64
3
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 matrix at fracture
5 V crit = Critical volume fraction of fibre
5 V min = Minimum volume fraction of fibre
If the matrix is stronger than the fibre, V min and V crit will be very
large. Hence, the resultant composite is strong. If a strain at a break
of the matrix is equal to a strain at a break of the fibre (i.e. ε m = ε f ),
the strength of the composite would be reduced, and the full potential of the fibre strength would not be realized. If the strain at a
break of the matrix is less than the strain at a break of the fibre (i.e.
ε m < ε f ), the strength of the composite would be reduced further,
and the matrix will fail at a very low stress [1, 4]. The load would not
be borne by the fibres, and the full potential of fibre strength would
not be realized. The composite will manifest the properties of the
matrix only. This will lead to the initiation of cracks and areas of
stress concentration. The different modes of failure in composites
are discussed in the following sections.
? Example 3.2 The strengths of a Kevlar fibre and matrix are
2350 MPa and 100 MPa, respectively. The experimental tensile
strength of this composite is 400 MPa. It has volume fractions
for the reinforcement and matrix 0.4 and 0.6, respectively. (a)
Calculate the critical volume fraction, minimum volume
fraction, and void contents of this composite. (b) Estimate the
possible nomenclature of the matrix.
v Answer
(a)
σ* c = 600 MPa
σ fu = 2350 MPa
σ mu = 100 MPa
V f = 0.4
V m = 0.6
σ
σ
σ
c
f f
m m
∗
∗
∗
=
+
V
V
σ
σ
σ σ
m
c
f f
m
m
MPa
∗
∗
∗
=
−
=
V
V
;
90
V
V
crit
mu
m
fu
m
crit
=
−
−
=
∗
∗
σ
σ
σ
σ
;
.
0 004
V
V
min
m in
;
.
=
−
+
−
=
∗
∗
σ
σ
σ
σ
σ
mu
m
fu
mu
m
0 004
(b) This matrix is expected to be a thermoplastic matrix such as
PEI or PEEK.
Chapter 3 · Micromechanics and Macromechanics of Polymeric Composites
3
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 matrix at fracture
5 V crit = Critical volume fraction of fibre
5 V min = Minimum volume fraction of fibre
If the matrix is stronger than the fibre, V min and V crit will be very
large. Hence, the resultant composite is strong. If a strain at a break
of the matrix is equal to a strain at a break of the fibre (i.e. ε m = ε f ),
the strength of the composite would be reduced, and the full potential of the fibre strength would not be realized. If the strain at a
break of the matrix is less than the strain at a break of the fibre (i.e.
ε m < ε f ), the strength of the composite would be reduced further,
and the matrix will fail at a very low stress [1, 4]. The load would not
be borne by the fibres, and the full potential of fibre strength would
not be realized. The composite will manifest the properties of the
matrix only. This will lead to the initiation of cracks and areas of
stress concentration. The different modes of failure in composites
are discussed in the following sections.
? Example 3.2 The strengths of a Kevlar fibre and matrix are
2350 MPa and 100 MPa, respectively. The experimental tensile
strength of this composite is 400 MPa. It has volume fractions
for the reinforcement and matrix 0.4 and 0.6, respectively. (a)
Calculate the critical volume fraction, minimum volume
fraction, and void contents of this composite. (b) Estimate the
possible nomenclature of the matrix.
v Answer
(a)
σ* c = 600 MPa
σ fu = 2350 MPa
σ mu = 100 MPa
V f = 0.4
V m = 0.6
σ
σ
σ
c
f f
m m
∗
∗
∗
=
+
V
V
σ
σ
σ σ
m
c
f f
m
m
MPa
∗
∗
∗
=
−
=
V
V
;
90
V
V
crit
mu
m
fu
m
crit
=
−
−
=
∗
∗
σ
σ
σ
σ
;
.
0 004
V
V
min
m in
;
.
=
−
+
−
=
∗
∗
σ
σ
σ
σ
σ
mu
m
fu
mu
m
0 004
(b) This matrix is expected to be a thermoplastic matrix such as
PEI or PEEK.
Chapter 3 · Micromechanics and Macromechanics of Polymeric Composites