11 F= SL SLe F-7 11 ST STe F6=0 Fu=Su 1 F22STSTe F66≠ 1 and F12 is a strength interaction term between ou and o22.Note that F1,F2, F,F22,and F66 can be calculated using the tensile,compressive,and shear strength properties in the principal material directions.Determination of F12 requires a suitable biaxial test [3].For a simple example,consider an equal biaxial tension test in which o=12 o at failure.Using Equation 6.4,we can write (F+F2)o+(Fi1+F22+2F2)o2=1, from which F12= -(+加-（点+） Since reliable biaxial tests are not always easy to perform,an approximate range of values for F12 has been recommended [4]: -)WP≤F2≤0. 1 (6.5) In the absence ofexperimental data,the lower limit of Equation 6.5 is frequently used for F12. Figure 6.3 shows a comparison of the maximum strain theory,the Azzi-Tsai-Hill Theory,and the Tsai-Wu theory with a set of experimental data for a carbon fiber-epoxy lamina.The Tsai-Wu theory appears to fit the data best,which can be attributed to the presence of the strength interaction terms in Equation 6.4.Note that,for a given value of T12,the failure envelope defined by the Tsai-Wu failure theory is a continuous ellipse in the (on,o22) plane.The inclination of the ellipse in the on,o22 plane and the lengths of its semi-axes are controlled by the value of F12.The ellipse intercepts the ou axis at SLt and -SLe,and the o22 axis at Srt and -STe. 2007 by Taylor&Francis Group.LLC
F1 ¼ 1 SLt 1 SLc F2 ¼ 1 STt 1 STc F6 ¼ 0 F11 ¼ 1 SLtSLc F22 ¼ 1 STtSTc F66 ¼ 1 S2 LTs and F12 is a strength interaction term between s11 and s22. Note that F1, F2, F11, F22, and F66 can be calculated using the tensile, compressive, and shear strength properties in the principal material directions. Determination of F12 requires a suitable biaxial test [3]. For a simple example, consider an equal biaxial tension test in which s11 ¼ s12 ¼ s at failure. Using Equation 6.4, we can write (F1 þ F2) s þ (F11 þ F22 þ 2F12) s2 ¼ 1, from which F12 ¼ 1 2s2 1 1 SLt 1 SLc þ 1 STt 1 STc s 1 SLtSLc þ 1 STtSTc s2 : Since reliable biaxial tests are not always easy to perform, an approximate range of values for F12 has been recommended [4]: 1 2 (F11F22) 1=2 F12 0: (6:5) In the absence of experimental data, the lower limit of Equation 6.5 is frequently used for F12. Figure 6.3 shows a comparison of the maximum strain theory, the Azzi–Tsai–Hill Theory, and the Tsai–Wu theory with a set of experimental data for a carbon fiber–epoxy lamina. The Tsai–Wu theory appears to fit the data best, which can be attributed to the presence of the strength interaction terms in Equation 6.4. Note that, for a given value of t12, the failure envelope defined by the Tsai–Wu failure theory is a continuous ellipse in the (s11, s22) plane. The inclination of the ellipse in the s11, s22 plane and the lengths of its semi-axes are controlled by the value of F12. The ellipse intercepts the s11 axis at SLt and SLc, and the s22 axis at STt and STc. 2007 by Taylor & Francis Group, LLC.
22 150 boo -500 500 1000 (a) 221 -150 0 -500 500 1000 0 00 (b) 22 150 1000 dp0-D00e011 -500 500 0 00 ●●● (c) FIGURE 6.3 Comparison of (a)Tsai-Wu,(b)maximum strain,and(c)Azzi-Tsai-Hill failure theories with biaxial strength data of a carbon fiber-reinforced epoxy composite (note that the stresses are in MPa).(After Tsai,S.W.and Hahn,H.T.,Inelastic Behavior of Composite Materials,C.T.Herakovich,ed.,American Society of Mechanical Engin- eers,New York,1975.) EXAMPLE 6.4 Estimate the failure strength of a unidirectional lamina in an off-axis tension test using the Tsai-Wu theory.Assume that all strength coefficients for the lamina are known. SOLUTION An off-axis tension test on a unidirectional lamina is performed at a fiber orien- tation angle 6 with the loading axis.The stress state in the gage section of the 2007 by Taylor Francis Group,LLC
EXAMPLE 6.4 Estimate the failure strength of a unidirectional lamina in an off-axis tension test using the Tsai–Wu theory. Assume that all strength coefficients for the lamina are known. SOLUTION An off-axis tension test on a unidirectional lamina is performed at a fiber orientation angle u with the loading axis. The stress state in the gage section of the (a) (b) (c) 150 150 150 −500 −500 −500 500 1000 1000 1000 500 500 s22 s22 s22 s11 s11 s11 FIGURE 6.3 Comparison of (a) Tsai–Wu, (b) maximum strain, and (c) Azzi–Tsai–Hill failure theories with biaxial strength data of a carbon fiber-reinforced epoxy composite (note that the stresses are in MPa). (After Tsai, S.W. and Hahn, H.T., Inelastic Behavior of Composite Materials, C.T. Herakovich, ed., American Society of Mechanical Engineers, New York, 1975.) 2007 by Taylor & Francis Group, LLC
lamina is shown in the figure.The stress o in the loading direction creates the following stresses in the principal material directions: m1=0ucos20=50x1+c0s20, 022=0 xx sin20= 50x(1-cos20, T12=-Cxx sin 0 cos 0=-Oxx sin 20. At failure,x=Se,where Se denotes the failure strength in the off-axis tension test.Substituting for ou,22,and Ti2 in Equation 6.4 gives S[(3FI+3F22+2F2+F66)+4FI-F22)cos29+(F11+F22-2F12-F66) cos40]+4Sa(F+F2)+(F1-F2)cos20-8=0. This represents a quadratic equation of the form AS号+BS#+C=0, which can be solved to calculate the failure strength Se. 6.1.2 FAILURE PREDICTION FOR UNNOTCHED LAMINATES Failure prediction for a laminate requires knowledge of the stresses and strains in each lamina,which are calculated using the lamination theory described in Chapter 3.The individual lamina stresses or strains in the loading directions are transformed into stresses or strains in the principal material directions for each lamina,which are then used in an appropriate failure theory to check whether the lamina has failed.After a lamina fails,the stresses and strains in the 2007 by Taylor&Francis Group.LLC
lamina is shown in the figure. The stress sxx in the loading direction creates the following stresses in the principal material directions: q s11 ¼ sxx cos2 u ¼ 1 2 sxx(1 þ cos 2u), s22 ¼ sxx sin2 u ¼ 1 2 sxx(1 cos 2u), t12 ¼ sxx sin u cos u ¼ 1 2 sxx sin 2u: At failure, sxx ¼ Su, where Su denotes the failure strength in the off-axis tension test. Substituting for s11, s22, and t12 in Equation 6.4 gives S2 u[(3F11 þ 3F22 þ 2F12 þ F66) þ 4(F11 F22) cos 2u þ (F11 þ F22 2F12 F66) cos 4u] þ 4Su[(F1 þ F2) þ (F1 F2) cos 2u] 8 ¼ 0: This represents a quadratic equation of the form AS2 u þ BSu þ C ¼ 0, which can be solved to calculate the failure strength Su. 6.1.2 FAILURE PREDICTION FOR UNNOTCHED LAMINATES Failure prediction for a laminate requires knowledge of the stresses and strains in each lamina, which are calculated using the lamination theory described in Chapter 3. The individual lamina stresses or strains in the loading directions are transformed into stresses or strains in the principal material directions for each lamina, which are then used in an appropriate failure theory to check whether the lamina has failed. After a lamina fails, the stresses and strains in the 2007 by Taylor & Francis Group, LLC.
remaining laminas increase and the laminate stiffness is reduced;however,the laminate may not fail immediately.Furthermore,the failed lamina may not cease to carry its share of load in all directions. 6.1.2.1 Consequence of Lamina Failure Several methods have been proposed to account for the failed lamina and the subsequent behavior of the laminate [5].Among them are the following: Total discount method:In this method,zero stiffness and strength are assigned to the failed lamina in all directions. Limited discount method:In this method,zero stiffness and strength are assigned to the failed lamina for the transverse and shear modes if the lamina failure is in the matrix material.If the lamina fails by fiber rupture,the total discount method is adopted. Residual property method:In this method,residual strength and stiffness are assigned to the failed lamina. EXAMPLE 6.5 A quasi-isotropic [0/+45/90]s laminate made from T-300 carbon-epoxy is sub- jected to an in-plane normal load Nex per unit width.With increasing values of Nxx,failure occurs first in the 90 layers owing to transverse cracks. Determine the stiffness matrices before and after the first ply failure(FPF). Assume that each ply has a thickness to.Use the same material properties as in Example 3.6. (1)0°Lamina (2)+45°Lamina (3)-45°Lamina (4)90°Lamina Mid-plane (5)90°Lamina (6)-45°Lamina (7)+45°Lamina (8)0°Lamina +2 SOLUTIONS Referring to the figure,we observe hs=-ho=4to,=-=3to,h6=-h2=2to, hs=h3=to,and h4=0.In addition,note that (2mh=(2mns=(②m, (2nm2=(2nmn)h=(0mm)+45, (2m)3=(2nmt)6=(0mn）-45, (2nn)4=(2n)5=(2mm)90. 2007 by Taylor Francis Group,LLC
remaining laminas increase and the laminate stiffness is reduced; however, the laminate may not fail immediately. Furthermore, the failed lamina may not cease to carry its share of load in all directions. 6.1.2.1 Consequence of Lamina Failure Several methods have been proposed to account for the failed lamina and the subsequent behavior of the laminate [5]. Among them are the following: Total discount method: In this method, zero stiffness and strength are assigned to the failed lamina in all directions. Limited discount method: In this method, zero stiffness and strength are assigned to the failed lamina for the transverse and shear modes if the lamina failure is in the matrix material. If the lamina fails by fiber rupture, the total discount method is adopted. Residual property method: In this method, residual strength and stiffness are assigned to the failed lamina. EXAMPLE 6.5 A quasi-isotropic [0=±45=90]S laminate made from T-300 carbon–epoxy is subjected to an in-plane normal load Nxx per unit width. With increasing values of Nxx, failure occurs first in the 908 layers owing to transverse cracks. Determine the stiffness matrices before and after the first ply failure (FPF). Assume that each ply has a thickness t0. Use the same material properties as in Example 3.6. h0 (1) 0o Lamina (8) 0 Lamina (2) +45 Lamina (7) +45 Lamina (3) −45 Lamina (6) −45 Lamina (4) 90 Lamina Mid-plane +z (5) 90 Lamina h8 t0 SOLUTIONS Referring to the figure, we observe h8 ¼ h0 ¼ 4t0, h7 ¼ h1 ¼ 3t0, h6 ¼ h2 ¼ 2t0, h5 ¼ h3 ¼ t0, and h4¼ 0. In addition, note that (Q mn)1 ¼ (Q mn)8 ¼ (Q mn)0 , (Q mn)2 ¼ (Q mn)7 ¼ (Q mn)þ45 , (Q mn)3 ¼ (Q mn)6 ¼ (Q mn)45 , (Q mn)4 ¼ (Q mn)5 ¼ (Q mn)90 : 2007 by Taylor & Francis Group, LLC.
Since the given laminate is symmetric about the midplane,[B][0].For in-plane loads,we need to determine the elements in the [A]matrix. Am=∑（②mnh--） =21ol(2mnoe+(②m)+45+(2m)-45+(2mm0l. Note that A does not depend on the stacking sequence,since (h-h1)=to regardless of where the ith lamina is located. Before the 90 layers fail:From Example 3.6,we tabulate the values of various as follows.The unit of m is GPa. 0° +45 -45 90° O 134.03 40.11 40.11 8.82 012 2.29 33.61 33.61 2.29 016 0 31.30 -31.30 0 O 8.82 40.11 40.11 134.03 026 0 31.30 -31.30 0 06 3.254 34.57 34.57 3.254 Therefore, 「446.141o143.60t0 0 [A]before= 143.601o 446.141o 0 0 0 151.301o After the 90 layers fail: 1.Total discount method:For the failed 90 layers,we assume Ou=012 016=022=226=266=0. Therefore, 428.501o139.0210 0 [Alafter 139.0210178.0810 0 0 0 144.791o 2.Limited discount method:Since the 90 layers failed by transverse crack- ing,we assume O11 =012=016 =026=266=0. However,O22 134.03 GPa.Therefore, 428.501o139.021o 0 [A]after= 139.0210446.141o 0 0 144.7910J 2007 by Taylor Francis Group.LLC
Since the given laminate is symmetric about the midplane, [B] ¼ [0]. For in-plane loads, we need to determine the elements in the [A] matrix. Amn ¼ X8 j¼1 (Q mn)j(hj hj1) ¼ 2t0[(Q mn)0 þ (Q mn)þ45 þ (Q mn)45 þ (Qmn)90 ]: Note that Amn does not depend on the stacking sequence, since (hj hj1) ¼ t0 regardless of where the jth lamina is located. Before the 908 layers fail: From Example 3.6, we tabulate the values of various Q mn as follows. The unit of Q mn is GPa. 08 1458 458 908 Q 11 134.03 40.11 40.11 8.82 Q 12 2.29 33.61 33.61 2.29 Q 16 0 31.30 31.30 0 Q 22 8.82 40.11 40.11 134.03 Q 26 0 31.30 31.30 0 Q 66 3.254 34.57 34.57 3.254 Therefore, [A]before ¼ 446:14t0 143:60t0 0 143:60t0 446:14t0 0 0 0 151:30t0 2 4 3 5: After the 908 layers fail: 1. Total discount method: For the failed 908 layers, we assume Q 11 ¼ Q 12 ¼ Q 16 ¼ Q 22 ¼ Q 26 ¼ Q 66 ¼ 0. Therefore, [A]after ¼ 428:50t0 139:02t0 0 139:02t0 178:08t0 0 0 0 144:79t0 2 4 3 5 2. Limited discount method: Since the 908 layers failed by transverse cracking, we assume Q 11 ¼ Q 12 ¼ Q 16 ¼ Q 26 ¼ Q 66 ¼ 0: However, Q 22 ¼ 134:03 GPa. Therefore, [A]after ¼ 428:50t0 139:02t0 0 139:02t0 446:14t0 0 0 0 144:79t0 2 4 3 5 2007 by Taylor & Francis Group, LLC.