69
3
where
5 Z = Mid-fibre length (i.e. l/2 where l = fibre length of short
fibre)
5 r = Fibre’s radius
If the length of the fibre is smaller than l c , the stress borne by the
fibre would be less than its actual stress bearing capacity. If length is
equal to l c , the entire length of the fibre would bear the stress, and
the maximum stress would be at the centre of the fibre. If the length
of fibre is more than l c , the composite will behave as a continuous
fibre-reinforced composite. When the force is applied on a short
fibre composite, the shear force is at a maximum at the ends of the
fibres and zero at the centre of the fibre [1, 7, 8]. However, the tensile force is at a maximum at the centre of fibre and zero at the ends
of fibre.The strength of short fibre-reinforced composites would follow the rule of mixtures as described in Eq. (3.1) if the fibre length l
> > l c . In the case where l < l c and l = l c , the ultimate tensile strength
of a composite is given as
σ
τ
σ
cu
y
f
m u m
c
for
=
+
>
l
d
V
V
l l
(3.17)
σ
σ
σ ε
cu
fu
c
f
m
f m
c
for
=
−





 + ( )
<
∗
1 2
l
l
V
V
l l
(3.18)
where
5 σ cu  = Ultimate tensile strength of composite
5 σ mu  = Ultimate tensile strength of matrix
5 σ fu  = Ultimate tensile strength of fibre
5 σ m  = Tensile strength of fibre
5 l c = Critical length for fibre
5 d = Diameter of fibre
5 τ y = Yield stress of matrix in shear
5 V f  = Volume fraction of fibre
5 ε ∗
f  = Fibre fracture strain
The modulus of the radome fibre composite may be calculated as
E
E
E
Random
L
T
=





 +






3
8
5
8
(3.19)
where
5 E L  = Longitudinal modulus
5 E T  = Transverse modulus
The Halpin-Tsai equation for longitudinal and transverse moduli of
short fibre-reinforced composite is given as
E
E
l
d
V
V
L
m
L f
L f
=
+
−
1
2
1
η
η
(3.20)
3.1 · Micromechanics of Polymeric Composites