VIEWS: 98 PAGES: 44 POSTED ON: 3/14/2011
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