Page 124 - Mechanics Analysis Composite Materials

Chapter 3. Mechanics of a unidirectionalply 109

Using traditional assumptions, i.e., taking that (dol/&) << I and 6 ez E and
substituting VI from Eqs. (3.108) we arrive at

Thus, Eq. (3.1 10) yields

IT2
A =-c,Pad(l +d) . (3.111)
44,
Strain energy consists of three parts, i.e.,

w=wf+w;+K, , (3.112)

where Wf is the energy of buckled fibers, while Wi and WG correspond to shear strain
and transverse extension of the matrix which supports the fibers. Strain energy of
fibers deformed in accordance with Eqs. (3.108) and shown in Fig. 3.61 has the
form

where Dfis the fiber bending stiffness. Substituting Eqs. (3.108) and calculating the
integrals we get

(3.113)

To determine the strain energy of the matrix, we assume that the matrix element
shown in Fig. 3.61 is in the plane state of stress (nonzero stresses are o.~,o!,and z.!,.),
and equilibrium equations, Eqs. (2.5) can be written as

(3.114)

To simplify the solution, we assume that longitudinal stress, G.~,acting in the matrix
can be neglected in comparison with the corresponding stress acting in the fibers.
Thus, we can put 0, = 0. Then, Eqs. (3.114) can be integrated and yield
z.T,. = z(x), Cr!, = B(X) - z’(x)y . (3.115)

Here, z(x) and CT(X) are arbitrary functions of integration and ()’ = d( )/dx.
Neglecting also the Poisson effects we can express the strains as follows