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

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

630 • Chapter 15 / Composites

(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

15.10a:

Ecl=(3.4 GPa)(0.6)+(69 GPa)(0.4)

=30 GPa (4. 3 × 106 psi)

(b)To solve this portion of the problem, first find the ratio of fiber load to

matrix load, using Equation 15.11; thus,

Ff

Fm

=

(69 GPa)(0.4)

(3.4 GPa)(0.6)

= 13. 5

orFf=13.5Fm.

In addition, the total force sustained by the compositeFcmay be com-

puted from the applied stressσand total composite cross-sectional areaAc

according to

Fc=Acσ=(250 mm^2 )(50 MPa)= 12 ,500 N (2900 lbf)

However, this total load is just the sum of the loads carried by fiber and

matrix phases; that is,

Fc=Ff+Fm= 12 ,500 N (2900 lbf)

Substitution forFffrom the above yields

13. 5 Fm+Fm= 12 ,500 N

or

Fm=860 N (200 lbf)

whereas

Ff=Fc−Fm= 12 ,500 N−860 N= 11 ,640 N (2700 lbf)

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:

Am=VmAc=(0.6)(250 mm^2 )=150 mm^2 (0.24 in.^2 )

and

Af=VfAc=(0.4)(250 mm^2 )=100 mm^2 (0.16 in.^2 )