Page 245 - Plastics Engineering

228 Mechanical Behaviour of Composites

Example 3.17 Short carbon fibres with a diameter of 10 pm are to be used
to reinforce nylon 66. If the design stress for the composite is 300 MN/m2
and the following data is available on the fibres and nylon, calculate the load
transfer length for the fibres and also the critical fibre length. The volume
fraction of the fibres is to be 0.3.

Modulus (CN/m*) Strength (GN/m2)

Carbon fibres 230 2.9
Nylon 66 2.8 -

The interfacial shear strength for carbonhylon may be taken as 4 MN/m2
Solution

EI = EfVf + EmVm

El = 230(0.3) + 2.8(0.7) = 71 GN/m2
Using (3.42)

300
= Ef (2) = 230 ( 11) 972 MN/m2
=
Using (3.43)

(af),,,=d - 972(10 x
e, = - = 1.2 mm
2TY 2x4
Using (3.44)
-
af,d
e --- - 2900(10 x
c- = 3.6
2rY 2x4
It may be seen from Fig. 3.29 that due to the ineffective end portions of
short fibres, the average stress in the fibre will be less than in a continuous
fibre. The exact value of the average stress will depend on the length of the
fibres. Using the stress distributions shown in Fig. 3.29(b) the fibre stresses
may be analysed as follows.