384 Mechanics of Composite Materials,Second Edition The Tsai-Wu failure theory applied to the top surface of the top 0 ply is applied as follows.The local stresses are o1=9.726×101Pa o2=1.313Pa T12=0 Using the parameters H,H2,Ho,Hu,H,He6,and Hi2 from Example 2.19, the Tsai-Wu failure theory Equation(2.152)gives the strength ratio as (0)(9.726×10)SR+(2.093×10-)(1.313)SR+(0×0)+ (4.4444×10-19)(9.726×102(SR)2+(1.0162×10-16)(1.313)2SR2 +(2.1626×1016)(0)2+2(-3.360×1018)(9.726×101)(1.313)SR)2=1 SR=1.339×107. The maximum strain failure theory can also be applied to the top surface of the top 0o ply as follows.The local strains are 5.353×10-10 E2 -2.297×10-11 Y12 0.000 Then,according to maximum strain failure theory(Equation 2.143),the strength ratio is given by SR=min{[(1500×10)/(181×10]/(5.353×10-10), [(246×10)/(10.3×10)]/(2.297×10-1)}=1.548×107. The strength ratios for all the plies in the laminate are summarized in Table 5.3 using the maximum strain and Tsai-Wu failure theories.The symbols in TABLE 5.3 Strength Ratios in Example 5.3 Ply no. Position Maximum strain Tsai-Wu 1(0) Top 1.548×10°(1T) 1.339×10 Middle 1.548×10(1T) 1.339×10 Bottom 1.548×10(1T) 1.339×10 2(90) Top 7.254×10(2T) 7.277×106 Middle 7.254×10(2T) 7.277×106 Bottom 7.254×10(2T) 7.277×10 3(0) Top 1.548×10°(1T) 1.339×10 Middle 1.548×10(1T) 1.339×10 Bottom 1.548×10(1T) 1.339×10 2006 by Taylor Francis Group,LLC
384 Mechanics of Composite Materials, Second Edition The Tsai–Wu failure theory applied to the top surface of the top 0° ply is applied as follows. The local stresses are σ1 = 9.726 × 101 Pa σ2 = 1.313 Pa τ12 = 0 Using the parameters H1, H2, H6, H11, H22, H66, and H12 from Example 2.19, the Tsai–Wu failure theory Equation (2.152) gives the strength ratio as (0) (9.726 × 101) SR + (2.093 × 10–8) (1.313) SR+ (0 × 0) + (4.4444 × 10–19) (9.726 × 101)2(SR)2 + (1.0162 × 10–16) (1.313)2(SR)2 + (2.1626 × 10–16) (0)2 + 2(–3.360 × 10–18) (9.726 × 101) (1.313)(SR)2=1 SR = 1.339 × 107. The maximum strain failure theory can also be applied to the top surface of the top 0° ply as follows. The local strains are . Then, according to maximum strain failure theory (Equation 2.143), the strength ratio is given by SR = min {[(1500 × 106)/(181 × 109)]/(5.353 × 10–10), [(246 × 106)/(10.3 × 109)]/(2.297 × 10–11)} = 1.548 × 107. The strength ratios for all the plies in the laminate are summarized in Table 5.3 using the maximum strain and Tsai–Wu failure theories. The symbols in TABLE 5.3 Strength Ratios in Example 5.3 Ply no. Position Maximum strain Tsai–Wu 1 (0°) Top Middle Bottom 1.548 × 107 (1T) 1.548 × 107 (1T) 1.548 × 107 (1T) 1.339 × 107 1.339 × 107 1.339 × 107 2 (90°) Top Middle Bottom 7.254 × 106 (2T) 7.254 × 106 (2T) 7.254 × 106 (2T) 7.277 × 106 7.277 × 106 7.277 × 106 3 (0°) Top Middle Bottom 1.548 × 107 (1T) 1.548 × 107 (1T) 1.548 × 107 (1T) 1.339 × 107 1.339 × 107 1.339 × 107 ε ε γ 1 2 12 10 11 5 353 10 2 297 10 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = × − × − − . . .000 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 1343_book.fm Page 384 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 385 the parentheses in the maximum strain failure theory column denote the mode of failure and are explained at the bottom of Table 2.3. From Table 5.3 and using the Tsai-Wu theory,the minimum strength ratio is found for the 90 ply.This strength ratio gives the maximum value of the allowable normal load as N,=7.277×105Y 11n and the maximum value of the allowable normal stress as N-7.277×10 h 0.015 =0.4851×109Pa where h=thickness of the laminate. The normal strain in the x-direction at this load is (eg)ply re(5.353×10r10)7.277×10) =3.895×10-3 Now,degrading the 90 ply completely involves assuming zero stiffnesses and strengths of the 90 lamina.Complete degradation of a ply does not allow further failure of that ply.For the undamaged plies,the [0/90],lami- nate has two reduced stiffness matrices as 181.8 2.897 0 [Q]= 2.897 10.35 0 GPa 0 0 7.17 and,for the damaged ply, 「000] [Q]=000GPa. 000 Using Equation(4.28a),the extensional stiffness matrix 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 385 the parentheses in the maximum strain failure theory column denote the mode of failure and are explained at the bottom of Table 2.3. From Table 5.3 and using the Tsai–Wu theory, the minimum strength ratio is found for the 90° ply. This strength ratio gives the maximum value of the allowable normal load as and the maximum value of the allowable normal stress as , where h = thickness of the laminate. The normal strain in the x-direction at this load is . Now, degrading the 90° ply completely involves assuming zero stiffnesses and strengths of the 90° lamina. Complete degradation of a ply does not allow further failure of that ply. For the undamaged plies, the laminate has two reduced stiffness matrices as and, for the damaged ply, . Using Equation (4.28a), the extensional stiffness matrix N N m x = × 7 277 106 . N h Pa x = × = × 7 277 10 0 015 0 4851 10 6 9 . . . ( ) ( . )( . εx 0 10 first ply failure 5 353 10 7 277 1 × × − 0 3 895 10 6 3 ) = × . − [ ] 0 90 / s [ ] . . . . . Q = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 GPa [ ] Q GPa = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 000 000 000 1343_book.fm Page 385 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
386 Mechanics of Composite Materials,Second Edition o,4--d Aj= k=1 181.8 2.897 [A]= 2.897 10.35 0 (10)0.005) 0 0 7.17 0 0 07 + 10 0 0(10°)0.005) 10 0 0 181.8 2.897 0 2.897 10.35 0 (10)0.005) 0 0 7.17 181.8 2.897 0 [A]= 2.897 10.35 0 (10)Pa-m. 0 0 7.17 Inverting the new extensional stiffness matrix [A],the new extensional compliance matrix is 5.525×10-10-1.547×10-10 0 [A']= -1.547×10-10 9.709×10-9 1 0 Pa-m 0 0 1.395×10-8 which gives midplane strains subjected to N,=1 N/m by Equation(5.1a) as e 5.525×10-10-1.547×10-10 0 -1.547×10-109.709×10-9 0 Yw 0 0 1.395×10-8 0 2006 by Taylor Francis Group,LLC
386 Mechanics of Composite Materials, Second Edition Inverting the new extensional stiffness matrix [A], the new extensional compliance matrix is , which gives midplane strains subjected to Nx = 1 N/m by Equation (5.1a) as A Q hh ij ij k k k k = − − = ∑[ ]( )1 1 3 [ ] . . . . . A = ( ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 10 0 005 9 )( . ) + ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 000 000 000 10 0 005 9 ( )( . ) + ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 . . . . . ( )( . ) 0 005 [ ] . . . . . A = ( ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 107 ) . Pa m- [ ] . . . . * A = × −× − × − − − 5 525 10 1 547 10 0 1 547 10 9 10 10 10 709 10 0 0 0 1 395 10 9 1 8 × × ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − − . Pa m- ε ε γ x y xy 0 0 10 5 525 10 1 547 10 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = × −× − − . . 10 10 9 8 0 1 547 10 9 709 10 0 0 0 1 395 10 −× × × ⎡ ⎣ ⎢ ⎢ − − − . . . ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 1 0 0 1343_book.fm Page 386 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 387 TABLE 5.4 Local Stresses after First Ply Failure in Example 5.3 Ply no. Position c 02 t2 1(0) Top 1.0000×102 0.0 0.0 Middle 1.0000×102 0.0 0.0 Bottom 1.0000×102 0.0 0.0 2(90) Top Middle Bottom 3(0) Top 1.0000×102 0.0 0.0 Middle 1.0000×102 0.0 0.0 Bottom 1.0000×102 0.0 0.0 TABLE 5.5 Local Strains after First Ply Failure in Example 5.3 Ply no. Position E1 E2 Ye 1(0) Top 5.25×1010 -1.547×10-10 0.0 Middle 5.525×10-10 -1.547×10-10 0.0 Bottom 5.525×10-10 -1.547×10-10 0.0 2(90) Top Middle - 一 Bottom 3(0) Top 5.525×10-10 -1.547×10-10 0.0 Middle 5.525×10-10 -1.547×10-10 0.0 Bottom 5.525×10-10 -1.547×10-10 0.0 5.525×10-10 -1.547×10-10 0 These strains are close to those obtained before the ply failure only because the 90 ply takes a small percentage of the load out of the normal load in the x-direction. The local stresses in each layer are found using earlier techniques given in this example and are shown in Table 5.4.The strength ratios in each layer are also found using methods given in this example and are shown in Table 5.5. From Table 5.6 and using Tsai-Wu failure theory,the minimum strength ratio is found in both the 0o plies.This strength ratio gives the maximum value of the normal load as N=1.5×107Y 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 387 . These strains are close to those obtained before the ply failure only because the 90° ply takes a small percentage of the load out of the normal load in the x-direction. The local stresses in each layer are found using earlier techniques given in this example and are shown in Table 5.4. The strength ratios in each layer are also found using methods given in this example and are shown in Table 5.5. From Table 5.6 and using Tsai–Wu failure theory, the minimum strength ratio is found in both the 0° plies. This strength ratio gives the maximum value of the normal load as TABLE 5.4 Local Stresses after First Ply Failure in Example 5.3 Ply no. Position σ1 σ2 τ12 1 (0°) Top Middle Bottom 1.0000 × 102 1.0000 × 102 1.0000 × 102 0.0 0.0 0.0 0.0 0.0 0.0 2 (90°) Top Middle Bottom — — — — — — — — — 3 (0°) Top Middle Bottom 1.0000 × 102 1.0000 × 102 1.0000 × 102 0.0 0.0 0.0 0.0 0.0 0.0 TABLE 5.5 Local Strains after First Ply Failure in Example 5.3 Ply no. Position ε1 ε2 γ12 1 (0°) Top Middle Bottom 5.25 × 10–10 5.525 × 10–10 5.525 × 10–10 –1.547 × 10–10 –1.547 × 10–10 –1.547 × 10–10 0.0 0.0 0.0 2 (90°) Top Middle Bottom — — — — — — — — — 3 (0°) Top Middle Bottom 5.525 × 10–10 5.525 × 10–10 5.525 × 10–10 –1.547 × 10–10 –1.547 × 10–10 –1.547 × 10–10 0.0 0.0 0.0 ε ε γ x y xy 0 0 0 10 5 525 10 1 547 10 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = × − × − . . − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 10 0 N N m x = × 1 5 107 . 1343_book.fm Page 387 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
388 Mechanics of Composite Materials,Second Edition TABLE 5.6 Strength Ratios after First Ply Failure in Example 5.3 Ply no. Position Max strain Tsai-Wu 1(0) Top 1.5000×10㎡(1T) 1.5000×10 Middle 1.5000×101T) 1.5000×107 Bottom 1.5000×10(1T) 1.5000×107 2(90) Top Middle 二 - Bottom 3(0) Top 1.5000×101T) 1.5000×109 Middle 1.5000×10(1T) 1.5000×107 Bottom 1.5000×10(1T) 1.5000×10 and the maximum value of the allowable normal stress as N=1.5×10 h0.015 =1.0×109Pa where h is the thickness of the laminate The normal strain in the x-direction at this load is (e)ap吵aure=(5.525×10-10)(1.5×10) =8.288×10-3 The preceding load is also the last ply failure(LPF)because none of the layers is left undamaged.Plotting the stress vs.strain curve for the laminate until last ply failure shows that the curve will consist of two linear curves, each ending at each ply failure.The slope of the two lines will be the Young's modulus inx direction for the undamaged laminate and for the FPF laminate that is,using Equation (4.35), 1 E.=0.0155.353×100), =124.5GPa until first ply failure,and 2006 by Taylor Francis Group,LLC
388 Mechanics of Composite Materials, Second Edition and the maximum value of the allowable normal stress as , where h is the thickness of the laminate. The normal strain in the x-direction at this load is . The preceding load is also the last ply failure (LPF) because none of the layers is left undamaged. Plotting the stress vs. strain curve for the laminate until last ply failure shows that the curve will consist of two linear curves, each ending at each ply failure. The slope of the two lines will be the Young’s modulus in x direction for the undamaged laminate and for the FPF laminate — that is, using Equation (4.35), , until first ply failure, and TABLE 5.6 Strength Ratios after First Ply Failure in Example 5.3 Ply no. Position Max strain Tsai–Wu 1 (0°) Top Middle Bottom 1.5000 × 107 (1T) 1.5000 × 107(1T) 1.5000 × 107(1T) 1.5000 × 107 1.5000 × 107 1.5000 × 107 2 (90°) Top Middle Bottom — — — — — — 3 (0°) Top Middle Bottom 1.5000 × 107(1T) 1.5000 × 107(1T) 1.5000 × 107(1T) 1.5000 × 107 1.5000 × 107 1.5000 × 107 N h Pa x = × = × 1 5 10 0 015 1 0 10 7 9 . . . ( ) ( . )( . εx o last ply failure =× × − 5 525 10 1 5 10 10 7 ) = × . − 8 288 10 3 E GPa x = × = − 1 0 015 5 353 10 124 5 10 ( . )( . ) . 1343_book.fm Page 388 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC