Manufacturing of Composites PDF

Manufacturing of composites MECH 415/6521 Suong V. Hoa Concordia Center for Composites Mechanical and Industrial Engineering Concordia University Introduction  General characteristics of manufacturing of composites Requirements for a good piece Goo...

Manufacturing of composites MECH 415/6521 Suong V. Hoa Concordia Center for Composites Mechanical and Industrial Engineering Concordia University Introduction  General characteristics of manufacturing of composites Requirements for a good piece Good bonding between matrix and fibers Proper orientation of fibers Good amount of volume fraction of fibers Uniform distribution of fibers within matrix material Proper curing or solidification of the resin Limited amount of voids and defects Good dimensional control of the final part. Good bonding between fibers and matrix Good bonding gives better strength and stiffness Poor bonding gives stress concentration. Dry spots. Partial bonding sometime can help to absorb impact energy Fiber orientation Fibers tend to be wavy- Microwaviness. Overlacing Misalignment due to flow of liquid during manufacturing Reinforcing along thickness direction- Z pinning, 3D weaving Hard pressing with stiff rollers Molding over complex geometry. Human errors during Hand Lay Up.  Fiber volume fraction Vf vf = Vc Ec = E f v f + vm E m 1 = v f + v m + vv 1 = v f + vm E c = (E f − E m )v f + E m  Uniform fiber distribution Resin rich area Fiber to fiber contact  Proper curing of the resin Curing for thermosets Consolidation for thermoplastics Limited amounts of voids and defects Less than 1% voids Good dimensional control of the part  Metal versus composites manufacturing Metals 1. Minerals 2. Liquid melts 3. Billets 4. Rods, plates 5. Cutting, Bending, Forming, Machining Transformation matrix Level 1: (liquid) Micromechanics Matrix (binder) Transformation Level 2: Macromechanics fiber Composites Lamina Structure bundle Laminate (polymer filaments Reinforcing fibers Curved matrix) laminated making structures fabric (a) (b) (c) (d) Functions of constituents Fibers Matrix Interface  Advantages of fiber form Stronger than bulk form More fabrication techniques Stretching Drawing Solvent removal Flexibility in forming d ε max = 2ρ  Fibers Disadvantages of fiber form 1. Requirements of large number of fibers Assembly issue Alignment issue  Fibers Disadvantages of fiber form 2. Fibers need to be bonded together to provide good mechanical properties 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 Fiber strength MPa x 10-2  Fibers Disadvantages of fiber form Fibers need to be bonded together to provide good mechanical properties P1 = 0.30 N P2 = 0.35 N Fiber P3 = 0.25 N P4 = 0.40 N P5 = 0.50 N P  Fiber bundle behavior Adhesive Fiber Crack P Stress distribution in the neighborhood of a crack  Fracture behavior of composites Metal fracture Composite fracture  Fibers Disadvantages of fiber form 3. The need for high fiber volume fraction Square array Hexagonal array a. Open packing b. Closed packing  Fibers Disadvantages of fiber form 4. Small inter-fiber spacing a. Stress concentration 2 0 la yers (w ater) 3 0 la yers (w ater) 2 0 la yers (o il) b. Fiber to fiber contact 1 0 00 K x’ oil = 0.35 Permeability Kx x 10- c. Large shear resistance of prepregs 100 d. Small permeability values 10 K x’ H 2 O = 0.6 8 10(cm2) e. Anisotropic behavior in 1 0 0.1 0.2 0.3 0.4 0.5 0.6 permeability Liquid volume fraction V R 20 layers (w ater) Fiber c o n tact Resin pocket 30 layers (w ater) 20 layers (oil) P 1000 K o z e n y- C a r m a n e q u a t i o n 1 Kz = 11 Permeability Kz x 10- 100 10 11(cm2) 1 0 0.1 0.2 0.3 0.4 0.5 0.6 P Li q u i d vo l u m e f r act i on V R  Matrix materials Functions of the matrix 1. Aligning the fibers 2. Transfer the load between fibers  Matrix materials Functions of the matrix 3. Assisting fibers in providing compression strength and modulus 4. Assisting fiber in providing shear strength and modulus 5. Protecting fibers from environmental attack.  Interface area 4vf Interface Composite volume ≈ d Two main requirements for good interface S ij ui = ∂p µ ∂x j  S11 0 0   S = 0 S  0  ij 22   0 0 S 33  S T dp uz = µ dz Compatibility: Depends on surface Availability: Depends on energy speed to be on site.  Volume fraction and weight fraction vf =Vf/ Vc wf =Wf/Wc Percent (%) and Parts per hundred (phr)  How are composite structures made? Transformation matrix Level 1: (liquid) Micromechanics Matrix (binder) Transformation Level 2: Macromechanics fiber Lamina Structure bundle Laminate Reinforcing filaments fibers Curved laminated making structures fabric (a) (b) (c) (d) Thank you Manufacturing of composites MECH 415/6521 Suong V. Hoa Concordia Center for Composites Mechanical and Industrial Engineering Concordia University Matrix Different types of matrix materials Polymer matrix composites – Carbon/epoxy – Glass/epoxy – Glass/polyester – Kevlar /epoxy Metal matrix composites – Silicon carbide/aluminum – Carbon fiber/aluminum Ceramic matrix composites – Carbon/carbon – Carbon/alumina There are more polymer matrix composites as compared to other types of composites due to the Compatibility condition. Different types of polymer matrix materials Thermoset matrix composites – Carbon/epoxy – Glass/epoxy – Glass/polyester – Kevlar /epoxy Thermoplastic matrix composites – Carbon/PEEK – Carbon/PPS – Glass/nylon There are more thermoset matrix composites as compared to thermoplastic matrix composites due to the Availability condition.  Material 20oC 25 oC ToC Air 0.0187 Water 1 Polyester 100-300 Vinyl ester 100-300 #10 Motor oil 500 Golden syrup 2,500 Epoxy (Shell Epon 828-14 600 phrMPDA, 15 phr BGE) Epoxy (Shell 826 – 16 phr 750 MPDA, 10phr BGE) Epoxy (Dow 332-16 phr MPDA, 500 10 phr BGE) Molasses 105 Epoxy 5208 100@177oC BMI 1000@150 oC Ryton (thermoplastic) 107 @313oC PEEK (thermoplastic) 106@400 oC Utem (thermoplastic) 108@305 oC Torlon (thermoplastic) 109@350 oC Table 2.1: Viscosity of a few thermoset and thermoplastic materials (in centipoise) 1 Pa-sec = 10 Poise = 1000 centi-Poise  (a) (b) (c) (d) (e) Figure 2.1 : Schematic of (a) the molecules in a thermoset resin, (b) the linking molecules, (c) the resin molecules and the linking molecules in a container before linking reactions, (d) the thermoset resin network after linking reactions, and (e) a partially linked network.  Thermoset matrix polymers Thermoset matrix composites – Carbon/epoxy – Glass/epoxy – Glass/polyester – Kevlar /epoxy Thermoplastic matrix composites – Carbon/PEEK – Carbon/PPS – Glass/nylon There are more thermoset matrix composites as compared to thermoplastic matrix composites due to the Availability condition. Figure 2.2: Schematic of the molecules in a thermoplastic resin. Polyester production  Example 2.1: It is desired to make a polyester resin using 100 g of maleic acid and a corresponding amount of ethylene glycol. A stoichiometric amount of ethylene glycol is used. After the condensate is removed, how many grams of polyester are obtained? Mass of different atoms: C = 12 g/mol, H = 1 g/mol, O = 16 g/mol, N = 14g/mol REPEATING UNIT Maleic acid: 4 C + 4 H + 4 O = 4 ( 12 + 1 + 16) =116 g/mole Ethylene glycol: 2 C + 6 H + 2 O = 24 + 6 + 32 = 62 g/mole The reaction takes place as shown in the following: It can be seen that one molecule of maleic acid reacts with one molecule of ethylene glycol to make a unit of ester and two water molecules. Mass of the water molecule: 2 H + O = 18 g/mole. Let Mp be the mass of the polyester made using 100 g of maleic acid, we have: (Mp/100) = (116 + 62 - 36)/116 = 1.224. Mp = 122.4 g.  Initiators Name of peroxide Chemical structure Hydrogen peroxide H-OO-H Hydroperoxides R-OO-H Dialkyl peroxides R-OO-R Diacyl peroxides R-C(O)-OO-C(O)-R Peroxyesters R-C(O)-OO-R Peroxy Acids R-C(O)-OO-H Peroxy Ketals R2-C-OO-R2 Peroxy Dicarbonates R-OC(O)-OO-C(O)O-R Table 2.2: Structures of commercial organic peroxides Polyester cross linking Example 2.2: Example on cross linking of polyester: It is desired to make a polyester using 100 g of maleic acid and ethylene glycol. A stoichiometric amount of ethylene glycol is used. Cross linking is done using styrene. Assume that one styrene monomer corresponds to one oligoester (this assumption is to simplify the calculation to illustrate the principle; in reality the crosslink between oligoester molecules can be in the range from 1 to 4 styrene monomers). From the cross linking process, how many C=C bonds are broken, how many C- C bonds are formed? Solution: Continuing from the same problem in Example 2.1. Number of bonds broken and formed: Therefore for each polyester unit, there are 2 C=C bonds broken and 4 C-C bonds formed. The chemical formula for a polyester unit is: O O The mass of a polyester unit is: C CH CH C O CH2 CH2 O n 6 C + 6 H + 4 O = 72 + 6 + 64 = 142 g/mole. Since there are 122.4 g of polyester made, the number of bonds involved is: C=C bonds: (122.4/142) 2 = 1.724 mole or 1.038 x 1024 bonds. (note that 1 mole = 0.602 x 10 24). C-C bonds: 2 x 1.724 = 3.446 moles or 2.076 x 1024 bonds. Atomic bond energy Bond Energy (kJ/mole) C-C 370 C=C 680 C-H 435 O-H 500 N-H 430 C-O 360 C=O 535 Example 2.3: Example on heat generation and temperature increase: It is desired to make a polyester using 100 g of maleic acid and ethylene glycol. A stoichiometric amount of ethylene glycol is used. Cross linking is done using styrene. Continuing from the same problem in Examples 2.1 and 2.2. The energy required to break a C=C bond is 680 kJ/mole and the energy created by the forming a C-C bond is 370 kJ/mole. How much energy is generated during the polymerization process? 1 mole = 0.602 x 1024. The heat capacity of polyester is 0.25 cal/g/oC. Assuming no heat loss, what is the increase in temperature of the polyester? Mass of carbon C= 12 g/mole, H = 1 g/mole, O = 16 g/mole Solution: a) Energy generated: Considering the energy in the bonds, one has: Energy inputted into the system to break the double bonds: (1.724mole) (680 kJ/mole) = 1172.32 kJ. Energy generated from the system to form the single bonds: (3.446 moles)(370 kJ/mole) = 1275.02 kJ. Net energy generated: 1275.2- 1172.32 = 102.7 kJ. b) Temperature increase: Heat stored in the material: Q = m c ∆T or ∆T = Q/(mc) For the mass of the material, apart from the polyester, there is also the styrene. Each unit of the polyester is corresponding to each unit of the styrene for complete crosslinking. The chemical formula for styrene is shown in the following: The mass of a styrene molecule is therefore: 8 C + 8 H = 96 + 8 = 104 g/mole.  Different types of reactants O HO CH2 CH2 OH Ethylene glycol Orthophthalic acid C OH HO CH CH2 OH (Ortho) C OH Propylene glycol CH3 O O O O Orthophthalic C Maleic acid (also HO C CH CH C OH anhydride O Fumeric acid) C O CH CH C O Maleic anhydride O C O O C OH Isophthalic acid (Iso) C O OH  Polyester use and Storage Container for shipping and storage Shelf life Pot life Inhibitors Accelerators Prepregs Coupling agents Fillers  Epoxy CH CH CH CH CH2 CH3 GROUPS O O C CH2 CH CH2 Cl + H O OH Epoxy Group Glycidyl O CH3 REACTANTS Epichlorohydrin Bisphenol-A CH3 CH2 CH CH2 O C O H + Cl CH2 CH CH2 O O CH3 1. Most commonly used resin for advanced composites 2. Good adhesive strength 3. Low shrinkage 4. Operating temperature up to 140 oC  Diglycidyl ether of bisphenol A (DGEBPA) PRODUCT CH3 CH2 CH CH2 O C O CH2 CH CH2 O CH3 O  Formation of epoxy resin CH3 CH2 CH CH2 Cl + H O C OH O Epichlorohydrin Bisphenol-A CH3 Add another Epichlorohydrin CH3 CH2 CH CH2Cl Epichlorohydrin CH2 CH CH2 O C O H O O CH3 CH3 CH3 CH CH2 O CH2 CH CH2 O C CH2 CH CH2 Add another Epichlorohydrin CH2 C O O O O CH3 OH CH3 CH2 CH CH2 Cl Epichlorohydrin O CH3 CH2 CH CH2 O C O CH2 CH CH2 O O CH3 Add another Bisphenol A CH3 HO C OH Bisphenol-A CH3 CH3 CH3 CH2 CH CH2 O C O CH2 CH CH2 O C OH O OH CH3 CH3  Specialty epoxy resins CH3 CH3 CH2 CH CH2 O C O CH2 CH CH2 O C O CH2 CH CH2 O n O CH3 OH CH3 Diglyci dyl Ether of Bisphenol A (DGEBPA) CH2 CH2 CH2 O O O CH CH CH CH2 CH2 CH2 O O O CH2 n CH2 Epoxy Novolac (Epoxidized Phenolic Resin) Example: DOW DEN 438 H CH2 CH CH2 O C O CH2 CH CH2 O O O O CH2 CH2 CH CH2 CH2 CH CH2 CH CH2 O C O CH2 CH CH2 O N CH2 N O O CH2 CH CH2 CH2 CH CH2 H O Tetraglycidylether of Tetrakis (Hydroxyphenyl) Ether Tetraglycidylmethylene Dianiline (TGMDA) Example: SHELL EPON 103 Example: CIBA MY-720  Diluents 1. To reduce the viscosity of the resin 2. To improve shelf or pot life 3. To lower the exotherm 4. Diluents may react with the resin and become an integral part of the cured resin system 5. Butyl glycidyl ether, Cresyl glycidyl ether, Phenyl glycidyl ether.  Curing systems for epoxies Amines Anhydrides Tertiary amines and accelerators  Amine curing agents AMINE HYDROGEN VISCOSITY @ 25°C AMINE CURING AGENTS EQUIVALENT WEIGHT (77°F), Pa.sec (cP) (g/eq) 0.0055-0.0085 Diethylenetriamine (DETA) 20 (5.5-8.5) H2N CH2 CH2 NH CH2 CH2 NH2 0.020-0.023 Triethylenetetramine (TETA) 24 (20-23) H2N ((CH2)2 NH)2 CH2 CH2 NH2 Diethylaminepropylamine 65 < 5.0 (DEAPA) CH3 CH2 N CH2 NH2 3 CH3 CH2  Amine curing agents OH NH2 NH2 + CH2 CH NH2 NH CH2 CH O OH OH NH2 N CH2 CH NH2 NH CH2 CH + CH2 CH O CH2 CH OH  Stoichiometric ratio Molecular weight of amine Parts by weight of amine to be Number of available amine hydrogens per molecule used with 100 parts by weight = X 100 Epoxy equivalent weight of epoxy resin (phr) Epoxy equivalent weight = (Molecular weight of epoxy molecule)/Number of epoxy groups in the molecule) Amine hydrogen: Hydrogen atom attached to Nitrogen Amine hydrogen equivalent weight = (Molecular weight of amine molecule)/(Number of amine hydrogen atoms in the amine molecule)  Anhydride curing agent O C Phthalic anhydride (PA) O C O O O R R CH C R R CH... C N + R O. R CH2 CH CH2 O + R N R R CH C R R CH C O O O O R R CH C... N + R R R CH C O CH2 CH2 CH R O O  Tertiary amines and accelerators N C CH3 C CH Boron Trifluoride Methyl CH3 CH2 N H EMI accelerator Amine (BF3-MEA) F F B NH2 CH2 CH3 F Homopolymerization Etherification  Nature of the reactions and heat generated Primary amine: RNH2 Secondary amine: RNH Tertiary amine: RN Energy associated with primary amine: 83 kJ/mole Energy associated with secondary amine: 131 kJ/mole Energy associated with tertiary amine: higher Example: 1. Assume that there are 100 primary amines to begin with. 2. In the first stage, 20 primary amines participate into the reaction. Heat comes out and raises the temperature of the material. 3. When the temperature is high enough, the 20 secondary amines can participate into the reaction, along with the remaining 80 primary amines. 4. As the temperature rises more, some of the tertiary amines may also participate into the reaction. From then on, at any instant of time, there can be three types of amines that participate into the reaction, giving rise to more heat coming out. 5. When most of the amines are consumed, there are less and less amines remaining for the reaction, heat generated becomes less and less. 6. The heat generated is there complex. 7. The rate of cure is also complex.  Example on relative concentration Question: It is desired to cross link a DGEBPA epoxy resin using an amine curing agent called DETA. The formula for the two materials are as shown in Figure 2.8 and Table 2.5. How many grams of DETA should be used if 100g of the epoxy resin are used? Solution: One repeat unit of the DGEBPA epoxy is shown at the bottom of Figure 2.7. Note that the symbol H H Diethylenetriamine (DETA) C C H2N CH2 CH2 NH CH2 CH2 NH2 means C C EPOXY PRODUCT C C CH3 CH2 CH CH2 O C O CH2 CH CH2 H H O CH3 O Counting the number of atoms, each unit of DGEBPA has 21 carbon atoms, 24 Hydrogen atoms, and 4 oxygen atoms. The mass of the epoxy unit is therefore: m1 = 21 (12) + 24 (1) + 4 (16) = 340 g/mole The DETA molecule (Table 2.5) has 4 carbon atoms, 3 Nitrogen atoms and 13 Hydrogen atoms. The mass of DETA is: m2 = 4 (12) + 3 (14) + 13 (1) =103 g/mole. In the epoxy molecule, there are two epoxy groups, the epoxy equivalent weight is therefore: m3 = 340/2 = 170 g/mole. There are 5 available hydrogens in the amine molecule. The value of molecular weight of amine over the number of available hydrogens per molecule is: m4 = 103/5 = 20.6 g/mole Parts by weight of amine to be used with 100 parts by weight of epoxy resin are:  Relative concentration 1. Large excess of epoxy 2. One reactive epoxy site for one reactive hardener site 3. One epoxy molecule for one hardener molecule 4. High excess of hardener  Cured epoxy resin systems 1. Aromatic rings 2. Cross link density 1. Number of cross links per volume of material 2. Lower cross link density improves toughness 3. Lower cross link density reduces shrinkage 4. Higher cross link density improves resistance to chemical attack 5. Higher cross link density increases heat distortion temperature Pot life Prepregs- Shelf life  Quality control At the liquid stage: Measure viscosity At the solid stage: degree of cure Chemical principle FTIR Acetone wipe  Quality control At the solid stage: degree of cure Electrical principle  Quality control At the solid stage: degree of cure Mechanical principle Barcol hardness test DMA Ultrasonic Shrinkage measurement  Quality control At the solid stage: degree of cure Thermal principle -DSC test Cycle 2 Cycle 3  Degree of cure and rate of cure Degree of cure α: 0< α < 1 dα Rate of cure dt dα = f (α , T ) dt dα  ∆E  m = A exp − α (1 − α ) n dt  RT  Vinyl ester resin O R R' O C C C  Vinyl ester resin O R' Heat O O catalyst C C C R C C C + 2HO C C C Resin formation R' O OH OH O R' C C C O C C C R C C C O C C C + coreactant + inhibitors C where R = O C O and R' = H or -CH3 C Addition of ethylenically unsaturated carboxylic acid molecules and epoxy molecules Properties: between epoxy and polyester Cost: between epoxy and polyester  Vinyl ester resin R' O OH OH O R' C C C O C C C R C C C O C C C + R'' C C * * * * * R (Free- Radical initiator) OH O R' Curing R C C C O C C C R'' C C C where R = O C O C R' = H or -CH3 R'' = * = Reactive sites  Polyimide O C N C O Temperature stability up to 350oC Epoxies stable up to 177oC Bismalimide (BMI) o Cured at 177 C o Post cured at 246 C to obtain full cure and properties significantly higher than epoxies  Phenolic matrix High temperature applications Made by reaction between phenol & formaldehyde Brittle and high shrinkage Fillers are used to reduce brittleness Used for electrical switches, auto molded parts, billiard balls Used as liner for rocket nozzles. Material ablate at high temperature.  Carbon matrix Made from carbon fibers reinforced phenolics Pyrolysis (charring) Porous material is re-impregnated with pitch, phenolics or carbon by vapor deposition Process may take up to 6 months. Higher temperature applications than phenolic Liners for rocket nozzles Tiles for space shuttle nose cones Aircraft, race car and truck brakes Carbon/carbon composites have the highests energy absorption (specific heat) than any known materials  Thermoplastic matrix No shelf life Short processing cycle Higher ductility than thermosets Repairable Recyclable Higher viscosity than thermosets Requires high temperature for processing Tapes are stiff and boardy  Two types of thermoplastic resins o Industrial thermoplastics (up to about 80 C) Polyethylene (PE) Polyvinyl chloride (PVC) Polymethymethacrylate (PMMA) Polypropylene (PP) Polystyrene (PS) Acrylonitrile Butadiene Styrene (ABS) o High performance thermoplastics (up to ~ 300 C) Poly-ether-ether-ketone (PEEK) Poly-ether-ketone-ketone (PEKK) Poly-ether-imide (PEI) Poly-phenylene sulfone (PPS) Poly sulfone Thermoplastic matrix Thermoplastic matrix  Thermoplastic matrix-Fabrication Difficult due to high viscosity- 106 centi-poise as compared to 1000 centi-poise for liquid epoxy Shear thinning Comingled fibers Power impregnation  Fillers Cost reduction Shrinkage reduction Improvement of mechanical properties Improvement of flame resistance Colorants, pigments  Metal matrix Aluminum, titanium, magnesium, copper Short fibers such as SiC- particles, whiskers, fibers Piston ring inserts, pistons, connecting rods High temperature applications Difficult to make Ceramic matrix Oxides, carbides, borides, nitrides Short fibers such as SiC- particles, whiskers Made by chemical vapor deposition High temperature applications Difficult to make  Polymer nanocomposites a = 1, R = 6 a = 0.5, R = 12 a = 0.1, R = 60 R = aspect ratio Structure of clay sheets Clay sheets with intercalating ions Different levels of clay structure Replacing positive ions with long omium ions  Dispersion techniques High pressure mixing technique X Ray pattern for samples mixed using different methods Minimum distance 0.3 microns, Maximum distance 3.2 microns, Mean distance 0.9 microns Distribution of sized of particles in epoxy samples mixed using High Pressure method- 2 wt% I 30E in TGDDM-DDS TEM of samples mixed using high pressure mixing method- Clay particle containing about 6 clay sheets  800 DMM HPM-A 600 G1C , J/m2 400 200 0 0 3 6 9 12 Clay Loading, phr Variation of strain energy release rate as a function of percent clays- Max increase: 5.8 times compared with no clay.  (b) (a) (c) High speed mixing (10,000 rpm)  5000 8.24 nm 3.74 nm 4000 3.84 nm Intensity (counts) 3000 1.85 nm 2000 C30B ENC-D230-6wt%C30B 1000 ENC-D2000-6wt%C30B EPON828-C30B 0 2 4 6 8 10 o 2θ ( ) X Ray Diffraction pattern for samples mixed using high speeds. Strain energy release rates of epoxy/clay samples mixed using high speed method. Procedure to incorporate modified epoxy into long continuous carbon fibers. a) Hand Lay Up carbon fibers with epoxy/clay. b) Bagging. c) Autoclave cure. d) Final sample.  450 0phr 400 350 2phr 300 4phr GIc, J/m 2 250 200 150 100 50 0 40 50 60 70 80 90 100 a, mm Improvement in fracture toughness of composite samples  1 region I QISF region II Direction of 0.8 QISU Normalized stiffness, E/E0 I crack growth 0.6 II region III 0.4 III 0.2 0 0 20000 40000 60000 80000 100000 120000 Number of cycles,N Improvement of fatigue lives of tapered composite sample (glass /epoxy) subjected to tensile-tensile fatigue loading. Broken curve for unfilled epoxy. Solid curve for epoxy with nanoclay.  Possible applications 1. Barrier properties: obstruction to diffusion of small particles: CO2, Flammability resistance. 2. Fracture resistance Carbon nanotube configurations Three-roll milling machine  102 101 100 Conductivity, S/m 10-1 Ref. in Ref. in , 10-2 10-3 10-4 NLIG NIKSW 10-5 0 1 2 3 4 5 6 7 8 9 CNT weight fraction, % Variation of electrical conductivity with amount of nanotubes  Possible applications 1. Enhance electrical conductivity: Lightning strike resistance, conductive adhesives. 2. Uses for strain measurement. 3. Uses to detect defect occurrence in composite structures. Aggregately conductive materials Electrical resistance between two points in a plate. a) copper plate, b) glass/epoxy containing CNTs.  R R ∆R 1  R R   1  Rtotal = Rtotal = =  − = N N −1 R R  N N − 1  N ( N − 1)   Hole 1 (1/16 inch) Hole 3 (3/16 inch) Hole 5 (5/16 inch) Hole 6 (6/16 inch) Hole 4 (4/16inch) Hole 2 (2/16 inch) 22×13 inch2 carbon fiber/epoxy/ composite plate Change in electrical resistance due to impact by projectile on carbon/epoxy with CNTs Thank you Manufacturing of composites MECH 415/6521 Suong V. Hoa Concordia Center for Composites Mechanical and Industrial Engineering Concordia University Fibers Tap e Woven fabric Braid Stack of many Knit layers (d) Mat (c)  Different types of fiber materials Glass fibers Carbon fibers Kevlar fibers Thermoplastic fibers  Glass fibers Made from sand plus oxides  Composition of glasses used in composite materials E glass range (%) S glass range C glass range (%) (%) Silicon Oxide 52-56 65 64-68 Calcium oxide 16-25 - 11-15 Aluminum oxide 12-16 25 3-5 Boric oxide 5-10 - 4-6 Magnesium oxide 0-5 10 2-4 Sodium oxide and potassium oxide 0-2 - 7-10 Titanium oxide 0-1.5 - - Iron 0-1 - - Iron oxide 0-0.8 - 0-0.8 Barium oxide - - 0-1  Properties of glasses Property Type of glass C E S Density (g/cm3) 2.49-2.50 2.54-2.62 2.48-2.50 Tensile strength (MPa) @22oC 3006-3280 3417 4544 @371 oC - 2597 3724-4408 @538 oC - 1708 2392 Tensile modulus (GPa) @22 oC 68.3 71.8 84.7 @538 oC - 80.6 88.2 Elongation 0.03 0.035 0.04 Coefficient of thermal 7.2 5.0 5.6 expansion (10-6m/m/oC) Heat capacity 800 800 800 (J/kg.C)@22 oC Softening point,oC 749-750 841-846 970 10 1 1 10 100  Matrix polymer R R R Si Si Si O O O O O O O O O Coupling Si Si Si Si Si Si Si Si Si agents Glass  Carbon fibers PAN based Pitch based Cellulosic based  PAN based carbon fibers 738-998oF (400-600oC) DEHYROGENATION N N N N N N N N N N N N 998-2350oF (600-1300oC) DENITROGENATION N N N N N N PAN and Pitch processes  Surface treatment of carbon fibers To improve interlaminar shear strength Liquid oxidative treatment (bath of oxidative agents- nitric acid, potassium permanganate, sodium hypochlorite) Non oxidative treatment (whiskerization, pyrolytic surface coating, polymer grafting) Surface coating (sizing) of carbon fibers Epoxy, polyvinyl alcohol, polyimide  Properties of carbon fibers Different types of carbon fibers and their properties PAN-BASED FIBERS LOW MODULUS HIGH MODULUS Tensile modulus (GPa) 226 383 Tensile strength (MPa) 3280 2392 Elongation (%) 1.4 0.6 Density (g/cc) 1.8 1.9 Carbon assay (%) 92-97 100 PITCH-BASED FIBERS Tensile modulus (GPa) 157 376 Tensile strength (MPa) 1367 1708 Elongation (%) 0.9 0.4 Density (g/cc) 1.9 2.0 Carbon assay (%) 97 99 Electrical conductivity about 104 S/cm-Pitch fibers more conductive than PAN fibers Aramid fiber (or Kevlar fibers) Poly Para-Phenyleneterephthalamide (Aramid) H O O N N C C N N H n H H O O O C N N C C n Intermolecular bonding in aramids Properties of aramid fibers Type of Kevlar 29 49 149 fiber Tensile modulus 83 131 186 (GPa) Tensile strength 3.6 3.6 3.4 (GPa) Elongation (%) 4 2.8 2 Density (g/cc) 1.44 1.44 1.47 Ultra high orientation polyethylene fibers Boron fibers, Silicon carbide (SiC)fibers Made by chemical vapor deposition  Fiber architecture 1. Fibers 2. Filaments 3. Strands 4. Tows 5. Yarn 6. Roving 7. Tape 8. Fabrics 9. Mat  Fabric forms Plain weave Basket weave 4_harness_satin 8_harness_satin Mats (random fiber orientation) Braids  Hybrids Glass and carbon fibers combined Deformation of fiber bundle  δf  Va   Space between fibers = −1 d  V f   -6 δf ~ 10 m  T Winding Pultrusion Molding Forming Mechanics of a fiber bundle  Deformation of a wavy fiber Slightly curved fiber with L/ a ≥ 100 and representative element for the fiber bundle  a 2πx  y = 1 − cos( ) 2 L   Beam mechanics Straight beam subjected to axial load ∆x Px = σ x A = Eε x A = E A L L ∆x = Px EA Curved beam subjected to axial load  L La 2  ∆x =  +  Px  EA 8EI  Curved beam subjected to bending load aL2 ∆x = − 2 Py 4π EI  a2L L − aL 2   + P  ∆x  8EI EA 4π 2 EI   x  ∆y  =     aL2 L3  Py  −   4π 2 EI 192EI  a2L L − aL 2  P   ∆x   8EI + EA 4π EI  x 2 ∆y  =  aL 2 L 3 P    −  y   4π 2 EI 192EI  Px Py ∆ ∆y σ1 = σb = e1 = x eb = L h h 2 hL a2L L aL2   Le1  =  8EI + EA − 4π 2 EI  σ h 2   1  heb  2 L 3 − aL  σ b hL   4π 2 EI 192EI  a h + h 2 2 2 aL2 h   e1  =  8EI EA − 4π 2 EI  σ 1    eb   − aL h 2 L 4 σ b   4π 2 EI 192EI   h − ho h Ao Vf eb = = 1− o eb = 1 − = 1− h h A Vo a h−d h π d2 /4 π  d 2 Va = = −1 Vf = =   h π /4 h = h 2 4h = Vf d d d d Vf d a Va = −1 d Vf L L a = β  Va − 1 β=L = a d ad  Vf    Va Fiber volume parameter ς= Vf  a h + h − 2 2 2aL 2 h   e1  =  8EI EA 4π 2 EI  σ 1     eb  2  − aL h L 4 σ b   4π 2 EI 192EI  (L= βa, a = d (ζ-1), h = dζ, A = (π /4)(d2), I = (π /64)(d4)  e1  =  F11 F1b σ 1       eb   Fb1 Fbb σ b  2 4 1 2 16 β F11 = ς [1 + 2(ς − 1) 2 ] F1b = Fb1 = − ς (ς −1)3 π E π3 E β 4 (ς −1) 4 Fbb = 3πE vo= Initial fiber volume fraction vf: Actual fiber volume fraction va: Maximum allowable fiber volume fraction Va Ao Vf ς= eb = 1 − A = 1− Vo Vf If σ1 and σb are given, use eb = Fb1σ 1 + Fbbσ b * * to find eb, then vf. Then use e1 = F11 (V f )σ 1 + F1b (V f )σ b  Axial extension (only extension load Px is applied) Vf 1− AoVo eb AV Vo P1 = Aσ 1 = P1 = − o o 3 V f Fb1 Vf 16 β 2 Va  Va  −1  π3 E Vf  Vf    Variation of fiber volume fraction and axial 500 stress Axial Stress (P1/A0) MPa Relation between axial 450 400 stress and fiber volume 350 300 fraction (Vo = 0.5, Va = 250 200 0.8, E = 24 GPa, β= 200) 150 100 50 0 0.5 0.55 0.6 0.65 0.7 Volume fraction Vf Bulk compressive stress (no axial stress) Vf 1− eb 3πE Vo σb = F = 4 4 bb β  V   a −1   Vf     Example 3.1: Consider the case of autoclave processing. A bed of fibers is being compressed. The initial fiber volume fraction is Vo = 0.50. The final fiber volume fraction is Vf = 0.68. Assume that the maximum allowable fiber volume fraction is Va = 0.785. Assume also that E = 234 GPa and L/a = 200. What would be the compressive stress to reach this final fiber volume fraction? Solution: Using equation (3.21), we have: Case a: β = L / a = 200 3πE = (3)(π )(234GPa) / 200 4 = 1378 Pa β4 0.5 0.5 Vf   0.68  1 −   = 1−   = −0.166  Vo   0.50  4 4  V  0.5   0.785  0.5   a  − 1 =   − 1 = 0.000031  V f    0.68       The required compressive stress is (1 psi= 6895 Pa): (1378 Pa)(−0.166) σb = = 7.38MPa = 1071 psi 0.000031 This stress is fairly high compared to normal autoclave pressure. It can be seen from equation (3.21) that apart from the volume fractions, the two parameters affecting the value of the compressive stress is the modulus E and the waviness ratio β = L/a  Case b: If one were to use the value of L/a = 350, and the modulus of E = 181 GPa, the result would be: 3πE = (3)(π )(181GPa) / 350 4 = 113.6 Pa β 4 The compressive stress now is: (113.6 Pa)(−0.166) σb = = 0.608MPa = 88.3 psi 0.000031 Case c: If L/a = 400 and the modulus is E = 181 GPa, one has: 3πE = (3)(π )(181GPa) / 400 4 = 66.6 Pa β 4 and the compressive stress is: (66.6 Pa)(−0.166) σb = = 0.357 MPa = 51.8 psi 0.000031  Maximum fiber volume fraction (Va) depends on 1. The state of lubrication of the fiber beds. Dry fibers have lower Va. 2. The rate of loading. Faster rate of loading produces lower Va. 3. Repeat loading. Va increases after first cycle of loading. This tapers off after 3 or 4 loadings. 4. Types of fabrics to be compressed. Mats, fabrics of different types. 5. Stacking sequence. 6. Fiber bed exhibits creep behavior. Thank you Manufacturing of composites MECH 415/6521 Suong V. Hoa Concordia Center for Composites Mechanical and Industrial Engineering Concordia University Hand Lay Up-Autoclave  Wet lay up Glass/polyester Glass/vinyl ester Early technique for composite manufacturing Low cost composite applications Boats Process by hand Air entrapped creating voids  Dry lay-up- Using prepregs Prepregs Tool preparation Laying up prepregs on the tool to make the part Curing of the part Removal of the part from the tool Inspection Finishing steps. Main steps in the Hand Lay UP- Autoclave process (a) Prepreg + (d) Vacuum Bag (b) Tool (c) Lay up (f) Final product (e) Curing in autoclave Prepreg making machine-Prepreg roll Tensioner Feeding Comb Resin Take up Heater creels spreader bath roll  Tools- Molds Reinforced polymers for low to intermediate temperature ranges. Metals for low to high temperatures. Ceramics and bulk graphite for very high temperatures.  Metal tools Aluminum Steel Invar Electro formed nickel Material Coefficient of thermal expansion (10-6/oC) Composite APC2/AS4 3.8 [45/0/-45/90]3s Bulk graphite 3.0 Ceramic 0.7 Metal-ceramic 7.2 Polyimide 4.7 Aluminum 26.2 0.2-0.5% carbon steel 13.2 18Cr+9%Ni steel 17.8 Cast iron 11.1  Metal tools Electro formed nickel  Graphite/epoxy tooling Similar coefficient of thermal expansion as the part Light weight Limited number of repeated use Elastomeric (rubber) tooling  Bulk graphite/Ceramic tooling High temperature applications (2000C) Very fragile  Release agent Chemical applied as a coating on the surface of the mold Helps to release the part from the mold after cure. Essential to prevent sticking and damage to the mold during part removal. Several coats may be applied (previous coat is allowed to harden before the next coat is applied.  Lay up 1. To deposit layers of prepregs to make the laminate 2. To facilitate the removal of the part after cure (without sticking problem) 3. To allow for compaction of prepregs using vacuum and pressure 4. To prevent excess resin from running within the plane of the fiber stack, which can distort the orientation of fibers 5. To provide an escape path for volatiles such as water vapor or gases that are generated during the curing process 6. To provide materials can can absorb excess resins that ooze out of the laminate during the curing and molding process. 7. To obtain good surface finish for the part Apart from just depositing the prepregs, many anciliary materials are used to achieve the above objectives. Lay up  Curing and consolidation Curing means making the molecules to cross link. Percentage of the bonds formed over the total number of bonds that can be formed is the degree of cure. Needs to make sure that all fibers are wetted before curing takes place. Needs to make sure that there are no (or limited) resin rich areas before curing takes place. Needs to understand the amount of heat coming out during curing. Consolidation means to assure that the layers are packed together with no voids or limited voids. Curing and consolidation need to be accomplished before cooling down.  Resin kinetics CH3 CH3 HO OH CH2 CH CH2 O O CH2 CH CH2 Epoxy molecule O O CH3 CH3 DETA molecule H2N CH2 CH2 NH CH2 CH2 NH2 Primary amine: 61.4 kJ/mole Secondary amine: 72 kJ/mole Etherification:101 kJ/mole  Degree of cure- Rate of cure Ht d 1 dH t   HT dt H T dt Differential Scanning Calorimetry (DSC) d d  f ( , T )  K m (1   ) n K  Ae  E / RT dt dt d  ( K 1  K 2 )(1   ) n1  K 3 (1   ) n 2 dt Two cases of cure rates, a) nth order, b) m, n ≠ 0  Typical DSC curve for epoxy 2.0 239.97°C pure TGDDM epoxy + DDS 1.5 Heat Flow (W/g) 1.0 0.5 0.0 198.40°C 581.0J/g -0.5 0 50 100 150 200 250 300 Exo Up Temperature (°C) Universal V4.2E TA Instruments Heat flow versus time curve for TGDDM/DDS epoxy resin  Heat Transfer and Energy Balance (  C pT )   T    T    T  dH  Kx    K y    K z  t x  x  y  y  z  z  dt k xx   m mn   k11  2 n2 k    n 2 m 2  mn    zz    k 33  k xz   mn mn m 2  n 2   k13  ρ (kg/m3) Cp (kJ/(W.oC) k33 (kW/(m.oC) k11/k33 Glass/polyester 1.89x103 1.26 2.16x10-4 2 Graphite/epoxy 1.52x103 0.94 4.46x10-4 5  Heat generated dH d  HT H T  wr H r  w f H f dt dt Glass/polyester Graphite/epoxy wr 46% 42% wf 54% 58% Hresin (kJ/kg) 168.6 473.6 HT(kJ/kg) 77.5 198.9  Kinetic equations d E m A = 3.7 x 1022 min-1  A exp(  ) (1   ) n ΔE = 1.674 x 105 J/mol dt RT m = 0.524 n = 1.476 d  ( K1  K 2 )(1   )( B   )   0.3 dt d  K 3 (1   )   0.3 dt A1  2.101x10 9 min 1  E  K 1  A1 exp   1  A2  2.014 x10 9 min 1  RT  A3  1.960 x10 5 min 1 B = 0.47  E 2  K 2  A2 exp    E1  8.07 x10 J / mol 4  RT   E3  E1  7.78 x10 4 J / mol K 3  A3 exp     RT  E1  5.66 x10 4 J / mol  0.2 1 f (a) a 0.9 0.18 (1-a)2.1 0.9 0.9 (1-a)2.1 a 0.16 0.8 0.14 0.7 0.1 0.6 0.10 f (a) = a 0.9 (1-a)2.1 0.5 0.08 0.4 0.06 0.3 0.04 0.2 0.02 0.1 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 a  E 63400 7629    g (T )  e RT e (8.31)T e T  Simplifying assumptions No variation along x,y directions. Only variation in z direction (  C pT )  T  d   z K  HT t z  z  dt (  C pT ) d Thin laminates  HT t dt  One example Consider a carbon/epoxy composite where the resin kinetic follows the equation: d  K m (1   ) n dt with A = 1.27x 105 sec-1 K  Ae  E / RT E = 63400 J/mol m = 0.9, n = 2.1 Values of the physical properties for many resins, fibers and composite systems are given in reference. Composite density:  = 1580 kg/m3 Composite specific heat: Cp = 870 J/(kg.K) Composite conductivity along the thickness direction: Kz = 0.69 W/m.K HT = 150 J/g (or 2.37 x 108 J/m3) R: Universal gas constant = 8.31 J/(mol.K) It is desired to manufacture a laminate of 8 layers with thickness of 0.150 mm per layer. The bottom of the laminate is an aluminum mold 12.7 mm thick and the top of the laminate has bagging materials (bleeder, breather layers etc). The whole assembly is placed inside an autoclave subject to a temperature cycle. Figure 4.14 shows the configuration.  One example Ta, Pa Bagging film Vacuum valve Breather Bleeder Release film Laminate Sealant tape Release material Tool Ta, Pa Two schedules of heating will be considered. Linear increase in temperature: In this schedule, the temperature of the autoclave is increased linearly from room temperature (20 oC) at a rate of 5 oC per minute. Two-step temperature increase: In this schedule, the autoclave temperature is increasing from room temperature (20 oC) at a rate of 5 oC per minute for 18 minutes. It is then held constant at 110 oC for 20 minutes. Then it is increased by 5 oC per minute for 14 minutes to a maximum of 180 oC and held there for 60 minutes. (Figure 4.15 shows the temperature schedule). KAl = 237 W/K.m L = 1.2 mm (laminate thickness) t1 = 12.7 mm (thickness of mold) t2 = 1 mm (thickness of bagging material on top of laminate) C p L =1.65x103 J/(K.m2 ) KAl = 237 W/m.K K glass cloth = 0.26 W/m.K Determine the development of degree of cure and rate of cure as functions of time.  Solution T d E  C p  HT  H T Ae RT ( ) 0.9 (1   ) 2,1 t dt dT C p L  net heat flux to the sample  Qc  Q1  Q2 dt d K K Qc  H T L  Ae  E / RT  0.9 (1   ) 2.1 H T L Q1  1 (T  T ) Q2  2 (T  T ) dt t1 t2 Governing equation C p L dT K K  Qc  1 (T  T )  2 (T  T ) dt t1 t2 dT a  bT  c  0 a  C p L dt K1 K 2 b  t1 t2 E  K1 K 2  c  [ Ae RT  0.9 (1   ) 2.1 H T L  T (  )] t1 t2 If we solve the equation by increments, and assuming that during each increment of time, the value of c is constant and it is calculated based on the temperature and degree of cure obtained from the previous increment. As such the value of c is constant for each increment. b m a T  c1e  mt  c2  mc1ae  mt  bc1e  mt  bc2  c  0 c c2   To  c1  c 2 b c mt c c T  (To  )e  c1  To  c 2  To  b b b  Final solution K1 K 2 K K Qc  T (  ) Qc  T ( 1  2 ) t1 t2  mt t1 t2 T  [To  ]e  K1 K 2 K1 K 2   t1 t2 t1 t2 d  7629 Qc  LHT  (3.6 x1010 J / sec.m 2 )e T  0.9 (1   ) 2.1 dt d K1 K AL 237 W / K.m  (1.27 x10 5 / sec)e 7629/ T ( ) 0.9 (1   ) 2.1    18661W /( m 2.K ) dt t1 t1 12.7 mm K 2 K glass cloth 0.26W / m.K    260W /( m 2.K ) Qc  1861T 11.47t Qc t2 1mm 0.001m T  T  [To  ]e  18661 18661 K1 K 2   18921W /(m 2 K )  d  n t1 t2  n 1   n  t  dt   Incremental solution To = 293 oK  o  0.1 Assume some degree of cure to start the solution Increment 1 (0 to 5 minutes): During this time increment, the temperature of the surrounding increases from 20C to 45C (293 K to 318 K ). The average of 305.5 K is used for T   0.1 Qc = 0.05 W/m2 0.05  18661(305.5) (11.47)( 300) 0.05 T  305.5  [293  ]e   2 x10 6 18661 18661 T  305.5 K d  (1.27 x105 / sec)e 7629/ 305.5  (0.1) 0.9 (1  0.1) 2.1  1.82 x10 7 / sec dt 1  0.1  (1.82 x10 7 )(300)  0.1  Incremental solution T1= 293 oK  o  0.1 Assume some degree of cure to start the solution Increment 2 (5 to 10 minutes): During this time increment, the temperature of the surrounding increases from 45C to 70C (318 K to 343 K ). The average of 330.5 K is used for T   0.1 Qc = 0.34 W/m2 d  (1.27 x105 / sec)e 7629/ 330.5  (0.1) 0.9 (1  0.1) 2.1  1.2 x10 6 / sec dt  2  0.1  (1.2x10 6 )(300)  0.1  Summary of results-Linear increase in T Time (minute) Ta(oC) (oC)  T (oK) Qc (W/m2) T(oK) Rate of cure sec-1 Degree of cure  0 20 20 293 0 293 1.2x10-8 0.1 5 45 32.5 305.5 0.05 305.5 1.2x10-7 0.1 10 70 57.5 330.5 0.34 330.5 1.2x10-6 0.1 15 95 82.5 355.5 1.72 355.5 6.1x10-6 0.102 20 120 107.5 380.5 7.21 380.5 2.55x10-5 0.110 25 145 132.5 405.5 26.1 405.5 9.17x10-5 0.138 30 170 157.5 430.5 89.2 430.5 2.46x10-4 0.211 35 180 175 448 217 448 7.64x10-4 0.440 40 180 180 453 247 453 8.71x10-4 0.702 45 180 180 453 99.8 453 3.70x10-4 0.813 50 180 180 453 42.9 453 1.89x10-4 0.941 55 180 180 453 4.33 453 1.53x10-5 0.946 60 180 180 453 3.6 453 1.27x10-5 0.950 Summary of results:Two-step increase in T Time (minute) Ta(oC) T  o T( C) (oK) Qc (W/m2) T(oK) Rate of cure sec-1 Degree of cure   0 20 20 293 0 293 1.2x10-8 0.1 5 45 32.5 305.5 0.05 305.5 1.2x10-7 0.1 10 70 57.5 330.5 0.34 330.5 1.2x10-6 0.1 15 95 82.5 355.5 1.72 355.5 6.1x10-6 0.102 18 110 102.5 375.5 5.52 375.5 1.95x10-5 0.106 23 110 110 383 8.39 383 2.96x10-5 0.114 28 110 110 383 8.86 383 3.13x10-5 0.123 33 110 110 383 9.27 383 3.27x10-5 0.132 38 110 110 383 9.62 383 3.40x10-5 0.142 40 120 115 388 13.0 388 4.6x10-5 0.148 45 145 132.5 405.5 31.1 405.5 1.1x10-4 0.181 50 170 157.5 430.5 102 430.5 3.6x10-4 0.289 52 180 175 448 231 448 8.2x10-4 0.387 60 180 180 453 265 453 9.4x10-4 0.837 70 180 180 453 32.7 453 1.15x10-4 0.906 80 180 180 453 11.1 453 3.92x10-5 0.930 90 180 180 453 6.14 453 2.17x10-5 0.942 100 180 189 453 4.2 453 1.47x10-5 0.951 Temperature and degree of cure in the part- Linear increase in autoclave temperature o a 1 200 T( C) 0.9 180 0.8 T 160 0.7 140 0.6 120 0.5 100 a 0.4 80 0.3 60 0.2 40 0.1 20 0 0 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 10 0 Time (min) Temperature and degree of cure in the part- Two step increase in autoclave temperature a 1 200 T(oC) 0.9 180 0.8 160 0.7 140 T 0.6 120 0.5 100 0.4 a 80 0.3 60 0.2 40 0.1 20 0 0 0 5 10 15 20 25 30 35 40 45 50 52 60 70 80 90 100 Time (min)  Thick laminates Temperature distribution in a thick laminate at 164 minutes (Glass/polyester laminate). Reproduced from Bogetti T.A. and Gillespie J.W. “Process induced stress and deformation in thick section thermoset composite laminates”, J. Composite Materials, Vol. 26, No. 5, 1992, pp.626-660,  Thick laminates Degree of cure distribution in a thick laminate at 164 minutes (Glass/polyester laminate). Reproduced from Bogetti T.A. and Gillespie J.W. “Process induced stress and deformation in thick section thermoset composite laminates”, J. Composite Materials, Vol. 26, No. 5, 1992, pp.626-660  U  Viscosities     exp   k   RT  Author System μo(Pa.s) E(J/mol) K(Pa.s) Range of validity Lee et al Hercules 3501- 7.93x10-14 9.08x104 14.11.2 α 900 pounds per hour (PPH) (IPM) lay up rate during production Minimum pieces and gaps:600 IPM part program execution. Off part motion: 6,000 IPM  Outline Thermoset matrix composites 1. Speed of deposition – Efficiency of machine use 2. Inspection 3. Fiber steering – Tow shearing 4. Dry fibers Thermoplastic matrix composites 1, Heating 2. Interlaminar shear strength 3. Distortion for structures with free edges. Conclusion  Defect detection In AFP, manual inspection of parts consumes a large portion of production time, and is susceptible to missing defects. Manual inspection can consume more than 25% of total production time, and many defects escape detection. Many organizations have projects to address this issue: Universities Companies http://concom.encs.concordia.ca  Thermographic system Denkena B., Schmidt C., Voltzer K., Hocke T. “Thermographic on line monitoring system for automated fiber placement processes”, Composites Part B. Eng., 2016, 97, 239-243. Can detect tow locations, gaps, laps, foreign bodies http://concom.encs.concordia.ca  Defect detection Marc Palardy-Sim et al, “Next generation inspection solution for automated fibre placement”, Proc. ACM4, Automated Composites Manufacturing, ed. S.V.Hoa, Destech 2019. Optical coherence tomography (OCT)-National Research Council of Canada, Fives Machining system Similar to ultrasound imaging. OCT imaging probes a material with light and detects reflection to map the position of the material interfaces. Interferometer with two optical paths. Sample arm emits and collects the light from the same point, like the end of an optical fiber. http://concom.encs.concordia.ca  Defect detection J. Cemenska, T. Rudberg, M. Henscheid, A. Lauletta, B. Dvis, “AFP automated Inspection system performance and expectations”, SAE technical paper 2018-01-2150. Electroimpact system Cameras, laser projectors, laser profilometers, & user interface. On large parts with >1,000 tow ends, a few minutes to inspect. Ply boundary inspection - Identify and measure 92% of tow ends in standard ply. Gap measurement has a mean error < 0.003” and a standard deviation of 0.005”. Overlap measurement data has a mean error < 0.010” and standard deviation of 0.013” http://concom.encs.concordia.ca  Electroimpact system Suite of software programs processes the data. Data is stored in a data base which acts as the interface between different software programs for interaction and data sharing. Extensive interface ties all the data together and displays the results. Inspection data consumes a large amount of hard drive space. Data can approach 1 TB/ part for complex parts. Data value and utility Reduces inspection time. Reduces human errors. Statistical process control can eliminate unnecessary fixes. Instead of reworking every out-of-tolerance error, operators can rework only conditions that make the population out-of-tolerance. Resource burden System cost: $1,000,000. Manufacturers need to make decisions about how to use the inspection data. http://concom.encs.concordia.ca  Defects Chevalier P.L., Kassapoglou C., Gurdal Z., “Fatigue behavior of composite laminates with automated fiber placement induced defects- a review”, Int. J. Fatigue, 2020, 140, 105775. Five perspectives: Anticipation: Predicting occurrence Existence: Inspection Significance: Effect on performance Progression: Potential evolution Disposition: Defect treatment http://concom.encs.concordia.ca  Effect of defects on performance Chevalier P.L., Kassapoglou C., Gurdal Z., “Fatigue behavior of composite laminates with automated fiber placement induced defects- a review”, Int. J. Fatigue, 2020, 140, 105775. Quasi-static loading in unnotched unidirectional laminates: Not significant effect. Fatigue loading: More influence More work required: Large variety of geometry configurations, large number of type of defects, loading conditions. http://concom.encs.concordia.ca  Outline Thermoset matrix composites 1. Speed of deposition – Efficiency of machine use 2. Inspection 3. Fiber steering 4. Dry fibers Thermoplastic matrix composites 1, Heating 2. Interlaminar shear strength 3. Distortion for structures with free edges. Future outlook Conclusion  Fibre steering Photos from Composites World magazine- Nov. 2020 AFP well suited to tow steering Laser heating highly focused, well controlled. Requires: Precise and focused heat Good compaction http://concom.encs.concordia.ca  Fiber steering Steering radius- Affecting factors: Tow width, Tow thickness, Resin type, Resin tackiness, Fiber type, Fiber format, Substrate contour complexity, Substrate quality, Arc length and tolerance parameters. http://concom.encs.concordia.ca  Fiber steering Possibilities: (Electroimpact, Coriolis, Ingersoll machines) 0.125” tow R = 750 mm (19.5”) 0.25” tow R = 280 mm (11”) to 1500 mm (59”) Such steering sees limited use, given the fact tha