Copyrighted Materials CpUPress o CHAPTER TEN Failure Criteria Failure of fiber-reinforced composites may be caused by fiber buckling,fiber breakage,matrix cracking,delamination,or by a combination of these factors (Fig.10.1).Local fiber buckling,or microbuckling,reduces the compressive stiff- ness and strength of the laminate.Microbuckling does not necessarily lead to immediate failure because the surrounding matrix supports the fibers.The prop- erties of the fibers and the matrix greatly affect the onset and magnitude of fiber buckling and the resulting losses in the compressive properties of the laminate. One of the main roles of the fibers is to carry tensile loads.When dry fibers (with no matrix surrounding them)break,they,of course,can no longer carry ten- sile loads.When the fibers are embedded in a matrix,the matrix acts as a bridge about the break and transmits the load across the gap created by the breakage as well from the broken to the adjacent fibers.Fiber bridging,as this phenomenon is called,is the main reason that the tensile strengths of unidirectional,continu- ous fiber-reinforced composites are higher than the tensile strengths of dry fiber bundles. Matrix cracking frequently occurs in composite laminates.In itself,matrix cracking generally does not result in ultimate failure of a laminate.Nonetheless, matrix cracks have many detrimental effects:they facilitate moisture absorbtion, reduce the matrix-dominated stiffnesses of the laminate and,last but not least,may propagate into the interface between adjacent layers,initiating delamination. Delamination is a separation of adjacent layers that may be introduced either during manufacture or subsequently by loads applied to the laminate.For example, loads due to transverse impact by an object on the laminate are a frequent cause of delamination.Delamination reduces the bending stiffness and strength as well as the load carrying capability of the laminate under compression.Significantly, under repeated loading the size of the delamination may increase to a critical point. Like the behavior of a crack in metal,once the critical size is reached,the growth of the delamination becomes unstable,leading to a rapid loss of compressive strength. 411
CHAPTER TEN Failure Criteria Failure of fiber-reinforced composites may be caused by fiber buckling, fiber breakage, matrix cracking, delamination, or by a combination of these factors (Fig. 10.1). Local fiber buckling, or microbuckling, reduces the compressive stiffness and strength of the laminate. Microbuckling does not necessarily lead to immediate failure because the surrounding matrix supports the fibers. The properties of the fibers and the matrix greatly affect the onset and magnitude of fiber buckling and the resulting losses in the compressive properties of the laminate. One of the main roles of the fibers is to carry tensile loads. When dry fibers (with no matrix surrounding them) break, they, of course, can no longer carry tensile loads. When the fibers are embedded in a matrix, the matrix acts as a bridge about the break and transmits the load across the gap created by the breakage as well from the broken to the adjacent fibers. Fiber bridging, as this phenomenon is called, is the main reason that the tensile strengths of unidirectional, continuous fiber-reinforced composites are higher than the tensile strengths of dry fiber bundles. Matrix cracking frequently occurs in composite laminates. In itself, matrix cracking generally does not result in ultimate failure of a laminate. Nonetheless, matrix cracks have many detrimental effects: they facilitate moisture absorbtion, reduce the matrix-dominated stiffnesses of the laminate and, last but not least, may propagate into the interface between adjacent layers, initiating delamination. Delamination is a separation of adjacent layers that may be introduced either during manufacture or subsequently by loads applied to the laminate. For example, loads due to transverse impact by an object on the laminate are a frequent cause of delamination. Delamination reduces the bending stiffness and strength as well as the load carrying capability of the laminate under compression. Significantly, under repeated loading the size of the delamination may increase to a critical point. Like the behavior of a crack in metal, once the critical size is reached, the growth of the delamination becomes unstable, leading to a rapid loss of compressive strength. 411
412 FAILURE CRITERIA Fiber buckling Fiber breakage Matrix cracking Delamination Figure 10.1:Typical failure modes of composites. Designers would be well served by mechanism-based (physical)theories that would indicate the load at which failure occurs as well as the mode of failure. Although such theories have been proposed,12 none is as yet at the stage where it could be applied in practical engineering design.Instead,frequently,ply-stress- based failure theories are used.3.4 According to these theories the criterion for failure in any one of the plies is 1 no failure f(o1,02,3,23,t13,t12,F,F, -1 failure limit, (10.1) >1 failure where o1,...,12 are the stresses in the ply and F1,F2,...are strength parameters. The criterion expressed by Eq.(10.1)is established in every ply,and failure is taken to occur when any one of the plies fails (first-ply failure). Here,we present three failure criteria for composites based on the aforemen- tioned concept:the quadratic,the maximum stress,and the maximum strain failure criteria.These criteria offer results that are sufficiently accurate for many(but by no means all)problems of practical interest.For this reason,in spite of their short- comings,they are relevant to the engineer.Nonetheless,the reader is warned to be cognizant of the following significant limits of the criteria listed above: Each criterion provides only the load at which first-ply failure occurs,that is, the load at which the linear load-displacement curve first changes(Fig.10.2). Under the load set that causes first-ply failure,the laminate does not necessarily fail because other undamaged plies can still carry load.As the applied loads 1 R.F.Gibson,Principles of Composite Material Mechanics.McGraw-Hill,New York,1994,pp.114- 126.244-249,and356-367. 2 S.R.Swanson,Advanced Composite Materials.Prentice-Hall,Upper Saddle River.New Jersey.1997. pp.91-120.123-147. 3 R.E.Rowlands,Strength(Failure)Theories and Their Experimental Correlation.In:Handbook of Composites,Vol.3.G.C.Sih and A.M.Skudra,eds.,Elsevier,Amsterdam,1985,pp.71-125. 4 M.N.Nahas,Survey of Failure and Post-Failure Theories of Laminated Fiber Reinforced Composites. Journal of Composites Technology and Research,Vol.8,138-153,1986
412 FAILURE CRITERIA Fiber buckling Fiber breakage Matrix cracking Delamination Figure 10.1: Typical failure modes of composites. Designers would be well served by mechanism-based (physical) theories that would indicate the load at which failure occurs as well as the mode of failure. Although such theories have been proposed,1,2 none is as yet at the stage where it could be applied in practical engineering design. Instead, frequently, ply–stressbased failure theories are used.3,4 According to these theories the criterion for failure in any one of the plies is f (σ1, σ2, σ3, τ23, τ13, τ12, F1, F2,...) 1 failure , (10.1) where σ1,...,τ12 are the stresses in the ply and F1, F2,... are strength parameters. The criterion expressed by Eq. (10.1) is established in every ply, and failure is taken to occur when any one of the plies fails (first-ply failure). Here, we present three failure criteria for composites based on the aforementioned concept: the quadratic, the maximum stress, and the maximum strain failure criteria. These criteria offer results that are sufficiently accurate for many (but by no means all) problems of practical interest. For this reason, in spite of their shortcomings, they are relevant to the engineer. Nonetheless, the reader is warned to be cognizant of the following significant limits of the criteria listed above: Each criterion provides only the load at which first-ply failure occurs, that is, the load at which the linear load-displacement curve first changes (Fig. 10.2). Under the load set that causes first-ply failure, the laminate does not necessarily fail because other undamaged plies can still carry load. As the applied loads 1 R. F. Gibson, Principles of Composite Material Mechanics. McGraw-Hill, New York, 1994, pp. 114– 126, 244–249, and 356–367. 2 S. R. Swanson, Advanced Composite Materials. Prentice-Hall, Upper Saddle River, New Jersey, 1997, pp. 91–120, 123–147. 3 R. E. Rowlands, Strength (Failure) Theories and Their Experimental Correlation. In: Handbook of Composites, Vol. 3. G. C. Sih and A. M. Skudra, eds., Elsevier, Amsterdam, 1985, pp. 71–125. 4 M. N. Nahas, Survey of Failure and Post-Failure Theories of Laminated Fiber Reinforced Composites. Journal of Composites Technology and Research, Vol. 8, 138–153, 1986
10.1 QUADRATIC FAILURE CRITERION 413 First-ply failure Load个 Displacement Figure 10.2:Load-displacement curve of a composite part. are increased beyond those at which first-ply failure occurs,there will be a sequence of ply failures until the load set is reached at which every ply has failed.The loads at ultimate failure may be considerably higher than at first- ply failure.Therefore,criteria based on first-ply failure are conservative. 。 None of the criteria sheds light on the failure mechanism or indicates the mode of failure. None of the criteria provides acceptable results for every condition of practical interest. Each criterion requires data,some of which are difficult to measure. Each criterion applies in regions inside the composite away from discontinu- ities such as holes,cracks,and edges.(Criteria applicable to plates containing a hole or a notch are given in Section 10.4.) 10.1 Quadratic Failure Criterion The quadratic failure criterion includes stresses up to the second power.In its most general form the quadratic failure criterion states that no failure occurs when the inequality below (Eq.10.2)is satisfied.This criterion and some of its simplified forms are variously referred to as Tsai-Wu,Hill,or Tsai-Hill failure criterion. F101+F202+F303+F4t23+Fst3+F6712+ F1o7+F2o+F3o+F442+F5+F66+ 2(F20102+F3O13+F4O123+F5O1t13+F16O1T12+ F30203+F242t23+F252T13+F26O2t12+F3403t23+ F353t13+f363t12+F45t23t3+F46t23t12+F56t13T2)<1, (10.2) where o1,02,...,ti2 are the components (in the x1,x2,x3 coordinate system)of the stress at the point of interest,that is,the stress that results from the applied loads,and the F's are strength parameters that depend on the material.No failure occurs when the left-hand side of Eq.(10.2)is less then unity.This means that the resultant stress is inside the failure surface (Fig.10.3,left).On the failure surface (Fig.10.3,middle),where the stress components are denoted by of
10.1 QUADRATIC FAILURE CRITERION 413 Load First-ply failure Displacement Figure 10.2: Load-displacement curve of a composite part. are increased beyond those at which first-ply failure occurs, there will be a sequence of ply failures until the load set is reached at which every ply has failed. The loads at ultimate failure may be considerably higher than at firstply failure. Therefore, criteria based on first-ply failure are conservative. None of the criteria sheds light on the failure mechanism or indicates the mode of failure. None of the criteria provides acceptable results for every condition of practical interest. Each criterion requires data, some of which are difficult to measure. Each criterion applies in regions inside the composite away from discontinuities such as holes, cracks, and edges. (Criteria applicable to plates containing a hole or a notch are given in Section 10.4.) 10.1 Quadratic Failure Criterion The quadratic failure criterion includes stresses up to the second power. In its most general form the quadratic failure criterion states that no failure occurs when the inequality below (Eq. 10.2) is satisfied. This criterion and some of its simplified forms are variously referred to as Tsai-Wu, Hill, or Tsai–Hill failure criterion. F1σ1 + F2σ2 + F3σ3 + F4τ23 + F5τ13 + F6τ12 + F11σ2 1 + F22σ2 2 + F33σ2 3 + F44τ 2 23 + F55τ 2 13 + F66τ 2 12 + 2(F12σ1σ2 + F13σ1σ3 + F14σ1τ23 + F15σ1τ13 + F16σ1τ12 + F23σ2σ3 + F24σ2τ23 + F25σ2τ13 + F26σ2τ12 + F34σ3τ23 + F35σ3τ13 + F36σ3τ12 + F45τ23τ13 + F46τ23τ12 + F56τ13τ12) < 1, (10.2) where σ1, σ2, ... , τ12 are the components (in the x1, x2, x3 coordinate system) of the stress at the point of interest, that is, the stress that results from the applied loads, and the F’s are strength parameters that depend on the material. No failure occurs when the left-hand side of Eq. (10.2) is less then unity. This means that the resultant stress is inside the failure surface (Fig. 10.3, left). On the failure surface (Fig. 10.3, middle), where the stress components are denoted by σf 1, σf 2,....,τ f 12
414 FAILURE CRITERIA No failure Failure Failure 02 01 01=0 01 Figure 10.3:Representation of the failure surface when only o1 and o2 stresses are applied. Eq.(10.2)is E+F吲+F+E吆+喵+F6h+1(o)2+ F2(2+F3()2+F4()2+s()2+ F6(t2)2+2(F2oo+Fi3oo+…+F6t2)=1. (10.3) The strength parameters must be determined by tests.For generally anisotropic and monoclinic materials,27 and 17 types of tests are required,respectively.This makes the use of the quadratic failure criterion impractical for structures made of generally anisotropic or monoclinic materials.The criterion becomes more manageable when the material is orthotropic or transversely isotropic.Therefore, in the following the criterion is presented only for these two types of materials. 10.1.1 Orthotropic Material An orthotropic material has three planes of symmetry (Figs.2.11 and 2.12).We select the x1,x2.x3 coordinate system with axes perpendicular to these symmetry planes. First,we consider only the shear-stress component t23 acting in the plane of symmetry (Fig.10.4).When only t23 acts,and it is in the positive direction,then Plane of symmetry Figure 10.4:The positive and negative shear stresses at failure acting in an orthotropic material; x1,x2,and x3 are perpendicular to the orthotropy planes
414 FAILURE CRITERIA σ1 σ2 No failure Failure Failure σ1 σ2 f = σσ 22 f = σσ 11 Figure 10.3: Representation of the failure surface when only σ1 and σ2 stresses are applied. Eq. (10.2) is F1σf 1 + F2σf 2 + F3σf 3 + F4τ f 23 + F5τ f 13 + F6τ f 12 + 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 +···+ F56τ f 13τ f 12 = 1. (10.3) The strength parameters must be determined by tests. For generally anisotropic and monoclinic materials, 27 and 17 types of tests are required, respectively. This makes the use of the quadratic failure criterion impractical for structures made of generally anisotropic or monoclinic materials. The criterion becomes more manageable when the material is orthotropic or transversely isotropic. Therefore, in the following the criterion is presented only for these two types of materials. 10.1.1 Orthotropic Material An orthotropic material has three planes of symmetry (Figs. 2.11 and 2.12). We select the x1, x2, x3 coordinate system with axes perpendicular to these symmetry planes. First, we consider only the shear–stress component τ23 acting in the plane of symmetry (Fig. 10.4). When only τ23 acts, and it is in the positive direction, then x3 x1 x2 Plane of symmetry f + 23 τ − − f 23 τ Figure 10.4: The positive and negative shear stresses at failure acting in an orthotropic material; x1, x2, and x3 are perpendicular to the orthotropy planes
10.1 QUADRATIC FAILURE CRITERION 415 at failure(2=)the quadratic failure criterion yields(Eq.10.3) E站+F4()2=1. (10.4) When only 723 acts in the negative direction,then,at failure(23=)the quadratic failure criterion becomes (Eq.10.3) -F站+F4(站)2=1. (10.5) Because of symmetry,the failure stress for positive shear is the same as for negative shear(=).The two preceding equations satisfy this condition only if F4 is zero. By similar argument it can be shown that Fs,F are zero.Thus,we have F=F=F6=0. (10.6) Next,we apply the normal stresses a1,02,o3;the shear stresses ti2,Ti3;and either a positive or a negative shear stress 2(Fig.10.5).For a positive shear stress at failure,ts=t∴，and we have(Eq.l0.3) o+Fi+Fo+F1(o'+F2()2+ Fs(o)2+F4(塔)2+Fs(6)2+6()2+ 2(E2ooi+3oo+）+·+ +2(F4o+F4o+F4o+F45+F46)站=1 (10.7) For a negative shear stress at failure,=,and Eq.(10.3)gives Fof+F2o+F3os+Fu(of)2+Fz2 ()2+ F3(o)2+F4(站)2+Fs(c)2+F6()2+ 2(2o+F3oog+…）+…+ -2(F4o+F4o+F4o吲+F4s+F46b)点=1 (10.8) Because of symmetry,the failure stress for positive out-of-plane shear is the same as for negative out-of-plane shear().This condition,together with 12 Plane of symmetry Figure 10.5:The stresses at failure acting on an orthotropic material
10.1 QUADRATIC FAILURE CRITERION 415 at failure (τ23 = τ f+ 23 ) the quadratic failure criterion yields (Eq. 10.3) F4τ f+ 23 + F44 τ f+ 23 2 = 1. (10.4) When only τ23 acts in the negative direction, then, at failure (τ23 = −τ f− 23 ) the quadratic failure criterion becomes (Eq. 10.3) −F4τ f− 23 + F44 τ f− 23 2 = 1. (10.5) Because of symmetry, the failure stress for positive shear is the same as for negative shear (τ f+ 23 = τ f− 23 ). The two preceding equations satisfy this condition only if F4 is zero. By similar argument it can be shown that F5, F6 are zero. Thus, we have F4 = F5 = F6 = 0. (10.6) Next, we apply the normal stresses σ1, σ2, σ3; the shear stresses τ12, τ13; and either a positive or a negative shear stress τ23 (Fig. 10.5). For a positive shear stress at failure, τ23 = τ f+ 23 , and we have (Eq. 10.3) F1σf 1 + F2σf 2 + F3σf 3 + F11 σf 1 2 + F22 σf 2 2 + F33 σf 3 2 + F44 τ f+ 23 2 + F55 τ f 132 + F66 τ f 122 + 2 F12σf 1σf 2 + F13σf 1σf 3 +··· +····+ +2 F14σf 1 + F24σf 2 + F34σf 3 + F45τ f 13 + F46τ f 12 τ f+ 23 = 1. (10.7) For a negative shear stress at failure, τ23 = −τ f− 23 , and Eq. (10.3) gives F1σf 1 + F2σf 2 + F3σf 3 + F11 σf 1 2 + F22 σf 2 2 + F33 σf 3 2 + F44 τ f− 23 2 + F55 τ f 132 + F66 τ f 122 + 2 F12σf 1σf 2 + F13σf 1σf 3 +··· +····+ −2 F14σf 1 + F24σf 2 + F34σf 3 + F45τ f 13 + F46τ f 12 τ f− 23 = 1. (10.8) Because of symmetry, the failure stress for positive out-of-plane shear is the same as for negative out-of-plane shear (τ f+ 23 = τ f− 23 ). This condition, together with x3 x1 x2 Plane of symmetry f + 23 τ − − f 23 τ f σ3 f σ3 f 13 τ f 13 τ f σ1 f f σ1 12 τ f 12 τ f σ2 f σ2 Figure 10.5: The stresses at failure acting on an orthotropic material
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)
420 FAILURE CRITERIA =-g T23=T23 Figure 10.8:Stresses in the x2-isotropy plane of a transversely isotropic material. fi are constants that may have any value within-1 and +1.Tsai and Hahn5 observed that the quadratic failure criterion for composites reduces to the von Mises quadratic failure criterion for isotropic materials (Section 10.1.3)when fij=-0.5.With this value the interaction strength parameters become =-3五 压=3m后 =3a (10.26) 10.1.2 Transversely Isotropic Material A transversely isotropic material has three planes of symmetry(Fig.2.15,page 19). Therefore,the stress parameters,which are zero for an orthotropic material(which also has three planes of symmetry),are also zero for transversely isotropic mate- rials (Egs.10.9 and 10.10).In addition,one of the symmetry planes is isotropic. We select this to be the x2-x3 plane(Fig.2.15).Because of isotropy,in this plane the subscripts 2 and 3 are interchangeable,and we can write F=F3F2=F33F13=F2 (10.27) For the same reason s12 =s13,and from Table 10.2(page 417)we have Fs5=F66. (10.28) We now apply tensile o2 and compressive o3 stresses of equal magnitude (02=-03)(Fig.10.8,top left).At failure,o2=a',o3 =-of,and Eq.(10.12) yields Fa-F3a+F2(o2+F3(a2-2F23(a52=1. (10.29) By combining Eqs.(10.27)-(10.29),we obtain 1 ()=2(F2-Fx3) (10.30) 5 S.W.Tsai and H.T.Hahn,Introduction to Composite Materials.Technomic,Lancester,Pennsylvania, 1980,P.286
420 FAILURE CRITERIA σ σ– 3 = f τ τ 45 = f 23 σ σ 2 = f τ τ 23 23 = f τ σ 45 f = o o Figure 10.8: Stresses in the x2–x3 isotropy plane of a transversely isotropic material. fi j are constants that may have any value within −1 and +1. Tsai and Hahn5 observed that the quadratic failure criterion for composites reduces to the von Mises quadratic failure criterion for isotropic materials (Section 10.1.3) when fi j = −0.5. With this value the interaction strength parameters become F12 = −1 2 5 F11F22 F13 = −1 2 5 F11F33 F23 = −1 2 5 F22F33. (10.26) 10.1.2 Transversely Isotropic Material A transversely isotropic material has three planes of symmetry (Fig. 2.15, page 19). Therefore, the stress parameters, which are zero for an orthotropic material (which also has three planes of symmetry), are also zero for transversely isotropic materials (Eqs. 10.9 and 10.10). In addition, one of the symmetry planes is isotropic. We select this to be the x2–x3 plane (Fig. 2.15). Because of isotropy, in this plane the subscripts 2 and 3 are interchangeable, and we can write F2 = F3 F22 = F33 F13 = F12. (10.27) For the same reason s12 = s13, and from Table 10.2 (page 417) we have F55 = F66. (10.28) We now apply tensile σ2 and compressive σ3 stresses of equal magnitude (σ2 = −σ3) (Fig. 10.8, top left). At failure, σ2 = σf , σ3 = −σf , and Eq. (10.12) yields F2σf − F3σf + F22(σf ) 2 + F33(σf ) 2 − 2F23(σf ) 2 = 1. (10.29) By combining Eqs. (10.27)–(10.29), we obtain (σf ) 2 = 1 2(F22 − F23) . (10.30) 5 S. W. Tsai and H. T. Hahn,Introduction to Composite Materials. Technomic, Lancester, Pennsylvania, 1980, p. 286