Page 219 - Mechanical Behavior of Materials

220 Chapter 5 Stress–Strain Relationships and Behavior

Substitution of Eq. 5.51 into Eq. 5.50, and also applying Eq. 5.52, yields the desired modulus of the
composite material:

E r A r + E m A m
E X = (5.53)
A
The ratios A r /A and A m /A are also the volume fractions of fiber and matrix, respectively,
denoted V r and V m :

A r A m
V r = , V m = 1 − V r = (5.54)
A A
Thus, Eq. 5.53 can also be written

E X = V r E r + V m E m (5.55)

This result confirms that, in this case, a simple rule of mixtures applies. Note that the same
relationship is also valid for a case where the reinforcement is in the form of well-bonded layers, as
in Fig. 5.15(b).

5.4.5 Elastic Modulus Transverse to Fibers
Now consider uniaxial loading in the other orthogonal in-plane direction, specifically, a stress σ Y as
shown in Fig. 5.15(c). An exact analysis of this case is more difficult, but analysis of a transversely
loaded layered composite as shown in (d) is a useful approximation. In fact, the E Y so obtained can
be shown by detailed analysis to provide a lower bound on the correct value for case (c). Therefore,
let us proceed to analyze case (d).
The stresses in reinforcement and matrix must now be the same and equal to the applied stress:

σ Y = σ r = σ m (5.56)
As before, we can use the definitions of the various elastic moduli:

σ Y = E Y ε Y , σ r = E r ε r , σ m = E m ε m (5.57)

The total length in the Y-direction is the sum of contributions from the layers of reinforcement and
the layers of matrix:
(5.58)
L = L r + L m
Also, the changes in these lengths give the strains in the overall composite material and in the
reinforcement and matrix portions. That is,

L L r L m
ε Y = , ε r = , ε m = (5.59)
L L r L m
where

L = L r + L m (5.60)