5 Failure,Analysis,and Design of Laminates Chapter Objectives Understand the significance of stiffness,and hygrothermal and mechanical response of special cases of laminates. Establish the failure criteria for laminates based on failure of indi- vidual lamina in a laminate. Design laminated structures such as plates,thin pressure vessels, and drive shafts subjected to in-plane and hygrothermal loads. Introduce other mechanical design issues in laminated composites. 5.1 Introduction The design of a laminated composite structure,such as a flat floor panel or a pressure vessel,starts with the building block of laminae,in which fiber and matrix are combined in a manufacturing process such as filament wind- ing or prepregs.The material of the fiber and matrix,processing factors such as packing arrangements,and fiber volume fraction determine the stiffness, strength,and hygrothermal response of a single lamina.These properties can be found by using the properties of the individual constituents of the lamina or by experiments,as explained in Chapter 3.Then the laminate can have variations in material systems and in stacking sequence of plies to tailor a composite for a particular application. In Chapter 4,we developed analysis to find the stresses and strains in a laminate under in-plane and hygrothermal loads.In this chapter,we will first use that analysis and failure theories studied in Chapter 2 to predict failure in a laminate.Then the fundamentals learned in Chapter 4 and the failure analysis discussed in this chapter will be used to design structures using laminated composites. 369 2006 by Taylor Francis Group,LLC
369 5 Failure, Analysis, and Design of Laminates Chapter Objectives • Understand the significance of stiffness, and hygrothermal and mechanical response of special cases of laminates. • Establish the failure criteria for laminates based on failure of individual lamina in a laminate. • Design laminated structures such as plates, thin pressure vessels, and drive shafts subjected to in-plane and hygrothermal loads. • Introduce other mechanical design issues in laminated composites. 5.1 Introduction The design of a laminated composite structure, such as a flat floor panel or a pressure vessel, starts with the building block of laminae, in which fiber and matrix are combined in a manufacturing process such as filament winding or prepregs. The material of the fiber and matrix, processing factors such as packing arrangements, and fiber volume fraction determine the stiffness, strength, and hygrothermal response of a single lamina. These properties can be found by using the properties of the individual constituents of the lamina or by experiments, as explained in Chapter 3. Then the laminate can have variations in material systems and in stacking sequence of plies to tailor a composite for a particular application. In Chapter 4, we developed analysis to find the stresses and strains in a laminate under in-plane and hygrothermal loads. In this chapter, we will first use that analysis and failure theories studied in Chapter 2 to predict failure in a laminate. Then the fundamentals learned in Chapter 4 and the failure analysis discussed in this chapter will be used to design structures using laminated composites. 1343_book.fm Page 369 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
370 Mechanics of Composite Materials,Second Edition First,special cases of laminates that are important in the design of laminated structures will be introduced.Then the failure criterion analysis will be shown for a laminate.Eventually,we will be designing laminates mainly on the basis of optimizing for cost,weight,strength,and stiffness. Other mechanical design issues are briefly introduced at the end of the chapter. 5.2 Special Cases of Laminates Based on angle,material,and thickness of plies,the symmetry or antisym- metry of a laminate may zero out some elements of the three stiffness matri- ces [A],[B],and [D].These are important to study because they may result in reducing or zeroing out the coupling of forces and bending moments, normal and shear forces,or bending and twisting moments.This not only simplifies the mechanical analysis of composites,but also gives desired mechanical performance.For example,as already shown in Chapter 4,the analysis of a symmetric laminate is simplified due to the zero coupling matrix [B].Mechanically,symmetric laminates result in no warpage in a flat panel due to temperature changes in processing. 5.2.1 Symmetric Laminates A laminate is called symmetric if the material,angle,and thickness of plies are the same above and below the midplane.An example of symmetric laminates is [0/30/60]: 0 30 60 30 0 For symmetric laminates from the definition of [B]matrix,it can be proved that [B]=0.Thus,Equation(4.29)can be decoupled to give N A12 A16 A12 A26 (5.1a) A16 A26 A6 2006 by Taylor Francis Group,LLC
370 Mechanics of Composite Materials, Second Edition First, special cases of laminates that are important in the design of laminated structures will be introduced. Then the failure criterion analysis will be shown for a laminate. Eventually, we will be designing laminates mainly on the basis of optimizing for cost, weight, strength, and stiffness. Other mechanical design issues are briefly introduced at the end of the chapter. 5.2 Special Cases of Laminates Based on angle, material, and thickness of plies, the symmetry or antisymmetry of a laminate may zero out some elements of the three stiffness matrices [A], [B], and [D]. These are important to study because they may result in reducing or zeroing out the coupling of forces and bending moments, normal and shear forces, or bending and twisting moments. This not only simplifies the mechanical analysis of composites, but also gives desired mechanical performance. For example, as already shown in Chapter 4, the analysis of a symmetric laminate is simplified due to the zero coupling matrix [B]. Mechanically, symmetric laminates result in no warpage in a flat panel due to temperature changes in processing. 5.2.1 Symmetric Laminates A laminate is called symmetric if the material, angle, and thickness of plies are the same above and below the midplane. An example of symmetric laminates is : For symmetric laminates from the definition of [B] matrix, it can be proved that [B] = 0. Thus, Equation (4.29) can be decoupled to give (5.1a) 0 30 60 30 0 [ /0 30/60]s N N N A A A A A A A A x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 11 12 16 12 22 26 16 26 66 0 0 A 0 x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ε ε γ 1343_book.fm Page 370 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 371 D12 D16 K My D12 D2 D26 (5.1b) D16 D26 D66」 This shows that the force and moment terms are uncoupled.Thus,if a laminate is subjected only to forces,it will have zero midplane curvatures. Similarly,if it is subjected only to moments,it will have zero midplane strains. The uncoupling between extension and bending in symmetric laminates makes analyzing such laminates simpler.It also prevents a laminate from twisting due to thermal loads,such as cooling down from processing temper- atures and temperature fluctuations during use such as in a space shuttle,etc. 5.2.2 Cross-Ply Laminates A laminate is called a cross-ply laminate(also called laminates with specially orthotropic layers)if only 0 and 90 plies were used to make a laminate.An example of a cross ply laminate is a [0/902/0/90]laminate: 0 90 90 0 90 For cross-ply laminates,A16=0,A26=0,Bi6=0,B26=0,D16=0,and D26 =0;thus,Equation(4.29)can be written as Nx An A12 0 B11 B12 0 Ny A An 0 B12 B22 0 e a 0 0 A66 0 0 Bos Ya (5.2) M B11 B12 0 D11 D12 0 Kx M, Bxz B 0 D12 Da 0 My」 0 0 Bos 0 0 Do K到 In these cases,uncoupling occurs between the normal and shear forces, as well as between the bending and twisting moments.If a cross-ply lami- nate is also symmetric,then in addition to the preceding uncoupling,the coupling matrix [B]=0 and no coupling takes place between the force and moment terms. 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 371 . (5.1b) This shows that the force and moment terms are uncoupled. Thus, if a laminate is subjected only to forces, it will have zero midplane curvatures. Similarly, if it is subjected only to moments, it will have zero midplane strains. The uncoupling between extension and bending in symmetric laminates makes analyzing such laminates simpler. It also prevents a laminate from twisting due to thermal loads, such as cooling down from processing temperatures and temperature fluctuations during use such as in a space shuttle, etc. 5.2.2 Cross-Ply Laminates A laminate is called a cross-ply laminate (also called laminates with specially orthotropic layers) if only 0 and 90° plies were used to make a laminate. An example of a cross ply laminate is a [0/902/0/90] laminate: For cross-ply laminates, A16 = 0, A26 = 0, B16 = 0, B26 = 0, D16 = 0, and D26 = 0; thus, Equation (4.29) can be written as . (5.2) In these cases, uncoupling occurs between the normal and shear forces, as well as between the bending and twisting moments. If a cross-ply laminate is also symmetric, then in addition to the preceding uncoupling, the coupling matrix [B] = 0 and no coupling takes place between the force and moment terms. 0 90 90 0 90 M M M D D D D D D D D x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 11 12 16 12 22 26 16 26 6 D 6 x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ κ κ κ N N N M M M x A A B y xy x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = 11 12 0 11 12 12 22 12 22 66 66 11 12 11 1 0 0 0 0 0 0 0 0 B A A B B A B B B D D 2 12 22 12 22 66 66 0 0 0 0 0 0 0 B B D D B D ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ε ε γ κ κ κ x y xy x y xy 0 0 0 ⎥ 1343_book.fm Page 371 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
372 Mechanics of Composite Materials,Second Edition 5.2.3 Angle Ply Laminates A laminate is called an angle ply laminate if it has plies of the same material and thickness and only oriented at +0 and-0 directions.An example of an angle ply laminate is [-40/40/-40/40]: -40 40 -40 40 If a laminate has an even number of plies,then A=A26=0.However,if the number of plies is odd and it consists of alternating +0 and-0 plies,then it is symmetric,giving [B]=0,and A1e,A26,Di6 and D26 also become small as the number of layers increases for the same laminate thickness.This behavior is similar to the symmetric cross-ply laminates.However,these angle ply laminates have higher shear stiffness and shear strength properties than cross-ply laminates. 5.2.4 Antisymmetric Laminates A laminate is called antisymmetric if the material and thickness of the plies are the same above and below the midplane,but the ply orientations at the same distance above and below the midplane are negative of each other.An example of an antisymmetric laminate is: 45 60 -60 -45 From Equation (4.28a)and Equation (4.28c),the coupling terms of the extensional stiffness matrix,Ai6=A26=0,and the coupling terms of the bending stiffness matrix,D16=D26=0: A12 0 B11 B12 B16 e A12 A2 B12 B2 B26 到 0 0 A66 B16 B26 Bos Y (5.3) M B11 B2 B16 D11 D12 0 K M, B12 B2 B26 D12 D2 0 Ky B16 B26 Bos 0 0 Doe 2006 by Taylor Francis Group,LLC
372 Mechanics of Composite Materials, Second Edition 5.2.3 Angle Ply Laminates A laminate is called an angle ply laminate if it has plies of the same material and thickness and only oriented at +θ and –θ directions. An example of an angle ply laminate is [–40/40/–40/40]: If a laminate has an even number of plies, then A16 = A26 = 0. However, if the number of plies is odd and it consists of alternating +θ and –θ plies, then it is symmetric, giving [B] = 0, and A16, A26, D16, and D26 also become small as the number of layers increases for the same laminate thickness. This behavior is similar to the symmetric cross-ply laminates. However, these angle ply laminates have higher shear stiffness and shear strength properties than cross-ply laminates. 5.2.4 Antisymmetric Laminates A laminate is called antisymmetric if the material and thickness of the plies are the same above and below the midplane, but the ply orientations at the same distance above and below the midplane are negative of each other. An example of an antisymmetric laminate is: From Equation (4.28a) and Equation (4.28c), the coupling terms of the extensional stiffness matrix, A16 = A26 =0, and the coupling terms of the bending stiffness matrix, D16 = D26 = 0: . (5.3) –40 40 –40 40 45 60 –60 –45 N N N M M M x A A B y xy x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = 11 12 0 11 12 16 12 22 12 22 26 66 16 26 66 11 0 0 0 B B A A B B B A B B B B B22 16 11 12 12 22 26 12 22 16 26 66 66 0 0 0 0 B D D B B B D D B B B D ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ε ε γ κ κ κ x y xy x y xy 0 0 0 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1343_book.fm Page 372 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 373 5.2.5 Balanced Laminate A laminate is balanced if layers at angles other than 0 and 90 occur only as plus and minus pairs of +0 and-0.The plus and minus pairs do not need to be adjacent to each other,but the thickness and material of the plus and minus pairs need to be the same.Here,the terms A16=A26=0.An example of a balanced laminate is [30/40/-30/30/-30/-40]: 30 40 -30 30 -30 -40 From Equation(4.28a), Nx An A12 0 B B12 B16 e Ny An An 0 B12 B22 B26 Na 0 0 A26 B16 B26 B66 M (5.4) Bu B12 B16 D Dv2 D16 Kx B1 B22 B26 D2 D2 D Ky May B16 B26 Bes Di6 D26 D66 5.2.6 Quasi-Isotropic Laminates For a plate of isotropic material with Young's modulus,E,Poisson's ratio, v,and thickness,h,the three stiffness matrices are E VE 1-v2 1-v2 [A]= vE E 1-v 1-v2 0 h, (5.5) E 0 0 2(1+v) [B= 0 0 (5.6) 0 0 0 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 373 5.2.5 Balanced Laminate A laminate is balanced if layers at angles other than 0 and 90° occur only as plus and minus pairs of +θ and –θ. The plus and minus pairs do not need to be adjacent to each other, but the thickness and material of the plus and minus pairs need to be the same. Here, the terms A16 = A26 = 0. An example of a balanced laminate is [30/40/–30/30/–30/–40]: From Equation (4.28a), . (5.4) 5.2.6 Quasi-Isotropic Laminates For a plate of isotropic material with Young’s modulus, E, Poisson’s ratio, ν, and thickness, h, the three stiffness matrices are , (5.5) , (5.6) 30 40 –30 30 –30 –40 N N N M M M x A A B y xy x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = 11 12 0 11 12 16 12 22 12 22 26 26 16 26 66 11 0 0 0 B B A A B B B A B B B B B12 16 11 12 16 12 22 26 12 22 26 16 26 66 B D D D B B B D D D B B B D16 26 66 0 0 0 D D x y xy x y ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ε ε γ κ κ κxy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ [ ] ( ) A E E E E E = − − − − + ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 1 1 0 1 1 0 0 0 2 1 2 2 2 2 ν ν ν ν ν ν ν ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ h [ ] B = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 000 000 000 1343_book.fm Page 373 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
374 Mechanics of Composite Materials,Second Edition E vE 0 12(1-v2) 12(1-v2) [D]= VE E 0 . 12(1-v2) (5.7) 12(1-v2) E 0 0 24(1+v) A laminate is called quasi-isotropic if its extensional stiffness matrix [A] behaves like that of an isotropic material.This implies not only that Au=Azz AA,and but also that these stiffnesses are indepen- 2 dent of the angle of rotation of the laminate.The reason for calling such a laminate quasi-isotropic and not isotropic is that the other stiffness matrices, [B]and [D],may not behave like isotropic materials.Examples of quasi- isotropic laminates include [0/+60],[0/+45/90],and [0/36/72/-36/-72]. Example 5.1 A [0/+60]graphite/epoxy laminate is quasi-isotropic.Find the three stiffness matrices [A],[B],and [D]and show that 1.A1=A2iA6=A6=0;A6=A1A2 2.[Bl≠0，unlike isotropic materials.. 3.[D]matrix is unlike isotropic materials. Use properties of unidirectional graphite/epoxy lamina from Table 2.1.Each lamina has a thickness of 5 mm. Solution From Example 2.6,the reduced stiffness matrix [Q]for the 0graphite/epoxy lamina is 181.8 2.897 0 [Q]= 2.897 10.35 0 (10)Pa. 0 0 7.17 From Equation(2.104),the transformed reduced stiffness matrices for the three plies are 2006 by Taylor Francis Group,LLC
374 Mechanics of Composite Materials, Second Edition . (5.7) A laminate is called quasi-isotropic if its extensional stiffness matrix [A] behaves like that of an isotropic material. This implies not only that A11 = A22, A16= A26 = 0, and , but also that these stiffnesses are independent of the angle of rotation of the laminate. The reason for calling such a laminate quasi-isotropic and not isotropic is that the other stiffness matrices, [B] and [D], may not behave like isotropic materials. Examples of quasiisotropic laminates include [0/±60], [0/±45/90]s, and [0/36/72/–36/–72]. Example 5.1 A [0/±60] graphite/epoxy laminate is quasi-isotropic. Find the three stiffness matrices [A], [B], and [D] and show that 1. . 2. [B] ≠ 0, unlike isotropic materials. 3. [D] matrix is unlike isotropic materials. Use properties of unidirectional graphite/epoxy lamina from Table 2.1. Each lamina has a thickness of 5 mm. Solution From Example 2.6, the reduced stiffness matrix [Q] for the 0° graphite/epoxy lamina is . From Equation (2.104), the transformed reduced stiffness matrices for the three plies are [ ] ( ) ( ) ( ) ( ) D E E E E = − − − − 12 1 12 1 0 12 1 12 1 2 2 2 2 ν ν ν ν ν ν 0 0 0 24 1 3 E h ( ) + ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ν A A A 66 11 12 2 = − A A A A A A A 11 22 16 26 66 11 12 0 2 = = = = − ; ; [ ] . . . . . Q = ( ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 ) Pa 1343_book.fm Page 374 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 375 [181.8 2.897 0 [顶= 2.897 10.35 0 (10)Pa, 0 0 7.17 23.65 32.46 20.05 [Ql4o= 32.46 109.4 54.19 (10)Pa, 20.05 54.19 36.74 23.65 32.46 -20.05 [@0= 32.46 109.4 -54.1910)Pa. -20.05 -54.19 36.74 The total thickness of the laminate is h=(0.005)(3)=0.015 m. The midplane is 0.0075 m from the top and bottom of the laminate.Thus, using Equation(4.20), h0=-0.0075m h1=-0.0025m h2=0.0025m h3=0.0075m Using Equation (4.28a)to Equation (4.28c),one can now calculate the stiffness matrices [A],[B],and [D],respectively,as shown in Example 4.2: 1.146 0.3391 0 [A]= 0.3391 1.146 0 (10)Pa-1, 0 0 0.4032 -3.954 0.7391 -0.5013 [B]= 0.7391 2.476 -1.355 (10)Pa-m2, -0.5013 -1.355 0.7391 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 375 , , The total thickness of the laminate is h = (0.005)(3) = 0.015 m. The midplane is 0.0075 m from the top and bottom of the laminate. Thus, using Equation (4.20), h0 = –0.0075 m h1 = –0.0025 m h2 = 0.0025 m h3 = 0.0075 m Using Equation (4.28a) to Equation (4.28c), one can now calculate the stiffness matrices [A], [B], and [D], respectively, as shown in Example 4.2: , , [ ] . . . . . Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ( ) 109 Pa [ ] . . . . . . . Q 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 = . . ( ) 19 36 74 109 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Pa [ ] . . . . . . . Q − = − − − 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 19 36 74 109 − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ . . ( ) Pa. [ ] . . . . . A = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ 1 146 0 3391 0 0 3391 1 146 0 0 0 0 4032 ⎥ ⎥ ⎥ ( ) 109 Pa-m [ ] . . . . . . . B = − − − − 3 954 0 7391 0 5013 0 7391 2 476 1 355 0 5013 1 355 0 7391 106 2 − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ . . ( ) Pa-m 1343_book.fm Page 375 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
376 Mechanics of Composite Materials,Second Edition 28.07 5.126 -2.507 [D]= 5.126 17.35 -6.774103)Pa-m3. -2.507 6.774 6.328 1.From the extensional stiffness matrix [A], A1=A22=1.146×109Pa-m A16=A26=0 A1-A2=1.146-0.3391×10 2 2 =0.4032×109Pa-m =A66 This behavior is similar to that of an isotropic material.However, a quasi-isotropic laminate should give the same [A]matrix,if a constant angle is added to each of the layers of the laminate.For example,adding 30 to each ply angle of the [0/+60]laminate gives a [30/90/-30]laminate,which has the same [A]matrix as the [0/ ±60]laminate. 2.Unlike isotropic materials,the coupling stiffness matrix [B]of the [0/+60]laminate is nonzero. 3.In an isotropic material, D1=D22, D16=D26=0, and Do=Du-Dg. 2 In this example,unlike isotropic materials,Du D22 because D11=28.07×103Pa-m3 2006 by Taylor Francis Group,LLC
376 Mechanics of Composite Materials, Second Edition . 1. From the extensional stiffness matrix [A], = 0.4032 × 109 Pa-m = A66. This behavior is similar to that of an isotropic material. However, a quasi-isotropic laminate should give the same [A] matrix, if a constant angle is added to each of the layers of the laminate. For example, adding 30° to each ply angle of the [0/±60] laminate gives a [30/90/–30] laminate, which has the same [A] matrix as the [0/ ±60] laminate. 2. Unlike isotropic materials, the coupling stiffness matrix [B] of the [0/±60] laminate is nonzero. 3. In an isotropic material, , , and . In this example, unlike isotropic materials, D11 ≠ D22 because D11 = 28.07 × 103 Pa-m3 [ ] .. . . .. . D = − − − − 28 07 5 126 2 507 5 126 17 35 6 774 2 507 6 774 6 328 103 3 . . ( ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Pa mA A Pa m 11 22 9 == × 1 146 10 . - A A 16 26 = = 0 A A 11 12 9 2 1 146 0 3391 2 10 − = − × . . D D 11 22 = D D 16 26 = = 0 D D D 66 11 12 2 = − 1343_book.fm Page 376 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 377 D22=17.35×103Pa-m3 D16≠0，D26≠0as D16=-2.507×103Pa-m3 D26=-6.774×103Pa-m3 D1-D2≠D6 2 because D1-D2=28.07×103-5.126×103 2 =11.47×103Pa-m3 D66=6.328×103Pa-m3. One can make a quasi-isotropic laminate by having a laminate with N(N >3)lamina of the same material and thickness,where each lamina is oriented at an angle of 180/N between each other. For example,a three-ply laminate will require the laminae to be oriented at 180/3=60 to each other.Thus,[0/60/-60],[30/90/ -30],and [45/-75/-15]are all quasi-isotropic laminates.One can make the preceding combinations symmetric or repeated to give quasi-isotropic laminates,such as [0/+60],[0/+60]s,and [0/+60]2 laminates.The symmetry of the laminates zeros out the coupling matrix [B]and makes its behavior closer (not same)to that of an isotropic material. Example 5.2 Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by U 0 [A]= U u, 0 h (5.8) 0 0 u1-U4 2 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 377 D22 = 17.35 × 103 Pa-m3 D16 ≠ 0, D26 ≠ 0 as D16 = –2.507 × 103 Pa-m3 D26 = –6.774 × 103 Pa-m3 because = 11.47 × 103 Pa-m3 D66 = 6.328 × 103 Pa-m3. One can make a quasi-isotropic laminate by having a laminate with N (N ≥ 3) lamina of the same material and thickness, where each lamina is oriented at an angle of 180°/N between each other. For example, a three-ply laminate will require the laminae to be oriented at 180°/3 = 60° to each other. Thus, [0/60/–60], [30/90/ –30], and [45/–75/–15] are all quasi-isotropic laminates. One can make the preceding combinations symmetric or repeated to give quasi-isotropic laminates, such as [0/±60]s, [0/±60]s, and [0/±60]2s laminates. The symmetry of the laminates zeros out the coupling matrix [B] and makes its behavior closer (not same) to that of an isotropic material. Example 5.2 Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by . (5.8) D D D 11 12 66 2 − ≠ D D 11 12 3 3 2 28 07 10 5 126 10 2 − = . . ×− × [ ] A U U U U U U = h − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ 1 4 4 1 1 4 0 0 0 0 2 1343_book.fm Page 377 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
378 Mechanics of Composite Materials,Second Edition where U and U are the stiffness invariants given by Equation(2.132)and h is the thickness of the laminate.Also,find the in-plane engineering stiffness constants of the laminate. Solution From Equation(2.131a),for a general angle ply with angle 0, u=L U2 Cos20 Us Cos40. (5.9) For the kth ply of the quasi-isotropic laminate with an angle0 (Q)=U U2 Cos20g+Us Cos40x (5.10) where 0=5=8=贤-N1,s= 2π From Equation (4.28a), N A1= (5.11) k=1 where f=thickness of kth lamina. Because the thickness of the laminate is h and all laminae are of the same thickness, Nk=1,2,,N, k= (5.12) and,substituting Equation(5.10)in Equation(5.11), A=02u+u,cos29+山，cas46) k=】 (5.13) =hu,+u0∑cos29+u0∑cos40 2006 by Taylor Francis Group,LLC
378 Mechanics of Composite Materials, Second Edition where U1 and U4 are the stiffness invariants given by Equation (2.132) and h is the thickness of the laminate. Also, find the in-plane engineering stiffness constants of the laminate. Solution From Equation (2.131a), for a general angle ply with angle θ, = U1 + U2 Cos2θ + U3 Cos4θ. (5.9) For the kth ply of the quasi-isotropic laminate with an angle θk, = U1 + U2 Cos2θk + U3 Cos4θk, (5.10) where From Equation (4.28a), , (5.11) where tk = thickness of kth lamina. Because the thickness of the laminate is h and all laminae are of the same thickness, (5.12) and, substituting Equation (5.10) in Equation (5.11), (5.13) Q11 ( ) Q11 k θ π θ π θ π θ π 12 1 θ π 2 1 = = …= … = − − = N N k N N N , ,, ,, kN N ( ) , . A tQk k k N 11 11 1 = = ∑ ( ) t h N k = = , , ,............, , k N 1 2 A h N UU U hU U k k k N 11 1 2 3 1 1 =+ + 2 4 = + = ∑( ) Cos Cos θ θ 2 3 1 1 2 4 h N U h N k k N k k N Cos Cos . θ θ + = = ∑ ∑ 1343_book.fm Page 378 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC