Manufacturing of Composites PDF

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...

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