86 Mechanics of Composite Materials,Second Edition ↑3 G D H: G 0 H 444 E B FIGURE 2.14 Deformation of a cubic element made of orthotropic material. Demonstrating the meaning of elastic symmetry for an orthotropic mate- rial is similar to the approach taken for a monoclinic material(Section 2.3.2). Consider a cubic element(Figure 2.14)taken out of the orthotropic material, where 1,2,and 3 are the principal directions or 1-2,2-3,and 3-1 are the three mutually orthogonal planes of symmetry.Apply a normal stress,o3, to the element.Then,using the Hooke's law Equation(2.26)and the com- pliance matrix(Equation 2.39)for the orthotropic material,one gets e1=S1303 e2=S2303 2006 by Taylor Francis Group,LLC
86 Mechanics of Composite Materials, Second Edition Demonstrating the meaning of elastic symmetry for an orthotropic material is similar to the approach taken for a monoclinic material (Section 2.3.2). Consider a cubic element (Figure 2.14) taken out of the orthotropic material, where 1, 2, and 3 are the principal directions or 1–2, 2–3, and 3–1 are the three mutually orthogonal planes of symmetry. Apply a normal stress, σ3, to the element. Then, using the Hooke’s law Equation (2.26) and the compliance matrix (Equation 2.39) for the orthotropic material, one gets FIGURE 2.14 Deformation of a cubic element made of orthotropic material. G σ3 σ3 D A H F C 3 2 1 E B G′ D′ F′ C′ E′ B′ H′ A′ ε σ 1 13 3 = S ε σ 2 23 3 = S 1343_book.fm Page 86 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 87 e3=S3303 Y23=0 (2.40a-f) Y31=0 Y12=0. The cube will deform in all directions as determined by the normal strain equations.However,the shear strains in all three planes(1-2,2-3,and 3-1) are zero,showing that the element will not change shape in those planes. Thus,the cube will not deform in shape under any normal load applied in the principal directions.This is unlike the monoclinic material,in which two out of the six faces of the cube changed shape. A cube made of isotropic material would not change its shape either; however,the normal strains,E and Ez,will be different in an orthotropic material and identical in an isotropic material. 2.3.4 Transversely Isotropic Material Consider a plane of material isotropy in one of the planes of an orthotropic body.If direction 1 is normal to that plane(2-3)of isotropy,then the stiffness matrix is given by Cn C12 C12 0 0 C12 Cz C23 0 0 0 C12 C23 C2 0 0 0 [C]= 0 0 0 C22-C23 0 0 (2.41) 3 0 0 0 0 C55 0 0 0 0 0 0 Css] Transverse isotropy results in the following relations: Ca=Cx.Cu=Cis.Css=Co6.Cu=C2-C2 2 Note the five independent elastic constants.An example of this is a thin unidirectional lamina in which the fibers are arranged in a square array or 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 87 (2.40a–f) The cube will deform in all directions as determined by the normal strain equations. However, the shear strains in all three planes (1–2, 2–3, and 3–1) are zero, showing that the element will not change shape in those planes. Thus, the cube will not deform in shape under any normal load applied in the principal directions. This is unlike the monoclinic material, in which two out of the six faces of the cube changed shape. A cube made of isotropic material would not change its shape either; however, the normal strains, ε1 and ε2, will be different in an orthotropic material and identical in an isotropic material. 2.3.4 Transversely Isotropic Material Consider a plane of material isotropy in one of the planes of an orthotropic body. If direction 1 is normal to that plane (2–3) of isotropy, then the stiffness matrix is given by . (2.41) Transverse isotropy results in the following relations: . Note the five independent elastic constants. An example of this is a thin unidirectional lamina in which the fibers are arranged in a square array or ε σ γ γ γ 3 33 3 23 31 12 0 0 0 = = = = S . [ ] C CCC CCC CCC = 11 12 12 12 22 23 12 23 22 000 000 000 0 0 0 2 0 0 000 0 0 000 00 22 23 55 55 C C C C − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ C CC CC CC C C 22 33 12 13 55 66 44 22 23 2 ==== − ,,, 1343_book.fm Page 87 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
88 Mechanics of Composite Materials,Second Edition 2 0 0 1 0 00 0 0 000 0 0 000 0 FIGURE 2.15 A unidirectional lamina as a transversely isotropic material with fibers arranged in a square array. a hexagonal array.One may consider the elastic properties in the two direc- tions perpendicular to the fibers to be the same.In Figure 2.15,the fibers are in direction 1,so plane 2-3 will be considered as the plane of isotropy. The compliance matrix reduces to Su S12 S12 0 0 01 S12 52m S23 0 0 0 0 [S= S12 S23 S2 0 0 (2.42) 0 0 0 2(S22-S2x) 0 0 0 0 0 0 Sss 0 0 0 0 0 0 Ss5] 2.3.5 Isotropic Material If all planes in an orthotropic body are identical,it is an isotropic material; then,the stiffness matrix is given by C11 C12 C12 0 0 01 C12 Cn C12 0 0 0 C12 C12 C11 0 0 0 0 0 C11-C12 0 0 [C]= 2 (2.43) 0 0 0 0 C11-C12 0 2 0 0 0 0 C11-C12 2 Isotropy results in the following additional relationships: 2006 by Taylor Francis Group,LLC
88 Mechanics of Composite Materials, Second Edition a hexagonal array. One may consider the elastic properties in the two directions perpendicular to the fibers to be the same. In Figure 2.15, the fibers are in direction 1, so plane 2–3 will be considered as the plane of isotropy. The compliance matrix reduces to . (2.42) 2.3.5 Isotropic Material If all planes in an orthotropic body are identical, it is an isotropic material; then, the stiffness matrix is given by . (2.43) Isotropy results in the following additional relationships: FIGURE 2.15 A unidirectional lamina as a transversely isotropic material with fibers arranged in a square array. 3 2 1 [ ] S SSS SSS SSS = 11 12 12 12 22 23 12 23 22 000 000 000 0 0 02 0 0 000 0 0 000 00 22 23 55 55 ( ) S S S S − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ [ ] C CCC CCC C C C = 11 12 12 12 11 12 12 12 11 000 000 000 0 0 0 2 0 0 000 0 2 0 000 0 0 2 11 12 11 12 11 12 C C C C C C − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1343_book.fm Page 88 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 89 Cu=Ca.CB-CaCu-Cx-Cn=Cu-CB. 2 2 This also implies infinite principal planes of symmetry.Note the two independent constants.This is the most common material symmetry avail- able.Examples of isotropic bodies include steel,iron,and aluminum.Relat- ing Equation(2.43)to Equation(2.18)shows that E(1-v) C1=0-2w1+W1 VE C2=1-2v1+ (2.44a-b) Note that C1-C2 2 1 E(1-v) VE =2a-2v1+w1-2vW1+W E 2(1+v) =G. The compliance matrix reduces to Su S12S12 0 0 01 S12 Su S12 0 0 0 [S]= 512 S12 S11 0 0 0 (2.45) 0 0 0 2(S1-S2) 0 0 0 0 0 0 2(S1-S12) 0 0 0 0 0 0 2(S1-S2)J We summarize the number of independent elastic constants for various types of materials: 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 89 . This also implies infinite principal planes of symmetry. Note the two independent constants. This is the most common material symmetry available. Examples of isotropic bodies include steel, iron, and aluminum. Relating Equation (2.43) to Equation (2.18) shows that (2.44a–b) Note that The compliance matrix reduces to . (2.45) We summarize the number of independent elastic constants for various types of materials: C CC CC CC CC 11 22 12 23 66 22 23 11 12 2 2 === − = − , , 11 1 12 1 C E = − − + ( ) ( )( ) , ν ν ν 12 12 1 C E = − + ν ( )( ) ν ν . C C 11 12 2 − = − − + − − + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 2 1 12 1 12 1 E E ( ) ( )( ) ( )( ) ν ν ν ν ν ν = + E 2 1( ) ν = G. [ ] S SSS SSS S S S = 11 12 12 12 11 12 12 12 11 000 000 000 0 0 02 0 0 0 0 0 02 0 0 0 0 0 02 11 12 11 12 11 1 ( ) ( ) ( S S S S S S − − − 2 ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1343_book.fm Page 89 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
90 Mechanics of Composite Materials,Second Edition ·Anisotropic:21 ·Monoclinic:13 ·Orthotropic:g Transversely isotropic:5 ·Isotropic:2 Example 2.3 Show the reduction of anisotropic material stress-strain Equation(2.25)to those of a monoclinic material stress-strain Equation(2.35). Solution Assume direction 3 is perpendicular to the plane of symmetry.Now in the coordinate system 1-2-3,Equation(2.25)with C=C from Equation(2.34)is 01 C11 C12 C13 C14 C15 C16 02 C12 C22 C23 C24 C25 C26 E2 3 C13 C23 C33 C34 C35 C36 E3 (2.46) T23 C14 C24 C34 Cu C45 C46 Yz T31 C15 C25 C35 C4s C55 C56 T12 C16 C26 C36 C46 C56 C66JY12] Also,in the coordinate system 1'-2'-3'(Figure 2.11), Or C11 C12 C13 C14 C15 C16 02 C12 C2 C23 C24 C25 C26 Ez 03 C13 C23 C33 C34 C35 C36 E3 (2.47 T23 C24 C34 C44 C45 C46 Y23 T31 C2 C35 C45 Css C56 Y3r C16 C26 C36 C46 C56 C66 Yr2 Because there is a plane of symmetry normal to direction 3,the stresses and strains in the 1-2-3 and 1'-2'-3'coordinate systems are related by 01=01,02=0x,03=03 T23-T23,T31=-T3r,T12=T1r2, (2.48a-f) 2006 by Taylor Francis Group,LLC
90 Mechanics of Composite Materials, Second Edition • Anisotropic: 21 • Monoclinic: 13 • Orthotropic: 9 • Transversely isotropic: 5 • Isotropic: 2 Example 2.3 Show the reduction of anisotropic material stress–strain Equation (2.25) to those of a monoclinic material stress–strain Equation (2.35). Solution Assume direction 3 is perpendicular to the plane of symmetry. Now in the coordinate system 1–2–3, Equation (2.25) with Cij = Cji from Equation (2.34) is (2.46) Also, in the coordinate system 1′–2′–3′ (Figure 2.11), (2.47) Because there is a plane of symmetry normal to direction 3, the stresses and strains in the 1–2–3 and 1′–2′–3′ coordinate systems are related by (2.48a–f) 1 2 3 23 31 12 σ 11 12 σ σ τ τ τ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = C CCCCC CCCCCC CCC 13 14 15 16 12 22 23 24 25 26 13 23 33 34 CCC CCCCCC CCCCC 35 36 14 24 34 44 45 46 15 25 35 45 55 56 C CCCCCC 16 26 36 46 56 66 ⎡ 1 ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ε 2 3 23 31 12 ε ε γ γ γ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ , ′ ′ ′ ′ ′ ′ ′ ′ ′ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1 2 3 2 3 3 1 1 2 σ σ σ τ τ τ ⎥ = CCCCCC CCCCCC 11 12 13 14 15 16 12 22 23 24 25 26 CCCCCC 13 CCCCCC C C 23 33 34 35 36 14 24 34 44 45 46 15 25 35 CCCC CCCCCC 45 55 56 16 26 36 46 56 66 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ′ ′ ′ ′ ′ ′ ′ ′ ′ 1 2 3 2 3 3 1 1 2 ε ε ε γ γ γ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ , σ σσ σσ σ 1 12 23 3 === ′′′ , , τ ττ ττ τ 23 2 3 31 3 1 12 1 2 − =− = ′′ ′′ ′′ , ,, 1343_book.fm Page 90 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC