1
Prof C. H. XU
School of Materials Science and EngineeringHenan University of Science and Technology
Chapter 9:Mechanical Properties of Composites
Subject: Composite MaterialsScience and Engineering
Subject code: 0210080060
2
Chapter 9: Mechanical Properties of Composites -Introduction
The methods of calculating the stiffness and strength of composites Based on the orientation of reinforcement, the properties of a
composite can be Isotropic ( 各向同性) :
composites with the random orientation of reinforcement, such as sort fibers or particles
Anisotropic ( 各向异性) : continuous fiber composites, Lamellar composites Skin-Core-Skin pattern Some composite with short fibers (pressure molded short fiber
composites)
This chapter introduces the calculation of stiffness and strength of the composite fiber-reinforced or lamellar composites at special directions.
3
Mechanical Property- Fiber-reinforced or lamellar composites
elastic modulus of composites
elastic modulus elastic modulus of reinforcement of matrix
Volume fraction of matrix
Volume fraction of reinforcement
4
Heterogeneity (异质 )Composites’ properties and structures vary from point to
point. Property relationship
The properties of a composite are determined largely by properties of the constituents, their relative concentration, their geometric arrangement and the nature of the interface between them.
Anisotropythe strength and stiffness are highest in the direction of
fibre orientation, but are week in the direction of the transverse direction.
Mechanical Property- Fiber-reinforced or lamellar composites
5
Deformation
The mechanical characteristics of afiber-reinforced composite dependon the properties of the fiber and on the degree of load transmittance to the matrix phase.
Important to the degree of this loadtransmittance is the magnitude ofthe interfacial bond between thefiber and matrix phases.
Under an applied stress, this fiber-matrix bond ceases at the fiber ends.
6
Stress can be transfer from matrix to fiber(or from fiber to matrix)
through the interface
Critical fiber length for transferring stress
Critical length:
c
fc
dl
2
dependent on the fiber diameter d the fibre ultimate (or tensile) strength
f
the fiber-matrix bond strength (or theshear yield strength of the matrix) c
7
Fiber-reinforced Composites
8
Continuous fiber reinforced composites:l >> lc (normally l > 15lc)
Discontinuous or short fibers: l < 15lc
For discontinuous fibers of lengthssignificantly less than lc the matrixdeforms around the fiber such thatthere is virtually no stresstransference and little reinforcementby the fiber.
Fiber-reinforced Composites
9
Influence of Fiber Orientation and Concentration
Two extremes: (a) a parallel alignment of the longitudinal axis of the fibers in a single direction; and (c) a totally random alignment.
Fiber-reinforced Composites
10
Continuous and aligned fiber compositesTensile stress-strain behavior depends on the fiber and matrix phases; the phase volume fractions; the direction of loading.
with longitudinal loading• in stage I, both fiber and matrix deform elastically.• in stage II, the fiber continues to deform elastically, but the matrix has yielded. • from stage I to II, the fiber picks up more load.• the onset of composite failure begins as the fibers start to fracture.• composite failure is not catastrophic.
11
Longitudinal Loading
The properties of a composite depend on thefibre direction.
Assuming the fiber-matrix interfacial bond isvery good, such that deformation of bothmatrix and fibers is the same (Isostrain).
Fiber-reinforced Composites
- Continuous and Aligned Fibre Composites
cfm
12
T o t a l l o a d s u b j e c t e d b y t h e c o m p o s i t e :
fmc FFF
F r o m t h e d e f i n i t i o n o f s t r e s s :
fmc AAAfmc
D i v i d i n g t h r o u g h b y t h e t o t a l c r o s s - s e c t i o n a la r e a o f t h e c o m p o s i t e :
c
f
c
m
A
A
AA
fmc
A m / A c : t h e a r e a f r a c t i o n s o f t h e m a t r i xA f / A c : t h e a r e a f r a c t i o n s o f t h e f i b e r
Fiber-reinforced Composites
13
Fiber-reinforced Composites
Using Volume Fraction: Vm = V vol,m /Vvol,c = Am/Ac: the volume fractions of the matrix Vf = V vol,f /Vvol,c=Af/Ac: the volume fractions of the fiber The composite stress becomes:
ffmmc VV
14
The previous assumption of an isostrain state means that
fmc Therefore, we can have:
ff
fm
m
m
c
c VV
Note that: Ec = c/c ; Em = m/m ; Ef = f/f
Fiber-reinforced Composites
15
The modulus of elasticity of a continuous and aligned fibrous composite in the direction of alignment (or longitudinal
direction): Ecl
E E V E Vcl m m f f
V Vm f 1
E E V E Vcl m f f f ( )1
Rule of Mixture
E cl is equal to the volume-fraction weighted average 加权平均值 ofthe moduli of elasticity of thefiber and matrix phases.
16
Other properties, including tensile strength, also have this dependence on volume fractions.
Students can give this equation.
(TS)lc= ?
It can also be shown, for longitudinal loading, that the ratio of the load carried by the fibers to that carried by the matrix is F
F
E V
E Vf
m
f f
m m
17
Transverse Loading A continuous and oriented fiber composite may be loaded in the transverse direction; that is, the load is applied at a 900 angle to the direction of fiber alignment. For this situation the stress to which the composite as well as both phases are exposed is the same, or
c m f
This is termed an isostress state. Also, the strain or deformation of the entire composite is
c m m f fV V
Fiber-reinforced Composites
18
since
E
ff
mmct
VE
VEE
where Ect is the modulus of elasticity in the transverse direction. Now, dividing through by yields
f
f
m
m
ct E
V
E
V
E1
Rule of Mixture
19
Fiber-reinforced Composites Discontinuous and aligned fiber composites
Moduli of elasticity and tensile strengths of short fiber composites are about 50 -90 % these of long fiber composites.
When l > lc, the longitudinal strength (TS)cd
Where (TS)f is the fracture strength of the fiber and (TS)’m is the stress in the matrix when the composite fails.
When l < lc, the longitudinal strength is
Where d is the fiber diameter and c is the shear yields strength of the matrix.
)1()()2
1()()( 'fm
cffcd VTS
l
lVTSTS
)1()()( '' fmf
ccd VTSV
d
lTS
20
Fiber-reinforced Composites Discontinuous and randomly oriented composites
The orientation of the short and discontinuous fibers is random in matrix.
A ‘rule-of-mixtures’ expression for the elastic modulus Ecd
where K is a fiber efficiency parameter depended on Vf and Ef/Em ratio (0.1 ~ 0.6)
mmffcd VEVKEE
21
Example Problem:A continuous and aligned glass-reinforced composite consists of
40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, display a modulus of 3.4 GPa.
a. Compute the modulus of elasticity of this composite in the longitudinal direction.
b. If the cross-sectional area is 250mm2 and a stress of 50 MPa is applied in this longitudinal direction, compute the magnitude of the load carries by each of the fiber and matrix phases.
c. Determine the strain that is sustained by each phase when the stress in part b is applied.
d. Compute the elastic modulus of the composite material, but assume that the stress is applied perpendicular to the direction of fiber alignment.
e. Assuming tensile strengths of 3.5 GPa and 69 MPa, respectively, for glass fibers and polyester resin, determine the longitudinal tensile strength of this fiber composite.
22
Equations on the calculation for long fiber reinforced matrix composite longitudinal direction:
Transverse direction
F
F
E V
E Vf
m
f f
m m
fmc
f
f
m
m
c E
V
E
V
E
1
ffmmc VV
ffmmc VEVEE
fmc FFF
23
GPaGPaGPaVEVEE ffmmcl 30)4.0()69()6.0()4.3(
Solution
We have Ef = 69 GPa
Em = 3.4 GPa
Vf = 0.4
Vm = 0.6
a) The modulus of elasticity of the composite is calculated using equation
24
b) To solve this portion of the problem, first find the ration of fiber load to matrix load, using equation
or Ff =13.5Fm
In addition, the total force sustained by the composite Fc may becomputed from the applied stress s and total composites cross
section area Ac according to
However, this total load is just the sum of the loads carried by
fiber and matrix phases, that is Fc=Ff+Fm = 12500 N.
Substitution for Ff from the above yields
13.5Fm + Fm =12500N, Fm=862N
Whereas Ff=Fc- Fm=12500N-860N=11640NThus, the fiber phase supports the majority of the applied load.
5.13)6.0()4.3(
)4.0()69(
GPa
GPa
VE
VE
F
F
mm
ff
m
f
NMPammAF cc 12500)50()250( 2
25
c) The stress for both fiber and matrix phases must be calculated. Then, by using the elastic modulus for each (from part a), the strain values my determined.
For stress calculations, phase cross-sectional areas are necessary:
Am = VmAc= (0.6)(250mm2)=150mm2
and Af =VfAc = (0.4)(250mm2)=100mm2
Thus,
Finally, strains are commutated as
Therefore, strains for both matrix and fiber phases are identical,
which they should be, according to in the previous development.
MPamm
N
A
F
m
mm 73.5
150
8602
33
1069.1104.3
73.5
MPa
MPa
Em
mm
33
1069.11069
4.116
MPa
MPa
E f
ff
MPamm
N
A
F
f
ff 4.116
100
116402
26
d) According to equation
This value for Ect is slightly greater than that of the matrix phase, but, only approximately one-fifth of the modulus of elasticity along the fiber direction (Ecl), which indicates the degree of anisotropy of continuous and oriented fiber composites.
e) For the tensile strength TS, (TS)cl = (TS)mVm +(TS)f Vf
(TS)cl=(69 MPa) (0.6) + (3.5 x 103MPa) (0.4)=1441 MPa
f
f
m
m
ct E
V
E
V
E
1
GPaGPaGPa
GPaGPaEct 5.5
)4.3)(4.0()69)(6.0(
)69)(4.3(