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 ∗ σ f Stress ∗ σ f ∗ σ f l/2 l/2 ∗ σ f Stress ∗ σ f ∗ σ f 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