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
n Figure 15.3 Interface between Two Phases 15.1.2.3 Stresses The stress vector=E(),where E represents the stress tensor,remains con- tinuous across an element of the interface with normal as: Txyny+Txznz Txyny+Txznz ① ① ① ① Oyny+Tyznz Oyny+Tyznz (15.6 ① ① ① ① Tyzny+Ozzn:Tyzny+Ozzn: 15.1.3 Equilibrium Relations Starting from the local equilibrium,in the absence of body forces,we have ∂oy=0 xj By integrating over the cross section: +(紧+是}s=0 where the normal stress resultant Nx appears as: N-Soud5. Then,transforming the second integral to an integral over the frontier aD of D': 袋+小+r=0 Note that equalitydsis made possible due to the conti- nuity of the expression (tn+tn)across the interfaces between the different phases (see Equation 15.6). 2003 by CRC Press LLC
15.1.2.3 Stresses The stress vector , where S represents the stress tensor, remains continuous across an element of the interface with normal as: (15.6) 15.1.3 Equilibrium Relations Starting from the local equilibrium, in the absence of body forces, we have By integrating over the cross section: where the normal stress resultant Nx appears as: Then, transforming the second integral to an integral over the frontier ∂D of D4 : Figure 15.3 Interface between Two Phases 4 Note that equality is made possible due to the continuity of the expression (txy ny + txz nz) across the interfaces between the different phases (see Equation 15.6). s = S( ) n n txyny + txznz = txyny + txznz ii jj syyny + t yznz = syyny + t yznz ii jj t yzny + szznz = t yzny + szznz i i jj ∂sij ∂xj --------- = 0 d dx------ sxxdS ∂txy ∂y --------- ∂txz ∂z + --------- Ë ¯ Ê ˆ dS = 0 DÚ + DÚ Nx sxx dS . DÚ = ÚD ∂t xy ∂y ------ ∂t xz ∂z ( ) + ------ dS Ú∂D t xyny t + xznz = ( )dG dNx dx --------- txyny + txznz ( )dG ∂D Ú + = 0 TX846_Frame_C15 Page 288 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
in which ny and n,are the cosines of the outward normal n,and dI represents element of frontier OD.If one assumes the absence of shear stresses applied over the lateral surface of the beam,then t n+tn=0 along the external frontier OD.Then for longitudinal equilibrium we have dNs= dx 引s+(+先s=0 where one recognizes the shear stress resultant: =n Then transforming the second integral into an integral over the external frontier OD of the domain D of the cross section°： 2张+no+=r=0 if one remarks that: ∫non4,+5.ar=jn-z动ar=jnar=p,Wm aD which is the transverse density of loading on the lateral surface of the beam, transverse equilibrium can be written as: +p,=0 dx 引。o.45+∫n-费+=0 5 We have neglected the body forces which appear in the local equation of equilibrium in the form of a function If these exist (inertia forces,centrifugal forces,or vibration inertia,for example),one obtains for the equilibrium:o in which Pds represents the longitudinal load density. 6 Note that the equalityds(dr is made possible due to the continuity of the expression n+-n.across the frontier lines between different phases (see Equation 15.6). 2003 by CRC Press LLC
in which ny and nz are the cosines of the outward normal , and d G represents element of frontier ∂D. If one assumes the absence of shear stresses applied over the lateral surface of the beam, then txy ny + txz nz = 0 along the external frontier ∂D. Then for longitudinal equilibrium we have5 where one recognizes the shear stress resultant: Then transforming the second integral into an integral over the external frontier ∂D of the domain D of the cross section6 : if one remarks that: which is the transverse density of loading on the lateral surface of the beam, transverse equilibrium can be written as: 5 We have neglected the body forces which appear in the local equation of equilibrium in the form of a function fx. If these exist (inertia forces, centrifugal forces, or vibration inertia, for example), one obtains for the equilibrium: in which represents the longitudinal load density. 6 Note that the equality is made possible due to the continuity of the expression (syy ny + tyz nz ) across the frontier lines between different phases (see Equation 15.6). n dNx dx ------ + px = 0 px ÚD f = xdS dNx dx --------- = 0 d dx------ txydS ∂syy ∂y ---------- ∂t yz ∂z + --------- Ë ¯ Ê ˆ dS = 0 DÚ + DÚ Ty txydS. DÚ = ÚD ∂syy ∂y ------- ∂t yz ∂z ( ) + ------ dS Ú∂D syyny t + yznz = ( )dG ∂Ty ∂x -------- syyny + t yznz ( )d G ∂D Ú + = 0 syyny + t yznz ( )dG ∂D Ú y S( ) n dG y sdG ∂D Ú ◊ = ◊ ∂D Ú = = py ( ) N/m dTy dx -------- + py = 0 d dx------ –ysxxdS y ∂txy ∂y --------- ∂txz ∂z + --------- Ë ¯ Ê ˆ – dS = 0 DÚ + DÚ TX846_Frame_C15 Page 289 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
where appears the moment resultant: M.=∫p-yos Then transforming the second integral': 0+∫n%+dr+ns=0 where one notes that: ∫n-pcg%,+.,)r=jn-定.2（动ar=jnar=hmNm which can be called a density moment on the beam.Then one obtains the equilibrium relation: M+T,+4:=0 dx The case where a density moment could exist in statics is practically nil,we therefore assume that u=0.In summary,one obtains for the equations of equilibrium: s= dx +p,=0 (15.7) dx dM: dx +T,=0 15.1.4 Constitutive Relations Taking into account the isotropic nature of the different phases,the constitutive relation can be written in tensor form for Phase i as: tr( E= (I=unity tensor) E Analogous note for the continuity of the expression (across the lines of internal interfaces (Equation 15.6). 2003 by CRC Press LLC
where appears the moment resultant: Then transforming the second integral7 : where one notes that: which can be called a density moment on the beam. Then one obtains the equilibrium relation: The case where a density moment could exist in statics is practically nil, we therefore assume that mz = 0. In summary, one obtains for the equations of equilibrium: (15.7) 15.1.4 Constitutive Relations Taking into account the isotropic nature of the different phases, the constitutive relation can be written in tensor form for Phase i as: 7 Analogous note for the continuity of the expression (txy ny + txz nz) across the lines of internal interfaces (Equation 15.6). Mz –ysxxdS. DÚ = dMz dx ---------- y txyny + txznz – ( )dG txydS DÚ + ∂D Ú + = 0 y txyny + txznz – ( )dG ∂D Ú –yx S( ) n dGº –y( ) s ◊ x dG ∂D Ú ◊ = ∂D Ú = = mz( ) mN/m dMz dx ---------- + + Ty mz = 0 dNx dx --------- = 0 dTy dx -------- + py = 0 dMz dx ---------- + Ty = 0 e 1 + ni Ei -------------S ni Ei = – ----tr( ) S I (I = unity tensor) TX846_Frame_C15 Page 290 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
One deduces,in integrating over the domain occupied by the cross section of the beam: eEds=∫noas-Jn Oy+:2)dS Taking into account the form of the displacements in Equation 15.3,one can write B=装=-影6+nE+别n which leads,with the notation in Equation 15.1,to the relation: N.=张+∫，vo+a.a西 (15.8) SyeEds-Jyods+Sv.ds Taking into account the form of the displacements in Equation 15.3 one can write e品=∫E-u-m. This leads,with the notation in Equation 15.1,to the relation: M=(喂-∫no+a n2eds=jnws (15.9) Taking into account the form of the displacements in Equation 15.3,one can write ∫n2a=j(等+装}=-ncds +∫cs+cs+n which gives with the notation in Equation 15.1,the relation: 五=(c偎-+∫，c映s (15.10) 2003 by CRC Press LLC
One deduces, in integrating over the domain occupied by the cross section of the beam: Taking into account the form of the displacements in Equation 15.3, one can write which leads, with the notation in Equation 15.1, to the relation: (15.8) Taking into account the form of the displacements in Equation 15.3 one can write This leads, with the notation in Equation 15.1, to the relation: (15.9) Taking into account the form of the displacements in Equation 15.3, one can write which gives with the notation in Equation 15.1, the relation: (15.10) exx EidS sxxdS ni syy + szz ( )dS DÚ – DÚ = DÚ exx Ei dS ∂ux ∂x --------Ei dS dqz dx –-------- yEi dS du dx ------ Ei dS ∂ ∂x ------ Eihx dS DÚ + DÚ + DÚ = DÚ = DÚ Nx · Ò ES du dx ------ ni syy + szz ( )dS DÚ = + –yexxEi dS y– sxx dS ni y syy + szz ( )dS DÚ + DÚ = DÚ –yexxEi dS dqz dx -------- Ei y 2 dS du dx ------ Ei y dS ∂ ∂x ------ Ei yhx dS DÚ – DÚ – DÚ = DÚ Mz EIz · Òdqz dx -------- ni y syy + szz ( )dS DÚ = – 2exyGidS txy dS DÚ = DÚ 2exyGidS ∂ux ∂y -------- ∂uy ∂x + -------- Ë ¯ Ê ˆ GidS –qz GidSº DÚ = DÚ = DÚ º Gi ∂hx ∂y --------dS DÚ dv dx------ GidS ∂ ∂x ------ hyGidS DÚ + DÚ + + Ty · Ò GS dv dx------ – qz Ë ¯ Ê ˆ Gi ∂hx ∂y --------dS DÚ = + TX846_Frame_C15 Page 291 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
15.1.5 Technical Formulation 15.1.5.1 Simplifications We extend to the composite beams the simplifications made for the homogeneous beams as: 1. and de. du dx (Equation 15.9)and N# (Equation 15.8): M2 ox=-B面'+B, (15.11) bending extension Remark:The continuity (=(at the interface between the Phases i and leads to E E This hypothesis is better verified by the fact that the Poisson coefficients of the different phases have similar values. 9 This hypothesis is known in the literature for the homogeneous beams as the generalized "Navier-Bernoulli"hypothesis. See Section 15.1.2. 2003 by CRC Press LLC
15.1.5 Technical Formulation 15.1.5.1 Simplifications We extend to the composite beams the simplifications made for the homogeneous beams as: 1. syy and szz -------- Nx # ES du dx ------ sxx Ei Mz EIz · Ò – ------------y Ei Nx · Ò ES = + ----------- bending extension sxx ( )i Ei ------------- sxx ( )j Ej = ------------- TX846_Frame_C15 Page 292 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
