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Stiffness of short fiber composites

Short fibres are less effective than long fibres

•  The load transfer mechanism results in end effects which may reduce the fibre stress.

• Difficult to control the alignment of short fibres.

• Randomly-oriented short fibres cannot be packed at such high volume fractions as continuous fibres.

And depending on the process they could become even shorter!

•  Short fibres are more often used with thermoplastic resins. Processes like injection moulding lead to considerable fibre damage and reduction in length:

σy 

σx 

Where is the average stress carried by the fibres

For any type of composite (short or long fibers) the applied load is shared by the fibers and the matrix Pc=Pm+Pf

If Ac = composite cross section Am=matrix equivalent cross section; Af= fibres equivalent cross section

σy 

σx 

The average load carried by the fibres depends on their length

L<Lc L=Lc

L>Lc L>>Lc

Fibre end effects

•  Because of the low stress at fibre ends, the average stress in the fiber will be lower than that in a continuous fiber, even if it is longer than the critical length.

•  For fibre length L = 5 Lc, the average fibre stress (σf

av) is about 90% of the maximum stress (σf

max) . •  It can be shown that (for L > Lc):

Stiffness of aligned short fibre composites

For aligned short fibre composites (difficult to achieve in polymers!), the rule of mixtures for modulus in the fibre direction is:

The length correction factor (ηL) can be derived theoretically. Provided L > 1 mm, ηL > 0.9

For composites in which fibres are not perfectly aligned the full rule of mixtures expression is used, incorporating both ηL and ηo.

€

E =ηLE fφ f + Em(1−φ f )

Theoretical length correction factor

Theore)cal length correc)on factor for glass fibre/epoxy, assuming inter‐fibre separa)on of 20 D. 

Again: Halpin-Tsai Equation

€

E1Em

=1+ηφ f

1−ηφ f

η =

E f

Em

−1

E f

Em

+ ξ

ξ = 2 Ld

€

E2

Em

=1+ 2ηφ f

1−φ fη

η =

E f

Em

−1

E f

Em

+ 2

ξ = 2

€

G12Gm

=1+ηφ f

1−φ fη

η =

Gf 12

Gm

−1

Gf 12

Gm

+1

ξ =1

Corrections for ξ

€

for ET ξ = 2 ab

+ 40φ f10

for EL ξ = 2 ld

+ 40φ f10

for GLT ξ = ab

1.732

+ 40φ f10

E1 secondo Halpin-Tsai

A di sotto di valori di ca. 3mm l’effetto rinforzante diventa pari a quello di un semplice riempitivo

Strenght of long fibers composites

Ultimate tensile strength in the fiber direction

Micromechanics 13 

The ul)mate tensile strenght can be predicted by the rule of mixtures as follows: 

σ

ε

ε*f 

fiber 

composite matrix 100 

3000 Important: The strength of the fiber is usually much higher than that one of the matrix, i.e., in the above equa)on the contribu)on of the second term is very small. Therefore, the above equa)on can be modified as follows: 

But: do the fibers reinforce the matrix, always?

Micromechanics  14 

This happens if: 

Where σmu is the ul)mate strength of the matrix 

A cri)cal fiber volume frac)on must be exceeded to produce strengthening 

Moreover…….

Micromechanics  15 

The composite will fail at a stress higher than the matrix strength only if: 

φmin and φcrit

Micromechanics  16 

σmu 

(σm)e*f 

σfu 

φcrit 

φmin 

Long fibres composites: Strength in the transversal direction

The transverse tensile strength of a composite is most likely less than the strength of the matrix alone. The fibers will act as stress concentrations, increasing the local stress in the material. Defects in fiber-matrix bond interface can be sufficiently large to constitute critical flaws. The transverse strength can be expressed as

€

σmu =σm

Swhere S is called the strength reduction factor.

The approach to determining the value of S uses either analytical or empirical estimates for the transverse composite failure strain, ε*T

Assuming brittle fracture: 

From the previous relations:

Also, it can be demonstrated that (eq. Nicolais-Narkis):

And (see slide before):

€

σTU =

ETσmu 1−4φ f

π

121− Em

E f

Em

€

εT* = εmu 1−φ f

13

€

σTU =ETσmu 1−φ f

1 3( )Em

By using Halpin-Tsai:

€

σTU = E f + ξEm + ξφ f E f −ξφ f Em( )σmu

φ f

13 −1

−E f −ξEm + φ f E f −φ f Em( )

€

σTU = Em

1+ ξ

E f

Em

−1

E f

Em

+ ξ

φ f

1−

E f

Em

−1

E f

Em

+ ξ

σmu

Em

1−φ f1 3( )

The above plots clearly indicate that it is impossible to have at the same time increase of transverse stiffness and strength.

Need to make laminates!

?

Calculate the volume fraction of fibres necessary to produce a transverse strength of 0.050 Gpa if the transverse Young’s modulus of the composite, ET, is 13.298 Gpa and the matrix has the following properties: Em = 3.1 Gpa and σmu = 0.072 GPa

Strenght of aligned short fiber composites

The micromechanics of strength are more complicated than for stiffness.

Strength depends on the relative failure strains of fibre and matrix (amongst other things).

Essen)ally, there is liQle difference between short and con)nuous fibre composite strengths once  L ≥ 10 Lc. 

•  The stress on a discontinuous fiber composite is given by the rule-of- mixtures based on the average fiber stress.

• When the composite breaks, assuming that the failure is governed by the fiber failure or pull out, the above equations can predict the composite strength if Lc replaces Lt and σmεfb replaces σm.

• 

•  For fibres longer of Lc:

€

σ cu =σ_fu φ f +σmε fb

1−φ f( )

Moduli for fibers randomly oriented

€

Young's modulus Erandom =58ET +

38EL

Shear Modulus Grandom =28ET +

18EL

Poisson's ratio ν random =Erandom

2Grandom

−1

Summary for Halpin-Tsai

•  P being the Modulus – E2 and G12 for long fibres, E1,

E2 and G12 for short fibers or other properties, such as the

thermal conductivity…

Depending on property and on geometry ξ assumes different values

€

PcPm

=1+ηφ f

1−ηφ f

η =

Pf

Pm−1

Pf

Pm+ ξ