Page 209 - Plastics Engineering

192 Mechanical Behaviour of Composites

Solution Let 8 = 25" and a, = 10 MN/m2 to illustrate the method of
solution.
The Stress Transformation Matrix is
2sc
c2 s2 -2sc ] 0.821 0.179 0.766
Tu = [ s2 c2 or Tu = ( 0.179 0.821 -0.766
-sc sc (2 -2) -0.383 0.383 0.643
In order to get the strains in the global directions it is necessary to determine
the overall compliance matrix [SI. This is obtained as indicated above, ie

[SI = @]-I where [D] = [Ta]-'[Q] [T,]
The local compliance matrix is
- 7
-v12
0
El
1
- 0
E2
1
0 -
Gl2 -
The Strain Transformation matrix is
2 s2 -sc ]
sc
Tc = [ s2 c2
-2sc 2sc (2 -2)

Then
1.12.10-~ -6.3.10-~ -1.87.10-~
s= [ -6.3. 2.65 . 5.89
-i.87.10-~ 5.89. io-6 4.35.10-~
Directly by matrix manipulation
E, = 1.126. cy = -6.309 * yxy = -1.878 -
or by multiplying out the terms

= [(Sii) * (ax) + $12) * (ay)] + (Si6) (txy) E, = 1.126 *
and similarly for the other two strains. Figure 3.13 illustrates how the strains
vary for different values of 8.
Example 3.8 A thin unidirectional carbon fibre composite is loaded as shown
in Fig. 3.14 and has the properties listed below. If the fibres are aligned at 35"
to the x-axis, calculate the stresses parallel and perpendicular to the fibres,