15 COMPOSITE BEAMS IN FLEXURE Due to their slenderness,a number of composite elements(mechanical components or structural pieces)can be considered as beams.A few typical examples are shown schematically in Figure 15.1.The study of the behavior under loading of these elements (evaluation of stresses and displacements)becomes a very complex problem when one gets into three-dimensional aspects.In this chapter,we propose a monodimensional approach to the problem in an original method.It consists of the definition of displacements corresponding to the traditional stress and moment resultants for the applied loads.This leads to a homogenized formulation for the flexure-and for torsion.This means that the equilibrium and behavior relations are formally identical to those that characterize the behavior of classical homogeneous beams.Utilization of these relations for the calculation of stresses and displacements then leads to expressions that are analogous to the common beams. We will limit ourselves to the composite beams with constant characteristics (geometry,materials)in any cross section,made of different materials-which we call phases-that are assumed to be perfectly bonded to each other. To clarify the procedure and for better simplicity in the calculations,we will limit ourselves in this chapter to the case of composite beams with isotropic phases. The extension to the transversely isotropic materials is immediate.When the phases are orthotropic,with eventually orthotropic directions that are changing from one point to another in the section,the study will be analogous,with a much more involved formulation. 15.1 FLEXURE OF SYMMETRIC BEAMS WITH ISOTROPIC PHASES In the following,D symbolizes the domain occupied by the cross section,in the y,z plane.The external frontier is denoted as OD.One distinguishes also (see Figure 15.2)the internal frontiers which limit the phases,denoted by l for two contiguous phases i and j.The area of the phase i is denoted as S;its moduli of elasticity are denoted by E,and G.The elastic displacement at any point of the beam has the components:ux (x,y,2);u (x,y,2);u (x,y,z). The beam is bending in the plane of symmetry xy under external loads which are also symmetric with respect to this plane. The only restrictive condition lies in the fact that one of the orthotropic directions is supposed to remain parallel to the longitudinal axis of the beam. 2003 by CRC Press LLC
15 COMPOSITE BEAMS IN FLEXURE Due to their slenderness, a number of composite elements (mechanical components or structural pieces) can be considered as beams. A few typical examples are shown schematically in Figure 15.1. The study of the behavior under loading of these elements (evaluation of stresses and displacements) becomes a very complex problem when one gets into three-dimensional aspects. In this chapter, we propose a monodimensional approach to the problem in an original method. It consists of the definition of displacements corresponding to the traditional stress and moment resultants for the applied loads. This leads to a homogenized formulation for the flexure—and for torsion. This means that the equilibrium and behavior relations are formally identical to those that characterize the behavior of classical homogeneous beams. Utilization of these relations for the calculation of stresses and displacements then leads to expressions that are analogous to the common beams. We will limit ourselves to the composite beams with constant characteristics (geometry, materials) in any cross section, made of different materials—which we call phases—that are assumed to be perfectly bonded to each other. To clarify the procedure and for better simplicity in the calculations, we will limit ourselves in this chapter to the case of composite beams with isotropic phases. The extension to the transversely isotropic materials is immediate. When the phases are orthotropic, with eventually orthotropic directions that are changing from one point to another in the section, the study will be analogous, with a much more involved formulation.1 15.1 FLEXURE OF SYMMETRIC BEAMS WITH ISOTROPIC PHASES In the following, D symbolizes the domain occupied by the cross section, in the y,z plane. The external frontier is denoted as ∂D. One distinguishes also (see Figure 15.2) the internal frontiers which limit the phases, denoted by lij for two contiguous phases i and j. The area of the phase i is denoted as Si ; its moduli of elasticity are denoted by Ei and Gi . The elastic displacement at any point of the beam has the components: ux (x,y,z); uy (x,y,z); uz (x,y,z). The beam is bending in the plane of symmetry x,y under external loads which are also symmetric with respect to this plane. 1 The only restrictive condition lies in the fact that one of the orthotropic directions is supposed to remain parallel to the longitudinal axis of the beam. TX846_Frame_C15 Page 283 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
laminated laminated unidirectional honey comb laminated foam foam blade box beam spar laminated steel aluminum foam unidirectional wood ski leaf spring transmission shaft Figure 15.1 Composite Beams S Figure 15.2 Composite Beam,with Plane of Symmetry 15.1.1 Degrees of Freedom 15.1.1.1 Equivalent Stiffness One writes in condensed form the following integrals taken over the total cross section: (E=Eds or=∑EBS number of phases (）=∫Esor=∑ (15.1) number of phases (G=∫Gids or=∑G number of phases is the quadratic moment of the Phase i with respect with z axis. 2003 by CRC Press LLC
15.1.1 Degrees of Freedom 15.1.1.1 Equivalent Stiffness One writes in condensed form the following integrals taken over the total cross section2 : (15.1) Figure 15.1 Composite Beams Figure 15.2 Composite Beam, with Plane of Symmetry 2 Izi is the quadratic moment of the Phase i with respect with z axis. · Ò ES EidS or EiSi i = ÚD = Â number of phases EIz · Ò Ei y 2 dS or EiIzi i = ÚD = Â number of phases · Ò GS GidS or GiSi i = ÚD = Â number of phases TX846_Frame_C15 Page 284 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
15.1.1.2 Longitudinal Displacement By definition,longitudinal displacement is denoted by u(x)and written as: u(0)=7 (.y.zy which consists of a mean displacement ux)and an incremental displacement △uxas: ux(x,y,z)=u(x)+Aux(x,y,z) where one notes that: E,△uxdS=0 (15.2) JD 15.1.1.3 Rotations of the Sections By definition,this is the fictitious rotation-or equivalent-given by the following expression: 0(x) 品 Eux(x,y,z)×ydS Or,with the above: 15.1.1.4 Elastic Center Origin 0 of the coordinate y is chosen such that the following integral is zero: Eyds =0 We call elastic center the corresponding point 0 that is located as in the expression above.Then Auy takes the form: Aux(x,y,z)=-ye(x)+nx(x,y,z) with: ∫and∫En.s=09 The displacement ux(,y,2)can then take the form: ux(x,y,2)=u(x)-ye(x)+n(x,y,z). 3 In what follows,the second property is the consequence of Equation 15.2. 2003 by CRC Press LLC
15.1.1.2 Longitudinal Displacement By definition, longitudinal displacement is denoted by u(x) and written as: which consists of a mean displacement u(x) and an incremental displacement Dux as: where one notes that: (15.2) 15.1.1.3 Rotations of the Sections By definition, this is the fictitious rotation—or equivalent—given by the following expression: Or, with the above: 15.1.1.4 Elastic Center Origin 0 of the coordinate y is chosen such that the following integral is zero: We call elastic center the corresponding point 0 that is located as in the expression above. Then Dux takes the form: with3 : The displacement ux (x,y,z) can then take the form: 3 In what follows, the second property is the consequence of Equation 15.2. u x( ) 1 · Ò ES ----------- Eiux( ) x, y, z dS DÚ = ux( ) x, y, z = u x() D + ux( ) x, y, z Ei Dux dS DÚ = 0 qz( ) x –1 EIz · Ò ------------ Eiux( ) x, y, z ¥ y dS DÚ = qz( ) x –1 EIz · Ò ------------ u x( ) Ei ydS DÚ Ei DÚ + Dux( ) x, y, z y dS Ó ˛ Ì ˝ Ï ¸ = Ei ydS DÚ = 0 Dux( ) x, y, z = –yqz( ) x + hx( ) x, y, z Eihx ydS DÚ and Eihx dS = 0 *( ) DÚ ux( ) x, y, z = u x( ) – yqz( ) x + hx( ) x, y, z . TX846_Frame_C15 Page 285 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
15.1.1.5 Transverse Displacement along y Direction By definition,this is v)that is given by the following expression: Gu(x,y.z)ds It follows from this definition that: u(x,y,z)=v(x)+n(x,y,z) where one notes that: ∫，Gns=0 15.1.1.6 Transverse Displacement along z Direction By definition,this is w(x)given by u)=高JGu.y2s 1 It follows from this definition and from the existence of the plane of symmetry x,y of the beam a zero average transverse displacement,as:w(x)=0. uz(x,y,z)=0+nz(x,y,z), with ∫nGn:s=0 In summary,we obtain the following elastic displacement field: (ux =u(x)-ye-(x)+nx(x,y,z) 4y=(x)+ n(x,y,z) (15.3) uz= n2(x,y,z) The origin of the axes is the elastic center such that: Eyds=0 (15.4) The three-dimensional incremental displacements nn n with respect to the unidimensional approximation u,v,0.verify the following: ∫En:s=n.=0 ∫nGn,d5=0 (15.5) ∫nGn:as=0 2003 by CRC Press LLC
15.1.1.5 Transverse Displacement along y Direction By definition, this is v(x) that is given by the following expression: It follows from this definition that: where one notes that: 15.1.1.6 Transverse Displacement along z Direction By definition, this is w(x) given by It follows from this definition and from the existence of the plane of symmetry x,y of the beam a zero average transverse displacement, as: w(x) = 0. In summary, we obtain the following elastic displacement field: (15.3) The origin of the axes is the elastic center such that: (15.4) The three-dimensional incremental displacements hx, hy, hz with respect to the unidimensional approximation u, v, qz verify the following: (15.5) v x( ) 1 · Ò GS ------------ Giuy( ) x, y, z dS DÚ = uy( ) x, y, z = v x( ) + hy( ) x, y, z Gihy dS DÚ = 0. w x( ) 1 · Ò GS ------------ Giuz( ) x, y, z dS DÚ = uz( ) x, y, z 0 hz( ) x, y, z , with Gihz dS DÚ = + = 0. ux = u x( ) – yqz( ) x + hx( ) x, y, z uy = v x( ) + hy( ) x, y, z uz Ë = hz( ) x, y, z Á Á Ê Ei y dS DÚ = 0 Eihx dS DÚ Eiyhx dS = 0 DÚ = Gihy dS = 0 DÚ Gihz dS = 0 DÚ TX846_Frame_C15 Page 286 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
Remarks: nx represents the longitudinal distortion of a cross section,that is,the quantity that this section displaces out of the plane which characterizes it if it moves truly as a rigid plane body. ny and n represent the displacements that characterize the variations of the form of the cross section in its initial plane. 15.1.2 Perfect Bonding between the Phases 15.1.2.1 Displacements The bonding is assumed to be perfect.Then the displacements are continuous when crossing through the interface between two phases in contact.For two phases in contact i and j,one has: 4x⑦=u.① 4,⑦ =4,① uz①=u,① 15.1.2.2 Strains For the phases i and j in Figure 15.3,in the plane of an elemental interface with a normal vector of n,the relations between the strain tensors g are: .e0=,erO 1.a T.an-i which can also be written as: Exx =Exx ①① -Exyn:+Exzny =-Exynz+Exzny ① ① ① ① Exn2E=Eyn-2Ey 2 ① ① ① 2003 by CRC Press LLC
Remarks: hx represents the longitudinal distortion of a cross section, that is, the quantity that this section displaces out of the plane which characterizes it if it moves truly as a rigid plane body. hy and hz represent the displacements that characterize the variations of the form of the cross section in its initial plane. 15.1.2 Perfect Bonding between the Phases 15.1.2.1 Displacements The bonding is assumed to be perfect. Then the displacements are continuous when crossing through the interface between two phases in contact. For two phases in contact i and j, one has: 15.1.2.2 Strains For the phases i and j in Figure 15.3, in the plane of an elemental interface with a normal vector of , the relations between the strain tensors e are: which can also be written as: ux i ux = j uy i uy = j uz i uz = j n x e( ) x i ◊ x e( ) x j = ◊ t e( ) x i ◊ t e( ) x j = ◊ t e( )t i ◊ t e( )t j = ◊ exx = exx i j – exynz + exzny = – exynz + exzny ii j j e yynz 2 – 2e yznynz e zzny 2 + e yynz 2 – 2e yznynz e zzny 2 = + ii i j j j TX846_Frame_C15 Page 287 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC