乐 4bh3, (4.11) where m is the initial slope of the load-deflection curve. Three-point flexural tests have received wide acceptance in the composite material industry because the specimen preparation and fixtures are very simple.However,the following limitations of three-point flexural tests should be recognized. 1.The maximum fiber stress may not always occur at the outermost layer in a composite laminate.An example is shown in Figure 4.24.Thus, Equation 4.10 gives only an apparent strength value.For more accurate values,lamination theory should be employed. 2.In the three-point bending mode,both normal stress oxx and shear stress Tx=are present throughout the beam span.If contributions from both stresses are taken into account,the total deflection at the midspan of the beam is PL3 3PL △= 4E6+ 10Gbh normal shear PL3 4Ebh3 (4.12) 90 90 90 0 000 90 0 0 90 90 0 (a) (b) FIGURE 4.24 Normal stress (oxx)distributions in various layers of (a)[90/0/ (90)6/0/90]and (b)[0/90/(0)6/90/0]laminates under flexural loading. 2007 by Taylor Francis Group,LLC
EF ¼ mL3 4bh3 , (4:11) where m is the initial slope of the load–deflection curve. Three-point flexural tests have received wide acceptance in the composite material industry because the specimen preparation and fixtures are very simple. However, the following limitations of three-point flexural tests should be recognized. 1. The maximum fiber stress may not always occur at the outermost layer in a composite laminate. An example is shown in Figure 4.24. Thus, Equation 4.10 gives only an apparent strength value. For more accurate values, lamination theory should be employed. 2. In the three-point bending mode, both normal stress sxx and shear stress txz are present throughout the beam span. If contributions from both stresses are taken into account, the total deflection at the midspan of the beam is D ¼ PL3 4Ebh3 |fflffl{zfflffl} normal þ 3PL 10Gbh |fflffl{zfflffl} shear ¼ PL3 4Ebh3 1 þ 12 10 E G h L 2 " #: (4:12) 90 0 90 90 90 90 90 90 0 90 (a) (b) 0 90 0 0 0 0 0 0 90 0 FIGURE 4.24 Normal stress (sxx) distributions in various layers of (a) [90=0= (90)6=0=90] and (b) [0=90=(0)6=90=0] laminates under flexural loading. 2007 by Taylor & Francis Group, LLC.
This equation shows that the shear deflection can be quite significant in a composite laminate,since the E/G ratio for fiber-reinforced compos- ites is often quite large.The shear deflection can be reduced employing a high span-thickness(L/h)ratio for the beam.Based on data of Zweben et al.[13],L/h ratios of 60:1 are recommended for the determination of flexural modulus. 3.Owing to large deflection at high L/h ratios,significant end forces are developed at the supports.This in turn affects the flexural stresses in a beam.Unless a lower L/h ratio,say 16:1,is used,Equation 4.10 must be corrected for these end forces in the following way: 2bh2 +-份 (4.13) where A is given by Equation 4.12. 4.Although the flexural strength value is based on the maximum tensile stress in the outer fiber,it does not reflect the true tensile strength of the material.The discrepancy arises owing to the difference in stress dis- tributions in flexural and tensile loadings.Flexural loads create a non- uniform stress distribution along the length,but a tensile load creates a uniform stress distribution.Using a two-parameter Weibull distribution for both tensile strength and flexural strength variations,the ratio of the median flexural strength to the median tensile strength can be written as *r% (4.14) OUT where a =shape parameter in the Weibull distribution function (assumed to be equal in both tests) Vr=volume of material stressed in a tension test Ve=volume of material stressed in a three-point flexural test Assuming Vr VF and using typical values of a =15 and 25 for 0 E-glass-epoxy and 0 carbon-epoxy laminates,respectively [12],Equation 4.14 shows that OUF =1.52our for 0 E-glass-epoxy laminates oUF 1.33our for 0 carbon-epoxy laminates 2007 by Taylor Francis Group.LLC
This equation shows that the shear deflection can be quite significant in a composite laminate, since the E=G ratio for fiber-reinforced composites is often quite large. The shear deflection can be reduced employing a high span–thickness (L=h) ratio for the beam. Based on data of Zweben et al. [13], L=h ratios of 60:1 are recommended for the determination of flexural modulus. 3. Owing to large deflection at high L=h ratios, significant end forces are developed at the supports. This in turn affects the flexural stresses in a beam. Unless a lower L=h ratio, say 16:1, is used, Equation 4.10 must be corrected for these end forces in the following way: max ¼ 3PmaxL 2bh2 1 þ 6 D L 2 4 h L D L " # , (4:13) where D is given by Equation 4.12. 4. Although the flexural strength value is based on the maximum tensile stress in the outer fiber, it does not reflect the true tensile strength of the material. The discrepancy arises owing to the difference in stress distributions in flexural and tensile loadings. Flexural loads create a nonuniform stress distribution along the length, but a tensile load creates a uniform stress distribution. Using a two-parameter Weibull distribution for both tensile strength and flexural strength variations, the ratio of the median flexural strength to the median tensile strength can be written as sUF sUT ¼ 2(1 þ a) 2 VT VF 1=a , (4:14) where a ¼ shape parameter in the Weibull distribution function (assumed to be equal in both tests) VT ¼ volume of material stressed in a tension test VF ¼ volume of material stressed in a three-point flexural test Assuming VT ¼ VF and using typical values of a ¼ 15 and 25 for 08 E-glass–epoxy and 08 carbon–epoxy laminates, respectively [12], Equation 4.14 shows that sUF ¼ 1.52sUT for 08 E-glass–epoxy laminates sUF ¼ 1.33sUT for 08 carbon–epoxy laminates 2007 by Taylor & Francis Group, LLC.
800 Three-point flexural test with L/h=11-16 unidirectional(0)laminates 600 T-300 Carbon-epoxy 超 GY-70 Carbon- 400 epoxy E-glass-epoxy peo Kevlar 49-epoxy 200 0.05 0.10 0.15 0.20 0.25 Deflection(in.) FIGURE 4.25 Load-deflection diagrams for various 0 unidirectional laminates in three-point flexural tests. Thus,the three-point flexural strength of a composite laminate can be signifi- cantly higher than its tensile strength.The experimental data presented by Bullock [14]as well as Whitney and Knight [15]verify this observation. Figure 4.25 shows the flexural load-deflection diagrams for four unidirec- tional 0 laminates.The materials of construction are an ultrahigh-modulus carbon(GY-70),a high-strength carbon(T-300),Kevlar 49,and E-glass fiber- reinforced epoxies.The difference in slope in their load-deflection diagrams reflects the difference in their respective fiber modulus.The GY-70 laminate exhibits a brittle behavior,but other laminates exhibit a progressive failure mode consisting of fiber failure,debonding (splitting),and delamination.The Kevlar 49 laminate has a highly nonlinear load-deflection curve due to com- pressive yielding.Fiber microbuckling damages are observed on the compres- sion side of both E-glass and T-300 laminates.Since high contact stresses just under the loading point create such damage,it is recommended that a large loading nose radius be used. The flexural modulus is a critical function of the lamina stacking sequence (Table 4.6),and therefore,it does not always correlate with the tensile modulus, which is less dependent on the stacking sequence.In angle-ply laminates, a bending moment creates both bending and twisting curvatures.Twisting curvature causes the opposite corners of a flexural specimen to lift off its supports.This also influences the measured flexural modulus.The twisting curvature is reduced with an increasing length-width(L/b)ratio and a decreasing degree of orthotropy (i.e.,decreasing E/E22). 2007 by Taylor Francis Group,LLC
Thus, the three-point flexural strength of a composite laminate can be significantly higher than its tensile strength. The experimental data presented by Bullock [14] as well as Whitney and Knight [15] verify this observation. Figure 4.25 shows the flexural load-deflection diagrams for four unidirectional 08 laminates. The materials of construction are an ultrahigh-modulus carbon (GY-70), a high-strength carbon (T-300), Kevlar 49, and E-glass fiberreinforced epoxies. The difference in slope in their load–deflection diagrams reflects the difference in their respective fiber modulus. The GY-70 laminate exhibits a brittle behavior, but other laminates exhibit a progressive failure mode consisting of fiber failure, debonding (splitting), and delamination. The Kevlar 49 laminate has a highly nonlinear load–deflection curve due to compressive yielding. Fiber microbuckling damages are observed on the compression side of both E-glass and T-300 laminates. Since high contact stresses just under the loading point create such damage, it is recommended that a large loading nose radius be used. The flexural modulus is a critical function of the lamina stacking sequence (Table 4.6), and therefore, it does not always correlate with the tensile modulus, which is less dependent on the stacking sequence. In angle-ply laminates, a bending moment creates both bending and twisting curvatures. Twisting curvature causes the opposite corners of a flexural specimen to lift off its supports. This also influences the measured flexural modulus. The twisting curvature is reduced with an increasing length–width (L=b) ratio and a decreasing degree of orthotropy (i.e., decreasing E11=E22). 800 600 400 200 0 0 0.05 0.10 0.15 0.20 0.25 Deflection (in.) Kevlar 49–epoxy E-glass–epoxy T-300 Carbon–epoxy GY-70 Carbon− epoxy Load (lbs) Three-point flexural test with L/h=11–16 unidirectional (0) laminates FIGURE 4.25 Load–deflection diagrams for various 08 unidirectional laminates in three-point flexural tests. 2007 by Taylor & Francis Group, LLC.
TABLE 4.6 Tensile and Flexural Properties of Quasi-Isotropic Laminates Tension Test Flexural Test Laminate Strength, Modulus, Strength, Modulus, Configurationb MPa (ksi) GPa(Msi) MPa(ksi) GPa(Msi) [0/±45/90s 506.4(73.5) 48.23(7) 1219.5(177) 68.9(10) [90/±45/0s 405.8(58.9) 45.47(6.6 141.2(20.5) 18.62.7) [45/0/-45/90s 460.9(66.9) 46.85(6.8) 263.9(38.3) 47.54(6.9) Source:Adapted from Whitney,J.M.,Browning.C.E.,and Mair,A.,Composite Materials:Testing and Design (Third Conference).ASTM STP,546,30,1974. Four-point flexural test with L/h =32 and L/b =4.8. bMaterial:AS carbon fiber-epoxy composite,vr=0.6,eight plies. 4.1.4 IN-PLANE SHEAR PROPERTIES A variety of test methods [16,17]have been used for measuring in-plane shear properties,such as the shear modulus Giz and the ultimate shear strength Tizu of unidirectional fiber-reinforced composites.Three common in-plane shear test methods for measuring these two properties are described as follows. +45 Shear test:The +45 shear test (ASTM D3518)involves uniaxial tensile testing of a [+45/-45]s symmetric laminate (Figure 4.26).The specimen dimensions,preparation,and test procedure are the same as those described in the tension test method ASTM D3039.A diagram of the shear stress T12 vs. the shear strain y12 is plotted using the following equations: T12=20, Y12=Exx -Eyy, (4.15) wherex,and s represent tensile stress,longitudinal strain,and trans- verse strain,respectively,in the [+45ls tensile specimen.A typical tensile stress- tensile strain response of a [+45]s boron-epoxy laminate and the corresponding shear stress-shear strain diagram are shown in Figure 4.27. 10 Off-axis test:The 10 off-axis test [18]involves uniaxial tensile testing of a unidirectional laminate with fibers oriented at 10 from the tensile loading direction(Figure 4.28).The shear stress 712 is calculated from the tensile stress xxusing the following expression: 1 712=i0rsin20l0=10=0.171or (4.16) 2007 by Taylor&Francis Group.LLC
4.1.4 IN-PLANE SHEAR PROPERTIES A variety of test methods [16,17] have been used for measuring in-plane shear properties, such as the shear modulus G12 and the ultimate shear strength t12U of unidirectional fiber-reinforced composites. Three common in-plane shear test methods for measuring these two properties are described as follows. ±45 Shear test: The ±45 shear test (ASTM D3518) involves uniaxial tensile testing of a [þ45=45]nS symmetric laminate (Figure 4.26). The specimen dimensions, preparation, and test procedure are the same as those described in the tension test method ASTM D3039. A diagram of the shear stress t12 vs. the shear strain g12 is plotted using the following equations: t12 ¼ 1 2 sxx, g12 ¼ «xx «yy, (4:15) where sxx, «xx, and «yy represent tensile stress, longitudinal strain, and transverse strain, respectively, in the [±45]nS tensile specimen. A typical tensile stress– tensile strain response of a [±45]S boron–epoxy laminate and the corresponding shear stress–shear strain diagram are shown in Figure 4.27. 108 Off-axis test: The 108 off-axis test [18] involves uniaxial tensile testing of a unidirectional laminate with fibers oriented at 108 from the tensile loading direction (Figure 4.28). The shear stress t12 is calculated from the tensile stress sxx using the following expression: t12 ¼ 1 2 sxx sin 2uj u¼10 ¼ 0:171sxx: (4:16) TABLE 4.6 Tensile and Flexural Properties of Quasi-Isotropic Laminates Tension Test Flexural Testa Laminate Configurationb Strength, MPa (ksi) Modulus, GPa (Msi) Strength, MPa (ksi) Modulus, GPa (Msi) [0=±45=90]S 506.4 (73.5) 48.23 (7) 1219.5 (177) 68.9 (10) [90=±45=0]S 405.8 (58.9) 45.47 (6.6) 141.2 (20.5) 18.6 (2.7) [45=0=45=90]S 460.9 (66.9) 46.85 (6.8) 263.9 (38.3) 47.54 (6.9) Source: Adapted from Whitney, J.M., Browning, C.E., and Mair, A., Composite Materials: Testing and Design (Third Conference), ASTM STP, 546, 30, 1974. a Four-point flexural test with L=h ¼ 32 and L=b ¼ 4.8. b Material: AS carbon fiber–epoxy composite, vf ¼ 0.6, eight plies. 2007 by Taylor & Francis Group, LLC.
25.4→ mm P End 38.5mm tab 45 45 Transverse 178mm strain gage Longitudinal strain gage 38.5mm End tab FIGURE 4.26 Test configuration for a [+45]s shear test. 43.75 [+45]s Tensile test data 35.00 20 夏2625 15 ě17.50 o 8.75 5 0 0 0.0050.0100.0150.0200.025 0 0.010.020.030.040.05 (a) Ex (b) Y12 FIGURE 4.27 (a)Tensile stress-strain diagram for a [+45]s boron-epoxy specimen and (b)the corresponding shear stress-shear strain diagram.(Adapted from the data in Rosen,B.M.,J.Compos.Mater.,6,552,1972.) 2007 by Taylor Francis Group,LLC
25.4 mm 45 45 38.5 mm 38.5 mm End tab End tab P 178 mm Transverse strain gage Longitudinal strain gage P FIGURE 4.26 Test configuration for a [±45]S shear test. 43.75 35.00 [±45]S Tensile test data 26.25 s 17.50 xx (ksi) t12 (ksi) exx 8.75 0 0 0 0.01 0.02 0.03 12 0.04 0.05 (a) (b) 20 15 10 5 0 0.005 0.010 0.015 0.020 0.025 FIGURE 4.27 (a) Tensile stress–strain diagram for a [±45]S boron–epoxy specimen and (b) the corresponding shear stress–shear strain diagram. (Adapted from the data in Rosen, B.M., J. Compos. Mater., 6, 552, 1972.) 2007 by Taylor & Francis Group, LLC.