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《纺织复合材料》课程参考文献(Experimental Characterization of Advanced Composite Materials, Third Edition)09 Lamina Off-Axis Tensile Response

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Lamina Off-Axis Tensile Response The off-axis tension test of unidirectional composites has received consider- able attention by the composites community."Off-axis"here refers to the material axes(1-2)being rotated through an angle 0 with respect to the specimen axis and direction of loading(Figure 9.1).The off-axis specimen is typically 230 mm long and between 12.7 and 25.4 mm wide.A thickness of eight plies is common(0.127 mm ply thickness). The off-axis tension test is rarely used to determine basic ply properties. Most commonly,the purpose of this test is to verify material properties determined in tension,compression,and shear,as discussed in Chapters 5-7, using the transformed constitutive relations discussed in Chapter 2.Testing of specimens at off-axis angles between 10 and 20 produces significant shear in the principal material system.Consequently,the 10 off-axis test has been proposed as a simple way to conduct a shear test [1].The test has been used FIGURE 9.1 Geometry of the off-axis tensile coupon. ©2003 by CRC Press LLC

9 Lamina Off-Axis Tensile Response The off-axis tension test of unidirectional composites has received consider￾able attention by the composites community. “Off-axis” here refers to the material axes (1-2) being rotated through an angle θ with respect to the specimen axis and direction of loading (Figure 9.1). The off-axis specimen is typically 230 mm long and between 12.7 and 25.4 mm wide. A thickness of eight plies is common (0.127 mm ply thickness). The off-axis tension test is rarely used to determine basic ply properties. Most commonly, the purpose of this test is to verify material properties determined in tension, compression, and shear, as discussed in Chapters 5–7, using the transformed constitutive relations discussed in Chapter 2. Testing of specimens at off-axis angles between 10 and 20° produces significant shear in the principal material system. Consequently, the 10° off-axis test has been proposed as a simple way to conduct a shear test [1]. The test has been used FIGURE 9.1 Geometry of the off-axis tensile coupon. w LT LT LG L I x y 2 θ TX001_ch09_Frame Page 131 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

also to verify biaxial strength criteria because,as will be discussed,uniaxial loading will lead to a combined state of stress in the principal material system. 9.1 Deformation and Stress in an Unconstrained Specimen Because of the off-axis configuration of the specimen,the in-plane response is characterized by a fully populated compliance matrix,as shown in Equation (2.16) Ex 52 516 Ox Ey S12 y (9.1) 56 526 where the x-y system is defined in Figure 9.1,and expressions for the trans- formed compliance elements Su are given in Appendix A. For an ideal,uniformly stressed off-axis tensile coupon,the only stress acting is ox,(oy=txy=0),and Equations (9.1)give the state of strain in the specimen, Ex Ey S (9.2) S.c] Consequently,the off-axis coupon subjected to a uniform uniaxial state of stress thus exhibits shear strain()in addition to the axial and transverse strains(Ex and )(Figure 9.2). A set of material properties may be evaluated based on measurement of axial stress (and axial,transverse,and shear strains (x yYy).It is customary to determine the axial Young's modulus(E)and Poisson's ratio (Vxy)of the off-axis specimen (9.3) Ex Vxy= y (9.4) Ex ©2003 by CRC Press LLC

also to verify biaxial strength criteria because, as will be discussed, uniaxial loading will lead to a combined state of stress in the principal material system. 9.1 Deformation and Stress in an Unconstrained Specimen Because of the off-axis configuration of the specimen, the in-plane response is characterized by a fully populated compliance matrix, as shown in Equation (2.16) (9.1) where the x-y system is defined in Figure 9.1, and expressions for the trans￾formed compliance elements Sij are given in Appendix A. For an ideal, uniformly stressed off-axis tensile coupon, the only stress acting is σx, (σy = τxy = 0), and Equations (9.1) give the state of strain in the specimen, (9.2) Consequently, the off-axis coupon subjected to a uniform uniaxial state of stress thus exhibits shear strain (γxy) in addition to the axial and transverse strains (εx and εy) (Figure 9.2). A set of material properties may be evaluated based on measurement of axial stress (σx) and axial, transverse, and shear strains (εx, εy, γxy). It is customary to determine the axial Young’s modulus (Ex) and Poisson’s ratio (νxy) of the off-axis specimen (9.3) (9.4) x y xy x y xy SSS SSS SSS ε ε γ σ σ τ               =                           11 12 16 12 22 26 16 26 66 x y xy x S S S ε ε γ σ               =               11 12 16 Ex x x = σ ε ν ε ε xy y x = − TX001_ch09_Frame Page 132 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

In addition,a ratio (n),which quantifies coupling between shear and axial strains,is defined according to (9.5) Ex The off-axis tension test may also be used to determine the in-plane shear modulus,G2,in the principal material coordinate system.This property is, according to Equation(2.9),defined by (9.6) Y12 Consequently,determination of G2 requires determination of shear stress and strain in the principal material coordinate system.Equations(2.12)and (2.14)yield T12=-7110x (9.7) where m=cos 0 and n=sin 0.The shear strain is obtained from Equations(2.14) Y2 2mn(y -E)+(m2-n2)Yy (9.8) where the strain(e)is directly measured,and the transverse strain ()and shear strain()are determined as subsequently explained. The properties EVyny and Gi2 can be evaluated from test data using procedures detailed later in this chapter.The mechanical properties so deter- mined can be compared to theoretical values calculated from the compliance relations,Equations (9.2),and the definitions in Equations (9.3-9.5), 1 (9.9a) Vxy=- 52 (9.9b) (9.9c) If the principal (basic)material properties(E,E2,Vi2,and G12)are known from previous tests (Chapters 5-7),it is possible to calculate the off-axis properties E,Vsy and nxy using Equations(A.1)(Appendix A),and compare those to the experimentally determined values.G12 may be compared to the modulus measured in the off-axis test(Equation (9.6)). ©2003 by CRC Press LLC

In addition, a ratio (ηxy), which quantifies coupling between shear and axial strains, is defined according to (9.5) The off-axis tension test may also be used to determine the in-plane shear modulus, G12, in the principal material coordinate system. This property is, according to Equation (2.9), defined by (9.6) Consequently, determination of G12 requires determination of shear stress and strain in the principal material coordinate system. Equations (2.12) and (2.14) yield τ12 = –mnσx (9.7) where m = cos θ and n = sin θ. The shear strain is obtained from Equations (2.14) γ12 = 2mn(εy – εx) + (m2 – n2)γxy (9.8) where the strain (εx) is directly measured, and the transverse strain (εy) and shear strain (γxy) are determined as subsequently explained. The properties Ex, νxy, ηxy, and G12 can be evaluated from test data using procedures detailed later in this chapter. The mechanical properties so deter￾mined can be compared to theoretical values calculated from the compliance relations, Equations (9.2), and the definitions in Equations (9.3–9.5), (9.9a) (9.9b) (9.9c) If the principal (basic) material properties (E1, E2, ν12, and G12) are known from previous tests (Chapters 5–7), it is possible to calculate the off-axis properties Ex, νxy, and ηxy using Equations (A.1) (Appendix A), and compare those to the experimentally determined values. G12 may be compared to the modulus measured in the off-axis test (Equation (9.6)). η γ ε xy xy x = G12 12 12 = τ γ E S x = 1 11 νxy S S = − 12 11 ηxy S S = 16 11 TX001_ch09_Frame Page 133 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

Unconstrained Ends P M Constrained Ends FIGURE 9.3 FIGURE 9.2 Influence of gripped end regions on deformation Off-axis coupon under uniform axial stress. of off-axis specimen [2]. 9.2 Influence of End Constraint As first pointed out by Halpin and Pagano [2],most test machines used in testing laboratories employ rigid grips that constrain the shear deformation illustrated in Figure 9.2.As a result,the specimen assumes a shape schematically illustrated in Figure 9.3 [2].To quantify the influence of gripping on the response of an off-axis tension specimen,Halpin and Pagano [2]performed a stress analysis of a constrained coupon and obtained the following expressions for the shear strain and longitudinal strain at the specimen centerline. Yxy=S1eC2-S66Cow2/4 (9.10a) ex=5C2-51Cow2/4 (9.10b) with 125160 C=3w25,56-5+25,民 (9.11a) C。-(356w2+52) C2=1256 (9.11b) where eo =AL/LG (elongation/gage length)and w and LG are specimen width and gage length,respectively(Figure 9.1). ©2003 by CRC Press LLC

9.2 Influence of End Constraint As first pointed out by Halpin and Pagano [2], most test machines used in testing laboratories employ rigid grips that constrain the shear deformation illustrated in Figure 9.2.As a result, the specimen assumes a shape schematically illustrated in Figure 9.3 [2]. To quantify the influence of gripping on the response of an off-axis tension specimen, Halpin and Pagano [2] performed a stress analysis of a constrained coupon and obtained the following expressions for the shear strain and longitudinal strain at the specimen centerline. (9.10a) (9.10b) with (9.11a) (9.11b) where ε0 = ∆L/LG (elongation/gage length) and w and LG are specimen width and gage length, respectively (Figure 9.1). FIGURE 9.2 Off-axis coupon under uniform axial stress. FIGURE 9.3 Influence of gripped end regions on deformation of off-axis specimen [2]. 2 y x σ x x σ w LG P P M V Constrained Ends Unconstrained Ends 1 γ xy = − S C S Cw 16 2 66 0 2 4 εx = − S C S Cw 11 2 16 0 2 4 C S w SS S SLG 0 16 0 2 11 66 16 2 11 2 12 3 2 = ( ) − + ε C C S 2 Sw SLG 0 16 66 2 11 2 12 = + ( ) 3 TX001_ch09_Frame Page 134 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

On the basis of this analysis,it is possible to derive an expression for the apparent axial Young's modulus including end constraint (Ex)a=x/Ex (9.12) where o,and are the stress and strain at the centerline of the constrained off-axis coupon.(E)a may be expressed as Ex (但)。=1 (9.13) in which Ex is the modulus for an unconstrained off-axis specimen.The parameterξis given by 35。 (9.14) Su 356+25.L6/w7 Examination of the above equations reveals that0and(E)E=1/Su when LG/w→o. Similarly,Pindera and Herakovich [3]derived an expression for the apparent Poisson's ratio,(Vxy)a 1- B ). (9.15) 1- 3 2 52 where B is given by [3], (9.16) The apparent shear coupling ratio of the specimen subjected to end con- straint is (nxy)a =Ysy/Ex (9.17) 2003 by CRC Press LLC

On the basis of this analysis, it is possible to derive an expression for the apparent axial Young’s modulus including end constraint (Ex)a = σx/εx (9.12) where σx and εx are the stress and strain at the centerline of the constrained off-axis coupon. (Ex)a may be expressed as (9.13) in which Ex is the modulus for an unconstrained off-axis specimen. The parameter ξ is given by (9.14) Examination of the above equations reveals that ξ → 0 and (Ex)a → when LG/w → ∞. Similarly, Pindera and Herakovich [3] derived an expression for the apparent Poisson’s ratio, (νxy)a (9.15) where β is given by [3], (9.16) The apparent shear coupling ratio of the specimen subjected to end con￾straint is (ηxy)a = γxy/εx (9.17) E E x a x ( ) = 1− ξ ξ = + ( )         1 S S 11 3S 2S L w G 16 2 66 11 2 3 E S x = 1 11 ν ν β β xy a xy S S S S ( ) = −       −       1 3 2 1 3 2 26 11 16 12 β =             +             w L S S w L S S G G 2 16 11 2 66 11 1 3 2 TX001_ch09_Frame Page 135 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

Substitution of Equations(9.10)and (9.11)into (9.17)yields (9.18) Note that when the length-to-width ratio,,Lc/w→o,∞(qy)a→Si6/Si,as given by Equation (9.9c). Pindera and Herakovich [3]examined the influence of end constraint on the evaluation of shear modulus,G2,from the off-axis tension specimen using the elasticity solution of Halpin and Pagano[2]and found that the procedure outlined in Section 9.1 leads to error in G2.The main source of error is the neglect of the shear stress txy in Equation (9.7).The proper transformation is [3] t2=-mnox +(m2-n2)txy (9.19) This equation,combined with the definition of G2(Equation(9.6)),yields an expression for the correct value of the shear modulus in terms of the apparent modulus(G2),evaluated using the procedure in Section 9.1 [3], 3m2-n2) 1+ Ga=(G2.1-B56/5 2mn (9.20) where m cos0,n sine,and B is defined in Equation (9.16).As (w/Lc) e∞,B→0,and the apparent shear modulus approaches Gi2: The above expressions for apparent off-axis properties(E)a,(vxy)a,and (nx)a may be used to correct measured values,or to design the off-axis specimen for minimum error resulting from end constraint.An obvious way to reduce the error is to use specimens with large aspect ratios,Lc/w.As mentioned early in this chapter,specimens are typically 230 mm long and between 12.7 and 25.4 mm wide.For 38-mm-long tabs at the ends,this corresponds to aspect ratios between 6 and 12.For a carbon/polyimide composite specimen with an aspect ratio of 10 and 10 off-axis angle,Pindera and Herakovich [3]found an error in E,of about 2 to 4%.The error in shear coupling ratio is larger,as will be discussed later. The error in shear modulus G2 for a 10 off-axis carbon/polyimide speci- men at LG/w=10 is approximately 12 to 15%[3].For proper determination of G12,Pindera and Herakovich [3]recommend use of coupons with an aspect ratio of 10 or more and an off-axis angle of 45. 2003 by CRC Press LLC

Substitution of Equations (9.10) and (9.11) into (9.17) yields (9.18) Note that when the length-to-width ratio, LG/w → ∞, (ηxy)a → , as given by Equation (9.9c). Pindera and Herakovich [3] examined the influence of end constraint on the evaluation of shear modulus, G12, from the off-axis tension specimen using the elasticity solution of Halpin and Pagano [2] and found that the procedure outlined in Section 9.1 leads to error in G12. The main source of error is the neglect of the shear stress τxy in Equation (9.7). The proper transformation is [3] τ12 = –mnσx + (m2 – n2)τxy (9.19) This equation, combined with the definition of G12 (Equation (9.6)), yields an expression for the correct value of the shear modulus in terms of the apparent modulus (G12)a, evaluated using the procedure in Section 9.1 [3], (9.20) where m = cosθ, n = sinθ, and β is defined in Equation (9.16). As (w/LG) → ∞, β → 0, and the apparent shear modulus approaches G12. The above expressions for apparent off-axis properties (Ex)a, (νxy)a, and (ηxy)a may be used to correct measured values, or to design the off-axis specimen for minimum error resulting from end constraint. An obvious way to reduce the error is to use specimens with large aspect ratios, LG/w. As mentioned early in this chapter, specimens are typically 230 mm long and between 12.7 and 25.4 mm wide. For 38-mm-long tabs at the ends, this corresponds to aspect ratios between 6 and 12. For a carbon/polyimide composite specimen with an aspect ratio of 10 and 10° off-axis angle, Pindera and Herakovich [3] found an error in Ex of about 2 to 4%. The error in shear coupling ratio is larger, as will be discussed later. The error in shear modulus G12 for a 10° off-axis carbon/polyimide speci￾men at LG/w = 10 is approximately 12 to 15% [3]. For proper determination of G12, Pindera and Herakovich [3] recommend use of coupons with an aspect ratio of 10 or more and an off-axis angle of 45°. ηxy a 1 2 G 2 S S 1 3 2 w L S S S S ( ) = +       −                       − 16 11 66 11 16 11 S S 16 11 G G m n mn S S a 12 12 2 2 16 11 1 3 2 1 = ( ) + ( ) − − ( ) β β / TX001_ch09_Frame Page 136 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

Specimen (x-y) Material (1-2) 2t1(<0j 6 FIGURE 9.4 State of stress in the specimen for an off-axis tension test. 9.3 Off-Axis Tensile Strength As mentioned earlier in this chapter,the off-axis tension test has been used to examine theories proposed for prediction failure of composites under com- bined stress.In such studies,slender specimens are used for strength mea- surements to avoid the complications of end constraint effects discussed above. This leads to a stress state in the on-axis system,as given by Equations(2.12) and(2.14: 01 m2 02 =6 n2 (9.21) T12 -mn Therefore,as illustrated in Figure 9.4,the state of stress in the principal material coordinate system is biaxial. Experimental studies conducted on on-axis and off-axis specimens,e.g., in References [4,5],show that the off-axis specimen under tension fails along planes parallel to the fibers except for zero and very small angles,where failure involves fiber fractures.To predict the failure stress of the off-axis tension specimen,the on-axis stresses given by Equations(9.21)are substi- tuted into the failure criterion of choice(see Section 2.5).The maximum stress and strain criteria(see Section 2.5),yield three equations for the ultimate stress,and the appropriate strength is identified by the least of the three values.Substitution of the stresses given by Equations(9.21)into the Tsai-Wu criterion,Equation(2.44),yields a quadratic equation in o,of the type Ao)+Bo-1=0 (9.22) ©2003 by CRC Press LLC

9.3 Off-Axis Tensile Strength As mentioned earlier in this chapter, the off-axis tension test has been used to examine theories proposed for prediction failure of composites under com￾bined stress. In such studies, slender specimens are used for strength mea￾surements to avoid the complications of end constraint effects discussed above. This leads to a stress state in the on-axis system, as given by Equations (2.12) and (2.14): (9.21) Therefore, as illustrated in Figure 9.4, the state of stress in the principal material coordinate system is biaxial. Experimental studies conducted on on-axis and off-axis specimens, e.g., in References [4,5], show that the off-axis specimen under tension fails along planes parallel to the fibers except for zero and very small angles, where failure involves fiber fractures. To predict the failure stress of the off-axis tension specimen, the on-axis stresses given by Equations (9.21) are substi￾tuted into the failure criterion of choice (see Section 2.5). The maximum stress and strain criteria (see Section 2.5), yield three equations for the ultimate stress, , and the appropriate strength is identified by the least of the three values. Substitution of the stresses given by Equations (9.21) into the Tsai-Wu criterion, Equation (2.44), yields a quadratic equation in , of the type (9.22) FIGURE 9.4 State of stress in the specimen for an off-axis tension test. 1 2 12 x 2 2 m n mn σ σ τ σ             = −             σx ult σx ult A B x ult x ult ( ) σ σ + −= 2 1 0 TX001_ch09_Frame Page 137 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

The solution of Equation(9.22)yields two roots,the positive associated with the tensile strength and the negative associated with the compressive strength of the off-axis specimen. As mentioned early in this chapter,the 10 off-axis tension test has been proposed for measuring the in-plane shear strength(S)of unidirectional composites [1].However,because failure occurs under the influence of nor- mal stresses o,and o,(Figure 9.4),which separate the specimen in two pieces, this test is not recommended for the generation of shear strength [6]. 9.4 Test Procedure 1.Prepare off-axis tension coupons from a unidirectional,six-to eight-ply-thick panel.The specimens should be about 230 mm long and between 12.5 and 25 mm wide.Select at least three different off-axis angles,e.g.,15,30,and 60.Use the same tolerances as for the tension specimen discussed in Chapter 5,and bond end tabs as described in Chapter 4. 2.The off-axis test specimen is instrumented with a three-element strain gage rosette with one of the elements aligned with the coupon axis(x-direction in Figure 9.1),one element at 45,and one element at45°. 3.Measure the specimen cross-sectional dimensions (average six measurements). 4.Mount the specimen in a properly aligned and calibrated test frame.Set the crosshead rate at about 0.5 to 1 mm/min. 5.Monitor the load-strain response of the specimen (all three elements). Take strain readings at small load intervals to collect at least 25 data points in the linear region.Load the specimen to failure. 9.5 Data Reduction 9.5.1 Elastic Properties Axial modulus,Poisson's ratio and shear coupling ratio may be determined from measured stress-strain data according to Equations(9.3)-(9.5).The axial strain,ex,is obtained directly from the axially oriented strain gage.Trans- verse strain and shear strain,y and Yy,are obtained from the +45 gages using Equations(2.13). ©2003 by CRC Press LLC

The solution of Equation (9.22) yields two roots, the positive associated with the tensile strength and the negative associated with the compressive strength of the off-axis specimen. As mentioned early in this chapter, the 10° off-axis tension test has been proposed for measuring the in-plane shear strength (S6) of unidirectional composites [1]. However, because failure occurs under the influence of nor￾mal stresses σ1 and σ2 (Figure 9.4), which separate the specimen in two pieces, this test is not recommended for the generation of shear strength [6]. 9.4 Test Procedure 1. Prepare off-axis tension coupons from a unidirectional, six- to eight-ply-thick panel. The specimens should be about 230 mm long and between 12.5 and 25 mm wide. Select at least three different off-axis angles, e.g., 15, 30, and 60°. Use the same tolerances as for the tension specimen discussed in Chapter 5, and bond end tabs as described in Chapter 4. 2. The off-axis test specimen is instrumented with a three-element strain gage rosette with one of the elements aligned with the coupon axis (x-direction in Figure 9.1), one element at 45°, and one element at –45°. 3. Measure the specimen cross-sectional dimensions (average six measurements). 4. Mount the specimen in a properly aligned and calibrated test frame. Set the crosshead rate at about 0.5 to 1 mm/min. 5. Monitor the load-strain response of the specimen (all three elements). Take strain readings at small load intervals to collect at least 25 data points in the linear region. Load the specimen to failure. 9.5 Data Reduction 9.5.1 Elastic Properties Axial modulus, Poisson’s ratio and shear coupling ratio may be determined from measured stress-strain data according to Equations (9.3)-(9.5). The axial strain, εx , is obtained directly from the axially oriented strain gage. Trans￾verse strain and shear strain, εy and γxy, are obtained from the ±45° gages using Equations (2.13). TX001_ch09_Frame Page 138 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

0.75 Carbon/Epoxy.[301s ny=-1.18 e/w=6.0 0.50 000..0.0 0.25 00-0000000 0 0.10.20.30.4 0.5 Axial Strain ()deg- FIGURE 9.5 Shear strain vs.axial strain for a [30l carbon/epoxy composite. e(45)=(ex+E,+Yy)/2 (9.23a) -45)=(e+8y-Yy)/2 (9.23b) Combining Equations (9.23)yields 8,=e(45)+c(-45)-e (9.24) Yy=e(45)+e(-45) (9.25) Young's modulus,E is determined from the initial slope of the curve ox vs.E Poisson's ratio,v is obtained by plotting the negative of the strain E vs. and determining the slope of the line.The shear coupling ratio,nxy is deter- mined by plotting shear strain vs.axial strain,as shown in Figure 9.5 for a 30 off-axis carbon/epoxy specimen.Note that the experimentally deter- mined properties are apparent because they may be influenced by the con- straints imposed by the grips. Figures 9.6-9.8 show experimentally obtained off-axis modulus,Poisson's ratio,and shear coupling ratio vs.off-axis angle for carbon/fiber compos- ites.Shown in these graphs are reduced data(apparent),theoretical curves calculated assuming the ends of the specimen are free to rotate(Equations (9.9)),and theoretical curves calculated using a correction for end constraint according to Section 9.2.It is observed that the apparent modulus and Poisson's ratio are larger than the unconstrained value,whereas the magni- tude of the apparent shear coupling ratio is reduced because of end constraints.The expressions in Section 9.2 incorporating end constraints due to finite aspect ratio bring the analytical results in close agreement with measured data. The apparent shear modulus,G2,determined for a carbon/polyimide com- posite using Equations (9.6)-(9.8)and corrected for shear stress and end constraint using Equation (9.20),is shown vs.off-axis angle in Figure 9.9. At off-axis angles up to 30 the end constraint will increase the apparent 2003 by CRC Press LLC

ε(45°) = (εx + εy + γxy)/2 (9.23a) ε(–45°) = (εx + εy – γxy)/2 (9.23b) Combining Equations (9.23) yields εy = ε(45°) + ε(–45°) –εx (9.24) γxy = ε(45°) + ε(–45°) (9.25) Young’s modulus, Ex, is determined from the initial slope of the curve σx vs. εx. Poisson’s ratio, νxy, is obtained by plotting the negative of the strain εy vs. εx and determining the slope of the line. The shear coupling ratio, ηxy, is deter￾mined by plotting shear strain vs. axial strain, as shown in Figure 9.5 for a 30° off-axis carbon/epoxy specimen. Note that the experimentally deter￾mined properties are apparent because they may be influenced by the con￾straints imposed by the grips. Figures 9.6–9.8 show experimentally obtained off-axis modulus, Poisson’s ratio, and shear coupling ratio vs. off-axis angle for carbon/fiber compos￾ites. Shown in these graphs are reduced data (apparent), theoretical curves calculated assuming the ends of the specimen are free to rotate (Equations (9.9)), and theoretical curves calculated using a correction for end constraint according to Section 9.2. It is observed that the apparent modulus and Poisson’s ratio are larger than the unconstrained value, whereas the magni￾tude of the apparent shear coupling ratio is reduced because of end constraints. The expressions in Section 9.2 incorporating end constraints due to finite aspect ratio bring the analytical results in close agreement with measured data. The apparent shear modulus, G12, determined for a carbon/polyimide com￾posite using Equations (9.6) – (9.8) and corrected for shear stress and end constraint using Equation (9.20), is shown vs. off-axis angle in Figure 9.9. At off-axis angles up to 30° the end constraint will increase the apparent FIGURE 9.5 Shear strain vs. axial strain for a [30]6 carbon/epoxy composite. TX001_ch09_Frame Page 139 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

140 Carbon/Epoxy,[ 120 w=6 △=Measured Volues 100 一=Theorelical [L/w=o ---Theoretical [Lo/w =6) 80 60 20 0 153045 6075 90 Off-Axis Angle(0),deg. FIGURE 9.6 Axial modulus vs.off-axis angle for a carbon/epoxy composite. Carbon/Polyimide,12 0.6 O =Measured values =Theoretical [Lg/w=co) =Theoretical (L/w=10) 0.4 0.2 0 15 3045607590 Off-Axis Angle (0),deg. FIGURE 9.7 Poisson's ratio vs.off-axis angle for a carbon/polyimide composite. -2.4 Carbon/Epox.I日l& -2.0 e/w=6 16 ---Theoretical (L/w=6) -1.2 0.8 -0.4 0 102030.405060708090 Off-Axis Angle 0,deg. FIGURE 9.8 Shear coupling ratio vs.off-axis angle for a carbon/epoxy composite. modulus.Correction of the apparent shear modulus for end constraint brings the value closer to the asymptotic value.This graph emphasizes the need to use off-axis specimens of angles 45 or greater when evaluating the shear modulus. ©2003 by CRC Press LLC

modulus. Correction of the apparent shear modulus for end constraint brings the value closer to the asymptotic value. This graph emphasizes the need to use off-axis specimens of angles 45° or greater when evaluating the shear modulus. FIGURE 9.6 Axial modulus vs. off-axis angle for a carbon/epoxy composite. FIGURE 9.7 Poisson’s ratio vs. off-axis angle for a carbon/polyimide composite. FIGURE 9.8 Shear coupling ratio vs. off-axis angle for a carbon/epoxy composite. TX001_ch09_Frame Page 140 Saturday, September 21, 2002 5:01 AM © 2003 by CRC Press LLC

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