《纺织复合材料》课程参考文献（Mechanics of Composite Structures）02 Displacements, Strains, and Stresses

Copyrighted Materials i0pUyPress rm CHAPTER TWO Displacements,Strains, and Stresses We consider composite materials consisting of continuous or discontinuous fibers embedded in a matrix.Such a composite is heterogeneous,and the properties vary from point to point.On a scale that is large with respect to the fiber diam- eter,the fiber and matrix properties may be averaged,and the material may be treated as homogeneous.This assumption,commonly employed in macromechan- ical analyses of composites,is adopted here.Hence,the material is considered to be quasi-homogeneous,which implies that the properties are taken to be the same at every point.These properties are not the same as the properties of either the fiber or the matrix but are a combination of the properties of the constituents. In this chapter,equations are presented for calculating the displacements, stresses,and strains when the structure undergoes only small deformations and the material behaves in a linearly elastic manner. Continuous fiber-reinforced composite materials(and structures made of such materials)often have easily identifiable preferred directions associated with fiber orientations or symmetry planes.It is therefore convenient to employ two co- ordinate systems:a local coordinate system aligned,at a point,either with the fibers or with axes of symmetry,and a global coordinate system attached to a fixed reference point(Fig.2.1).In this book the local and global Cartesian coordinate systems are designated respectively by xi,x2,x3 and the x,y,z axes.In the x,y,z directions the displacements at a point Aare denoted by u,v,w,and in the x1,x2, x3 directions by u1,u2,u3 (Fig.2.2). In the x,y,z coordinate system the normal stresses are denoted by ox,oy,and o:and the shear stresses by tyz,txz,and try(Fig.2.3).The corresponding normal and shear strains are ,y and yy,respectively. In the xi,x2,x3 coordinate system the normal stresses are denoted by a1,o2, and o3 and the shear stresses by t23,t13,and 7i2(Fig.2.3).The corresponding normal and shear strains are e1,e2,e3,and y23,y13,yi2,respectively.The symbol y represents engineeringshear strain that is twice the tensorial shear strain,Yij=2eij (i,j=x,y,zori,j=1,2,3). 3

CHAPTER TWO Displacements, Strains, and Stresses We consider composite materials consisting of continuous or discontinuous fibers embedded in a matrix. Such a composite is heterogeneous, and the properties vary from point to point. On a scale that is large with respect to the fiber diam￾eter, the fiber and matrix properties may be averaged, and the material may be treated as homogeneous. This assumption, commonly employed in macromechan￾ical analyses of composites, is adopted here. Hence, the material is considered to be quasi-homogeneous, which implies that the properties are taken to be the same at every point. These properties are not the same as the properties of either the fiber or the matrix but are a combination of the properties of the constituents. In this chapter, equations are presented for calculating the displacements, stresses, and strains when the structure undergoes only small deformations and the material behaves in a linearly elastic manner. Continuous fiber-reinforced composite materials (and structures made of such materials) often have easily identifiable preferred directions associated with fiber orientations or symmetry planes. It is therefore convenient to employ two co￾ordinate systems: a local coordinate system aligned, at a point, either with the fibers or with axes of symmetry, and a global coordinate system attached to a fixed reference point (Fig. 2.1). In this book the local and global Cartesian coordinate systems are designated respectively by x1, x2, x3 and the x, y, z axes. In the x, y, z directions the displacements at a point Aare denoted by u, v, w, and in the x1, x2, x3 directions by u1, u2, u3 (Fig. 2.2). In the x, y, z coordinate system the normal stresses are denoted by σx, σy, and σz and the shear stresses by τyz, τxz, and τxy (Fig. 2.3). The corresponding normal and shear strains are x, y, z and γyz, γxz, γxy, respectively. In the x1, x2, x3 coordinate system the normal stresses are denoted by σ1, σ2, and σ3 and the shear stresses by τ23, τ13, and τ12 (Fig. 2.3). The corresponding normal and shear strains are 1, 2, 3, and γ23, γ13, γ12, respectively. The symbol γ represents engineering shear strain that is twice the tensorial shear strain, γi j = 2i j (i, j = x, y, z or i, j = 1, 2, 3). 3

4 DISPLACEMENTS,STRAINS,AND STRESSES 2,E3个 Figure 2.1:The global x,y,z and local xi.x2.x3 coordinate systems. A stress is taken to be positive when it acts on a positive face in the positive direction.According to this definition,all the stresses shown in Figure 2.3 are positive. The preceding stress and strain notations,referred to as engineering notations, are used throughout this book.Other notations,most notably tensorial and con- tracted notations,can frequently be found in the literature.The stresses and strains in different notations are summarized in Tables 2.1 and 2.2. 2.1 Strain-Displacement Relations We consider a Ax long segment that undergoes a change in length,the new length being denoted by Ax'.From Figure 2.4 it is seen that u+△x'=△x+ u+ (2.1) where uandux are the displacements of points Aand B.respectively,in the x direction.Accordingly,the normal strain in the x direction is △x'-△xau Ex= (2.2) △x ax Similarly,in the y and z directions the normal strains are av y≠8y (2.3) aw e红202 (2.4) where v and w are the displacements in the y and z directions,respectively. Figure 2.2:The x,y,z and x,x2,x3 coordinate systems and the corresponding displacements

4 DISPLACEMENTS, STRAINS, AND STRESSES x1 x2 z x, 3 y x Figure 2.1: The global x, y, z and local x1, x2, x3 coordinate systems. A stress is taken to be positive when it acts on a positive face in the positive direction. According to this definition, all the stresses shown in Figure 2.3 are positive. The preceding stress and strain notations, referred to as engineering notations, are used throughout this book. Other notations, most notably tensorial and con￾tracted notations, can frequently be found in the literature. The stresses and strains in different notations are summarized in Tables 2.1 and 2.2. 2.1 Strain–Displacement Relations We consider a x long segment that undergoes a change in length, the new length being denoted by x . From Figure 2.4 it is seen that u + x = x +  u + ∂u ∂x x  , (2.1) where u and u + ∂u ∂x x are the displacements of points Aand B, respectively, in the x direction. Accordingly, the normal strain in the x direction is x = x − x x = ∂u ∂x . (2.2) Similarly, in the y and z directions the normal strains are y = ∂v ∂y (2.3) z = ∂w ∂z , (2.4) where v and w are the displacements in the y and z directions, respectively. A w v u z y x x3 x2 x1 A u2 u1 u3 A' A' Figure 2.2: The x, y, z and x1, x2, x3 coordinate systems and the corresponding displacements.

2.1 STRAIN-DISPLACEMENT RELATIONS 5 0 x3 037 个 2 T2 2 Figure 2.3:The stresses in the global x,y,z and the local x,x2,x3 coordinate systems. For angular(shear)deformation the tensorial shear strain is the average change in the angle between two mutually perpendicular lines(Fig.2.5) +B Exy=- 2 (2.5) For small deformations we have a≈tana= o+器△x)-v-u (2.6) △x ax Similarly B =au/ay,and the xy component of the tensorial shear strain is 1 /au Exy (2.7) In a similar manner we obtain the following expressions for the eyz and ex components of the tensorial shear strains: du d (2.8) Table 2.1.Stress notations Normal stress Shear stress x,y,z coordinate system Tensorial stress Oyy Gy: Oxy Engineering stress Ox Oy 0: Tyz try Contracted notation Gg Os x1,x2,x3 coordinate system Tensorial stress 011022 033 023013012 Engineering stress 01 02 03 T23 T13 T12 Contracted notation 01 02 0304 0506

2.1 STRAIN–DISPLACEMENT RELATIONS 5 x3 x2 x1 σ3 σ2 σ1 τ13 τ12 τ23 τ21 τ32 τ31 z y x σz σy σx τxz τxy τyz τyx τ τzx zy Figure 2.3: The stresses in the global x, y, z and the local x1, x2, x3 coordinate systems. For angular (shear) deformation the tensorial shear strain is the average change in the angle between two mutually perpendicular lines (Fig. 2.5) xy = α + β 2 . (2.5) For small deformations we have α ≈ tan α =  v + ∂v ∂x x  − v x = ∂v ∂x . (2.6) Similarly β = ∂u/∂y, and the xy component of the tensorial shear strain is xy = 1 2 ∂u ∂y + ∂v ∂x  . (2.7) In a similar manner we obtain the following expressions for the yz and xz components of the tensorial shear strains: yz = 1 2 ∂v ∂z + ∂w ∂y  xz = 1 2 ∂u ∂z + ∂w ∂x  . (2.8) Table 2.1. Stress notations Normal stress Shear stress x, y, z coordinate system Tensorial stress σxx σyy σzz σyz σxz σxy Engineering stress σx σy σz τyz τxz τxy Contracted notation σx σy σz σq σr σs x1, x2, x3 coordinate system Tensorial stress σ11 σ22 σ33 σ23 σ13 σ12 Engineering stress σ1 σ2 σ3 τ23 τ13 τ12 Contracted notation σ1 σ2 σ3 σ4 σ5 σ6

6 DISPLACEMENTS,STRAINS,AND STRESSES Table 2.2.Strain notations (the engineering and contracted notation shear strains are twice the tensorial shear strain) Normal strain Shear strain x,y,z coordinate system Tensorial strain Exx Eyz Ex2 Exy Engineering strain Ex Ey e Yyt Yus Yry Contracted notation Ex Ey Eg Es x1,X2,X3 coordinate system Tensorial strain e11e22633E23 613 e12 Engineering strain E2 E3 23M3 M12 Contracted notation 1 E2 E3 EA ES E6 The engineering shear strains are twice the tensorial shear strains: av aw Yy =2Ey= ay (2.9) 8z yit 2ex au aw (2.10) 8z'ax du dv y=2=)+x (2.11) In the x1,x2,x3 coordinate system the strain-displacement relationships are also given by Eqs.(2.2)-(2.4)and (2.9)-(2.11)with x,y,z replaced by x1,x2,x3, the subscripts x,y,z by 1,2,3,and u,v,w by u,u2,u3. 2.2 Equilibrium Equations The equilibrium equations at a point O are obtained by considering force and moment balances on a small AxAyAz cubic element located at point O.(The point O is at the center of the element,Fig.2.6.)We relate the stresses at one face to those at the opposite face by the Taylor series.By using only the first term of the Taylor series,force balance in the x direction gives -Ox△zAy-tx△x△y-tyr△x△z+ 0x+8x △z△y +(+股axa+(.+ayA x△z+f△x△y△z=0, (2.12) △x y B u+ 0u△正 Figure 2.4:Displacement of the AB line segment

6 DISPLACEMENTS, STRAINS, AND STRESSES Table 2.2. Strain notations (the engineering and contracted notation shear strains are twice the tensorial shear strain) Normal strain Shear strain x, y, z coordinate system Tensorial strain xx yy zz yz xz xy Engineering strain x y z γyz γxz γxy Contracted notation x y z q r s x1, x2, x3 coordinate system Tensorial strain 11 22 33 23 13 12 Engineering strain 1 2 3 γ23 γ13 γ12 Contracted notation 1 2 3 4 5 6 The engineering shear strains are twice the tensorial shear strains: γyz = 2yz = ∂v ∂z + ∂w ∂y (2.9) γxz = 2xz = ∂u ∂z + ∂w ∂x (2.10) γxy = 2xy = ∂u ∂y + ∂v ∂x . (2.11) In the x1, x2, x3 coordinate system the strain–displacement relationships are also given by Eqs. (2.2)–(2.4) and (2.9)–(2.11) with x, y, z replaced by x1, x2, x3, the subscripts x, y, z by 1, 2, 3, and u, v, w by u1, u2, u3. 2.2 Equilibrium Equations The equilibrium equations at a point O are obtained by considering force and moment balances on a small xyz cubic element located at point O. (The point O is at the center of the element, Fig. 2.6.) We relate the stresses at one face to those at the opposite face by the Taylor series. By using only the first term of the Taylor series, force balance in the x direction gives −σxzy − τzxxy − τyxxz +  σx + ∂σx ∂x x  zy +  τzx + ∂τzx ∂z z  xy +  τyx + ∂τyx ∂y y  xz + fxxyz= 0, (2.12) u A B ∆x y x x∆x u u ∂ ∂ + ∆x′ A′ B′ Figure 2.4: Displacement of the AB line segment

2.2 EQUILIBRIUM EQUATIONS 7 个 + Figure 2.5:Displacement of the ABC segment. where f is the body force per unit volume in the x direction.After simplification, this equation becomes ++ 8题+f=0. (2.13) ax ay az By similar arguments,the equilibrium equations in the y and z directions are 0xy+2++fv=0, (2.14) ax ay +g+o+5=0. (2.15) ax ay az where f,and f:are the body forces per unit volume in the y and z directions. A moment balance about an axis parallel to x and passing through the center (point O)gives(Fig.2.7) gAra号-rAay +(+aya-(+aAa (2.16) 0:+ 00Ξ△z 0z T+ 0r型△2 8z 0r丝△z T江+ 8z 0 Ty+ rg△y T:+ 0r丝△r by Ox 0g+ do1△yy 0x+ 0c王△士 8 8 △x 0r型△E Tw+ arg△y by Ox I △y Figure 2.6:Stresses on the AxAyAz cubic element

2.2 EQUILIBRIUM EQUATIONS 7 y x ∆x v A B β α C x∆x v v ∂ ∂ B′ A′ C′ + Figure 2.5: Displacement of the ABC segment. where fx is the body force per unit volume in the x direction. After simplification, this equation becomes ∂σx ∂x + ∂τyx ∂y + ∂τzx ∂z + fx = 0. (2.13) By similar arguments, the equilibrium equations in the y and z directions are ∂τxy ∂x + ∂σy ∂y + ∂τzy ∂z + fy = 0, (2.14) ∂τxz ∂x + ∂τyz ∂y + ∂σz ∂z + fz = 0, (2.15) where fy and fz are the body forces per unit volume in the y and z directions. A moment balance about an axis parallel to x and passing through the center (point O) gives (Fig. 2.7) τyzxz y 2 − τzyxy z 2 +  τyz + ∂τyz ∂y y  xz y 2 −  τzy + ∂τzy ∂z z  xy z 2 = 0. (2.16) σz σy σx τxz τxy τyz τyx τzy τzx z y x O y∆ x∆ x∆ x τ τ x∆ x σ σ x∆ x τ τ xy xy x x xz xz ∂ ∂ + ∂ ∂ + ∂ ∂ + z∆ z τ τ z∆ z τ τ z∆ z σ σ zx zx zy zy z z ∂ ∂ + ∂ ∂ + ∂ ∂ + y∆ y τ τ y∆ y σ σ y∆ y τ τ yx yx y y yz yz ∂ ∂ + ∂ ∂ + ∂ ∂ + z∆ Figure 2.6: Stresses on the xyz cubic element

8 DISPLACEMENTS,STRAINS,AND STRESSES 0ry△z 0z 0 Tw+ 0rg△y 业 by ← △y Figure 2.7:Stresses on the AxAyAz cubic element that appear in the moment balance about an axis parallel to x and passing through the center(point O). By omitting higher order terms,.which vanish in the limit△x→O,△y→O,△z→O, this equation becomes Tyz Tay. (2.17) Similarly,we obtain the following equalities: Txz Tx Txy Tyx. (2.18) By virtue of Egs.(2.17)and(2.18),the three equilibrium equations(Egs.2.13- 2.15)contain six unknowns,namely,the three normal stresses (ox,oy,o)and the three shear stresses (Tyz,txz,txy). In the x1,x2,x3 coordinate system the equilibrium equations are also given by Egs.(2.13)-(2.15)with x,y,z replaced by x1,x2,x3 and the subscripts x,y,z by 1.2,3 2.3 Stress-Strain Relationships In a composite material the fibers may be oriented in an arbitrary manner.De- pending on the arrangements of the fibers,the material may behave differently in different directions.According to their behavior,composites may be charac- terized as generally anisotropic,monoclinic,orthotropic,transversely isotropic, or isotropic.In the following,we present the stress-strain relationships for these types of materials under linearly elastic conditions. 2.3.1 Generally Anisotropic Material When there are no symmetry planes with respect to the alignment of the fibers the material is referred to as generally anisotropic.A fiber-reinforced composite material is,for example,generally anisotropic when the fibers are aligned in three nonorthogonal directions(Fig.2.8)

8 DISPLACEMENTS, STRAINS, AND STRESSES z y x τyz τzy O z∆ y∆ x∆ z∆ z τ τ zy zy ∂ ∂ + y∆ y τ τ yz yz ∂ ∂ + Figure 2.7: Stresses on the xyz cubic element that appear in the moment balance about an axis parallel to x and passing through the center (point O). By omitting higher order terms, which vanish in the limit x →0, y→0, z→0, this equation becomes τyz = τzy. (2.17) Similarly, we obtain the following equalities: τxz = τzx τxy = τyx. (2.18) By virtue of Eqs. (2.17) and (2.18), the three equilibrium equations (Eqs. 2.13– 2.15) contain six unknowns, namely, the three normal stresses (σx, σy, σz) and the three shear stresses (τyz, τxz, τxy). In the x1, x2, x3 coordinate system the equilibrium equations are also given by Eqs. (2.13)–(2.15) with x, y, z replaced by x1, x2, x3 and the subscripts x, y, z by 1, 2, 3. 2.3 Stress–Strain Relationships In a composite material the fibers may be oriented in an arbitrary manner. De￾pending on the arrangements of the fibers, the material may behave differently in different directions. According to their behavior, composites may be charac￾terized as generally anisotropic, monoclinic, orthotropic, transversely isotropic, or isotropic. In the following, we present the stress–strain relationships for these types of materials under linearly elastic conditions. 2.3.1 Generally Anisotropic Material When there are no symmetry planes with respect to the alignment of the fibers the material is referred to as generally anisotropic. A fiber-reinforced composite material is, for example, generally anisotropic when the fibers are aligned in three nonorthogonal directions (Fig. 2.8)

2.3 STRESS-STRAIN RELATIONSHIPS Figure 2.8:Example of a generally anisotropic material For a generally anisotropic linearly elastic material,in the x,y,z global coor- dinate system,the stress-strain relationships are Ox CHex C12ey +C13e:+C14Yyz+Cisyxz +C16Yxy y C2Ex +Czey +C23e:+C24Yy:+C2sYsz +C26Ysy O:C31Ex +C32Ey +C336:+C34Yy:+C35Yx:+C36Yxy (2.19) Tyz=C41∈x+C42∈y+C43e:+C44yz+C4sh:+C46hy Tx:CsLEx +Cs2ey Cs3e:+Cs4Yy:+Cssyxz+C56Yty Txy =C6LEx +C62Ey +C63e:+C64Yy:+C65Yx:+C66Yxy. Equation(2.19)may be written in the form Ox C11 C12 C13 C14 Cis C16 Ca C22 C23 C24 C25 C26 ∈y 31 C C33 C34 C35 C36 E2 Ca C42 C43 C44 C45 C46 (2.20) Yyz 【x 1 C52 C53 C54 C55 Cs6 Yxz C61 C62 C63 C64 ℃65 C66 Yxy where Cij are the elements of the stiffness matrix [C]in the x,y,z coordinate system. Inversion of Eq.(2.20)results in the following strain-stress relationships: ∈x 下12 S13 下i4 s S167 Ox 21 下 S23 下24 526 2 下38 4 下36 541 下42 下4 下45 下46 (2.21) Tyz Yxz 1 s2 下35 6 1 S62 563 564 S65 56 where Si;are the elements of the compliance matrix [S]in the x,y,z coordinate system and are defined in Table 2.3 (page 10).In this table tests are illustrated that,in principle,could provide means of determining the different compliance matrix elements

2.3 STRESS–STRAIN RELATIONSHIPS 9 Figure 2.8: Example of a generally anisotropic material. For a generally anisotropic linearly elastic material, in the x, y, z global coor￾dinate system, the stress–strain relationships are σx = C11x + C12y + C13z + C14γyz + C15γxz + C16γxy σy = C21x + C22y + C23z + C24γyz + C25γxz + C26γxy σz = C31x + C32y + C33z + C34γyz + C35γxz + C36γxy τyz = C41x + C42y + C43z + C44γyz + C45γxz + C46γxy τxz = C51x + C52y + C53z + C54γyz + C55γxz + C56γxy τxy = C61x + C62y + C63z + C64γyz + C65γxz + C66γxy. (2.19) Equation (2.19) may be written in the form    σx σy σz τyz τxz τxy    =          C11 C12 C13 C14 C15 C16 C21 C22 C23 C24 C25 C26 C31 C32 C33 C34 C35 C36 C41 C42 C43 C44 C45 C46 C51 C52 C53 C54 C55 C56 C61 C62 C63 C64 C65 C66             x y z γyz γxz γxy    , (2.20) where Ci j are the elements of the stiffness matrix [C ] in the x, y, z coordinate system. Inversion of Eq. (2.20) results in the following strain–stress relationships:    x y z γyz γxz γxy    =          S11 S12 S13 S14 S15 S16 S21 S22 S23 S24 S25 S26 S31 S32 S33 S34 S35 S36 S41 S42 S43 S44 S45 S46 S51 S52 S53 S54 S55 S56 S61 S62 S63 S64 S65 S66             σx σy σz τyz τxz τxy    , (2.21) where Si j are the elements of the compliance matrix [S ] in the x, y, z coordinate system and are defined in Table 2.3 (page 10). In this table tests are illustrated that, in principle, could provide means of determining the different compliance matrix elements

10 DISPLACEMENTS,STRAINS,AND STRESSES Table 2.3.The elements of the compliance matrix [S]in the x,y,z coordinate system.The elements S(without bar)in the为，x2,为 coordinate system are obtained by replacing x,y,zby 1,2,3 on the right-hand sides of the expressions. Test Elements of the compliance matrix O: S11=Ex/ox Sa =Yyzlax 521=ey/o Ss1=Yxzlax S31 =E:/ax S61 Yxy/ax S12=Exlay 下42=z/oy Sm=Eylay 下52=k:/oy 下2=e:/oy 下62=yxy/oy 个0= 513=ex/a: 下43=y:/a 523=e,/oz S53=Yxzla: 33=e:/o S63 Yxy/o: 514=ex/y 下4=y:/: 524=ey/r 下4=hx:/ry 534=e:/tu 下64=yxy/ 下15=ex/txt 下4s=:/x: 52s=∈y/tx 下s5=x:/tx 下35=e:/tx 下6s=hx/tx: S16=Ex/Txy 下6=Yy:/ty 下26=ey/xy 下s6=k:/txy 下36=e:/txy 下6=yzy/xy In thex,x2,x3 coordinate system the stress-strain relationships are 01 C11 C12 C13 C14 C15 C16 E1 02 C21 C22C23 C24 C25 C26 2 03 C31 C32 C33 C34 C35 C36 E3 (2.22) t23 C41 C42 C43 C44 C45 C46 23 t13 C51 C52C53 C54 Cs5 C56 T12 C61 C62C6 C64 C65 C66 12 where Ci;are the elements of the stiffness matrix [C]in the x1,x2,x3 coordinate system. By inverting Eq.(2.22)we obtain the following strain-stress relationships: S11 S12 S13 S14 S15S16 6 62 S21S22S23S24S25 S26 2 S31 S32S33 S34S35 S36 03 S41 S42S43 S45 (2.23) 23 S44 S46 T23 13 S51 S52S53 S54S55 Ss6 T13 S61 S62 So3 S64S65 S66 T12 where Sij are the elements of the compliance matrix [S]in the x1,x2,x3 coordinate system

10 DISPLACEMENTS, STRAINS, AND STRESSES Table 2.3. The elements of the compliance matrix [S] in the x, y, z coordinate system. The elements Sij (without bar) in the x1, x2, x3 coordinate system are obtained by replacing x, y, z by 1, 2, 3 on the right-hand sides of the expressions. Test Elements of the compliance matrix σx σx S11 = x/σx S41 = γyz/σx S21 = y/σx S51 = γxz/σx S31 = z/σx S61 = γxy/σx σ S12 = x/σy S42 = γyz/σy σy y S22 = y/σy S52 = γxz/σy S32 = z/σy S62 = γxy/σy σz σz S13 = x/σz S43 = γyz/σz S23 = y/σz S53 = γxz/σz S33 = z/σz S63 = γxy/σz τyz S14 = x/τyz S44 = γyz/τyz S24 = y/τyz S54 = γxz/τyz S34 = z/τyx S64 = γxy/τyz τxz S15 = x/τxz S45 = γyz/τxz S25 = y/τxz S55 = γxz/τxz S35 = z/τxz S65 = γxy/τxz τxy S16 = x/τxy S46 = γyz/τxy S26 = y/τxy S56 = γxz/τxy S36 = z/τxy S66 = γxy/τxy In the x1, x2, x3 coordinate system the stress–strain relationships are    σ1 σ2 σ3 τ23 τ13 τ12    =          C11 C12 C13 C14 C15 C16 C21 C22 C23 C24 C25 C26 C31 C32 C33 C34 C35 C36 C41 C42 C43 C44 C45 C46 C51 C52 C53 C54 C55 C56 C61 C62 C63 C64 C65 C66             1 2 3 γ23 γ13 γ12    , (2.22) where Ci j are the elements of the stiffness matrix [C] in the x1, x2, x3 coordinate system. By inverting Eq. (2.22) we obtain the following strain–stress relationships:    1 2 3 γ23 γ13 γ12    =          S11 S12 S13 S14 S15 S16 S21 S22 S23 S24 S25 S26 S31 S32 S33 S34 S35 S36 S41 S42 S43 S44 S45 S46 S51 S52 S53 S54 S55 S56 S61 S62 S63 S64 S65 S66             σ1 σ2 σ3 τ23 τ13 τ12    , (2.23) where Si j are the elements of the compliance matrix [S] in the x1, x2, x3 coordinate system

2.3 STRESS-STRAIN RELATIONSHIPS 11 Plane of symmetry Figure 2.9:Illustrations of fiber-reinforced monoclinic materials.The fibers are only in planes parallel to the x-x2 plane of symmetry (top),only perpendicular to the plane of symmetry (middle),and in the plane of symmetry and perpendicular to the plane of symmetry (bottom). It is evident from Eqs.(2.20)-(2.23)that the compliance matrix [S]is the inverse of the stiffness matrix [C]: [同=@]-1[g=[C-1. (2.24) It can be shown(see Section 2.11.1)that for an elastic material the stiffness and compliance matrices are symmetrical in both the x,y,z and x,x2,x3 coordinate systems as follows: Sii=Sii Sij=Sji Cij=Ci C=Cii,j=1,2,.,6. (2.25) Because of this symmetry,in both the [S]and the [C]matrices only 21 of the 36 elements are independent. 2.3.2 Monoclinic Material When there is a symmetry plane with respect to the alignment of the fibers,the material is referred to as monoclinic.Examples of monoclinic fiber-reinforced composites are shown in Figure 2.9. For a monoclinic material we specify the compliance [S]and stiffness [C] matrices in an x1,x2,x3 coordinate system chosen in such a way thatx and x2 are in the plane of symmetry,whereas x3 is perpendicular to this plane

2.3 STRESS–STRAIN RELATIONSHIPS 11 x1 x3 x2 Plane of symmetry x1 x3 x2 x1 x3 x2 Figure 2.9: Illustrations of fiber-reinforced monoclinic materials. The fibers are only in planes parallel to the x1–x2 plane of symmetry (top), only perpendicular to the plane of symmetry (middle), and in the plane of symmetry and perpendicular to the plane of symmetry (bottom). It is evident from Eqs. (2.20)–(2.23) that the compliance matrix [S]is the inverse of the stiffness matrix [C]: [S] = [C] −1 [S] = [C] −1 . (2.24) It can be shown (see Section 2.11.1) that for an elastic material the stiffness and compliance matrices are symmetrical in both the x, y, z and x1, x2, x3 coordinate systems as follows: Si j = Sji Si j = Sji Ci j = C ji Ci j = Cji i, j = 1, 2,..., 6. (2.25) Because of this symmetry, in both the [S] and the [C] matrices only 21 of the 36 elements are independent. 2.3.2 Monoclinic Material When there is a symmetry plane with respect to the alignment of the fibers, the material is referred to as monoclinic. Examples of monoclinic fiber-reinforced composites are shown in Figure 2.9. For a monoclinic material we specify the compliance [S] and stiffness [C] matrices in an x1, x2, x3 coordinate system chosen in such a way that x1 and x2 are in the plane of symmetry, whereas x3 is perpendicular to this plane

12 DISPLACEMENTS,STRAINS,AND STRESSES Generally anisotropic Monoclinic 3 Plane of symmetry 个工3 Figure 2.10:The normal stress o causes shear strain y3 in a generally anisotropic material(left) and no shear strain in a monoclinic material(right). The elements of the compliance matrix for a monoclinic material are ob- tained by modifying the compliance matrix of a generally anisotropic material. We observe that in a generally anisotropic material a normal stress o1 causes an out-of-plane shear strain y3(Fig.2.10 left),but in a monoclinic material subjected to a normal stress o1(o being in the plane of symmetry)the out-of-plane shear strain yi3 is zero(Fig.2.10 right).Consequently,for a monoclinic material the Ssi element of the compliance matrix is zero.By similar arguments it can be shown that Table 2.4.The elements of the compliance matrix for monoclinic materials.For orthotropic,transversely isotropic,and isotropic materials S16=S61=0,S26=S62=0,S36=S63=0,S45=S4=0 Test Elements of the compliance matrix 01 S1=1/o S1=3/o1=0 S21=e2/a1 S1=3/o1=0 S31=e3/o1 S61=2/o1 S12=e1/o2 S42=y3/o2=0 Sp=E2/02 S2=h3/o2=0 S2=e3/o2 S2=h2/o2 个03 S3=e1/o3 S3=ys/o3=0 S23=e2/o3 S3=h3/o3=0 S3=e3/o3 S3=12/o3 「23 Si4=e1/t23=0 S44=ys/23 S24=e2/T3=0 S4=13/3 S4=e3/23=0 S64=h2/t23=0 Si5=e1/t13=0 Sas =y/t13 S2s=e2/t13=0 Sss =y13/t13 S3s=e3/B=0 S6s=12/T13=0 T12 S6=e1/t12 S46=y3/t12=0 S26=e2/t12 S6=13/t2=0 S36=e3/t12 S66=y12/t12

12 DISPLACEMENTS, STRAINS, AND STRESSES Plane of symmetry σ1 σ1 σ1 σ1 σ1 σ1 x1 x1 x1 x2 x3 x3 x3 γ13 Generally anisotropic Monoclinic σ1 σ1 x1 x2 x3 Figure 2.10: The normal stress σ1 causes shear strain γ13 in a generally anisotropic material (left) and no shear strain in a monoclinic material (right). The elements of the compliance matrix for a monoclinic material are ob￾tained by modifying the compliance matrix of a generally anisotropic material. We observe that in a generally anisotropic material a normal stress σ1 causes an out-of-plane shear strain γ13 (Fig. 2.10 left), but in a monoclinic material subjected to a normal stress σ1 (σ1 being in the plane of symmetry) the out-of-plane shear strain γ13 is zero (Fig. 2.10 right). Consequently, for a monoclinic material the S51 element of the compliance matrix is zero. By similar arguments it can be shown that Table 2.4. The elements of the compliance matrix for monoclinic materials. For orthotropic, transversely isotropic, and isotropic materials S16 = S61 = 0, S26 = S62 = 0, S36 = S63 = 0, S45 = S54 = 0 Test Elements of the compliance matrix σ1 σ1 S11 = 1/σ1 S41 = γ23/σ1 = 0 S21 = 2/σ1 S51 = γ13/σ1 = 0 S31 = 3/σ1 S61 = γ12/σ1 σ S12 = 1/σ2 S42 = γ23/σ2 = 0 σ2 2 S22 = 2/σ2 S52 = γ13/σ2 = 0 S32 = 3/σ2 S62 = γ12/σ2 σ3 σ3 S13 = 1/σ3 S43 = γ23/σ3 = 0 S23 = 2/σ3 S53 = γ13/σ3 = 0 S33 = 3/σ3 S63 = γ12/σ3 τ23 S14 = 1/τ23 = 0 S44 = γ23/τ23 S24 = 2/τ23 = 0 S54 = γ13/τ23 S34 = 3/τ23 = 0 S64 = γ12/τ23 = 0 τ13 S15 = 1/τ13 = 0 S45 = γ23/τ13 S25 = 2/τ13 = 0 S55 = γ13/τ13 S35 = 3/τ13 = 0 S65 = γ12/τ13 = 0 τ12 S16 = 1/τ12 S46 = γ23/τ12 = 0 S26 = 2/τ12 S56 = γ13/τ12 = 0 S36 = 3/τ12 S66 = γ12/τ12