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