2.3 STRESS-STRAIN RELATIONSHIPS 13 for a monoclinic material the S41,S22,S52,S43,S53,S64,S6s elements are also zero. (Since the compliance matrix is symmetrical the elements S14,$24,S2s,S34,S35,S46, S56 are also zero.)The elements of the compliance matrix are listed in Table 2.4. The elements of the compliance matrix may be expressed in terms of the engineering constants defined in Table 2.5.In Tables 2.4 and 2.5 the types of tests are also illustrated that,at least in principle,could provide the elements of the compliance matrix and the engineering constants.The relationships between the elements of the compliance matrix and the engineering constants are shown in Tables 2.6 and 2.7. The nonzero and zero elements of the compliance matrix can best be seen when the matrix is written in the form S11 S12S130 0 S16 S12 S22 S23 0 0 S26 0 0 [S= S3 S23 S33 S36 (2.26) 0 0 S44 S45 0 0 0 0 S45 S55 0 S26 S36 0 0 S66 Table 2.5.The engineering constants for monoclinic materials.For orthotropic,transversely isotropic,and isotropic materials v16=v61=0,v26=v62=0,v36=V63=0,v45=154=0 Test Engineering constants Young's modulus in the x direction E=o1/ Poisson's ratio in the x-x2 plane 12=-62/e1 Poisson's ratio in the xi-x3 plane 13=-e3/e1 1 Poisson parameter 16=h12/e1 Young's modulus in the x2 direction E2=2/e2 02 Poisson's ratio in the x2-xi plane 21=-∈1/e2 Poisson's ratio in the x2-x3 plane 23=-e3/e2 Poisson parameter 26=Y12/e2 个03 Young's modulus in the x3 direction E=o3/3 Poisson's ratio in the x3-xi plane 31=-61/e3 Poisson's ratio in the x3-x2 plane 2=-2/e3 Poisson parameter '36=h2/3 123 Shear modulus in the x2-x3 plane G =t/yn Poisson parameter V45=n3/23 Shear modulus in the xi-x3 plane G13=t13/h3 Poisson parameter V54=y3/h3 Shear modulus in the xi-x2 plane G12=t12/h2 12 Poisson parameter 61=∈1/h2 Poisson parameter v62=e2/h2 Poisson parameter V6s=3/h2 = 器=器 管=器6j=1.23)
2.3 STRESS–STRAIN RELATIONSHIPS 13 for a monoclinic material the S41, S42, S52, S43, S53, S64, S65 elements are also zero. (Since the compliance matrix is symmetrical the elements S14, S24, S25, S34, S35, S46, S56 are also zero.) The elements of the compliance matrix are listed in Table 2.4. The elements of the compliance matrix may be expressed in terms of the engineering constants defined in Table 2.5. In Tables 2.4 and 2.5 the types of tests are also illustrated that, at least in principle, could provide the elements of the compliance matrix and the engineering constants. The relationships between the elements of the compliance matrix and the engineering constants are shown in Tables 2.6 and 2.7. The nonzero and zero elements of the compliance matrix can best be seen when the matrix is written in the form [S] = S11 S12 S13 0 0 S16 S12 S22 S23 0 0 S26 S13 S23 S33 0 0 S36 000 S44 S45 0 000 S45 S55 0 S16 S26 S36 0 0 S66 . (2.26) Table 2.5. The engineering constants for monoclinic materials. For orthotropic, transversely isotropic, and isotropic materials ν16 = ν61 = 0, ν26 = ν62 = 0, ν36 = ν63 = 0, ν45 = ν54 = 0 Test Engineering constants σ1 σ1 Young’s modulus in the x1 direction E1 = σ1/1 Poisson’s ratio in the x1–x2 plane ν12 = −2/1 Poisson’s ratio in the x1–x3 plane ν13 = −3/1 Poisson parameter ν16 = γ12/1 σ2 σ2 Young’s modulus in the x2 direction E2 = σ2/2 Poisson’s ratio in the x2–x1 plane ν21 = −1/2 Poisson’s ratio in the x2–x3 plane ν23 = −3/2 Poisson parameter ν26 = γ12/2 σ3 σ3 Young’s modulus in the x3 direction E3 = σ3/3 Poisson’s ratio in the x3–x1 plane ν31 = −1/3 Poisson’s ratio in the x3–x2 plane ν32 = −2/3 Poisson parameter ν36 = γ12/3 τ23 Shear modulus in the x2–x3 plane G23 = τ23/γ23 Poisson parameter ν45 = γ13/γ23 τ13 Shear modulus in the x1–x3 plane G13 = τ13/γ13 Poisson parameter ν54 = γ23/γ13 τ12 Shear modulus in the x1–x2 plane G12 = τ12/γ12 Poisson parameter ν61 = 1/γ12 Poisson parameter ν62 = 2/γ12 Poisson parameter ν63 = 3/γ12 νi j Ei = ν ji Ej , ν45 G23 = ν54 G13 , νi6 Ei = ν6i G12 (i, j = 1, 2, 3)
14 DISPLACEMENTS,STRAINS,AND STRESSES Table 2.6.Elements of the compliance matrix in terms of the engineering constants for monoclinic materials.The expressions are also valid for orthotropic,transversely isotropic,and isotropic materials with Si6=S1=0, S26=S62=0,S36=S63=0,S45=S4=0. S1=/a1=/(Ee)=言 S1=0 =a/o1=aEi)=-置 S51=0 s1=3/am=(E)=-晋 S61=h2/a1=h2/(E)=曾 S=61/om=646)=- S42=0 5m=2/2=/Ee2)=a S2=0 2=s/2=(B)=-詈 562=h2/o2=h2/(Ee2)=爱 Ss=/o3=1/(Ee)=- S43=0 53=61a3=(E)=-詈 S3=0 S3=3/a3=/(Ee3)=言 S3=h2/a3=2/(E3)= S4=0 S4=2/23=s/(Gs)= S24=0 S4=s/3=h/(G2s)=是 45 S34=0 S64=0 S5=0 Ss=a/s=hs/GeMs)=器 S25=0 Ss=h/13=h/(G1sn)=品 S5=0 S6s=0 Si6=61/t12=G1/(G12h2)= V61 G12 S46=0 S26=e2/t12=2/(G12h2)= G13 S56=0 S6=e3/t12=3/(G12h2)= v6图 S66=h2/2=h2/(G22)=品 The stiffness matrix is obtained by inverting the compliance matrix as follows: C11C12C13 0 0 C16 C12C22 C2 0 0 C26 C13C23C33 0 0 [C]= C36 0 0 (2.27) 0 C44 C45 0 000 C45 C55 0 LC16 C26 C3 00 C66 Because the [S]and [C]matrices are symmetrical(Eq.2.25)only 13 of the elements are independent (Table 2.8). 2.3.3 Orthotropic Material When there are three mutually perpendicular symmetry planes with respect to the alignment of the fibers the material is referred to as orthotropic (Fig.2.11). Examples of orthotropic fiber-reinforced composites are shown in Figure 2.12.For an orthotropic material we specify the stiffness and compliance matrices in the x, x2,x3 coordinate system defined in such a way that the axes are perpendicular to the three planes of symmetry(Fig.2.11). We apply a normal stress (Fig.2.13).Because o is in the xi-x2 symmetry (orthotropy)plane the out-of-plane shear strains are zero,(y13=y23=0);and
14 DISPLACEMENTS, STRAINS, AND STRESSES Table 2.6. Elements of the compliance matrix in terms of the engineering constants for monoclinic materials. The expressions are also valid for orthotropic, transversely isotropic, and isotropic materials with S16 = S61 = 0, S26 = S62 = 0, S36 = S63 = 0, S45 = S54 = 0. S11 = 1/σ1 = 1/(E11) = 1 E1 S41 = 0 S21 = 2/σ1 = 2/(E11) = −ν12 E1 S51 = 0 S31 = 3/σ1 = 3/(E11) = −ν13 E1 S61 = γ12/σ1 = γ12/(E11) = ν16 E1 S12 = 1/σ2 = 1/(E22) = −ν21 E2 S42 = 0 S22 = 2/σ2 = 2/(E22) = 1 E2 S52 = 0 S32 = 3/σ2 = 3/(E22) = −ν23 E2 S62 = γ12/σ2 = γ12/(E22) = ν26 E2 S13 = 1/σ3 = 1/(E33) = −ν31 E3 S43 = 0 S23 = 2/σ3 = 2/(E33) = −ν32 E3 S53 = 0 S33 = 3/σ3 = 3/(E33) = 1 E3 S63 = γ12/σ3 = γ12/(E33) = ν36 E3 S14 = 0 S44 = γ23/τ23 = γ23/(G23γ23) = 1 G23 S24 = 0 S54 = γ13/τ23 = γ13/(G23γ23) = ν45 G23 S34 = 0 S64 = 0 S15 = 0 S45 = γ23/τ13 = γ23/(G13γ13) = ν54 G13 S25 = 0 S55 = γ13/τ13 = γ13/(G13γ13) = 1 G13 S35 = 0 S65 = 0 S16 = 1/τ12 = 1/(G12γ12) = ν61 G12 S46 = 0 S26 = 2/τ12 = 2/(G12γ12) = ν62 G12 S56 = 0 S36 = 3/τ12 = 3/(G12γ12) = ν63 G12 S66 = γ12/τ12 = γ12/(G12γ12) = 1 G12 The stiffness matrix is obtained by inverting the compliance matrix as follows: [C] = C11 C12 C13 0 0 C16 C12 C22 C23 0 0 C26 C13 C23 C33 0 0 C36 000 C44 C45 0 000 C45 C55 0 C16 C26 C36 0 0 C66 . (2.27) Because the [S] and [C] matrices are symmetrical (Eq. 2.25) only 13 of the elements are independent (Table 2.8). 2.3.3 Orthotropic Material When there are three mutually perpendicular symmetry planes with respect to the alignment of the fibers the material is referred to as orthotropic (Fig. 2.11). Examples of orthotropic fiber-reinforced composites are shown in Figure 2.12. For an orthotropic material we specify the stiffness and compliance matrices in the x1, x2, x3 coordinate system defined in such a way that the axes are perpendicular to the three planes of symmetry (Fig. 2.11). We apply a normal stress σ1 (Fig. 2.13). Because σ1 is in the x1–x2 symmetry (orthotropy) plane the out-of-plane shear strains are zero, (γ13 = γ23 = 0); and
2.3 STRESS-STRAIN RELATIONSHIPS 15 Table 2.7.The compliance matrices in terms of the engineering constants for monoclinic,orthotropic,transversely isotropic,and isotropic materials 0 0 室 51550 0 0 0 63 [S= 0 0 0 monoclinic 0 0 0 爱 0 1 0 0 0 1 2. 0 0 07 [S= 1百号0 0 0 0 0 0 0 0 orthotropic 0 0 00 0 0 0 G 0 0 0 1百号 0 0 0 0 0 0 0 0 0 [S= 0 2(1+23)】 transversely isotropic 0 0 0 0 0 0 1 0 0 0 0 0 品」 0 0 0 [9= 0 0 E1E0 0 0 0 0 0 0 21+ 0 isotropic E 0 0 0 0 21+w】 0 0 0 0 0 0 241+】 E because o1 is also in the xi-x3 symmetry plane,the yi2=0 shear strain is zero. This implies that S14,Ss,Si6 are zero.By similar arguments it can be shown that for an orthotropic material the S24,S25,S26,S34,S35,$36,S45,S46,S56 elements are also zero.Accordingly,the compliance matrix is S11 S12 S13 0 0 0 S12S2 S23 0 0 0 0 0 [S= S1 S23 S33 0 0 0 0 (2.28) S44 0 0 0 0 0 S55 0 0 0 0 0 0 S66 The elements of the compliance matrix are listed in Table 2.4.In terms of the engineering constants,the compliance matrix is given in Table 2.7.The stiffness
2.3 STRESS–STRAIN RELATIONSHIPS 15 Table 2.7. The compliance matrices in terms of the engineering constants for monoclinic, orthotropic, transversely isotropic, and isotropic materials [S] = 1 E1 −ν21 E2 −ν31 E3 0 0 ν61 G12 −ν12 E1 1 E2 −ν32 E3 0 0 ν62 G12 −ν13 E1 −ν23 E2 1 E3 0 0 ν63 G12 000 1 G23 ν54 G13 0 000 ν45 G23 1 G13 0 ν16 E1 ν26 E2 ν36 E3 0 0 1 G12 monoclinic [S] = 1 E1 −ν21 E2 −ν31 E3 000 −ν12 E1 1 E2 −ν32 E3 000 −ν13 E1 −ν23 E2 1 E3 000 000 1 G23 0 0 0 0 00 1 G13 0 0 0 0 00 1 G12 orthotropic [S] = 1 E1 −ν21 E2 −ν21 E2 0 00 −ν12 E1 1 E2 −ν32 E2 0 00 −ν12 E1 −ν23 E2 1 E2 0 00 000 2(1+ν23) E2 0 0 000 0 1 G13 0 000 0 0 1 G13 transversely isotropic [S] = 1 E − ν E − ν E 000 − ν E 1 E − ν E 000 − ν E − ν E 1 E 000 000 2(1+ν) E 0 0 000 0 2(1+ν) E 0 000 0 0 2(1+ν) E isotropic because σ1 is also in the x1–x3 symmetry plane, the γ12 = 0 shear strain is zero. This implies that S14, S15, S16 are zero. By similar arguments it can be shown that for an orthotropic material the S24, S25, S26, S34, S35, S36, S45, S46, S56 elements are also zero. Accordingly, the compliance matrix is [S] = S11 S12 S13 000 S12 S22 S23 000 S13 S23 S33 000 000 S44 0 0 0000 S55 0 00000 S66 . (2.28) The elements of the compliance matrix are listed in Table 2.4. In terms of the engineering constants, the compliance matrix is given in Table 2.7. The stiffness
16 DISPLACEMENTS,STRAINS,AND STRESSES Table 2.8.The nonzero engineering constants for monoclinic,orthotropic, transverselysc,andser(学=号，seTable2.7刀 Nonzero engineering constants Material Independent Dependent Monoclinic E1,E,Es G23,G13,G12 12,13,23 h6,26,V45,6 Orthotropic E.E,Es G23,G13,G12 12,13,23 Transversely E1,E2 E=E2,G13=G12 isotropic G12 E G23=20+2 12,123 13=12 Isotropic E(=E) E2=E3=E,h3=23=V 2(=) G3=G13=G12=2+ 个E3 Figure 2.11:Material with three planes of symmetry. 个3 +⊙ Figure 2.12:Illustrations of fiber-reinforced orthotropic composites.The fibers are oriented in three mutually perpendicular directions(left);the fibers are distributed equally in the and directions in planes parallel to the xi-x2 plane(right)
16 DISPLACEMENTS, STRAINS, AND STRESSES Table 2.8. The nonzero engineering constants for monoclinic, orthotropic, transversely isotropic, and isotropic materials ( νi j E i = ν j i E j , see Table 2.7) Nonzero engineering constants Material Independent Dependent Monoclinic E1, E2, E3 G23, G13, G12 ν12, ν13, ν23 ν16, ν26, ν45, ν36 Orthotropic E1, E2, E3 G23, G13, G12 ν12, ν13, ν23 Transversely E1, E2 E3 = E2, G13 = G12 isotropic G12 G23 = E2 2(1+ν23) ν12, ν23 ν13 = ν12 Isotropic E1 (= E) E2 = E3 = E, ν13 = ν23 = ν ν12 (= ν) G23 = G13 = G12 = E 2(1+ν) x3 x3 x2 x2 x1 x1 Figure 2.11: Material with three planes of symmetry. x3 x2 x2 x1 x1 x3 + Θ −Θ Figure 2.12: Illustrations of fiber-reinforced orthotropic composites.The fibers are oriented in three mutually perpendicular directions (left); the fibers are distributed equally in the + and − directions in planes parallel to the x1–x2 plane (right)
2.3 STRESS-STRAIN RELATIONSHIPS 17 12=0 13=0 23=0 Figure 2.13:A normal stress1 applied in the xi-x2 and xi-x3 symmetry planes of an orthotropic material. matrix is obtained by inverting the compliance matrix.The nonzero terms of the stiffness matrix are C11 C12 C13 0 0 07 C12 C22 C23 0 0 0 0 0 [C]= C13 C23 C33 0 0 0 0 0 0 (2.29) C44 0 0 0 0 C55 0 0 0 0 0 0 C66 In the [S]and [C]matrices of the 12 nonzero elements only 9 are independent (Table 2.8).Equation(2.29)can be written in the form 0 0 [可 0 [C= 0 0 0 0 00 (2.30) 0 0 0 M 0 0 0 The submatrices [L]and [M]are given in Tables 2.9 and 2.10 in terms of the en- gineering constants. With the compliance matrix given by Eq.(2.28),the strain-stress relationships (Eq.2.23)become E1 TSu S12 S13 0 0 07 S12 S22 S23 0 0 0 3 S1 S23 S33 0 0 0 y23 0 0 0 0 (2.31) S44 0 T23 Y13 0 0 0 0 S55 0 9 12 0 0 0 0 0 S66」 T12 This equation shows an important feature of orthotropic materials,namely, that normal stresses do not produce shear deformations when these stresses are in the x1,x2,x3 orthotropy directions.Note,however,that normal stresses applied in the x,y,z directions (which do not coincide with thex,x2,x3 orthotropy directions)result in shear deformations,as illustrated in Figure 2.14.In this case (in the x,y,z coordinate system)none of the elements of the compliance and stiffness matrices is zero
2.3 STRESS–STRAIN RELATIONSHIPS 17 x2 γ12 = 0 σ1 x1 σ1 x3 γ13 = 0 σ1 x1 σ1 x3 γ23 = 0 x1 Figure 2.13: A normal stress σ1 applied in the x1–x2 and x1–x3 symmetry planes of an orthotropic material. matrix is obtained by inverting the compliance matrix. The nonzero terms of the stiffness matrix are [C] = C11 C12 C13 000 C12 C22 C23 000 C13 C23 C33 000 000 C44 0 0 0000 C55 0 00000 C66 . (2.29) In the [S] and [C] matrices of the 12 nonzero elements only 9 are independent (Table 2.8). Equation (2.29) can be written in the form [C] = [L] 000 000 000 000 000 000 [M] . (2.30) The submatrices [L] and [M] are given in Tables 2.9 and 2.10 in terms of the engineering constants. With the compliance matrix given by Eq. (2.28), the strain–stress relationships (Eq. 2.23) become 1 2 3 γ23 γ13 γ12 = S11 S12 S13 000 S12 S22 S23 000 S13 S23 S33 000 000 S44 0 0 0000 S55 0 00000 S66 σ1 σ2 σ3 τ23 τ13 τ12 . (2.31) This equation shows an important feature of orthotropic materials, namely, that normal stresses do not produce shear deformations when these stresses are in the x1, x2, x3 orthotropy directions. Note, however, that normal stresses applied in the x, y, z directions (which do not coincide with the x1, x2, x3 orthotropy directions) result in shear deformations, as illustrated in Figure 2.14. In this case (in the x, y, z coordinate system) none of the elements of the compliance and stiffness matrices is zero