Eindhoven University of Technology MASTER. Finite element methods for geometrically linear curved beams. Kinyanjui, T.W.

Transcription

1 Eindhoven University of Technology MASTER Finite element methods for geometrically linear curved beams Kinyanjui, T.W. Award date: 2012 Link to publication Disclaimer This document contains a student thesis (bachelor's or master's), as authored by a student at Eindhoven University of Technology. Student theses are made available in the TU/e repository upon obtaining the required degree. The grade received is not published on the document as presented in the repository. The required complexity or quality of research of student theses may vary by program, and the required minimum study period may vary in duration. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Users may download and print one copy of any publication from the public portal for the purpose of private study or research. You may not further distribute the material or use it for any profit-making activity or commercial gain

2 TECHNISCHE UNIVERSITÄT KAISERLSLAUTERN DEPARTMENT OF MATHEMATICS MASTER S THESIS Finite element methods for geometrically linear curved beams by: Tabitha Wangari Kinyanjui Supervisors: Dr - ing. Joachim Linn (ITWM) Dr. Joseph M. Maubach (TU/e) Kaiserslautern, Germany November 22, 2007

3 The Faculty members appointed to examine the thesis by Tabitha W. Kinyanjui find it satisfactory and recommend that it be accepted. TU-Kaiserslautern Date: TU-Eindhoven Date:

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5 Contents 1 Introduction 8 2 Straight Beam Theory Geometry Stresses Euler - Bernoulli Beam Theory Total potential energy Timoshenko Beam Theory Total potential energy Finite Element Method Formulation of the Finite Element Method Bubnov-Galerkin s Approximation Method The Stiffness Matrix and the Load vector Assembly of global stiffness matrix and force vector Linear Curved Beams The Equilibrium Equations Normal stresses in Curved Beams Deflection of a Curved Beam Shear Deformation Finite Element Implementation Stiffness Matrix and Load Vector

6 Contents Timoshenko beam stiffness matrix Euler-bernoulli beam stiffness matrix Numerical Integration Displacement Transformation Matrix Common Boundary Conditions Numerical results of beams under different boundary conditions Shear and Membrane Locking Numerical Test Examples Example 1: An arc for various subtended angles Castigliano s Energy Method Example 2 :A quarter-circular Cantilever Beam Results for Timoshenko curved Beam Results for the Euler-Bernoulli curved beam Points to Note: Timoshenko curved beam strains definition Summary and Conclusion Appendix 64.1 Matlab Programmes Common programmes and Parameters for both beam theories Curved Beam - Euler-Bernoulli beam theory Curved Beam - Timoshenko Beam Theory Declaration 79 2

7 List of Figures 2.1 Geometry of a simple Beam Deformation of cross-section of an Euler - Bernoulli beam Deformation of the cross-section of a Timoshenko beam Bending stress in curved beams Forces and displacements on an infinitesimal curved beam element Geometry of a beam s element vertical displacement of magnitude y (a.)tangential displacement (b.)radial displacement(c.)cross-sectional displacements and rotations of the cross-sectional plane around the y- axis (d)about the z-axis Geometry of a curved beam element with three degrees of freedom at each node curved beam element related to local and global displacements straight beam element displacement Relation of curved beam element in local and global coordinates system Different beam supports and their reactions Tangential displacement and Radial displacements for Euler -Bernoulli and Timoshenko beams fixed at both ends Cross sectional Rotation for Euler -Bernoulli and Timoshenko beams fixed at both ends Tangential displacement and Radial displacements for cantilever Euler -Bernoulli and Timoshenko beams subjected to external load of Cross sectional Rotation for cantilever beam Euler -Bernoulli and Timoshenko beams,subjected to external load of

8 List of figures 4.14 Tangential Displacement and Rotational Displacement for increased external load from of a beam clamped at both ends Cross sectional Rotation for increased external load from of a beam clamped at both ends Displacement of an Euler- Bernoulli Beam and a Timoshenko Beam that are both simply supported at both ends,subjected to external load from Cross sectional Rotation of an Euler- Bernoulli Beam and a Timoshenko Beam that are both simply supported at both ends, subjected to external load from Effects of reduced membrane energy term integration for Euler-Bernoulli beam, and reduced shear and membrane energy terms integration for Timoshenko beam An arc fixed on one end and supported on rollers and subjected to point load on the other end Radial displacement (a)with reduced integration, (b) Without reduced integration for an arc fixed on one end and supported on rollers and subjected to point load on the other end A quarter-circular cantilever beam subjected to radial point load Q, at the free end. (Lee, Sin [22]) Free body diagram of a quarter-circular cantilever beam subjected to radial point load Q, and we introduce axial dummy force Θ and a dummy moment - m at the free end Maximum Absolute Error for Tangential, Radial displacements and Cross sectional Rotation when the conventional strains of a Timoshenko curved beam are used

9 List of Tables 4.1 Parameters used for a curved beam subtended by angle π Comparison of the finite element radial displacement with reduced integration for the Timoshenko beam and Euler-Bernoulli beam with, the method by Saffari and Tabatabaei [2] Comparison of the two different finite element solution models loaded with tip radial force, of a quarter circular cantilever beam with Castigliano s energy solutions for Timoshenko beam Comparison of the two different finite element solution models loaded with tip radial force, of a quarter circular cantilever beam with Castigliano s energy solutions for Euler-Bernoulli Beam Comparison of the two different strain definition for fem solution models loaded with tip radial force, of a quarter circular cantilever beam with Castigliano s energy solutions for Timoshenko Beam

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11 Preface I would like to express my sincere thanks and appreciations to my supervisor Dr - ing. Joachim Linn for his great effort and a lot of time spent in introducing me into this area of study and guiding me through. Without his undoubted advice and guidance, it would have been impossible to get through this work. I would like to thank my lecturers at TU - Kaiserslautern and TU - Eindhoven. They kept us very busy, we barely noticed days passing by, but they tireless helped and assisted us all the way. I would like to thank the consortium for industrial mathematics for giving me this chance to participate in the Erasmus Mundus programme. Indeed it is a privilege to be a participant and the knowledge, and skills acquired during my two years of study are without doubt enormous. Many thanks to my friends in Eindhoven, and in Kaiserslautern. I wish to mention especially Changgih, Henry and Vincent. Thanks for keeping the long hours in the discussion groups. Lastly, i thank my parents and my siblings, for their tireless and relentless love, continuous support, and countless sacrifices they ve had to make on my behalf. Thank you for believing in me. Tabitha W. Kinyanjui November 22, 2007 Kaiserslautern

12 Chapter 1 Introduction Beams are the most common structural components found in most mechanical structures. They can be defined as thin long structures that are capable of carrying loads in flexure, and are characterized by (a) The longitudinal dimension, which defines the axial direction of the beam is considerably larger than the transverse dimensions, the dimension defining the directions normal to the axial directions. (b) If the beam is restricted to deform in plane, then it resists primarily loading applied in one plane and has a cross section that is symmetric with respect to that plane, hence it can be analyzed with two-dimensional idealizations [1]. A variety of practical engineering problems for example some of the structures such as tyres, pipelines, rings, arc-like structures, just to mention a few however are modeled by curved finite elements. A lot can be learned by studying the performance of simple (essentially one-dimensional) models so as to gain insight into the more general behavior of the overall structures. The circular ring serves as a simple yet realistic model in addition for analyzing the effects of symmetry in structural systems elements. Since the deformations are symmetric, a closed ring can be regarded as an example of a curved beam with restrained sections or ends. From conditions of symmetry, the distribution of stress in one quadrant is know to be the same as in another. Thus one quadrant may be considered to be the curved beam in which the behavior of the ring under load, at any section is to be found. Furthermore, curved beams are known to be more efficient in transfer of loads than straight beams because the transfer is affected by bending, shear, and membrane action (Saffari, Tabatabaei [2]). Therefore the main objective in this thesis is 8

13 (a) Give a detailed derivation of equations of a curved beam in three dimesions, with reference to the quoted literature, having first introduced a straight beam model as a starting ground. (b) To investigate the behavior of a plane curved beam, resulting from loading applied to the beam with cross-sections symmetric about that plane. (c) The behavior of the beam under the assumption that there is a linear variation of displacement over the cross-section, the Euler-Bernoulli beam theory. This implies that the cross-sections which are normal to the centroid axis of the beam before deformation remain plane, undeformed and perpendicular to the deformed axis.this model accounts for bending moment effects on stresses and deformations. This implies that transverse shear forces are recovered from equilibrium but their effect on beam deformations is neglected. (d) Account for shear deformation by constraining the normal to remain straight, but not necessarily normal, to the centroidal axis after shearing and bending; Timoshenko beam theory. (e) Implement the finite element method to model the numerical analysis of the behavior of the displacement of curved beams. (f) To check how accurate the finite element we are going to use here, we compare the finite element approximation solution of the analytical solution, and this comparison is carried out with already existing finite element results in literature which have been proven to be very accurate in their approximation. The examples we are going to use are the typical examples of the shear and membrane locking. They are frequently dealt with to check the capability of analysis of the elements developed and to see if the locking phenomena is property overcome. Assumptions The main assumptions are that 1 (a) The beam deformation is geometrically linear. Therefore the displacement gradients are small compared to unity 1, implying that the difference between the deformed and the undeformed beam is very small. G = U X =: ε 1 where U is the displacement vector, and X is position of material point p in a reference configuration with respect to a given origin O. 9

14 (b) The beam is prismatic with constant circular cross section and constant radius of curvature. (c) We are neglecting changes in dimensions of the cross sections as the beam is displaced. 10

15 Chapter 2 Straight Beam Theory To set ground for curved beams, we first give an overview of the deformation of a straight beam with a square cross section, while adopting definitions according to [3] Geometry Y q(x) Y X h Z L b Figure 2.1: Geometry of a simple Beam Cross Section (a) The beam is long and thin, implying that L >> b and h. (b) Loading is in the y direction to obtain the bending moment. (c) Constant cross section. (d) The beam does not experience torsion Stresses (a) No load applied in the z- direction implying the stresses are zero in that direction. (b) The only significant stress are σ xx and τ xy. 11

16 2.1 Euler - Bernoulli Beam Theory We define v the axial deflection and u as the transverse deflection of midplane. After deformation, (a) Plane sections remain plane and perpendicular to the midplane. For this to be true, we require that (i). The beam should be bent only with bending moments, therefore there are no shear on transverse planes, as we will explain below; (ii). The applied loads are such that no twisting occurs. (b) In addition to this, we use the assumptions that the axis of the rod is inextensible. (c) The Euler-Bernoulli beam theory assumes that the internal energy of a beam is Therefore entirely due to bending strains and stresses, therefore the predominant stresses and strains are σ xx and ε xx. ε yy = u y = 0 (2.1) ε xy = 1 2 ( u x + v ) = 0 (2.2) y Thus from (2.1) u is not a function of y, thus u = u(x) (2.3) This implies that every point of the cross section at a given location along the x axis has the same vertical displacement.(timoshenko, Goodier [4]) Therefore Thus ε xy = 0 v y = u x = du dx (2.4) v = y du = yθ (2.5) dx implying that the axial displacement vary linearly with the y distance from the neutral axis. We also note that the equation (2.5) implies that the cross-sections rotational (bending) angle directly depend on the gradient of deformation, i.e θ = du dx 12 (2.6)

17 which along with (2.3) confirms the assumption (b) above that any cross-sectional plane stays a plane and normal to the deformed longitudinal axis, though it displaces and rotates a bit from its original position. y,u θ deformed state Normal to the reference beam Normal to deformed beam axis, which is also the direction of the deformed crosssections. du/dx u undeformed state x, v Figure 2.2: Deformation of cross-section of an Euler - Bernoulli beam Total potential energy For a given loaded elastic body under geometrical constrains or boundary conditions, the potential energy of the deformed body assumes a stationary value, and it attains an absolute minimum when the displacement of the body are those of equilibrium configuration. From (2.5), the axial strain distribution is ε xx = v x = y 2 u x 2 = yd2 u dx 2 (2.7) Using the Hooke s law, (Timoshenko,Goodier, [4]), the constitutive equation is thus σ xx = Eε xx = Ey d2 u dx 2 (2.8) the stress distribution in terms of displacement field, where E is the Young s modulus of elasticity. The resulting bending moment M is defined as M = yσ xx dx = E d2 u A dx 2 y 2 da = EIκ A for κ = d2 u dx 2 (2.9) Where I is the area moment of inertia A y2 da a property of the beam that is used to predict its resistance to bending and deflection, and κ is the curvature of the beam. Since the internal energy for the Euler-Bernoulli beam accounts only for the bending 13

18 moment deformation, the total potential energy of the beam, subjected to force q acting in the y direction is W i = 1 2 W e = V σ xx ε xx dv = 1 2 L 0 qudx Π = W i +W e = 1 2 L 0 EI L 0 EI ( d 2 ) 2 u dx 2 dx (2.10) ( d 2 ) 2 u L dx 2 qudx 0 Where W i and W e are internal and external virtual work respectively and L is the length of the beam.(oden, [5]) 2.2 Timoshenko Beam Theory As cited by Felippa [1] the Timoshenko beam corrects the classical beam theory with first-order shear deformation effects. Therefore the effect of the shear stresses on the deformation are taken into account. After deformation, (a) Plane sections are assumed to remain plane and rotate about the same neutral axis, with the assumption under (a) - (ii) in section 2.1 also holding for this case too. (b) Unlike the euler-bernoulli beam, the cross sections do not remain perpendicular to the deformed longitudinal axis. Therefore ε xx = v x = ydθ dx (2.11) ε xy = 1 2 ( v x + u ) = 1 ( θ + du ) = 1 x 2 dx 2 ϒ (2.12) Thus the deviation from the normal direction which is the difference between the normal to the longitudinal axis and the plane section rotation is the shear angle ϒ. It consists of the contributions from the gradient of the deflection (the rotation of the longitudinal axis), and the cross sectional rotation bending angle θ. This is further illustrated in figure (2.3). For Timoshenko beam the cross section deformation is the sum of two contributions that is one is due to bending, and the other is the shear deformation. 14

19 γ Shear angle y, u θ du/dx Normal to the reference beam Normal to defromed beam axis du / dx Direction of deformed cross section u x, v Figure 2.3: Deformation of the cross-section of a Timoshenko beam Total potential energy From equations (2.11) and (2.12), the axial and shear strains for a Timoshenko beam are ε = y dθ dx ϒ = θ + du dx (2.13) (2.14) Therefore Π = 1 2 = EI 2 L (σ xx ε xx + σ xy ε xy )dv V L 0 ( dθ dx ) 2 dx + κga 2 0 L 0 qudx (2.15) ( θ + du ) 2 L dx qudx dx 0 (2.16) where κ is the shear correction factor, G the shear modulus,a is the cross sectional area, and q is the external load applied to the beam, as has been explained by (Kwon, Bang [6]). 15

20 Chapter 3 Finite Element Method In this chapter we introduce the finite element numerical methods, that will be adopted for the problems to follow. Most numerical methods involve an approximation to an unknown function U by a new function U, which is a linear combination of the basis functions U ˆ = n i=1 U i W i where U are the unknown coefficients, W i are the chosen weight functions. While classical methods employ global basis functions of various frequencies, approximation of this form will work adequately for simple geometries but yield poor approximations for more complex geometries, as global basis functions cannot reflect properties of the solution that are induced by the geometry. Instead of employing global basis functions, the finite element method approximates a solution for problems which are described by partial differential equations or formulated as functional minimization, based on the concept of discretization of the solution domain into sub-domains called finite elements. As summarized by (Nikishkov, [7]), this can be achieved with the following general steps. (a) Discretize the domain, that is to divide the solution region into finite elements. The solution domain is divided into several simpler finite elements, where each element has a simple geometry, so appropriate assumed solutions can easily be written for the element. (b) Selection of interpolation functions to describe solution over an element is the next step. 16

21 If we use a polynomial function to interpolate U within elements from nodal values of U, then for U to approach the exact values as the finite element mesh is repeatedly refined and if α is the highest derivative of U, then (i) Within each element, the assumed field for U must contain a complete polynomial of degree α or higher, (ii) Across the elements boundaries, there must be a continuity of order α 1 for U and its derivatives. (c) Establish the matrix equation for the finite element which relates the nodal values of the unknown function to other parameters.different approaches used are for example the most the variational approach and the Galerkin method. (d) To find the global equation system for the whole solution, all element equations must be assembled. This involves the combination local element equations for all elements used for discretization. Element connectivities are used for the assembly process. Before solution, boundary conditions (which are not accounted in element equations) should be imposed. (e) Solving the global equation system. Since the finite element global equation system is typically sparse, symmetric and positive definite, direct and iterative methods can be used for solution. The nodal values of the sought function are produced as a result of the solution,since approximating functions are determined in terms of nodal values of as physical field which is sought Formulation of the Finite Element Method As we have mentioned in (c) above, several approaches can be used to transform the physical formulation of the problem to its finite element discrete state. If the physical formulation of the problem is described by a differential equation then the most popular method of its finite element formulation is the Galerkin method. If the physical problem can be formulated as minimization of a functional then variational formulation of the finite element equations is usually used. 3.1 Bubnov-Galerkin s Approximation Method The formulation given here is standard, and can be found in many finite element resources. The construct of the finite element given below is in reference mainly from (Hughes, [8]); Also we refereed to approach used by (Nikishkov, [7]); ( 17

22 Strang et al., [9]); With a general example, let the problem to be solved in one dimensional be defined by the equation L u = f in Ω (3.1) u Γ1 = u ρ on Γ 1 u α Γ2 = u β on Γ 2 for L a given operator acting on functions u L p (Ω), such that the functions satisfy in some sense the boundary conditions. The fundamental question is then: To match such a space of functions u with a class of inhomogeneous terms f, in such a way that to each f, there corresponds one and only one solution u. Let the space δ = {u u H s,u Γ1 = u ρ } (3.2) be defined as a function space for trail solutions u. Define a second collection of functions know as test functions such that we require that they vanish at boundaries corresponding to the essential boundary condition u ρ. That is and V = {ω ω H s,ω Γ1 = 0} (3.3) L : H s (H s ), and f (H s ) the dual space of H s. (3.4) Then the weak formulation of the problem is to find u δ, such that for all ω V, a(u,ω) = ω T L u = (ω, f ) (3.5) Where a : δ V R is a bilinear operator linear in its second argument. The u α for any α with α s exist in the weak sense, and derivatives of small order needn t exist, like in the classical sense. The Galerkin s method with piecewise polynomial subspaces δ h and V h is the finite element method. Let the approximations of sets δ and V, denoted by δ h and V h respectively, be such that δ h in δ (i.e.; if u δ h,then u h δ) (3.6a) V h in V (i.e.; if ω V h,then ω h V ) (3.6b) 18

23 Consequently from (3.1) u h Γ1 = u ξ and ω h Γ1 = 0 (3.6c) since this is an essential boundary condition then it must be satisfied by every function ω h Implying that the functions ω h, should be zero at that boundary. Given the collection V h, then to each member ν h V h, we construct a function u h δ h, by u h = ν h + u h ρ (3.7) where u h ρ = u ρ. Therefore δ h is all functions of the form (3.6a). The problem is now defined as: Given f,u ρ, and u β, find u h = ν h +uρ, h where ν h V such that for all ω h V h, a(ω h,ν h ) = (ω h, f ) + ω h Γ2 u β a(ω h,u h ρ) (3.8) u h ρ and V h have to be explicitly defined. Note that the approximate solution is required to fulfil the essential boundary conditions, but not the natural boundary condition, which doesn t prevent the approximations form converging in the H s norm, to a solution u, which does satisfy u α Γ2 = u β The Stiffness Matrix and the Load vector The Galerkin method leads to a coupled system of linear algebraic equations. Every test function ν h is determined by its nodal parameters, which are the unknowns c i of the discrete problem. Each of these nodal parameters is the value at a given node, of either the function or its derivatives. Let V h consist of all linear combinations of given functions which are denoted by φ i : = Ω R, for i = 1,2,...,n. Then if ω h V h, then there exists constants c i such that ω h (x) = n i=1 c i φ i (x) (3.9) for c i = w h (x i ) for c i = w h (x i ), the nodal values for w h (x), { 1 if i = j φ j (x i ) = 0 if i j i, j = 1,...,n 19

24 and (φ i ) Γ1 = 0, i = 1,2,...,n Thus V h is a linear space of dimensions n with the basis {φ i } n i=1. To define members of δ, we need to specify u ξ. Defining another shape function as φ n+1 : = Ω R, with the property that (φ n+1 ) Γ1 = 1, φ n+1 / V h, then and thus u h ρ = uφ n+1 u h ρ Γ1 = u ρ For u h δ h, and constanst d i, i = 1,2,...,n u h = ν h + u ρ (3.10) = n i=1 d i φ i + u ρ φ n+1 Substituting (3.9) and (3.10), into (3.8), then a ( n i=1 c i φ i, n j=1 which simplifies to d j φ j ) = ( n c i φ i, f i=1 ) + [ n i=1c i φ i Γ1 ] u β a ( n i=1c i φ i,u ρ φ n+1 ) (3.11) For n i=1 D i = c i D i = 0 (3.12) n j=1 a(φ i,φ j )d j (φ i, f ) φ i Γ1 u β + a(φ i,φ n+1 )u ρ (3.13) Since the solution hold ω h V h, and {c i } n i, i = 1,...,n, arbitrary in (3.12), then n j=1 for each D i, a(φ i,φ j )d j = (φ i, f ) + φ i Γ1 u β a(φ i,φ n+1 )u ρ (3.14) i = 1,..,n. Since everything is know besides d j, then (3.14) constitute a system of n equations, in n unknowns. Let K i j = a(φ i,φ j ) (3.15a) 20

25 F i = (φ i, f ) + φ i Γ1 u β a(φ i,φ n+1 )u ρ (3.15b) Then (3.14) while adopting a matrix notation can compactly be restated as solving for d in the algebraic equations defined as Kd = F (3.16) 3.2 Assembly of global stiffness matrix and force vector The elements stiffness must now be assembled into the beam s stiffness, implying now that to add values of element stiffnesses in global axes, we need to adjust the numbering into a numbering system of the beam as a whole. This particular task is accomplished by the location matrix.the location matrix (if we denote it by loc) dimensions are the number of element nodes by the number of elements. Thus give a particular degree of freedom, and an element number, for example i and j respectively, then the value returned by the location matrix array is the corresponding global equation number à such that j if i = 1 j + 1 if i = 2 à = loc(i, j) = j + 2 if i = 3 for j = 1,2,...,n el.. j + n el 1 if i = n el Therefore the contribution of an element to the beams global stiffness is obtained by adding to the beams global stiffness matrix term in row loc(i), and column loc( j), the term K loc(i),k loc( j), the element stiffness term K (e) i, j. This applies to the force vector too, where the representation of the global force vector is now F loc(i) = F loc(i) + F (e) loc(i) (3.17) In the next two subsections, we are going to discuss a plane two-dimensional, linear Euler-Bernoulli beam finite element, and Timoshenko beam element. From the definitions of the potential energy for each beam,we derive the displacement formulation on the basis of the galerkin method already described. 21

26 Chapter 4 Linear Curved Beams In our discussion in the previous chapters, the deformation of the beam subjected to a load, has been restricted to beams with the assumption that the longitudinal elements have the same length. The xy- longitudinal plane has been assumed to be the plane of symmetry with the load applied in this plane. This has restricted the beam theory formulated above to initially straight beams of constant cross section. Although considerable deviations from this restriction can be tolerated in real problems, when the initial curvature of the beams becomes significant, the linear variations of strain over the cross section is no longer valid, even though the assumption of plane cross sections remaining plane is valid. The cross section of part of an initially curved beam is described by the figure (4.1), with xy plane the plane of symmetry. The radii R references the location of the centroid of the cross sectional area; R n references the location of the neutral axis; and r references some arbitrary point p of area element da on the cross section. For curved beams, the longitudinal deformation of any element will be proportional to the distance of the element from the neutral surface, implying that the total deformations are proportional to the distance from the neutral axis. This implies then that the linear variation of the strain over the cross section is no longer valid as the beam s elements are not of equal length, even though the assumptions of the plane cross section remaining plane after deformation is valid, as we assumed for the straight beam case. This citation can be found at university of Washington s department of mechanical engineering website [13]. Since the stress during deformation is proportional to the strain of the beam, then 22

27 Stress distribution Centroidal Axis A A R R n r A p M Neutral Axis R n r p A θ θ R = Radius of curvature R n = Radius of Neutral axis r = Radius of general fiber in the beam M = Bending moments θ = Cross sectional rotations Figure 4.1: Bending stress in curved beams the elastic stress of a curved beam is not promotional to the distance of the element from the neutral axis. For the same reason, the neutral axis in a curved beam does not pass through the centroid section. We define the geometry of the curved beam with reference to (Oden, [5]). Consider a beam, having a constant cross section and a constant radius of curvature in the plane of bending. The geometry of the curved beam is defined by establishing two orthogonal coordinates systems: one fixed coordinate system (x, y, z), with its origin located on a convenient cross-section, and another curvilinear system (s, y, z) in which s is tangent to the curve, y is a radial coordinate directed toward the center of the geometric axis, and z is directed normal to the plane of the beam. The assumptions taken are that, (a) The beam material is homogeneous, isotropic and linearly elastic. (b) The beam axis always lies in the plane of bending. (c) The Radius of curvature is constant and the cross section is constant and has symmetry about the plane of curvature, that is xy-plane. 23

28 (d) External loads act in the plane of curvature, therefore the plane of bending and the plane of curvature are the same. (e) Plane sections remain plane, and the cross section does not deform. We note that these assumptions are almost the same as the ones used for straight beams only that the beam is initially curved in this case. s s M y Vy y R V y + V y M s + Ms M N s s N s + M s My + M y z Mz + Mz V z + V s z V z M z Figure 4.2: Forces and displacements on an infinitesimal curved beam element 4.1 The Equilibrium Equations With reference to figure (4.2), the equilibrium equations at a point p in the deformed beam element have been derived and found to be dn s ds V y R + q 1 = 0 (4.1) dv y ds + N s R + q 2 = 0 dv z ds + q 3 = 0 For the equilibrium of the moments we get dm s ds M y R + m 1 = 0 (4.2) dm y ds + M s R V z + m 2 = 0 dm z ds +V y + m 3 = 0 24

29 Where N s,v y,v z are Normal force along s, and shear forces in in y and z direction, while M y,m z are bending moments around y and z axis, and M s is the twisting moment around the longitudinal axis, and {q i,m i, i = 1,...,3} are external loads and moments applied in the tangential, radial direction and z direction respectively. Detailed derivation are derived by (Oden, [5]), (Blake [14], pg. 575 ) and ( Lebeck, Knowlton, [15]) derives the equilibrium equations in polar coordinates. 4.2 Normal stresses in Curved Beams We follow the derivation given by (Oden, [5], pg. 64). Following similar derivation is (Wallerstein, [16], pg. 99). To be able to find the displacement of the beam, we find the stress-strain relations,assuming that the material is elastic and obeys the Hooke s law. First the derivation of the pure bending of the curved beam is presented, implying that we use the assumption that plane sections normal to the beam s axis before deformation remain plane and normal to this axis after deformation. With this assumption,the components of displacement in a direction normal to a cross section must satisfy the equation of a plane u = ĉ 0 + ĉ 1 y + ĉ 2 z (4.3) where ĉ 0,ĉ 1,ĉ 2 are constants. Let the geometry of element between the cross-sections of the rod be as in the diagram below. s s y y R Figure 4.3: Geometry of a beam s element vertical displacement of magnitude y Materials located at a radial distance y from the axis are of different length s y. Thus taking limit as s approaches zeros, ds 1 = ds y 1 y/r 25

30 From strain-displacement formula, ε s = v s y (4.4) Substituting this to (4.3), then for ε s = c 0 ds ds y + c 1 ds ds y y + c 2 ds ds y z = ds ds y (c 0 + c 1 y + c 2 z) c 0 = dĉ 0 ds, c 1 = dĉ 1 ds, c 2 = dĉ 2 ds Implying that ε s = 1 1 y (c 0 + c 1 y + c 2 z) (4.5) R Thus the normal stress is σ s = Eε s = E 1 y (c 0 + c 1 y + c 2 z) (4.6) R To determine c 0,c 1,c 2 then: N s = M z = M y = A A A σ s da = c 0 E σ s yda = c 0 E A σ s zda = c 0 E A 1 y z 1 y da + c 1 E R A 1 y da + c 2 E R A 1 y (4.7) da R y y 2 yz A 1 y da + c 1 E R A 1 y da + c 2 E R A 1 y da R z yz z 2 1 y da + c 1 E R A 1 y da + c 2 E R A 1 y da R The inertia terms, also derived by (A.O.Lebeck, J.S.Knowlton, [15]) can be defined as I y = I yz = I z = A A A z 2 1 y da (4.8) R yz 1 y da R y 2 1 y da R We note here when the dimensions of the cross section are small compared to the radius of curvature of the longitudinal axis, that is y << R, then the term y/r is negligibly small, and therefore the inertia terms are the same as for a straight 26

31 beam. Simplifying further, 1 A 1 y da = da + 1 yda + 1 y 2 R A R A R 2 A 1 y da (4.9) R = A + 1 yda + 1 R A R 2 I z z A 1 y da = zda + 1 yz R A R A 1 y da = zda + 1 R A R I yz y A 1 y da = yda + 1 y 2 R A R A 1 y da = yda + 1 R A R I y Since the origin of the coordinate system is at the centroid of the section, A yda = zda = 0 A Therefore equations (4.7) reduces to N s E M z E M y E = Solving the equations, ( A + I ) z R 2 c 0 + I z R c 1 + I yz R c 2 (4.10) = I yz R c 0 + I yz c 1 + I y c 2 = I z R c 0 + I z c 1 + I yz c 2 Ec 0 = N s A M z AR Ec 1 = M zi y M y I yz I y I z I 2 yz Ec 2 = M zi y M y I yz I y I z I 2 yz N s RA + M z AR 2 (4.11) Therefore we get the normal stress as σ s = N s A M z RA + M zi y M y I yz I y I z I 2 yz y 1 y R + M zi y M z I yz I y I z I 2 yz z 1 y R (4.12) The term N s A M z RA represents a uniform normal stress over the section, while the rest of the terms represent a non uniform distribution of stress. Lack of symmetry is represented by the terms I yz, which we will equate to zero in this case. (Oden, [5], Wallerstein, [16]). We refereed to 27

32 4.3 Deflection of a Curved Beam Having noted down the formulation of the normal stress, the corresponding strains, are formulated, so as to calculate the displacements of the beam after deformation through the strain-displacement relations. Let any point on the cross-section of the beam be displaced from its unstrained position through small components of displacement v and u representing the tangential and radial displacements of points on the cross section, and w represent the displacement in a direction normal to the plane of the rod, with the previous assumptions that plane cross sections remain plane after deformation still holding. From (4.4), ε s = v s y (4.13) The displacements can be described as in the following figure (4.4) described as in (Oden, [5], pg 132) (a.)tangential displacement (b)radial displacement (c.)cross-sectional displacements and rotations of the cross-sectional plane around the y-axis (d)about the z-axis. Using the principle of superposition we can get the overall tangential strain (4.13) as follows. From (a), we can see that the strain as a result of tangential displacement of the centroidal axis, while ignoring higher order terms is just the change of v, along s dv ds (4.14) The radial strain is as a result of the difference between the original length of material at the neutral axis to a distance y below the axis. see (Timoshenko,Goodier, [4]). Therefore the radial strain is s[1 y/(r u)] s(1 y/r) s(1 y/r) = y ( 1 1 y/r R u 1 ) u R R (4.15) 28

33 s s s u R u u v (a) v + dv/ds s (b) du/ds 2 2 du/ds + d u/ds s z dw/ds y s z s y du/ds (c) dw/ds (d) 2 2 dw/ds + dw/d s s Figure 4.4: (a.)tangential displacement (b.)radial displacement(c.)cross-sectional displacements and rotations of the cross-sectional plane around the y-axis (d)about the z-axis. The strains from (c) and (d) are as a result of the rotation of sections owing to change in u and w, thus from the diagrams (c)and(d), the rotations of the crosssections is y du ds zdw ds Where the strain is calculated to be ( du ) ds s ds zdw = ds ds y 1 1 y/r ( ) y d2 u ds 2 w zd2 ds 2 (4.16) Thus adding (4.14), (4.15) and (4.16), then we get dv ds = σ s E u ( d 2 R u ds 2 + u ) y R 2 1 y R = N s AE M z RAE + M zi y M y I yz E ( ) I y I z Iyz 2 d2 w ds 2 y 1 y R z 1 y R + M yi z M z I yz E ( ) I y I z Iyz 2 z 1 y R (4.17) Equating the like y/(1 y/r) and (z/(1 y/r)) on each side we get the equations of the elastic curve dv ds u R = N s AE M z RAE (4.18) 29

34 d2 u ds 2 u R 2 = M zi y M y I yz E ( ) I y I z Iyz 2 (4.19) d2 w ds 2 = M yi z M z I yz E ( ) I y I z Iyz 2 (4.20) Note The last equation describing the out of plane deformation is not considered here as we are dealing with plane deformation, as we will see below. 4.4 Shear Deformation Shearing stress for a symmetric circular curved bar is given in (Oden, [5]) as τ ys = ( ) V y ˆQ z b(1 y R ) A I z AR (4.21) From Hooke s law, we have the relation γ ys = τ ( ) ys G = V y ˆQ z b(1 y R )G A I z AR (4.22) Where y ˆQ = A 1 y/r da A, element cross sectional area, with dimensions b of A parallel to the z- axis. With reference to figure (4.4) the normal displacement due to the rotations of the cross-sections, owing to change in u and w, about the y axis, as a result of change in curvature, produces the displacement y du ds zdw ds Since we assumed that there are no out-of -plane displacement (in the z - direction), then w = 0. Therefore the displacement is v = y du b ds for u b the deflection due to bending. Thus ( ) V y ˆQ z b(1 y R )G A = y 2 u b I z AR s y u b s + u s (4.23) 30

35 The first term on the right side represent a slight warping of the cross section due to shear deformation. On the assumption that planes remain plane, for which τ ys was derived, then this term is small and can be neglected. ( ) V y ˆQ z b(1 y R )G A = du b I z AR ds + du ds (4.24) But the left side of the equation is a function of both s and y. This contradicts the assumption that planes remain normal after deformation, as u b is a function of s only. To remedy this we replace τ sy by uniform stress distribution and account for its variation with y by introducing a shear correction factor κ = β ˆβ. Thus (4.24) gives the shear deformation. That is Where V y du b ds + du ds = κ AG(1 y R ) (4.25) β = ˆQA I z b and ˆβ = A Rb θ + du ds = κ ( ) Vy AG 1 y R (4.26) If we assume that the cross section is symmetric with respect to y, then I yz. M y, and w are equal to zero, because of our choice of external loading to be in the xy plane. The strain-displacement equations describing the deformation of a shearable curved beam reduces to ( dv EA ds u ) = N s M z R R (4.27a) ( EI z d2 u ds 2 u ) R 2 = M z (4.27b) ( AG θ + du ) ( ) Vy = κ ds 1 y R (4.27c) Where EI z the bending rigidity, G is the shear modulus of elasticity, A is the cross sectional area of the curved element, and κ the shear correction factor. The correction factor arise from the fact that by admitting a nonzero shearing strain, and with the assumption of plane sections remaining plane leads to a 31

36 constant shear stress through the curved beam thickness. Therefore a shear correction factor is used to account for the difference in the constant state of shear stress and the more realistic parabolic variation of shear stress. We notice the difference between the displacement for a straight beam, and a curved beam is that (a) Due to the initial curvature of the beam, the terms are multiplied by 1/R, where R is the radius of curvature. (b) The tangential deformation, the bending and the rotation are intrinsically coupled, unlike the case for straight beams, hence efficient transfer of loads than for a straight beam because the transfer is affected by bending, shear, and membrane action. (a) However, since we are considering beams, whose longitudinal direction is much larger compare to transverse directions, then the term y/r found in the inertia terms is negligible, thus the inertia terms are the same as for the straight beams. 4.5 Finite Element Implementation Once again, the element kinematics of a plane beam is completely defined if the following functions are given: the axial displacement v 1, the transverse displacement u 2 and the cross section rotation θ. The geometry of the curved beam element in figure (4.5) is as described by (Friedman, Kosmatka, [17]) At a typical node, three variables d i = {v i,u i,θ i } are used to define the element behavior. The corresponding tangential ε, bending χ and shear strains γ respectively are in the form: ε = dv ds u R (4.28a) χ = d2 u ds 2 dv Rds (4.28b) γ = θ + du ds (4.28c) 1 The use of axial and tangential displacements are used interchangeably, and is assumed to mean the same displacement 2 The use of the transverse and Radial displacements are used interchangeably, and is assumed to refer to the same displacement 32

37 v1 s, v θ1 θ 2 y, u u 1 R u 2 v 2 Figure 4.5: Geometry of a curved beam element with three degrees of freedom at each node. For a given loaded elastic body under geometrical constrains or boundary conditions, the potential energy of the deformed body assumes a stationary value, and it attains an absolute minimum when the displacement of the body are those of equilibrium configuration. We can find the total potential energy of the element by thus L (e) Π = U M +U B +U S (q 1 v + q 2 u + mθ)ds (4.29) 0 where U M is the membrane energy, U B the bending strain energy, and U S the shear strain energy, and q 1, q 2 and m the external tangential and radial loads respectively, and the distributed moment, while L (e) is the length of the element. see (Seok-Soon Lee, [18]). U M,U B, and U S are expressed by U M = L (e) 0 ( 1 dv 2 EA ds u ) 2 ds (4.30a) R U B = L (e) 0 ( 1 2 EI z d2 u ds 2 dv ) 2 ds (4.30b) Rds U S = L (e) 0 ( 1 2 κga θ + du ) 2 ds (4.30c) ds 33

38 4.5.1 Stiffness Matrix and Load Vector If we let the displacement vector be U, then following the Galerkin method already introduced in section one, then we can define the finite element approximation of the displacements and the rotation of the cross sections with the finite element shape functions φ j as follows Ui h n nd = j=1 θi h n nd = j=1 φ j U h i j φ j θ h i j where n nd is the number of nodes in the discretization Timoshenko beam stiffness matrix With reference to (2.13), we saw that the cross sectional rotation for a Timoshenko beam are independent of the vertical displacement u, therefore they can be interpolated independently,implying that the degrees of freedom are v, u and θ. Thus the axial, bending and shear strains are ε = dv ds u R (4.31a) χ = dθ ds dv Rds (4.31b) γ = θ + du ds (4.31c) The highest derivative in the displacements for the total potential energy functional in (4.29) with the strains as defined in (4.31) is C(1) continuous, therefore we only need C(0) continuous interpolation functions of the Timoshenko curved beam, Thus for the shape functions φ = φ, v u = θ φ φ φ φ φ φ 2 v 1 u 1 θ 1 v 2 u 2 θ 2 34

39 Introducing the non-dimensional coordinate system formulation 1 ξ 1 3 such that ξ = x L (4.32) then φ 1 = 1 (1 ξ ) (4.33) 2 φ 2 = 1 2 (1 + ξ ) Defining the vector of the unknown values as d, and since we have three degrees of freedom per node, that is v,u and θ then we can write the axial, bending, shear strains as ε = B a d χ = B b d γ = B s d respectively, for the axial, bending and shear - displacement matrices [ ] B a = φ 1 φ 1 R 0 φ 2 φ 2 R 0 [ ] B b = φ 1 R 0 φ 1 φ 2 0 R φ 1 [ ] B s = 0 φ 1 φ 1 0 φ 2 φ 2 Therefore the element stiffness matrix is a contribution from the membrane, bending, and the shearing stiffness such that where [K e ] = [K a ] + [K b ] + [K s ] (4.34) [K a ] = dt 2 [ K b] = dt 2 [K s ] = dt B at (EA)B a d L(e) 2 dξ B bt (EI)B b d L(e) 2 dξ B st (κga)b s d L(e) 2 dξ 3 The global and local domains description can be related by the mapping ξ : [x i,x i+1 ] [ξ 1,ξ 2 ] : ξ (x i ) = ξ 1, and ξ (x i+1 ) = ξ 2, for example for a linear two noded element. Thus in general, for ξ 1 ξ, and an n-noded element, the lagrangian shape functions of n 1 degree is φ i (ξ ) (e) n ( ) ξ ξ j = j=1, j i ξ i ξ j 35

40 The load vector is d T +1 1 q L i φ (e) j 2 dξ t = 3 j 3 + i F (e) = f t = d T +1 1 m L φ (e) j dξ t = 3 j for 1 j 2 and d is d = v 1 u 1 θ 1 v 2 u 2 θ 2 where q i represents external tangential and radial loads q 1,q 2, and m the external moment The stiffness matrix (4.34) represents the Timoshenko curved beam, with the inclusion of shear strain Euler-bernoulli beam stiffness matrix For the Euler-Bernoulli curved beam, shear strain is neglected, and therefore, since we are assuming the beam is inextensional, the relevant strains in this case are the axial strain ε, and the bending strain, χ, where the bending strain now depends only on the radial displacement, u. ε = dv ds u R χ = d2 u ds 2 u R 2 (4.35a) (4.35b) In choosing the shape for the distribution of the unknown displacement, the geometry of the element is defined using the nodal coordinates and the same shape functions which are used to interpolate the displacement. Since in Euler- Bernoulli theory we have the normality assumption, cross-sectional rotations directly depend on the gradient of the deflection, the vertical displacement of a beam must have continuous slope as well as continues deflection at any neighboring beam elements. The bending of the beam has a second derivative term in the bending strain, thus we must have C(1) continuity of u and its derivative, hence C(1) continuous elements. For the tangential displacement v, the highest derivative is first order, so we only 36

41 need C(0) continuous interpolation functions. Therefore the simplest approximation of u and v is v(s) = b 0 + b 1 s (4.36) u(s) = c 0 + c 1 s + c 2 s 2 + c 3 s 3 du ds (s) = c 1 + 2c 2 s + 3c 3 s 2 for constants b 0,b 1,c 0,c 1,c 2,c 3. The Euler-bernoulli beam element thus has v, and u as degrees of freedom. Cross sectional rotations depend on the gradient of deflection, therefore θ is not a free parameter, and hence we evaluate the deflection u and its slope du/ds at both nodes, u(0) = c 0 = u 1 (4.37) du ds (0) = c 1 = du 1 ds u(l) = c 0 + c 1 L + c 2 L 2 + c 3 L 3 = u 2 du ds (L) = c 1 + 2c 2 L + 3c 3 L 2 = du 2 ds Solving (4.37) for the constants c i in terms of u i and du i /ds and substituting into (4.36), and for shape functions φ i = φ i we get du 1 u(s) = φ 1 u 1 + φ 2 ds + du 2 φ 3 u 2 + φ 4 ds (4.38) Re-writing in non-dimensional coordinate system formulation, then we have u(ξ ) (e) = φ 1 (ξ )u (e) 1 + φ 2 (ξ ) L(e) 2 where 4 φ i i=1 are cubic hermitian shape functions, as below: ( ) du (e) + φ 3 (ξ )u (e) 3 ds + φ 4 (ξ ) L(e) 1 2 ( ) du (e) (4.39) ds 2 φ 1 = 1 ( 2 3ξ + ξ 3 ) 4 (4.40a) φ 2 = 1 ( 1 ξ ξ 2 ξ 3) 4 (4.40b) φ 3 = 1 ( 2 + 3ξ ξ 3 ) 4 (4.40c) φ 4 = 1 ( 1 ξ + ξ 2 + ξ 3) 4 (4.40d) 37

42 Thus for functions φ i = φ i defined in (4.33), and hermitian shape functions φ i = φ i defined in (4.40) respectively, we can represent v,u and du/dx as v u du/ds = φ φ φ 1 L φ φ 3 L φ φ 1 φ 2 L 2 0 φ 3 Therefore the axial strain displacement matrix is [ ] B a = φ 1 φ 1 R φ 2 R φ 2 φ 3 R φ 4 R [K (e) a ] = dt φ 4 L 2 v 1 u 1 (du/ds) 1 v 2 u 2 (du/ds) 2 B T a (EA)B a d L(e) dξ (4.41) 2 and the bending strain displacement matrix is [ ] B b = φ 1 R φ 1 φ 2 φ 2 R φ 3 φ 4 [K (e) b ] = dt Hence the element stiffness matrix is thus B T b (EI)B b d L(e) dξ (4.42) 2 [K e ] = [K a ] + [K b ] (4.43) = dt While the load vector is now φ [F e ] = d T q φ where d in this case is v 1 u 1 d = du 1 /ds v 2 u 2 du 2 /ds B T a (EA)B a d L(e) 2 dξ + dt 2 L (e) +1 dξ + dt q 2 1 B T a (EI)B a d L(e) 2 dξ 0 φ 1 φ 2 0 φ 3 φ 4 L (e) 2 dξ 38

43 4.6 Numerical Integration Evaluation of the integrals needed to determine the element characteristics can be tedious and possibility of having so many errors is high, therefore we have adopted Gauss-Legendre numerical integration rules through out.(owen, [10]) This can be in short defined as follows: If we have a function f (ξ ), and we wish to calculate the integral +1 Ī = f (ξ )dξ (4.44) 1 for pre-determined sampling position ξ = ξ, we sample the function f (ξ ), and multiply the results by some predetermined weights W i. Therefore for a p - point rule, Ī p = W 1 f (ξ ) + W 2 f (ξ ) W p f (ξ ) (4.45) We note that an n- point rule can integrate polynomial functions of degree 2n 1 or less exactly. Tables of sampling positions and weights can be found in many literatures. see for example (Stewart, pg. 169 [11]). 4.7 Displacement Transformation Matrix Since we have used a rectangular cartesian coordinates system, it is necessary to relate the displacements defined along the local element axes to a global coordinate system common to every element in the beam. 4 Let the displacements of the beam in the local and global coordinate systems be v, u, θ and X, Y, θ respectively, as shown in the figure (4.6) Then to determine the transformation matrix for the curved beam element, we first consider a transformation matrix in two dimensions of straight beam element, of length L. If the lateral and normal displacements are ν and υ, and if the angle that the element makes with the X direction is α, as shown in figure (4.7), then the relationship between the displacements ν,υ, θ and X, Y, θ is known 4 If elements matrices are derived based on the local polar coordinate system instead of the local Cartesian one, then the matrices coefficients are invariant for any curved beam element with constant radius of curvature and subtended angle. Therefore there is no need for transformation. 39