Page 84 - Mechanics Analysis Composite Materials

Chapter 3. Mechanics of a unidirectionalply 69

-1 -I
Fig. 3.17. Shear strain in the matrix layer.

is the fiber strain expressed in terms of the displacement in the x-direction. Shear
strain in the matrix follows from Fig. 3.17, Le.,

(3.21)

Thus obtained set of equations, Eqs. (3.18)-(3.21) is complete - it includes 10k + 3
equations and contains the same number of unknown stresses, strains and
displacements.
Consider the boundary conditions. If there is no crack in the central fiber, the
solution of the problem is evident and has the form a,, = a, z,, = 0. Assuming that
the perturbation of the stressed state induced by the crack vanishes at a distance
from the crack we arrive at

a,& + m) = 0: z,(x + 00) = 0 . (3.22)

Because of the crack in the central fiber we have

O(](X = 0) = 0 . (3.23)

For the rest fibers, symmetry conditions yield

u,,(x=O)=O (n= 1,2,3.....k) . (3.24)

To solve the problem, we use the stress formulation and, in accordance with Section
2. IO, should consider equilibrium equations in conjunction with compatibility
equations expressed in terms of stresses.
First, transform equilibrium equations introducing the stress function, F(x) such
that

T,~=Fi, E,(x+ 03) = 0 . (3.25)

Substituting Eqs. (3.25) into equilibrium equations, Eqs. (3.18), integrating them
from x to XI and taking into account Eqs. (3.22) and (3.25) we get