Callister - An introduction - 8th edition

640 • Chapter 16 / Composites
Substitution for F f from the preceding equation yields
or
whereas
Thus, the fiber phase supports the vast majority of the applied load.
(c) The stress for both fiber and matrix phases must first be calculated. Then,
by using the elastic modulus for each [from part (a)], the strain values may be
determined.
For stress calculations, phase cross-sectional areas are necessary:
and
Thus,
s m F m
A m
Finally, strains are computed as
s f F f
11,640 N 116.4 MPa 116,875 psi2
A
2 f 100 mm
m s m
E m
f s f
E f
13.5 F m F m 12,500 N
F m 860 N 1200 lb f 2
F f F c F m 12,500 N 860 N 11,640 N 12700 lb f 2
A m V m A c 10.621250 mm 2 2 150 mm 2 10.24 in. 2 2
A f V f A c 10.421250 mm 2 2 100 mm 2 10.16 in. 2 2
860 N 5.73 MPa 1833 psi2
2
150 mm
5.73 MPa
1.69
3.4 10 3 103
MPa
116.4 MPa
1.69
69 10 3 103
MPa
Therefore, strains for both matrix and fiber phases are identical, which
they should be, according to Equation 16.8 in the previous development.
transverse direction
Elastic Behavior—Transverse Loading
A continuous and oriented fiber composite may be loaded in the transverse direction;
that is, the load is applied at a 90 angle to the direction of fiber alignment as shown
in Figure 16.8a. For this situation the stress to which the composite as well as both
phases are exposed is the same, or
(16.12)
This is termed an isostress state. Also, the strain or deformation of the entire composite
is
c
s c s m s f s
c m V m f V f
(16.13)