1' INTRODUCTION GENERAL PRINCIPLES AND BASIC CONCEPTS

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• 1. INTRODUCTION GENERAL PRINCIPLES AND BASIC
  CONCEPTS
• Composites in the real world Classification of
  composites scale effects the role of
  interfacial area and adhesion three simple
  models for a-priori materials selection the role
  of defects Stress and strain thermodynamics of
  deformation and Hookes law anisotropy and
  elastic constants micromechanics models for
  elastic constants Lectures 1-2
• 2. MATERIALS FOR COMPOSITES FIBERS, MATRICES
• Types and physical properties of fibers
  flexibility and compressive behavior stochastic
  variability of strength types and physical
  properties of matrices combining the phases
  residual thermal stresses Lectures 3-5
• 3. THE PRINCIPLES OF FIBER REINFORCEMENT
• Stress transfer The model of Cox The model of
  Kelly Tyson Other model Lectures 6-7
• 4. INTERFACES IN COMPOSITES
• Basic issues, wetting and contact angles,
  interfacial adhesion, the fragmentation
• phenomenon, microRaman spectroscopy,
  transcrystalline interfaces, Lectures 8-9
• 5. FRACTURE PHYSICS OF COMPOSITES
• Griffith theory of fracture, current models for
  idealized composites, stress concentration,
  simple mechanics of materials, micromechanics of
  composite strength, composite toughness Lectures
  10-11
• 6. DESIGN EXAMPLE
• A composite flywheel Lecture 12

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Types and physical properties of matrices
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Thermosets

• Heavily crosslinked polymers consisting of a
  rigid 3d molecular network
• Upon the application of heat, they degrade rather
  than melt
• Initial uncured state low viscosity
• Curing ( polymerization) involves chain
  extension, branching, crosslinking. This leads to
  rigidity, strength, solvent resistance, good
  thermal/oxidative stability
• Sensitivity to moisture use temperature limit
  200 C
• Polyesters, epoxies

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Preparation of epoxy resin
STEP 1
GROUP
epoxy
glycidyl

REACTANTS
epichlorohydrin
Bisphenol A
epoxy
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STEP 2
Reaction of an epoxide group with a diethylene
triamine (DETA) molecule ( the curing agent, or
hardener)
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STEP 3
Formation of crosslinks

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TYPES OF HARDENERS Anhydrides provide good
electrical insulating properties, thermal
resistance, environmental stability Aromatic
amines provide higher thermal resistance but
require higher cure T Aliphatic amines lead to
fast cure, suited to room T curing of epoxy ROLE
OF HARDENERS Reduce curing time, achieve
desirable properties
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EXAMPLES OF EPOXIES
DGEBA
NOVALAC (Dow DEN 438)
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SHELL EPON 1031
Ciba Geigy MY-720
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• Curing can be achieved at room temperature but
  is usually performed at higher temperatures (for
  shorter times). A postcure at relatively high
  temperature is usually added to avoid further
  (unwanted) property change during service.
• Shrinkage during cure, thermal contraction upon
  cooling after cure lead to built-in residual
  stresses in composites.
• Thermosets are usually isotropic, homogeneous
  solids.
• Thermosets loose their stiffness at the Heat
  Distorsion Temperature (HDT) upper structural
  limit.
• High-temperature thermosets include polyimides
  (PIs), bismaleimides (BMI), etc. due to
  aromatic structures

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Types and physical properties of matrices
Not treated here, limited use in composites in
general (tires)
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THERMOPLASTICS

• Advantages easier to use, faster to prepare,
  cheaper, longer shelf life, tougher, higher
  temperature use
• Not crosslinked. Origin of their
  strength/stiffness is the inherent properties of
  monomer unit, and the high molecular weight.
• Amorphous thermoplastics high concentration of
  molecular entanglements (act like crosslinks).
  Heating leads to disentanglement (rigid to
  viscous)
• Crystalline thermoplastics high degree of
  molecular order, alignment. Heating leads to
  melting (sharp melting point)
• Examples Polypropylene (semi-cryst), nylon
  (semi-cryst), polycarbonate (amorphous)
• Creep under load
• Phenomenon of transcrystallinity

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TRANSCRYSTALLINITY
HM graphite fiber in PP. The fiber acts as a
nucleating agent, heterogeneous nucleation arises.
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RESIDUAL THERMAL STRESSES

• First, some experimental observations
• Metal-matrix composites Large difference
  between TEC (a) of fiber and matrix, leading to
  interface or matrix cracking. And DT is very
  large, inducing fracture. Example SiC/Ti
• Thermoplastic polymer matrix composites Same as
  above, very high Da, DT. Moreover, fibers that
  are weak in compression break following sample
  preparation (spontaneous fragmentation). Example
  HM graphite/polypropylene

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Compressive breaks in HM graphite fiber embedded
in isotactic PP
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Compressive shrinkage arises in thermoset
matrices as well
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These compressive phenomena are attributed to
residual thermal stresses during specimen
preparation.

• QUESTIONS
• What is the magnitude of these stresses?
• Under which conditions and for which materials do
  such stresses become a critical problem for a
  structure?

Theoretical models schemes exist for (1)
one-dimensional models, and (2) isotropic
homogeneous shrink-fit concentric cylinders
with perfect interface.
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One-dimensional model

• Main hypothesis only z (longitudinal) (residual)
  stress components exist in the fibers and matrix
• Secondary hypothesis E and a are not
  temperature-dependent

Eq. 1
where a coefficient of thermal expansion E
Youngs modulus T temperature (Tref
stress-free reference temperature) f volume
fraction
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3D models

• Composite (concentric) cylinders model
• Reference papers
• J.A. Nairn, Polymer Composites 6(2) (1985),
  123-130
• H.D. Wagner, Physical Review B, 53 (9) (1 March
  1996-I), 5055-5058.
• H.D. Wagner, J.A. Nairn, Composites Science and
  Technology, 57 (9-10) (1997), 1289-1302.

History (1) 1934 fabrication of guns
(Poritski) (2) 1962 Glass/epoxy composites
(Haslett McGarry) (3) 3 cylinder solution,
anisotropic fiber, isotropic interlayer and
matrix (Nairn).
Schematic overview of the (simple version of the)
model for 2 cylinders only There are 3 steps
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One hollow cylinder Classical solution exists
for a linear elastic, isotropic cylinder
subjected to internal and external pressure
(P)i designates internal or in, o
designates external or out
HOLE
CYLINDER
For the radial, hoop and longitudinal stresses in
the cylinder. The parameter x is a constant.
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2. Next, we consider 2 concentric cylinders
internal cylinder is a solid fiber external
cylinder is hollow matrix.
Both cylinders are transversally isotropic rather
than isotropic ! In cylindrical coordinates
2 indept TECs
4 indept elastic constants
n Poisson ratio, E Youngs modulus, DT T -
Tref
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It can be shown stresses in transversally
isotropic cylinders are the same as for the
isotropic case (which were just presented).
For the internal cylinder (the fiber), internal
pressure Pi 0, internal radius Ri 0. Thus,
the equations (of step 1 above) become for the
fiber
where Afiber and Cfiber are constants.
For the external cylinder (the matrix), external
pressure Po 0. Thus, the equations (of step 1
above) become for the matrix
where Amatrix , Bmatrix, Cmatrix are constants.
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• Therefore, five unknows Af, Cf, Am, Bm, Cm.
• Radial stress boundary conditions

_at_ r R2
_at_ r R1

• Force balance in longitudinal direction

This leads to
So there are only 2 unknowns left
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To determine the 2 unknowns (AM and CM), combine
stress-strain relations with interfacial no-slip
conditions
Both _at_ r R1 (and u displacement)
But since ur r eqq, the above is equivalent to
_at_ r R1
These no-slip conditions combined with
stress-strain relations yield 2 simultaneous eqs
with 2 unknowns
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With
and
(These reduce to Nairns elements (1985) if the
matrix is isotropic)
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Solving the system of 2 equations with 2 unknowns
leads to
By inserting these into the earlier expressions
for BM, AF, CF, the residual thermal stresses in
both the fiber and the matrix may be determined.
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• Finally, we must consider 3 concentric cylinders
  internal cylinder is a solid fiber middle
  cylinder is hollow interphase external
  cylinder is hollow matrix. The cylinders are
  transversally isotropic. The solution is similar
  as for the 2 cylinder case, but more complex (see
  referenced papers).

• Illustrative examples using HM graphite/polypropyl
  ene composites
• Key questions
• What is the effect of volume fraction on residual
  stresses?
• Can we predict the amount of fiber compressive
  breaks generated by residual stresses as a
  function of DT?

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MICROCOMPOSITE (very small volume
fraction) Large compressive stresses are induced
in the fiber, as the materials cools.
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MACROCOMPOSITE (large volume fraction) Small
compressive stresses are induced in the fiber, as
the materials cools.
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Comparison between volume fractions on a single
plot
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Microcomposite
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Macrocomposite
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SUMMARY IN TABULAR FORM, FOR DIFFERENT COMPOSITES
CONCLUSIONS Criticality in micromechanical
testing mostly
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predictive tool for the effect of large
compressive stresses in fibers (of the order of
100,000 Atmospheres!)
Can we also predict the number of fiber breaks as
a function of DT?
Assumption The compressive fiber strength
follows a Weibull distribution. Then the ratio of
average compressive strengths of two fiber
populations with lengths L1 and L2 is
To predict the strength for any other length, we
need to know one pair of values (Li, ltsgtLi), and
the shape parameter b.
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ltsgtLi may be viewed as the average compressive
strength of a fiber fragment of length Li. The
average fragment length at stress level ltsgtLi is
equal to L0/Ni where L0 is the original
(unbroken) fiber length and Ni is the average
number of breaks. Therefore, for two populations
with fragment lengths L1 and L2, one has
In particular, when L1 L0, we have N1 1 and
the average number of fiber breaks up to ltsgtLi is
()
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• For a given value of the undercooling DT, one
  calculates the residual stress level ltsgtLi in the
  fiber using the models developed earlier. From
  this, the number of breaks can be derived via the
  expression () above.
• EXAMPLE Literature data for HM graphite/PP
  suggests a compressive strength of 1.08 GPa for
  a 10 mm gauge length fiber. Using () we find

This leads to the following predictive plot for 2
values of b
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How do we know which curve is correct? By
comparison with experiments in which the number
of breaks per unit length is found to vary
between 5 and 8.5 breaks per mm under DT -130
C, we see that the curve for b 2 gives a
fairly good fit. Alternatively, the fiber
volume fraction may be varied and the number of
spontaneous breaks counted per unit length. This
gives the following plot
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As predicted, there are many more breaks at low
fiber content than at high fiber content.
Calculated values of N provide the following plot
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Again, the best-fit value for b is in the range
1.5 to 2.