Page 236 - Plastics Engineering

Mechanical Behaviour of Composites 219

If the diameter of the bottle is 160 mm, calculate the hoop and axial strains
in the bottle wall when an internal pressure of 200 kN/m2 is applied. Calculate
also the stresses in the individual layers.
Solution Although the composite wall is curved, the (r/h) value is large
and so it can be analysed using the method illustrated for laminates. In this
case, each ply is isotropic and so the properties do not vary with 6. It is thus
necessary to get Q for each ply relative to the centre line of the wall thickness

h~ = 0.6, hl = -0.2, h2 =O, h3 =0.2, h4 = 0.6
ie the wall section is treated as if it has four plies
[material A (0.4 mm)/material B (0.2 mm)],

Also
pr 0.2 x 80
hoop stress, cy = - = = 13.33 MN/m2
h 1.2
(or N, = cyh = 16 N/mm)
pr 0.2 x 80
axial stress, a, = - = = 6.67 MN/m2
2h 2 x 1.2
(or NR = a~h = 8 N/mm)
- [3,mOx io3 1.244 103 0
= 1.244 x io3 3.444 x io3 0 1 N/mm2,
]
- [ 956.5 376.7 ; 1.103 x lo3
0
Q2 = 376.7 956.5 N/I-M12
0 0 285.8
Also,
and
Then the Extensional Stiffness Matrix is given by
4
A = ef(hf - hf-,), u = A-' (since [BI = 0).
f =1
The strains are related to the forces (or stresses) by
[;!yl Nx

Therefore the axial and hoop strains are
Ex = 7.932 Ey = 4.809 10-~(~,, = 01