Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-15 GTBL042-Callister-v2 August 29, 2007 8:52

15.5 Influence of Fiber Orientation and Concentration • 631

Thus,

σm=

Fm

Am

=

860 N

150 mm^2

= 5 .73 MPa (833 psi)

σf=

Ff

Af

=

11,640 N

100 mm^2

= 116 .4 MPa (16,875 psi)

Finally, strains are computed as

m=

σm

Em

=

5 .73 MPa

3. 4 × 103 MPa

= 1. 69 × 10 −^3

f=

σf

Ef

=

116 .4 MPa

69 × 103 MPa

= 1. 69 × 10 −^3

Therefore, strains for both matrix and fiber phases are identical,

which they should be, according to Equation 15.8 in the previous devel-

opment.

Elastic Behavior—Transverse Loading

transverse direction A continuous and oriented fiber composite may be loaded in thetransverse direction;

that is, the load is applied at a 90◦angle to the direction of fiber alignment as shown

in Figure 15.8a. For this situation the stressσto which the composite as well as both

phases are exposed is the same, or

σc=σm=σf=σ (15.12)

This is termed anisostressstate. Also, the strain or deformation of the entire com-

positecis

c=mVm+fVf (15.13)

but, since=σ/E,

σ

Ect

=

σ

Em

Vm+

σ

Ef

Vf (15.14)

whereEctis the modulus of elasticity in the transverse direction. Now, dividing

through byσyields

1

Ect

=

Vm

Em

+

Vf

Ef

(15.15)

which reduces to

Ect=

EmEf

VmEf+VfEm

=

EmEf

(1−Vf)Ef+VfEm

(15.16)

For a continuous and

aligned fiber-

reinforced

composite, modulus

of elasticity in the

transverse direction

Equation 15.16 is analogous to the lower-bound expression for particulate com-

posites, Equation 15.2.