Naturally Occurring Composites 材料早论 Wood: cellulose fibers in a lignin matrix. Bone: short and soft collagen fibers embedded in a mineral matrix called apatite 第四章复合料 Glass fiber reinforced resins have been in use since about the 1940s Comparison between conventional monolithic materials and composite materials 4.1概述 A mixture of two or more materials that are LIu distinct in composition and form, each being present in significant quantities(e.g, >5%) 两种或多种不同组成、不同存在形式材料的 混合物,各以显奢的量存在 41概述 41概述 定义 高分子材料 The union of two or more diverse materials to attain synergestic or superior qualities to those 两种或多种不同材料的结合体,可获得协 的或优于个别材料的质量 陶瓷材料 金属材料
1 材料导论 第四章 复合材料 Wood: cellulose fibers in a lignin matrix. Bone: short and soft collagen fibers embedded in a mineral matrix called apatite. Glass fiber reinforced resins have been in use since about the 1940s. Naturally Occurring Composites Comparison between conventional monolithic materials and composite materials. Weight Thermal expansion Stiffness Strength Fatigue resistance Steel Aluminum Composites 4.1 概述 定义1 A mixture of two or more materials that are distinct in composition and form, each being present in significant quantities (e.g., >5%) 。 两种或多种不同组成、不同存在形式材料的 混合物，各以显著的量存在 4.1 概述 定义2 The union of two or more diverse materials to attain synergestic or superior qualities to those exhibited by individual members 两种或多种不同材料的结合体，可获得协同 的或优于个别材料的质量 复 合 材 料 按 基 体 分 类 陶瓷材料 增 强 材 料 金属材料 高分子材料 CMC MMC 4.1 概述 PMC
41概述 复合材料按结构分类 42混合原理 迭层型 粒型 连续织物型 42混合原理 42.1应力平行于纤维,等应变 基本假定 应力符合混合规律 今纤维与基体必须紧密结合。 今纤维必须是连续的或在长度方向上搭接的。 ☆存在一个临界纤维体积分数Vrcm高于此值方 能发生纤维增强 v体积分数,∝应力 m分别代表纤维与基体 令存在一个临界纤维长度,高于此值方能发生 模量加和规律 求受力比 =E2E2Ef代表应变 由等应变假定 E v+E 故有
2 4.1 概述 复合材料按结构分类 4.2 混合原理 4.2 混合原理 基本假定 纤维与基体必须紧密结合。 纤维必须是连续的或在长度方向上搭接的。 存在一个临界纤维体积分数V f crit,高于此值方 能发生纤维增强。 存在一个临界纤维长度，高于此值方能发生 增强。 应力符合混合规律： σc = Vf σf + Vm σm V:体积分数，σ:应力， f与m分别代表纤维与基体。 4.2.1 应力平行于纤维，等应变 σf = Ef εf ，σm = Em εm ，σc = Ec εc ε代表应变 Ec εc = Vf Ef εf + Vm Em εm Ec = Ef Vf + Em Vm 由等应变假定 σc = Vf σf + Vm σm 模量加和规律 σf = Ef εf ，σm = Em εm m m f f m m f f m m f f m f V V V l V l A A F F σ σ σ σ σ σ = = = / / 求受力比 由 m f m f E E = σ σ 故有 m m f f m f E V E V F F =
EXAMPLE PROBLEM 4.1 复合材料模量预测(1) A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having 纤维沿外力方向 modulus of elasticity of 69 GPa resin that, when hardened, displays a modulus of 3. 4 GPa. (a) Compute the modulus of elasticity of this composite in E=VmEm+ ret the longitudinal direction (b)If the cross-sectional area is 250 mmand a stress of 50 MPa is applied in this longitudinal direction, compute the magnitude of the load carried by each of the fiber and (c)Determine the strain that is sustained by each phase when the stress in part b is applied consists of 40 vol% of glass fibers having a modulus of elasticity consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, of 69 GPa and 60 vol% of a polyester resin that, when hardened displays a modulus of 3.4 GPa. fthe cross-sectional area is 250 mm2 and a stress of 50 MPa Compute the modulus of elasticity of this composite in the plied in this longitudinal direction, compute the magnitude of rried by each of the fiber and matrix pha Solution: First find the ratio of fiber load to matrix load SOLUTION (a) The modulus of elasticity of the composite is calculated using E=(3.4GPa)0.6)+(69GPa)0.4)=30GPa 13.5or Fm(34GPa)0.6) A continuous and aligned glass fiber-reinforced consists of 40 vol%of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened consists of 40 vol%o ter resin that. when hardened displays a modulus of 3. 4 GPa. The cross-sectional area is 250 mm?2 (b)If the cross-sectional area is 250 mm?and a stress of 50 MPaI and a stress of 50 MPa is applied is applied in this longitudinal direction, compute the magnitude the load carried by each of the fiber and matrix phases (c)Determine the strain that is sustained by each phase when the stress in part b is applied. The total force sustained by the composite F F2=A2a=(250mm250MPa)=12,500N For stress calculations, phase cross-sectional areas are This total load is just the sum of the loads carried by fiber and necessary An=V42=(060250mm2)=150mn 4=V=(0.4)250mm2)=100mm2 whereas FrFe-F.=12, 500N-860N=11, 640N
3 复合材料模量预测（1） A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa. (a) Compute the modulus of elasticity of this composite in the longitudinal direction. (b) If the cross-sectional area is 250 mm2 and a stress of 50 MPa is applied in this longitudinal direction, compute the magnitude of the load carried by each of the fiber and matrix phases. (c) Determine the strain that is sustained by each phase when the stress in part b is applied. EXAMPLE PROBLEM 4.1 (a) The modulus of elasticity of the composite is calculated using Equation Ec = EmVm + Ef Vf : Ec = (3.4 GPa)(0.6) + (69 GPa)(0.4) = 30 GPa SOLUTION A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa. (a) Compute the modulus of elasticity of this composite in the longitudinal direction. Solution: First find the ratio of fiber load to matrix load, using Equation ， m m f f m f E V E V F F = f m m f or F F F F 13.5 13.5 (3.4 GPa)(0.6) (69 GPa)(0.4) = = = A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa. (b) If the cross-sectional area is 250 mm2 and a stress of 50 MPa is applied in this longitudinal direction, compute the magnitude of the load carried by each of the fiber and matrix phases. The total force sustained by the composite Fc: Fc = Acσ = (250 mm2)(50 MPa) = 12,500 N This total load is just the sum of the loads carried by fiber and matrix phases, that is 13.5 Fm + Fm = 12,500 N or Fm = 860 N whereas Ff = Fc - Fm = 12,500 N - 860 N = 11,640 N A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa. (b) If the cross-sectional area is 250 mm2 and a stress of 50 MPa is applied in this longitudinal direction, compute the magnitude of the load carried by each of the fiber and matrix phases. A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa. The cross-sectional area is 250 mm2 and a stress of 50 MPa is applied. (c) Determine the strain that is sustained by each phase when the stress in part b is applied. For stress calculations, phase cross-sectional areas are necessary: Am = VmAc = (0.6)(250 mm2) = 150 mm2 Af = Vf Ac = (0.4)(250 mm2) = 100 mm2
422外力垂直于纤维:等应力 4150mnx=5.73MPa 11,640N 116.4MPa 复合材料的应变可表示 Finally, strains are computed as 5.73MPa 代入虎克定律 En34×103MPa 1164Ma=169×103 复合材料模量预测(2) 由等应力条件: 纤单垂直于外力方向 我们得到 EXAMPLE PROBLEM 4.2 42.3基体的塑性流动 e material described in Example Problem 4.1, but assume that the stress is applied 基体发生塑性流动时,复合材料的极限强度可表示为 perpendicular to the direction of fiber alignment. SOLUTION According to Equation 17 16, 其中σ。是纤维的极限拉伸强度, 是应变硬化基体的流动应力 (3.4 GPa)69 GPa) =5.5GPa 复合材料的极限强度¤必然高于基体的极限强度 (06(69GPa)+(0.4)34GPa)
4 Thus, MPa mm N A F MPa mm N A F f f f m m m 116.4 100 11,640 5.73 150 860 2 2 = = = = = = σ σ 3 3 3 3 1.69 10 69 10 116.4 1.69 10 3.4 10 5.73 − − = × × = = = × × = = MPa MPa E MPa MPa E f f f m m m σ ε σ ε Finally, strains are computed as 复合材料的应变可表示 为： 代入虎克定律： εc = εf Vf + εm Vm m m m f f f c c V E V E E σ σ σ = + 4.2.2 外力垂直于纤维：等应力 由等应力条件： σc = σf = σm 我们得到： m m f f c E V E V E = + 1 复合材料模量预测（2） Compute the elastic modulus of the composite material described in Example Problem 4.1 , but assume that the stress is applied perpendicular to the direction of fiber alignment. SOLUTION According to Equation 17.16, EXAMPLE PROBLEM 4.2 Ec 5.5GPa (0.6)(69 GPa) (0.4)(3.4 GPa) (3.4 GPa)(69 GPa) = + = 基体发生塑性流动时，复合材料的极限强度可表示为： σcu = σfu Vf + σ’m Vm 其中σfu 是纤维的极限拉伸强度， σ’m 是应变硬化基体的流动应力。 复合材料的极限强度σcu必然高于基体的极限强度： σfu Vf + σ’m Vm ≥ σmu 4.2.3 基体的塑性流动
423基体的塑性流动 42.4应力传递 可导出发生增强的临界纤帷体积分vrn fcrit O :Ⅲ axImum 42.4应力传递 Stress 临界纤维长度:1 do 临界长度 临界长径比 非连续纤维整齐排列,复合材料的强度可以按下式修 43聚合物基体 非等长纤維:只要<L2,以上分析仍然适用 对比
5 4.2.3 基体的塑性流动 可导出发生增强的临界纤维体积分数Vf crit： fu m mu m Vfcrit ' ' σ σ σ σ − − = 4.2.4 应力传递 τ σ Stress Position 4.2.4 应力传递 4 2 2 l d d f τπ π σ = c f c c d l τ σ 2 = c fu c c d l τ σ 2 = 临界纤维长度： 临界长径比： τ σ Stress Position 临界长度 l/2 l/2 l = lc Maximum applied load ∗ σ f ∗ σ f ∗ σ f Stress l lc 非连续纤维整齐排列，复合材料的强度可以按下式修 正： 非等长纤维：只要l < lc ，以上分析仍然适用。 m m c cu fu f V l l V ' 2 σ σ 1 +σ = − σcu = σfu Vf + σ’m Vm 对比 4.3 聚合物基体
43聚合物基体 43聚合物基体 聚合物基体 聚苯硫隆(PPS) 热塑性基体 热固性基体 聚砜(PES) 聚酰胶、聚碳酸酚醛树脂 环氧树脂 不饱和聚酯 酯、聚砜、聚苯 硫醚、聚醚醚 聚PE◎ooc Important characteristics 44增强纤维 3. A very high degree of flexibility Flexibility The flexibility of a material is determined by shape, size of the cross section, and its radius of curvature We can use the inverse of the product of bending moment(M) and the radius of curvature (R)as a measure of flexibility. MR End Fiber diameter of materials with flexibility equal to that of
6 热塑性基体 酚酚酚酚 环环酚酚 不不不不不 热固性基体 不聚聚聚聚 聚酰胺、聚碳酸 酯、聚砜、聚苯 硫醚、聚醚醚 酮、氟塑料 4.3 聚合物基体 聚苯硫醚 (PPS) 聚砜 (PES) 聚醚醚酮 (PEEK) － S－ n O S = O = O －－ n － O C = O －－ n － O 4.3 聚合物基体 4.4 增强纤维 1. A small diameter 2. A high aspect ratio 3. A very high degree of flexibility Important characteristics Flexibility 4 1 64 MR Eπd = The flexibility of a material is determined by shape, size of the cross section, and its radius of curvature. We can use the inverse of the product of bending moment (M) and the radius of curvature (R) as a measure of flexibility. Fiber diameter of materials with flexibility equal to that of a 25-µm-diameter nylon fiber. d (µm) E (GPa) 100 200 300 400 500 25 20 15 10 5 0 nylon glassmullite Si3N4 C ZrO2 Al2O3 B SiC
44.1 Boron fibers硼纤维 A boron filament production facility Chemical vapor deposition(cVD)化学气相沉 2BX3+3H2 +6HX X: Cl, Br, or I BC/H2气相沉积 44.1硼纤维 44.2 Carbon fibers碳纤维 比重:2.34-2.6 拉伸强度:2.1-5lGPa 模量:-400GPa 碳合量在92-95%之间,桃量在34GPa以下的为碳纤维, 碳含量在9%以上,横量在J4GPa以上的为石曼纤维 Reaction zone 在1300°C左右热解的称碳纤维 在1900°以上热解的称石纤维 44.2碳纤維与石纤维 Ladder molecule z2座oe polyacrylonitrile 聚丙烯睛法 C≡N 石墨化
7 2BX3+3H2 2B+6HX X : C1, Br, or I 4.4.1 Boron Fibers 硼纤维 Chemical vapor deposition (CVD) 化学气相沉 积法 钨丝 卷取 加热 加热 预热 H2 BCl3/H2 气相沉积 A boron filament production facility 4.4.1 硼纤维 硼纤维的结构 比重：2.34 ~ 2.6 拉伸强度：2.1 ~ 5.1GPa 模量： ~ 400GPa Reaction zone W2B5+WB4 W SiC B 4.4.2 Carbon Fibers 碳纤维 碳含量在92~95%之间，模量在344GPa以下的为碳纤维， 碳含量在99%以上，模量在344GPa以上的为石墨纤维。 在1300°C左右热解的称碳纤维 在1900°C以上热解的称石墨纤维 4.4.2 碳纤维与石墨纤维 Oxidation up to 250°C Carbonization 250~1500°C Graphitization 1500~2500°C C C H H H C N C C H H H C N C C H H H C N C C H H H C N C C H H H C N C C H H H C N C C H H H C N C C H H H C N ∆ Ladder molecule polyacrylonitrile
Properties of different carbon fibers EX-Pitch Carbon fibers Density Sources of pitch Polyvinyl chloride(PvC) Pitch( Kureha Petroleum asphalt Mesophase pitch Coal tar Single-crystal graphite 2.25 4.4.3 Organic Fibers 44增强纤维 44.3有机纤维 单维增强环氧树脂 Kevlar⑥ 石墨纤锥 拉伸强度3.GPa 1000 玻璃纤维 Kevlar49:模量130GPa Kevlar1491模量180GPa 005101520253.0 4.4.3有机纤维:聚乙烯纤维 Spectra聚乙烯纤维 聚乙烯溶液 烘 CH
8 Ex-Pitch Carbon Fibers Sources of pitch Polyvinyl chloride (PVC) Petroleum asphalt Coal tar Density Young's Electrical Precursor (g/m3) modulus resistivity (GPa) (10-4 Ω−cm) Rayon 1.66 390 10 Polyacrylonitrile 1.74 230 18 Pitch (Kureha) LT 1.6 41 100 HT 1.6 41 50 Mesophase pitch LT 2.1 340 9 HT 2.2 690 1.8 Single-crystal graphite 2.25 l000 0.40 Properties of different carbon fibers Kevlar 4.4.3 Organic Fibers N N C O C O H H 4.4 增强纤维 4.4.3 有机纤维 Kevlar29：高韧性， 拉伸强度3.4GPa Kevlar49：模量130GPa Kevlar149：模量180GPa 比重：1.44 Spectra CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 聚乙烯纤维 4.4.3 有机纤维：聚乙烯纤维 烘箱 冷却 聚乙烯溶液
44增强纤维 444 Ceramic Fibers陶瓷纤维 443有机纤维 Three fabrication methods 聚乙烯纤维的性质 1. Chemical vapor deposition 性质 Spectra900 Spectra I0001 Kevlar149 密度(g/em)0.97 2. Polymer pyrolysis 直径(μ 拉伸强度(GPa)2 3.0 3. Sol-gel techniques 拉伸模量(GPa)119 The 3M Co. process of making AlO, fiber Precursor fib ymer and alkyl silicate Calcination solvent The Sumitomo process of making Draw wheels Composition and properties of Nextel series fibers 312Al2O262,SiO24,B2O314 440Al2O370,SO228,B2O3210-123.052000186 550A12O373,S0227 10-123.03 610A12O39+,(siO,Fe2O3)10-12 720A2O385.S0215 2130 Optical micrograph of Nextel 312(AL, O +SiO,+B,O3)
9 4.4 增强纤维 4.4.3 有机纤维 聚乙烯纤维的性质 性质 Spectra900 Spectra 1000 Kevlar 149 密度(g/cm3) 0.97 0.97 1.44 直径（µ） 38 27 拉伸强度(GPa) 2.7 3.0 拉伸模量(GPa) 119 175 180 4.4.4 Ceramic Fibers 陶瓷纤维 Three fabrication methods 1. Chemical vapor deposition 2. Polymer pyrolysis 3. Sol-gel techniques The Sumitomo process of making a silica- stabilized Al2O3 fiber Organoalumino compound Alkyl aluminum or alkoxy aluminum (AlR3) Polymerization AlR3 + H2O －Al－O－ － R Organic solvent + Si-containing compound (alkyl silicate) Dry spinning Precursor fiber (organoaluminum polymer and alkyl silicate) Calcination Inorganic fiber Al2O3: 70-100% SiO2: 30-0% Filter Sol reservior Pump Draw wheels Pyrolysis furnace Spinneret High temperature straightening furnace Takeup spool The 3M Co. process of making Al2O3 fiber Composition Diameter Density Tensile Young’s No. (wt%) (µm) (g/cm3) strength modulus (MPa) (GPa) 312 Al2O362, SiO224, B2O314 10-12 2.7 1700 152 440 Al2O3 70, SiO2 28, B2O3 2 10-12 3.05 2000 186 550 Al2O3 73, SiO2 27 10-12 3.03 2000 193 610 Al2O3 99+, (SiO2, Fe2O3) 10-12 3.75 1900 370 720 Al2O3 85, SiO2 15 10-12 3.4 2130 260 Composition and properties of Nextel series fibers Optical micrograph of Nextel 312(Al2O3+SiO2+B2O3 )
CVD Silicon Carbide Fibers 4. 4.5 Glass Fiber CH3 13(g) SiC(s)+3 HCI(g) 55%二氧化硅,20%氧化钙,15%氧化铝,10%氧化 roperties 拉伸强度;5GPa( Electrical application) Diameter Density Tensile Y 6氧化,25%氧化铝,65%二氧化硅 拉伸强度:7GPa( High Strength) 140 3500430 ch mixing 445玻璃纤维 melting area Y Filament collecting Collet and size applicator Strand traverse Glass fiber manufacture 44增强纤维 44.6 Wiskers晶须 44.6晶须:谷法 单晶 焦化 F反应器 接近理论强度(1~2万MPa) 直径:几至几mm 长径比:50-10000 缺点:尺寸不均一 一并匀一余 性能不均 烘干—脱碳碳化硅晶须
10 CH3SiC13(g) SiC(s) + 3 HCl(g) H2 CVD Silicon Carbide Fibers Diameter Density Tensile Young’s (µm) (g/cm3) strength modulus (MPa) (GPa) 140 3.3 3500 430 Properties E玻璃： 55%二氧化硅，20%氧化钙，15%氧化铝，10%氧化硼 拉伸强度：5GPa（Electrical application） S玻璃 10%氧化镁，25%氧化铝，65%二氧化硅 拉伸强度：7GPa （High Strength） 4.4.5 Glass Fiber Batch mixing Batch hopper at furnace Furnace melting area Platinum bushing Collet Filament collecting and size applicator Strand traverse motion Winding head unit Glass fiber manufacture 4.4.5 玻璃纤维 Roving Fabric Chopped strand continuous yarn 单晶 接近理论强度（1~2万MPa ） 直径：几µ至几mm 长径比：50~10000 缺点：尺寸不均一 性能不均一 4.4.6 Wiskers 晶须 4.4 增强纤维 4.4.6 晶须：谷糠法 糠 研磨 焦化 碳管反应器 ＞1600 °C 破碎 碳化硅晶须 拌匀 分离多余碳 分离残糠 烘干 脱碳