Page 221 - Mechanical Behavior of Materials

222 Chapter 5 Stress–Strain Relationships and Behavior

Example 5.5
A composite material is to be made of tungsten wire aligned in a single direction in a copper
matrix. The elastic modulus parallel to the fibers must be at least 250 GPa, and the elastic
modulus perpendicular to the fibers must be at least 200 GPa.

(a) What is the smallest volume fraction of wire that can be used?
(b) For the volume fraction of wire chosen in (a), what are the major Poisson’s ratio and
the shear modulus of the composite material?

Solution (a) We need to determine the volume fractions of fibers V r needed to meet each
requirement. The larger of the two different V r values obtained is then chosen as meeting or
exceeding both requirements. The matrix and reinforcement properties are:

Tungsten reinforcement: E = 411 GPa, ν = 0.280, G = 160.5GPa
Copper matrix: E = 130 GPa, ν = 0.343, G = 48.4GPa

where E and ν are from Table 5.2, and each G is calculated from Eq. 5.28 under the assumption
that each material is isotropic.
First consider the E X = 250 GPa requirement, where X is the fiber direction. Equation 5.55
can be solved to obtain the needed V r . Substituting the appropriate preceding values and solving
for V r gives

E X = V r E r + V m E m , V m = 1 − V r

250 GPa = V r (411 GPa) + (1 − V r )(130 GPa), V r = 0.427

Then, similarly, consider the E Y = 200 GPa requirement, and employ Eq. 5.64 to obtain

1 V r V m
= + , V m = 1 − V r
E Y E r E m
1 V r 1 − V r
= + , V r = 0.512
200 GPa 411 GPa 130 GPa
Hence, the required volume fraction is the larger value, V r = 0.512. Ans.
(b) The major Poisson’s ratio and the shear modulus for the composite material with
V r = 0.512 can be estimated from Eqs. 5.65 and 5.66:

ν XY = V r ν r + V m ν m = 0.512(0.280) + (1 − 0.512)(0.343) = 0.311 Ans.

G r G m (160.5GPa)(48.4GPa)
G XY = = = 75.3GPa Ans.
V r G m + V m G r 0.512(48.4GPa) + (1 − 0.512)(160.5GPa)