Subject: Composite Materials Science and Engineering Subject code: 0210080060

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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

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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.

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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

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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

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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.

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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

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Fiber-reinforced Composites

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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

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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

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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.

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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

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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

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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

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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

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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.

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     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

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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(