61
3
or
E E V E
V
c
f f
m
f
=
+
−
(
)
1
where
5 E c = Young’s modulus of composite
5 E f = Young’s modulus of fibre
5 E m = Young’s modulus of matrix
The transverse modulus, E T , may be given as
1
E
V
E
V
E
T
f
f
m
m
=
+
(3.3)
The shear modulus of composite may be given as
G
G G
G V G V
c
f m
f m
m f
=
+
(3.4)
where
5 G c = Shear modulus of composite
5 G f = Shear modulus of fibre
5 G m = Shear modulus of matrix
However, the theoretically calculated values are higher than that of
the experimental values. The reasons to this discrepancy are many.
The most important reasons are:
(a) Fibres of nonuniform strength
(b) Discontinuity in the fibres
(c) Interfacial conditions
(d) Disorientation of fibres
(e) Residual stresses
(f) Presence of voids
Interfacial conditions may deteriorate due to either mishandling or
improper storage of the fibres. Nonuniformity of strength and discontinuity of fibres occur due to mishandling of the fibres. The role
of the interface and the interphase becomes very important in
deciding the overall properties of composites. The molecular structure and surface treatments on both reinforcement and matrix
define the compatibility of these interfaces, thus the properties of
interphase. The disorientation of fibres, residual stress, and the
presence of voids occur during processing of composites. The volume fraction of voids may be calculated as
V V
ct
ce
ct
=
−
ρ
ρ
ρ
(3.5)
where
5 Vv = Volume fraction of voids
5 ρ ct = Theoretical density of composite
5 ρ ce = Experimentally determined density of composite
Point to Ponder…
It is virtually impossible to make
composites free from voids or
defects. In metals also, defects
in the form of slip or screw
dislocations are present, but they
make machinability of metals
possible.
3.1 · Micromechanics of Polymeric Composites
3
or
E E V E
V
c
f f
m
f
=
+
−
(
)
1
where
5 E c = Young’s modulus of composite
5 E f = Young’s modulus of fibre
5 E m = Young’s modulus of matrix
The transverse modulus, E T , may be given as
1
E
V
E
V
E
T
f
f
m
m
=
+
(3.3)
The shear modulus of composite may be given as
G
G G
G V G V
c
f m
f m
m f
=
+
(3.4)
where
5 G c = Shear modulus of composite
5 G f = Shear modulus of fibre
5 G m = Shear modulus of matrix
However, the theoretically calculated values are higher than that of
the experimental values. The reasons to this discrepancy are many.
The most important reasons are:
(a) Fibres of nonuniform strength
(b) Discontinuity in the fibres
(c) Interfacial conditions
(d) Disorientation of fibres
(e) Residual stresses
(f) Presence of voids
Interfacial conditions may deteriorate due to either mishandling or
improper storage of the fibres. Nonuniformity of strength and discontinuity of fibres occur due to mishandling of the fibres. The role
of the interface and the interphase becomes very important in
deciding the overall properties of composites. The molecular structure and surface treatments on both reinforcement and matrix
define the compatibility of these interfaces, thus the properties of
interphase. The disorientation of fibres, residual stress, and the
presence of voids occur during processing of composites. The volume fraction of voids may be calculated as
V V
ct
ce
ct
=
−
ρ
ρ
ρ
(3.5)
where
5 Vv = Volume fraction of voids
5 ρ ct = Theoretical density of composite
5 ρ ce = Experimentally determined density of composite
Point to Ponder…
It is virtually impossible to make
composites free from voids or
defects. In metals also, defects
in the form of slip or screw
dislocations are present, but they
make machinability of metals
possible.
3.1 · Micromechanics of Polymeric Composites