7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 323 0x(x) △L 0.(x+△L) Figure 7.5:The loads applied to a thin-walled,open-section beam. are evaluated in the o reference plane defined in Figure 6.26.When this expression is used,Eq.(7.43)becomes (7.44) The work done by the external forces due to shear deformation is W- y:△L + 2 -2 due to displacement due to displacement due to the twist- in the y direction in the z direction induced displacement (7.45) alx+△L due to warping(neglected) We neglect the work done by the axial stresses on the shear-force-induced warping. The law of conservation of energy gives U=W. (7.46) By introducing Eqs.(7.44),(7.45),(7.37),and (7.41)into Eq.(7.46),and by performing the algebra,we obtain 登w+x+T己+，时：+，T如+T加= Vs ooc(q)dn+V2s (q:)dn+T2s ci(q)dn VyV:h(s angig:dn+VyTlis ansgyqodn+V:This agiqodn. (7.47) where S indicates integration around the entire circumference.This equation is
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 323 z y σ ∆L) x ( +x ∆L x Vz Vz Vy Vy T T σ ) x (x Figure 7.5: The loads applied to a thin-walled, open-section beam. are evaluated in the reference plane defined in Figure 6.26. When this expression is used, Eq. (7.43) becomes U = 1 2 ) αν 66q2 dη L. (7.44) The work done by the external forces due to shear deformation is W = 1 2 V yγy L % &' ( due to displacement in the y direction + 1 2 V zγz L % &' ( due to displacement in the z direction + 1 2 T ωϑS L % &' ( due to the twistinduced displacement + 1 2 ) at x σxuxdA − ) at x+ L σxuxdA % &' ( due to warping (neglected) . (7.45) We neglect the work done by the axial stresses on the shear–force-induced warping. The law of conservation of energy gives U = W. (7.46) By introducing Eqs. (7.44), (7.45), (7.37), and (7.41) into Eq. (7.46), and by performing the algebra, we obtain V 2 y 1 2 syy + V 2 z 1 2 szz + T 2 ω 1 2 sωω + V yV zsyz + V yT ωsyω + V zT ωszω = V 2 y 1 2 (S) αν 66(q∗ y )2dη + V 2 z 1 2 (S) αν 66(q∗ z )2dη + T 2 ω 1 2 (S) αν 66(q∗ ω)2dη + V yV z (S) αν 66q∗ yq∗ zdη + V yT ω (S) αν 66q∗ yq∗ ωdη + V zT ω (S) αν 66q∗ zq∗ ωdη, (7.47) where S indicates integration around the entire circumference. This equation is
324 BEAMS WITH SHEAR DEFORMATION .=1个 3 91= 2b11+8) =1 9:2 9w2= 2b2(1+ 2dbg2 q q如 Figure 7.6:I-beam and the shear flows from unit shear forces in the z and y directions and from a unit torque. valid for arbitrary values of V,V,and T.Hence,the elements of the shear compliance matrix are (q)dn x= a(g:)dn a(qdn (S 了= 6qyqodn g= a669-qydn (s J(S (S) (7.48) The stiffness is given by Eq.(7.38). l-beam.We consider an l-beam symmetrical about the z-axis (Fig.7.6).The layup of the web is symmetrical,although the layups of the top and bottom flanges may be unsymmetrical.Unit forces V:and V and a unit torque T are applied at the shear center. The shear flows,resulting from unit shear forces=1 and=1 acting at the shear center and from a unit torque T=T=1,are shown in Figure 7.6. Details of the calculations are not presented here.We merely note that the shear flow along the length of the web due to V is taken to be constant.We now define Y1,Y2,8c,and &sc as follows: 1(a11)m d h=1+3akbm1+家) d=d-2 Ze ,1(a11)nd d-e h=1+31+8】 8sc=9 (7.49) e where e is the location of the shear center,ze is the location of the centroid,d is the distance between the midplanes of the flanges,and ba and bp are the widths of the flanges (Fig.7.6.) To calculate Syy we apply V,andperform the integration indicatedin Eq.(7.48) for the web and for the top and bottom flanges.When only V acts,there is no shear flow in the web(Fig.7.6),and hence the integration needs to be performed
324 BEAMS WITH SHEAR DEFORMATION bf2 z e y SC bw d bf1 Vz = 1 Vy = 1 zc * z q * y q * wq T = 1 ω 1 1 2 1 dγ q z = d qz 1 w = )+1(2 3 = 1f sc 1 b δ qy 1f 1 2 3 = db qω 2 2 2 1 dγ q z = ) 1 +1(2 3 = sc 2f 2 δ b q y 2f 2 2 3 = db qω Figure 7.6: I-beam and the shear flows from unit shear forces in the z and y directions and from a unit torque. valid for arbitrary values of V y, V z, and T ω. Hence, the elements of the shear compliance matrix are syy = ) (S) αν 66(q∗ y ) 2 dη szz = ) (S) αν 66(q∗ z ) 2 dη sωω = ) (S) αν 66(q∗ ω) 2 dη szω = ) (S) αν 66q∗ zq∗ ωdη syω = ) (S) αν 66q∗ yq∗ ωdη syz = ) (S) αν 66q∗ zq∗ ydη. (7.48) The stiffness is given by Eq. (7.38). I-beam. We consider an I-beam symmetrical about the z-axis (Fig. 7.6). The layup of the web is symmetrical, although the layups of the top and bottom flanges may be unsymmetrical. Unit forces V z and V y and a unit torque T ω are applied at the shear center. The shear flows, resulting from unit shear forces V z = 1 and V y = 1 acting at the shear center and from a unit torque T = T ω = 1, are shown in Figure 7.6. Details of the calculations are not presented here. We merely note that the shear flow along the length of the web due to V z is taken to be constant. We now define γ1, γ2, δc, and δsc as follows: γ1 = 1 + 1 3 (α11)f1 (α11)w d bf1 1 + 1 δc δc = d − zc zc γ2 = 1 + 1 3 (α11)f2 (α11)w d bf2(1 + δc) δsc = d − e e , (7.49) where e is the location of the shear center, zc is the location of the centroid, d is the distance between the midplanes of the flanges, and bf1 and bf2 are the widths of the flanges (Fig. 7.6.) To calculate syy we apply V y and perform the integration indicated in Eq. (7.48) for the web and for the top and bottom flanges. When only V y acts, there is no shear flow in the web (Fig. 7.6), and hence the integration needs to be performed
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 325 only for the two flanges as follows: -a=广{e[+品(- fe[ag6-h (7.50) By performing the integrations,we obtain (n =网1+子 (7.51) where the subscripts fl and f2 refer to the top and bottom flanges.The other elements of the shear compliance matrix are calculated similarly from Eq.(7.48) with the shear flows shown in Figure 7.6.The results are 3u= 66(q)2dn= +1anb如+1wn加 (S) 12d2y7 12dy3 (7.52) s=)o e(qo)dn= 1.2 (7.53) bp 1.2 %649d= ()n (c62 (7.54) b1(1+8sc) b(1+)】 := :dn =0 S0= ai6qtq。dn=0. (7.55) (S) The stiffness is given by Eq.(7.38). Arbitrary cross-section beam.The shear compliances of selected thin-walled composite beams?are presented in Tables A.8 and A.9.The shear stiffness matrix is determined by Eq.(7.38). 7.2.2 Shear Stiffnesses and Compliances of Thin-Walled Closed-Section Beams When there is no restrained warping (T=0,and consequentlys=0,see Eq.7.28),the shear-force-strain relationships(Eq.7.36)become -[度] (7.56) 5 L.P.Kollar,Flexural-Torsional Vibration of Open Section Composite Beams with Shear Deforma- tion.International Journal of Solids and Structures,Vol.38.7543-7558.2001
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 325 only for the two flanges as follows: syy = ) (S) αν 66(q∗ y ) 2 dη = ) bf1 0 1 (αν 66)f1 3 2bf1 (1 + δsc) 4 η bf1 1 − η bf1 !2 6 dη + ) bf2 0 αν 66 f2 3 2bf2 1 + 1 δsc 4 η bf2 1 − η bf2 2 dη. (7.50) By performing the integrations, we obtain syy = 1.2 αν 66 f1 bf1 (1 + δsc) 2 + αν 66 f2 bf2 1 + 1 δsc 2 , (7.51) where the subscripts f1 and f2 refer to the top and bottom flanges. The other elements of the shear compliance matrix are calculated similarly from Eq. (7.48) with the shear flows shown in Figure 7.6. The results are szz = ) (S) αν 66(q∗ z ) 2 dη = αν 66 w d + 1 12 αν 66 f1 bf1 d2γ 2 1 + 1 12 αν 66 f2 bf2 d2γ 2 2 (7.52) sωω = ) (S) αν 66(q∗ ω) 2 dη = 1.2 d2 - αν 66 f1 bf1 + αν 66 f2 bf2 . (7.53) syω = ) (S) αν 66q∗ yq∗ ωdη = 1.2 d − αν 66 f1 bf1 (1 + δsc) + αν 66 f2 bf2 1 + 1 δsc (7.54) syz = ) (S) αν 66q∗ yq∗ zdη = 0 szω = ) αν 66q∗ zq∗ ωdη = 0. (7.55) The stiffness is given by Eq. (7.38). Arbitrary cross-section beam. The shear compliances of selected thin-walled composite beams5 are presented in Tables A.8 and A.9. The shear stiffness matrix is determined by Eq. (7.38). 7.2.2 Shear Stiffnesses and Compliances of Thin-Walled Closed-Section Beams When there is no restrained warping (T ω = 0, and consequently ϑS = 0, see Eq. 7.28), the shear-force–strain relationships (Eq. 7.36) become V y V z = Syy Syz Syz Szz! γy γz . (7.56) 5 L. P. Koll´ar, Flexural-Torsional Vibration of Open Section Composite Beams with Shear Deformation. International Journal of Solids and Structures, Vol. 38, 7543–7558, 2001
326 BEAMS WITH SHEAR DEFORMATION The shear compliance matrix is (see Eq.7.38) 原-原 (7.57) The elements of the shear stiffness matrix of orthotropic thin-walled closed- section beams are obtained by reasoning similar to that used for open-section beams.Thus,from Eq.(7.48)we have $v=∮%(g}dx=∮a%(g}d砌r=f 6q2q,dn. (7.58) where g andare shear flows(per unit force)calculated by setting either=1 andV.=0orV=0and V.=1.These shear flows are calculated according to the analysis of thin-walled closed-section beams(Section 6.7).Shear compliances of rectangular and circular cross-section thin-walled beams are given in Table A.7. The shear stiffness are given by Eq.(7.38). 7.2.3 Stiffnesses of Sandwich Beams Like sandwich plates(page 169),sandwich beams consist of a foam or honeycomb core covered by facesheets (Fig.7.7).The core acts much like the web in an I-beam;it separates the two facesheets,resulting in high bending stiffness. In this section we consider orthotropic sandwich beams consisting of an ortho- tropic core and orthotropic facesheets(Fig.7.7). To determine the shear stiffness in the x-plane,we apply a shear force V.. For an orthotropic sandwich plate the relationships between the shear forces(per unit length)and the shear strains are (Egs.5.15,5.34) -[] sandwich (7.59) plate. The total shear force V.corresponds to bV in the plate(where b is the width). and V=0(only V acts).Thus,we can apply the laminate plate theory expressions b Figure 7.7:Sandwich beam
326 BEAMS WITH SHEAR DEFORMATION The shear compliance matrix is (see Eq. 7.38) syy syz syz szz! = Syy Syz Syz Szz!−1 . (7.57) The elements of the shear stiffness matrix of orthotropic thin-walled closedsection beams are obtained by reasoning similar to that used for open-section beams. Thus, from Eq. (7.48) we have syy = @ αν 66(q∗ y ) 2 dη szz = @ αν 66 (q∗ z ) 2 dη syz = @ αν 66q∗ zq∗ ydη, (7.58) where q∗ y and q∗ z are shear flows (per unit force) calculated by setting either V y = 1 and V z = 0 or V y = 0 and V z = 1. These shear flows are calculated according to the analysis of thin-walled closed-section beams (Section 6.7). Shear compliances of rectangular and circular cross-section thin-walled beams are given in Table A.7. The shear stiffness are given by Eq. (7.38). 7.2.3 Stiffnesses of Sandwich Beams Like sandwich plates (page 169), sandwich beams consist of a foam or honeycomb core covered by facesheets (Fig. 7.7). The core acts much like the web in an I-beam; it separates the two facesheets, resulting in high bending stiffness. In this section we consider orthotropic sandwich beams consisting of an orthotropic core and orthotropic facesheets (Fig. 7.7). To determine the shear stiffness in the x–z plane, we apply a shear force V z. For an orthotropic sandwich plate the relationships between the shear forces (per unit length) and the shear strains are (Eqs. 5.15, 5.34) Vx Vy = S 11 0 0 S 22! γxz γyz sandwich plate. (7.59) The total shear force V z corresponds to bVx in the plate (where b is the width), and Vy = 0 (only V z acts). Thus, we can apply the laminate plate theory expressions z x y b z x y Vz b Vy N T My Mz Figure 7.7: Sandwich beam.
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 327 to sandwich beams by making the following substitutions: 4= V=0, (7.60) where b is the width of the beam.The shear strain in the plate yx corresponds to the transverse shear strain y:in the beam(Fig.7.1).Thus,Egs.(7.59)and(7.60) yield sandwich (7.61) beam. The term in parentheses is the shear compliance of a sandwich beam in the x-z plane (see Eq.7.37) 1 Sa=Sub (7.62) To determine the bending stiffness in the x-z plane,we again refer to a sandwich plate for which Eqs.(5.11)and(5.12)yield N M =[al +[] M (7.63) Ny Mey N g- =1 +[] M (7.64) ay where M,My,My,N,N,Nry are bending moments and in-plane forces per unit length and [5],[B],and [are evaluated in the x,y,z coordinate system. We now apply only a bending moment My to the sandwich beam(Fig.7.7).The relationship between the bending moment in the plate Mr and the beam My is (Eq.6.20) Mx= My b (7.65) Since only M,acts,we have My=Mty=N=Ny=Nty=0. (7.66) Equations(7.64),(7.65),and (7.66)yield (xx2=xx) -（） (7.67) We compare Eq.(7.67)with Eq.(7.32)and note that an orthotropic sandwich beam is symmetrical with respect to the z-axis and hence Ely:is zero.Thus,the bending stiffness in the x-z plane is 6 y=81 (7.68)
7.2 STIFFNESSES AND COMPLIANCES OF BEAMS 327 to sandwich beams by making the following substitutions: Vx = V z b Vy = 0, (7.60) where b is the width of the beam. The shear strain in the plate γxz corresponds to the transverse shear strain γz in the beam (Fig. 7.1). Thus, Eqs. (7.59) and (7.60) yield γz = 1 S 11 1 b V z sandwich beam. (7.61) The term in parentheses is the shear compliance of a sandwich beam in the x–z plane (see Eq. 7.37) szz = 1 S 11b . (7.62) To determine the bending stiffness in the x–z plane, we again refer to a sandwich plate for which Eqs. (5.11) and (5.12) yield o x o y γ o xy = [α] Nx Ny Nxy + [β] Mx My Mxy (7.63) −∂χxz ∂x −∂χyz ∂y −∂χxz ∂y − ∂χyz ∂x = [β] T Nx Ny Nxy + [δ] Mx My Mxy , (7.64) where Mx, My, Mxy, Nx, Ny, Nxy are bending moments and in-plane forces per unit length and [δ], [β] , and [δ] are evaluated in the x, y, z coordinate system. We now apply only a bending moment My to the sandwich beam (Fig. 7.7). The relationship between the bending moment in the plate Mx and the beam My is (Eq. 6.20) Mx = My b . (7.65) Since only My acts, we have My = Mxy = Nx = Ny = Nxy = 0. (7.66) Equations (7.64), (7.65), and (7.66) yield (χxz = χx) −∂χx ∂x = δ11 b My. (7.67) We compare Eq. (7.67) with Eq. (7.32) and note that an orthotropic sandwich beam is symmetrical with respect to the z-axis and hence EI yz is zero. Thus, the bending stiffness in the x–z plane is EI yy = b δ11 . (7.68)