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Fracture physics of composite materials (PowerPoint)

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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 Hooke’s 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}

7. THE FUTURE:
    Composites based on nanoreinforcement, composites based on biology, ribbon- and
    platelet-reinforced materials, biomimetic concepts {Lectures 13-15}
                                                                                                    1
   Types and physical properties of matrices



                              POLYMERS


    THERMOPLASTICS        ELASTOMERS (RUBBERS)   THERMOSETS


Crystalline   Amorphous




                                                              2
                          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




                                                                        3
               Preparation of epoxy resin

STEP 1
                     GROUP
                                epoxy            glycidyl



         REACTANTS                        +
                       epichlorohydrin
                                                 Bisphenol A




                                         epoxy




                                                               4
STEP 2   Reaction of an epoxide group with a diethylene triamine
           (DETA) molecule (= the curing agent, or hardener)




                                                                   5
STEP 3   Formation of crosslinks




         +




                                   6
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

                                                       7
EXAMPLES OF EPOXIES




                           DGEBA




                        NOVALAC
                      (Dow DEN 438)



                                      8
 SHELL EPON 1031




Ciba Geigy MY-720



                    9
• 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




                                                               10
   Types and physical properties of matrices



                              POLYMERS


    THERMOPLASTICS        ELASTOMERS (RUBBERS)         THERMOSETS


Crystalline   Amorphous



                                Not treated here, limited
                                use in composites in general
                                (tires)



                                                                    11
                     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

                                                                  12
TRANSCRYSTALLINITY




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




                                         13
14
15
            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




                                                               16
17
Compressive breaks in HM graphite fiber embedded in isotactic PP

                                                                   18
Compressive shrinkage arises in thermoset matrices as well




                                                             19
These compressive phenomena are attributed to residual
thermal stresses during specimen preparation.



 QUESTIONS:
 1. What is the magnitude of these stresses?
 2. 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.




                                                                20
                     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 – Young’s modulus
    T – temperature (Tref = stress-free reference temperature)
    f – volume fraction                                          21
                               3D models
Composite (concentric) cylinders model

Reference papers:
1.  J.A. Nairn, Polymer Composites 6(2) (1985), 123-130
2. H.D. Wagner, Physical Review B, 53 (9) (1 March 1996-I),
    5055-5058.
3. 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
                                                                          22
1.      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.



                                                                               23
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 indep’t TECs



                          4 indep’t elastic
                          constants

        n – Poisson ratio, E – Young’s modulus, DT = T - Tref
                                                                           24
[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 P i = 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.




                                                                          25
•   Therefore, five unknows Af, Cf, Am, Bm, Cm.

•   Radial stress boundary conditions:
                                                                @ r = R2

                                                                @ r = R1


• Force balance in longitudinal direction:




    This leads to:

                                                          So:
                                             there are only 2 unknowns left

                                                                              26
To determine the 2 unknowns (A M and CM), combine stress-strain
relations with interfacial no-slip conditions:


                   Both @ r = R1 (and u = displacement)



  But since ur = r eqq, the above is equivalent to:

                                                          @ r = R1


These no-slip conditions combined with stress-strain
relations yield 2 simultaneous eqs with 2 unknowns:




                                                                     27
With:




and
        (These reduce to
        Nairn’s elements
        (1985) if the matrix
        is isotropic)


                          28
Solving the system of 2 equations with 2 unknowns leads to:




   By inserting these into the earlier expressions for B M, AF, CF, the
   residual thermal stresses in both the fiber and the matrix may be
   determined.




                                                                          29
•     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/polypropylene composites
    Key questions:
    1. What is the effect of volume fraction on residual stresses?
    2. Can we predict the amount of fiber compressive breaks
       generated by residual stresses as a function of DT?




                                                                       30
31
MICROCOMPOSITE
(very small volume
fraction):
Large compressive
stresses are induced in
the fiber, as the
materials cools.




                     32
MACROCOMPOSITE
(large volume fraction):
Small compressive
stresses are induced in
the fiber, as the
materials cools.




                    33
Comparison between volume fractions on a single plot:




                                                        34
Microcomposite




                 35
Macrocomposite




                 36
SUMMARY IN TABULAR FORM, FOR DIFFERENT COMPOSITES




   CONCLUSIONS: Criticality in micromechanical testing mostly
                                                                37
= 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, <s>Li), and the shape parameter b.
                                                                    38
<s>Li may be viewed as the average compressive strength of a fiber
fragment of length Li. The average fragment length at stress level <s>Li
is equal to L0/Ni where L0 is the original (unbroken) fiber length and N i
is the average number of breaks.
Therefore, for two populations with fragment lengths L 1 and L2, one has:




In particular, when L1 = L0, we have N1 = 1 and the average number of
fiber breaks up to <s>Li is


                                                     (#)




                                                                         39
   For a given value of the undercooling DT, one calculates the residual
   stress level <s>Li 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:




                                                                         40
41
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:




                                                              42
As predicted, there are many more breaks at low fiber
content than at high fiber content. Calculated values of N
provide the following plot:
                                                             43
Again, the best-fit value for b is in the range 1.5 to 2.   44