318 BEAMS WITH SHEAR DEFORMATION The last two equations give 2 dx (7.23) where the term in the first parentheses EE)is the warping stiffness, which for an isotropic beam is given by Eq.(6.238)and for an orthotropic l-beam byEq.(6.244) The shear force in the flange is related to Mr by =d dx (7.24) By referring to Eq.(7.13),we write Vi as =(y)h(y), (7.25) where (Syy)is the shear stiffness of the flange in the x-y plane.The warping- induced torque is(Eq.6.235) T。=d. (7.26) Equations (7.22),(7.24),and (7.26)result in 立= dx (7.27) From Eqs.(7.26),(7.25),and (7.20)we obtain To= d2 restrained-warping- (7.28) induced torque, where S is the rotational shear stiffness defined as 及=区，h号 (7.29) Equations (7.12),(7.13),(7.15),(7.23),and (7.28)are the force-strain relation- ships for an orthotropic l-beam with doubly symmetrical cross section subjected to a bending moment M,a shear force acting in the plane of symmetry V,and a torque T(=Tsv +T). Arbitrary cross-section beams.We now consider orthotropic beams of arbi- trary cross sections with internal forces N,My,M,V.V,andM.T. The relationship between the axial force(acting at the centroid)and the axial strain is (Egs.6.7 and 6.8) N=EAe. (7.30)
318 BEAMS WITH SHEAR DEFORMATION The last two equations give Mω = d2 2 EI f % &' ( EIω −dϑB dx % &' ( , (7.23) where the term in the first parentheses EI ω(= d2 2 EI f) is the warping stiffness, which for an isotropic beam is given by Eq. (6.238) and for an orthotropic I-beam by Eq. (6.244). The shear force in the flange is related to Mf by V f = dMf dx . (7.24) By referring to Eq. (7.13), we write V f as V f = (Syy)f (γ )f , (7.25) where (Syy)f is the shear stiffness of the flange in the x–y plane. The warpinginduced torque is (Eq. 6.235) T ω = V fd. (7.26) Equations (7.22), (7.24), and (7.26) result in T ω = dMω dx . (7.27) From Eqs. (7.26), (7.25), and (7.20) we obtain T ω = (Syy)f d2 2 ! % &' ( S ωω ϑS restrained–warpinginduced torque, (7.28) where Sωω is the rotational shear stiffness defined as Sωω = (Syy)f d2 2 . (7.29) Equations (7.12), (7.13), (7.15), (7.23), and (7.28) are the force–strain relationships for an orthotropic I-beam with doubly symmetrical cross section subjected to a bending moment My, a shear force acting in the plane of symmetry V z, and a torque T (= T sv + T ω). Arbitrary cross-section beams. We now consider orthotropic beams of arbitrary cross sections with internal forces N, My, Mz, V y, V z, and Mω, T . The relationship between the axial force (acting at the centroid) and the axial strain is (Eqs. 6.7 and 6.8) N = EAo x . (7.30)
7.1 GOVERNING EQUATIONS 319 The bending moments and bimoment for a (symmetrical)l-beam are (see Eqs.7.12and7.23) M. dXy 0 0 d M 0 x: (7.31) 0 El d dx where the equation forM.is analogous to the equation for My.For an arbitrary cross-section orthotropic beam,the 12 and 21 elements are not zero.Hence,we write M. dxy 0 El 0 d (7.32) dx 0 0 dx By utilizing the definitions given in Eq.(7.6),these equations may be written as M. 1 Elyy 0 (7.33) P 0 The relationship between Tsv and the rate of twist is(Eq.7.15) Tsv =GIt. (7.34) The shear forces and the restrained-warping-induced torque for a(doubly symmetrical)l-beam is (see Egs.7.13 and 7.28) 0 (7.35) 0 0 "s where the first equality is analogous to the second one.Shear is introduced by T and by V and V.Therefore,the force-strain relationships are coupled. We now extend the preceding equations to include these couplings and(for arbi- trary cross-section beams)write the force-strain relationships as 屋美到周 (7.36)
7.1 GOVERNING EQUATIONS 319 The bending moments and bimoment for a (symmetrical) I-beam are (see Eqs. 7.12 and 7.23) Mz My Mω = EI zz 0 0 0 EI yy 0 0 0 EI ω −dχy dx −dχz dx −dϑB dx , (7.31) where the equation for Mz is analogous to the equation for My. For an arbitrary cross-section orthotropic beam, the 12 and 21 elements are not zero. Hence, we write Mz My Mω = EI zz EI yz 0 EI yz EI yy 0 0 0 EI ω −dχy dx −dχz dx −dϑB dx . (7.32) By utilizing the definitions given in Eq. (7.6), these equations may be written as Mz My Mω = EI zz EI yz 0 EI yz EI yy 0 0 0 EI ω 1 ρz 1 ρy . (7.33) The relationship between T sv and the rate of twist ϑ is (Eq. 7.15) T sv = GI tϑ. (7.34) The shear forces and the restrained–warping-induced torque for a (doubly symmetrical) I-beam is (see Eqs. 7.13 and 7.28) V y V z T ω = Syy 0 0 0 Szz 0 0 0 Sωω γy γz ϑS , (7.35) where the first equality is analogous to the second one. Shear is introduced by T ω and by V y and V z. Therefore, the force–strain relationships are coupled. We now extend the preceding equations to include these couplings and (for arbitrary cross-section beams) write the force–strain relationships as V y V z T ω = Syy Syz Syω Syz Szz Szω Syω Szω Sωω γy γz ϑS = " Si j# γy γz ϑS , (7.36)
320 BEAMS WITH SHEAR DEFORMATION where [S]is the shear stiffness matrix of the beam.The inverse of Eq.(7.36)is (7.37) where [;is the shear compliance matrix (7.38) Syo y Equations (7.30),(7.33),(7.34),and (7.36)are the force-strain relationships for a beam with shear deformation.The way in which the elements of the stiffness and compliance matrices are determined is discussed in Section 7.2. 7.1.3 Equilibrium Equations In the presence of shear deformation the equilibrium equations for a straight beam subjected to the loads shown in Figure 6.1 are identical to those of a beam without shear deformation (see Eq.6.3): aN a(Tsv+T.) =- 8x ax a av. =-Py =-Pa (7.39) ax ax aM=立 ax aM:=Vy 0x When the beam is axially constrained,we have the additional equation (Eq.7.27)as follows: dMo =To. (7.40) dx 7.1.4 Summary of Equations In summary,we have 23 unknowns:7 generalized displacements(u,v,w,v,xz, Xy,),8 generalized strain components (e,1/p:,1/p,r,ry,y,,).and 8 generalized internal forces (N,M:,My,M.V,V:,T,Tsv). The equations that yield the solution are as follows: Number of equations Equation number strain-displacement 8 7.6.7.7 force-strain 8 7.30,7.33,7.34,7.36 equilibrium 7 7.39,7.40
320 BEAMS WITH SHEAR DEFORMATION where " Si j# is the shear stiffness matrix of the beam. The inverse of Eq. (7.36) is γy γz ϑS = [si j] V y V z T ω , (7.37) where [si j] is the shear compliance matrix Syy Syz Syω Syz Szz Szω Syω Szω Sωω = syy syz syω syz szz szω syω szω sωω −1 . (7.38) Equations (7.30), (7.33), (7.34), and (7.36) are the force–strain relationships for a beam with shear deformation. The way in which the elements of the stiffness and compliance matrices are determined is discussed in Section 7.2. 7.1.3 Equilibrium Equations In the presence of shear deformation the equilibrium equations for a straight beam subjected to the loads shown in Figure 6.1 are identical to those of a beam without shear deformation (see Eq. 6.3): ∂N ∂x = −px ∂(T sv + T ω) ∂x = −t ∂V y ∂x = −py ∂V z ∂x = −pz ∂My ∂x = V z ∂Mz ∂x = V y. (7.39) When the beam is axially constrained, we have the additional equation (Eq. 7.27) as follows: dMω dx = T ω. (7.40) 7.1.4 Summary of Equations In summary, we have 23 unknowns: 7 generalized displacements (u, v, w, ψ, χz, χy, ϑB), 8 generalized strain components (o x , 1/ρz, 1/ρy, , γy, γz, ϑS, ϑ), and 8 generalized internal forces (N, Mz, My, Mω, V y, V z, T ω, T sv). The equations that yield the solution are as follows: Number of equations Equation number strain–displacement 8 7.6, 7.7 force–strain 8 7.30, 7.33, 7.34, 7.36 equilibrium 7 7.39, 7.40
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 321 Table 7.1.Boundary conditions for beams with shear stiffnesses x-z plane x-y plane Built-in 0=0 X:=0 =0 Xy=0 Simply supported 0=0 应，=0 v=0 应=0 Free 立=0M,=0可=0应=0 Axially restrained 4=0 unrestrained N=0 Warping restrained 0B=0 unrestrained M.=0 Rotationally restrained 业=0 unrestrained f=0 In general the strain-displacement,force-strain,and equilibrium equations must be solved simultaneously to determine the forces in the beam.The analysis becomes simpler when the beam is orthotropic.When the beam is orthotropic the internal forces can be determined by substituting the replacement stiffnesses into the corresponding isotropic beam expressions.The replacement stiffnesses are discussed in Section 7.2. 7.1.5 Boundary Conditions At a built-in end the displacement and the rotations of the cross section are zero. At a simply supported end the displacement and the moment are zero.At a free end the moments and the shear force are zero. When the end of the beam is restrained so that it cannot move axially,the axial displacement is zero.When the end is not restrained axially,the axial force is zero.When warping of the cross section is restrained,the distortion is zero.When warping is not restrained,the bimoment is zero. When the end may rotate,the torque is zero.When the end is rotationally restrained,the rate of twist is zero. The preceding boundary conditions are summarized in Table 7.1. 7.2 Stiffnesses and Compliances of Beams The stiffnesses that characterize orthotropic beams with shear deformation are shown in Table 7.2.For thin-walled beams these stiffnesses are given in Sec- tions 6.2-6.6 except for the shear stiffnesses derived below in Sections 7.2.1 and 7.2.2.The stiffnesses of sandwich beams are presented in Section 7.2.3. The shear stiffness of thick solid-section beams are not discussed here;they are given by Whitney.3 3 J.M.Whitney,Structural Analysis of Laminated Anisotropic Plates.Technomic.Lancaster, Pennsylvania,1987,p.270
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 321 Table 7.1. Boundary conditions for beams with shear stiffnesses x−z plane x−y plane Built-in w = 0 χz = 0 v = 0 χy = 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 Warping restrained ϑB = 0 unrestrained Mω = 0 Rotationally restrained ψ = 0 unrestrained T = 0 In general the strain–displacement, force–strain, and equilibrium equations must be solved simultaneously to determine the forces in the beam. The analysis becomes simpler when the beam is orthotropic. When the beam is orthotropic the internal forces can be determined by substituting the replacement stiffnesses into the corresponding isotropic beam expressions. The replacement stiffnesses are discussed in Section 7.2. 7.1.5 Boundary Conditions At a built-in end the displacement and the rotations of the cross section are zero. At a simply supported end the displacement and the moment are zero. At a free end the moments and the shear force are zero. When the end of the beam is restrained so that it cannot move axially, the axial displacement is zero. When the end is not restrained axially, the axial force is zero. When warping of the cross section is restrained, the distortion is zero. When warping is not restrained, the bimoment is zero. When the end may rotate, the torque is zero. When the end is rotationally restrained, the rate of twist is zero. The preceding boundary conditions are summarized in Table 7.1. 7.2 Stiffnesses and Compliances of Beams The stiffnesses that characterize orthotropic beams with shear deformation are shown in Table 7.2. For thin-walled beams these stiffnesses are given in Sections 6.2–6.6 except for the shear stiffnesses derived below in Sections 7.2.1 and 7.2.2. The stiffnesses of sandwich beams are presented in Section 7.2.3. The shear stiffness of thick solid-section beams are not discussed here; they are given by Whitney.3 3 J. M. Whitney, Structural Analysis of Laminated Anisotropic Plates. Technomic, Lancaster, Pennsylvania, 1987, p. 270
322 BEAMS WITH SHEAR DEFORMATION Table 7.2.Stiffnesses of orthotropic beams with shear deformation Tensile stiffness EA Sections 6.2-6.4 Bending stiffnesses El,Elyy.Ely Sections 6.2-6.4 Torsional stiffness Gi Sections 6.5.1-6.6.2 Warping stiffness E Section 6.6.4 Shear stiffnesses n,33,s Sections 7.2.1-7.2.3 7.2.1 Shear Stiffnesses and Compliances of Thin-Walled Open-Section Beams To determine the shear stiffness and shear compliance matrices of thin-walled open-section beams we consider an element of length AL of a thin-walled open- section beam(Fig.7.5).On the two faces of the element there are equal and opposite shear forces equal and opposite shear forcesequal and opposite torques T(=+T),and axial stresseso,which are different at the two faces. We neglect the Saint-Venant torque;hence,=T. The shear stresses are represented by the shear flow g, 9=Vyqy +V:q:+Toqo (7.41) where g and g are shear flows(per unit force)and are calculated by setting either V=1 and V.=0or V =0 and V.=1 in Eq.(6.281)and go is the shear flow(per unit torque)introduced by a unit torque T;q is evaluated by replacing Eh by 1/(see page 264)in the expression of givend for the corresponding isotropic beams. In the following we take into account only shear deformation caused by shear. Thus,the strain energy is(Eq.2.200) U-i (csnn）ddnd5, (7.42) where is the coordinate perpendicular to the wall,n is the coordinate along the circumference of the cross section,and=x.We assume that the shear flow g(=frd)does not vary along the ALlong element and the shear strain y is uniform across the thickness of the wall.Thus,the strain energy is U= )d△L (7.43) where is the shear strain in the wall's reference surface and is related to the shear flow in the composite wall byr=q(see Eq.6.195),whereis given by 6=i6- (see Eq.6.196).The superscript indicates that the compliances 88 4 T.H.G.Megson,Aircraft Structures for Engineering Students.3rd edition.Halsted Press,John Wiley Sons,New York,1999,pp.465-472
322 BEAMS WITH SHEAR DEFORMATION Table 7.2. Stiffnesses of orthotropic beams with shear deformation Tensile stiffness EA Sections 6.2–6.4 Bending stiffnesses EI zz, EI yy, EI yz Sections 6.2–6.4 Torsional stiffness GI t Sections 6.5.1–6.6.2 Warping stiffness EI ω Section 6.6.4 Shear stiffnesses Syy, Szz, Sωω, Syz, Syω, Szω Sections 7.2.1–7.2.3 7.2.1 Shear Stiffnesses and Compliances of Thin-Walled Open-Section Beams To determine the shear stiffness and shear compliance matrices of thin-walled open-section beams we consider an element of length L of a thin-walled opensection beam (Fig. 7.5). On the two faces of the element there are equal and opposite shear forces V y, equal and opposite shear forces V z, equal and opposite torques T (= T sv + T ω), and axial stresses σx, which are different at the two faces. We neglect the Saint-Venant torque; hence, T = T ω. The shear stresses are represented by the shear flow q, q = V yq∗ y + V zq∗ z + T ωq∗ ω, (7.41) where q∗ y and q∗ z are shear flows (per unit force) and are calculated by setting either V y = 1 and V z = 0 or V y = 0 and V z = 1 in Eq. (6.281) and q∗ ω is the shear flow (per unit torque) introduced by a unit torque T ω; q∗ ω is evaluated by replacing Eh by 1/α 11 (see page 264) in the expression of q∗ ω given4 for the corresponding isotropic beams. In the following we take into account only shear deformation caused by shear. Thus, the strain energy is (Eq. 2.200) U = 1 2 ))) (τξ ηγξ η) dζdηdξ , (7.42) where ζ is the coordinate perpendicular to the wall, η is the coordinate along the circumference of the cross section, and ξ = x. We assume that the shear flow q(= τξ ηdζ ) does not vary along the L long element and the shear strain γξ η is uniform across the thickness of the wall. Thus, the strain energy is U = 1 2 ) γ o ξ ηq dη L, (7.43) where γ o ξ η is the shear strain in the wall’s reference surface and is related to the shear flow in the composite wall by γ o ξ η = αν 66q (see Eq. 6.195), where αν 66 is given by αν 66 = α 66 − (β 66)2 δ 66 (see Eq. 6.196). The superscript indicates that the compliances 4 T. H. G. Megson, Aircraft Structures for Engineering Students. 3rd edition. Halsted Press, John Wiley & Sons, New York, 1999, pp. 465–472.