TABLE 3.2 Important Energy Absorption Mechanisms During Longitudinal Tensile Loading of a Unidirectional Continuous Fiber Lamina Stress relaxation energy (energy dissipated ￡- 6Er owing to reduction in stresses at the ends of a broken fiber [6]) Stored elastic energy in a partially E(where y=debonded length of debonded fiber[7] 4Er the fiber when it breaks) Fiber pullout energy 8] v for ke l Epo= 12h veo for k<le 121 Energy absorption by matrix deformation End=(1-vr)(mude between parallel matrix cracks [9] Vr (where Um=energy required in deforming unit volume of the matrix to rupture) Notes. 1.All energy expressions are on the basis of unit fracture surface area. 2.Debonding of fibers ahead of a crack tip or behind a crack tip is an important energy absorption mechanism.However.no suitable energy expression is available for this mechanism. 3.Energy absorption may also occur because of yielding of fibers or matrix if either of these constituents is ductile in nature. may develop either at the fiber-matrix interface or in the matrix at 0=90 (Figure 3.16c). In a lamina containing a high volume fraction of fibers,there will be interactions of stress fields from neighboring fibers.Adams and Doner [10] used a finite difference method to calculate the stresses in unidirectional com- posites under transverse loading.A rectangular packing arrangement of paral- lel fibers was assumed,and solutions were obtained for various interfiber spacings representing different fiber volume fractions.Radial stresses at the fiber-matrix interface for 55%and 75%fiber volume fractions are shown in Figure 3.17.The maximum principal stress increases with increasing Er/Em ratio and fiber volume fraction,as indicated in Figure 3.18.The transverse modulus of the composite has a similar trend.Although an increased transverse modulus is desirable in many applications,an increase in local stress concen- trations at high volume fractions and high fiber modulus may reduce the transverse strength of the composite (Table 3.3). The simplest model used for deriving the equation for the transverse modulus of a unidirectional continuous fiber-reinforced composite is shown in Figure 3.19 2007 by Taylor&Francis Group.LLC
may develop either at the fiber–matrix interface or in the matrix at u ¼ 908 (Figure 3.16c). In a lamina containing a high volume fraction of fibers, there will be interactions of stress fields from neighboring fibers. Adams and Doner [10] used a finite difference method to calculate the stresses in unidirectional composites under transverse loading. A rectangular packing arrangement of parallel fibers was assumed, and solutions were obtained for various interfiber spacings representing different fiber volume fractions. Radial stresses at the fiber–matrix interface for 55% and 75% fiber volume fractions are shown in Figure 3.17. The maximum principal stress increases with increasing Ef=Em ratio and fiber volume fraction, as indicated in Figure 3.18. The transverse modulus of the composite has a similar trend. Although an increased transverse modulus is desirable in many applications, an increase in local stress concentrations at high volume fractions and high fiber modulus may reduce the transverse strength of the composite (Table 3.3). The simplest model used for deriving the equation for the transverse modulus of a unidirectional continuous fiber-reinforced composite is shown in Figure 3.19 TABLE 3.2 Important Energy Absorption Mechanisms During Longitudinal Tensile Loading of a Unidirectional Continuous Fiber Lamina Stress relaxation energy (energy dissipated owing to reduction in stresses at the ends of a broken fiber [6]) Er ¼ vfs2 fulc 6Ef Stored elastic energy in a partially debonded fiber [7] Es ¼ vfs2 fuy 4Ef (where y ¼ debonded length of the fiber when it breaks) Fiber pullout energy ½8 Epo ¼ vfsful 2 c 12lf for lf > lc ¼ vfsful 2 f 12lc for lf < lc Energy absorption by matrix deformation between parallel matrix cracks [9] Emd ¼ (1 vf) 2 vf smudf 4ti Um (where Um ¼ energy required in deforming unit volume of the matrix to rupture) Notes: 1. All energy expressions are on the basis of unit fracture surface area. 2. Debonding of fibers ahead of a crack tip or behind a crack tip is an important energy absorption mechanism. However, no suitable energy expression is available for this mechanism. 3. Energy absorption may also occur because of yielding of fibers or matrix if either of these constituents is ductile in nature. 2007 by Taylor & Francis Group, LLC.
6 At0=90° 个 4 Goela Onla 0.4 0 Fiber Stress 1.6 ratio 1.6 0 0.4 d 42 Matrix 4 6 At0=-90° (a) () Matrix cracking Fiber Fiber-matrix interfacial cracking (c) FIGURE 3.16 (a)Transverse tensile loading on a lamina containing a single cylindrical fiber,(b)stress distribution around a single fiber due to transverse tensile loading,and (c)possible microfailure modes. in which the fibers and the matrix are replaced by their respective"equivalent" volumes and are depicted as two structural elements (slabs)with strong bond- ing across their interface.The tensile load is acting normal to the fiber direc- tion.The other assumptions made in this simple slab model are as follows. 2007 by Taylor Francis Group,LLC
in which the fibers and the matrix are replaced by their respective ‘‘equivalent’’ volumes and are depicted as two structural elements (slabs) with strong bonding across their interface. The tensile load is acting normal to the fiber direction. The other assumptions made in this simple slab model are as follows. s srr r Fiber q sqq s (a) Matrix df At q=908 At q=−908 6 4 r df 2 0 0 2 4 6 1.6 Stress ratio 1.6 0.4 0.4 srr /s sqq /s r df (b) Matrix cracking Fiber–matrix interfacial cracking Fiber (c) FIGURE 3.16 (a) Transverse tensile loading on a lamina containing a single cylindrical fiber, (b) stress distribution around a single fiber due to transverse tensile loading, and (c) possible microfailure modes. 2007 by Taylor & Francis Group, LLC
Tensile normal stress Fiber Shear stress -1 0 (a) 2 Shear stress Tensile normal stress -ibe (b) FIGURE 3.17 Variation of shear stress T and normal stress or at the surface of a circular fiber in a square array subjected to an average tensile stress o transverse to the fiber directions:(a)v =55%and (b)vr 75%.(After Adams,D.F.and Doner,D.R., J.Compos.Mater.,1,152,1967.) 1.Total deformation in the transverse direction is the sum of the total fiber deformation and the total matrix deformation,that is,AWe= △Wr+△Wm. 2.Tensile stress in the fibers and the tensile stress in the matrix are both equal to the tensile stress in the composite,that is,or=om=oc. Since&c=,&鲜=，and&a=,the deformtionqn△W.=△W:+ △W can be written as Sc We st Wi Sm Wm. (3.22) 2007 by Taylor&Francis Group.LLC
1. Total deformation in the transverse direction is the sum of the total fiber deformation and the total matrix deformation, that is, DWc ¼ DWf þ DWm. 2. Tensile stress in the fibers and the tensile stress in the matrix are both equal to the tensile stress in the composite, that is, sf ¼ sm ¼ sc. Since «c ¼ DWc Wc , «f ¼ DWf Wf , and «m ¼ DWm Wm , the deformation equation DWc¼ DWfþ DWm can be written as «cWc ¼ «fWf þ «mWm: (3:22) (a) 0 123 4 (b) −1 1 Shear stress 0 −1 Tensile normal stress Fiber 1 0 −10 1 2 3 −1 Tensile normal stress Shear stress Fiber FIGURE 3.17 Variation of shear stress tru and normal stress srr at the surface of a circular fiber in a square array subjected to an average tensile stress s transverse to the fiber directions: (a) vf ¼ 55% and (b) vf ¼ 75%. (After Adams, D.F. and Doner, D.R., J. Compos. Mater., 1, 152, 1967.) 2007 by Taylor & Francis Group, LLC
Circular fibers arranged in a square array 3.0 4=0.78 Denotes locations of maximum principal 0.75 stress 0.70 2.0 0.55 0.40 0.04 1.0 46810 20 4060100200 4006001000 Ratio of fiber modulus to matrix modulus,E/Em FIGURE 3.18 Ratio of the maximum principal stress in the matrix to the applied transverse stress on the composite for various fiber volume fractions.(After Adams, D.F.and Doner,D.R.,J.Compos.Mater.,1,152,1967.) TABLE 3.3 Effect of Transverse Loading in a Unidirectional Composite Transverse Transverse Modulus, Strength, Composite Material 会 (%) GPa(Msi) MPa (ksi) E-glass-epoxy 20 39 8.61(1.25) 47.2(6.85) 67 18.89(2.74) 30.87(4.48) E-glass-epoxy 24 6 8.96(1.30) 69.1(10.03) 13.23(1.92) 77.92(11.31) 68 21.91(3.18) 67.93(9.86 7 25.9(3.76 41.27(5.99) Boron-epoxy 120 65 23.43(3.4) 41.96(6.09) Source:Adapted from Adams.D.F.and Doner.D.R..J.Compos.Mater..1.152,1967. Dividing both sides by Wand noting that-vr and Vm,we can rewrite Equation 3.22 as Sc =8fVf EmVm. (3.23) 2007 by Taylor Francis Group,LLC
Dividing both sides by Wc and noting that Wf Wc ¼ vf and Wm Wc ¼ vm, we can rewrite Equation 3.22 as «c ¼ «fvf þ «mvm: (3:23) 1.0 1 2 4 6 8 10 20 40 Circular fibers arranged in a square array Denotes locations of maximum principal stress r d s s vf = 0.78 0.75 0.70 0.55 0.40 0.04 60 100 200 400 600 1000 2.0 Ratio of maximum principal stress to applied stress Ratio of fiber modulus to matrix modulus, Ef/Em 3.0 d FIGURE 3.18 Ratio of the maximum principal stress in the matrix to the applied transverse stress on the composite for various fiber volume fractions. (After Adams, D.F. and Doner, D.R., J. Compos. Mater., 1, 152, 1967.) TABLE 3.3 Effect of Transverse Loading in a Unidirectional Composite Composite Material Ef Em vf (%) Transverse Modulus, GPa (Msi) Transverse Strength, MPa (ksi) E-glass–epoxy 20 39 8.61 (1.25) 47.2 (6.85) 67 18.89 (2.74) 30.87 (4.48) E-glass–epoxy 24 46 8.96 (1.30) 69.1 (10.03) 57 13.23 (1.92) 77.92 (11.31) 68 21.91 (3.18) 67.93 (9.86) 73 25.9 (3.76) 41.27 (5.99) Boron–epoxy 120 65 23.43 (3.4) 41.96 (6.09) Source: Adapted from Adams, D.F. and Doner, D.R., J. Compos. Mater., 1, 152, 1967. 2007 by Taylor & Francis Group, LLC
Fiber Matrix slab slab Transverse direction W -Wm W FIGURE 3.19 Transverse loading of a unidirectional continuous fiber lamina and the equivalent slab model. Since sc= e 6= E and em= mEquation 3.23 can be written as of Om vm (3.24) In Equation 3.24,Er is the transverse modulus of the unidirectional continuous fiber composite. Finally,since it is assumed that or=om=de,Equation 3.24 becomes 日营+2 (3.25) Rearranging Equation 3.25,the expression for the transverse modulus Er becomes EfEm EfEm B=ENn+EN--(G-E面 (3.26 Equation 3.26 shows that the transverse modulus increases nonlinearly with increasing fiber volume fraction.By comparing Equations 3.7 and 3.26,it can be seen that the transverse modulus is lower than the longitudinal modulus and is influenced more by the matrix modulus than by the fiber modulus. A simple equation for predicting the transverse tensile strength of a unidir- ectional continuous fiber lamina [11]is Omu OTwu= (3.27) 2007 by Taylor Francis Group.LLC
Since «c ¼ sc ET , «f ¼ sf Ef , and «m ¼ sm Em , Equation 3.23 can be written as sc ET ¼ sf Ef vf þ sm Em vm: (3:24) In Equation 3.24, ET is the transverse modulus of the unidirectional continuous fiber composite. Finally, since it is assumed that sf ¼ sm ¼ sc, Equation 3.24 becomes 1 ET ¼ vf Ef þ vm Em : (3:25) Rearranging Equation 3.25, the expression for the transverse modulus ET becomes ET ¼ EfEm Efvm þ Emvf ¼ EfEm Ef vf(Ef Em) : (3:26) Equation 3.26 shows that the transverse modulus increases nonlinearly with increasing fiber volume fraction. By comparing Equations 3.7 and 3.26, it can be seen that the transverse modulus is lower than the longitudinal modulus and is influenced more by the matrix modulus than by the fiber modulus. A simple equation for predicting the transverse tensile strength of a unidirectional continuous fiber lamina [11] is sTtu ¼ smu Ks , (3:27) Fiber slab Matrix slab Transverse direction Wc Wf Wm FIGURE 3.19 Transverse loading of a unidirectional continuous fiber lamina and the equivalent slab model. 2007 by Taylor & Francis Group, LLC.