CONTINUOUS FIBER COMPOSITE STRENGTH Composite stress is given by the simple rule of mixtures:

1 (1 )f f fV V mσ σ σ= + − 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 uσ σ= + − σ u 1 (1 )u f fu f mfV Vσ σ σ= + −

' ' muf

fu fmu mu

V f σσ σ σ

= =− +

' ' mu mfuf

fu mfu m

V fu

σ σσ σ σ

−= =

− +

Short-Fiber Composites 1 CE 537 - Washington State University

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 TransferThe 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:

σc

dz

σf

σf + dσf

τ

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:

( ) ( ) ( ) ( )π σ π τ π σ σr r dz rf f2 22+ = + d f

1 Agarwal, B.D. and L.J. Broutman. 1990. Analysis and Performance of Fiber Composites; 2nd edition. John Wiley and Sons, Inc. NY, NY. 449 pp.

Short-Fiber Composites 2 CE 537 - Washington State University

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

ddz r

fσ τ=

2

This relation implies that the change in fiber stress is proportional to the shear stress at the 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.

σ σ τf fo

z

rdz= + ∫

2

0

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

σ τf

z

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 (τy). The normal stresses in the fiber can then be given as:

στ

fyz

r=

2

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

στ

fy lr

max =

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

σ σff

cc

EE

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:

Short-Fiber Composites 3 CE 537 - Washington State University

( )ld

E Et f

y

f c

y

= =στ

στ

max /2 2

c

If we consider that the maximum fiber stress can be equated to that required to fail the fiber (σfu), then we can imagine that extremely short fibers will never develop enough stress to fail. This concept leads to the theory that a critical fiber length (lc) exists for effective reinforcement:

ldc fu

y=στ2

Semi-Empirical Equations Halpin-Tsai EquationsHalpin 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

EE

υηυξη

−

+=

11

where: ( )( ) imf

mfi EE

EEξ

η+

−=

/1/

i indicates the two principal material directions; longitudinal (L) or transverse (T) to the fiber direction

The reinforcement shape parameter, ξi depends on the direction of loading and the filler shape:

2,2 == TL xL ξξ where: x equals the fiber diameter or platelet thickness

For random filler orientation the modulus is approximated as3:

85

83 TL

rEEE += OR

48TL

rEEG +=

2 Halpin, J.C. and S.W. Tsai. 1969. Effects of Environmental Factors on Composite Materials. AFML-TR 67-423. 3 Agarwal, BD and JL Broutman. 1990. Analysis and Performance of Fiber Composites. John Wiley and Sons, NY, NY. 449pp.

Short-Fiber Composites 4 CE 537 - Washington State University

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 υσυσσ +=

And peak stress at the center of fiber = max 2f

ldτσ = because force acting on half a fiber is:

x

F3

Assuming a linear stress variation in the fiber

21

22

3

/ 4

[ ]f

f f

F D

F d D

F Ddx

σ π

σ σ π

τπ

=

= +

=

/ 4

D

For equilibrium, F1 + F3 = F2 and by integration and simplification, we get

4 ( / 2 ) /f l xσ τ= − Note, x is measured from the mid-length of the fiber and τ 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

Short-Fiber Composites 5 CE 537 - Washington State University

σfu

The average fiber stress, fσ , 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 Notesl < lt mmffc υσυσσ += max

21 Triangular stress over lt

l > lt ( ) mmfftc ll υσυσσ +−= max21

l >> lt mmffc υσυσσ += max 121 ≈− llt Considering both the ultimate fiber stress (σfu) and matrix stress (σmu) we can compute the composite material strength (σcu)… Length Governing Equation Notesl < lc ( ) mmufyc dl υσυτσ += Matrix failure or fiber pull-

out l > lt ( ) mmffutc ll υσυσσ ′+−= 21 σ’m = σm at εfu

l >> lt mmffuc υσυσσ ′+= 121 ≈− llt

l lc/2

lc

l

2 clD

τ

2 clD

τ

2 lD

τ

σfu

σfu

fσ

fσ

fσ

Short-Fiber Composites 6 CE 537 - Washington State University

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

is often )

where υf’ is 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 andincluded in the fiber volume fraction. Therefore, an effective fiber volume fraction (υf

eff

and fiber modulus (Efeff ) is substituted in the equations above for υ and E respectively.

These terms can be computed as:

f f

ifefff υυυ +=

iiffefff EEE υυζ += '

A common

effffi EEE

modified Halpin-Tsai equations to the reinforced composite modulus at varying levels υf and solving for υi

4 Jancar, J. 1999. Engineered Interfaces in Polypropylene Composites. In: Handbook of Polypropylene and Polypropylene Composites, HG Karian ed. Marcel Dekker, NY, NY. 559pp.

Short-Fiber Composites 7 CE 537 - Washington State University

Equations Relating to Weight and Volume Fractions

y Definition B

n Volume Fractio cmmcff VVandVV == υυ Weight Fraction cmmcff WWandWW == ωω Invoking the definition of density:

fc

ffff VWυ

cccf VW ρ

ρρρ

ω === and mc

m

cc

mm

c

mm V

VWW

υρρ

ρρ

ω ===

y identity:B

m 1=+=+ fmf ωωυυ

ssuming that the material is void free:

voking the definition of density:

Amfcmfc VVVandWWW +=+=

Inmmffcc VVV ρρρ +=

ividing each side by the composite’s volume leads to: D

mmffc ρυρυρ += r using similar paths for weight fraction: o

m

mf ω

fc ρρω

ρ

+=1

Short-Fiber Composites 8 CE 537 - Washington State University