4 Performance The performance of an engineering material is judged by its properties and behavior under tensile,compressive,shear,and other static or dynamic loading conditions in both normal and adverse test environments.This information is essential for selecting the proper material in a given application as well as designing a structure with the selected material.In this chapter,we describe the performance of fiber-reinforced polymer composites with an emphasis on the general trends observed in their properties and behavior.A wealth of property data for continuous fiber thermoset matrix composites exists in the published literature.Continuous fiber-reinforced thermoplastic matrix com- posites are not as widely used as continuous fiber-reinforced thermoset matrix composites and lack a wide database. Material properties are usually determined by conducting mechanical and physical tests under controlled laboratory conditions.The orthotropic nature of fiber-reinforced composites has led to the development of standard test methods that are often different from those used for traditional isotropic materials.These unique test methods and their limitations are discussed in relation to many of the properties considered in this chapter.The effects of environmental conditions,such as elevated temperature or humidity,on the physical and mechanical properties of composite laminates are presented near the end of the chapter.Finally,long-term behavior,such as creep and stress rupture,and damage tolerance are also discussed. 4.1 STATIC MECHANICAL PROPERTIES Static mechanical properties,such as tensile,compressive,flexural,and shear properties,of a material are the basic design data in many,if not most, applications.Typical mechanical property values for a number of 0 laminates and sheet-molding compound(SMC)laminates are given in Appendix A.5 and Appendix A.6,respectively. 4.1.1 TENSILE PROPERTIES 4.1.1.1 Test Method and Analysis Tensile properties,such as tensile strength,tensile modulus,and Poisson's ratio of flat composite laminates,are determined by static tension tests in accordance 2007 by Taylor Francis Group.LLC
4 Performance The performance of an engineering material is judged by its properties and behavior under tensile, compressive, shear, and other static or dynamic loading conditions in both normal and adverse test environments. This information is essential for selecting the proper material in a given application as well as designing a structure with the selected material. In this chapter, we describe the performance of fiber-reinforced polymer composites with an emphasis on the general trends observed in their properties and behavior. A wealth of property data for continuous fiber thermoset matrix composites exists in the published literature. Continuous fiber-reinforced thermoplastic matrix composites are not as widely used as continuous fiber-reinforced thermoset matrix composites and lack a wide database. Material properties are usually determined by conducting mechanical and physical tests under controlled laboratory conditions. The orthotropic nature of fiber-reinforced composites has led to the development of standard test methods that are often different from those used for traditional isotropic materials. These unique test methods and their limitations are discussed in relation to many of the properties considered in this chapter. The effects of environmental conditions, such as elevated temperature or humidity, on the physical and mechanical properties of composite laminates are presented near the end of the chapter. Finally, long-term behavior, such as creep and stress rupture, and damage tolerance are also discussed. 4.1 STATIC MECHANICAL PROPERTIES Static mechanical properties, such as tensile, compressive, flexural, and shear properties, of a material are the basic design data in many, if not most, applications. Typical mechanical property values for a number of 08 laminates and sheet-molding compound (SMC) laminates are given in Appendix A.5 and Appendix A.6, respectively. 4.1.1 TENSILE PROPERTIES 4.1.1.1 Test Method and Analysis Tensile properties, such as tensile strength, tensile modulus, and Poisson’s ratio of flat composite laminates, are determined by static tension tests in accordance 2007 by Taylor & Francis Group, LLC
-38 mm (1.5 in.)Gage length +2(width) +中-38mm(15in.)+ End tab End tab Width Tab 25 thickness LSpecimen thickness Tab thickness FIGURE 4.1 Tensile test specimen configuration. with ASTM D3039.The tensile specimen is straight-sided and has a constant cross section with beveled tabs adhesively bonded at its ends (Figure 4.1). A compliant and strain-compatible material is used for the end tabs to reduce stress concentrations in the gripped area and thereby promote tensile failure in the gage section.Balanced [0/90]cross-ply tabs of nonwoven E-glass-epoxy have shown satisfactory results.Any high-elongation (tough)adhesive system can be used for mounting the end tabs to the test specimen. The tensile specimen is held in a testing machine by wedge action grips and pulled at a recommended cross-head speed of 2 mm/min (0.08 in./min). Longitudinal and transverse strains are measured employing electrical resistance strain gages that are bonded in the gage section of the specimen.Longitudinal tensile modulus Eu and the major Poisson's ratio v12 are determined from the tension test data of o unidirectional laminates.The transverse modulus E22 and the minor Poisson's ratio vz are determined from the tension test data of 90 unidirectional laminates. For an off-axis unidirectional specimen (0<0<90),a tensile load creates both extension and shear deformations (since 416 and 4260).Since the specimen ends are constrained by the grips,shear forces and bending couples are induced that create a nonuniform S-shaped deformation in the specimen (Figure 4.2).For this reason,the experimentally determined modulus of an off- axis specimen is corrected to obtain its true modulus [1]: Etrue =(1-m)Eexperimental, where 356 7=3,356/5）+2L/wT' (4.1) 2007 by Taylor Francis Group,LLC
with ASTM D3039. The tensile specimen is straight-sided and has a constant cross section with beveled tabs adhesively bonded at its ends (Figure 4.1). A compliant and strain-compatible material is used for the end tabs to reduce stress concentrations in the gripped area and thereby promote tensile failure in the gage section. Balanced [0=90] cross-ply tabs of nonwoven E-glass–epoxy have shown satisfactory results. Any high-elongation (tough) adhesive system can be used for mounting the end tabs to the test specimen. The tensile specimen is held in a testing machine by wedge action grips and pulled at a recommended cross-head speed of 2 mm=min (0.08 in.=min). Longitudinal and transverse strains are measured employing electrical resistance strain gages that are bonded in the gage section of the specimen. Longitudinal tensile modulus E11 and the major Poisson’s ratio n12 are determined from the tension test data of 08 unidirectional laminates. The transverse modulus E22 and the minor Poisson’s ratio n21 are determined from the tension test data of 908 unidirectional laminates. For an off-axis unidirectional specimen (08 < u < 908), a tensile load creates both extension and shear deformations (since A16 and A26 6¼ 0). Since the specimen ends are constrained by the grips, shear forces and bending couples are induced that create a nonuniform S-shaped deformation in the specimen (Figure 4.2). For this reason, the experimentally determined modulus of an offaxis specimen is corrected to obtain its true modulus [1]: Etrue ¼ (1 h) Eexperimental, where h ¼ 3S2 16 S2 11[3(S66=S11) þ 2(L=w) 2 ] , (4:1) ≥58 Specimen thickness Tab thickness Tab thickness End tab Width 38 mm (1.5 in.) Gage length + 2 (width) 38 mm (1.5 in.) End tab FIGURE 4.1 Tensile test specimen configuration. 2007 by Taylor & Francis Group, LLC.
FIGURE 4.2 Nonuniform deformation in a gripped off-axis tension specimen. where L is the specimen length between grips w is the specimen width Su,S16,and S66 are elements in the compliance matrix (see Chapter 3) The value of n approaches zero for large values of L/w.Based on the investigation performed by Rizzo [2],L/w ratios >10 are recommended for the tensile testing of off-axis specimens. The inhomogeneity of a composite laminate and the statistical nature of its constituent properties often lead to large variation in its tensile strength. Assuming a normal distribution,the average strength,standard deviation, and coefficient of variation are usually reported as Average strength Cave= Standard deviation =d= (n-1) 100d Coefficient of variation= (4.2) Jave where n is the number of specimens tested o;is the tensile strength of the ith specimen Instead of a normal distribution,a more realistic representation of the tensile strength variation of a composite laminate is the Weibull distribution. Using two-parameter Weibull statistics,the cumulative density function for the composite laminate strength is F(o)=Probability of surviving stress o=e exp (4.3) 2007 by Taylor&Francis Group.LLC
where L is the specimen length between grips w is the specimen width S11,S16, and S66 are elements in the compliance matrix (see Chapter 3) The value of h approaches zero for large values of L=w. Based on the investigation performed by Rizzo [2], L=w ratios >10 are recommended for the tensile testing of off-axis specimens. The inhomogeneity of a composite laminate and the statistical nature of its constituent properties often lead to large variation in its tensile strength. Assuming a normal distribution, the average strength, standard deviation, and coefficient of variation are usually reported as Average strength ¼ save ¼ Xsi n , Standard deviation ¼ d ¼ Pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð Þ si save 2 (n 1) s , Coefficient of variation ¼ 100d save , (4:2) where n is the number of specimens tested si is the tensile strength of the ith specimen Instead of a normal distribution, a more realistic representation of the tensile strength variation of a composite laminate is the Weibull distribution. Using two-parameter Weibull statistics, the cumulative density function for the composite laminate strength is F(s) ¼ Probability of surviving stress s ¼ exp s s0 a , (4:3) FIGURE 4.2 Nonuniform deformation in a gripped off-axis tension specimen. 2007 by Taylor & Francis Group, LLC.
1.0 0.8 [90] 0.6 [0/性45/90] 0.4 0.2 0 7.48.29.0 9.810.6Y5567 79148181214 Ultimate stress(ksi) FIGURE 4.3 Tensile strength distribution in various carbon fiber-epoxy laminates. (Adapted from Kaminski,B.E.,Analysis of the Test Methods for High Modulus Fibers and Composites,ASTM STP,521,181,1973.) where a is a dimensionless shape parameter oo is the location parameter(MPa or psi) The mean tensile strength and variance of the laminates are 0= =(告9）-r(t (4.4) where I represents a gamma function. Figure 4.3 shows typical strength distributions for various composite lamin- ates.Typical values of a and oo are shown in Table 4.1.Note that the decreasing value of the shape parameter a is an indication of greater scatter in the tensile strength data. EXAMPLE 4.1 Static tension test results of 22 specimens of a 0 carbon-epoxy laminate shows the following variations in its longitudinal tensile strength (in MPa):57.54,49.34, 68.67,50.89,53.20,46.15,71.49,72.84,58.10,47.14,67.64,67.10,72.95,50.78 63.59,54.87,55.96,65.13,47.93,60.67,57.42,and67.51.Plot the Weibull distribution curve,and determine the Weibull parameters a and oo for this distribution. 2007 by Taylor Francis Group,LLC
where a is a dimensionless shape parameter s0 is the location parameter (MPa or psi) The mean tensile strength and variance of the laminates are s ¼ s0G 1 þ a a , s 2 ¼ s2 0 G 2 þ a a G2 1 þ a a , (4:4) where G represents a gamma function. Figure 4.3 shows typical strength distributions for various composite laminates. Typical values of a and s0 are shown in Table 4.1. Note that the decreasing value of the shape parameter a is an indication of greater scatter in the tensile strength data. EXAMPLE 4.1 Static tension test results of 22 specimens of a 08 carbon–epoxy laminate shows the following variations in its longitudinal tensile strength (in MPa): 57.54, 49.34, 68.67, 50.89, 53.20, 46.15, 71.49, 72.84, 58.10, 47.14, 67.64, 67.10, 72.95, 50.78, 63.59, 54.87, 55.96, 65.13, 47.93, 60.67, 57.42, and 67.51. Plot the Weibull distribution curve, and determine the Weibull parameters a and s0 for this distribution. 1.0 0.8 [90] [0] [0/±45/90] 0.6 0.4 0.2 0 7.4 8.2 9.0 9.8 10.6 55 67 Ultimate stress (ksi) Probability of survival 79 148 181 214 FIGURE 4.3 Tensile strength distribution in various carbon fiber–epoxy laminates. (Adapted from Kaminski, B.E., Analysis of the Test Methods for High Modulus Fibers and Composites, ASTM STP, 521, 181, 1973.) 2007 by Taylor & Francis Group, LLC.
TABLE 4.1 Typical Weibull Parameters for Composite Laminates Shape Location Parameter, Material Laminate Parameter,o oo MPa (ksi) Boron-epoxy O 24.3 1324.2 (192.0) [90 15.2 66.1 (9.6 [02/±45s 18.7 734.5 (106.6 [0/±45/90s 19.8 419.6 (60.9) [902/45]s 19.8 111.9 (16.1) T-300 Carbon-epoxy5 [Os] 17.7 1784.5 (259) [016d 18.5 1660.5 (241) E-glass-polyester SMCe SMC-R25 7.6 74.2 (10.8) SMC-R50 8.7 150.7 (21.9) a From B.E.Kaminski,Analysis of the Test Methods for High Modulus Fibers and Composites, ASTM STP,521,18L,1973. bFrom R.E.Bullock,J.Composite Mater.,8,200.1974. e From C.D.Shirrell,Polym.Compos..4.172.1983. SOLUTION Step 1:Starting with the smallest number,arrange the observed strength values in ascending order and assign the following probability of failure value for each strength. i P= n+l: where i=1,2,3,,n n=total number of specimens tested P 46.15 1/23=0.0435 2 47.14 2/23=0.0869 2 47.94 3/23=0.1304 21 72.84 21/23=0.9130 22 72.95 22/23=0.9565 Step 2:Plot P vs.tensile strength o to obtain the Weibull distribution plot (see the following figure). 2007 by Taylor&Francis Group.LLC
SOLUTION Step 1: Starting with the smallest number, arrange the observed strength values in ascending order and assign the following probability of failure value for each strength. P ¼ i n þ 1 , where i ¼ 1, 2, 3, . . . , n n ¼ total number of specimens tested i s P 1 46.15 1=23 ¼ 0.0435 2 47.14 2=23 ¼ 0.0869 3 47.94 3=23 ¼ 0.1304 ... ... ... ... ... ... 21 72.84 21=23 ¼ 0.9130 22 72.95 22=23 ¼ 0.9565 Step 2: Plot P vs. tensile strength s to obtain the Weibull distribution plot (see the following figure). TABLE 4.1 Typical Weibull Parameters for Composite Laminates Material Laminate Shape Parameter, a Location Parameter, s0 MPa (ksi) Boron–epoxya [0] 24.3 1324.2 (192.0) [90] 15.2 66.1 (9.6) [02=±45]S 18.7 734.5 (106.6) [0=±45=90]S 19.8 419.6 (60.9) [902=45]S 19.8 111.9 (16.1) T-300 Carbon–epoxyb [08] 17.7 1784.5 (259) [016] 18.5 1660.5 (241) E-glass–polyester SMCc SMC-R25 7.6 74.2 (10.8) SMC-R50 8.7 150.7 (21.9) a From B.E. Kaminski, Analysis of the Test Methods for High Modulus Fibers and Composites, ASTM STP, 521, 181, 1973. b From R.E. Bullock, J. Composite Mater., 8, 200, 1974. c From C.D. Shirrell, Polym. Compos., 4, 172, 1983. 2007 by Taylor & Francis Group, LLC
1.0 0 0.8 0.2 einjle] 0.6 可⊙ 0.4 eAlns 0.4 ⊙-⊙⊙0 0.6 ⊙ 0.2 0.8 ⊙ o 0 0 40 50 60 70 801.0 Tensile strength(MPa) Step 3:Calculate Ye In(In[1/(1-P)])for each strength value,and plot Yp vs.Ino. Use a linear least-squares method to fit a straight line to the data.The slope of this line is equal to a,and its intersection with the Ino axis is equal to In oo.In our example,a 7.62 and In oo 4.13,which gives oo 62.1 MPa. 4.1.1.2 Unidirectional Laminates For unidirectional polymer matrix laminates containing fibers parallel to the tensile loading direction(i.e.,=0),the tensile stress-strain curve is linear up to the point of failure(Figure 4.4).These specimens fail by tensile rupture of fibers, which is followed or accompanied by longitudinal splitting(debonding along the fiber-matrix interface)parallel to the fibers.This gives a typical broom-type appearance in the failed area of0specimens(Figure 4.5a).For off-axis specimens with 0°<0<go°，the tensile stress-strain curves may exhibit nonlinearity.For9o° specimens in which the fibers are 90 to the tensile loading direction,tensile rupture of the matrix or the fiber-matrix interface causes the ultimate failure. For intermediate angles,failure may occur by a combination of fiber-matrix interfacial shear failure,matrix shear failure,and matrix tensile rupture.For many of these off-axis specimens (including 90),matrix craze marks parallel to the fiber direction may appear throughout the gage length at low loads.Repre- sentative failure profiles for these specimens are shown in Figure 4.5b and c. Both tensile strength and modulus for unidirectional specimens depend strongly on the fiber orientation angle 0(Figure 4.6).The maximum tensile strength and modulus are at 0=0.With increasing fiber orientation angle, both tensile strength and modulus are reduced.The maximum reduction is observed near 0=0 orientations. 2007 by Taylor Francis Group,LLC
1.0 0.8 0.6 0.4 Probability of failure Probability of survival Tensile strength (MPa) 0.2 0 30 40 50 60 70 80 0 0.2 0.4 0.6 0.8 1.0 Step 3: Calculate YP ¼ ln{ln[1=(1 P)]} for each strength value, and plotYP vs. lns. Use a linearleast-squares method to fit a straight line to the data. The slope of thisline is equal to a, and itsintersection with the lns axisis equal to lns0.In our example, a ¼ 7.62 and lns0 ¼ 4.13, which gives s0 ¼ 62.1 MPa. 4.1.1.2 Unidirectional Laminates For unidirectional polymer matrix laminates containing fibers parallel to the tensile loading direction (i.e., u ¼ 08),the tensile stress–strain curve islinear up to the point of failure (Figure 4.4). These specimens fail by tensile rupture of fibers, which is followed or accompanied by longitudinal splitting (debonding along the fiber–matrix interface) parallel to the fibers. This gives a typical broom-type appearance in the failed area of 08specimens(Figure 4.5a).For off-axisspecimens with 08 < u < 908,the tensile stress–strain curves may exhibit nonlinearity.For 908 specimens in which the fibers are 908 to the tensile loading direction, tensile rupture of the matrix or the fiber–matrix interface causes the ultimate failure. For intermediate angles, failure may occur by a combination of fiber–matrix interfacial shear failure, matrix shear failure, and matrix tensile rupture. For many of these off-axis specimens (including 908), matrix craze marks parallel to the fiber direction may appear throughout the gage length at low loads. Representative failure profiles for these specimens are shown in Figure 4.5b and c. Both tensile strength and modulus for unidirectional specimens depend strongly on the fiber orientation angle u (Figure 4.6). The maximum tensile strength and modulus are at u ¼ 08. With increasing fiber orientation angle, both tensile strength and modulus are reduced. The maximum reduction is observed near u ¼ 08 orientations. 2007 by Taylor & Francis Group, LLC.
25 1,725 Fiber volume fraction (v)=0.6 A 20 Boron- HT Carbon- S-Glass- 1,380 epoxy epoxy Kevlar 49- epoxy epoxy 15 1.035 10 690 Aluminum 5 (7075-T6) 345 0 0 .0 0.5 1.0 1.52.0 2.5 3.0 Tensile strain (% FIGURE 4.4 Tensile stress-strain curves for various 0 laminates. 4.1.1.3 Cross-Ply Laminates The tensile stress-strain curve for a cross-ply [0/90]s laminate tested at 0=0 direction is slightly nonlinear;however,it is commonly approximated as a bilinear curve(Figure 4.7).The point at which the two linear sections intersect is called the knee and represents the failure of 90 plies.Ultimate failure of the Craze marks Longitudinal splitting (a) (b) (c) FIGURE4.5 Schematic failure modes in unidirectional laminates::(a)0=0°，(b)0=90°， and(c)0<0<90°. 2007 by Taylor&Francis Group.LLC
4.1.1.3 Cross-Ply Laminates The tensile stress–strain curve for a cross-ply [0=90]S laminate tested at u ¼ 08 direction is slightly nonlinear; however, it is commonly approximated as a bilinear curve (Figure 4.7). The point at which the two linear sections intersect is called the knee and represents the failure of 908 plies. Ultimate failure of the 1,725 1,380 1,035 690 345 0 25 Fiber volume fraction (vf ) = 0.6 Boron− epoxy HT Carbon− epoxy Kevlar 49− epoxy S-Glass− epoxy Aluminum (7075-T6) Tensile strain (%) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 20 15 10 Tensile stress (10,000 psi) 5 0 FIGURE 4.4 Tensile stress–strain curves for various 08 laminates. Longitudinal splitting Craze marks (a) (b) (c) FIGURE 4.5 Schematic failure modes in unidirectional laminates: (a) u ¼ 08, (b) u ¼ 908, and (c) 0 < u < 908. 2007 by Taylor & Francis Group, LLC
300 60 40 80 50 1 200 30 20 (sW)sninpow 0 60 卫 30 名 100 10 20 0 0 90 0 999 0 30 60 30 60 90 0 (a) 6(degrees) (b) 6(degrees) FIGURE 4.6 Variations of tensile modulus and tensile strength of a unidirectional carbon fiber-epoxy laminate with fiber orientation angle.(After Chamis,C.C.and Sinclair,J.H.,Mechanical behavior and fracture characteristics of off-axis fiber com- posites,II-Theory and comparisons,NASA Technical Paper 1082,1978.) laminate occurs at the fracture strain of 0 plies.The change in slope of the stress-strain curve at the knee can be reasonably predicted by assuming that all 90 plies have failed at the knee and can no longer contribute to the laminate modulus. Denoting the moduli of the 0 and 90 plies as E and E22,respectively,the initial (primary)modulus of the cross-ply laminate can be approximated as E=40 Eu+4 A A90 E22 (4.5) [o] [0/9o]ls Es Knee [90j ETt ELt FIGURE 4.7 Schematic tensile stress-strain diagram for a [0/90]s cross-plied laminate tested at =0 direction. 2007 by Taylor Francis Group,LLC
laminate occurs at the fracture strain of 08 plies. The change in slope of the stress–strain curve at the knee can be reasonably predicted by assuming that all 908 plies have failed at the knee and can no longer contribute to the laminate modulus. Denoting the moduli of the 08 and 908 plies as E11 and E22, respectively, the initial (primary) modulus of the cross-ply laminate can be approximated as E ¼ A0 A E11 þ A90 A E22, (4:5) 80 60 40 Strength (ksi) 20 0 40 30 20 10 0 0 0 30 60 90 100 200 300 60 90 q (degrees) q (degrees) 0 30 (a) (b) 0 10 20 Strength (MPa) Modulus (GPa) Modulus (Msi) 30 40 50 s q s 60 FIGURE 4.6 Variations of tensile modulus and tensile strength of a unidirectional carbon fiber–epoxy laminate with fiber orientation angle. (After Chamis, C.C. and Sinclair, J.H., Mechanical behavior and fracture characteristics of off-axis fiber composites, II—Theory and comparisons, NASA Technical Paper 1082, 1978.) [90] [0] s sF sk [0/90]S Knee E eTt e Lt e ES FIGURE 4.7 Schematic tensile stress–strain diagram for a [0=90]S cross-plied laminate tested at u ¼ 08 direction. 2007 by Taylor & Francis Group, LLC
where Ao =net cross-sectional area of the 0 plies 490=net cross-sectional area of the 90 plies A=A0+A90 At the knee,the laminate strain is equal to the ultimate tensile strain sru of the 90 plies.Therefore,the corresponding stress level in the laminate is ok ESTU, (4.6 where ok is the laminate stress at the knee. If90 plies are assumed to be completely ineffective after they fail,the second- ary modulus(slope after the knee)Es of the laminate can be approximated as E= A0 En. (4.7) A Failure of the laminate occurs at the ultimate tensile strain sLu of the 0 plies. Therefore,the laminate failure stress oF is CF =Ok +Es(SLU -STU). (4.8) Unloading of the cross-ply laminate from a stress level oL above the knee follows a path AB (Figure 4.8)and leaves a small residual strain in the laminate.Reloading takes place along the same path until the stress level oL is recovered.If the load is increased further,the slope before unloading is also A OL k BD Strain FIGURE 4.8 Unloading and reloading of a [0/90]s laminate. 2007 by Taylor&Francis Group.LLC
where A0 ¼ net cross-sectional area of the 08 plies A90 ¼ net cross-sectional area of the 908 plies A ¼ A0 þ A90 At the knee, the laminate strain is equal to the ultimate tensile strain «TU of the 908 plies. Therefore, the corresponding stress level in the laminate is sk ¼ E«TU, (4:6) where sk is the laminate stress at the knee. If 908 plies are assumed to be completely ineffective after they fail, the secondary modulus (slope after the knee) Es of the laminate can be approximated as Es ¼ A0 A E11: (4:7) Failure of the laminate occurs at the ultimate tensile strain «LU of the 08 plies. Therefore, the laminate failure stress sF is sF ¼ sk þ Esð Þ «LU «TU : (4:8) Unloading of the cross-ply laminate from a stress level sL above the knee follows a path AB (Figure 4.8) and leaves a small residual strain in the laminate. Reloading takes place along the same path until the stress level sL is recovered. If the load is increased further, the slope before unloading is also A sL sk C B D Stress Strain FIGURE 4.8 Unloading and reloading of a [0=90]S laminate. 2007 by Taylor & Francis Group, LLC.
recovered.Unloading from a higher stress level follows a path CD,which has a smaller slope than AB.The difference in slope between the two unloading paths AB and CD is evidence that the 90 plies fail in a progressive manner.Neglect- ing the small residual strains after unloading,Hahn and Tsai [6]predicted the elastic modulus Ep of the damaged laminate as E ED=1+[(AE/AoE)-11(1-ox/o) (4.9) 4.1.1.4 Multidirectional Laminates Tensile stress-strain curves for laminates containing different fiber orientations in different laminas are in general nonlinear.A few examples are shown in Figure 4.9.For the purposes of analysis,these curves are approximated by a number of linear portions that have different slopes.When these linear portions are extended,a number of knees,similar to that observed in a cross-ply laminate,can be identified.The first knee in these diagrams is called the first ply failure (FPF)point.Many laminates retain a significant load-carrying capacity beyond the FPF point,but for some laminates with high notch sensitivity,failure occurs just after FPF (Table 4.2).Furthermore,cracks appearing at the FPF may increase the possibility of environmental damage (such as moisture pickup)as well as fatigue failure.For all these reasons,the FPF point has special importance in many laminate designs Angle-ply laminates containing [te]layups exhibit two kinds of stress- strain nonlinearity (Figure 4.10).At values of 0 closer to 0,a stiffening effect 140 1104 120 828 [0/±451s 80 552 苏 ssenS 40 [0/±45/90]s 276 [±45/90]s 0 0 0 2 6 Strain(10-5) FIGURE 4.9 Typical tensile stress-strain diagrams for multidirectional laminates. 2007 by Taylor Francis Group,LLC
recovered. Unloading from a higher stress level follows a path CD, which has a smaller slope than AB. The difference in slope between the two unloading paths AB and CD is evidence that the 908 plies fail in a progressive manner. Neglecting the small residual strains after unloading, Hahn and Tsai [6] predicted the elastic modulus ED of the damaged laminate as ED ¼ E 1 þ [(AE=A0E11) 1](1 sk=sL) : (4:9) 4.1.1.4 Multidirectional Laminates Tensile stress–strain curves for laminates containing different fiber orientations in different laminas are in general nonlinear. A few examples are shown in Figure 4.9. For the purposes of analysis, these curves are approximated by a number of linear portions that have different slopes. When these linear portions are extended, a number of knees, similar to that observed in a cross-ply laminate, can be identified. The first knee in these diagrams is called the first ply failure (FPF) point. Many laminates retain a significant load-carrying capacity beyond the FPF point, but for some laminates with high notch sensitivity, failure occurs just after FPF (Table 4.2). Furthermore, cracks appearing at the FPF may increase the possibility of environmental damage (such as moisture pickup) as well as fatigue failure. For all these reasons, the FPF point has special importance in many laminate designs. Angle-ply laminates containing [±u] layups exhibit two kinds of stress– strain nonlinearity (Figure 4.10). At values of u closer to 08, a stiffening effect 140 120 80 40 0 0 2 4 Strain (10−6) [0/±45]S [0/±45/90]S [±45/90]S Stress (MPa) Stress (ksi) 6 1104 828 552 276 0 FIGURE 4.9 Typical tensile stress–strain diagrams for multidirectional laminates. 2007 by Taylor & Francis Group, LLC.