Copyrighted Materials CpUyPress o CHAPTER SIX Beams The response of composite beams to loading is more complex than that of isotropic beams,and the analyses of composite beams must take these complexities into account.This requires analyses that are,by necessity,more involved than for isotropic beams but which,nonetheless,result in expressions readily amenable to numerical computations. In this chapter we treat rectangular solid cross sections as well as thin-walled beams that undergo small deformations and in which the material behaves in a linearly elastic manner.We neglect shear deformations and adopt the Bernoulli- Navier hypothesis,according to which the originally plane cross sections of a beam undergoing bending remain plane and perpendicular to the axis of the beam. Axial,transverse,and torque loads may be applied to the beam(Fig.6.1),re- sulting in the following internal forces:normal force N;bending moments My.M; torque T:and the transverse shear forcesVV,(Fig.6.2). 6.1 Governing Equations The response of a beam to the applied forces is described by the strain- displacement,force-strain,and equilibrium equations.These equations are given in this section for conditions in which there is no restrained warping.The effect of restrained warping is discussed in Sections 6.5.5,6.5.6,and 6.6.4. Here,as well as in the following analyses,we employ an x-y-z coordinate system with the origin at the centroid.The centroid is defined such that an axial load acting at the centroid does not change the curvature of the axis passing through the centroid.As a consequence of this definition,a bending moment acting on the beam does not introduce an axial strain along this axis.Unlike for isotropic beams,for composite beams the centroid does not necessarily coincide with the center of gravity of the cross section. There are four independent displacements(Fig.6.3):the axial displacement u, the transverse displacements v and w in the y and z directions,respectively,and the twist of the cross section.The corresponding axial strain e,curvatures 1/Py 203
CHAPTER SIX Beams The response of composite beams to loading is more complex than that of isotropic beams, and the analyses of composite beams must take these complexities into account. This requires analyses that are, by necessity, more involved than for isotropic beams but which, nonetheless, result in expressions readily amenable to numerical computations. In this chapter we treat rectangular solid cross sections as well as thin-walled beams that undergo small deformations and in which the material behaves in a linearly elastic manner. We neglect shear deformations and adopt the Bernoulli– Navier hypothesis, according to which the originally plane cross sections of a beam undergoing bending remain plane and perpendicular to the axis of the beam. Axial, transverse, and torque loads may be applied to the beam (Fig. 6.1), resulting in the following internal forces: normal force N; bending moments My, Mz; torque T ; and the transverse shear forces V z, V y (Fig. 6.2). 6.1 Governing Equations The response of a beam to the applied forces is described by the strain– displacement, force–strain, and equilibrium equations. These equations are given in this section for conditions in which there is no restrained warping. The effect of restrained warping is discussed in Sections 6.5.5, 6.5.6, and 6.6.4. Here, as well as in the following analyses, we employ an x–y–z coordinate system with the origin at the centroid. The centroid is defined such that an axial load acting at the centroid does not change the curvature of the axis passing through the centroid. As a consequence of this definition, a bending moment acting on the beam does not introduce an axial strain along this axis. Unlike for isotropic beams, for composite beams the centroid does not necessarily coincide with the center of gravity of the cross section. There are four independent displacements (Fig. 6.3): the axial displacement u, the transverse displacements v and w in the y and z directions, respectively, and the twist of the cross section ψ. The corresponding axial strain o x , curvatures 1/ρy 203
204 BEAMS 与与与 Figure 6.1:Axial,transverse,and torque loads acting on a section of a beam. and 1/p,in the x-z and x-y planes,and the rate of twist are defined through the strain-displacement relationships Bu 12v 82w a业 = ax 8x2 ax (6.1) Pz Py The generalized force-strain relationship is defined as N P P2 Pi3 P2 (6.2) Pi3 P33 4 P2A P4 where Pi are the elements of the stiffness matrix. The equilibrium equations for a straight beam subjected to the loads shown in Figure 6.1 are2 aN aT ax =一Px ax =一Py ap ax (6.3) ax =-P aMy -V: aM: = ax ax The preceding three sets of equations,(together with the appropriate bound- ary conditions)completely describe the displacements of,and the forces in,a composite beam. The internal forces N,My,M,V,V,and T are determined by the simulta- neous solution of Eqs.(6.1)-(6.3)together with the appropriate boundary condi- tions given below.When a beam is statically determinate,the internal forces can be obtained from the equilibrium equations.When a composite beam is statically indeterminate,the internal forces can be obtained with the use of replacement stiffnesses in the relevant isotropic beam expressions provided that either the beam is orthotropic or the cross section is symmetrical and the load is applied in the plane of symmetry.The concepts of orthotropic beam and replacement stiffnesses are discussed in Section 6.1.2. 1 T.H.G.Megson,Aircraft Structures for Engineering Students.3rd edition.Halsted Press,John Wiley Sons,New York,1999,p.284. 2 B.K.Donaldson,Analysis of Aircraft Structures.An Introduction.McGraw-Hill.New York.1993. Pp.277-278
204 BEAMS z x y pz z py z t x y x y z y x px Figure 6.1: Axial, transverse, and torque loads acting on a section of a beam. and 1/ρz in the x–z and x–y planes,1 and the rate of twist ϑ are defined through the strain–displacement relationships o x = ∂u ∂x 1 ρz = −∂2v ∂x2 1 ρy = −∂2w ∂x2 ϑ = ∂ψ ∂x . (6.1) The generalized force–strain relationship is defined as N My Mz T = P11 P12 P13 P14 P12 P22 P23 P24 P13 P23 P33 P34 P14 P24 P34 P44 o x 1 ρy 1 ρz ϑ , (6.2) where Pij are the elements of the stiffness matrix. The equilibrium equations for a straight beam subjected to the loads shown in Figure 6.1 are2 ∂N ∂x = −px ∂T ∂x = −t ∂V y ∂x = −py ∂V z ∂x = −pz ∂My ∂x = V z ∂Mz ∂x = V y. (6.3) The preceding three sets of equations, (together with the appropriate boundary conditions) completely describe the displacements of, and the forces in, a composite beam. The internal forces N, My, Mz, V y, V z, and T are determined by the simultaneous solution of Eqs. (6.1)–(6.3) together with the appropriate boundary conditions given below. When a beam is statically determinate, the internal forces can be obtained from the equilibrium equations. When a composite beam is statically indeterminate, the internal forces can be obtained with the use of replacement stiffnesses in the relevant isotropic beam expressions provided that either the beam is orthotropic or the cross section is symmetrical and the load is applied in the plane of symmetry. The concepts of orthotropic beam and replacement stiffnesses are discussed in Section 6.1.2. 1 T. H. G. Megson, Aircraft Structures for Engineering Students. 3rd edition. Halsted Press, John Wiley & Sons, New York, 1999, p. 284. 2 B. K. Donaldson, Analysis of Aircraft Structures. An Introduction. McGraw-Hill, New York, 1993, pp. 277–278
6.1 GOVERNING EQUATIONS 205 个 M. N, Figure 6.2:The normal force N:the bending moments My.M:;the torque T;and the transverse shear forces V,V.inside a beam. 6.1.1 Boundary Conditions At a built-in end,the in-plane displacements and the slopes are zero.At a simply supported end,the in-plane displacements and the moments are zero.At a free end,the moments and the transverse shear forces are zero. When the end of the beam is restrained axially,the axial displacement is zero. When the end is not restrained axially,the axial force is zero. When the end may rotate,the torque is zero.When the end is rotationally restrained,the twist is zero. The preceding boundary conditions are summarized in Table 6.1. 6.1.2 Stiffness Matrix The stiffness matrix depends on the geometry of the cross section and on the type of material used in the construction of the beam.The geometry (i.e.,the shape) of the cross section changes when the beam is loaded.We neglect the effects of this change in shape on the stiffness and evaluate the stiffness matrix for the cross section of the unloaded beam. Figure 6.3:Displacements of a beam
6.1 GOVERNING EQUATIONS 205 Vz N Vy z y x Mz My T Figure 6.2: The normal force N; the bending moments My, Mz; the torque T ; and the transverse shear forces V y, V z inside a beam. 6.1.1 Boundary Conditions At a built-in end, the in-plane displacements and the slopes are zero. At a simply supported end, the in-plane displacements and the moments are zero. At a free end, the moments and the transverse shear forces are zero. When the end of the beam is restrained axially, the axial displacement is zero. When the end is not restrained axially, the axial force is zero. When the end may rotate, the torque is zero. When the end is rotationally restrained, the twist is zero. The preceding boundary conditions are summarized in Table 6.1. 6.1.2 Stiffness Matrix The stiffness matrix depends on the geometry of the cross section and on the type of material used in the construction of the beam. The geometry (i.e., the shape) of the cross section changes when the beam is loaded. We neglect the effects of this change in shape on the stiffness and evaluate the stiffness matrix for the cross section of the unloaded beam. ρy z y x u w ρz v ψ z y x z y x z y x Figure 6.3: Displacements of a beam
206 BEAMS Table 6.1.Boundary conditions for beams. x-z plane x-yplane Built-in w=0 =0 v=0 =0 Simply supported w=0 M,=0 v=0 M2=0 Free 2=0 M,=0 ,=0 M:=0 Axially restrained u=0 unrestrained N=0 Rotationally restrained 女=0 unrestrained f=0 For a beam made of an isotropic material("isotropic beam")the force-strain relationships are3 N 「(EA) 0 0 0 M, 0 (EIyy) (EIyz) 0 1 M isotropic. (6.4) 0 (EIvz)(EI) 0 1% 0 0 (G) The terms in parentheses are the tensile EAbending Elyy,EI,Ely(=Ely), and torsional GI stiffnesses. We observe that for an isotropic beam there is no coupling between tension(or compression),bending,and torsion.On the other hand,for a beam made of com- posite materials,in general,none of the elements of the stiffness matrix is zero, and there is coupling between tension,bending,and torsion.Accordingly,ten- sion may cause bending and torsion,torsion may cause tension and bending,and bending may cause tension and torsion (see Eq.6.2).The displacements resulting from these couplings are often unexpected and are most of the time undesirable. Fortunately for the designer,some of the couplings and the corresponding dis- placements are not present when either the beam's cross section is symmetrical or when the beam is orthotropic. Symmetrical cross-section beams.First,we consider an isotropic beam whose cross section is symmetrical about the z-axis.An axial load N and a bending moment My(acting in the x-symmetry plane)are applied to the beam.For this beam the force-strain relationships(Eq.6.4)reduce to =[wE2 isotropic (6.5) symmetrical cross section. Next,we consider a composite beam whose cross section is symmetrical about the z-axis (Fig.6.4).As a result of the symmetry,an axial load N acting at the centroid does not introduce either bending or twisting of the beam,whereas a 3 T.H.G.Megson,Aircraft Structures for Engineering Students.3rd edition.Halsted Press,John Wiley Sons,New York,1999,pp.56 and 285
206 BEAMS Table 6.1. Boundary conditions for beams. x–z plane x–y plane Built-in w = 0 ∂w ∂x = 0 v = 0 ∂v ∂x = 0 Simply supported w = 0 My = 0 v = 0 Mz = 0 Free V z = 0 My = 0 V y = 0 Mz = 0 Axially restrained u = 0 unrestrained N = 0 Rotationally restrained ψ = 0 unrestrained T = 0 For a beam made of an isotropic material (“isotropic beam”) the force–strain relationships are3 N My Mz T = (EA) 0 00 0 (EIyy) (EIyz) 0 0 (EIyz) (EIzz) 0 00 0 (GIt) o x 1 ρy 1 ρz ϑ isotropic. (6.4) The terms in parentheses are the tensile EA, bending EIyy, EIzz, EIyz (= EIzy), and torsional GIt stiffnesses. We observe that for an isotropic beam there is no coupling between tension (or compression), bending, and torsion. On the other hand, for a beam made of composite materials, in general, none of the elements of the stiffness matrix is zero, and there is coupling between tension, bending, and torsion. Accordingly, tension may cause bending and torsion, torsion may cause tension and bending, and bending may cause tension and torsion (see Eq. 6.2). The displacements resulting from these couplings are often unexpected and are most of the time undesirable. Fortunately for the designer, some of the couplings and the corresponding displacements are not present when either the beam’s cross section is symmetrical or when the beam is orthotropic. Symmetrical cross-section beams. First, we consider an isotropic beam whose cross section is symmetrical about the z-axis. An axial load N and a bending moment My (acting in the x–z symmetry plane) are applied to the beam. For this beam the force–strain relationships (Eq. 6.4) reduce to N My = (EA) 0 0 (EIyy) ! o x 1 ρy isotropic symmetrical cross section. (6.5) Next, we consider a composite beam whose cross section is symmetrical about the z-axis (Fig. 6.4). As a result of the symmetry, an axial load N acting at the centroid does not introduce either bending or twisting of the beam, whereas a 3 T. H. G. Megson, Aircraft Structures for Engineering Students. 3rd edition. Halsted Press, John Wiley & Sons, New York, 1999, pp. 56 and 285
6.1 GOVERNING EQUATIONS 207 30°90°0°0°90°30° Figure 6.4:Illustrations of composite beams with symmetrical cross sections subjected to a trans- verse load in the x-z symmetry plane. moment My acting in the x-zsymmetry plane introduces only bending in this plane.We designate the elements of the stiffness matrix by EAand E and write the stress-strain relationships as {-[] composite symmetrical cross section. (6.6) Orthotropic beams.A beam is orthotropic when its wall is made of an or- thotropic laminate and one of the orthotropy axes is aligned with the axis of the beam.A laminate is orthotropic when every layer is made of either an isotropic material or a fiber-reinforced composite (page 75).In the latter case,a layer may consist of plies made either of woven fabric or of unidirectional fibers(Fig.6.5) Woven fabric plies must be arranged such that one of the ply symmetry axes is aligned with the longitudinal x-axis of the beam.Unidirectional plies must be Figure 6.5:Layups that result in no coupling between tension,bending,and torsion.Unidirec- tional ply (left):woven fabric (middle):two-ply layer(right).For each configuration,one of the symmetry axes must be parallel to the beam's longitudinal x-axis
6.1 GOVERNING EQUATIONS 207 30° 90° 0° 0° 90° 30° z z x x Figure 6.4: Illustrations of composite beams with symmetrical cross sections subjected to a transverse load in the x–z symmetry plane. moment My acting in the x–z symmetry plane introduces only bending in this plane. We designate the elements of the stiffness matrix by EA and EI yy and write the stress–strain relationships as 1 N My 6 = / EA 0 0 EI yy0 1o x 1 ρy 6 composite symmetrical cross section. (6.6) Orthotropic beams. A beam is orthotropic when its wall is made of an orthotropic laminate and one of the orthotropy axes is aligned with the axis of the beam. A laminate is orthotropic when every layer is made of either an isotropic material or a fiber-reinforced composite (page 75). In the latter case, a layer may consist of plies made either of woven fabric or of unidirectional fibers (Fig. 6.5). Woven fabric plies must be arranged such that one of the ply symmetry axes is aligned with the longitudinal x-axis of the beam. Unidirectional plies must be x x x x x Figure 6.5: Layups that result in no coupling between tension, bending, and torsion. Unidirectional ply (left); woven fabric (middle); two-ply layer (right). For each configuration, one of the symmetry axes must be parallel to the beam’s longitudinal x-axis.