416 FAILURE CRITERIA the two preceding equations (Egs.10.7 and 10.8),give Fi4=F24=F34=F45=F46=0, (10.9) By applying the procedure just given to the other two symmetry planes we find that the following strength parameters are also zero: Fi5=F5=F35=F45=F56=0 (10.10) F6=F26=F36=F46=F6=0. For an orthotropic material the quadratic failure criterion(Eq.10.2)becomes FO1+F2o2+F303+ F110+F2202+F3303+F44+Fssti3+F66ti2+ 2(F20102+f13O103+f30203)<1. (10.11) At failure,where the stress components are designated by the superscript f, Eq.(10.11)is +F1+时+F1(o02+F2(o)2+ F3(o)2+F4()2+Fss()2+F6()2+ 2(F12ai02+Fi3aio3+F230203)=1. (10.12) Noninteraction strength parameters.The noninteraction strength parameters are denoted by F1,F2,F3,F11,F22,F33,F44,Fss,F66.The values of these parameters are obtained from uniaxial and from shear tests. To obtain Fi and Fu we subject the material to uniaxial tension and compres- sion in the xi orthotropy direction(Table 10.1,top left and middle left).At fail- ure,the stresses are of =st and of=-s,wheres is the strength of the material and the superscripts (+)and (-)refer to tension and compression.With these stresses,the quadratic failure criterion gives(Eq.10.12) Fist+Fu(s )2=1 -Fis+Fu(s )2=1. (10.13) Solution of these two equations yields 5s1、1 1 F1= (10.14) stst The strength parameters F2,F3,F22,F33 are obtained in a similar manner and are given in Table 10.2.The tests to determine s?,s2,s3,s3 are illustrated in Table 10.1. To obtain the strength parameter F44 we subject the material to shear t23 in the x2-x3 orthotropy plane (Table 10.1,bottom left).Because of material symmetry the failure stress is independent of the direction of the shear stress,and at failure =523(=s=s3).With this stress,Eq.(10.12)gives F44(S23)2=1, (10.15)
416 FAILURE CRITERIA the two preceding equations (Eqs. 10.7 and 10.8), give F14 = F24 = F34 = F45 = F46 = 0. (10.9) By applying the procedure just given to the other two symmetry planes we find that the following strength parameters are also zero: F15 = F25 = F35 = F45 = F56 = 0 F16 = F26 = F36 = F46 = F56 = 0. (10.10) For an orthotropic material the quadratic failure criterion (Eq. 10.2) becomes F1σ1 + F2σ2 + F3σ3+ F11σ2 1 + F22σ2 2 + F33σ2 3 + F44τ 2 23 + F55τ 2 13 + F66τ 2 12+ 2(F12σ1σ2 + F13σ1σ3 + F23σ2σ3) < 1. (10.11) At failure, where the stress components are designated by the superscript f, Eq. (10.11) is F1σf 1 + F2σf 2 + F3σf 3 + F11 σf 1 2 + F22 σf 2 2 + F33 σf 3 2 + F44 τ f 232 + F55 τ f 132 + F66 τ f 122 + 2 F12σf 1σf 2 + F13σf 1σf 3 + F23σf 2σf 3 = 1. (10.12) Noninteraction strength parameters. The noninteraction strength parameters are denoted by F1, F2, F3, F11, F22, F33, F44, F55, F66. The values of these parameters are obtained from uniaxial and from shear tests. To obtain F1 and F11 we subject the material to uniaxial tension and compression in the x1 orthotropy direction (Table 10.1, top left and middle left). At failure, the stresses are σf 1 = s+ 1 and σf 1 = −s− 1 , where s is the strength of the material and the superscripts (+) and (−) refer to tension and compression. With these stresses, the quadratic failure criterion gives (Eq. 10.12) F1s+ 1 + F11(s+ 1 ) 2 = 1 − F1s− 1 + F11(s− 1 ) 2 = 1. (10.13) Solution of these two equations yields F1 = 1 s+ 1 − 1 s− 1 F11 = 1 s+ 1 s− 1 . (10.14) The strength parameters F2, F3, F22, F33 are obtained in a similar manner and are given in Table 10.2. The tests to determine s+ 2 ,s− 2 , s+ 3 ,s− 3 are illustrated in Table 10.1. To obtain the strength parameter F44 we subject the material to shear τ23 in the x2–x3 orthotropy plane (Table 10.1, bottom left). Because of material symmetry the failure stress is independent of the direction of the shear stress, and at failure τ f 23 = s23(= s+ 23 = s− 23). With this stress, Eq. (10.12) gives F44 (s23) 2 = 1, (10.15)
10.1 QUADRATIC FAILURE CRITERION 417 Table 10.1.Tests to determine the strengths of orthotropic materials;,2, 3 are perpendicular to the planes of orthotropy. 8 3个 1 82 83 01 s15 贴 which results in 1 F44= 23F (10.16) The strength parameters Fss,F66 are obtained in a similar manner.These,as well as the other noninteraction strength parameters,are given in Table 10.2. Interaction strength parameters.The interaction strength parameters F12,F3, and F23 can be determined from tests that result in two or more nonzero stress components inside the material.Off-axis uniaxial tests and biaxial tests offer pos- sible means for determining the interaction strength parameters. Off-axis uniaxial tests.When the interaction strength parameters of orthotropic materials are to be determined by off-axis tests,we take test coupons from each (x1-x2,x2-x3,and x1-x3)orthotropy plane (Fig.10.6). Table 10.2.The noninteraction strength parameters in terms of strengths F=容 Fn F=密 F4= 1 （2sP s=高 f6二m子
10.1 QUADRATIC FAILURE CRITERION 417 Table 10.1. Tests to determine the strengths of orthotropic materials; σ1, σ2, σ3 are perpendicular to the planes of orthotropy. σ1 σ1 + 1 s σ2 σ2 + 2 s σ3 σ3 + 3 s – σ1 – σ1 − 1s – σ2 – σ2 − 2 s – σ3 – σ3 − 3 s τ23 + 23 s τ13 + 13 s τ12 + 12 s which results in F44 = 1 (s23) 2 . (10.16) The strength parameters F55, F66 are obtained in a similar manner. These, as well as the other noninteraction strength parameters, are given in Table 10.2. Interaction strength parameters. The interaction strength parameters F12, F13, and F23 can be determined from tests that result in two or more nonzero stress components inside the material. Off-axis uniaxial tests and biaxial tests offer possible means for determining the interaction strength parameters. Off-axis uniaxial tests. When the interaction strength parameters of orthotropic materials are to be determined by off-axis tests, we take test coupons from each (x1–x2, x2–x3, and x1–x3) orthotropy plane (Fig. 10.6). Table 10.2. The noninteraction strength parameters in terms of strengths F1 = 1 s+ 1 − 1 s− 1 F2 = 1 s+ 2 − 1 s− 2 F3 = 1 s+ 3 − 1 s− 3 F11 = 1 s+ 1 s− 1 F22 = 1 s+ 2 s− 2 F33 = 1 s+ 3 s− 3 F44 = 1 (s23) 2 F55 = 1 (s13) 2 F66 = 1 (s12) 2
418 FAILURE CRITERIA Coupon 3 Coupon 2 Coupon 1 Figure 10.6:Test coupons in the xi-x2,xi-x3.x2-t3 orthotropy planes. Coupon 1,taken from the xi-x2 orthotropy plane,is subjected either to an axial tensile or to an axial compressive stress.At failure the stress is designated by p.Superscript 1 indicates Coupon 1.The corresponding stresses in thex,x2 coordinate system are (Eg.2.182) of pl cos21 of=pl sin21 1t9=pcos⊙1sin⊙1.(10.17) The angle is shown in Figure 10.6.Substitution of the preceding stresses into Eq.(10.12)results in the expression p(Ecos2⊙1+Fsin2⊙1)+ (pl)2(Fi1cos4⊙1+fF2sin4⊙1+F66cos2⊙1sin2Θ1)+ (p)22F12 cos2 1sin2 1=1. (10.18) This equation can be solved for F12.The result is given in Table 10.3.The Fi3, F23 interaction strength parameters are obtained in a similar manner and are also included in Table 10.3. Biaxial tests.When the interaction strength parameter Fi2 is to be determined from biaxial tests,the specimen is loaded in biaxial tension,resulting in stresses o and o2(Fig.10.7).The load is then increased proportionally such that the ratio of the two stresses remains constant.At failure the stresses are denoted by 01=g1-2 2=31-2 (10.19) Table 10.3.The interaction strength parameters obtained from uniaxial tests (orthotropic material).The angles⊙1，⊙2，ande3 are shown in Figure10.6. F12=2sin2 @1 co62 01 1 (pa o2o+5m2o-cos01-2sinO4）-警 F23= 1 2sin2 82 cos2 e2 (p2 o2o+5m2@-F2cos4o2-Fasin2）-婴 F13= 1 1 2sin2 @3 c0s203 (pB)2 FacoinF3 cos 3-Fn sin 3
418 FAILURE CRITERIA x1 x3 x2 Θ1 Θ3 Θ2 Coupon 3 Coupon 2 Coupon 1 x3 x2 x1 Figure 10.6: Test coupons in the x1–x2, x1–x3, x2–x3 orthotropy planes. Coupon 1, taken from the x1–x2 orthotropy plane, is subjected either to an axial tensile or to an axial compressive stress. At failure the stress is designated by pf1. Superscript 1 indicates Coupon 1. The corresponding stresses in the x1, x2 coordinate system are (Eq. 2.182) σf1 1 = pf1 cos2 1 σf1 2 = pf1 sin2 1 τ f1 12 = pf1 cos 1 sin 1. (10.17) The angle 1 is shown in Figure 10.6. Substitution of the preceding stresses into Eq. (10.12) results in the expression pf1(F1 cos2 1 + F2 sin2 1) + (pf1) 2 (F11 cos4 1 + F22 sin4 1 + F66 cos2 1 sin2 1) + (pf1) 2 2F12 cos2 1 sin2 1 = 1. (10.18) This equation can be solved for F12. The result is given in Table 10.3. The F13, F23 interaction strength parameters are obtained in a similar manner and are also included in Table 10.3. Biaxial tests. When the interaction strength parameter F12 is to be determined from biaxial tests, the specimen is loaded in biaxial tension, resulting in stresses σ1 and σ2 (Fig. 10.7). The load is then increased proportionally such that the ratio of the two stresses remains constant. At failure the stresses are denoted by σ1 = σf(1–2) 1 σ2 = σf(1–2) 2 . (10.19) Table 10.3. The interaction strength parameters obtained from uniaxial tests (orthotropic material). The angles Θ1, Θ2, and Θ3 are shown in Figure 10.6. F12 = 1 2 sin2 1 cos2 1 1 (pf1)2 − F1 cos2 1+F2 sin2 1 pf1 − F11 cos4 1 − F22 sin4 1 − F66 2 F23 = 1 2 sin2 2 cos2 2 1 (pf2 )2 − F2 cos2 2+F3 sin2 2 pf2 − F22 cos4 2 − F33 sin4 2 − F44 2 F13 = 1 2 sin2 3 cos2 3 1 (pf3)2 − F3 cos2 3+F1 sin2 3 pf3 − F33 cos4 3 − F11 sin4 3 − F55 2
10.1 QUADRATIC FAILURE CRITERION 419 (-2) →0f1-2 Figure 10.7:Test coupon for biaxial testing in the xi-x2 orthotropy plane. The superscript f refers to the stresses at failure,and the superscript 1-2 identi- fies the applied biaxial stress components in the xi-x2 orthotropy plane.By substi- tutingandintoEq.(10.12).and by setting all other stress components equal to zero,we obtain 五o1-2+F1-2+F1-22+z(o1-22+2F2o1-2a1-2=1. (10.20) Equation (10.20)results in fi=1F-)F()fno) 2a0-2g1-2 (10.21) The other interaction strength parameters,obtained in a similar manner,are 1-Fa-0-Eo-》-F(o--Fo-2 Fij= 2o,0-o-7 ij=1,2,3. (10.22) Approximate expressions for the interaction strength parameters.In practice,it is difficult to perform the tests needed to generate the interaction strength param- eters Fi;.To eliminate the need for these tests,numerous approximate expressions have been proposed for Fij.One of these expressions is obtained by observing that a quadratic equation is characterized by its discriminants.When all but any two of the normal stresses are zero in the failure criterion(Eq.10.12),the discriminants are △=F:Fi-Fi,j=1,2,3i≠. (10.23) For the quadratic equation to represent a closed domain(which in our case is necessary to ensure that the stresses remain finite),every discriminant must be positive.Thus,from Eq.(10.23)we have -VFi Fii Fii <VFii Fij. (10.24) For convenience,we write E=i√EmFi (10.25)
10.1 QUADRATIC FAILURE CRITERION 419 f (1--2) σ1 (1--2)f σ1 f (1--2) σ2 (1--2)f σ2 Figure 10.7: Test coupon for biaxial testing in the x1–x2 orthotropy plane. The superscript f refers to the stresses at failure, and the superscript 1–2 identi- fies the applied biaxial stress components in the x1–x2 orthotropy plane. By substituting σf(1–2) 1 and σf(1–2) 2 into Eq. (10.12), and by setting all other stress components equal to zero, we obtain F1σf(1–2) 1 + F2σf(1–2) 2 + F11 σf(1–2) 1 2 + F22 σf(1–2) 2 2 + 2F12σf(1–2) 1 σf(1–2) 2 = 1. (10.20) Equation (10.20) results in F12 = 1 − F1σf(1–2) 1 − F2σf(1–2) 2 − F11 σf(1–2) 1 2 − F22 σf(1–2) 2 2 2σf(1–2) 1 σf(1–2) 2 . (10.21) The other interaction strength parameters, obtained in a similar manner, are Fi j = 1 − Fiσf(i– j) i − Fjσf(i– j) j − Fii σf(i– j) i 2 − Fj j σf(i– j) j 2 2σf(i– j) i σf(i– j) j i, j = 1, 2, 3. (10.22) Approximate expressions for the interaction strength parameters. In practice, it is difficult to perform the tests needed to generate the interaction strength parameters Fi j . To eliminate the need for these tests, numerous approximate expressions have been proposed for Fi j . One of these expressions is obtained by observing that a quadratic equation is characterized by its discriminants. When all but any two of the normal stresses are zero in the failure criterion (Eq. 10.12), the discriminants are i j = Fii Fj j − F2 i j i, j = 1, 2, 3 i = j. (10.23) For the quadratic equation to represent a closed domain (which in our case is necessary to ensure that the stresses remain finite), every discriminant must be positive. Thus, from Eq. (10.23) we have − 5Fii Fj j < Fi j < 5Fii Fj j . (10.24) For convenience, we write Fi j = fi j5Fii Fj j . (10.25)