy = iit and cr! = ii: we get

The results of calculation for composites listed in Table 3.5 are presented in
Table 4.2. As can be seen, ii~;') >> ii?). This means that the first failure occurs in the
transverse ply under stress

(4.109)

This stress does not cause the failure of the whole laminate because the longitudinal
plies can carry the load, but the material behavior becomes nonlinear. Actually, the
effect under consideration is the result of the difference between the ultimate
elongations of the unidirectional plies along and across the fibers. From the data
presented in Table 4.2 we can see that for all the materials listed in the table EI >> 6.
As a result, transverse plies following under tension longitudinal plies that have

Table 4.2
Ultimate stresses causing the failure of longitudinal (3;'))or transverse (a?') plies and deformation
characteristics of typical advanced composites.

o (MPa); Glass- Carbon- Carbon- Aramid- Boron- Boron-AI Carbon- Al@-AI
e (%) epoxy epoxy PEEK epoxy epoxy carbon
~~ ~
*:I! 2190 2160 2250 2630 1420 2000 890 1100
a?l 225 690 1125 590 840 400 100 520
c:l 3 1.43 1.5 2.63 0.62 0.50 0.47 0.27
82 0.31 0.45 0.75 0.2 0.37 0.1 0.05 0.13
El /E2 9.7 3.2 2 13.1 1.68 5 94 2.1