Page 229 - Plastics Engineering

212 Mechanical Behaviour of Composites

when f =3, f =4, f =7 and f = 8
a, = -10.8 MNIm’, a, = -19.2 MNlm2, txy = -11.8 MNlm2

when f =5and6

a, = 62.6 MNlm2, a,, = -9 MNlm2, txy = -0.9 MNIm’.
Note that in the original question, the applied force per unit width in the
x-direction was 100 Nlmm (ie a, (10)). As each ply is 1 mm thick, then the
above stresses are also equal to the forces per unit width for each ply. If we
add the above values for all 10 plies, then it will be seen that the answer is
100 Nlmm as it should be for equilibrium. Similarly, if we add N, and N,, for
each ply, these come to -140 Nlmm and -50 Nlmm which also agree with
the applied forces in these directions.
In the local (1 -2) directions we can obtain the stresses and strains by using
the transformation matrices. Hence, for the tf’th ply
[ ::I = T, [ ;]

t12 T*Y
[E:] =TEf [ 21

Y12 YXY
So that for f = 1, 2, 9 and 10
a1 = -0.44 MNlm2, a2 = -6.4 MNlm2, txy = 7.3 MNlm2

c1 = 1.38 x g2 = -8.15 x y12 = 1.67 10-~
For f = 3,4,7 and 8

01 = -0.96 MN/m2, a2 = -5.8 MN/m2, q2 = -7.5 MNlm2
= 1.82 E2 = -6.2 x y12 = -1.8 10-~

For f =5and6
a1 = 4.5 MNIm’, a2 = -11.3 MNlm2, t12 = -0.28 MN/m2
g1 = 5.25 x g2 = -1.33 x y12 = -2.09 10-~

When the moment M, = 10o0 Ndm is added, the curvatures, K, can be
calculated from