Callister - An introduction - 8th edition

16.5 Influence of Fiber Orientation and Concentration • 639
this dependence on volume fractions. Equation 16.10a is the fiber analogue of Equation
16.1, the upper bound for particle-reinforced composites.
It can also be shown, for longitudinal loading, that the ratio of the load carried
by the fibers to that carried by the matrix is
Ratio of load carried
by fibers and the
matrix phase, for
longitudinal loading
F f
F m
E fV f
E m V m
The demonstration is left as a homework problem.
(16.11)
EXAMPLE PROBLEM 16.1
Property Determinations for a Glass Fiber–Reinforced
Composite—Longitudinal Direction
A continuous and aligned glass fiber–reinforced composite consists of 40 vol%
of glass fibers having a modulus of elasticity of 69 GPa (10 10 6 psi) and
60 vol% of a polyester resin that, when hardened, displays a modulus of
3.4 GPa (0.5 10 6 psi).
(a) Compute the modulus of elasticity of this composite in the longitudinal
direction.
(b) If the cross-sectional area is 250 mm 2 (0.4 in. 2 ) and a stress of 50 MPa
(7250 psi) is applied in this longitudinal direction, compute the magnitude of
the load carried by each of the fiber and matrix phases.
(c) Determine the strain that is sustained by each phase when the stress in
part (b) is applied.
Solution
(a) The modulus of elasticity of the composite is calculated using Equation
16.10a:
E cl 13.4 GPa210.62 169 GPa210.42
30 GPa 14.3 10 6 psi2
(b) To solve this portion of the problem, first find the ratio of fiber load to
matrix load, using Equation 16.11; thus,
F f 169 GPa210.42
F m 13.4 GPa210.62 13.5
or F f 13.5 F m .
In addition, the total force sustained by the composite F c may be computed
from the applied stress and total composite cross-sectional area A c
according to
F c A c s 1250 mm 2 2150 MPa2 12,500 N 12900 lb f 2
However, this total load is just the sum of the loads carried by fiber and matrix
phases; that is,
F c F f F m 12,500 N 12900 lb f 2