Elastic Beams in Three Dimensions

Transcription

1 Elastic Beams in Three Dimensions Lars Andersen and Søren R.K. Nielsen ISSN DCE Lecture Notes No. 23 Department of Civil Engineering

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3 Aalborg Universit Department of Civil Engineering Structural Mechanics DCE Lecture Notes No. 23 Elastic Beams in Three Dimensions b Lars Andersen and Søren R.K. Nielsen August 28 c Aalborg Universit

4 Scientific Publications at the Department of Civil Engineering Technical Reports are published for timel dissemination of research results and scientific work carried out at the Department of Civil Engineering (DCE) at Aalborg Universit. This medium allows publication of more detailed eplanations and results than tpicall allowed in scientific journals. Technical Memoranda are produced to enable the preliminar dissemination of scientific work b the personnel of the DCE where such release is deemed to be appropriate. Documents of this kind ma be incomplete or temporar versions of papers or part of continuing work. This should be kept in mind when references are given to publications of this kind. Contract Reports are produced to report scientific work carried out under contract. Publications of this kind contain confidential matter and are reserved for the sponsors and the DCE. Therefore, Contract Reports are generall not available for public circulation. Lecture Notes contain material produced b the lecturers at the DCE for educational purposes. This ma be scientific notes, lecture books, eample problems or manuals for laborator work, or computer programs developed at the DCE. Theses are monograms or collections of papers published to report the scientific work carried out at the DCE to obtain a degree as either PhD or Doctor of Technolog. The thesis is publicl available after the defence of the degree. Latest News is published to enable rapid communication of information about scientific work carried out at the DCE. This includes the status of research projects, developments in the laboratories, information about collaborative work and recent research results. Published 28 b Aalborg Universit Department of Civil Engineering Sohngaardsholmsvej 57, DK-9 Aalborg, Denmark Printed in Denmark at Aalborg Universit ISSN DCE Lecture Notes No. 23

5 Preface This tetbook has been written for the course Statics IV on spatial elastic beam structures given at the 5th semester of the undergraduate programme in Civil Engineering at Aalborg Universit. The book provides a theoretical basis for the understanding of the structural behaviour of beams in three-dimensional structures. In the course, the tet is supplemented with laborator work and hands-on eercises in commercial structural finite-element programs as well as MATLAB. The course presumes basic knowledge of ordinar differential equations and structural mechanics. A prior knowledge about plane frame structures is an advantage though not mandator. The authors would like to thank Mrs. Solveig Hesselvang for tping the manuscript. Aalborg, August 28 Lars Andersen and Søren R.K. Nielsen i

6 ii DCE Lecture Notes No. 23

7 Contents 1 Beams in three dimensions Introduction Equations of equilibrium for spatial beams Section forces and stresses in a beam Kinematics and deformations of a beam Constitutive relations for an elastic beam Differential equations of equilibrium for beams Governing equations for a Timoshenko beam Governing equations for a Bernoulli-Euler beam Uncoupling of aial and bending deformations Determination of the bending centre Determination of the principal aes Equations of equilibrium in principal aes coordinates Normal stresses in beams The principle of virtual forces Elastic beam elements A plane Timoshenko beam element A three-dimensional Timoshenko beam element Summar Shear stresses in beams due to torsion and bending Introduction Homogeneous torsion (St. Venant torsion) Basic assumptions Solution of the homogeneous torsion problem Homogeneous torsion of open thin-walled cross-sections Homogeneous torsion of closed thin-walled cross-sections Shear stresses from bending Shear stresses in open thin-walled cross-sections Determination of the shear centre Shear stresses in closed thin-walled sections Summar References 93 iii

8 iv Contents DCE Lecture Notes No. 23

9 CHAPTER 1 Beams in three dimensions This chapter gives an introduction is given to elastic beams in three dimensions. Firstl, the equations of equilibrium are presented and then the classical beam theories based on Bernoulli- Euler and Timoshenko beam kinematics are derived. The focus of the chapter is the fleural deformations of three-dimensional beams and their coupling with aial deformations. Onl a short introduction is given to torsional deformations, or twist, of beams in three dimensions. A full description of torsion and shear stresses is given in the net chapters. At the end of this chapter, a stiffness matri is formulated for a three-dimensional Timosheko beam element. This element can be used for finite-element analsis of elastic spatial frame structures. 1.1 Introduction In what follows, the theor of three-dimensional beams is outlined. 1.2 Equations of equilibrium for spatial beams An initiall straight beam is considered. When the beam is free of eternal loads, the beam occupies a so-called referential state. In the referential state the beam is clindrical with the length l, i.e. the cross-sections are everwhere identical. The displacement and rotation of the beam is described in a referential (,, )-coordinate sstem with base unit vectors {i, j, k}, the origin O placed on the left end-section, and the -ais parallel with the clinder and orientated into the beam, see Fig For the time being, the position of O and the orientation of the - and -aes ma be chosen freel. The beam is loaded b a distributed load per unit length of the referential scale defined b the vector field q = q() and a distributed moment load vector per unit length m = m(). A differential beam element of the length d is then loaded b the eternal force vector qd and eternal moment vector md as shown in Fig The length of the differential beam element ma change during deformations due to aial strains. However, this does not affect the indicated load vectors which have been defined per unit length of the referential state. Measured in the (,, )-coordinate sstem, q and m have the components q = q q q, m = m m m. (1 1) 1

10 2 Chapter 1 Beams in three dimensions qd md F + df M + dm M id F j i k d l Figure 1 1 Beam in referential state. As a consequence of the eternal loads, the beam is deformed into the so-called current state where the eternal loads are balanced b an internal section force vector F = F() and an internal section moment vector M = M(). These vectors act on the cross-section with the base unit vector i of the -ais as outward directed normal vector. With reference to Fig. 1 2, the components of F and M in the (,, )-coordinate sstem are: F = N Q Q, M = M M M (1 2) Here, N = N() is the aial force, whereas the components Q = Q () and Q = Q () signif the shear force components in the - and -directions. The aial componentm = M () of the section moment vector is denoted the torsional moment. The components M = M () and M = M () in the - and -directions represent the bending moments. The torsional moment is not included in two-dimensional beam theor. However, in the design of three-dimensional frame structures, a good understanding of the torsional behaviour of beams is crucial. Assuming that the displacements remain small, the equation of static equilibrium can be established in the referential state. With reference to Fig. 1 1, the left end-section of the element is loaded with the section force vector F and the section moment vector M. At the right end-section, these vectors are changed differentiall into F + df and M + dm, respectivel. Force equilibrium and moment equilibrium formulated at the point of attack of the section force vector F at the left end-section then provides the following equations of force and moment DCE Lecture Notes No. 23

11 1.2 Equations of equilibrium for spatial beams 3 Q M M N M Q Figure 1 2 Components of the section force vector and the section moment vector. equilibrium of the differential beam element: F + F + df + qd = df d + q = M + M + dm + id (F + df) + md = (1 3a) dm d + i F + m = (1 3b) From Eqs. (1 1) and (1 2) follows that Eqs. (1 3a) and (1 3b) are equivalent to the following component relations: dn d + q =, dq d + q =, dq d + q =, (1 4a) dm d + m dm =, d Q dm + m =, d + Q + m =. At the derivation of Eq. (1 4b), it has been utilised that (1 4b) i F = i (Ni + Q j + Q k) = Ni i + Q i jq i k = i Q j + Q k. (1 5) Hence, i F has the components {, Q, Q }. It is noted that a non-ero normal-force component is achieved when the moment equilibrium equations are formulated in the deformed state. This ma lead to coupled lateral-fleural instabilit as discussed in a later chapter Section forces and stresses in a beam On the cross-section with the outward directed unit vector co-directional to the -ais, the normal stress σ and the shear stresses σ and σ act as shown in Fig These stresses must be Elastic Beams in Three Dimensions

12 4 Chapter 1 Beams in three dimensions staticall equivalent to the components of the force vector F and the section moment vector M as indicated b the following relations: N = A σ da, Q = A σ da, Q = M = (σ σ )da, M = A A σ da, A σ da, M = σ da. A (1 6a) (1 6b) Q M da σ M N σ M σ Q Figure 1 3 Stresses and stress resultant on a cross-section of the beam. σ σ σ σ d σ σ σ σ σ d d Figure 1 4 Components of the stress tensor. On sections orthogonal to the - and -aes, the stresses {σ, σ, σ } and {σ, σ, σ } act as shown in Fig The first inde indicates the coordinate ais co-directional to the outward normal vector of the section, whereas the second inde specifies the direction of action of the stress component. The stresses shown in Fig.1 4 form the components of the stress tensor DCE Lecture Notes No. 23

13 1.2 Equations of equilibrium for spatial beams 5 σ in the (,, )-coordinate sstem given as σ = σ σ σ σ σ σ. σ σ σ (1 7) Moment equilibrium of the cube shown in Fig. 1 4 requires that σ = σ, σ = σ, σ = σ. (1 8) Hence, σ is a smmetric tensor Kinematics and deformations of a beam The basic assumption in the classical beam theor is that a cross-section orthogonal to the -ais at the coordinate remains plane and keeps its shape during deformation. In other words, the cross-section translates and rotates as a rigid bod. Especiall, this means that Poisson contractions in the transverse direction due to aial strains are ignored. Hence, the deformed position of the cross-section is uniquel described b a position vector w = w() and a rotation vector θ = θ() with the following components in the (,, )-coordinate sstem: w = w w w, θ = θ θ θ. (1 9) Further, onl linear beam theor will be considered. This means that the displacement components w, w and w in Eq. (1 9) all small compared to the beam length l. Further the rotation components θ, θ and θ are all small. Especiall, this means that sin θ tan θ θ, (1 1) where θ represents an of the indicated rotation components measured in radians. The various displacement and rotation components have been illustrated in Fig The rotation component around the -ais is known as the twist of the beam. Now, a material point on the cross-section with the coordinates (,, ) in the referential state achieves a displacement vector u = u(,, ) with the components {u, u, u } in the (,, )-coordinate sstem given as (see Fig. 1 5): u (,, ) = w () + θ () θ (), (1 11a) u (,, ) = w () θ (), u (,, ) = w () + θ (). (1 11b) (1 11c) It follows that the displacement of an material point is determined if onl the 6 components of w() and θ() are known at the beam coordinate. Hence, the indicated kinematic constraint reduces the determination of the continuous displacement field u = u(,, ) to the determination of the 6 deformation components w = w (), w = w (), w = w (), θ = θ (), θ = θ () and θ = θ () of a single spatial coordinate along the beam ais. Elastic Beams in Three Dimensions

14 6 Chapter 1 Beams in three dimensions θ θ dw d dw d w w w w θ Figure 1 5 Deformation components in beam theor. The strains conjugated to σ, σ and σ are the aial strain ε and the angular strains γ = 2ε and γ = 2ε. The are related to displacement components as follows: ε = u = dw d + dθ d dθ d, γ = u + u = dw d dθ d θ (), (1 12a) (1 12b) γ = u + u = dw d + dθ d + θ (). (1 12c) From Eq. (1 12) follows that γ = γ (, ) is independent of as a consequence of the presumed plane deformation of the cross-section. Then, the shear stress σ = σ (, ) must also be constant over the cross-section. Especiall, σ at the upper and lower edge of the cross-section as illustrated in Fig. 1 6a. However, if the clindrical surface is free of surface shear tractions, then σ = at the edge. Hence, σ σ in contradiction to Eq. (1 8). In realit σ = at the edges, corresponding to γ =. This means that the deformed cross-section forms a right angle to the clindrical surface as shown in Fig. 1 6b. The displacement fields Eq. (1 11) are onl correct for beams with cross-sections which are circular smmetric around the -ais. In all other cross-sections, the torsional moment M will DCE Lecture Notes No. 23

15 1.2 Equations of equilibrium for spatial beams 7 σ γ σ dw d σ dw d w w w w (a) Figure 1 6 Shear stresses on deformed beam section: (a) Deformation of cross-section in beam theor and (b) real deformation of cross-section. (b) induce an additional non-planar displacement in the -ais, which generall can be written in the form u (,, ) = ω(, )dθ /d. This is illustrated in Fig Hence, the final epression for the aial displacement reads u (,, ) = w () + θ () θ () + ω(, ) dθ d. (1 13) The epressions for u and u in Eq. (1 11) remain unchanged, and ω(, ) is called the warping function. Whereas and in Eq. (1 13) ma be considered as shape functions for the deformations caused b the rotations θ () and θ (), the warping function is a shape function defining the aial deformation of the cross-section from the rotation component. The definition and determination of the warping function is considered in a subsequent section. A Deformed top flange Undeformed state B Section A A A Deformed bottom flange B Section B B Figure 1 7 Warping deformations in an I-beam induced b homogeneous torsion. The cross sections A A and B B are shown with the top flange on the left and the bottom flange on the right. As a consequence of the inclusion of the warping, the strain components in Eq. (1 12) are modified as follows: ε = u = dw d + dθ d dθ d + θ ωd2 d 2, γ = u + u = dw ( ) ω d θ + dθ d, γ = u + u = d d + θ + ( ω + ) dθ d. (1 14a) (1 14b) (1 14c) Elastic Beams in Three Dimensions

16 8 Chapter 1 Beams in three dimensions θ θ dw d dw d w w w w Figure 1 8 Kinematics of Bernoulli-Euler beam theor. Bernoulli-Euler beam kinematics presumes that the rotated cross-section is alwas orthogonal to the deformed beam ais. This involves the following additional kinematical constraints on the deformation of the cross-section (see Fig. 1 8): θ = dw d, θ = dw d. (1 15) Assuming temporaril that θ in bending deformations, i.e. disregarding the twist of the beam, Eqs. (1 14) and (1 15) then provide: γ = γ =. (1 16) Equation (1 16) implies that the shear stresses are σ = σ =, and in turn that the shear forces become Q = Q =, cf. Eq. (1 6). However, non-ero shear forces are indeed present in bending of Bernoulli-Euler beams. The apparent parado is dissolved b noting that the shear forces in Bernoulli-Euler beam theor cannot be derived from the kinematic condition, but has to be determined from the static equations. The development of the classical beam theor is associated with names like Galilei ( ), Mariotte ( ), Leibner ( ), Jacob Bernoulli ( ), Euler ( ), Coulomb ( ) and Navier ( ), leading to the mentioned Bernoulli-Euler beam based on the indicated kinematic constraint. The inclusion of transverse shear deformation was proposed in 1859 b Bresse ( ) and etended to dnamics in 1921 b Timoshenko ( ). Due to this contribution, the resulting beam theor based on the strain relations Eq. (1 12), is referred to as Timoshenko beam theor (Timoshenko 1921). The first correct analsis of torsion in beams was given b St. Venant (1855). The underling assumption was that dθ /d in Eq. (1 13) was constant, so the warping in all cross-sections become identical. Then, the aial strain ε from torsion vanishes and the distribution of the shear strains γ and γ are identical in all sections. Because of this, St. Venant torsion is also referred to as homogeneous torsion. DCE Lecture Notes No. 23

17 1.2 Equations of equilibrium for spatial beams 9 Whenever the twist or the warping is prevented at one or more cross-sections, dθ /d is no longer constant as a function of. Hence, aial strains occur and, as a consequence of this, aial stresses arise and the shear strains and shear stresses are varing along the beam. These phenomena were sstematicall analsed b Vlasov (1961) for thin-walled beams, for which reason the resulting theor is referred to as Vlasov torsion or non-homogeneous torsion. Notice that the shear stresses from Vlasov torsion have not been included in the present formulation. These will be considered in a subsequent chapter. Seen from an engineering point of view, the primar advantage of Vlasov torsion theor is that it eplains a basic feature of beams, namel that prevention of warping leads to a much stiffer structural elements than achieved in the case of homogeneous warping, i.e. a given torsional moment will induce a smaller twist. Warping of the cross-section ma, for eample, be counteracted b the inclusion of a thick plate orthogonal to the beam ais and welded to the flanges and the web. The prevention of torsion in this manner is particularl useful in the case of slender beams with open thin-walled cross-sections that are prone to coupled fleural torsional buckling. Obviousl, Vlasov torsion theor must be applied for the analsis of such problems as discussed later in the book. Net, the deformation of the cross-section ma be decomposed into bending and shear components. The bending components are caused b the bending moments M and M and deform as a Bernoulli-Euler beam. Hence, the bending components are causing the rotations θ and θ of the cross-section. The shear components are caused b the shear forces Q and Q. These cause the angular shear strains γ and γ without rotating the cross-section. Further, the displacement of the beam ais in shear takes place without curvature. Hence, the curvature of the beam ais is strictl related to the bending components, see Fig With reference to Fig. 1 1, the radii of curvatures r and r are related to the rotation increments dθ and dθ of the end-sections in the bending deformations of a differential beam element of the length d as follows r dθ = d r dθ = d κ = 1/r = dθ /d κ = 1/r = dθ /d (1 17) Here, κ and κ denote the components of the curvature vector κ of the -ais. Especiall, for a Bernoulli-Euler beam the curvature components become, cf. Eq. (1 15), κ = d2 w d 2, κ = d2 w d 2. (1 18) From Eqs. (1 14) and (1 17) follows that the aial strain ma be written as ε (,, ) = ε() + κ () κ () + ω(, ) d2 θ d 2, (1 19) where ε() denotes the aial strain of the beam along the -ais given as ε() = dw d. (1 2) Here, ε(), κ () and κ () define the aial strain and curvatures of the beam ais, i.e. the ()-ais. Elastic Beams in Three Dimensions

18 1 Chapter 1 Beams in three dimensions = M + γ Q γ Figure 1 9 Decomposition of cross-section deformation into bending and shear components. r dθ r dθ w ds d w ds d Figure 1 1 Definition of curvature Constitutive relations for an elastic beam In what follows we shall refer to N(), Q (), Q (), M (), M () and M () as generalised stresses. These are stored in the column matri σ() = N() Q () Q () M () M () M (). (1 21) DCE Lecture Notes No. 23

19 1.2 Equations of equilibrium for spatial beams 11 The internal virtual work of these quantities per unit length of the beam is given as δω = Nδε + Q δγ + Q δγ + M δθ + M δκ + M δκ = σ T δε, (1 22) where ε() = ε() γ () γ () θ () κ () κ (). (1 23) The components of ε() are referred to as the generalised strains. The components of σ() and ε() are said to be virtual work conjugated because these quantities define the internal virtual work per unit length of the beam. Let E and G denote the elasticit modulus and the shear modulus. Then, the normal stress σ and the shear stresses σ and σ ma be calculated from Eq. (1 14) as follows: ( dw σ = Eε = E d + dθ d dθ d + ω(, θ )d2 d 2 ( ( ) dw ω σ = Gγ = G d θ + dθ d ( ( ) dw2 ω σ = Gγ = G d + θ + + dθ d B integration over the cross-sectional area, it then follows that ( N = E A dw d + S ( ( dw Q = G A ( ( dw Q = G A ( ( dw M = G S ( dw M = E S d θ d + θ d + θ dθ d S dθ 2 d + S ω ) dθ + R d ) dθ + R d d 2 θ d 2 ) S ( dw d θ dθ d I dθ d + I ω d + I ( dw M = E S d I dθ d + I dθ d I ω ), (1 24a) ), (1 24b) ). (1 24c) ), (1 25a) ), (1 25b) ), (1 25c) ) + K dθ d ), (1 25d) ) dθ, (1 25e) d dθ d ), (1 25f) where A, A, R, R, S, S, S ω, K, I, I, I = I, I ω and I ω are cross-sectional (or geometrical) constants identified as: A = da, A = α A, A = α A, (1 26a) A Elastic Beams in Three Dimensions

20 12 Chapter 1 Beams in three dimensions R = S = A A I = I = K = A A A ( ) ( ) ω ω da, R = A + da, da, S = da, S ω = ωda, A 2 da, I = da, I ω = A A 2 da, ( ω ω ) da. A ωda, I ω = A ωda, (1 26b) (1 26c) (1 26d) (1 26e) (1 26f) Here, A is the cross-sectional area, whereas A and A signif the so-called shear areas. Beam theor presumes a constant variation of the shear stresses in bending, whereas the actual variation is at least quadratic. The constant variation results in an overestimation of the stiffness against shear deformations, which is compensated b the indicated shear reduction factors α and α. If the actual distribution of the shear stresses is parabolic, these factors become α = α = 5/6. For an I-profile, the shear area is approimatel equal to the web area. For GA we have γ = Q /(GA ) =. Bernoulli-Euler beam theor is characterised b γ =. Hence, Timoshenko theor must converge towards Bernoulli-Euler theor for the shear areas passing towards infinit. The magnitude of the shear deformations in proportion to the bending deformations depends on the quantit (h/l) 2, where h is the height and l is the length of the beam. This relation is illustrated in Eample 1-3. R and R are section constants which depend on the warping mode shape ω(, ) as well as the bending modes via and. Further, the section constants S and S are denoted the static moments around the - and -aes. S ω specifies a corresponding static moment of the warping shape function. I and I signif the bending moments of inertia around the - and -aes, respectivel. I is denoted the centrifugal moment of inertia, whereas I ω and I ω are the corresponding centrifugal moments of the warping shape function and the bending mode shapes. K is the so-called torsion constant. This defines merel the torsional stiffness in St. Venant torsion. As mentioned above, the additional contribution to M from Vlasov torsion will be considered in a subsequent section. 1.3 Differential equations of equilibrium for beams In what follows, the governing differential equations for Timoshenko and Bernoulli-Euler beams are derived. At this stage, the twist θ and the torsional moment M are ignored. With no further assumptions and simplifications, Eq. (1 25) reduces to N M M Q Q = EA ES ES ES EI EI ES EI EI GA GA dw /d dθ /d dθ /d dw /d θ dw /d + θ. (1 27) DCE Lecture Notes No. 23

21 1.3 Differential equations of equilibrium for beams 13 The coefficient matri of Eq. (1 27) is smmetric. When formulated in a similar matri format, the corresponding matri in Eq. (1 25) is not smmetric. This is a consequence of the ignorance of the Vlasov torsion in M Governing equations for a Timoshenko beam Net, Eq. (1 27) is inserted into the equilibrium equations (1 4a) and (1 4b), which results in the following sstem of coupled ordinar differential equations for the determination of w, w, w, θ and θ : d d dn/d dm /d dm /d dq /d dq /d = = Q Q q m m q q EA ES ES ES EI EI ES EI EI GA GA GA GA dw /d dθ /d dθ /d dθ /d θ dw /d + θ dw /d dθ /d dθ /d dw /d θ dw /d + θ q m m q q. (1 28) Equation (1 28) specifies the differential equations for Timoshenko beam theor. These should be solved with proper boundar condition at the end-sections of the beam. Let denote the abscissa of an of the two end-sections, i.e. = or = l, where l is the length of the beam. At = either kinematical or mechanical boundar conditions ma be prescribed. Kinematical boundar conditions mean that values of w, w, w, θ and θ are prescribed, w ( ) = w, w ( ) = w, w ( ) = w, θ ( ) = θ, θ ( ) = θ,, =, l, (1 29) whereas mechanical boundar conditions impl the prescription of N, Q, Q, M and M, N( ) = N Q ( ) = Q, Q ( ) = Q,, =, l. (1 3) M ( ) = M, M ( ) = M, Elastic Beams in Three Dimensions

22 14 Chapter 1 Beams in three dimensions In Eq. (1 3), the left-hand sides are epressed in kinematical quantities b means of Eq. (1 27). Of the 1 possible boundar conditions at = specified b Eqs. (1 35) and (1 3), onl 5 can be specified. The 5 boundar conditions at = and = l can be selected independentl from Eq. (1 35) and Eq. (1 3). With given boundar conditions Eq. (1 28) can be solved uniquel for the 5 kinematic quantities w, w, w, θ, θ, which make up the degrees of freedom of the cross-section. Although an analtical solution ma be cumbersome, a numerical integration is alwas within reach Governing equations for a Bernoulli-Euler beam Net, similar differential equations are specified for a Bernoulli-Euler beam. At first the shear forces Q and Q in the equations of equilibrium for M and M in Eq. (1 4b) are eliminated b means of the 2nd and 3rd equations in Eq. (1 4a): d 2 M /d 2 dq /d + dm /d = d 2 M /d 2 + dq /d + dm /d = d 2 M /d 2 + q + dm /d = d 2 M /d 2 q + dm /d =. (1 31) Using the Bernoulli-Euler kinematical constraint Eq. (1 15), the constitutive equations for the resulting section forces ma be written as N M M = EA ES ES ES EI EI ES EI EI dw /d d 2 w /d 2 d 2 w /d 2. (1 32) Then, the equations of equilibrium Eq. (1 4a) and Eq. (1 31) ma be recasted as the following sstem of coupled ordinar differential equations ( d EA dw d d ES d 2 w d 2 ES d 2 ) w d 2 + q =, (1 33a) d 2 ( dw d 2 ES d EI d 2 w d 2 EI d 2 ) w d 2 + q + dm d =, d 2 ( dw d 2 ES d + EI d 2 w d 2 + EI d 2 ) w d 2 + q + dm d =. (1 33b) (1 33c) The governing equations (1 33) should be solved with 5 of the same boundar conditions as indicated b Eqs. (1 35) and (1 3). The difference is that θ ( ), θ ( ), Q ( ) and Q ( ) are represented as, cf. Eqs. (1 4b) and (1 15), dw ( ) d = θ,, dw ( ) d = θ,, (1 34a) dm ( ) d m ( ) = Q,, dm ( ) d + m ( ) = Q,. (1 34b) DCE Lecture Notes No. 23

23 1.4 Uncoupling of aial and bending deformations 15 With this in mind, the kinematic boundar conditions for Bernoulli-Euler beams are given in the form w ( ) = w, w ( ) = w, w ( ) = w,, =, l, (1 35) dw ( )/d = θ, dw ( )/d = θ, whereas the mechanical boundar conditions defined in Eq. (1 3) are still valid. 1.4 Uncoupling of aial and bending deformations Up to now the position of the origin O and the orientation of the - and -aes in the crosssection have been chosen arbitraril. As a consequence of this, the deformations from the aial force and the deformation from the bending moments M and M will generall be coupled. This means that the aial force N referred to the origin O will not merel induce a uniform displacement w of the cross-section, but also non-ero displacements w and w of O as well as rotations θ and θ. Similarl, the bending moment M will not merel cause a displacement w and a rotation θ of the cross-section, but also a non-ero displacement w and a rotation θ in the orthogonal direction in addition to an aial displacement w of the origin. The indicated mechanical couplings are the reason for the couplings in the differential equations (1 28) and (1 33). The couplings ma have a significant impact on the structural behaviour and stabilit of an engineering structure and the position of the origin for a given beam element as well as the orientation of the coordinate aes must be implemented correctl in a computational model. In this section, two coordinate transformations will be indicated, in which the aial force referred to the new origin B, called the bending centre, onl induces a uniform aial displacement over the cross-section. Similarl, the bending moments M and M around the new rotated - and -aes, referred to as the principal aes, will onl induce the non-ero deformation components (w, θ ) and (w, θ ), respectivel. Especiall, the moments will induce the displacement w = of the bending centre, B Determination of the bending centre The position of the bending centre B is given b the position vector r B with the components {, B, B } in the (,, )-coordinate sstem. In order to determine the components B and B, a translation of the (,, )-coordinate sstem to a new (,, )-coordinate sstem with origin in the et unknown bending centre is performed (see Fig. 1 11). The relations between the new and the old coordinates read =, = + B, = + B, (1 36) In the new coordinate sstem, the displacement of B (the new origin) in the -direction (the new beam ais) becomes (see Fig. 1 11): w = w + B θ B θ. (1 37) Elastic Beams in Three Dimensions

24 16 Chapter 1 Beams in three dimensions θ, M θ, M B B N θ, M r B O θ, M N B Figure 1 11 Translation of coordinate sstem. The aial strain of fibres placed on the new beam ais becomes ε ( ) = dw d = dw d = ε + Bκ B κ, (1 38) where Eq. (1 17), Eq. (1 2) and Eq. (1 37) have been used. The components of the rotation vector θ of the cross-section are identical, i.e. θ = θ, θ = θ, θ = θ. (1 39) In turn, this means that the components of the curvature vector κ in the two coordinate sstems are identical as well κ = dθ d = dθ d = κ, κ = dθ d = dθ d = κ. (1 4) Further, the components of the section force vector F in the two coordinate sstems become identical, i.e. N = N, Q = Q Q = Q. (1 41) As a consequence of referring the aial force N = N to the new origin B, the components of the section vector in the (,, )-coordinate sstem are related to the components in the (,, )-coordinate as follows: M = M, M = M B N, M = M + B N. (1 42) Equations (1 38) and (1 4) provide the following relation for ε in terms of ε, κ and κ : ε = ε B κ + B κ = ε B κ + B κ. (1 43) DCE Lecture Notes No. 23

25 1.4 Uncoupling of aial and bending deformations 17 Then the relation between {N, M, M } and {N, M, M } and {ε, κ, κ } and {ε, κ, κ } ma be specified in the following matri formulation: σ = A T σ, (1 44a) ε = Aε, where σ = σ = A = N M M N M M, ε =, ε = 1 B B 1 1. ε κ κ ε κ κ,, (1 44b) (1 45a) (1 45b) (1 45c) The components {N, M, M } and {ε, κ, κ } of σ and ε ma be interpreted as work conjugated generalised stresses and strains. With reference to Eq. (1 32), the constitutive relation between σ and ε is given as σ = Cε, (1 46) where C denotes the constitutive matri, A S S C = E S I I. (1 47) S I I Likewise, the constitutive relation in the (,, )-coordinate sstem reads σ = C ε (1 48) where the constitutive matri has the form A S S C = E S I I. (1 49) S I I Obviousl, as given b Eqs. (1 47) and (1 49), the cross-sectional area A is invariant to a rotation of the cross-section about the -ais and a translation in the - and -directions. From Eqs. (1 44a), (1 44b) and (1 46) follows that σ = A T Cε = A T CAε 1 C = A T CA = E B 1 B 1 A S S S I I S I I 1 B B 1 1 Elastic Beams in Three Dimensions

26 18 Chapter 1 Beams in three dimensions A S B A (S B A) C = E S B A I 2 B S +B 2 A I + B S + B (S B A). (S B A) I + B S + B (S B A) I 2 B S +B 2 A (1 5) The idea is now to use the translational coordinate transformation to uncouple the aial deformations from the bending deformations. This requires that S = S =. Upon comparison of Eq. (1 49) and Eq. (1 5), this provides the following relations for the deformation of the coordinates of the bending centre: B = S A, B = S A. (1 51) With B and B given b Eq. (1 51), the bending moments of inertia, I and I, and the centrifugal moment of inertia, I, in the new coordinate sstem can be epressed in terms of the corresponding quantities in the old coordinate sstem as follows: I = I 2 B (A B ) + B 2 A = I B 2 A, I = I 2 B (A B ) + BA 2 = I BA, 2 I = I B (A B ) B (A B ) + B B A = I B B A. The final results in Eq. (1 52) are known as König s theorem. (1 52a) (1 52b) (1 52c) O B Figure 1 12 Single-smmetric cross-section. If the cross-section is smmetric around a single line, and the -ais is placed so that it coincides with this line of smmetr, then the static moment S vanishes, i.e. S = A da = (1 53) As a result of this, the bending centre B will alwas be located on the line of smmetr in a single-smmetric cross-section, see Fig Obviousl, if the cross-section is double smmetric, then the position of B is found at the intersection of the two lines of smmetr. DCE Lecture Notes No. 23

27 1.4 Uncoupling of aial and bending deformations 19 Eample 1.1 Determination of bending and centrifugal moments of inertia of non-smmetric thin-walled cross-section The position of the bending centre of the cross-section shown in Fig. A is determined along with the bending moments of inertia I and I and the centrifugal moment of inertia I. O 2a 2t 2a t t a Figure A Thin-walled cross-section. The (,,)-coordinate sstem is placed as shown in Fig. A. Then, the following cross-sectional constants are calculated: A = 2a 2t + 2a t + a t = 7at, S = 2a 2t a + 2a t t 2 + a t a 2 = 1 (2t + 9a) + ta, (b) 2 ( S = 2a 2t t + 2a t (2t + a) + a t 2t + 2a + t ) = 1 (21t + 8a)ta, (c) 2 2 I = 1 3 2t (2a) a t t a3 = 1 ( 2t a 2) ta, (d) 3 I = 1 3 2a 2a (2t) (2a)3 t + 2a t (2t + a) ( 12 a t3 + a t 2t + 2a + t ) 2 2 = 1 ( 59t ta + 2a 2) ta, (e) 3 I = 2a 2t t a + 2a t (2t + a) 1 ( 2 + a t 2t + 2a + t ) a 2 2 = 1 ( 8t ta + 4a 2) ta. (f) 4 Here, use has been made of König s theorem at the calculation of contributions to I, I and I from the three rectangles forming the cross-section. The coordinates of the bending centre follow from Eq. (1 51) and Eqs. (a) to (c). Thus, (a) B = S A = 3 2 t a, S B = A = 1 7 t a. (g) (continued) Elastic Beams in Three Dimensions

28 2 Chapter 1 Beams in three dimensions 3 2 t a 1 7 t a B Figure B Position of bending centre in the thin-walled cross-section. Subsequentl, the moments of inertia around the aes of the (,, )-coordinate sstem follow from Eq. (1 52), Eq. (1 53) and Eqs. (d) to (f): I = 1 ( 1 3 (2t2 + 17a 2 )ta 7 t + 9 ) 2 14 a 7ta = 1 84 (44t2 18ta + 233a 2 )ta, (h) I = 1 ( 3 3 (59t2 + 54ta + 2a 2 )ta I = 1 ( 1 4 (8t2 + 25ta + 4a 2 )ta 7 t a) 2 t + 4 ) 2 7 a 7ta = 1 84 (329t2 + 54ta + 368a 2 )ta, (i) ) ( 3 2 t + 4 ) 7 a 7ta = 1 14 (7t2 15ta 22a 2 )ta. (j) Now, for a thin-walled cross-section the thickness of the flanges and the web is much smaller than the widths of the flanges and the height of the web. In the present case this means that t a. With this in mind, Eqs. (g) to (j) reduce to B 4 7 a, B 9 14 a (k) and I ta3, I ta3, I 11 7 ta3. (l) It is noted that the error on I estimated b Eq. (l) increases rapidl with increasing values of t/a. Thus, for t/a =.1, the error is about 13%. The errors related to the estimated values of I and I are somewhat smaller, i.e. about 5% and 7%, respectivel. From now on, the origin of the (,, )-coordinate sstem is placed at the bending centre. Then, the constitutive matri given b Eq. (1 47) takes the form C = E A I I. (1 54) I I DCE Lecture Notes No. 23

29 1.4 Uncoupling of aial and bending deformations 21 As a result of this, an aial force N no longer induces deformations in the - and -directions, and the bending moments M and M do not induce aial displacements. However, the bending moment M will still induce displacements in the -direction in addition to the epected displacements in the -direction. Similarl, the bending moment M induces displacements in both the - and -directions Determination of the principal aes In order to uncouple the bending deformations, so that M will onl induce deformations in the -direction, and M onl deformations in the -direction, a new (,, )-coordinate sstem is introduced with origin in B and rotated the angle ϕ around the -ais as shown in Fig κ, M κ, M κ, M B ϕ N = N κ, M Figure 1 13 Rotation of coordinate sstem. Let {N, M, M } and {N, M, M } denote the components of the generalised stresses in the (,, )-coordinate sstem and the (,, )-coordinate sstem, respectivel. The two sets of generalised stresses are related as σ = Bσ, (1 55a) where σ = N M M, σ = N M M, B = 1 cosϕ sinϕ sinϕ cosϕ. (1 55b) Likewise, the components of the generalised strains in the two coordinate sstems are denoted as {ε, κ, κ } and {ε, κ, κ }, respectivel. These are related as ε = Bε, (1 56a) Elastic Beams in Three Dimensions

30 22 Chapter 1 Beams in three dimensions where ε = ε κ κ, ε = ε κ κ, B = 1 cosϕ sin ϕ sin ϕ cosϕ. (1 56b) The constitutive relation in the (,, )-coordinate sstem reads σ = C ε, C = E A I. (1 57) I I The corresponding constitutive relation in the (,, )-coordinate sstem is given b Eq. (1 46) with C given b Eq. (1 54). Use of Eqs. (1 55a) and (1 56a) in Eq. (1 46) provides Bσ = Cε = CBε σ = B T CBε, (1 58) where it has been utilised that B 1 = B T. Comparison of Eq. (1 57) and Eq. (1 58) leads the following relation between the constitutive matrices C = B T CB = E = E 1 cosϕ sin ϕ sinϕ cosϕ A C 22 C 23 C 32 C 33. A I I I I 1 cosϕ sin ϕ sin ϕ cosϕ (1 59a) where C 22 = cos 2 ϕi 2 cosϕsin ϕi + sin 2 ϕi, C 23 = C 32 = sinϕcos ϕ(i I ) (cos 2 ϕ sin 2 ϕ)i, C 33 = cos2 ϕi + 2 cosϕsin ϕi + sin 2 ϕi. From Eq. (1 57) and Eq. (1 59) follows that I = 1 2 (I + I ) (I I )cos(2ϕ) I sin(2ϕ), (1 59b) (1 59c) (1 59d) (1 6a) I = 1 2 sin(2ϕ)(i I ) cos(2ϕ)i, I = 1 2 (I + I ) 1 2 (I I )cos(2ϕ) + I sin(2ϕ), (1 6b) (1 6c) where use has been made of the relations sin(2ϕ) = 2 sin ϕcosϕ, cos(2ϕ) = cos 2 ϕ sin 2 ϕ, cos 2 ϕ = 1 (1 + cos2ϕ), 2 sin2 ϕ = 1 2 (1 cos2 ϕ). DCE Lecture Notes No. 23

31 1.4 Uncoupling of aial and bending deformations 23 Uncoupling of bending deformations in the (,, )-coordinate sstem requires that I =. This provides the following relation for the determination of the rotation angle ϕ: 1 2 sin(2ϕ)(i I ) cos(2ϕ)i = tan(2ϕ) = 2I I I for I I, (1 61) cos(2ϕ) = for I = I. Note that cos(2ϕ) = implies that sin(2ϕ) = ±1. The sign of sin(2ϕ) is chosen as follows: sin(2ϕ) = 1 ϕ = 1 4 π for I < sin(2ϕ) = 1 ϕ = 3 4 π for I. (1 62) > Then, Eq. (1 6) provides the following solutions for I and I : I = 1 2 (I + I )+ I, I = 1 2 (I + I ) I. (1 63) For I I the solution for tan(2ϕ) is fulfilled for the following two alternative solutions for sin(2ϕ) and cos(2ϕ): sin(2ϕ) = 2I J, cos(2ϕ) = I I, (1 64a) J sin(2ϕ) = 2I J, where J = (I I ) 2 + 4I. 2 cos(2ϕ) = I I, (1 64b) J The sign definition in Eq. (1 64a) is chosen. This implies that (1 65) 2ϕ [, π] for I < and 2ϕ [π, 2π] for I >. (1 66) Insertion of the solution for sin(2ϕ) and cos(2ϕ) into Eq. (1 6) provides the following results for I and I : I = 1 2 (I + I ) (I I ) 2 + 4I 2, (1 67a) J I = 1 2 (I + I ) 1 (I I ) 2 + 4I 2, (1 67b) 2 J or, b insertion of J, cf. Eq. (1 65), I = 1 2 (I + I ) I = 1 2 (I + I ) 1 2 (I I ) 2 + 4I 2, (1 68a) (I I ) 2 + 4I 2. (1 68b) Elastic Beams in Three Dimensions

32 24 Chapter 1 Beams in three dimensions ϕ B = ϕ = B (a) (b) Figure 1 14 Position of principal aes: (a) I < and (b) I >. The coordinate aes and are known as the principal aes of the cross-section, whereas I and I are called the principal moments of inertia. It follows from Eqs. (1 63) and (1 67) that the choices of signs for sin(2ϕ) implies that I becomes the larger of the principal moments of inertia and I is the smaller principal moment of inertia. It is emphasised that this choice is performed merel to have a unique determination of ϕ. Three other choices of ϕ are possible obtained b additional rotations of the magnitudes π 2, π and 3 2π relative to the indicated. If the cross-section has a smmetr line, and the -ais is placed along this line, then I =. Hence, a smmetr line is alwas a principal ais. Since the principal aes are orthogonal, the -ais is also a principal ais even if the cross-section is not smmetric around the ais. Eample 1.2 Determination of principal aes coordinate sstem The cross-section analsed in Eample 1.1 is reconsidered. The thin-wall approimation is used, so the moments of inertia are given b Eq. (l) in Eample 1.1 and repeated here (without the primes): I ta ta 3, I ta ta 3, I 11 7 ta ta 3.(a) The position of the bending centre relativel to the top-left corner of the cross-section is provided in Fig. A. From Eq. (a) and Eq. (1 65) follows that (233 J = 84 ta3 92 ) 2 ( 21 ta ) 2 7 ta3 = ta , (b) 28 which b insertion into Eq. (1 64) provides: sin(2ϕ) = 2 ( 11 7 ta3) 28 ta = , (c) cos(2ϕ) = ( ta ta3) 28 ta = (d) 84 (continued) DCE Lecture Notes No. 23

33 1.4 Uncoupling of aial and bending deformations 25 B 4 7 a 2a B 9 14 a 2a 2t B = ϕ t t a Figure A Position of principal aes coordinated sstem for the thin-walled cross-section. From Eqs. (c) and (d) it is found that ϕ = radians corresponding to ϕ = Hence, ϕ [, π ] in agreement with 2 I = 11 7 ta3 <. Finall, the moments of inertia in the principal aes coordinate sstem follow from Eq. (1 67), i.e. I I = 1 2 ( ) (233 ± ) ( ) 2 11 ta 3 = ta 3, ta 3. Clearl, I is greater than an of I or I, whereas I is smaller than the bending moments of inertia defined with respect to the original - and -aes. (e) Equations of equilibrium in principal aes coordinates From now on it will be assumed that the (,, )-coordinate sstem forms a principal aes coordinate sstem with origin at the bending centre. In this case, the sstem of differential equations (1 28) for a Timoshenko beam uncouples into three differential subsstems. Thus, the aial deformation is governed b the equation d d ( EA dw d ) + q =, (1 69) whereas bending deformation in the -direction is defined b the coupled equations ( ) ( ) d dθ dw EI + GA d d d θ + m =, (1 7a) d d ( GA ( dw d θ )) + q =, (1 7b) where the double inde on the bending moment of inertia has been replaces b a single inde in order to indicate that the principal-aes coordinates are utilised. Elastic Beams in Three Dimensions

34 26 Chapter 1 Beams in three dimensions d d Similarl, the fleural deformations in the -directions are determined b ( ) ( dθ dw EI GA d d + θ ( GA ( dw ) + m =, (1 71a) )) + q =, (1 71b) d d d + θ where again the double inde on the bending moment of inertia has been replaced b a single inde. As seen from Eqs. (1 7) and (1 71), {w, θ } and {w, θ } are still determined b pairwise coupled ordinar differential equations of the second order. For a Bernoulli-Euler beam, the sstem of ordinar differential equations (1 33) uncouples completel into the following differential equations for the determination of w, w and w : d d d 2 d 2 d 2 d 2 ( EA dw d ( d 2 w EI d 2 ( EI d 2 w d 2 ) + q =, ) q + dm d =, ) q dm d =. (1 72a) (1 72b) (1 72c) Eample 1.3 Plane, fied Timoshenko beam with constant load per unit length Figure A shows a plane Timoshenko beam of the length l with constant bending stiffness EI and shear stiffness GA. The beam is fied at both end-sections and is loaded with a constant load q and a constant moment load m. The displacement w (), the rotation θ (), the shear force Q () and the bending moment are to be determined. δq m EI, GA l Figure A Fied beam with constant load per unit length. The differential equations for determination of w () and θ () follow from Eq. (1 7). Thus, ( ) d 2 θ EI d + dw 2 GA d θ + m =, According to Eq. (1 35), the boundar conditions are: ( ( )) d dw GA d d θ + q =. (a) w () = w (l) =, θ () = θ (l) =. (b) DCE Lecture Notes No. 23 (continued)

35 1.4 Uncoupling of aial and bending deformations 27 Integration of the second equation in Eq. (a) provides: ( ) dw Q = GA d θ = q + c 1 (c) Then, the following solution is obtained for θ () from the first equation in Eq. (a): EI d 2 θ d 2 = q (c1 + m) EIθ() = 1 6 q3 1 2 (c1 + m)2 + c 2 + c 3. (d) Further, the boundar conditions θ () = θ (l) = provide c 3 =, c 2 = 1 6 ql2 + 1 (c1 + m)l. (e) 2 Hence, the following reduced form is obtained for θ (): θ () = 1 6EI ( q( 3 l 2 ) 3(c 1 + m )( 2 l) ). (f) Net, Eq. (f) is inserted into Eq. (c) which is subsequentl integrated with respect to, leading to the following solution for w (): GA dw d GA ( = q + c1 + q( 3 l 2 ) 3(c 1 + m ) 2 l) ) 6EI GA w () = 1 2 q2 + c 1 + c 4 + GA 6EI ( 1 4 q(4 2 2 l 2 ) 1 2 (c1 + m)( l) The boundar conditions w () = w (l) = provide the integration constants ). (g) c 4 =, c 1 = 1 2 ql 1 Φ + 1 m, (h) where Φ = 12 EI GA l 2 (i) Then, Eq. (a) and Eq. (f) provide the following solutions: w () = w () = q 2GA (l ) + q (l ) 2 2 m 24EI GA q 24EI ( (l ) Φ l 2 (l ) ) m 12EI Φ + 1 m 12EI Φ Φ + 1 ( l) Φ Φ + 1 ( l + l 2 ), (j) θ () = q ( l + l 2 ) + m Φ (l ). 12EI 2EI Φ + 1 (k) The non-dimensional parameter Φ is a measure of the influence of the shear deformations. For a rectangular cross-section with the height h we have I = 1 12 h2 A and A = 5 A. Then Φ becomes 6 Φ = 72 5 h2 l 2 E G. Hence, shear deformations are primaril of importance for short and high beams. On the other hand, for long beams with a small height of the cross-section, shear deformations are of little importance, i.e. onl the bending deformation is significant. (continued) (l) Elastic Beams in Three Dimensions

36 28 Chapter 1 Beams in three dimensions For Bernoulli-Euler beams we have γ =, corresponding to GA =, cf. Eqs. (1 16) and (1 27). Then, Φ = and Eqs. (j) and (k) reduce to w () = q 24EI (l ) 2 2, (m) q θ () = ( l + l). (n) 12EI It is remarkable that the distributed moment load m does not induce an displacements or rotations in the considered beam with Bernoulli-Euler kinematics. The shear force Q () and bending moment M () follow from Eq. (c), Eq. (h) and Eq. (k), respectivel, i.e. Q () = q + c 1 = 1 2 q(l 2) 1 Φ + 1 m, (o) dθ M () = EI d = q 12 (62 6l + l 2 ) + m Φ (l 2). 2 Φ + 1 (p) For a Bernoulli-Euler beam these results reduce to Q () = 1 q(l 2) m, 2 (q) M () = q 12 (62 6l + l 2 ). (r) The constant moment load m onl induces a constant shear force of magnitude m, whereas M () is not affected b this load. Especiall, for = and = l, Eq. (o) and Eq. (p) provide Q () = 1 2 ql m, M() = 1 12 ql Φ + 1 ml, Q ( l) = 1 2 ql m, M(l) = 1 12 ql2 1 2 Φ + 1 ml. The displacement at the midpoint = l/2 follows from Eq. (k): Φ Φ (s) (t) w (l/2) = ql4 384EI (1 + 4Φ ). The first and second terms within the parenthesis specif the contributions from bending and shear, respectivel. Again the parameter Φ reveals itself as a measure of the relative contribution from shear deformations. (u) 1.5 Normal stresses in beams For at beam without warping, the normal stress σ (,, ) in terms of the generalised strains follows from Eqs. (1 17), (1 2) and (1 24): σ = E(ε κ + κ ). (1 73) In the principal aes coordinate sstem, where S = S = I =, the generalised strains {ε, κ, κ } are related to the conjugated generalised stresses {N, M, M } as determined from DCE Lecture Notes No. 23

37 1.6 The principle of virtual forces 29 Eqs. (1 45a), (1 45b), (1 46) and (1 47) for S = S = I =, i.e. ε = N EA, κ = M EI, κ = M EI. (1 74) Insertion of Eq. (1 74) into Eqs. (1 19) and (1 24) provides the result for the aial stress in terms of the generalised stresses, σ = N A M I + M I. (1 75) Equation (1 75) is due to Navier, and is therefore referred to as Navier s formula. It should be noticed that Eq. (1 75) presumes that the stresses are formulated in a principal aes coordinate sstem, so I and I indicate the principal moments of inertia. The relation is valid for both Timoshenko and Bernoulli-Euler beams. This is so because onl the relation (1 15), but not the relation (1 18) has been utilised. Hence, Eq. (1 75) is based on the assumption that plane cross-sections remain plane, but not that the remain orthogonal to the beam ais. The so-called ero line specifies the line in the (, )-plane on which σ =. The analtical epression for the ero line becomes N A M I + M I =. (1 76) It is finall noted that warping introduces displacements in the aial direction in addition to those provided b bending. However, if the torsion is homogeneous, these displacements will not introduce an normal strains and therefore no normal stresses. Hence, Navier s formula is also valid in the case of St. Venant torsion, but in the case of Vlasov torsion, or inhomogeneous torsion, additional terms must be included in Eq. (1 75). 1.6 The principle of virtual forces In this section the principle of virtual forces is derived for a plane Timoshenko beam of the length l. The deformation of the beam is taking place in the (, )-plane. In the referential state, the left end-section is placed at the origin of the coordinate sstem and the -ais is placed along the bending centres of the cross-sections, see Fig The principle of virtual forces is the dual to the principle of virtual displacements. In the principle of virtual displacements the actual sectional forces and sectional moments are assumed to be in equilibrium with the loads and the reaction forces applied at the end sections. The virtual displacements and rotations are considered as arbitrar increments to the actual displacements and the onl need to fulfil homogeneous kinematic boundar conditions, so that the combined field made up b the actual and the virtual fields alwas fulfils the actual non-homogeneous boundar conditions as given b Eq. (1 35). Further, the generalised virtual strains defining the internal virtual work must be derived from the virtual displacement and rotation fields. In contrast, the principle of virtual forces presumes that the displacements and rotations of the beam are fulfilling the kinematic boundar conditions, and that the generalised internal strains are compatible to these fields. The actual loads on the beam are superimposed with the virtual incremental loads per unit length δq and δq, the virtual moment load per unit length δm, Elastic Beams in Three Dimensions

38 3 Chapter 1 Beams in three dimensions δq δq δm δm,1 δn δn 2 δn 1 δm δm,2 δq δq,2 δq,1 l Figure 1 15 Virtual internal and eternal forces. the virtual reaction forces δn j and δq,j along the - and -directions, and the virtual reaction moments δm in the -directions, where j = 1 and j = 2 indicate the left and right end-sections, respectivel. Due to the load increments, the internal section forces and section moment achieve increments δn, δq and δm, see Fig These variational fields are assumed to be in equilibrium with the variational load fields δq, δq and δm, and to compl with the variations δn j, δq,j and δm,j of the reaction forces and reaction moments. In what follows, δn, δq and δm will be referred to as the virtual internal forces, whereas δq, δq, δm, δn j, δq,j and δm,j are called the virtual eternal forces. The starting point is taken in the kinematical conditions provided b Eqs. (1 12) and (1 17) and rewritten in the form dw d ε =, dw d θ γ =, dθ d κ =. (1 77) The virtual internal forces are related to the virtual eternal loads per unit length δq, δq and δm via the following equations of equilibrium, cf. Eqs. (1 4a) and (1 4b): d(δn) d + δq =, d(δq ) d + δq =, d(δm ) d + δq + δm =. (1 78) The first equation in Eq. (1 78) is multiplied with δn(), the second equation is multiplied with δq (), and the third equation with δm(). Net, the equations are integrated from = to = l, and the three resulting equations are added, leading to the identit l [ ( ) ( ) ( )] dw δn d ε dw + δq d θ dθ γ + δm d κ d = (1 79) Integration b parts is carried out on the first terms within the innermost parentheses, leading to [ ] l l ( d(δn) δnw + δq w + δm Q d w + d(δq ) d w + d(δm ) θ δq θ )d d l ) = (δnε + δq γ + δm κ d. (1 8) DCE Lecture Notes No. 23

39 1.6 The principle of virtual forces 31 Upon utilisation of Eq. (1 78), this is reduced to [δn w + δq w + δm θ ] l + l = l (δq w + δq w + δm θ ) d (δn ε + δq γ + δm κ ) d. (1 81) The generalised strains on the right-hand side of Eq. (1 81) are now epressed in mechanical quantities b means of Eq. (1 74). Further, δn, δq and δm fulfil the following boundar conditions at = and = l, cf. Fig. 1 15, δn() = δn 1, δq () = δq,1, δm () = δm,1, (1 82a) δn(l) = δn 2, δq (l) = δq,2, δm (l) = δm,2. (1 82b) Equation (1 81) then obtains the following final form: 2 j=1 = (δn j w,j + δq,j w,j + δm,j θ,j ) + l ( δn N EA + δq Q GA l (δq w + δq w + δm θ ) d + δm ) M d, (1 83) EI where w,j, w,j and θ,j denote the displacements in the - and -directions and the rotation in the -direction at the end-sections, respectivel. Equation (1 83) represents the principle of virtual forces. The left- and right-hand sides represent the eternal and internal virtual work, respectivel. The use of Eq. (1 83) in determining the displacements and rotations of a Timoshenko beam is demonstrated in Eamples 1.4 and 1.5 below. Furthermore, the principle of virtual forces ma be used to derive a stiffness matri for a Timoshenko beam element as shown later. Eample 1.4 End-displacement of cantilevered beam loaded with a force at the free end Figure A shows a plane Timoshenko beam of the length l with constant aial stiffness EA, shear stiffness GA and bending stiffness EI. The beam is fied at the left end-section and free at the right end-section, where it is loaded with a concentrated force Q,2 in the -direction. The displacement w,2 at the free end is searched. The principle of virtual forces Eq. (1 83) is applied with the following eternal virtual loads: δq = δq = δm =, δn 1 = δq,1 = δm,1 = δn 2 = δm,2 = and δq,2 = 1. Further N() =. Then, Eq. (1 83) reduces to l ( ) δq Q δm M 1 w,2 = + d. (a) GA EI (continued) Elastic Beams in Three Dimensions

40 32 Chapter 1 Beams in three dimensions Q,2 δq,2 = 1 EA, GA, EI l l M δm Q,2 l l Q δq Q,2 1 Figure A Fied plane Timoshenko beam loaded with a concentrated force at the free end: Actual force and section forces (left) and virtual force and section forces (right). The variation of the bending moment M () and the shear force Q () from the actual load Q,2 has been shown in Fig. A on the left. The corresponding variational moment field δm () and shear force δq () from δq,2 = 1 are shown in Fig. A on the right. Insertion of these distributions in Eq. (b) provides the solution w,2 = Q,2l + 1 Q,2l 3 = 1 Q,2l3 (4 + Φ), (b) GA 3 EI 12 EI where Φ is given b Eq. (i) in Eample 1.3. The deformation contributions from shear and bending are additive. This is a consequence of the additive nature of the fleibilities indicated b Eq. (1 83) in contribution to the fact that the beam is staticall determinate, which provides the fields M () and Q () as well as δm () and δq () directl. Eample 1.5 End-deformations of fied beam loaded with a moment at the free end The beam described in Eample 1.4 is considered again. However, now the free end is loaded with a concentrated moment M,2. The displacement w,2 and the rotation θ,2 of the end-section is to be found. At the determination of w,2 from M,2, the principle of virtual forces given b Eq. (1 83) is again applied with δq,2 = 1 and all other eternal variational loads equal to ero, leading to Eq. (a) in Eample 1.4. However, M () and Q () are now caused b M,2, and are given as shown in Fig. A, whereas δm () and δq () are as shown in Fig. A of Eample 1.4. Then, w,2 becomes w,2 = l δm M d = 1 M,2l 2. (a) EI 2 EI (continued) DCE Lecture Notes No. 23

41 1.7 Elastic beam elements 33 EA, GA, EI l M,2 l δm,2 = 1 M δm M,2 1 Q = δq = Figure A Fied plane Timoshenko beam loaded with a moment at the free end: Actual moment and section forces (left) and virtual moment and section forces (right). At the determination of θ,2 the principle of virtual forces Eq. (1 83) is applied with the following eternal virtual loads δq = δq = δm =, δn 1 = δq,1 = δm,1 = δn 2 = δq,2 = and δm,2 = 1. Then, Eq. (1 83) reduces to l ( ) δq Q δm M 1 θ,2 = + d. (b) GA EI The variation of Q () and M () from M,2 has been shown in Fig. A on the left, and the variation of δq () and δm () from δm,2 = 1 is shown in Fig. A on the right. Then θ,2 becomes θ,2 = l 1 M,2 d = M,2l. EI EI In the present load case, the shear force is given as Q () =. Consequentl it will not induce an contributions in Eq. (a) and Eq. (c). (c) 1.7 Elastic beam elements When frame structures consisting of multiple beams are to be analsed, the establishment of analtical solutions is not straightforward and instead a numerical solution must be carried out. For this purpose, a discretiation of the frame structure into a number of so-called beam elements is necessar, eventuall leading to a finite-element model. The aim of the present section is not to provide a full introduction to the finite-element method for the analsis of frame structures, e.g. tower blocks with a steel frame as the load-carring structure. However, a formulation is given for a single beam element to be applied in such analses. Both the Timoshenko and Bernoulli-Euler beam theories are discussed in this contet, and plane as well as threedimensional beams are touched upon. Elastic Beams in Three Dimensions

42 34 Chapter 1 Beams in three dimensions A plane Timoshenko beam element Firstl, the stiffness matri and element load vector is derived for a plane Timoshenko beam element with constant aial stiffness EA, shear stiffness GA and bending stiffness EI, cf. Fig The stiffness relation is described in an (, )-coordinate sstem with origin at the left endsection and the -ais along the bending centres. N 1, w,1 M,1, θ,1 Q,1, w,1 EA, GA, EI 1 2 Q,2, w,2 N 2, w,2 M,2, θ,2 l Figure 1 16 Plane Timoshenko beam element with definition of degrees of freedom and nodal reaction forces. At the end-nodes, nodal reaction forces N j and Q,j are acting along the - and -directions, respectivel, and reaction moments M,j are applied around the -ais. Here, j = 1 and j = 2 stand for the left-end and right-end nodes of the beam element, respectivel, and the reaction forces and moments are in equilibrium with the remaining eternal loads on the element for arbitrar deformations of the beam. The element has 6 degrees of freedom defining the displacements and rotations of the endsections, cf. Fig These are organised in the column vector [ ] we1 w e = = [ ] T w w,1 w,1 θ,1 w,2 w,2 θ,2 (1 84) e2 The sub-vector w ej defines the degrees of freedom related to element node j. Similarl, the reaction forces N j, Q,j and M,j, j = 1, 2, at the end-sections, work conjugated to w,j, w,j and θ,j, are stored in the column vector [ ] re1 r e = = [ ] T N r 1 Q,1 M,1 N 2 Q,2 M,2 (1 85) e2 p p m M,1 N 1 Q,1 Q,2 M,2 N 2 Figure 1 17 Eternal loads and reaction forces from eternal loads on a plane beam element. The equilibrium of the beam element relating the nodal reaction forces to the degrees of freedom of the element ma be derived b the principle of virtual displacements as demonstrated DCE Lecture Notes No. 23

43 1.7 Elastic beam elements 35 in a subsequent paper. The resulting equilibrium equations on matri form ma be written on the form r e = K e w e + f e. (1 86) The vector f e in Eq. (1 86) represents the nodal reaction forces from the eternal element loads when w e =, i.e. when the beam is fied at both ends as shown in Fig We shall merel consider constant element loads q and q per unit length in the - and -directions, and a constant moment load per unit length m in the -direction, see Fig The reaction forces and reaction moments follow from Eqs. (s) and (t) in Eample 1.3: 1 2 q l 1 2 q l + m 1 12 f e = q l 2 1 Φ 2 Φ m +1 l 1 2 q. (1 87) l 1 2 q l m 1 12 q l 2 1 Φ 2 Φ m +1 l The matri K e in Eq. (1 86) denotes the stiffness matri in the local (,, )-coordinate sstem. Let w i denote the ith component of w e. Then, the ith column in K e represents the nodal reaction forces for f e =, and with w i = 1 and w j =, j i. These forces are obtained following the derivations in Eample 1.3 from Eq. (a) to Eq. (t) with q = m = and with the boundar condition in Eq. (b) replaced b the indicated conditions. Because of the smmetr of the problem, onl two such analses need to be performed. Still, this is a rather tedious approach. Partl because of this, and partl in order to demonstrate an alternative approach, the stiffness matri will be derived based on the principle of virtual forces. Undeformed state Undeformed state w θ w (a) (b) Figure 1 18 Rigid-bod modes of a plane beam element: (a) Translation and (b) rotation. The beam element has 6 degrees of freedom, b which a total of 6 linear independent modes of deformation ma be defined. These consist of 3 linear independent rigid bod modes and 3 linear independent elastic modes. The rigid modes ma be chosen as a translation in the -direction, a translation in the -direction and a rotation around the -direction as shown in Fig An rigid bod motion of the beam element ma be obtained as a linear combination of these component modes of deformation. Obviousl, the rigid bod motions do not introduce stresses in the beam. Hence, the aial force N, the shear force Q and the bending moment M are all ero during such motions. Since, aial elongations are uncoupled from bending deformations, the elastic elongation mode is uniquel defined as shown in Fig. 1 19a. The two bending deformation modes ma be Elastic Beams in Three Dimensions

44 36 Chapter 1 Beams in three dimensions N N u 2 N u 2 N (a) 2M a l M s M s M a θ a θ s θ s θ a 2M a l Q M a Q = 2M a l M M a M s M a M (b) Figure 1 19 Elastic modes and related section forces in a plane beam element: (a) Aial elongation; (b) smmetric bending and (c) antismmetric bending. (c) chosen in arbitraril man was. Tpicall, these are chosen b prescribing an angle of rotation at the other end-section. Following an idea b Krenk (21), a more convenient formulation ma be obtained b choice of two bending modes smmetric and anti-smmetric around the mid-point of the beam element as shown in Figs. 1 19b and 1 19c. It should be noticed that these modes also appl if the material properties of the beam are not smmetrical around the mid-point. The aial elongation and conjugated aial force related to the aial elongation mode are denoted u and N, respectivel. The smmetric and anti-smmetric bending modes are described b the end-section rotations θ s and θ a defined in Figs. 1 19b and 1 19c, respectivel. The conjugated moments are denoted M s and M a, respectivel. The related distributions of the shear force Q () and the bending moment M () are shown in Figs. 1 19b and 1 19c. The shear force is equal to Q = in smmetric bending, because the bending moment is constant. Then, no shear deformations are related to this mode. In contrast, a constant shear force appears in the anti-smmetric bending mode. Hence, the deformations occurring in this mode are affected b bending as well as shear contributions. At first the constitutive relations between the deformation measures and the conjugated generalised strains for the indicated elastic modes are found b means of the principle of virtual forces Eq. (1 83). In all cases, the beam element is unloaded, so δq = δq = δm =. For the aial elongation mode N = N, Q = M =, and δn = 1, δn 1 = 1, δn 2 = 1. Further, DCE Lecture Notes No. 23

45 1.7 Elastic beam elements 37 w,1 = 1 2 u, w,2 = 1 2 u, w,j = θ,j =. Then, Eq. (1 83) reduces to ( 1) ( 1 2 u ) u = l 1 N EA d u = N l d EA = l EA N. (1 88) The last statement holds for a beam element with constant aial stiffness EA. If EA varies, the integral in the middlemost statement must be evaluated analticall or numericall. For the smmetric bending mode N = Q =, M = M s, δm = 1, δm,1 = 1, and δm,2 = 1. Further, θ,1 = θ s, θ,2 = θ s and w,j = w,j =. Then, Eq. (1 83) provides 1 θ s + ( 1) ( θ s ) = l ( 1)( M s ) EI d 2θ s = M s l d = l M s. (1 89) EI EI Again, the last statement onl applies for a homogeneous beam, whereas the middlemost statement applies for an variation of the bending stiffness EI along the beam. For the anti-smmetric bending mode N =, Q = 2 Ma l, M () = ( 1 + 2/l)M a, and δq = 2δM ()/l = ( 1 + 2/l). Further, θ,1 = θ a, θ,2 = θ a, w,j = w,j =. Then, Eq. (1 83) provides ( l 2 l 1 θ a + 1 θ a = 2Ma l + ( l )( l )M ) a d GA EI ( 4 l ) d l ( l 2θ a = M a l 2 + )2 d GA EI 1 2θ a = M a (4l GA l ) l = 1 l (1 + Φ )M a, 3 EI 3 EI Φ = 12 EI GA l 2. (1 9) As discussed in Eample 1.3, the non-dimensional parameter Φ defines the contribution of shear fleibilit relativel to the bending fleibilit. The fleibilit relations provided b Eqs. (1 88), (1 89) and (1 9) ma be written in the following equivalent stiffness matri formulation: r = K w, where r = N M s M a, w = u 2θ s 2θ a, K = (1 91) EA l EI l. (1 92) 3 EI 1+Φ l The nodal reaction forces r e and r in Eq. (1 85) and Eq. (1 92) are related via the transformation r e = Sr = SK w, where S = 1 2/l /l 1 1. (1 93) (1 94) Elastic Beams in Three Dimensions

46 θ,1 θ,2 38 Chapter 1 Beams in three dimensions Similarl, the elastic deformation measures stored in w can be epressed b the degrees of freedom of the element stored in w e as follows (see Fig. 1 2): u = w,2 w,1, (1 95) θ,1 = θ a + θ s + 1 l (w,2 w,1 ) θ,2 = θ a θ s + 1 l (w,2 w,1 ) 2θ a = θ,1 + θ,2 2 l (w,2 w,1 ) 2θ s = θ,1 θ,2. (1 96) w,1 w,2 θ a + θ s θ a θ s Figure 1 2 Connection between elastic deformation measures and element degrees of freedom. Equations (1 95) and (1 96) ma be rewritten in the common matri form w = S T w e, (1 97) where S is given b Eq. (1 94). Insertion of Eq. (1 97) into Eq. (1 93) provides upon comparison with Eq. (1 86): r e = SK S T w e K e = SK S T. (1 98) Insertion of K and S as given b Eq. (1 92) and Eq. (1 94) provides the following eplicit solution for K e : K e = E l 3 Al 2 Al Φ I 6 1+Φ I l 12 1+Φ I 6 1+Φ I l 6 1+Φ I l 4+Φ 1+Φ I l Φ I l 2 Φ 1+Φ I l 2 Al 2 Al Φ I 6 1+Φ I l 12 1+Φ I 6 1+Φ I l 6 1+Φ I l 2 Φ 1+Φ I l Φ I l 4+Φ 1+Φ I l 2. (1 99) The corresponding result for a plane Bernoulli-Euler beam element is obtained simpl b setting Φ =. The equivalent element relations for a three-dimensional beam formulated in a (,, ) principal aes coordinate sstem are given in the net section. DCE Lecture Notes No. 23

47 1.7 Elastic beam elements 39 Eample 1.6 Deformations of a plane Timoshenko beam structure Figure A shows a plane beam structure ABC consisting of two Timoshenko beam elements AB and BC of the lengths l and l/2, respectivel. The loads on the structure and the resulting displacements are described in the indicated (,)-coordinate sstem. The shear stiffness GA and the bending stiffness EI are constant and the same in both beam elements. The structure is fied at point A, free at point C, and simpl supported at point B. Both beam elements are loaded with a constant load per unit length q. Additionall, beam BC is loaded with a concentrated load P = q l at the free end C. q P = q l A 1 B 2 C GA, EI l l/2 Figure A Plane Timoshenko beam structure consisting of two beam elements. We want to determine the displacement of point C in the -direction, and the reaction forces and moments at the support points A and B. The calculations are performed with the shear stiffness given as GA = 12 EI l 2. We shall refer to the beam elements AB and BC with the inde e = 1 and e = 2, respectivel. With reference to Eq. (i) in Eample 1.3, the shear fleibilit parameters for the two beam elements become: (a) Φ 1 = 12EI 12EI =.1, Φ2 = GA l2 GA =.4. (l/2) 2 (b) θ 1, M 1 θ 2 θ 2 θ w 1 w 2 w 2 w 3 Q 1 Q 2 Figure B Global degrees of freedom and reaction forces in the plane Timoshenko beam structure. Since, the aial deformations are disregarded, each element has 4 degrees of freedom out of which two are common, namel the displacement and rotation at point B. The related global degrees of freedom have been defined in Fig. B. The element stiffness matrices follow from Eq. (1 99) and Eq. (b): 12 6l 12 6l K 1 = EI 6l 4.1l 2 6l 1.9l 2 1.1l l 12 6l, 6l 1.9l 2 6l 4.1l l 12 3l 8EI K2 = 3l 1.1l 2 3l.4l 2 1.4l l 12 3l 3l.4l 2 3l 1.1l 2. (c) (continued) Elastic Beams in Three Dimensions

48 4 Chapter 1 Beams in three dimensions The corresponding element loads become, cf. Eq. (1 87): 1 ql 6 2 f 1 = 1 12 ql2 1 ql = ql l 6, f2 = 1 12 ql2 l 1 2 q l q( l 2 )2 1 2 q l 2 + P 1 12 q( l 2 )2 = 1 48 ql The global equilibrium equation, made up of contributions from both elements, has the structure r = Kw + f, 12 l 6 l. (d) (e) where r 1 r = = r 2 Q 1 M 1 Q 2, w = w 1 θ 1 w 2 θ 2 w 3 θ 3 f 1, f = = 1 48 ql f l 36 3l 6 l, (f) K 1 K = K 2 = EI l l l l2 6 l l l ( l l 1.1 ( 1.1 l2 6 + ) l ) l l 2 24 l l2 ) l ( l l l l l l2. (g) The details in the derivation of the matri equation Eq. (e), including the formation of the column vectors r and f and the global stiffness matri K b adding contribution from element components, will be eplained in a later chapter. The displacement degrees of freedom w 1 and w 2 at the nodes A and C are both equal to ero. Similarl, the rotation θ 1 at the fied support at A is ero. When these values are introduced in w, Eq. (e) provides the values of the reaction components Q 1, M 1 and Q 2 if the remaining unconstrained degrees of freedom θ 2, w 3 and θ 3 are inserted. These are determined from the corresponding equations in Eq. (e) using w 1 = θ 1 = w 2 =, leading to EI l 3 θ 2l w 3 θ 3l ( )l l l2 24 l l l2 24 l l2 = ql4 EI θ 2 w 3 θ ql 3l 6 l = Insertion of Eq. (h) along with w 1 = θ 1 = w 2 = into the remaining equation (e) provides the following solution for the reaction components Q 1, M 1 and Q 2: (continued) (h) DCE Lecture Notes No. 23

49 1.7 Elastic beam elements 41 Q 1 M 1 Q 2 Q 1l M 1 O 2l = EI l 3 = q l 2 6 l l2 ( )l l ql4 EI.1453/l /l 1 48 ql 24 4l 36 (i) In case of Bernoulli-Euler kinematics, corresponding to Φ 1 = Φ 2 =, the corresponding solutions become: θ 1l.1354 w 3 = ql (j) EI θ 2l.2812 Q 1l M 1 Q 2l = q l Comparison of Eqs. (h) and (j) reveals that the displacement w 3 as well as the rotations θ 1 and θ 3 are increased b the shear fleibilit. This is so because bending and shear deformations in general are coupled for statical indeterminate structures. At the same time, a comparison of Eqs. (i) and (k) shows that Bernoulli-Euler beam kinematics lead to higher stresses than Timoshenko beam theor, which is due to the fact that the shear stiffness in the Bernoulli-Euler beam is infinite. (k) A three-dimensional Timoshenko beam element The formulation of the beam-element stiffness matri is now etended to three dimensions. This involves fleural displacements in two directions, aial displacements and, in addition to this, twist of the beam. It is assumed that the beam element is straight, of the length l and with constant cross-section. The element relation is described in a principal aes (,, )-coordinate sstem with origin at the left end-section and the -ais placed along the bending centres of the cross-sections. Onl St. Venant torsion is taken into consideration. The aial stiffness EA, the shear stiffnesses GA and GA in the - and -directions, respectivel, the torsional stiffness GK, and the principal inertial bending stiffnesses EI and EI around the - and -aes are all constant along the beam element. The degrees of freedom of the element are made up b the 6 componentsw,j, w,j and w,j, j = 1, 2, providing the displacements of the bending centres, and the 6 components θ,j, θ,j and θ,j defining the rotation of the end-sections. Again, j = 1 and j = 2 refer to the left and right end-sections, respectivel. The nodal reaction forces conjugated to the indicated degrees of freedom consist of the aial forces Q,j and Q,j in the - and -directions, respectivel, the torsional moments M,j and the bending moment components M,j and M,j in the - and -directions. Finall, the element loadings consist of constant loads per unit length {q, q, q } and constant moment loads per unit length {m, m, m } in the -, - and -directions. No concentrated element forces or moments are considered. The loads q and q as well as the shear forces Q,j and Q,j are assumed to act through the so-called shear centre, leading to an uncoupling of the fleural and torsional deformations. The definition of the shear centre and further details about uncoupling of torsional and fleural displacements are given in Chapter 2. Elastic Beams in Three Dimensions

50 A,GA 42 Chapter 1 Beams in three dimensions q M,1 m N 1 N 2 M,2 θ,1 w,1 w EA,GK,2 θ,2 q m θ,1, M,1 θ,2, M,2 GA, EI w,1, Q,1 w,2, Q,2 q m θ,1, M,1 θ,2, M,2 GA, EI w,1, Q,1 w,2, Q,2 Figure 1 21 Three-dimensional Timoshenko beam element with definition of degrees of freedom, nodal reaction forces, element loads and sectional properties. The element equilibrium equations ma be epressed on the matri form, cf. Eq. (1 86), r e = K e w e + f e (1 1) Here, r e and w e are 12-dimensional column vectors storing the reaction forces and the element degrees of freedom, respectivel, cf. Eqs. (1 84) and (1 85), N 1 w,1 Q,1 w,1 Q,1 w,1 M,1 θ,1 [ ] M,1 [ ] θ,1 re1 r e = = M,1 we1 r e2 N 2, w e = = θ,1 w e2 w,2. (1 11) Q,2 w,2 Q,2 w,2 M,2 θ,2 M,2 θ,2 M,2 θ,2 Likewise, f e is 12-dimensional column vector storing the contributions to the reaction forces DCE Lecture Notes No. 23

51 1.7 Elastic beam elements 43 from the element loads, given as, cf. Eq. (1 87), f e = 1 2 q l 1 2 q l + m 1 2 q l m 1 2 m l 1 ] 12 q l 2 1 Φ 2 1+Φ m l 1 12 = q l 2 1 Φ 2 1+Φ m l f e2 1 2 q, (1 12) l 1 2 q l m 1 2 q l + m 1 2 m l 1 12 q l 2 1 Φ 2 1+Φ m l 1 12 q l 2 1 Φ 2 1+Φ m l [ fe1 where Φ and Φ are given as, cf. Eq. (i) in Eample 1.3, Φ = 12 EI GA l 2, Φ = 12 EI GA l2. (1 13) Finall, K e specifies the element stiffness given as, cf. Eq. (1 99), K e = where EA l EA l k11 k12 k13 k14 k 11 k 12 k 13 k 14 GK l GK l k 12 k 22 k 23 k 24 k12 k22 k23 k24 EA EA l l k13 k23 k33 k34 k 13 k 23 k 33 k 34 GK l GK l k 14 k 24 k 34 k 44 k 14 k 24 k 34 k 44 k11 k12 k13 k14 k22 k23 k24 k33 k34 k44 k 11 k 12 k 13 k 14 k 22 k 23 k 24 k 33 k 34 k 44 = = EI (1 + Φ )l 2 EI (1 + Φ )l l 12 6l (4 + Φ )l 2 6l (2 Φ )l l (4 + Φ )l l 12 6l (4 + Φ )l 2 6l (2 Φ )l l (4 + Φ )l 2, (1 14), (1 15). (1 16) Elastic Beams in Three Dimensions

52 44 Chapter 1 Beams in three dimensions The element equilibrium relation provided b Eq. (1 1) presumes that onl St. Venant torsion is taken into consideration, corresponding to the torsional equilibrium equations [ ] M,1 = GK [ ] [ ] 1 1 θ,1 1 [ ] 1 M,2 l 1 1 θ,2 2 m l. (1 17) 1 The torsional constant K is determined in the net chapter. It will be shown that the inclusion of Vlasov torsion requires the introduction of two etra degrees of freedom dθ,1 d and dθ,2 d. The conjugated generalised stresses are the so-called bimoments. Hence, a Timoshenko beam element, where both St. Venant and Vlasov torsion are taken into consideration, is described b a total of 14 degrees of freedom. 1.8 Summar In this chapter, the basic theor of Timoshenko and Bernoulli-Euler beams in three-dimensional space has been presented. Some of the main topics covered are summarised below. Beams are one-dimensional structures that ma carr loads in three dimensions including aial forces, shear forces in two orthogonal directions and moments around three directions. Bernoulli-Euler beam kinematics assume that cross-sections remain orthogonal to the beam ais during deformation. Hence, no shear deformation occurs. Timoshenko beam kinematics include shear fleibilit, but still a cross-section remains plane during deformation. Hence, shear strains and stresses are homogeneous over the beam height. The bending centre of a beam cross-section is defined as the point of attack of an aial force not producing a bending moment. The principal aes of a beam cross-section are defined as the aes around which a bending moment will neither produce an aial force nor fleural displacements in the other direction. The principle of virtual forces can b applied to the analsis of deformations in a beam. In the case of Timoshenko beam theor, both shear and bending deformation occurs, whereas onl bending deformation is present in a Bernoulli-Euler beam. Plane beam elements have si degrees of freedom with three at either end, i.e. two displacements and one in-plane rotation. In the general case, the rotations and the aial displacements are coupled, but in a principle-aes description, the become uncoupled. Spatial beam elements have 12 degrees of freedom, that is three displacements and three rotations at either end. Generall, the displacements and rotations are coupled, but an uncoupling can be achieved b a proper choice of coordinate sstem. Thus, a detailed description has been given of the lateral and fleural deformations in a beam. However, in this chapter onl a brief introduction has been given to twist and torsion of a beam. A thorough eplanation and analsis of these phenomena will be the focus of the net chapter. DCE Lecture Notes No. 23

53 CHAPTER 2 Shear stresses in beams due to torsion and bending In this chapter, a theoretical eplanation is given for the shear forces in beams stemming from bending as well as torsion. In this regard, the coupling of torsion and bending is discussed, and the so-called shear centre is introduced. The derivations are confined to homogeneous torsion, or St. Venant torsion. Later, the strains and stresses provided in the case of non-homogeneous torsion, or Vlasov torsion, will be dealt with. 2.1 Introduction When the beam is eposed to the loads per unit length q and q, the beam will generall deform with bending deformations {u, θ } and {u, θ }, respectivel. Depending on the line of action of these loads, the bending deformations will be combined with a torsional rotation θ around the -ais as illustrated in Fig. 2 1a and Fig. 2 1c. Under certain conditions, the bending of the beam is not associated with a torsional deformation. This happens if the loads per unit length q and q, the reaction forces at the ends of the beam as well as the shear forces Q and Q are acting through a special point S, known as the shear centre, as illustrated in Fig. 2 1b. When this is the case, torsion is caused solel b the moment load m per unit length, which in part includes contributions from the translation of q and q to S. These torsional deformations take place without bending deformation as illustrated in Fig. 2 1d. Hence, bending and torsion can be analsed independentl. The position of the shear centre depends on the geometr of the cross-section and is generall different from the position of the bending centre B. However, for double-smmetric crosssections, the positions of the bending and torsion centres will coincide. The shear forces Q and Q as well as the torsional moment M bring about shear stresses σ and σ on the beam section. In what follows these will be determined independentl for the two deformation mechanisms. Hence, q and q are presumed to be referred to the shear centre. The shear stresses caused b the torsional moment M are staticall equivalent to the shear forces Q = Q =. The position of the shear centre has no influence on the distribution of shear stresses in this case. Likewise, in the decoupled bending problem, in which the cross-section is eposed to the shear forces Q and Q, the position of the shear centre is determined from the requirement that the resulting shear stresses are staticall equivalent to Q and Q, and produce the torsional moment M = around S. 45

54 46 Chapter 2 Shear stresses in beams due to torsion and bending Q, q Q, q S B S B (a) (b) S B Q, q M, m S B (c) Figure 2 1 Coupled and uncoupled bending and torsion. Coupling eists in cases (a) and (c), whereas cases (b) and (d) involve no coupling. (d) 2.2 Homogeneous torsion (St. Venant torsion) It is assumed that the torsional moment M and the incremental twist per unit length dθ /d and the warping of the cross-sections remain unchanged along the beam. Then all cross-sections of the beam are eposed to the same distribution of the shear stresses σ and σ. For this reason, this case is referred to as homogeneous torsion. Since the solution of the problem was given b St. Venant (ref.), the case is also called St. Venant torsion. Inhomogeneous torsion refers to the case, where either M or the material properties var along the beam. Then dθ /d or the warping will var as well. Figure 2 2a shows a cross-section of a clindrical beam of the length l. The cross-sectional area is A. The curve along the outer peripher is denoted Γ. The cross-section ma have a number N of holes determined b the boundar curves Γ j, j = 1, 2,..., N. At the boundar curves arc-length coordinates s, s 1,..., s N are defined. The arc-length coordinate s along Γ is orientated in the anti clock-wise direction, whereas s 1, s 2,..., s N, related to the interior boundaries Γ 1, Γ 2,..., Γ N, are orientated in the clock-wise direction. The outward directed unit DCE Lecture Notes No. 23

55 2.2 Homogeneous torsion (St. Venant torsion) 47 n Γ N Γ s s n s B S s j Γ j M n j S S n j Γ j (a) Figure 2 2 Cross-section with holes: (a) Interior and eterior edges; (b) definition of local (, n j, s j )-coordinate sstems. (b) s j s j Γ vector on a point of the eterior or interior boundaries Γ j is denoted n j, j =, 1,..., N. The unit tangential vector to a boundar curve is denoted s j and is co-directional to the arc-length coordinate s j, see Fig. 2 2b. Thus, a local (, n j, s j )-coordinate sstem ma be defined with the base unit vectors {i,n j,s j }. The indicated orientation of the eterior and interior arc-length coordinates s j, j =, 1,..., N, insures that the related (, n j, s j )-coordinate sstem forms a right-hand coordinate sstem. The beam material is assumed to be homogeneous, isotropic linear elastic with the shear modulus G. In homogeneous torsion, onl shear stresses are present for which reason G is the onl needed elasticit constant Basic assumptions For convenience the inde is omitted on the twist θ (the rotation angle around the -ais), i.e. θ θ. Figure 2 3 shows a differential beam element of the length d. Both end-sections of the element are eposed to the torsional moment M, so the element is automaticall in equilibrium. On the left and right end-sections the twists are θ and θ + dθ, respectivel. The increment dθ ma be written as dθ = dθ d d. (2 1) Since M and the material properties are the same in all cross-sections, dθ/d must be constant along the beam. Further, the warping must be the same in all cross-sections, i.e. u = u (, ). This implies that the warping in homogeneous torsion does not induce normal strains, i.e. ε = u =. (2 2) In turn this means that the normal stress becomes σ = Eε =. Hence, onl the shear stresses σ and σ are present on a cross-section in homogeneous torsion. Elastic Beams in Three Dimensions

56 48 Chapter 2 Shear stresses in beams due to torsion and bending M l M d θ M θ + dθ d M Figure 2 3 Beam and differential beam element subjected to homogeneous torsion. The onl deformation measure of the problem is the twist gradient dθ/d. Then, due to the linearit assumptions, the torsional moment M must depend linearl on dθ/d. Further, σ and σ (and hence M ) depend linearl on G. This implies the following relation: M = GK dθ d. (2 3) The proportionalit constant K with the dimension [unit of length] 4 is denoted the torsional constant. The determination of this constant is a part of the solution of the torsion problem Solution of the homogeneous torsion problem As for all beam theories, the shape of the cross-section is assumed to be preserved during the deformation. Then, the displacements in the (, )-plane are caused merel b the rotation θ around the shear centre S. The warping displacements u must also be linearl dependent on the strain measure dθ/d, corresponding to the last term in Eq. (1 13). This implies the displacement components u = u (, ) = ω(, ) dθ d, u = ( s )θ, u = ( s )θ. (2 4) Here ω(, ) is the so-called warping function as discussed in Section 1.2.2, and its determination constitutes the basic part of the solution of the homogeneous torsion problem. Similarl to Eq. (1 14), it follows from Eq. (2 4) that the components of the strain tensor become: ε = u =, ε = u =, ε = u =, (2 5a) DCE Lecture Notes No. 23

57 2.2 Homogeneous torsion (St. Venant torsion) 49 ε = 1 ( u 2 + u ) = 1 ( θ + θ) =, (2 5b) 2 ε = 1 ( u 2 + u ) = 1 ( ) ω dθ 2 ( s) d, (2 5c) ε = 1 ( u 2 + u ) = 1 ( ) ω dθ 2 + ( s) d. (2 5d) As seen onl ε and ε are non-vanishing. Correspondingl, all components of the Cauch stress tensor become equal to ero, save the shear stresses σ and σ. These are given as ( ) ω dθ σ = σ (, ) = 2Gε = G ( s) d, ( ) ω dθ σ = σ (, ) = 2Gε = G + ( s) d. Ignoring the volume loads, the equilibrium equations read σ + σ + σ =, σ + σ + σ =, (2 6a) (2 6b) (2 7a) (2 7b) σ + σ + σ =. (2 7c) With σ = σ = σ = σ =, and σ and σ onl dependent on and, the two last equations are identicall fulfilled, and the first equation reduces to σ + σ =. (2 8) Equation (2 8) ma be formulated at a point on the boundar curve Γ j, j =, 1,..., N, in the related local (, n j, s j )-coordinate sstem. Ignoring the inde j, the non-trivial equilibrium equation then reads σ n n + σ s s =, (2 9) where σ n and σ s denote the shear stresses along the local n- and s-aes. According to Cauch s boundar condition (ref.), σ n can be epressed in terms of the shear stress components σ and σ as σ n = σ n + σ n. (2 1) Here n and n denote the components of the unit normal vector n along the - and -aes. The smmetr of the stress tensor implies that σ n = σ n. Further, since the eterior and all interior surfaces are free of surface traditions, corresponding to σ n =, it follows that σ n = (see Fig. 2 4). Then, Eq. (2 1) reduces to σ n + σ n =. (2 11) Elastic Beams in Three Dimensions

58 5 Chapter 2 Shear stresses in beams due to torsion and bending σ n σ n = n Figure 2 4 Shear stress in the normal direction at an eterior or interior boundar. Finall, insertion of Eq. (2 6) into Eq. (2 11) provides the following boundar condition formulated in the warping function ω n + ω n ( s )n + ( s )n = ω n = ( s)n ( s )n, (2 12) where ω/ n denotes the partial derivative of ω in the direction of the outward directed unit normal. Equation (2 12) must be fulfilled at the eterior and all interior boundaries. The partial differential for ω to be fulfilled in the interior A of the profile follows from insertion of Eq. (2 6) into the equilibrium equation (2 8), leading to 2 ω ω 2 =. (2 13) If a solution to Eq. (2 13) with the boundar conditions (2 12) is obtained, the shear stresses are subsequentl determined from Eq. (2 1). Equation (2 13) is Laplace s differential equation, and the boundar conditions Eq. (2 13) are classified as the so-called Neumann boundar conditions. Notice that the solution to Eqs. (2 12) and (2 13) is not unique. Actuall, if ω(, ) is a solution, then ω(, )+ω will be a solution as well, where ω is an arbitrar constant. Since the shear stresses are determined b partial differentiation of the of the warping function, all these solutions lead to the same stresses. The boundar value problem for the warping function has been summarised in Bo 2.1. DCE Lecture Notes No. 23

59 2.2 Homogeneous torsion (St. Venant torsion) 51 Bo 2.1 Boundar value problem for the warping function The differential equation, representing the non-trivial equation of equilibrium, reads: 2 ω + 2 ω =, (,) A. (2 14a) 2 2 The Neumann boundar condition, representing the relevant Cauch boundar condition, reads: ω = ( s)n ( s)n, (,) Γ Γ1 ΓN. (2 14b) n An alternative formulation for the solution of the problem can be obtained b the introduction of the so-called Prandtl s stress function S with the defining properties σ = S, σ = S. (2 15) Upon insertion of Eq. (2 15) into Eq. (2 9), the equilibrium equation is seen to be identical fulfilled, i.e. σ + σ From Eq. (2 6) follows σ σ = = 2 S 2 S. (2 16) ( 2 ω 1 ) G dθ ( 2 ) d ω + 1 G dθ d = 2Gdθ d. (2 17) Then, the differential equation for S is obtained b insertion of Eq. (2 15) on the left-hand side of Eq. (2 17), leading to 2 S S 2 = 2Gdθ d. (2 18) Equation (2 18) is a compatibilit condition for S in order that the kinematical conditions (2 6) are fulfilled. The boundar condition for S follow upon insertion of Eq. (2 15) into Eq. (2 11), i.e. S n S n = (2 19) The tangential unit vector is given as s T = [s, s ] = [ n, n ], cf. Fig. 2 2b, where {n, n } denotes the Cartesian components of the outward directed unit normal vector at an of the boundar curves Γ j, j =, 1,...,N, cf. Fig. 2 2b. Then, Eq. (2 19) ma be written as S s + S s = S s =. (2 2) where S/ s denotes the directional derivative of S in the direction of the tangential unit vector s. Equation (2 2) implies that S is constant along the eterior and the interior boundar curves, i.e. S = S j, j =, 1,...,N. (2 21) Elastic Beams in Three Dimensions

60 52 Chapter 2 Shear stresses in beams due to torsion and bending Equation (2 18) is a Poisson differential equation (inhomogeneous Laplace equation), and the boundar conditions Eq. (2 21) are classified as the so-called Dirichlet boundar conditions. In principle, the solution to the indicated boundar value problem is unique. The problem is that the constant values S j of the stress function along the boundar curves are unknown. The determination of these is a part of the problem. For profiles with interior holes this is onl possible b the introduction of additional geometric conditions. The boundar value problem for the Prandtl stress function has been summarised in Bo 2.2. The formulation in terms of Prandtl s stress function is especiall useful in relation to homogeneous torsion of thin-walled profiles and will be utilised in a number of eamples below. Bo 2.2 Boundar value problem for the Prandtl stress function The equilibrium equation is automaticall fulfilled. The compatibilit condition is represented b the following differential equation: 2 S + 2 S dθ = 2G, (, ) A. (2 22a) 2 2 d The Dirichlet boundar condition, representing the Cauch boundar condition, reads: S = S j, j =,1,..., N, (,) Γ Γ 1 Γ N. (2 22b) The shear stresses σ and σ must be staticall equivalent to the shear forces Q = Q = and the torsional moment M. Application of Gauss s theorem on the vector field v T = [v, v ] = [, S] provides, Q = A σ da = A ( + S ) da = N j= Γ j ( d S d) = N S j d, (2 23) Γ j where the circulation is taken anticlockwise along Γ and clockwise along Γ j, j = 1,, N. Further it has been used that S j is constant along the boundar curve, and hence ma be transferred outside the circulation integral. Now, Γ j d =. Hence, it follows that an solution to the boundar value problem defined b Eqs. (2 22a) and (2 22b) automaticall provides a solution fulfilling Q =. Using v T = [S, ] it can in the same wa be shown that Q = Q = A σ da = N j= A ( S + ) da N S j d =. (2 24) Γ j Γ j (S d d) = j= The torsional moment M must be staticall equivalent to the moment of the shear stresses around S, i.e. M = A (( S )σ ( S )σ )da = DCE Lecture Notes No. 23 A j= (σ σ )da, (2 25)

61 2.2 Homogeneous torsion (St. Venant torsion) 53 B S da σ σ S M S Figure 2 5 Static equivalence of torsional moment to shear stresses. where it has been used that A σ da = σ da =. Net, insertion of Eq. (2 15) provides ( M = S A + S ) ( da = A (S) + ) (S) da + 2 SdA. (2 26) A The divergence theorem with v T = [S, S] provides A ( (S) + ) (S) da = N j= Γ j (Sd Sd) = N S j (d d). (2 27) Γ j Let A and A j denote the area inside the boundar curves Γ and Γ j. Then, use of Eq. (1 4a) in Eq. (2 27) provides A ( (S) + ) (S) da = 2A S j= N 2A j S j. (2 28) j=1 The negative sign of the last term is because the circulation on the interior boundaries is taken clockwise, cf. the discussion subsequent to Eq. (1 4a). Insertion of Eq. (2 28) gives the following final result: M = 2A S + 2 A SdA + 2 N A j S j. (2 29) j=1 The shear stresses remain unchanged if an arbitrar constant is added to S. Then, without restriction one can choose S =, which is assumed in what follows. Let the domain of definition for S(, ) be etended to the interior of the holes, where S(, ) is given the same value S j as the boundar value along the holes. With S =, Eq. (2 29) determines M as twice the volume below S(, ). Further, with S(, ) determined along with Elastic Beams in Three Dimensions

62 54 Chapter 2 Shear stresses in beams due to torsion and bending Γ, S = B Γ 1 S(, ) S 1 Figure 2 6 Variation of Prandtl s stress function over a cross-section with a hole. the boundar values S j, j = 1, 2,..., N, the torsional constant K can net be determined upon comparison of Eqs. (2 3) and (2 29). This is illustrated b eamples below. It is remarkable that no reference is made to the position of the shear centre, neither in the boundar value problem (2 22), for the Prandtl stress function, nor in the epression (2 25) for the torsional moment. In contrast, the coordinates of the shear centre enter the boundar value problem (2 14) for the warping function. Defining the same homogeneous torsion problem, the warping function and Prandtl s stress function cannot be independent function. The relation follows from a comparison of Eqs. (2 6) and (2 15): ( ) S ω = ( s) G dθ d, (2 3a) S ( ) ω = + ( s) G dθ d, (2 3b) where it is recalled that θ θ. Eample 2.1 Homogeneous torsion of infinitel long rectangular cross-section Figure A shows an infinitel long rectangular cross-section with the thickness t eposed to a torsional moment M. The torsional moment is carried b shear stresses uniforml distributed in the -direction. Then, S = S() is independent of, and the boundar value problem (2 22) reduces to d 2 S dθ = 2G, S( t/2) = S(t/2) = (a) d2 d with the solution S() = 1 4 (42 t 2 )G dθ d. (b) (continued) DCE Lecture Notes No. 23

63 2.2 Homogeneous torsion (St. Venant torsion) 55 The shear stresses follow from Eq. (2 15): σ = S dθ = 2G, σ =. (c) d The shear stresses are linearl distributed in the thickness direction, and has been illustrated in Fig. A. t σ s = σ τ = G dθ d t dm d d Figure A Torsion of a infinitel long rectangular cross-section: Distribution of shear stresses (left); torsional moment on differential cross-sectional segment (right). Due to the independence of the shear stresses on, the torsional problem can be analsed b merel considering a differential cross-sectional segment of the length d eposed to the torsional moment dm, see Fig. A. The increment dm is related to the stress function b Eq. (2 28), i.e. t/2 dm = 2 S()dd = 1 t/2 2 dgdθ d, d/2 t/2 (4 2 t 2 )d = 1 3 t3 dg dθ d. (d) At the same time dm = GdKdθ/d, where dk denotes the torsional constant related to the differential segment. As seen from Eq. (d), this is given as dk = 1 3 d3 d. The value dk of the torsional constant for the differential segment will be applied below. (e) Eample 2.2 Homogeneous torsion of a solid ellipsoidal cross-section Figure A shows an ellipsoidal cross-section without holes with semi-aes a and b, eposed to a torsional moment M. At first, it is verified that the Prandtl s stress-function of this problem is given as ( ) S(, ) = a2 b 2 2 a 2 + b 2 a b 1 G dθ 2 d. (a) (continued) Elastic Beams in Three Dimensions

64 56 Chapter 2 Shear stresses in beams due to torsion and bending Γ b M a Figure A Ellipsoidal cross-section. The boundar curve Γ is described b the ellipsis 2 a b 2 = 1. Hence, S(,) = for (,) Γ, so the boundar condition (2 22b) is fulfilled b Eq. (a). The Laplacian of s(, ) becomes ( 2 S + 2 S 2 = a2 b a 2 + b 2 a + 2 ) G dθ dθ = 2G 2 b 2 d d. (c) Then, also the differential equation (2 22a) is fulfilled, from which is concluded that Eq. (a) is indeed the solution to the homogeneous torsion problem. From Eq. (2 29) follows that M = 2 SdA = 2 a2 b 2 A a 2 + b G dθ 2 d A ( 2 where the following result has been utilised: ( ) 2 a b 1 da = π 2 2 ab. A a b 2 1 ) da = π a3 b 3 a 2 + b 2 G dθ d, From Eq. (2 3) and Eq. (d) follows that the torsional constant for an ellipsoidal cross-section becomes: (b) (d) (e) K = π a3 b 3 a 2 + b 2. (f) Gdθ/d follows from Eq. (d), i.e. G dθ d = 1 a 2 + b 2 π a 3 b M. 3 Then, the shear stresses become, cf. Eqs. (2 15) and (a): (g) σ = S = 2 a 2 b 2 b 2 a 2 + b 1 a 2 + b 2 M 2 π a 3 b 3 = 2 M π ab, 3 (h) σ = S = 2 a 2 b 2 a 2 a 2 + b 1 a 2 + b 2 M 2 π a 3 b 3 = 2 M π a 3 b. (i) (continued) DCE Lecture Notes No. 23

65 2.2 Homogeneous torsion (St. Venant torsion) 57 The warping of the cross-section is given b Eq. (2 4). Due to the smmetr of the cross-section, it is observed that s = s =, i.e. the shear centre coincides with the bending centre. Then, the warping function is determined from, cf. Eq. (2 3): ω = 1 S Gdθ/d + = 2a2 a 2 + b + = b2 a 2 2 b 2 + a, 2 ω = 1 S Gdθ/d = 2b 2 a 2 + b = b2 a 2 2 b 2 + a. 2 The solution to Eqs. (j) and (k) is given as ω(,) = b2 a 2 b 2 + a 2. It is left as an eercise to prove that Eq. (l) fulfils the boundar value problem (2 14). The warping follows from Eq. (2 4), Eq. (g) and Eq. (l) (j) (k) (l) u (, ) = ω(,) dθ d = b2 a 2 a 2 + b 1 a 2 + b 2 M 1 b 2 a 2 2 π a 3 b 3 G π a 3 b 3 M G. The solution (l) has been chosen so that the warping from torsion provided b Eq. (m) is ero at the bending centre. This will generall be presumed in what follows. Then, the displacement of the bending centre in the -direction is caused entirel b the aial force N. Finall, it is noted that, especiall for a circular profile with a = b, the warping vanishes everwhere on the profile. (m) Homogeneous torsion of open thin-walled cross-sections Figure 2 7 shows an open cross-section of a clindrical beam. An arc-length coordinate s is defined along the midpoints of the profile wall, where s = is chosen at one of the free ends, and s = L at the other free. Further, L specifies the total length of the profile wall, and the wall thickness at arc-length coordinate s is denoted t(s). τ σ s s s = L M, m S ds n t(s) s s = Figure 2 7 Open thin-walled cross-section of clindrical beam eposed to homogeneous torsion. Elastic Beams in Three Dimensions

66 58 Chapter 2 Shear stresses in beams due to torsion and bending For thin-walled cross-sections it is assumed that t(s) L. Then, the profile can be considered as built-up of differential rectangles of the length ds, similar to those considered in Eample 2.2. Each has the torsional constant dk = 1 3 t3 (s)ds, cf. Eq. (e) in Eample 2.2. Hence, the torsional constant for the whole profile is given as K = 1 t 3 (s)ds, 3 L (2 33) where the inde L indicates that the line integral is etended over the whole length of the profile measured along the profile wall. The shear stresses are specified in a local (, n, s)-coordinate sstem as shown in Fig These coordinates follow from Eq. (c) upon replacing with s and with w. Then, σ s = 2nG dθ d, σ n =. (2 34) Finall, using Gdθ/d = M /K, the maimum shear stresses for n = t(s)/2 becomes τ = G dθ d t(s) = M K t(s). (2 35) The computation of the shear stress in a U-profile is considered in the eample below. Eample 2.3 Homogeneous torsion of a U-profile Figure A shows a U-profile eposed to homogeneous torsion from the torsional moment M. With the thin-wall approimation t a, the position of the bending centre is as shown in the figure. Further, the cross-sectional area and the bending moments of inertia around the - and -aes become A = 1at, I = a3 t, I = 26 3 a3 t, (a) where the single indices and indicate that the corresponding aes are principal aes. The crosssectional constants given in Eq. (a) have been calculated for later use. With reference to Eq. (2 33), the torsional constant becomes: K = 2 3 2a(2t) at3 = 34 5 a3 t. (b) (continued) DCE Lecture Notes No. 23

67 2.2 Homogeneous torsion (St. Venant torsion) 59 2τ 4 5 a 2a 2t 2τ a B M M a t 2t τ τ 2τ 2τ Figure A Homogeneous torsion of U-profile: Dimensions (left) and shear stresses (right). The distribution of shear stresses follows from Eq. (2 35). The maimum shear becomes τ and 2τ, respectivel, where τ is given as τ = M K t = 3 M 34 at. 2 Hence, in the present case, the shear stresses in the flanges are higher than those in the web. (c) Homogeneous torsion of closed thin-walled cross-sections The boundar value problem for the warping function ω(, ), defined b Eq. (2 14), has a unique solution (save an arbitrar additive constant) no matter if the profile has an interior hole or not. Although an analtical solution is seldom obtainable, a numerical solution can alwas be achieved b a discretiation of the Laplace operator b a finite-difference or a finite-element approach. In contrast to this, the boundar value problem defined b Eq. (2 22) for Prandtl s stress function cannot immediatel be solved, because the boundar values S j, j = 1, 2,...,N, are not known. The determination of these values requires the formulation of additional geometrical conditions which epress that the warping function ω(, ) shall be continuous along the Elastic Beams in Three Dimensions

68 6 Chapter 2 Shear stresses in beams due to torsion and bending boundar curves Γ j. This ma be formulated as ( ) ω ω dω = d + Γ j d =, j = 1, 2,..., N. (2 37) Γ j Γ j Equation (2 37) can be epressed in the stress function b use of Eq. (2 3): ( ) S S d d + G dθ (( s )d ( s )d) =. (2 38) d Γ j It can be shown that the integrand of the first integral can be rewritten in terms of the coordinates s and n b the substitution S S d d = S n ds. (2 39) The second integral in Eq. (2 38) ma be recast as (( s )d ( s )d) = (d d) = 2A j. (2 4) Γ j Γ j The change of sign in Eq. (2 4) is because all circulations are taken clockwise. Further, it has utilised that Γ j s d = s Γ j d = and Γ j s d =. Insertion of Eqs. (2 39) and (2 4) into Eq. (2 38) provides the following form for the geometrical conditions: S Γ j n ds = 2A jg dθ d. (2 41) Onl thin-walled cross-sections are considered. At first a cross-section with a single cell is considered as shown in Fig An arc-length coordinate is introduced, orientated in the clockwise direction. Correspondingl, a local (, n, s)-coordinate sstem is defined at each point of the boundar curve with the n-ais orientated inward into the cavit, whereas the s-ais is tangential to the boundar curve and unidirectional to the arc-length coordinate. Then, the shear stresses along the n- and s-directions become, cf. Eq. (2 15), σ n = S s, σ s = S n. (2 42) Here, S is constant along the eterior and interior boundar curve of the wall, given as S = S = and S = S 1, respectivel. Hence, σ n = along these boundaries. If the thickness t(s) of the wall is small compared to a characteristic diameter of the profile, it then follows from continuit that σ n is ignorable in the interior of the wall, i.e. σ n. (2 43) The stress function decreases from S = S 1 at the inner side of the wall to S = S = at the outer side. If t(s) is small, the variation of S(n, s) must var approimatel linearl over the wall thickness t(s), see Fig This implies the following approimation σ s = S n S 1 S t(s) DCE Lecture Notes No. 23 = S 1 t(s). (2 44)

69 2.2 Homogeneous torsion (St. Venant torsion) 61 σ n σ s s s B n n s S M t S(, ) S 1 Figure 2 8 Closed thin-walled cross section of a clindrical beam eposed to homogeneous torsion. Then, the geometrical condition Eq. (2 41) ma be written as S 1 Γ 1 ds t(s) = 2A 1G dθ d, (2 45) where A 1 is area of the cavit. The torsional moment is given b Eqs. (2 3) and (2 29): M = GK dθ d = 2A 1S 1 (2 46) Combining Eq. (2 45) and Eq. (2 46) provides the following result for the torsional constant: S 1 K = 2A 1 Gdθ/d = 4A2 1 J, J = ds Γ 1 t(s). (2 47) Equation (2 47) is known as Bredt s formula. The shear stresses follow from Eqs. (2 44) and (2 46): M σ s (s) = S 1 t(s) = 2A 1 t(s). (2 48) Notice, that the shear stress σ s (s) is uniforml distributed over the wall-thickness as shown in Fig This is in contrast to an open section, where a linear variation was obtained, as given b Eq. (2 34). Elastic Beams in Three Dimensions

70 62 Chapter 2 Shear stresses in beams due to torsion and bending Eample 2.4 Homogeneous torsion of a closed thin-walled cross-section Figure A shows the cross-section of a beam with a single cell eposed to homogeneous torsion. The thin-wall approimation t a is assumed to be valid. t 1 6 τ s a t t M 1 6 τ 1 6 τ n M 1 12 τ a 2t 2t 2t 1 12 τ 1 12 τ a a Figure A Homogeneous torsion of a closed thin-walled cross-section: Geometr (left) and distribution of shear stresses with τ = M /(a 2 t) (right). The area of the cavit and the line integral entering Eqs. (2 45) and (2 47) become A 1 = 3a 2 ds, Γ 1 t(s) = 4a t + 4a 2t = 6a t. Then, the torsional constant becomes, cf. Eq. (2 47), (a) K = 4(3a2 ) 2 6a/t = 6a 3 t. (b) The shear stresses follow from Eq. (2 48): M σ s = 6a 2 t(s). (c) The distribution has been shown in Fig. A along with the direction of action. Note that τ = M /(a 2 t) has been introduced as a normalisation quantit. Eample 2.5 Comparison of homogeneous torsion of open and closed profiles Figure A shows two clindrical beams, both with a clindrical thin-walled cross-section with the side length a and the thickness t. In one case, the cross-section is closed, whereas the cross-section in the other case has been made open b a cut along a developer. The rotation gradient dθ/d and the shear stresses in two beams due to homogeneous torsion are compared below. (continued) DCE Lecture Notes No. 23

71 2.2 Homogeneous torsion (St. Venant torsion) 63 τ o a a M,o, θ o a τ c a M,c, θ c Figure A Homogeneous torsion of open and closed cross-sections. According to Eqs. (2 33) and (2 47), the torsional constants K o and K c for the beams with open and closed cross-sections are given as K o = 1 t 3 (s)ds = at3, (a) K c = 4A2 ds t = 4a4 4a/t = a3 t. If the two beams are eposed to the same torsional moment M, it follows from Eq. (2 3) that the rotation gradients dθ o/d and dθ c/d for the open and closed sections are related as (b) dθ o/d dθ = Kc = 4 a 2 c/d K o 3 t. 2 (c) Net, assume that both cross-sections are full stressed, i.e. the maimum shear stress τ o = τ c is in both cases equal to the ield shear stress τ. The corresponding torsional moments that can be carried b the profiles are denoted M,o and M,c, respectivel. With reference to Eqs. (2 35) and (2 48), these torsional moment are related as M,o = Koτ/t = 2 a M,c 2a 2 tτ 3 t. Hence, whereas the deformations of the two beams depend on the fraction a 2 /t 2, the torsional moment which can be carried depends on a/t. In conclusion, open sections are etremel ill-conditioned to carr torsional moments in homogeneous torsion compared to closed sections. However, in man cases another mechanics is active, which involves inhomogeneous torsion. This substantiall increases the torsional properties of open sections. (d) Elastic Beams in Three Dimensions

72 64 Chapter 2 Shear stresses in beams due to torsion and bending As illustrated b Eample 2.5, closed thin-walled cross-sections, e.g. tubes and bo girders, have a much higher torsional strength and stiffness than open thin-walled cross-section. A mechanical eplanation for the highl increased stiffness obtained for a closed cross-section compared with the open cross-section is the fact that warping is hindered b the fact that the displacements in the aial direction must be continuous along the s-direction, i.e. along the wall. Especiall, for a circular tube no warping is achieved in homogeneous torsion, making this profile particularl useful for structural elements with the primar function of carring torsional loads. Net, consider a thin-walled cellular cross-section with a total of N cavities as illustrated in Fig Each cavit has the cross-sectional area A j, j = 1, 2,...,N, and the boundar of the cell is denoted Γ j. In each cell, a local (, n j, s j )-coordinate sstem is defined with the n j -ais oriented into the cavit as illustrated in Fig The local arc-length coordinate s j is defined along the cell boundar in the clockwise direction, and the wall-thickness is t(s j ). σ s(s) = S N t(s) Γ n s 1 s 1 n s N s k B N s M n k s n Γ j s j j s σ s(s) = S j S k t(s) Figure 2 9 Cellular thin-walled cross section of a clindrical beam eposed to homogeneous torsion. As discussed in Subsection and illustrated in Fig. 2 6, Prandtl s stress function S is constant along each interior boundar. Hence, on the cell boundar Γ j, S has the constant value S j, j = 1, 2,...,N. However, different values of Prandtl s stress function are generall present on either side of a common wall between two adjacent cells, j and k. Because of the thin-wall assumption the variation of S over the wall thickness must be approimatel linear. Hence, with reference to Eq. (2 42), σ s = S S j S k. n j t(s j ) (2 51) Thus, σ s is uniforml distributed over the wall-thickness as illustrated in Fig The shear stress component σ n vanishes along Γ j as well as Γ k. It then follows from continuit that σ n is ignorable in the interior of the wall as well, i.e. σ n. (2 52) DCE Lecture Notes No. 23

73 2.2 Homogeneous torsion (St. Venant torsion) 65 Insertion of Eq. (2 51) into Eq. (2 41) provides: ds j (S j S k ) t(s j ) = 2A jg dθ, j = 1, 2,..., N (2 53) d k Γ jk where the summation on the left-hand side is etended over all cavities adjacent to cell j. Notice that S k = at the part of cell j adjacent to the outer peripher. Then, Eq. (2 53) represents N coupled linear equations for the determination of S j, j = 1, 2,..., N. Subsequentl the shear stress σ s in all interior and eterior walls is determined from Eq. (2 51). According to Eq. (2 29), the torsional moment is given as N M = 2 A j S j, j=1 (2 54) where the contribution S da can be ignored due to the thin-wall approimation. Then, the A torsional constant follows from Eq. (2 3): K = M N Gdθ/d = 2 A j j=1 S j Gdθ/d. (2 55) Since S j turns out to be proportional to Gdθ/d, the right-hand side of Eq. (2 55) will be independent of this quantit. Eample 2.6 Homogeneous torsion of a rectangular thin-wall profile with two cells Figure A shows a rectangular thin-walled cross-section with two cells eposed to homogeneous torsion from the torsional moment M. The wall-thickness is everwhere t a. σ s = 8 52 τ σs = 1 52 τ σs = 9 52 τ a t 1 M 2 n n s 1 s s 2 s a Figure A Homogeneous torsion of a rectangular thin-walled cross-section with two cells. Definition of local coordinate sstems and distribution of shear stresses with τ = M /(a 2 t). (continued) 2a Elastic Beams in Three Dimensions

74 66 Chapter 2 Shear stresses in beams due to torsion and bending For the two cells, Eq. (2 53) takes the form: (S 1 ) a t + (S1 )a t + (S1 )a t + (S1 S2)a t = 2a2 G dθ d, (S 2 ) 2a t + (S2 S1)a t + (S2 )a t + (S2 )2a t 4S 1 S 2 = 2atG dθ d, S1 + 6S2 = 4atGdθ d. The solution of Eq. (c) reads S 1 = atgdθ 18, S2 = d 23 atgdθ d, and b Eq. (2 54), the torsional moment is derived as = 2 2a 2 G dθ d. (a) (b) (c) (d) M = 2a 2 S a 2 S 2 = a3 tg dθ d G dθ d = 23 M 14 a 3 t. (e) Then, S 1 and S 2 ma be written as S 1 = 8 M 52 a, 2 S2 = 9 M 52 a. 2 The shear stresses σ s follow from Eq. (2 51) and Eq. (f). The distribution has been shown in Fig. A with their direction of action. The quantit τ = M /(a 2 t) represents a normalisation quantit for the shear stresses. The torsional constant follows from Eq. (2 55) and Eq. (d) K = 2 a at + 2 2a at = a3 t. (g) If the interior wall is skipped, the corresponding quantities become: S 1 = 1 M 6 a, 2 σs = 1 M 6 a 2 t, K = 9 2 a3 t. (h) Hence, the interior wall increases the shear stress in the eterior cell walls of the right cell from 1 τ to 6 9 τ (1.3%). The torsional constant is increased merel from K = a3 t to K = a3 t (2.2%). (f) As indicated b Eample 2.6, onl a small increase of the torsional stiffness is achieved b the inclusion of internal walls in a cross-section. Hence, from an engineering point of view, the benefits of appling structural members with a cellular cross-sections are insignificant in relation to torsion. Since the advantages regarding fleural deformations are also limited, and profiles with two or more cavities are not easil manufactured, it ma be concluded that open cross-sections are generall preferred for beams loaded in unidirectional bending. Likewise, closed cross-sections with a single cavit (pipes or bo girders) are preferable when the beam is primaril subjected to torsion and/or bending in two directions. Finall it is noted that the solution method based on Prandtl s stress-function will onl be used in relation to hand calculation for thin-walled cross-sections with at most two or three cells. Otherwise, for thick-walled sections or multi-cell problems, the homogeneous torsion problem will be solved numericall b a finite-element approach or another spatial discretiation method based on the Neumann boundar problem specified in Bo 2.1. DCE Lecture Notes No. 23

75 2.3 Shear stresses from bending Shear stresses from bending In the previous section, the shear stresses occurring due to homogeneous torsion of a beam have been analsed. A second contribution to the shear stresses stem from bending or fleural deformations of a beam and with a proper choice of coordinates, the two contributions decouple. Figure 2 1 shows a cross-section eposed to bending without torsion. Correspondingl, the loads per unit length q and q as well as the shear forces Q and Q have been referred to the shear centre S, the position of which is discussed in this section. s Γ N Γ n n s σ Γ 1 σ s N σ s B S M s 1 s n S S Q q Q M q Figure 2 1 Cross-section eposed to bending. The cross-section has the eternal boundar Γ and ma have a number of cavities, N, bounded b the interior boundar curves Γ j, j = 1, 2,...,N. Again, a local arc-length coordinate s j is defined along each boundar, orientated clockwise for all interior boundaries, j = 1, 2,..., N, whereas the arc-length coordinate s along the outer boundar curve is orientated anti clockwise. At an point along the outer and inner boundar curves, local right-handed (, n j, s j )-coordinate sstems are defined as shown in Fig The (,, )-coordinate sstem with origin at the bending centre B is assumed to be a principal-aes coordinate sstem. On the cross-section, the normal stress σ as well as the shear stresses σ and σ are acting. With reference to Eq. (2 7), these stresses fulfil the equilibrium equation σ + σ + σ =, where σ is determined from Navier s formula (1 75), i.e. σ (,, ) = N() A It then follows that σ = dn d 1 A + dm d (2 57) + M () M (). (2 58) I I dm I d I = q A + (Q + m ) I + (Q m ) I, (2 59) Elastic Beams in Three Dimensions

76 68 Chapter 2 Shear stresses in beams due to torsion and bending where the following equilibrium equations have been used in the statement: dn d + q =, dm d Q + m =, dm d + Q + m =. (2 6) A stress function T = T(,, ) is introduced with the defining properties σ = T, σ = T. (2 61) Hence, T is defined differentl from the somewhat similar Prandtl s stress function, cf. Eq. (2 15). Insertion of Eqs. (2 59) and (2 61) into Eq. (2 57) provides the following Poisson partial differential equation for T : 2 T T 2 = q A (Q + m ) (Q m ). (2 62) I I With reference to Eqs. (2 1) and (2 11), the boundar conditions read σ n = σ n + σ n + σ n = σ n + σ n =, (2 63) where n = because the beam is clindrical. Insertion of Eq. (2 61) into this equation provides the following homogeneous Neumann boundar conditions to be fulfilled on all eterior and interior boundar curves: T n = T n + T n =. (2 64) The boundar value problems defined b Eqs. (2 14) and (2 22) for the warping function and Prandtl s stress function, respectivel, are independent of, i.e. the solution applies for all cross-sections of the beam. In contrast, the corresponding boundar value problem defined b Eqs. (2 62) and (2 64) for T = T(,, ) must be solved at each cross-section defined b, where the shear stresses are determined. This is so, because q, m, m, Q and Q entering the right-hand side of Eq. (2 62) ma var along the beam. The solution to Eq. (2 62) with boundar conditions given b Eq. (2 64) is unique save an arbitrar function T = T () which has no influence on the shear stresses. The method can be applied to thick-walled or thin-walled cross-sections with or without interior cavities. Normall the boundar value problem can onl be solved numericall based on a discretiation of the Laplace operator entering Eq. (2 62). The boundar value problem for the stress function T has been summarised in Bo 2.3. Bo 2.3 Boundar value problem for the stress function determining shear stresses in bending For a given cross-section determined b the abscissa the stress-function T = T(,,) is obtained from the Poisson partial differential equation 2 T T 2 = q() A (Q() + m()) I (Q () m ()) I, (, ) A (2 65a) with the homogeneous Neumann boundar conditions T =, (,) Γ Γ1 ΓN. (2 65b) n DCE Lecture Notes No. 23

77 2.3 Shear stresses from bending 69 The indicated method is not applicable for hand-calculations. For this purpose a method will be devised in the following sub-sections, which solves the problem for thin-walled open cross-sections and closed thin-walled sections with few cells Shear stresses in open thin-walled cross-sections Figure 2 12 shows the same open thin-walled cross-section as shown in Fig. 2 7, when eposed to homogeneous torsion. Now the shear stresses σ s and σ n defined in local (, n, s)- coordinates are requested, caused b the shear forces Q and Q acting through the shear centre S. We shall return to the definition and determination of the shear centre later in this chapter. σ s s s = L n S Q M Q B t(s) M Ω(s) s s = Figure 2 11 Cross-section eposed to bending. The shear stress component σ n still vanishes at the surfaces n = ± 1 2t(s), cf. Eq. (2 34). For continuit reasons, Eq. (2 43) remains valid in the interior of the wall, i.e. σ n (, n, s). (2 66) Then, with reference to Eq. (2 57), the equilibrium equation of stress components in the (, n, s)- coordinate sstem reduces to σ + σ s s. (2 67) From Eq. (2 58) follows that σ varies linearl over the cross-section. Thus it must be almost constant over the thin wall and can be replaced b its value at the midst of the wall. From Eq. (2 67) it then follows that σ s must also be approimatel constant in the n-direction, i.e. σ s (, n, s) σ s (, s). (2 68) The constanc of the shear stress component σ s in the thickness direction has been illustrated in Fig This variation should be compared to Eq. (2 34) for homogeneous torsion of an open cross-section, where a linear variation with n is obtained as illustrated in Fig Elastic Beams in Three Dimensions

78 7 Chapter 2 Shear stresses in beams due to torsion and bending ds σ s σ t(s) s H + dh, σ s + dσ s d H, σ s σ + dσ σ s + dσ s Figure 2 12 Differential element of a thin-walled cross-section. A differential element with the side lengths d and ds is cut free from the wall at the aial coordinate and at the arc-length coordinate s, see Fig On the sections with the arc-length coordinates s and s + ds, the shear forces per unit length H and H + dh are acting, where H(, s) = t/2 t/2 σ s (, n, s)dn σ s (, s)t(s). (2 69) Here it has been eploited that σ s = σ s, and Eq. (2 68) has been utilised. On the sections with the aial coordinates and + d, the normal stresses σ and σ + dσ are acting (see Fig. 2 12). Equilibrium in the -direction then provides (σ + dσ )t(s)ds σ t(s)ds + (H + dh)d Hd = H s + t(s) σ =, (2 7) where it has been utilised that σ has an ignorable variation in the thickness direction. It has therefore been replaced with a constant value equal to the value present at the midst of the wall. Alternativel, Eq. (2 7) ma be obtained simpl b multiplication of Eq. (2 67) with t(s) and use of Eq. (2 69). With H () representing an integration constant, integration of Eq. (2 7) provides the solution H(, s) = H () s σ t(s)ds. (2 71) Especiall, in the open thin-walled section the boundar condition σ s = applies at the ends of the profile, corresponding to the arc-length coordinates s = and s = L, i.e. H(, ) = H(, L) =. (2 72) Thus, for an open thin-walled cross-section, the integration constant in Eq. (2 71) is H () =. The partial derivative σ / is given b Eq. (2 59). In what follows it is for ease assumed that q = m =. Equation (2 59) then reduces to σ (, s) = Q () (s) + Q () (s), (2 73) I I DCE Lecture Notes No. 23

79 2.3 Shear stresses from bending 71 where ((s), (s)) denotes the principal-aes coordinate of a position on the wall determined b the arc-length coordinate s. Then, insertion in Eq. (2 71) and use of Eq. (2 69) provide the following solution for σ s : σ s (, s) = 1 t(s) H(, s) = Q () S Ω (s) Q S Ω (s), t(s)i t(s)i where it has been utilised that H () = for the open section, and s s S Ω (s) = (s)t(s)ds = da, S Ω (s) = (s)t(s)ds = Ω Ω da. (2 74a) (2 74b) The quantities S Ω (s) and S Ω (s) denote the statical moment around the - and -aes of the area segment Ω(s), shown with a dark gre signature in Fig and defined as s Ω(s) = t(s)ds = da. (2 75) Ω Equation (2 74) is known as Grashof s formula. The formula is valid for Timoshenko as well as Bernoulli-Euler beams with a thin-walled open cross-section. In this contet it is noted that Bernoulli-Euler beam theor is based on the kinematic constraint that plane cross-sections orthogonal to the beam ais in the referential state remain plane and orthogonal to the deformed beam ais. In turn this implies that the angular strains γ and γ vanish, and hence that σ = σ =. Hence, the shear stresses in bending cannot be determined from the beam theor itself. Instead, these are determined from Eqs. (2 66) and (2 74) which are derived from static equations alone and, hence, are independent of an kinematic constraints. The shear strain caused b the shear stress σ s is given as ε s = 1 2G σ s = 1 H(, s). (2 76) 2Gt(s) The shear strain ε s implies a warping u (, ) of the cross-section, additional to the displacement in the -direction caused b the aial force and the bending moments. The latter is described b the kinematic conditions defined b Eqs. (1 11a) and (1 15) for Bernoulli-Euler beam theor. Hence, the displacements of the cross-section relative to the principal-aes coordinate sstem can be written as u (, s) = w () dw () d (s) dw () d (s) + u (, s), (2 77a) u (, s) = w (), u (, s) = w (). (2 77b) (2 77c) The component u s (, s) of the displacement vector u(, s) in the tangential s-direction in the local (, n, s)-coordinate sstem shown in Fig is given as u s (, s) = s T u(, s) = [ d(s) ds d(s) ds ] T [ u (, s) u (, s) ] = d(s) ds u (, s) + d(s) ds u (, s), (2 78) Elastic Beams in Three Dimensions

80 72 Chapter 2 Shear stresses in beams due to torsion and bending where s T = s T (s) = [d/ds, d/ds] signifies the unit tangential vector. Then, the strain ε s (, S) follows from Eqs. (2 77) and (2 78): ε s = 1 ( us 2 + u ) = 1 s 2 ε s = 1 u 2 s. ( d dw ds d + d(s) ds dw d d dw ds d d dw ds d + u ) s (2 79) The warping u (, s) is net determined b integration of Eq. (2 79): u (, s) = 2 s ε s (, s)ds + u (). (2 8) The arbitrar function u () is adjusted, so that u = at the bending centre B. When ε s is determined from Eq. (2 76), the warping of the cross-section additional to displacements in the -direction predicted b Bernoulli-Euler beam theor can be determined b Eq. (2 8). Eample 2.7 Shear stresses and warping due to bending of a rectangular cross-section Figure A shows a rectangular beam of height h and thickness t eposed to a shear force Q. The bending moment of inertia, the area segment Ω(s) and static moment S Ω(s) around the -ais become I = 1 ( h 12 h3 t, Ω(s) = st, S Ω(s) = st 2 s ). (a) 2 where the arc-length parameter is defined from the upper edge of the profile, i.e. s = h 2 σ s = σ is determined from Eq. (2 74), that is ( ) σ = Q s S Ω(s) = 6 ti h s2 Q h 2 ht = 3 ) (1 4 2 Q 2 h 2 ht. +. Then, (b) Equation (b) specifies a parabolic distribution over the cross-section, where σ = at the edges = ± h 2 in agreement with the boundar conditions, and the maimum value 3 Q is achieved at the bending 2 ht centre B at =. Net, the warping is determined from Eqs. (2 76) and (2 8) together with Eq. (b): u (,s) = u + 1 s s ( ) σ sds = u + 6 Q s G Ght h s2 ds = (3 Q s2 h 2 Gt h 2 s3 2 h 1 ), (c) 3 2 where u is adjusted, so u = at the bending centre, i.e. at s = h. With s = h +, the warping 2 2 displacements become ( u () = Q 3 2 ( ) ) 3 2 4Gt h. (d) h Equation (d) specifies a cubic polnomial variation, leading to an S-shape of the warping function. The distribution of σ () and u () has been shown in Fig. A. (continued) DCE Lecture Notes No. 23

81 2.3 Shear stresses from bending 73 t s 1 Q 2G t h 2 Ω(s) 3 Q 2 th B = S u h 2 Q σ Figure A (right). 1 Q 2G t Rectangular cross-section subjected to bending: Area segment (left); shear stresses (centre); warping Eample 2.8 Shear stresses due to bending of a double-smmetric I-profile Figure A (left) shows a double-smmetric cross-section eposed to the shear forces Q and Q. The wall thickness is everwhere t a, i.e. the thin-wall assumption applies. The cross-sectional area and the principal bending moments of inertia become A = 3at, I = a3 t = 1 ( a ) a3 t, I = 2 at a3 t = 7 12 a3 t. (a) Due to the smmetr, the bending and shear centres are coinciding. t n s 1 s s 2 a 2 B = S t Q n n s 5 s s a 2 Q t n s 3 s 4 a 2 a 2 Figure A Double-smmetric I-profile: Geometr (left); definition of arc-length coordinates and local coordinate sstems (right). At first the shear stresses caused b Q are considered. For each of the four branches and the web of the profile, arc-length coordinates s j and local right-handed (,n j, s j)-coordinate sstems are defined as shown in Fig. A (right). At the free edge, the shear stresses are σ s =. Hence, the shear force per unit length H(,s j) vanishes at this point, and Eq. (2 71) becomes valid for each branch. (continued) n s s Elastic Beams in Three Dimensions

82 74 Chapter 2 Shear stresses in beams due to torsion and bending S Ω (s 1 ) 1 8 a2 t S Ω (s 2 ) σ s(s 1 ) 3 Q 4 at σ s(s 2 ) B = S Q B = S Q S Ω (s 3 ) 1 S Ω (s 4 ) σ s(s 3 ) σ s(s 4 ) 8 a2 3 Q t 4 at Figure B Double-smmetric I-profile: Distribution of statical moment of the area segment Ω(s j ) around the -ais (left); distribution of the shear stresses σ s(s j ) from Q (right). On the five branches, the area segments become Ω(s j) = s jt. These have been shown with a dark gre signature in Fig. A (right). For the flanges, the statical moments of Ω(s j) around the -ais become S Ω(s 1) = 1 2 ts1(s1 a), SΩ(s2) = 1 ts2(s2 a), (b) 2 S Ω(s 3) = 1 2 ts3(s3 a), SΩ(s4) = 1 ts4(s4 a). (c) 2 Further, S Ω = along the web. The distribution has been shown in Fig. B (left). S Ω (s 1 ) 1 4 a2 t S Ω (s 2 ) σ s(s 1 ) 3 Q 7 at σ s(s 2 ) B = S 1 8 a2 t B = S 3 Q 14 at Q Q 1 2 a2 t 6 Q 7 at S Ω (s 3 ) S Ω (s 4 ) σ s(s 3 ) σ s(s 4 ) 1 4 a2 3 Q t 7 at Figure C Double-smmetric I-profile: Distribution of statical moment of the area segment Ω(s j ) around the -ais (left); distribution of the shear stresses σ s(s j ) from Q (right). (continued) DCE Lecture Notes No. 23

83 2.3 Shear stresses from bending 75 The shear stresses σ s(s j) follow from Eq. (2 74): σ s(s 1) = Q S Ω(s 1) = 3 Q s1(a s1), ti a 3 t Q σs(s2) = 3 s2(a s2), a 3 t (d) σ s(s 3) = 3 Q Q s3(a s3), σs(s4) = 3 s4(a s4), σs(s5) =. (e) a 3 t a 3 t The distribution has been shown in Fig. B (right). The sign refers to the local,n j, s j)-coordinate sstem. Hence, a negative sign implies that the shear stress is acting in the negative s j-direction and, hence, is co-directional to Q. Then, the shear stresses in the flanges are distributed parabolicall in the same wa as stresses in the rectangular cross-section considered in Eample 2.7. Each flange carries the shear force Q /2, and no shear force is carried b the web in the present case. Net, the shear stresses caused b Q are analsed. The statical moment of the area segment Ω(s j) around the -ais becomes: S Ω(s 1) = 1 2 ats1, SΩ(s2) = 1 ats2, (f) 2 S Ω(s 3) = 1 2 ats3, SΩ(s4) = 1 2 ats4, SΩ(s5) = a 2 at 1 ts5(a s5), (g) 2 and b Eq. (2 74) the shear stresses σ s(s j) are determined as Q σ s(s 1) = Q S Ω(s 1) = 6 ti 7 a 2 t s1, σs(s2) = 6 7 a 2 t s2, Q Q σ s(s 3) = 6 7 a 2 t s3, σs(s4) = 6 7 a 2 t s4, σs(s5) = 6 7 (a2 + s 5a s 2 5) Q at. The distribution of Ω(s j) and σ s(s j) has been shown in Fig. C. Q (h) (i) Determination of the shear centre So far in this section we have onl considered double-smmetric cross-sections for which the shear centre S coincides with the bending centre B. However, in the general case, the shear centre will be different from the bending centre as suggested b Fig In an case, the shear forces Q and Q must be staticall equivalent to the shear stresses integrated over the crosssection, i.e. Q = σ da, Q = σ da. (2 83) A A In the present case, we are concerned with bending uncoupled from torsion. Hence, the torsional moment M stemming from the shear stresses on the cross-section should vanish. According to Eq. (2 25), this implies the identit (( s )σ ( s )σ )da =. (2 84) A Combining Eqs. (2 83) and (2 84) provides the relation for the coordinates of the shear centre: Q s + Q s = ( σ + σ )da. (2 85) A Elastic Beams in Three Dimensions

84 76 Chapter 2 Shear stresses in beams due to torsion and bending For a thin-walled cross-section, this identit ma be recast in terms of the local (, s, n)-coordinates, providing Q s + Q s = h(s)σ s (, s)t(s)ds, (2 86) L where h(s) is the moment arm of the differential shear force σ s (, s)t(s)ds, h(s) = (s) d(s) ds (s) d(s) ds. (2 87) It is noted that, with the given definition, h(s) ma have positive as well as negative values. Insertion of Eq. (2 74) into Eq. (2 86) provides the identit Q s + Q s = Q S Ω (s)h(s)ds Q S Ω h(s)ds (2 88) I I L which holds for arbitrar Q and Q. This implies the following solutions for the coordinates of the shear centre: s = 1 S Ω (s)h(s)ds, s = 1 S Ω (s)h(s)ds. (2 89) I I L In particular, for a thin-walled section without branching, S Ω (s) and S Ω (s) become S Ω (s) = s (s)t(s)ds, S Ω (s) = L s L (s)t(s)ds. (2 9) Hence, Eq. (2 89) involves a double integration with respect to the arc-length parameter over the length L. We shall arrange the calculation in a wa such that onl a single line integral needs to be evaluated. Ω(s) s = S Q Q s ds B h(s) ω(s) t(s) Figure 2 13 Cross-section eposed to bending. s = L DCE Lecture Notes No. 23

85 2.3 Shear stresses from bending 77 At first the so-called sector coordinate ω(s) with pole B is introduced. The sector coordinate ω(s) is equal to the area shown in Fig. 2 13, delimited b the bending centre B and the area segment Ω(s). Thus, ω(s) is given as ω(s) = s h(s)ds. (2 91) It follows that ω(s) ma be considered as an integral of h(s) with respect to s, i.e. d dsω(s) = h(s). Hence, from Eq. (2 9) follows that the derivation of S Ω (s) and S Ω (s) with respect to s become d ds S Ω(s) = (s)t(s), d ds S Ω(s) = (s)t(s). (2 92) Then, integration b parts of Eq. (2 89) and use of Eq. (2 92) provides ( s = 1 [ ] L ) L S Ω (s)ω(s) I (s)ω(s)t(s)ds (2 93) Now, S Ω (L) = S =, because the (,, )-coordinate sstem has origin in the bending centre. Further, both S Ω () = and ω() =, so the first term within the parentheses vanishes in both limits and, hence, s = I ω I, I ω = L Similarl, it can be shown that s = I ω I, I ω = L ω(s)(s)t(s)ds = ω(s)(s)t(s)ds = A A ωda. ωda. (2 94a) (2 94b) The quantities I ω and I ω are denoted sector centrifugal moments. An arbitrar constant ω can be added to ω(s) in Eq. (2 94) without changing the value of I ω and I ω. This is so because A ω da = ω A da = and A ω da =. Hence, the sector coordinate is determined within an arbitrar constant. As a third method, the coordinates of the shear centre ma be determined from Eq. (2 86), if the shear stress σ s (, s) is calculated from Q and Q separatel. Thus, s = 1 σ s (, s)h(s)t()ds for Q =, (2 95a) Q s = 1 Q L L σ s (, s)h(s)t(s)ds for Q =. (2 95b) Obviousl, if the profile has a line of smmetr, then the shear centre is placed on this line. Finall, a note is made regarding the notation in this chapter. Previousl, ω(, ) has denoted the normalised warping function or, simpl, the warping function. However, we shall later see that for open thin-walled sections the warping function is identical to the sector coordinate, which motivates the naming of the latter quantit. Elastic Beams in Three Dimensions

86 78 Chapter 2 Shear stresses in beams due to torsion and bending Eample 2.9 Determination of the shear centre for an I-profile with a single line of smmetr The thicknesses of the flanges and the web of the profile shown in Fig. A (left) are all t. Due to the smmetr around the -ais, the -ais as well as the -ais become principal aes. The bending centre is placed as shown in the figure and the principal moments of inertia become: I = a3 t, I = a3 t. The profile is eposed to a horiontal shear force Q, and the position of the shear centre will be determined both b Eqs. (2 89), (2 94) and (2 95). Due to the smmetr, the shear centre is placed on the -ais, i.e. s =. (a) 4a t 2a 2 t S Ω (s 1 ) S Ω (s 2 ) n 2 Q a n s 1 s 2 1 6a S B t t a s 3 s 4 n 4 3a n 3 S Ω (s 3 ) S Ω (s 4 ) 9 8 a2 t Figure A Single-smmetric I-profile: Geometr (left); definition of arc-length coordinates and distribution of static moments (right). For each of the four branches indicated in Fig. A (right), an arc-length parameter s j and a local (, n j, s j) coordinate sstem are defined. The statical moments S Ω(s j) = (sj)t(sj)dsj become: Ω S Ω(s 1) = S Ω(s 3) = s1 s3 (2a s)tds 1 = 1 2 (4a s1)s1t, SΩ(s2) = 1 (4a s2)s2t, (b) 2 ( ) 3 2 a s3 tds 3 = 1 2 (3a s3)s3t, SΩ(s4) = 1 (3a s4)st. (c) 2 The shear stresses σ s(, s j) are positive, when acting in the direction of the arc-length parameter s j. These follow from Grashof s formula (2 74): σ s(s j) = Q S Ω(s j) = 12 Q ti 91 a 3 t SΩ(sj). 2 (d) The distribution of shear stresses has been shown in Fig. B (left). (continued) DCE Lecture Notes No. 23

87 2.3 Shear stresses from bending 79 Q τ a2 n 2 S B Q a B s ω(s 2) a a a Q τ a2 Figure B Single-smmetric I-profile: Shear stresses from the shear force Q with τ = Q /(at) (left); distribution of the sector coordinate ω(s) (right). The shear stress resultants Q 1 and Q 2 in the top and bottom flanges become: Q 1 = Q 4at 182 at = Q, Q2 = Q 3at 182 at = 27 Q. (e) 91 The shear centre then follows from Eq. (2 95): 36 Q s = Q 1 13 a + 42 Q a s = 9a. (f) Net, emploing another approach, the moment arm h(s j) defined b Eq. (2 87) is negative on the branches described b arc length parameters s 1 and s 2, and positive along the arc-length parameters s 2 and s 3. Then, from Eq. (b) it follows that L S Ω(s)h(s)ds = + 2a 3a/2 1 ( 2 (4a s1)s1t 36 ) 2a 13 a ds 1 + ) 1 2 (3a s3)s3t ( a ds a/2 ( (4a s2)s2t (3a s4)s4t 2 ) 13 a ds 2 ( a ) ds 4 = s4 t a4 t a4 t a4 t = 15 2 a4 t. (g) Then, from Eqs. (2 89) and (a) it follows that s = a3 t 15 2 a4 t = 9 91 a. This result is identical to the result achieved in Eq. (f). However, it is noted that the result in Eq. (h) has been achieved without the determination of the shear stresses. Hence, the second approach ma appear to be simpler than the first approach, but it does not provide an information about the stress distribution. (continued) (h) Elastic Beams in Three Dimensions

88 8 Chapter 2 Shear stresses in beams due to torsion and bending Finall, the same calculation is performed b means of Eq. (2 94). The distribution of the sector coordinate ω(s j) = s j h(s)ds with the pole B for each of the four branches becomes: ω(s 1) = a ω(s 4) = a s1 s4 ds 1 = ds 4 = as1, ω(s2) = as2, (i) as4, ω(s3) = as4. (j) 13 The sector centrifugal moment I ω = ωda = ω(s)(s)t(s)ds becomes: A L I ω = 2 2a (2a s 1) 36 as1 2tds a ( 32 a + s4 ) as4 2tds4 = 15 2 a4 t, (k) where the smmetr of the integrand ω(s)(s) has been eploited. Then, from Eqs. (2 94) and (a) the shear centre is determined as s = a3 t 2 a4 t = 9 91 a. Hence, the three different approaches lead to the same result. (l) Eample 2.1 Determination of the shear centre for a smmetric U-profile The bending centre of the U-profile has the position shown in Fig. A. Since the profile is smmetric, the indicated (,, )-coordinate sstem is a principal ais sstem. The moments of inertia with respect to the principal aes become: I = a3 t, I = 26 3 a3 t. 2a 4 5 a 4 2t 5 a (a) a Q 2 S Q a t B 2t S B Q 1 Q 2 Q Q a Figure A Smmetric U-profile: Geometr (left); position of the shear centre (right). Local arc length coordinates s 1, s 2 and s 3 are introduced as indicated in Fig. B. The distributions of the static moments S Ω(s j) and S Ω(s j) are shown in the figure. The analtic epressions are given as (continued) DCE Lecture Notes No. 23

89 2.3 Shear stresses from bending 81 S Ω(s 1) = s1 S Ω(s 2) = 4 s2 5 a2 t + S Ω(s 3) = 4 5 a2 t + S Ω(s 1) = s1 S Ω(s 2) = 4a 2 t + S Ω(s 3) = 4a 2 t + ( ) ( ) a s1 2tds 1 = 5 as1 s2 1 t, (b) s3 ( 45 ) a tds 2 = 4 at(a s2), 5 (c) ( 45 ) a + s3 2tds 3 = 4 5 a2 t 8 5 as3t + s2 3t, ( a)2tds 1 = 2as 1t, s2 s3 ( a + s 2)tds 2 = 4a 2 t as 2t s2 2t, a 2tds 3 = 4a 2 t + 2as 3t. (d) (e) (f) (g) 4 5 a2 t 26 5 a2 t S Ω (s 1 ) 4a 2 t S Ω (s 1 ) S Ω (s 2 ) s 1 S Ω (s 2 ) s 1 s 2 s 2 B B 1 2 a2 t s 3 s a2 t 26 5 a2 t S Ω (s 3 ) 4a 2 t S Ω (s 3 ) Figure B Smmetric U-profile: Distribution of the static moments S Ω (s j ) (left) and S Ω (s j ) (right) τ τ σ s(s 1 ) 6 26 τ σ s(s 1 ) σ s(s 2 ) σ s(s 2 ) B 3 52 τ B τ 16 τ σ s(s 3 ) 6 26 τ σ s(s 3 ) Figure C Smmetric U-profile: Distribution of the shear stresses from the shear forces Q with τ = Q /(at) (left) and Q with τ = Q /(at) (right). (continued) Elastic Beams in Three Dimensions

90 82 Chapter 2 Shear stresses in beams due to torsion and bending The shear stresses follow from Grashof s formula (2 74). Figure B (left) shows the distribution of shear stresses from Q, and Fig. B (right) shows the distribution from Q. The position of the shear centre is given b Eq. (2 95). Since the shear stresses σ s from Q are distributed smmetricall around the -ais it follows that s =, reflecting the fact that the -ais is a line of smmetr. The resulting shear forces Q 1 and Q 2 in Fig. A then become Q 1 = 2t 1 2 2a 6 26 τ = 6 Q, Q2 = t 13 Finall, from Eq. (2 95) follows that s = 1 Q ( Q 1 a a Q2 + Q1a ) ( ) 2aτ = Q. = 112 a. (i) 65 (h) Hence, for a U-profile the shear centre lies at a considerable distance outside the profile Shear stresses in closed thin-walled sections Figure 2 14 shows a closed single-cell section of a thin-walled beam. The shear force H(, s) acting within the wall per unit length of the beam (see Fig. 2 12) in a corresponding open section is uniquel determined b Eq. (2 71) due to the condition H(, ) = H () = at the boundar s =. However, in a closed section H(, s) is not a priori known in one or more points of the peripher Γ 1, for which reason the shear stresses cannot be determined from static equations alone. Hence, a geometrical condition must be introduced, from which the initial value H () at s = can be determined. The derivation of this geometric condition will be considered at first. σ s s s s n Ω(s) B S n Q Γ 1 s = L Q t(s) H s = Figure 2 14 Closed single-cell section eposed to bending. DCE Lecture Notes No. 23

91 2.3 Shear stresses from bending 83 The arc-length coordinate s is orientated clock-wise with an arbitraril selected origin O. Then, the local (, n, s)-coordinate sstem is orientated with the n-ais orientated toward the interior of the cell. The shear stress σ s (, s) follows from Eqs. (2 71) and (2 73): σ s (, s) = 1 t(s) H(, s) = H () Q () S Ω (, s) Q () S Ω (, s). t(s) t(s)i t(s)i (2 98a) where the static moments of the area segment Ω(s) are again given as s s S Ω (s) = (s)t(s)ds = da, S Ω (s) = (s)t(s)ds = Ω Ω da. (2 98b) Equation (2 98a) is the equivalence of Grashof s formula (2 74) for a closed thin-walled section. Unlike the case of the open section, H () is generall different from ero and the determination of H () is part of the problem. The shear strain ε s in the local (, n, s)-coordinate sstem is given b Eq. (2 76) which is valid for open as well as closed thin-walled sections. This implies the warping u (, s) defined b Eq. (2 8), where it is recalled that u (, s) = in the bending centre B. This is so, because w () has been defined as the total displacement in the -direction of B. Now, for the closed section, the displacement u (, s) must be continuous as a function of s along the peripher Γ 1. The aial and bending contributions, i.e. the first three contributions on the right-hand side of Eq. (2 77), are alwas continuous. A possible discontinuit then stems from the warping. If the profile is open, as illustrated in Fig. 2 12, the ends at s = and s = L can move freel relativel to each other. Hence, a warping discontinuit develops between these two endpoints. However, in a closed section, the warping of these points must be identical. Hence, the geometrical condition can be formulated as u (, ) = u(, L) = u (). (2 99) Insertion of Eqs. (2 76) and (2 99) into Eq. (2 8) provides u () = u () + 2 L L H(s) t(s) ds = H(s) Γ 1 t(s) ε s (, s)ds = u () + 1 G L H(, s) ds t(s) ds =. (2 1) Insertion of Eq. (2 98a) gives the following formulation of the geometrical condition from which the initial condition H can be determined: ds H () Γ 1 t(s) Q S Ω I Γ 1 t(s) ds Q S Ω ds =. (2 11) I Γ 1 t(s) With H () determined, H(, s) and the shear stress σ s (, s) can net be determined from Eq. (2 98a). Eample 2.11 Shear stresses in a smmetric thin-walled single-cell section Figure A shows a double-smmetric thin-walled single-cell profile. The thickness is everwhere t. We want to determine the shear stresses from a shear force Q acting at the shear centre. (continued) Elastic Beams in Three Dimensions

92 84 Chapter 2 Shear stresses in beams due to torsion and bending The cross-sectional area and moment of inertia around the -ais become A = 8at, I = 5 3 a3 t. (a) The origin of the arc length coordinate is chosen at the lower left corner. The corresponding distribution of the static moment S Ω(s) = s (s)t(s)ds has been shown in Fig. B. a B = S t(s) Q 2 3 a 2 3 a Figure A Geometr of the smmetric rectangular thin-walled single-cell section. 3 2 a2 t B = S 1 8 a2 t s 1 8 a2 t 3 2 a2 t Figure B Distribution of the static moment S Ω (s) in the smmetric rectangular thin-walled single-cell section. Net, the following line integrals are calculated: ds Γ 2 t = 1 t (a + 3a + a + 3a) = 8a t, (b) S Ω(s) ds Γ 1 t = 1 ( 2 t a2 t a 1 2 3a 3 2 a2 t 3 2 a2 t a a2 t a 1 2 3a 3 ) 2 a2 t = 6a 3. (c) (continued) DCE Lecture Notes No. 23

93 2.3 Shear stresses from bending 85 From Eq. (2 11) then follows that H 8a t Q 5 3 a3 t ( 6a3 ) = H = 9 Q 2 at. (d) 18τ 18τ 3τ B = S Q 3τ 18τ 18τ Figure C Shear stresses in the smmetric rectangular thin-walled single-cell section with τ = Q /(4at). The distribution of shear stresses σ s(, s) follows from Eq. (2 98a) and has been indicated in Fig. C. The arrows indicate the positive direction of the shear stresses. These will be referred to as the shear flow in what follows, due to the analogous behaviour of water flowing down through a sstem of pipes. As shown in Eample 2.11, a smmetric thin-walled single-celled profile eposed to a shear force acting along the line of smmetr will have a smmetric distribution of shear stresses. Especiall, the shear stress at the line of smmetr vanishes. Hence, if the origin of arc-length coordinate is placed at the line of smmetr we must have H () =, entailing that a singlecelled smmetric profile can be analsed from the static equations alone. For a thin-walled profile with N cells, arc-length parameters s j are introduced for all cells, orientated clock-wise as shown in Fig The origins O 1, O 2,..., O N for the arc-length coordinates are chosen arbitraril. The shear forces per unit length H j () at the origins are unknown and must be determined from geometrical conditions in addition to the static shear flow equations. Similarl to Eq. (2 99) these are determined b the conditions that the warping must be continuous along all the peripheries Γ j of the cells. At first, the static moments S Ω and S Ω for the open profile in Fig are determined. Especiall, the variation of S Ω (, s j ) and S Ω (, s j ) along Γ j as a function of s j are registered. Net, the calculation of the shear force per unit length H j (, s j ) within each cell can be arranged as illustrated in Fig Firstl, the shear force per unit length H sj (s j ) of the open, branched profile is calculated, orientated in the direction of the arc-length coordinate s j. This is given b Grashof s formula (2 74), i.e. H sj (s j ) = Q I S Ω (s j ) Q I S Ω (s j ). (2 13) Elastic Beams in Three Dimensions

94 86 Chapter 2 Shear stresses in beams due to torsion and bending σ s H N Γ S Q O N Q H 1 O 1 s N B N Γ j s j j s 1 1 O j s k k H j O k H k Figure 2 15 Closed multi-cell section eposed to bending. N H N j H 1 1 Hk H j k H s1 1 N H sn Hsk k j H sj H 1 1 N H N Hk k j H j Figure 2 16 Arrangement of calculations of shear forces per unit length within cells. DCE Lecture Notes No. 23

95 2.3 Shear stresses from bending 87 In each cell, the initial conditions induce a constant shear force per unit length H j. Due to the chosen sign convention, the net shear force on the peripher segment Γ jk between cell j and cell k becomes H sj H k. Hence, on this segment the total shear force per unit length becomes H j (s j ) = H sj (s j ) + H j H k. (2 14) Continuit requires that Eq. (2 11) is fulfilled for all N cells, i.e. Γ j H j (s j ) t(s j ) ds j =, j = 1, 2,..., N. (2 15) Insertion of Eq. (2 14) leads to the following N linear equations for the determination of H 1, H 2,..., H N : N (H j H k ) k=1 Γ jk ds j t(s j ) = H sj(s j ) Γ j t(s j ) ds j, j = 1, 2,..., N. (2 16) The coefficient matri in Eq. (2 16) is identical to this for the determination of Prandtl s stress function S j in the cells in the corresponding St. Venant torsion problem. With H j determined from Eq. (2 16), H j (s j ) can be calculated from Eq. (2 14). Finall, the shear stresses σ s (s j ) follow from Eq. (2 98a) with positive sign when acting in the direction of s j. Eample 2.12 Shear stresses in a single-smmetric double-cell section The profile shown in Fig. A is identical to that considered in Eample However, a partition with the wall thickness t has been included as shown in the figure. The bending centre B is placed as indicated. Because the section is smmetric around the -ais the indicated coordinate sstem with origin in B is a principal ais coordinate sstem. The section is still eposed to a shear force Q acting through the shear centre S. Due to the smmetr the shear centre is placed on the -ais ( S = ). The s will be determined as a part of the solution. The cross-sectional area and inertial around the -ais become A = 9at, I = 7 4 a3 t. (a) 13 9 a a t(s) t(s) t(s) S B Q a 2 a 2 a 2a Figure A Geometr of the rectangular thin-walled double-cell section. (continued) Elastic Beams in Three Dimensions

96 88 Chapter 2 Shear stresses in beams due to torsion and bending O 1 O 2 1 H s 1 s H 2 Γ 1 Γ 2 Figure B Definition of the origins of the arc-length coordinates in the rectangular thin-walled double-cell section. 1 2 a2 t a 2 t S Ω (s 1 ) S Ω (s 2 ) 1 8 a2 t a2 t 8 a2 t 1 8 a2 t 1 2 a2 t a 2 t Figure C Distribution of the static moments S Ω (s1) and S Ω (s2) in the rectangular thin-walled double-cell section. The origin of the arc-length coordinates are chosen as shown in Fig. B. The static moments S Ω(s 1) and S Ω(s 2) are net calculated with the sign and magnitude as indicated in Fig. C. Especiall, it is seen that S Ω(s 1) = S Ω(s 2). The shear force per unit length in the open profile is given b Eq. (2 13): H sj(s j) = Q I S Ω(s j), j = 1, 2. (b) Net, Eq. (2 16) provides (H 1 H 2) a t + (H1 )a t + (H1 )a t + (H1 )a t = (H 2 ) 2a t + (H2 )a t + (H2 )2a t + (H2 H1)a t = Γ 1 H sj(s 1) t(s 1) ds1, Γ 2 H sj(s 2) t(s 2) ds2. The right-hand sides are calculated b insertion of Eq. (b) with the distribution of S Ω(s j) shown in Fig. C. The results become H sj(s 1) ds 1 = 4 ( Q 2 Γ 1 t(s 1) 7 a 3 t 2 3 a a2 t 8 a 2 a2 t 2 t aa2 2 2 t 3 aa2 8 s ) a 2 t = 4 Q t, (e) ( H sj(s 2) Γ 2 t(s 2) ds2 = 4 Q 2a 7 a 3 t 2 2 a2 t + a a 2 t a a2 t + 2a 2 a2 t 2 ) t 3 aa2 = 12 Q 8 7 t. (f) (c) (d) (continued) DCE Lecture Notes No. 23

97 2.3 Shear stresses from bending 89 Then, the following solutions are obtained for the initial values of the shear forces per unit length: 4H 1 H 2 = 4 Q 7 a, H1 + 6H2 = 12 7 which implies that H 1 = Q 88 Q, H2 = a 322 a. Net, H j(s j) is calculated from Eqs. (2 13) and (2 14) along both peripheries Γ 1 and Γ 2. Finall, the shear stresses σ s(s j) are computed from Eq. (2 98a). Due to the constant wall thickness this reduces to σ s(s j) = 1 Hj(sj), j = 1, 2. (i) t The flow of the shear stresses σ s has been shown in Fig. E. Q a, (g) (h) 68τ 24τ 96τ 23τ 88τ 88τ 23τ 88τ 88τ 23τ 68τ 24τ 96τ Figure D Distribution of the shear stresses in the rectangular thin-walled double-cell section with τ = Q /(322at) a Q 5 Q 4 Q 4 Q 3 Q S B Q 1 Q 1 Q 2 a 2 a Figure E Shear forces in the wall segments of the rectangular thin-walled double-cell section. (continued) Elastic Beams in Three Dimensions

98 9 Chapter 2 Shear stresses in beams due to torsion and bending Net, the shear forces Q 1, Q 2, Q 3, Q 4 and Q 5 in the wall segments shown in Fig. E are calculated. These become: Q 1 = 1 2 (96 88) 2a t 1 Q 322 at = 12 Q, (j) 483 Q 2 = ( ) a t 1 Q 322 at = 167 Q, 483 (k) Q 3 = ( ) a t 1 Q 322 at = 191 Q, 483 (l) Q 4 = 1 2 (68 24) a t 1 Q 322 at = 33 Q, 483 (m) Q 5 = ( ) a t 1 Q 1288 at = 125 Q. 483 (n) It is seen that Q 2 + Q 3 + Q 5 = Q. The static equivalence (2 83) of the shear stresses is then fulfilled. The position s of the shear centre follows from Eq. (2 95): s = 1 ( (Q 4 Q 1)a Q 2 14 Q 9 a + Q3 4 ) 9 a + 13 Q5 9 a s = 8 a. (o) 1449 The position of the shear centre has been illustrated in Fig. E. Eample 2.13 Distribution of shear stresses in open and closed sections due to bending and St. Venant torsion Q Q Q Q Q M M Figure A Shear flow in open and closed sections due to transverse shear forces and torsional moments. Fig. A shows the shear flow in open and closed thin-walled sections eposed to a transverse shear force Q or a torsional moment M. For closed sections the shear stresses σ s(, s) are uniforml distributed over the wall thickness for both loadings. For open section this is onl the case in the case for the shear force loading, whereas the shear stresses for St. Venant torsion is linearl varing in the thickness direction around the mid-line of the wall. DCE Lecture Notes No. 23

99 2.4 Summar Summar The computation of shear stresses in beams has been the focus of this chapter with detailed eplanation of the theor for thin-walled sections subjected to torsion and/or bending. A brief summar of the main findings is given in the following. Uncoupling of bending and torsion requires that the shear force acts through the so-called shear centre. The shear centre alwas lies on a line of smmetr within a smmetric cross-section. Hence, the position of the shear centre coincides with that of the bending centre for doublesmmetric sections. St. Venant torsion is characterised b the fact that the beam is allows to warp freel, leading to a homogeneous twist of the cross-section along the beam ais. The homogenous torsion problem can be defined in terms of the warping function or, alternativel, in terms of Prandtl s stress function. Shear stresses due to homogeneous torsion var linearl over the thickness of the wall in open thin-walled profiles. However, in a closed thin-walled section with one or more cells, the shear stresses due homogeneous torsion is homogeneous over the thickness. Homogeneous torsion induces warping in a beam. The warping increases linearl with the torsional moment. However, in a circular profile (a clinder as well as a tube) there is no warping. The torsional stiffnesses of open and closed cross-sections are different. Thus, a closed crosssection has a much higher torsional stiffness than an open cross-section with a similar crosssectional area. Likewise, the ultimate strength of a closed section is higher than that of a similar open section. Shear stresses from bending can generall be analsed b means of the so-called stress function. However, this requires the use of a numerical scheme, e.g. the finite-element method. Grashof s formula defines the shear stresses from bending in open thin-walled sections. For closed thin-walled sections, Grashof s formula must be adjusted be an additional term that Warping takes place due to bending since the shear stresses are accompanied b shear strains. For a rectangular cross-section, the beam warps into an S-shape. Shear flow is a graphical interpretation of the direction in which the shear stresses are acting on the cross-section of a beam. In the case of bending, this forms an analog to water streaming down a sstem of pipes. Internal walls in a section will not increase the torsional stiffness and strength significantl. However, the inclusion of an internal wall oriented in the direction of the shear force will provide an increase of the shear strength and stiffness. The theor presented in this chapter can be used for the determination of the shear stresses and warping deformations in beams subjected to an combination of bending and homogeneous torsion. However, if torsion is prevented, e.g. at one end of the beam, another theor must be applied as described in the net chapter. Elastic Beams in Three Dimensions

100 92 Chapter 2 Shear stresses in beams due to torsion and bending DCE Lecture Notes No. 23

101 References Krenk, S. (21). Mechanics and analsis of beams, columns and cables (2nd ed.). Springer Verlag. SaintVenant, B. (1855). Memoire sur la torsion des prismes. Mem. Acad. Sci. Savants Etrangers 14, Timoshenko, SP (1921). On the correction for shear of the differential equation for transverse vibrations of static bars. Philosophical Magaine, Series 6 41, Vlasov, V.Z. (1961). Thin-walled elastic beams. Israel Program for Scientific Translations. 93

102 94 References DCE Lecture Notes No. 23

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104 ISSN DCE Lecture Notes No. 23