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.