Page 239 - Plastics Engineering

222 Mechanical Behaviour of Composites

[ 9.69; 16 3.89 x 16
D = 3.89 x 16 9.69 x Id x ] Nmm,
0 2.89 x 105
im0X 10-6 -4.94 x 10-7 0
d = -4.94 x 1.23 x
0 3.45 x O 10-6 I
Then the curvatures, K, are given by

so,
K, = 3.44 x 10-~ mm-l, K~ = 1.72 x 10-~ mm-', K,~ = o

The strains will be directly proportional to the distance across the thickness, ie
E=K*Z

At top surface, Z = -11 mm, so E~ = (3.44 x 10-4)(-11) = -3.78 x
stress, a, = EIE, = 3500(3.78 x = -13.25 MN/m2

At interface (skin side), Z = -10 mm, so E, = (3.44 x 10-4)(-10) =
-3.44 x 10-3:

stress, a, = E~E, = 3500(3.44 x = -12.04 m/m2

At interface (core side), E, = -3.44 x
stress, a, = 60(-3.44 x = -0.21 MN/m2
Similarly in the y-direction,
At top surface

cy = -1.89 x ay = -6.6 MN/m2

At interface (skin)
= -1.72 x ay = -6 MN/m2
At interface (core)

= -1.72 x ay = -0.1 m/m2

In both cases the stresses and strains will be mirrored on the bottom section of
the beam (ie positive instead of negative). These are illustrated in Fig. 3.26.