18 DISPLACEMENTS,STRAINS,AND STRESSES Table 2.9.The [L]submatrix in Eq.(2.30)for orthotropic,transversely isotropic, and isotropic materials Orthotropic E（1-号喝) E(2+号s）〉 E(13+23) [叫=六 互（啦+看吃） 6(1-要6) E(2+号) E3(13+23) 后(2+会2we) E(1-号哈) D= E1E-2E号-2号马-2123E号-2E号 E1 E2E3 Transversely isotropic E(1-喝) E2(1+23) E12(1+23) [四= Ev12(1+s) （1-号哈) 五(%+异品) E22(1+z) 五(+鲁哈) (1-鲁哈) D=1-喝-21+)会哈 Isotropic 1-v [=0+-2可 E Table 2.10.The [M]submatrix in Eq.(2.30)for orthotropic, transversely isotropic,and isotropic materials Orthotropic Gy 0 0 [M= 0 G13 0 0 0 G12 Transversely isotropic E 21+23) 0 0 [M= 0 G12 0 0 0 L G12 Isotropic 21+) 0 0 [M= 0 E 21+ 0 0 E 21+」 6 个3 0 Figure 2.14:Orthotropic material subjected to a normal stress.There is no shear strain when the stress is applied in one of the orthotropy directions(left),but there is shear strain when the stress is not along an orthotropy direction(right)
18 DISPLACEMENTS, STRAINS, AND STRESSES Table 2.9. The [L] submatrix in Eq. (2.30) for orthotropic, transversely isotropic, and isotropic materials Orthotropic [L] = 1 D E1 1 − E3 E2 ν2 23 E2 ν12 + E3 E2 ν13ν23 E3 (ν13 + ν12ν23) E2 ν12 + E3 E2 ν13ν23 E2 1 − E3 E1 ν2 13 E3 ν23 + E2 E1 ν12ν13 E3 (ν13 + ν12ν23) E3 ν23 + E2 E1 ν12ν13 E3 1 − E2 E1 ν2 12 D = E1 E2 E3−ν2 23 E1 E2 3−ν2 12 E2 2 E3−2ν12ν13ν23 E2 E2 3−ν2 13 E2 E2 3 E1 E2 E3 Transversely isotropic [L] = 1 D E1 1 − ν2 23 E2ν12 (1 + ν23) E2ν12 (1 + ν23) E2ν12 (1 + ν23) E2 1 − E2 E1 ν2 12 E2 ν23 + E2 E1 ν2 12 E2ν12 (1 + ν23) E2 ν23 + E2 E1 ν2 12 E2 1 − E2 E1 ν2 12 D = 1 − ν2 23 − 2 (1 + ν23) E2 E1 ν2 12 Isotropic [L] = E (1+ν)(1−2ν) 1 − νν ν ν 1 − ν ν ν ν 1 − ν Table 2.10. The [M ] submatrix in Eq. (2.30) for orthotropic, transversely isotropic, and isotropic materials Orthotropic [M] = G23 0 0 0 G13 0 0 0 G12 Transversely isotropic [M] = E2 2(1+ν23) 0 0 0 G12 0 0 0 G12 Isotropic [M] = E 2(1+ν) 0 0 0 E 2(1+ν) 0 0 0 E 2(1+ν) σx σ1 σx σ1 x x1 x z 3 Figure 2.14: Orthotropic material subjected to a normal stress. There is no shear strain when the stress is applied in one of the orthotropy directions (left), but there is shear strain when the stress is not along an orthotropy direction (right)
2.3 STRESS-STRAIN RELATIONSHIPS 19 Plane of isotropy Figure 2.15:Example of a fiber-reinforced,transversely isotropic composite material. 2.3.4 Transversely Isotropic Material A transversely isotropic material has three planes of symmetry(Fig.2.11)and,as such,it is orthotropic.In one of the planes of symmetry the material is treated as isotropic.An example of transversely isotropic material is a composite reinforced with continuous unidirectional fibers with all the fibers aligned in the x direction (Fig.2.15).In this case the material in the plane perpendicular to the fibers(x2-x3 plane)is treated as isotropic. For a transversely isotropic material we specify the stiffness and compliance matrices in an x1,x2,x3 coordinate system chosen in such a way that the axes are perpendicular to the planes of symmetry and x1 is perpendicular to the plane of isotropy(Fig.2.15).In this coordinate system,because of material symmetry four of the Poisson ratios are zero (v6 v26 v36=v45 =0).Furthermore,because of isotropy the following engineering constants are related: E3=E2, G13=G12, V13=12 (2.32) For an isotropic material the shear modulus is! E G=21+) (2.33) Correspondingly,for a material that is isotropic in the x2-x3 plane we write E2 G3=2(1+z) (2.34) Equations (2.32)and (2.34),together with the expressions in Table 2.6 (page 14),yield the compliance matrix in terms of the engineering constants.The 1 E.P.Popov,Engineering Mechanics of Solids.Prentice-Hall,Englewood Cliffs,New Jersey,1990. p.151
2.3 STRESS–STRAIN RELATIONSHIPS 19 Plane of isotropy x3 x2 x1 Figure 2.15: Example of a fiber-reinforced, transversely isotropic composite material. 2.3.4 Transversely Isotropic Material A transversely isotropic material has three planes of symmetry (Fig. 2.11) and, as such, it is orthotropic. In one of the planes of symmetry the material is treated as isotropic. An example of transversely isotropic material is a composite reinforced with continuous unidirectional fibers with all the fibers aligned in the x1 direction (Fig. 2.15). In this case the material in the plane perpendicular to the fibers (x2–x3 plane) is treated as isotropic. For a transversely isotropic material we specify the stiffness and compliance matrices in an x1, x2, x3 coordinate system chosen in such a way that the axes are perpendicular to the planes of symmetry and x1 is perpendicular to the plane of isotropy (Fig. 2.15). In this coordinate system, because of material symmetry four of the Poisson ratios are zero (ν16 = ν26 = ν36 = ν45 = 0). Furthermore, because of isotropy the following engineering constants are related: E3 = E2, G13 = G12, ν13 = ν12. (2.32) For an isotropic material the shear modulus is1 G = E 2 (1 + ν) . (2.33) Correspondingly, for a material that is isotropic in the x2–x3 plane we write G23 = E2 2 (1 + ν23) . (2.34) Equations (2.32) and (2.34), together with the expressions in Table 2.6 (page 14), yield the compliance matrix in terms of the engineering constants. The 1 E. P. Popov, Engineering Mechanics of Solids. Prentice-Hall, Englewood Cliffs, New Jersey, 1990, p. 151
20 DISPLACEMENTS,STRAINS,AND STRESSES results are given in Table 2.7 (page 15).The zero and nonzero elements of the compliance matrix are T Su S12 S12 0 0 07 S12 S22 S23 0 0 0 S12 S23 S22 0 0 [= 0 0 (2.35) 0 0 0 2(S22-23) 0 0 0 0 0 S66 0 0 0 0 0 0 S66 The stiffness matrix is obtained by inverting the compliance matrix.The zero and nonzero elements of the stiffness matrix are C11 C12 C12 0 0 0 C12 C22 C2 0 0 0 [C= C12 C23 Cn 0 0 0 0 0 Cm-C≌ 0 0 (2.36) 0 0 0 0 0 C66 0 0 0 0 0 C66」 In terms of the engineering constants,the elements of the stiffness matrix are given by Eq.(2.30). In both the compliance and stiffness matrices,of the 12 nonzero elements only 5 are independent (Table 2.8,page 16). 2.3.5 Isotropic Material In an isotropic material there are no preferred directions and every plane is a plane of symmetry.For example,a composite containing a large number of ran- domly oriented fibers behaves in an isotropic manner.For an isotropic material the coordinate system may be chosen arbitrarily.Here,we present the compliance and the stiffness matrices in the x,x2,and x3 coordinate system. Because of material symmetry four of the Poisson ratios are zero(vi6=v26 v36=v4s=0).Also,because of isotropy some of the engineering constants are related as follows: E=E=E=E G23=G13=G12=G (2.37) 23=13=12=V E G=21+) (2.38) Equations (2.37)and (2.38),together with the expressions in Table 2.6 (page 14),give the compliance matrix in terms of the engineering constants.The
20 DISPLACEMENTS, STRAINS, AND STRESSES results are given in Table 2.7 (page 15). The zero and nonzero elements of the compliance matrix are [S] = S11 S12 S12 0 00 S12 S22 S23 0 00 S12 S23 S22 0 00 0 0 02 (S22 − S23) 0 0 000 0 S66 0 000 0 0 S66 . (2.35) The stiffness matrix is obtained by inverting the compliance matrix. The zero and nonzero elements of the stiffness matrix are [C] = C11 C12 C12 0 00 C12 C22 C23 0 00 C12 C23 C22 0 00 000 C22−C23 2 0 0 000 0 C66 0 000 0 0 C66 . (2.36) In terms of the engineering constants, the elements of the stiffness matrix are given by Eq. (2.30). In both the compliance and stiffness matrices, of the 12 nonzero elements only 5 are independent (Table 2.8, page 16). 2.3.5 Isotropic Material In an isotropic material there are no preferred directions and every plane is a plane of symmetry. For example, a composite containing a large number of randomly oriented fibers behaves in an isotropic manner. For an isotropic material the coordinate system may be chosen arbitrarily. Here, we present the compliance and the stiffness matrices in the x1, x2, and x3 coordinate system. Because of material symmetry four of the Poisson ratios are zero (ν16 = ν26 = ν36 = ν45 = 0). Also, because of isotropy some of the engineering constants are related as follows: E1 = E2 = E3 = E G23 = G13 = G12 = G ν23 = ν13 = ν12 = ν (2.37) G = E 2 (1 + ν) . (2.38) Equations (2.37) and (2.38), together with the expressions in Table 2.6 (page 14), give the compliance matrix in terms of the engineering constants. The
2.3 STRESS-STRAIN RELATIONSHIPS 21 results are in Table 2.7.The elements of the compliance matrix are Su S12 S12 0 0 0 S12 S11 S12 0 0 0 0 0 0 [S= S12 S12 S11 0 0 0 2(S1-S2) 0 0 0 0 0 0 2(S1-S2) 0 0 0 0 0 0 2(S1-S2)J (2.39) The stiffness matrix is obtained by inverting the compliance matrix.The ele- ments of the stiffness matrix are Cn C12 C12 0 0 0 C12 C11 C12 0 0 0 [C]= C12 C12 C11 0 0 0 0 0 (2.40) 0 C11-C12 0 0 0 0 0 0 Cu-CR 2 0 0 0 0 0 0 C1-C卫 2 In terms of the engineering constants,the elements of the stiffness matrix are given by Eq.(2.30). In both the compliance and stiffness matrices,of the 12 nonzero elements only 2 are independent (Table 2.8,page 16). 2.1 Example.Calculate the elements of the stiffness and compliance matrices of a graphite epoxy unidirectional ply.The engineering constants are given as E= 148×109N/hm2,E2=9.65×109N/m2,G12=4.55×109NW/m2,12=0.3,and 23=0.6. Solution.For a transversely isotropic material the compliance matrix is given in Table 2.7(page 15,third row).By substituting the engineering constants into the expression in Table 2.7,and by using the condition that vj/E=v/Ej(see Table 2.8,page 16)we obtain 6.76 -2.03 -2.03 0 0 0 -2.03 103.63 -62.18 0 0 0 -2.03 -62.18 103.63 0 0 [S]= 0 0 0 1012m2 0 331.61 0 0 N 0 0 0 0 219.78 0 0 0 0 0 0 219.78 (2.41)
2.3 STRESS–STRAIN RELATIONSHIPS 21 results are in Table 2.7. The elements of the compliance matrix are [S] = S11 S12 S12 000 S12 S11 S12 000 S12 S12 S11 000 0 0 02 (S11 − S12) 0 0 000 0 2 (S11 − S12) 0 000 0 0 2 (S11 − S12) . (2.39) The stiffness matrix is obtained by inverting the compliance matrix. The elements of the stiffness matrix are [C] = C11 C12 C12 000 C12 C11 C12 000 C12 C12 C11 000 000 C11−C12 2 0 0 000 0 C11−C12 2 0 000 0 0 C11−C12 2 . (2.40) In terms of the engineering constants, the elements of the stiffness matrix are given by Eq. (2.30). In both the compliance and stiffness matrices, of the 12 nonzero elements only 2 are independent (Table 2.8, page 16). 2.1 Example. Calculate the elements of the stiffness and compliance matrices of a graphite epoxy unidirectional ply. The engineering constants are given as E1 = 148 × 109 N/m2, E2 = 9.65 × 109 N/m2, G12 = 4.55 × 109 N/m2, ν12 = 0.3, and ν23 = 0.6. Solution. For a transversely isotropic material the compliance matrix is given in Table 2.7 (page 15, third row). By substituting the engineering constants into the expression in Table 2.7, and by using the condition that νi j /Ei = νji /Ej (see Table 2.8, page 16) we obtain [S] = 6.76 −2.03 −2.03 0 0 0 −2.03 103.63 −62.18 0 0 0 −2.03 −62.18 103.63 0 0 0 0 0 0 331.61 0 0 0 0 0 0 219.78 0 0 0 0 0 0 219.78 10−12 m2 N . (2.41)
22 DISPLACEMENTS,STRAINS,AND STRESSES Figure 2.16:The x,y.z and the xi,x2.x3 coordinate systems. The elements of the stiffness matrix are obtained by inverting the compliance matrix 「152.47 7.467.46 0 0 0 7.46 15.44 9.41 0 0 0 [C]=[g1= 7.46 9.4115.44 0 0 0 0 0 0 109W 0 0 3.016 m2 0 0 0 0 4.55 0 0 0 0 0 0 4.55 (2.42) 2.4 Plane-Strain Condition There are circumstances when the stresses and strains do not vary in a certain direction.This direction is designated by either the x3 or the z axis(Fig.2.16). Although the stresses and strains do not vary along x3(or z),they may vary in planes perpendicular to the x3(or z)axis.This condition is referred to as plane- strain condition. When plane-strain condition exists in a body made of an isotropic material the xi-x2(or x-y)planes of the cross section remain plane and perpendicular to the x3(or z)axis.In a body made of an anisotropic material these planes do not necessarily remain plane. Plane-strain condition may exist far from the edges in a long body with constant cross section when both the material properties and the applied loads are uniform →个 个←- 个 个<- 个个 →个 Body 个← Force 个← Body Force 7777777777 777 Figure 2.17:Surface and body forces that may be applied under plane-strain condition.The applied forces must be uniform along the longitudinal axis and must be in equilibrium for each segment
22 DISPLACEMENTS, STRAINS, AND STRESSES x3 x1 x2 x y z Figure 2.16: The x, y, z and the x1, x2, x3 coordinate systems. The elements of the stiffness matrix are obtained by inverting the compliance matrix [C] = [S] −1 = 152.47 7.46 7.46 0 0 0 7.46 15.44 9.41 0 0 0 7.46 9.41 15.44 0 0 0 0 0 0 3.016 0 0 0 0 0 0 4.55 0 0 0 0 0 0 4.55 109 N m2 . (2.42) 2.4 Plane-Strain Condition There are circumstances when the stresses and strains do not vary in a certain direction. This direction is designated by either the x3 or the z axis (Fig. 2.16). Although the stresses and strains do not vary along x3 (or z), they may vary in planes perpendicular to the x3 (or z) axis. This condition is referred to as plane– strain condition. When plane-strain condition exists in a body made of an isotropic material the x1–x2 (or x–y) planes of the cross section remain plane and perpendicular to the x3 (or z) axis. In a body made of an anisotropic material these planes do not necessarily remain plane. Plane-strain condition may exist far from the edges in a long body with constant cross section when both the material properties and the applied loads are uniform Body Force Body Force Figure 2.17: Surface and body forces that may be applied under plane-strain condition. The applied forces must be uniform along the longitudinal axis and must be in equilibrium for each segment