Design In the preceding chapters,we have discussed various aspects of fiber-reinforced polymers,including the constituent materials,mechanics,performance,and manufacturing methods.A number of unique characteristics of fiber-reinforced polymers that have emerged in these chapters are listed in Table 6.1.Many of these characteristics are due to the orthotropic nature of fiber-reinforced com- posites,which has also necessitated the development of new design approaches that are different from the design approaches traditionally used for isotropic materials,such as steel or aluminum alloys.This chapter describes some of the design methods and practices currently used for fiber-reinforced polymers including the failure prediction methods,the laminate design procedures,and the joint design considerations.A number of design examples are also included. 6.1 FAILURE PREDICTION Design analysis of a structure or a component is performed by comparing stresses (or strains)due to applied loads with the allowable strength (or strain capacity)of the material.In the case of biaxial or multiaxial stress fields,a suitable failure theory is used for this comparison.For an isotropic material that exhibits yielding,such as a mild steel or an aluminum alloy,either the maximum shear stress theory or the distortional energy theory(von Mises yield criterion)is commonly used for designing against yielding.Fiber-reinforced polymers are not isotropic,nor do they exhibit gross yielding.Thus,failure theories developed for metals or other isotropic materials are not applicable to composite materials.Instead,many new failure theories have been proposed for fiber-reinforced composites,some of which are discussed in this section. 6.1.1 FAILURE PREDICTION IN A UNIDIRECTIONAL LAMINA We consider the plane stress condition of a general orthotropic lamina contain- ing unidirectional fibers at a fiber orientation angle of a with respect to the x axis(Figure 6.1).In Chapter 3,we saw that four independent elastic constants, namely,E1,E22,Gi2,and vi2,are required to define its elastic characteristics Its strength properties are characterized by five independent strength values: SLt longitudinal tensile strength Srt transverse tensile strength 2007 by Taylor&Francis Group.LLC
6 Design In the preceding chapters, we have discussed various aspects of fiber-reinforced polymers, including the constituent materials, mechanics, performance, and manufacturing methods. A number of unique characteristics of fiber-reinforced polymers that have emerged in these chapters are listed in Table 6.1. Many of these characteristics are due to the orthotropic nature of fiber-reinforced composites, which has also necessitated the development of new design approaches that are different from the design approaches traditionally used for isotropic materials, such as steel or aluminum alloys. This chapter describes some of the design methods and practices currently used for fiber-reinforced polymers including the failure prediction methods, the laminate design procedures, and the joint design considerations. A number of design examples are also included. 6.1 FAILURE PREDICTION Design analysis of a structure or a component is performed by comparing stresses (or strains) due to applied loads with the allowable strength (or strain capacity) of the material. In the case of biaxial or multiaxial stress fields, a suitable failure theory is used for this comparison. For an isotropic material that exhibits yielding, such as a mild steel or an aluminum alloy, either the maximum shear stress theory or the distortional energy theory (von Mises yield criterion) is commonly used for designing against yielding. Fiber-reinforced polymers are not isotropic, nor do they exhibit gross yielding. Thus, failure theories developed for metals or other isotropic materials are not applicable to composite materials. Instead, many new failure theories have been proposed for fiber-reinforced composites, some of which are discussed in this section. 6.1.1 FAILURE PREDICTION IN A UNIDIRECTIONAL LAMINA We consider the plane stress condition of a general orthotropic lamina containing unidirectional fibers at a fiber orientation angle of u with respect to the x axis (Figure 6.1). In Chapter 3, we saw that four independent elastic constants, namely, E11, E22, G12, and n12, are required to define its elastic characteristics. Its strength properties are characterized by five independent strength values: SLt ¼ longitudinal tensile strength STt ¼ transverse tensile strength 2007 by Taylor & Francis Group, LLC
TABLE 6.1 Unique Characteristics of Fiber-Reinforced Polymer Composites Nonisotropic Orthotropic Directional properties Four independent elastic constants instead of two Principal stresses and principal strains not in the same direction Coupling between extensional and shear deformations Nonhomogeneous More than one macroscopic constituent Local variation in properties due to resin-rich areas,voids,fiber misorientation,etc. Laminated structure Laminated structure Extensional-bending coupling Planes of weakness between layers Interlaminar stresses Properties depend on the laminate type Properties may depend on stacking sequence Properties can be tailored according to requirements Poisson's ratio can be greater than 0.5 Nonductile behavior Lack of plastic yielding Nearly elastic or slightly nonelastic stress-strain behavior Stresses are not locally redistributed around bolted or riveted holes by yielding Low strains-to-failure in tension Noncatastrophic failure modes Delamination Localized damage (fiber breakage,matrix cracking.debonding,fiber pullout,etc.) Less notch sensitivity Progressive loss in stiffness during cyclic loading Interlaminar shear failure in bending Low coefficient of thermal expansion Dimensional stability Zero coefficient of thermal expansion possible Attachment problem with metals due to thermal mismatch High internal damping:High attenuation of vibration and noise Noncorroding SLe =longitudinal compressive strength Sre transverse compressive strength SLTs in-plane shear strength Experimental techniques for determining these strength properties have been presented in Chapter 4.Note that the in-plane shear strength SLrs in the principal material directions does not depend on the direction of the shear stress although both the longitudinal and transverse strengths may depend on the direction of the normal stress,namely,tensile or compressive. 2007 by Taylor Francis Group,LLC
SLc ¼ longitudinal compressive strength STc ¼ transverse compressive strength SLTs ¼ in-plane shear strength Experimental techniques for determining these strength properties have been presented in Chapter 4. Note that the in-plane shear strength SLTs in the principal material directions does not depend on the direction of the shear stress although both the longitudinal and transverse strengths may depend on the direction of the normal stress, namely, tensile or compressive. TABLE 6.1 Unique Characteristics of Fiber-Reinforced Polymer Composites Nonisotropic Orthotropic Directional properties Four independent elastic constants instead of two Principal stresses and principal strains not in the same direction Coupling between extensional and shear deformations Nonhomogeneous More than one macroscopic constituent Local variation in properties due to resin-rich areas, voids, fiber misorientation, etc. Laminated structure Laminated structure Extensional–bending coupling Planes of weakness between layers Interlaminar stresses Properties depend on the laminate type Properties may depend on stacking sequence Properties can be tailored according to requirements Poisson’s ratio can be greater than 0.5 Nonductile behavior Lack of plastic yielding Nearly elastic or slightly nonelastic stress–strain behavior Stresses are not locally redistributed around bolted or riveted holes by yielding Low strains-to-failure in tension Noncatastrophic failure modes Delamination Localized damage (fiber breakage, matrix cracking, debonding, fiber pullout, etc.) Less notch sensitivity Progressive loss in stiffness during cyclic loading Interlaminar shear failure in bending Low coefficient of thermal expansion Dimensional stability Zero coefficient of thermal expansion possible Attachment problem with metals due to thermal mismatch High internal damping: High attenuation of vibration and noise Noncorroding 2007 by Taylor & Francis Group, LLC
y y FIGURE 6.1 Two-dimensional stress state in a thin orthotropic lamina. Many phenomenological theories have been proposed to predict failure in a unidirectional lamina under plane stress conditions.Among these,the simplest theory is known as the maximum stress theory;however,the more commonly used failure theories are the maximum strain theory and the Azzi-Tsai-Hill failure theory.We discuss these three theories as well as a more generalized theory,known as the Tsai-Wu theory.To use them,applied stresses(or strains) are first transformed into principal material directions using Equation 3.30. The transformed stresses are denoted o11,022,and T12,and the applied stresses are denoted oxx,oy,and Tx. 6.1.1.1 Maximum Stress Theory According to the maximum stress theory,failure occurs when any stress in the principal material directions is equal to or greater than the corresponding ultimate strength.Thus to avoid failure, -SLe <011<SLt, -STe <22 STL -SLTs TI2<SLTs. (6.1) For the simple case of uniaxial tensile loading in the x direction,only oxx is present and oyy =Tx=0.Using Equation 3.30,the transformed stresses are 2007 by Taylor&Francis Group.LLC
Many phenomenological theories have been proposed to predict failure in a unidirectional lamina under plane stress conditions. Among these, the simplest theory is known as the maximum stress theory; however, the more commonly used failure theories are the maximum strain theory and the Azzi–Tsai–Hill failure theory. We discuss these three theories as well as a more generalized theory, known as the Tsai–Wu theory. To use them, applied stresses (or strains) are first transformed into principal material directions using Equation 3.30. The transformed stresses are denoted s11, s22, and t12, and the applied stresses are denoted sxx, syy, and txy. 6.1.1.1 Maximum Stress Theory According to the maximum stress theory, failure occurs when any stress in the principal material directions is equal to or greater than the corresponding ultimate strength. Thus to avoid failure, SLc < s11 < SLt, STc < s22 < STt, SLTs < t12 < SLTs: (6:1) For the simple case of uniaxial tensile loading in the x direction, only sxx is present and syy ¼ txy ¼ 0. Using Equation 3.30, the transformed stresses are y syy syy sxx q sxx txy txy x FIGURE 6.1 Two-dimensional stress state in a thin orthotropic lamina. 2007 by Taylor & Francis Group, LLC.
011=0xc0s20, 022=0 xx sin20, T12 =-Oxx sin0 cos 0. Thus,using the maximum stress theory,failure of the lamina is predicted if the applied stress oxx exceeds the smallest of (SLt/cos),(Srt/sin 0),and (SLTs/sine cose).Thus the safe value of ox depends on the fiber orientation angle 6,as illustrated in Figure 6.2.At small values of 6,longitudinal tensile failure is expected,and the lamina strength is calculated from (SL/cos )At high values of 0,transverse tensile failure is expected,and the lamina strength is calculated from(Sr/sin20).At intermediate values of 0,in-plane shear failure of the lamina is expected and the lamina strength is calculated from(SLTs/sine cose).The change from longitudinal tensile failure to in-plane shear failure occurs at =01=tan-1 SLTs/SL and the change from in-plane shear failure to 200 100 60 40 20 Maximum strain theory 10 Maximum stress theory 6 Azzi-Tsai- Hill theory 2 15 30 45 60 75 90 Fiber orientation angle,(deg) FIGURE 6.2 Comparison of maximum stress,maximum strain,and Azzi-Tsai-Hill theories with uniaxial strength data of a glass fiber-reinforced epoxy composite.(After Azzi,V.D.and Tsai,S.W.,Exp.Mech.,5,283,1965.) 2007 by Taylor Francis Group,LLC
s11 ¼ sxx cos2 u, s22 ¼ sxx sin2 u, t12 ¼ sxx sin u cos u: Thus, using the maximum stress theory, failure of the lamina is predicted if the applied stress sxx exceeds the smallest of (SLt=cos2 u), (STt=sin2 u), and (SLTs=sinu cosu). Thus the safe value of sxx depends on the fiber orientation angle u, as illustrated in Figure 6.2. At small values of u, longitudinal tensile failure is expected, and the lamina strength is calculated from (SLt=cos2 u). At high values of u, transverse tensile failure is expected, and the lamina strength is calculated from (STt=sin2 u). At intermediate values of u, in-plane shear failure of the lamina is expected and the lamina strength is calculated from (SLTs=sinu cosu). The change from longitudinal tensile failure to in-plane shear failure occurs at u ¼ u1 ¼ tan1 SLTs=SLt and the change from in-plane shear failure to Maximum stress theory Maximum strain theory Azzi–Tsai− Hill theory Uniaxial tensile strength (ksi) Fiber orientation angle, q (deg) 2 0 15 30 45 60 75 90 200 100 60 40 20 10 8 6 4 q FIGURE 6.2 Comparison of maximum stress, maximum strain, and Azzi–Tsai–Hill theories with uniaxial strength data of a glass fiber-reinforced epoxy composite. (After Azzi, V.D. and Tsai, S.W., Exp. Mech., 5, 283, 1965.) 2007 by Taylor & Francis Group, LLC.
transverse tensile failure occurs at =02=tan-Sn/SLTs.For example,for an E-glass fiber-epoxy composite with SLt =1100 MPa,Sn=96.5 MPa,and SLTs 83 MPa,01=4.3 and 02 =49.3.Thus,according to the maximum stress theory,longitudinal tensile failure of this composite lamina will occur for0°≤0<4.3°，in-plane shear failure will occur for4.3°≤0≤49.3oand transverse tensile failure will occur for 49.3<0<90. EXAMPLE 6.1 A unidirectional continuous T-300 carbon fiber-reinforced epoxy laminate is subjected to a uniaxial tensile load P in the x direction.The laminate width and thickness are 50 and 2 mm,respectively.The following strength properties are known: SL SLe =1447.5 MPa,ST =44.8 MPa,and SLTs =62 MPa Determine the maximum value of P for each of the following cases:(a)0=0, (b)0=30°，and(c0=60°. SOLUTION The laminate is subjected to a uniaxial tensile stress oxx due to the tensile load applied in thexdirection.In all three cases,where A is the cross-sectional area of the laminate. L.Since0=0°，11=gxx,o22=0,andT12=0. Therefore,in this case the laminate failure occurs whenou=oxx =Su 1447.5MPa. Since the tensile load p at which failure occurs is 144.75 kN.The mode of failure is the longitudinal tensile failure of the lamina. 2.Since0=30°，using Equation3.30, 11=0xc0s230°=0.750x 22=0asin230°=0.25rxx, T12=oxx sin30°cos30°=0.433ax. According to Equation 6.1,the maximum values of ou,o22,and Ti2 are (1)o =0.750xx SLt 1447.5 MPa,which gives oxx 1930 MPa (2)022 =0.250xx STt =44.8 MPa,which gives oxx 179.2 MPa (3)T12 =0.4330xx SLTs 62 MPa,which gives oxx =143.2 MPa Laminate failure occurs at the lowest value ofx In this case,the lowest value is 143.2 MPa.Using x==143.2 MPa,P=14.32 kN.The mode of failure is the in-plane shear failure of the lamina. 2007 by Taylor Francis Group.LLC
transverse tensile failure occurs at u ¼ u2 ¼ tan1 STt=SLTs. For example, for an E-glass fiber–epoxy composite with SLt ¼ 1100 MPa, STt ¼ 96:5 MPa, and SLTs ¼ 83 MPa, u1 ¼ 4:3 and u2 ¼ 49:3. Thus, according to the maximum stress theory, longitudinal tensile failure of this composite lamina will occur for 0 u < 4:3, in-plane shear failure will occur for 4:3 u 49:3 and transverse tensile failure will occur for 49:3 < u 90. EXAMPLE 6.1 A unidirectional continuous T-300 carbon fiber-reinforced epoxy laminate is subjected to a uniaxial tensile load P in the x direction. The laminate width and thickness are 50 and 2 mm, respectively. The following strength properties are known: SLt ¼ SLc ¼ 1447:5 MPa, STt ¼ 44:8 MPa, and SLTs ¼ 62 MPa: Determine the maximum value of P for each of the following cases: (a) u ¼ 08, (b) u ¼ 308, and (c) u ¼ 608. SOLUTION The laminate is subjected to a uniaxial tensile stress sxx due to the tensile load applied in the x direction. In all three cases, sxx ¼ P A, where A is the cross-sectional area of the laminate. 1. Since u ¼ 0, s11 ¼ sxx, s22 ¼ 0, and t12 ¼ 0. Therefore, in this case the laminate failure occurs when s11 ¼ sxx ¼ SLt ¼ 1447:5 MPa. Since sxx ¼ P A ¼ P (0:05 m)(0:002 m) , the tensile load P at which failure occurs is 144.75 kN. The mode of failure is the longitudinal tensile failure of the lamina. 2. Since u ¼ 308, using Equation 3.30, s11 ¼ sxx cos2 30 ¼ 0:75 sxx, s22 ¼ sxx sin2 30 ¼ 0:25 sxx, t12 ¼ sxx sin 30 cos 30 ¼ 0:433 sxx: According to Equation 6.1, the maximum values of s11, s22, and t12 are (1) s11 ¼ 0:75sxx ¼ SLt ¼ 1447:5 MPa, which gives sxx ¼ 1930 MPa (2) s22 ¼ 0:25sxx ¼ STt ¼ 44:8 MPa, which gives sxx ¼ 179:2 MPa (3) t12 ¼ 0:433sxx ¼ SLTs ¼ 62 MPa, which gives sxx ¼ 143:2 MPa Laminate failure occurs at the lowest value of sxx. In this case, the lowest value is 143.2 MPa. Using sxx ¼ P A ¼ 143:2 MPa, P ¼ 14.32 kN. The mode of failure is the in-plane shear failure of the lamina. 2007 by Taylor & Francis Group, LLC.
3.Since0=60°，using Equation3.30, 11=xxc0s260°=0.25xx, 022 =Oxx sin2 60=0.75 Oxx, T12=xx sin60°cos60°=0.433gxx. According to Equation 6.1,the maximum values of u,o22,and Ti2 are (1)o11 =0.250xx=SL 1447.5 MPa,which gives oxx 5790 MPa (2)o22 =0.750 xx=STt =44.8 MPa,which gives oxx 59.7 MPa (3)TI2 =0.4330xx SLTs 62 MPa,which gives axx 143.2 MPa Laminate failure occurs at the lowest value of xx.In this case,the lowest value is 59.7 MPa.Using oxx==59.7 MPa,P 5.97 kN.The mode of failure is transverse tensile failure of the lamina. 6.1.1.2 Maximum Strain Theory According to the maximum strain theory,failure occurs when any strain in the principal material directions is equal to or greater than the corresponding ultimate strain.Thus to avoid failure, -8L 811 SLt, -STe 822 STt, -YLTs Y12 YLTs. (6.2) Returning to the simple case of uniaxial tensile loading in which a stress ox is applied to the lamina,the safe value of this stress is calculated in the following way. 1.Using the strain-stress relationship,Equation 3.72,and the transformed stresses,the strains in the principal material directions are 811=S11o11+S12o22=(S1cos20+S12sin20)oxx, E2=S12011+S202=(S12cos20+S22sin2)0x, Y12 S66T22=-S66 sin 0 cos 0 Oxx, where 1 Su=Eu S2=-2=- 21 E11 E22 1 S2n≠E2 S6=G12 2007 by Taylor Francis Group,LLC
3. Since u ¼ 608, using Equation 3.30, s11 ¼ sxx cos2 60 ¼ 0:25 sxx, s22 ¼ sxx sin2 60 ¼ 0:75 sxx, t12 ¼ sxx sin 60 cos 60 ¼ 0:433 sxx: According to Equation 6.1, the maximum values of s11, s22, and t12 are (1) s11 ¼ 0:25sxx ¼ SLt ¼ 1447:5 MPa, which gives sxx ¼ 5790 MPa (2) s22 ¼ 0:75sxx ¼ STt ¼ 44:8 MPa, which gives sxx ¼ 59:7 MPa (3) t12 ¼ 0:433sxx ¼ SLTs ¼ 62 MPa, which gives sxx ¼ 143:2 MPa Laminate failure occurs at the lowest value of sxx. In this case, the lowest value is 59.7 MPa. Using sxx ¼ P A ¼ 59:7 MPa, P ¼ 5.97 kN. The mode of failure is transverse tensile failure of the lamina. 6.1.1.2 Maximum Strain Theory According to the maximum strain theory, failure occurs when any strain in the principal material directions is equal to or greater than the corresponding ultimate strain. Thus to avoid failure, «Lc < «11 < «Lt, «Tc < «22 < «Tt, gLTs < g12 < gLTs: (6:2) Returning to the simple case of uniaxial tensile loading in which a stress sxx is applied to the lamina, the safe value of this stress is calculated in the following way. 1. Using the strain–stress relationship, Equation 3.72, and the transformed stresses, the strains in the principal material directions are «11 ¼ S11s11 þ S12s22 ¼ (S11 cos2 u þ S12 sin2 u) sxx, «22 ¼ S12s11 þ S22s22 ¼ (S12 cos2 u þ S22 sin2 u) sxx, g12 ¼ S66t22 ¼ S66 sin u cos u sxx, where S11 ¼ 1 E11 S12 ¼ n12 E11 ¼ n21 E22 S22 ¼ 1 E22 S66 ¼ 1 G12 2007 by Taylor & Francis Group, LLC.
2.Using the maximum strain theory,failure of the lamina is predicted if the applied stress oxx exceeds the smallest of ELt (1) E11Su SL Su cos20+S12sin2 0 cos20-v12 sin20 cos20-v12 sin2 0 (2) STt E228Tt ST S12 cos20+S22 sin20 sin20-v2I Cos2 0 sin20-v21 cos20 (3) YLTs G12YLTs SLTs S66 sin0 cos sin cos0 sin0 cos The safe value of oxx for various fiber orientation angles is also shown in Figure 6.2.It can be seen that the maximum strain theory is similar to the maximum stress theory for 0 approaching 0.Both theories are operationally simple;however,no interaction between strengths in different directions is accounted for in either theory. EXAMPLE 6.2 A T-300 carbon fiber-reinforced epoxy lamina containing fibers at a +10 angle is subjected to the biaxial stress condition shown in the figure.The following material properties are known: 10,000psi 2 21 10=10° 20,000psi 2007 by Taylor Francis Group.LLC
2. Using the maximum strain theory, failure of the lamina is predicted if the applied stress sxx exceeds the smallest of (1) «Lt S11 cos2 u þ S12 sin2 u ¼ E11«Lt cos2 u n12 sin2 u ¼ SLt cos2 u n12 sin2 u (2) «Tt S12 cos2 u þ S22 sin2 u ¼ E22«Tt sin2 u n21 cos2 u ¼ STt sin2 u n21 cos2 u (3) gLTs S66 sin u cos u ¼ G12gLTs sin u cos u ¼ SLTs sin u cos u The safe value of sxx for various fiber orientation angles is also shown in Figure 6.2. It can be seen that the maximum strain theory is similar to the maximum stress theory for u approaching 08. Both theories are operationally simple; however, no interaction between strengths in different directions is accounted for in either theory. EXAMPLE 6.2 A T-300 carbon fiber-reinforced epoxy lamina containing fibers at a þ 108 angle is subjected to the biaxial stress condition shown in the figure. The following material properties are known: 10,000 psi 20,000 psi x 1 2 θ=10 y 2007 by Taylor & Francis Group, LLC.
E11t=Ei1e=21×10psi E221=1.4×106psi E22c=2×106psi G12=0.85×106psi 121=0.25 12c=0.52 a:=9,500μin./in. 8mt=5,100μin./in. c=11,000μin./in. eTc=14,000μin./in. hrs=22,000μin./in. Using the maximum strain theory,determine whether the lamina would fail. SOLUTION Step 1:Transform oxx and oyy into o1,o22,and 712. 11=20,000cos210°+(-10,000)sin210°=19,095.9psi, 022=20,000sin210°+(-10,000)cos210°=-9,095.9psi, T12=(-20,000-10,000)sin10°cos10°=-5,130psi. Step 2:Calculate su,822,and Yi2. L一vae Ere 81二E 22=1134.3×10-6in./im, 1+2=-4774.75×10-6in./in, 22=-12mE1i.E22 =T2=-6035.3×10-6in./im. y12=G2 Step 3:Compare u,822,and y12 with the respective ultimate strains to determine whether the lamina has failed.For the given stress system in this example problem, 811 8Lt, -8Te<822, -YLTs Y12. Thus,the lamina has not failed. 2007 by Taylor Francis Group,LLC
E11t ¼ E11c ¼ 21 106 psi E22t ¼ 1:4 106 psi E22c ¼ 2 106 psi G12 ¼ 0:85 106 psi n12t ¼ 0:25 n12c ¼ 0:52 «Lt ¼ 9,500 min:=in: «Tt ¼ 5,100 min:=in: «Lc ¼ 11,000 min:=in: «Tc ¼ 14,000 min:=in: gLTs ¼ 22,000 min:=in: Using the maximum strain theory, determine whether the lamina would fail. SOLUTION Step 1: Transform sxx and syy into s11, s22, and t12. s11 ¼ 20,000 cos2 10 þ (10,000) sin2 10 ¼ 19,095:9 psi, s22 ¼ 20,000 sin2 10 þ (10,000) cos2 10 ¼ 9,095:9 psi, t12 ¼ (20,000 10,000) sin 10 cos 10 ¼ 5,130 psi: Step 2: Calculate «11, «22, and g12. «11 ¼ s11 E11t n21c s22 E22c ¼ 1134:3 106 in:=in:, «22 ¼ n12t s11 E11t þ s22 E22c ¼ 4774:75 106 in:=in:, g12 ¼ t12 G12 ¼ 6035:3 106 in:=in: Step 3: Compare «11, «22, and g12 with the respective ultimate strains to determine whether the lamina has failed. For the given stress system in this example problem, «11 < «Lt, «Tc < «22, gLTs < g12: Thus, the lamina has not failed. 2007 by Taylor & Francis Group, LLC.
6.1.1.3 Azzi-Tsai-Hill Theory Following Hill's anisotropic yield criterion for metals,Azzi and Tsai [1] proposed that failure occurs in an orthotropic lamina if and when the following equality is satisfied: 011022 2+ 十 (6.3) where ou and o22 are both tensile (positive)stresses.When ou and o22 are compressive,the corresponding compressive strengths are used in Equation 6.3. For the uniaxial tensile stress situation considered earlier,failure is predicted if Uxx 2 cos40 sin2 6 cos26 sin6 sin26 cos20 1/2 Sia Sir SLE This equation,plotted in Figure 6.2,indicates a better match with experimental data than the maximum stress or the maximum strain theories. EXAMPLE 6.3 Determine and draw the failure envelope for a general orthotropic lamina using Azzi-Tsai-Hill theory. SOLUTION A failure envelope is a graphic representation of failure theory in the stress coordinate system and forms a boundary between the safe and unsafe design spaces.Selecting ou and o22 as the coordinate axes and rearranging Equation 6.3,we can represent the Azzi-Tsai-Hill failure theory by the following equations. 1.In the +o/+022 quadrant,both ou and o22 are tensile stresses.The corre- sponding strengths to consider are SL and Sr. -2+=1-五 S2S兄 S 2.In the +o/-22 quadrant,ou is tensile and o22 is compressive.The corres- ponding strengths to consider are SLt and Sre. +112+2=1-2 2007 by Taylor Francis Group.LLC
6.1.1.3 Azzi–Tsai–Hill Theory Following Hill’s anisotropic yield criterion for metals, Azzi and Tsai [1] proposed that failure occurs in an orthotropic lamina if and when the following equality is satisfied: s2 11 S2 Lt s11s22 S2 Lt þ s2 22 S2 Tt þ t2 12 S2 LTs ¼ 1, (6:3) where s11 and s22 are both tensile (positive) stresses. When s11 and s22 are compressive, the corresponding compressive strengths are used in Equation 6.3. For the uniaxial tensile stress situation considered earlier, failure is predicted if sxx 1 cos4 u S2 Lt sin2 u cos2 u S2 Lt þ sin4 u S2 Tt þ sin2 u cos2 u S2 Lts 1=2 : This equation, plotted in Figure 6.2, indicates a better match with experimental data than the maximum stress or the maximum strain theories. EXAMPLE 6.3 Determine and draw the failure envelope for a general orthotropic lamina using Azzi–Tsai–Hill theory. SOLUTION A failure envelope is a graphic representation of failure theory in the stress coordinate system and forms a boundary between the safe and unsafe design spaces. Selecting s11 and s22 as the coordinate axes and rearranging Equation 6.3, we can represent the Azzi–Tsai–Hill failure theory by the following equations. 1. In the þs11=þs22 quadrant, both s11 and s22 are tensile stresses. The corresponding strengths to consider are SLt and STt. s2 11 S2 Lt s11s22 S2 Lt þ s2 22 S2 Tt ¼ 1 t2 12 S2 LTs 2. In the þs11=s22 quadrant, s11 is tensile and s22 is compressive. The corresponding strengths to consider are SLt and STc. s2 11 S2 Lt þ s11s22 S2 Lt þ s2 22 S2 Tc ¼ 1 t2 12 S2 LTs 2007 by Taylor & Francis Group, LLC.
3.In the-o/+022 quadrant,ou is compressive and o22 is tensile.The corres- ponding strengths to consider are SLe and Sr. 2 +2+亭=1- S 4.In the-1/-022 quadrant,both ou and o22 are compressive stresses.The corresponding strengths to consider are SLe and STe. _2+2=1- 221 Increasing t12 011 A failure envelope based on these equations is drawn in the figure for various values of the Ti2/SLTs ratio.Note that,owing to the anisotropic strength charac- teristics of a fiber-reinforced composite lamina,the Azzi-Tsai-Hill failure envel- ope is not continuous in the stress space. 6.1.1.4 Tsai-Wu Failure Theory Under plane stress conditions,the Tsai-Wu failure theory [2]predicts failure in an orthotropic lamina if and when the following equality is satisfied: F1o11+F022+F6T12+F10+F22022+F66Ti2+2F2011022=1,(6.4) where F1,F2,and so on are called the strength coefficients and are given by 2007 by Taylor Francis Group,LLC
3. In the s11=þs22 quadrant, s11 is compressive and s22 is tensile. The corresponding strengths to consider are SLc and STt. s2 11 S2 Lc þ s11s22 S2 Lc þ s2 22 S2 Tt ¼ 1 t2 12 S2 LTs 4. In the s11=s22 quadrant, both s11 and s22 are compressive stresses. The corresponding strengths to consider are SLc and STc. s2 11 S2 Lc s11s22 S2 Lc þ s2 22 S2 Tc ¼ 1 t2 12 S2 LTs s22 s11 Increasing t12 A failure envelope based on these equations is drawn in the figure for various values of the t12=SLTs ratio. Note that, owing to the anisotropic strength characteristics of a fiber-reinforced composite lamina, the Azzi–Tsai–Hill failure envelope is not continuous in the stress space. 6.1.1.4 Tsai–Wu Failure Theory Under plane stress conditions, the Tsai–Wu failure theory [2] predicts failure in an orthotropic lamina if and when the following equality is satisfied: F1s11 þ F2s22 þ F6t12 þ F11s2 11 þ F22s2 22 þ F66t2 12 þ 2F12s11s22 ¼ 1, (6:4) where F1, F2, and so on are called the strength coefficients and are given by 2007 by Taylor & Francis Group, LLC.
