Page 251 - Plastics Engineering

234 Mechanical Behaviour of Composites
In practice the second term in the above equation is found to be small relative
to the others and so it is often ignored and the reduced form of the Tsai-Hill
Criterion becomes

(3.54)

3.16.1 Strength of Single Plies
These failure criteria can be applied to single ply composites as illustrated in
the following Examples.

Example 3.19 A single ply Kevlar 49/epoxy composite has the following
properties.

E1 = 79 GN/m2, E2 = 4.1 GN/m2, G12 = 1.5 GN/m2, u12 = 0.43

=
&IT = 1.5 GN/m2, 62~ 0.027 GN/m2, ?12 = 0.047 GN/m2
= 0.24 GN/m2, 62~ 0.09 GN/m2.
=
If the fibres are aligned at 15" to the x-direction, calculate what tensile value
of a, will cause failure according to (i) the Maximum Stress Criterion (ii) the
Maximum Strain Criterion and (iii) the Tsai-Hill Criterion. The thickness of
the composite is 1 mm.
Solution
(i) Maximum Stress Criterion
Consider the situation where a, = 1 MN/m2.
The stresses on the local (1-2) axes are given by
[:+.["]

r12 tXY
Hence, 01 = 0.93 MN/m2, 02 = 0.067 MN/m2, t12 = -0.25 MN/m2
so

31 T 62T $12
=
=
- 1608, - 402, - = 188
01 02 tl2
Hence, a stress of a, = 1608 MN/m2 would cause failure in the local
1-direction. A stress of a, = 402 MN/m2 would cause failure in the local
2-direction and a stress of a, = 188 MN/m2 would cause shear failure in the
local 1-2 directions. Clearly the latter is the limiting condition since it will
occur first.