Aligned Short Fibres
In many applications it is inconvenient to use continuous fibres,
for example, composites that are injection moulded or where the form
is laid up by spraying - just look at the back of any fibre glass
chair and you can readily see the short lengths of fibre (these are
actuallty quite large, about 3cm or so) in comparison with the short
fibres used in injection moulded parts. One interesting feature of
composites containing chopped fibres is that they are almost as
strong as those containing continuous fibres; providing the fibres
exceed a critical length. Fibres shorter than the critical length
will not carry their maximum load are thus unable to function
effectively. Beyond the critical length the fibres will carry an
increasing fraction of the applied load and may fracture before the
matrix especially if the matrix material has some ductility eg. a
thermoplastic such as peek or a metal matrix. It is therefore
necessary to determine what the CRITICAL FIBRE LENGTH is.
To evaluate the critical fibre length we need to look at the
process of load transfer to the fibre - since we are grabbing hold of
the fibre through the matrix then we are relying on the shear
strength of the matrix and/or matrix fibre interface to carry the
load to the fibre - in effect a frictional loading by SHEAR
LAG. Just try holding your index finger in the fist of your other
hand and then try to pull the finger ouit from the clenched fist -
you can feel the shear resistance in the palm of your fist and you
can also feel the tensile stress in the index finger near your
knuckles put not down at the tip of your index finger. Hence a shear
stress is used to transfer the applied load to the fibre - so that
the fibre can do its job and the tensile stress that results from
this in the fibre is not the same along the length of the fibre - in
fact it increases from zero at the free end to some arbitary value in
th emiddle of the fibre then decreases as we move towards the other
free end.
First we shall work out what that shear stress in the matrix,
tm, does. Look at the small
shaded element in red,
The force acting normal to that small element df must be equal to
the shear force acting on the edge of the element or else we would
see the disc bulge at the centre. So the force on the edge is just
the product of the stress (tm)
and the area it acts on ie. the circumference multiplied by
dx
The total force acting along the fibre is simply calculated by
integrating the above equation with respect to x, the distance along
the fibre.
(We have substituted sm/2
for tm where sm
is the normal stress in the matrix - this is ok because the Tresca
criteria says that the shear stress is half the difference between
the two principal stresses - since there is only one stress,
sm, applied parallel to the
fibres and none applied perpendicualr tothe fibres (principal
stresses are always at right angles to each other) then tm
= sm/2. We can get an equation
for the stress at any point along the fibre - measured from either of
the free ends by dividing the force by the cross-sectional area of
the fibre.
So the stress would appear to increase linearly from zero at the
end to a maximum at the centre of the fibre. However, we must be
careful, as the fibre gets longer, the stress at the mid point could
rise beyond the fracture strength of the fibre - before that though
we reach a situation where it is necessary to impart some
compatability between the fibre and the matrix namely, that the
strain in the fibre (measured parallel to the fibre) cannot exceed
the strain in the matrix adjacent to it - In other words, the stress
in the fibre can increase only to a value that is equal to the strain
in the same fibre if it were a continuous fibre (isostrain rule).
Shows how the stress varies along the lengthof a fibre when the
fibre is shorter than the critical length (l1) and longer
than the critical length (l2)
for a linear elastic matrix. Similarly, the maximum fibre stress
cannot exceed the fracture strength of the fibre.
The fibre is used most efficiently when the fibre length,
lc, is such that the matrix and the fibre fail at the same
strain. This is the critical fibre length is
When l < lc then it is impossible for the fibre to
fail and the stress will increase linearly with distance from each
end to a maximum s(max) = l/d.sm.
The average stress in the fibre is then obtained by integrating the
stress function over the length of the fibre.
When the fibre exceeds the critical length, the isostrain criteria
sets a maximum strain that the fibre cannot exceed (ie. the strain in
the matrix) and so the stress in the centre section of the fibre will
be that determined by the isostrain rule. The average stress in the
fibre is then
where sf(e) is the stress in
a 'continuous' fibre at the current strain.
Now that we know the average stress in the fibre for both the case
of fibres shorter than lc and longer than lc,
we can work out the stress in the composite which is simply given by
the volume average of the stress on the two consituents and the
elastic modulus parallel to the short fibres follows from the
isostrain rule.
For l < lc, the stress on the composite is
For l > lc, the stress on the composite is
Stiffness of Aligned Short Fibre Composites
We can use the same methodology to obtain the elastic modulii of
the short fibre composite as we did for the continuous fibre
composite. Using Hookes' Law we can substitute Ee
for the stress in each component. In the matrix, the strain is
uniform and equal to that in the composite, in th eshort fibres, the
strain varies with position along the fibre but averages out, along
each fibre, to the strain in the matrix. Hence after canceling the
strain terms we end up with.
- The stifness when l < lc
- The stifness when l > lc
The stiffness measured perpendicular to
the fibre axis is just the same as the continuous fibre case that
we looked at earlier. But what about where the fibres are oriented in
random directions ? Since random orientations occur quite commonly in
composites which are made by spraying a combination of chopped fibres
and resin onto a mould form we need a different method for estimating
the elastic modulus.
Orientation Distributions
How do we calculate the elastic properties of a short fibre composite when the fibres are not aligned but have a distribution of orientations. Click here to find out!
Strength of Short Fibre Composites
In short fibre composites - such as dough and sheet moulding
compounds, or even chopped fibre reinforced mmc castings, the
strength of the composite will depend on the length of the fibres and
the orientation as well as the volume fraction. It is perhaps one of
the great advanages of the injection moulding - squeeze casting
fabrication route (possibly forging) that the dies can be made such
that the flow of matrix is approximately parallel to the direction
which will experience the greatest tensile stress. Given the fibre
morphology, the bulk of the fibres will align in the flow direction
and the effective volume fraction of fibres can range from 50 to
almost 100% of the nominal fibre loading.
If the fibres are of optimum length then the matrix and fibre will
fail simultaneously, the fibre experiencing its fracture stress at
the midpoint of the fibre, with the average stress carried in the
fibre being one half that value. Using the isostrain rule we find
where f, in this case, is the actual fibre loading parallel to the
tensile stress. If the fibres are longer than the critical length
then the strength in the composite will depend on whether the matrix
or fibres fail first and the arguments developed in the previous two
sections can be followed if 
is substituted for sf
How do we calculate the strength of a short fibre composite when the fibres are not aligned but have a distribution of orientations. Click here to find out!
