Short Fiber Composite Theory

8/13/2019 Short Fiber Composite Theory

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CONTINUOUS FIBER COMPOSITE STRENGTH

Composite stress is given by the simple rule of mixtures:

1 (1 )f f fV V m = +

for strains before onset of matrix cracking.

Two Cases

Case 1: Fiber has a higher strain to failure, and

Case 2: Matrix has the higher strain to failure

In the figure above (Hull, D. and Clyne, T. W. 1996.An Introduction to Composite

Materials.Cambridge Univ. Press):

1 (1 )u f fmu f mV V u = + u 1 (1 )u f fu f mf V V = +

' ' muf

fu fmu mu

V f

= =

+ ' '

mu mfu

f

fu mfu m

V fu

= =

+

Short-Fiber Composites 1

CE 537 - Washington State University


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SHORT FIBER COMPOSITES

Cost effective manufacturing often drives the use of discontinuous or short reinforcing

elements in composites. These materials differ from continuous reinforced composites

because the fibers do not extend throughout the entire material. Therefore, load is not

directly applied to each reinforcing element.

Stress Transfer

The most commonly quoted theory of stress transfer in discontinuous element composites

is the shear-lag analysis.1 Classically, this element is considered as a discontinuous

cylindrical fiber embedded in a continuous matrix.

Differences in longitudinal strain in the matrix and adjacent fiber (matrix >> fiber) will

result in shear stresses at the interface. Tensile load is transferred to the fibers by a

shearing mechanism between fibers and matrix. The stresses acting on this body are:

c

dz

f

+ d

Ignoring stress transfer at the fiber end cross sections and interaction between

neighboring fibers, we can calculate normal stress distribution by a simple equilibrium

analysis. Consider the force equilibrium on the infinitesimal length, dz:

( ) ( ) ( )( ) r r dz r f f2 22+ = + d f

1Agarwal, B.D. and L.J. Broutman. 1990. Analysis and Performance of Fiber Composites; 2ndedition.

John Wiley and Sons, Inc. NY, NY. 449 pp.




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Reducing this relation, we find that the change in fiber stress with length is:

d

dz r

f

=2

This relation implies that the change in fiber stress is proportional to the shear stress atthe interface between the fiber and matrix. The fiber stress at any point along the fiber is

obtained by integrating this equation with respect to z.

f fo

z

rdz= +

2

0

The stress at the fiber end, fo, is often negligible and can be neglected in theformulation:

fz

rdz= 2

0

This formulation requires knowledge for the shear stresses along the fiber length.

Although these are seldom known, one frequently used assumption is that the matrix

material is perfectly rigid and perfectly plastic. In this case, the shear stress would

develop instantly at the end of the fiber, reaching a maximum of the yield stress (y). Thenormal stresses in the fiber can then be given as:

f

yz

r=

2

The maximum normal stress in the fiber will occur a mid-length (z = l/2):

f

yl

r

max =

Assuming continuity of strains between the fiber, matrix, and composite (i.e. f= m=c), the maximum fiber stress can be given as:

ff

c

cEE

max =

Because the normal stress in the fiber changes along the fiber length, we can assume that

a minimum fiber length (lt) exists to transfer the maximum fiber stress. This minimum

fiber length can be calculated by combining the previous two equations and rearranging:




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( )ld

E Et f

y

f c

y

= =

max /

2 2

c

If we consider that the maximum fiber stress can be equated to that required to fail the

fiber (fu), then we can imagine that extremely short fibers will never develop enoughstress to fail. This concept leads to the theory that a critical fiber length (lc)exists foreffective reinforcement:

l

d

c fu

y

=

2

Semi-Empirical Equations

Halpin-Tsai Equations

Halpin and Tsai2 performed a more exact micro-mechanics analysis on unidirectional

composites with short fibers. For simplicity, approximate equations were produced that

afford a more precise prediction of properties.

The basic form of the relations is given as:

fi

fii

m

i

E

E

+=

1

1

where:

( ) imf

mf

iEE

EE

+

=

/

1/

iindicates the two principal material directions; longitudinal (L) or transverse (T) to the

fiber direction

The reinforcement shape parameter, idepends on the direction of loading and the fillershape:

2,2 == TLx

L where: x equals the fiber diameter or platelet thickness

For random filler orientation the modulus is approximated as3:

8

5

8

3 TLr

EEE += OR

48

TLr

EEG +=

2Halpin, J.C. and S.W. Tsai. 1969. Effects of Environmental Factors on Composite Materials. AFML-

TR 67-423.3Agarwal, BD and JL Broutman. 1990. Analysis and Performance of Fiber Composites. John Wiley and

Sons, NY, NY. 449pp.




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Discontinuous Fiber Composite Strength

Assuming (1) an aligned short fiber composite with (2) the load placed parallel to the

fiber direction, the overall stress in the composite can be calculated as:

mmffc +=

And peak stress at the center of fiber = max2

f

l

d

= because force acting on half a fiber is:

x

F3

Assuming a linear stress variation in the fiber

2

1

2

2

3

/ 4

[ ]

f

f f

F D

F d DF Ddx

=

= +=

/ 4

D

For equilibrium, F1+ F3= F2and by integration and simplification, we get

4 ( / 2 ) /f

l x =

Note, x is measured from the mid-length of the fiber and is assumed to be a constant.

In solving these equations, recall that the normal stress in a fiber changes along the

discontinuous fiber length.

dx

F1 F2

0

l/2




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fu

The average fiber stress,f , is determined by taking the area under the stress-fiber

length graph and dividing by the fiber length.

Now consider the following three cases:

Length Governing Equation Notes

l < ltmmffc +=

max

21 Triangular stress over lt

l > lt ( ) mmfftc ll +=

max

21 l >> lt

mmffc += max 121 llt

Considering both the ultimate fiber stress (fu) and matrix stress (mu) we can computethe composite material strength (cu)

Length Governing Equation Notes

l < lc ( ) mmufyc dl += Matrix failure or fiber pull-out

l > lt ( ) mmffutc ll += 21 m= mat fu

l >> lt mmffuc += 121 llt

llc/2

lc

l

2clD

2 clD

2 lD

fu

fu

f

f

f




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Interphase Volume

For composites with engineered filler-matrix interfaces, a third phase must be

is often

)

where fis the filler volume fraction in the fiber-interphase

complex.

simplification assumes:

==

Using this assumption, the effective fiber volume fraction can be solved by fitting the

of

considered.4 The material interphase has properties higher than the matrix and

included in the fiber volume fraction. Therefore, an effective fiber volume fraction (feff

and fiber modulus (Efeff

) is substituted in the equations above for and E respectively.These terms can be computed as:f f

if

eff

f +=

iiff

eff

f EEE += '

A common

eff

ffi EEE

modified Halpin-Tsai equations to the reinforced composite modulus at varying levelsfand solving for i

4Jancar, J. 1999. Engineered Interfaces in Polypropylene Composites. In:Handbook of Polypropylene

and Polypropylene Composites, HG Karian ed. Marcel Dekker, NY, NY. 559pp.




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Equations Relating to Weight and Volume Fractions

y Definition

B

nVolume Fractiocmmcff VVandVV ==

Weight Fractioncmmcff WWandWW ==

Invoking the definition of density:

f

c

ffff VW

ccc

fVW

=== and m

c

m

cc

mm

c

m

mV

V

W

W

===

y identity:B

m 1=+=+ fmf

ssuming that the material is void free:

voking the definition of density:

A

mfcmfcVVVandWWW +=+=

In

mmffcc VVV +=

ividing each side by the composites volume leads to:D

mmffc +=

r using similar paths for weight fraction:o

m

mf

fc

+=1



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Short Fiber Composite Theory