第四章复合材料(共4学时)(彩色版)

Weight
Thermal expansion
Stiffness
Strength
Fatigue resistance
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 CMC 陶瓷材料
σfu Vf + σ’m Vm ≥ σmu
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： Position
4.2.4 应力传递
τ σ
V fcrit
σ − σ 'm = mu σ fu − σ 'm
Stress
Stress
σ
Position
σf
πd 2
τ
Stress
l = τπd 4 2
4.2.3 基体的塑性流动
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：
Ec =
(3.4 GPa)(69 GPa) = 5.5GPa (0.6)(69 GPa) + (0.4)(3.4 GPa)
Fm 860 N = = 5.73MPa Am 150mm 2 Ff Af = 11,640 N = 116.4 MPa 100mm 2
4.2.2 外力垂直于纤维：等应力
复合材料的应变可表示 为：
Finally, strains are computed as
εc = εf Vf + εm Vm
代入虎克定律：
σ f Ef = σ m Em
Ec εc = Vf Ef εf + Vm Em εm
由等应变假定
σf = Ef εf ，σm = Em εm
Ec = Ef Vf + Em Vm
故有
Ff Fm
=
E fVf EmVm
复合材料模量预测（1）
EXAMPLE PROBLEM 4.1
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.
εm =
σm
Em
=
5.73MPa = 1.69 ×10 −3 3.4 ×103 MPa
116.4 MPa = = 1.69 ×10 −3 εf = E f 69 ×103 MPa
σf
σc
Ec
=
σf
Ef
Vf +
σm
Em
Vm
复合材料模量预测（2） 由等应力条件：
σc = σf = σm
我们得到：
1 V f Vm = + Ec E f Em
4.1 概述
复合材料按基体分类
高分子材料 PMC 增 强材 料 MMC 金属材料
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。
Ff Fm
=
E fVf E mVm
，
Ff Fm
=
(69 GPa)(0.4) = 13 .5 or F f = 13 .5 Fm (3.4 GPa)(0.6)
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 whereas or Fm = 860 N 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. Solution: First find the ratio of fiber load to matrix load,
EXAMPLE PRห้องสมุดไป่ตู้BLEM 4.2
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,
SOLUTION
(a) The modulus of elasticity of the composite is calculated using Equation Ec = EmVm + EfVf : Ec = (3.4 GPa)(0.6) + (69 GPa)(0.4) = 30 GPa using Equation
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.
Naturally Occurring Composites
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.
Comparison between conventional monolithic materials and composite materials.
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%) 。
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 = VfAc = (0.4)(250 mm2) = 100 mm2
Thus,
σm = σf =
4.2.1 应力平行于纤维，等应变
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。
模量加和规律
σc = Vf σf + Vm σm σf = Ef εf ，σm = Em εm ，σc = Ec εc
ε代表应变
求受力比
Ff Fm
由
=
σ f Af σ f V f / l σ f V f = = σ m Am σ mVm / l σ mVm
σ d lc = f c 2τ c
l/2