Page 191 - Plastics Engineering

174 Mechanical Behaviour of Composites

are known to be 2540 kg/m3 and 1300 kg/m3, calculate the weight fraction of
fibres in the composite.

Solution From the rule of mixtures
~c = PfVf + Pmvm = PfVf + Pm(l- Vf)
1950 = 2540Vf + 1300(1 - Vf)

Vf = 0.52
Pf 2540
Using equation (3.1). Wf = -Vf = -(0.52) = 0.68
PC 1950
Example 3.2 PEEK is to be reinforced with 30% by volume of unidirec-
tional carbon fibres and the properties of the individual materials are given
below. Calculate the density, modulus and strength of the composite in the
fibre direction.

Density Tensile strength Modulus
Material (kdm3) (GN/m*) (GN/m*)

PEEK 1300 0.0d 3.8
Carbon fibre (HM) 1800 2.1 400

Note that as shown below, this must be the matrix stress at the fibre fracture strain.
Solution From the rule of mixtures

~c = PfVf + PmVm = 0.13(1800) + 0.7(1300) = 1450 kg/m3
= CfuVf + CmVm = 0.3(2.1) + 0.7(0.058) = 0.67 GN/m2
E,L = EfVf + EmVm = 0.3(400) + 0.7(3.8) = 122.7 GN/m2

Example 3.3 Calculate the fraction of the applied force which will be taken
by the fibres in the composite referred to in Example 3.2

Solution From equations (3.2), (3.3) and (3.4), the force in the fibres is
given by
Ff = UfAf = Ef &f Vf vC
Similarly the force in the composite is given by

F1 = EfEfVfVc + EmCmVmVc

Hence.