Page 211 - Plastics Engineering

194 Mechanical Behaviour of Composites

when the stresses a,, ay and txy are applied. Calculate also the strains in the
global X - Y directions.

E1 = 125O0O1 MN/m2, E2 = 7800 MN/m2, G12 = 4400 MN/m2,
u12 = 0.34 a, = 10 MN/m2, ay = -14 MN/m2, txy = -5 MN/m2
Solution The Stress Transformation Matrix is 0.94 )
c2 s2 -2sc ] 0.671 0.329
2sc
T, = [ s2 c2 or T, = ( 0.329 0.671 -0.94
-sc sc (2 - 2) -0.47 0.47 0.342
The stresses parallel and perpendicular to the fibres are then given by
[E:] =To. [ 21

t12 =xY
so,
a1 = -2.59 MN/m2 02 = -1.4 MN/m2 t12 = -12.98 MN/m2

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 = [a]-' where [a] = [To]-' [e]. [T,]
The local compliance matrix is
-
0

0 and Q=S-'
1
-

The Strain Transformation matrix is
c2 s2 -sc ]
SC
T, = [ s2 c2
-2sc 2sc (2 -2)
so, -
Q=T;'.Q.T, and s=a-'
Then
6.64 x -2.16 x -7.02 x
["]=3.["] - s = [ -2.16 x 10-~ 1.07 x 10-~ -4.27 x 10-~
YXY TXY -7.02 x 10-~ -4.27 x 10-~ 1.51 x 10-~