The TL Plane Beam Element: Formulation

Transcription

1 11 The T Plane Beam Element: Formulation 11 1

2 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION TABE OF CONTENTS Page 11.1 Introduction Beam Models Basic Concepts and Terminology Beam Mathematical Models Finite Element Models Shear ocking X-Aligned Reference Configuration Element Description Motion Displacement Interpolation Strain-Displacement Relations *Consistent inearization *Rigid Body Motion Check Arbitrary Reference Configuration Strain-Displacement Matrix Constitutive Equations Strain Energy Internal Force Vector Tangent Stiffness Matrix Material Stiffness Matrix Eliminating Shear ocking by RBF Geometric Stiffness Matrix Commentary on Element Performance Derivation Summary Notes and Bibliography Exercises

3 11.2 BEAM MODES Introduction In this Chapter the finite element equations of a geometrically nonlinear, two-node Timoshenko plane beam element are formulated using the Total agrangian (T) kinematic description. The derivation is more typical of the general case. It is still short, however, of the enormous complexity involved, for instance, in the FEM analysis of nonlinear three-dimensional beams or shells. In fact the latter are still doctoral thesis topics. In the formulation of the bar element in Chapter 9, advantage was taken of the direct expression of the axial strain in terms of reference and current element lengths. That shortcut bypasses the use of displacement gradients, and allows the reference configuration to be arbitrarily oriented. The simplification works equally well for straight bars in three-dimensional space. A more systematic but lengthier procedure is required with more complicated elements. The procedure requires going through the displacement gradients to construct a strain measure. Sometimes this measure is too complex and must be simplified while retaining physical correctness. Then work-conjugate stresses are introduced and paired with strains to form the strain energy function of the element. Repeated differentiations with respect to node displacements yield the expressions of the internal force vector and tangent stiffness matrix. Finally, a transformation to the global coordinate system may be required. In addition to giving a better picture of the general procedure just outlined, the beam element illustrates the treatment of rotational degrees of freedom Beam Models Basic Concepts and Terminology Beams represent the most common structural component found in civil and mechanical structures. Because of their ubiquity they are extensively studied, from an analytical viewpoint, in Mechanics of Materials courses. Such a basic knowledge is assumed here. The following material recapitulates definitions and concepts that are needed in the finite element formulation. A beam is a rod-like structural member that can resist transverse loading applied between its supports. By rod-like it is meant that one of the dimensions is considerably larger than the other two. This one is called the longitudinal dimension and defines the longitudinal direction or axial direction. Directions normal to the longitudinal direction are called transverse. The intersection of planes normal to the longitudinal direction with the beam are called cross sections, just as for bar elements. The beam longitudinal axis is directed along the longitudinal direction and passes through the centroid of the cross sections. 1. Beams may be used as isolated structures. But they can also be combined to form framework structures. This is actually the most common form of high-rise building construction. Individual beam components of a framework are called members, which are connected at joints. Frameworks can be distinguished from trusses by the fact that their joints are sufficiently rigid to transmit bending moments between members. 1 If the beam is built of several materials, as in the case of reinforced concrete, the longitudinal axis passes through the centroid of a modified cross section. The modified-area technique is explained in elementary courses of Mechanics of Materials 11 3

4 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION reference configuration motion current configuration Figure A geometrically nonlinear plane framework structure. In practical structures beam members can take up a great variety of loads, including biaxial bending, bidirectional shears, axial forces and even torsion. Such complicated actions are typical of spatial beams, which are used in three-dimensional frameworks and are subject to forces applied along arbitrary directions. A plane beam resists primarily loading applied in one plane and has a cross section that is symmetric with respect to that plane. Plane frameworks, such as the one illustrated in Figure 11.1, are assemblies of plane beams that share that symmetry. Those structures can be analyzed with twodimensional idealizations. A beam is straight if the longitudinal direction is a straight line. A beam is prismatic if its cross section is uniform. Only straight, prismatic, plane beams will be considered in this Chapter Beam Mathematical Models Beams are actually three-dimensional solids. One-dimensional mathematical models of plane beams are constructed on the basis of beam theories. All such theories involve some form of approximation that describes the behavior of the cross sections in terms of quantities evaluated on the longitudinal axis. More precisely, the element kinematics of a plane beam is completely defined by the two models described below if the following functions are given: Axial displacement u X (X) Transverse displacement u Y (X) (also called lateral displacement) Cross section rotation θ Z (X) θ(x): angle by which cross section rotates. 2 Here X denotes the longitudinal coordinate in the reference configuration. See Figure In the sequel we will drop the subscript Z from θ for brevity. Two beam mathematical models are commonly used in structural mechanics: Bernoulli-Euler (BE) Model. This model is also called classical beam theory or engineering beam theory, and is the one covered in elementary treatments of Mechanics of Materials. It accounts for bending moment effects on stresses and deformations. Transverse shear forces are recovered from equilibrium but their effect on beam deformations is ignored (more precisely: the strain energy due to shear stresses is neglected). The fundamental kinematic assumption is that cross sections 2 This angle suffices if the cross section remains plane, an assumption verified by both models described here. 11 4

5 11.2 BEAM MODES current configuration Current cross section θ Z(X) θ(x) motion Y, y X, x u (X) Y u (X) X reference configuration X Reference cross section Figure Definition of beam kinematics in terms of the three displacement functions: u X (X), u Y (X), and θ(x). The figure actually depicts the BE model kinematics. In the Timoshenko model, θ(x) is not constrained by normality, as illustrated in Figure current configuration finite element idealization of current configuration motion reference configuration finite element idealization of reference configuration Figure Idealization of a geometrically nonlinear beam member (as taken, for example, from a plane framework structure like the one in Figure 11.1) as an assembly of finite elements. remain plane and normal to the deformed longitudinal axis. This rotation occurs about a neutral axis parallel to Z that passes through the centroid of the cross section. Timoshenko Model. This model corrects the BE theory with first-order shear deformation effects. The key assumption is that cross sections remain plane and rotate about the same neutral axis, but do not remain normal to the deformed longitudinal axis. The deviation from normality is produced by a transverse shear stress that is assumed to be constant over the cross section. Both the BE and Timoshenko models rest on the assumptions of small deformations and linearelastic isotropic material behavior. In addition both models neglect any change in dimensions of the cross sections as the beam deforms. Either theory can account for geometrically nonlinear behavior due to large displacements and rotations, as long as the other assumptions hold. 11 5

6 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION 1 θ 2 (a) C (BE) model (b) C (Timoshenko) model θ 2 θ 1 θ 1 Y, y X, x u X1 u Y1 u X2 u Y Figure Two-node beam elements have six DOF, regardless of which model is used. u X1 u Y1 u X2 u Y Finite Element Models To carry out the geometrically nonlinear finite element analysis of a framework structure, beam members are idealized as an assembly of finite elements, as illustrated in Figure Beam elements used in practice have usually two end nodes. The i th node has three DOF: two node displacements u Xi and u Yi, and one nodal rotation θ i, positive counterclockwise in radians, about the Z axis. See Figure The cross section rotation from the reference to the current configuration is called θ in both models. In the BE model this is the same as the rotation ψ of the longitudinal axis. In the Timoshenko model, the shear distortion angle is γ = θ ψ, as shown in Figure The mean shear strain has the opposite sign: γ = γ = ψ θ, so as to make the shear strain e XY positive, as per the usual conventions of structural mechanics. Either the BE or the Timoshenko model may be used as the basis for the T beam element formulation. Superficially it appears that one should select the latter only when shear effects are to be considered, as in deep beams whereas the former is used for ordinary beams. But here a twist appear because of FEM considerations. This twist is one that has caused significant confusion among users over the past 25 years. See Appendix S for the tragicomical story of the topic. Although the Timoshenko beam model appears to be more complex because of the inclusion of shear deformation, finite elements based on this model are in fact simpler to construct! Here are the two key reasons: (i) Separate kinematic assumptions on the variation of cross-section rotations are possible, as made evident by Figure Mathematically: θ(x) may be assumed independently of u X (X) and u Y (X). As a consequence, two-node Timoshenko elements may use linear variations in both displacement and rotations. On the other hand, a two-node BE model requires a cubic polynomial for u Y (X) because the rotation θ(x) is not independent. (ii) The linear transverse displacement variation matches that commonly assumed for the axial deformation (bar-like behavior). The transverse and axial displacement assumptions are then said to be consistent. 11 6

7 11.2 BEAM MODES Angles are positive as shown. Note that _ γ = θ ψ whereas γ = γ = ψ θ direction of deformed cross section θ _ γ γ ψ 9 normal to reference beam axis X normal to deformed beam axis ψ _ // X (X = X) ds Figure Illustrates total and BE section rotations θ and ψ, respectively, in the plane beam Timoshenko model. The mean shear distortion angle is γ = θ ψ, but we take γ = γ = θ ψ as shear strain measure,to match usual sign conventions of structural mechanics. For elastic deformations of engineering materials γ << 1. Typical values for γ would be O(1 3 ) radians whereas rotations ψ and θ may be much larger, say 1-2 radians. The magnitude of γ is grossly exaggerated in the Figure for visualization convenience. This simplicity is even more important in geometrically nonlinear analysis, as strikingly illustrated by the elements contrasted in Figure Although as pictured there both elements have six degrees of freedom, the internal kinematics of the Timoshenko model is far simpler Shear ocking In the FEM literature, a BE-based model such as the one shown in Figure 11.4(a) is called a C 1 beam because this is the kind of mathematical continuity achieved in the longitudinal direction when a beam member is divided into several elements (cf. Figure 11.3). On the other hand, the Timoshenko-based element pictured in Figure 11.4(b) is called a C beam because both transverse displacements, as well as the rotation angle θ, preserve only C continuity. What would be the first reaction of an experienced but old-fashioned (i.e, never heard about FEM ) structural engineer on looking at Figure 11.6? The engineer would pronounce the C element unsuitable for practical use. And indeed the kinematics looks way wrong. The shear distortion grossly violates the basic assumptions of beam behavior. And indeed, a huge amount of shear energy would be required to keep the element straight as pictured. The engineer would be both right and wrong. If the two-node element of Figure 11.6(b) were constructed with actual shear properties and exact integration, an overstiff model results. This phenomenom is well known in the FEM literature and receives the name of shear locking. To avoid locking while retaining the element simplicity it is necessary to use certain computational devices that have nothing to do with physics. The most common are: 1. Selective integration for the shear energy. 2. Residual energy balancing. The second device will be used with only a cursory explanation at the end of this Chapter. For explanatory literature references see Notes and Bibliography and Appendix S. In this Chapter the C model will be used to illustrated the T formulation of a two-node, geometrically nonlinear beam element. 11 7

8 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION 2-node C 1 (cubic) element for Euler-Bernoulli beam model: plane sections remain plane and normal to deformed longitudinal axis (a) 2-node C linear displacement & rotations element for Timoshenko beam model: plane sections remain plane but not normal to deformed longitudinal axis (b) C 1 element with same DOFs Figure Contrasting kinematics of 2-node beam FEM models based on (a) BE beam theory, and (b) Timoshenko beam theory. These are called C 1 and C beam elements, respectively, in the FEM literature. Remark As a result of the application of the aforementioned devices the beam element behaves like a BE beam although the underlying model is Timoshenko s. This represents a curious paradox: shear deformation is used to simplify the kinematics, but then most of the shear will be removed to restore the correct physical behavior. 3 As a result, the name C element is more appropriate than Timoshenko element because capturing the actual shear deformation is not the main objective. Remark The two-node C 1 beam element is used primarily in linear structural mechanics. It is in fact the beam model used in the IFEM course [?, Chapter 12.) This is because some of the easier-construction advantages cited for the C element are less noticeable, while no artificial devices to eliminate locking are needed. The C 1 element is also called the Hermitian beam element because the shape functions are cubic polynomials specified by Hermite interpolation formulas X-Aligned Reference Configuration Element Description We consider a two-node, straight, prismatic C plane beam element moving in the (X, Y ) plane, as depicted in Figure 11.7(a). For simplicity in the following derivation the X axis system is initially aligned with the longitudinal direction in the reference configuration, with origin at node 1. This assumption is relaxed in the next section, once invariant strain measured are obtained. The reference element length is. The cross section area A and second moment of inertia I zz with respect to the neutral axis 4 are defined by the area integrals A = da, YdA=, I = Y 2 da, (11.1) A A A 3 The FEM analysis of plates and shells is also rife with such two wrongs make a right paradoxes. 4 For a plane prismatic beam, the neutral axis at a particular section is the intersection of the cross section plane X = constant with the plane Y =. 11 8

9 11.3 X -AIGNED REFERENCE CONFIGURATION (a) y y P y C Section rotation is θ=ψ+γ=ψ γ _ θ 1 γ 1 P(x,y) Y, y 1 x x Y, y P x C(X+u,u ) X Y X X u X(X) = u P X X C u P (X,Y) XP u XC C (X) Y Y P 1 C (X,) 2 X, x 1 2 ψ C θ 2 γ2 ψ ψ 2 //X C(x,y ) C C reduction to 1D C (b) 1 C θ(x) XC Y 2 ψ u (X) = u X, x YC Figure agrangian kinematics of C beam element with X-aligned reference configuration: (a) plane beam moving as a 2D body; (b) reduction of motion description to 1D as measured by coordinate X. In the current configuration those quantities become A, I zz and, respectively, but only the current length is frequently used in the T formulation. The material remains linearly elastic, with Young s modulus E relating the stress and strain measures defined below. As in the previous Chapters the element identification superscript e will be omitted to reduce clutter until it is necessary to distinguish elements within structural assemblies. The element has the six degrees of freedom depicted in Figure These degrees of freedom and the associated node forces are collected in the node displacement and node force 6-vectors u = [ u X1 u Y 1 θ 1 u X2 u Y 2 θ 2 T, f = [ f X1 f Y 1 f θ1 f X2 f Y 2 f θ2 T. (11.2) The external loads applied to the nodes will be assumed to be conservative Motion The kinematic assumptions of the Timoshenko model element have been outlined in Basically they state that cross sections remain plane upon deformation, but not necessarily normal to the deformed longitudinal axis. In addition, changes in cross section geometry are neglected. To work out the agrangian kinematics of the element shown in Figure 11.6(a), we study the motion of a generic particle located at P (X, Y ) in the reference configuration, which moves to P(x, y) in the current configuration. The projections of P and P on the neutral axis, along the cross sections at C and C are called C (X, ) and C(x C, y C ), respectively. The exact kinematics is [ [ [ x xc Y (sin ψ+ sin γ cos ψ) xc Y [sin θ + (1 cos γ)sin ψ = = y y C + Y (cos ψ sin γ sin ψ) y C + Y [cos θ + (1 cos γ)cos ψ. (11.3) in which ψ + γ has been replaced by θ. But x C = X + u XC and y C = u YC. From now we denote u XC and u YC simply as u X and u Y, respectively. Assume for simplicity that γ is constant over the 11 9

10 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION element. Expand (11.3) in Taylor series in γ up to O(γ 4 ): [ [ x u X + X + 1 = 2 Y γ 3 cos θ Y ( γ γ 4 ) sin θ y u Y + Y ( γ γ 4 ) cos θ Y γ 3 sin θ (11.4) If we assume the shear distortion is very small, setting γ in (11.4) gives the simplified agrangian description of the motion [ [ x X + u = X Y sin θ, (11.5) y u Y + Y cos θ in which u X, u Y and θ are functions of X only. This form, which is correct up to O(γ 2 ), can be directly obtained from the last of (11.3) on replacing 1 cos γ by. This concludes the reduction to a one-dimensional model, as sketched in Figure 11.7(b). For future use it is convenient to define an extended internal displacement vector w, and its X derivative: [ [ [ u X (X) w = u Y (X), w = dw dux /dx u X θ(x) dx = du Y /dx = u Y, (11.6) dθ/dx θ in which primes denote derivatives with respect to X. The derivative θ will be also denoted as κ, which has the meaning of beam curvature in the current configuration. Also useful are the following differential relations 1 + u X = s cos ψ, u Y = s sin ψ, s = ds dx = (1 + u X )2 + (u Y )2, (11.7) in which ds is the differential arclength in the current configuration; see Figure If u X and u Y are constant over the element, s = /, 1 + u X = cos ψ/, u Y = sin ψ/. (11.8) Remark Replacing 1 cos γ by zero in (11.3) is equivalent to saying that Y (1 cos γ)can be neglected in comparison to other cross section dimensions. This is consistent with uncertainties in cross section changes. Such changes would depend on the Y -normal stress as well as Poisson s ratio effects. The motion (11.5) has the virtue of being purely kinematic (i.e., no material properties appear) and of leading to (exactly) e YY = Displacement Interpolation For a 2-node C element it is natural to express the displacements and rotation functions as linear in the node displacements: w = [ [ u X (X) 1 ξ 1+ ξ u Y (X) = ξ 1+ ξ θ(x) 1 ξ 1+ ξ 11 1 u X1 u Y 1 θ 1 u X2 u Y 2 θ 2 = Nu, (11.9)

11 11.3 X -AIGNED REFERENCE CONFIGURATION in which ξ = 2X/ is the isoparametric coordinate that varies from ξ = 1 at node 1 to ξ = 1 at node 2, and N as usual denotes the shape function matrix. Differentiating this expression with respect to X yields the displacement gradient interpolation: [ u X w = u Y θ = 1 [ u X1 u Y 1 θ 1 u X2 u Y 2 θ 2 = N u. (11.1) Strain-Displacement Relations The deformation gradient matrix of the motion (11.5) is [ x x [ F = X Y 1 + u = X Y κ cos θ sin θ y y u Y X Y Y κ sin θ cos θ, (11.11) in which primes denote derivatives with respect to X, and κ = θ is the curvature. The displacement gradient matrix is [ u G = F I = X Y κ cos θ sin θ u, (11.12) Y Y κ sin θ cos θ 1 From the definition (8.1), the plane portion of the Green-agrange (G) strain tensor follows as [ exx e e = XY = 1 2 (FT F I) = 1 2 (G + GT ) GT G. (11.13) e YX e YY This expands to the component form [ e = 1 2(u X Y κ cos θ)+ (u X Y κ cos θ)2 + (u Y Y κ sin θ)2 (1 + u X ) sin θ + u Y cos θ 2 (1 + u X ) sin θ + u Y cos θ (11.14) It is seen that the only nonzero strains are the axial strain e XX and the shear strain e XY +e YX = 2e XY, whereas e YY vanishes. Through the consistent-linearization techniques described in below, it can be shown that under the small-strain assumptions made precise therein, the axial strain e XX can be replaced by the simpler form e XX = (1 + u X ) cos θ + u Y sin θ Y κ 1, (11.15) in which all quantities appear linearly except θ. The nonzero axial and shear strains will be arranged in the strain vector e = [ e1 e 2 = [ exx 2e XY = [ (1 + u X ) cos θ + u Y sin θ Y θ [ 1 e Y κ (1 + u X ) sin θ + u Y cos θ =. γ (11.16)

12 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION The three strain quantities introduced in (11.16) e = (1 + u X ) cos θ + u Y sin θ 1, γ = (1 + u X ) sin θ + u Y cos θ, κ = θ, (11.17) characterize axial strains, shear strains and curvatures, respectively. These are collected in the following generalized strain vector: h = [ e γ κ T. (11.18) Because of the assumed linear variation in X of u X (X), u Y (X) and θ(x), e and γ only depend on θ whereas κ is constant over the element. Making use of (11.8) one can express e and γ in the geometrically invariant form 1 + e = s cos(θ ψ) = cos γ, γ = s sin(θ ψ) = sin γ (11.19) In theory one could further reduce e to / and γ to γ/, but those simplifications would actually complicate the strain variations taken in the following Section *Consistent inearization The derivation of the consistent linearization (11.15) is based on the polar decomposition analysis of the deformation gradient. Introduce the matrix [ cos α sin α Ω(α) =, (11.2) sin α cos α which represents a two-dimensional rotation (about Z) through an angle α. Since Ω is an orthogonal matrix, Ω T = Ω 1. The deformation gradient (11.11) can be written [ s [ Y θ F = Ω(ψ) + Ω(θ). (11.21) 1 where s is defined in (11.7) Premultiplying both sides of (11.21) by Ω( θ) gives the modified deformation gradient [ s F = Ω( θ)f = cos(θ ψ) Y θ s (11.22) sin(θ ψ) 1 Now the G strain tensor 2e = F T F I does not change if F is premultiplied by the orthogonal matrix Ω because F T Ω T ΩF = F T F. Consequently 2e = F T F I. But if the strains remain small, as it is assumed in the Timoshenko model, the following are small quantities: (i) s 1 = (/ ) 1 because the axial strains are small; (ii) Y θ = Y κ because the curvature κ is of order 1/R, R being the radius of curvature, and R >> Y according to beam theory because Y can vary only up to the cross section in-plane dimension; (iii) γ = ψ θ, which is the mean angular shear deformation. Then F = I + + higher order terms, where is a first-order linearization in the small quantities s 1, Y θ and γ = ψ θ. It follows that e = 1 2 ( + T ) + higher order terms. (11.23) Carrying out this linearization one finds that e XY and e YY do not change, but that e XX simplifies to (11.15). It can also be shown that 2e XY = γ. = γ within the order of approximation of (11.23)

13 *Rigid Body Motion Check 11.4ARBITRARY REFERENCE CONFIGURATION It is instructive to check whether the foregoing displacements and strain fields can correctly represent rigid body motions (RBM). An arbitrary RBD of the element can be represented by a rigid translation {u XR, u YR }, along X and Y, resectively, and a rotation of angle θ R about the translated position. Denoting the current coordinates by {x R, y R } the motion is given by [ [ [ xr cos θr sin θ = R X + u XR. (11.24) y R sin θ R cos θ R Y + u YR The displacements are u X = x R X R and u Y = y R Y. The X derivatives are u X = cos θ R 1, u Y = sin θ, θ R = κ R =. (11.25) This motion is inserted into the following strain measures: e (1) XX = (u X Y κ cos θ)+ 1 2 (u X Y κ cos θ) (u Y Y κ sin θ)2, e (2) XX = (1 + u X ) cos θ + u Y sin θ Y κ 1, e (1) XY = (1 + u X ) sin θ + u Y cos θ e (2) XY = e(1) XY 2 cos θ(u Y βy κ sin θ). (11.26) in which β is arbitrary. In (11.27) e (1) XX and e(1) XY are those in the G strain tensor (11.13). Normal strain e(2) XX is the result of the linearization (11.15), while the shear strain e (2) XX includes a correction not derivable from the G formulas. Inserting the RBM described by (11.24) and (11.25) into the shear and normal strain measures (11.27) one obtains e (1) XX =, e(2) XX = 2 sin2 θ R, e (1) XY = 1 2 sin(2θ R), e (2) XY =. (11.27) Conclusion: the strain measures e = e (2) XX and γ = e(1) XY used in the strain vector (11.18) do not capture exactly RBM as the rotation angle θ R gets large. For that to be achieved, it would be necessary to pick e = e (1) XX and γ = e (2) XY. There is no displacement field, however, that produces both of these as components of the G strain tensor. See also Exercise Arbitrary Reference Configuration In the general case the reference configuration C of the element is not aligned with X. The longitudinal axis X forms an angle ϕ with X, as illustrated in Figure 11.8(a). The six degrees of freedom of the element are indicated in that Figure. Note that the section rotation angle θ is measured from the direction Ȳ, normal to X, and no longer from Y as in Figure Given the node coordinates (X 1, Y 1 ) and (X 2, Y 2 ), the reference angle ϕ is determined by cos ϕ = X 21 /, sin ϕ = Y 21 /, X 21 = X 2 X 1, Y 21 = Y 2 Y 1, 2 = X Y (11.28) The angle φ = ψ +ϕ formed by the current longitudinal axis with X (see Figure 11.5) is determined by cos φ = cos(ψ + ϕ) = x 21 sin φ = sin(ψ + ϕ) = y 21, (11.29) in which x 21 = x 2 x 1, y 21 = y 2 y 1 and may be obtained from the motion as x 21 = X 21 + u X2 u X1, y 21 = Y 21 + u Y 2 u Y 1, 2 = x y2 21. (11.3) 11 13

14 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION (a) θ 2 2(x,y ) 2 2 // γy _ φ = ψ+ϕ ψ // γx _ ϕ // γx (b) M V N // γy _ C C u Y2 Y, y θ 1 X, x Y _ γ 1(x,y ) 1 1 C u Y1 u X2 2(X,Y ) 2 2 ϕ X _ γ // γx C V M N 1(X,Y ) 1 1 u X1 Figure Plane beam element with arbitrarily oriented reference configuration: (a) kinematics, (b) internal stress resultants. Solving the foregoing trigonometric relations for ψ gives cos ψ = X 21x 21 + Y 21 y 21 = X 21(X 21 + u X2 u X1 ) + Y 21 (Y 21 + u Y 2 u Y 1 ), sin ψ = X 21y 21 Y 21 x 21 = X 21(Y 21 + u Y 2 u Y 1 ) Y 21 (X 21 + u X2 u X1 ). (11.31) It follows that sin ψ and cos ψ are exactly linear in the translational node displacements. This property simplifies considerably the calculations that follow Strain-Displacement Matrix For the generalized strains it is convenient to use the invariant form (11.19), which does not depend on ϕ. The variations δe, δγ and δκ with respect to nodal displacement variations are required in the formation of the strain displacement relation δh = B δu. To form B we take partial derivatives of e, γ and κ with respect to node displacements. Here is a sample of the kind of calculations involved: e = [ cos(θ ψ)/ 1 = [(cos θ cos ψ + sin θ sin ψ)/ 1 u X1 u X1 u X1 = X 21 cos θ + Y 21 sin θ cos ϕ cos θ + sin ϕ sin θ 2 = = cos ω (11.32), in which ω = θ + ϕ. Use was made of (11.31) in a key step. These derivatives were checked with Mathematica. Collecting all of them into matrix B: [ B = 1 cos ω sin ω N 1 γ cos ω sin ω N 2 γ sin ω cos ω N 1 (1 + e) sin ω cos ω N 2 (1 + e). (11.33)

15 11.6 INTERNA FORCE VECTOR Here N 1 = (1 ξ)/2 and N 2 = (1 + ξ)/2 are abbreviations for the element shape functions (caligraphic symbols are used to lessen the chance of clash against axial force symbols) Constitutive Equations Because the beam material is assumed to be homogeneous and isotropic, the only nonzero PK2 stresses are the axial stress s XX and the shear stress s XY. These are collected in a stress vector s related to the G strains by the linear elastic relations s = [ sxx s XY = [ s1 s 2 [ [ s = 1 + Ee 1 s s2 + Ge = s 2 [ [ E e1 = s + Ee, (11.34) G e 2 in which E is the modulus of elasticity and G is the shear modulus. We introduce the initial stress (prestress) resultants N = s1 da, V = s2 da, M = Ys1 da. (11.35) A A A These define the axial forces, transverse shear forces and bending moments, respectively, in the reference configuration. We also define the stress resultants N = N + EA e, V = V + GA γ, M = M + EI κ. (11.36) These represent axial forces, tranverse shear forces and bending moments in the current configuration, respectively, defined in terms of PK2 stresses. See Figure 11.8(b) for signs. These are collected in the stress-resultant vector z = [ N V M T. (11.37) Strain Energy As in the case of the T bar element, the total potential energy = U P is separable because P = λq T u. The strain (internal) energy is given by [ U = (s ) T e et Ee [ dv = (s 1 e 1 + s2 e 2) (Ee2 1 + Ge2 2 ) dad X. (11.38) V A Carrying out the area integrals while making use of (11.34) through (11.37), U can be written as the sum of three length integrals: U = (N e EA e 2 ) d X + (V γ GA γ 2 ) d X + (M κ EI κ 2 ) d X, (11.39) The three terms in (11.39) define the energy stored through bar-like axial deformations, shear distortion and pure bending, respectively

16 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION Internal Force Vector The internal force vector can be obtained by taking the first variation of the internal energy with respect to the node displacements. This can be compactly expressed as ( ) δu = N δe + V δγ + M δκ d X = z T δh d X = z T B d X δu. (11.4) Here h and B are defined in equations (11.18) and (11.33), respectively, whereas z collects the stress resultants in C as defined in (11.35) through (11.37). Because δu = p T δu, weget p = B T z d X. (11.41) This expression may be evaluated by a one point Gauss integration rule with the sample point at ξ = (beam midpoint). et θ m = (θ 1 + θ 2 )/2, ω m = θ m + ϕ, c m = cos ω m, s m = sin ω m, e m = cos(θ m ψ)/ 1, γ m = sin(ψ θ m )/, and B m = B ξ= = 1 c m s m 1 2 γ m c m s m 1 2 γ m 1 s m c m 2 1 (1 + e m ) s m c m 2 (1 + e m ) (11.42) 1 1 where subscript m stands for at beam midpoint. Then p = Bm T z = Tangent Stiffness Matrix c m s m 1 2 γ m c m s m 1 2 γ m s m c m 1 2 (1 + e m ) s m c m 1 2 (1 + e m ) 1 1 T [ N V M (11.43) The first variation of the internal force vector (11.41) defines the tangent stiffness matrix ( δp = B T δz + δb T z ) d X = (K M + K G )δu = K δu. (11.44) This is again the sum of the material stifness K M and the geometric stiffness K G Material Stiffness Matrix The material stiffness comes from the variation δz of the stress resultants while keeping B fixed. This is easily obtained by noting that [ δn [ EA [ δe δz = δv = GA δγ = S δh, (11.45) δm EI δκ where S is the diagonal constitutive matrix with entries EA, GA and EI. Because δh = B δu, the term B T δz becomes B T SB δu = K M δu, whence the material matrix is K M = B T SB d X. (11.46) 11 16

17 11.7 TANGENT STIFFNESS MATRIX This integral is evaluated by the one-point Gauss rule at ξ =. Denoting again by B m the matrix (11.43), we find K M = Bm T SB m d X = K a M + Kb M + Ks M (11.47) where K a M, Kb M and Ks M are due to axial (bar), bending, and shear stiffness, respectively: cm 2 c m s m c m γ m /2 c 2 m c m s m c m γ m /2 c K a M = EA m s m sm 2 γ m s m /2 c m s m sm 2 γ m s m /2 c m γ m /2 γ m s m /2 γm 2 2 /4 c mγ m /2 γ m s m /2 γm 2 2 /4 cm 2 c m s m c m γ m /2 cm 2 c m s m c m γ m /2 c m s m sm 2 γ m s m /2 c m s m sm 2 γ m s m /2 c m γ m /2 γ m s m /2 γm 2 2 /4 c mγ m /2 γ m s m /2 γm 2 2 (11.48) /4 K b M = EI 1 1 (11.49) 1 1 sm 2 c m s m a 1 s m /2 sm 2 c m s m a 1 s m /2 c m s m cm 2 c m a 1 /2 c m s m cm 2 c m a 1 /2 K s M = GA a 1 s m /2 c m a 1 /2 a1 22 /4 a 1 s m /2 c m a 1 /2 a1 22 /4 sm 2 c m s m a 1 s m /2 sm 2 c m s m a 1 s m /2 c m s m cm 2 c m a 1 /2 c m s m cm 2 c m a 1 /2 a 1 s m /2 c m a 1 /2 a1 22 /4 a 1 s m /2 c m a 1 /2 a1 22 (11.5) /4 in which a 1 = 1 + e m Eliminating Shear ocking by RBF How good is the nonlinear material stiffness given by (11.49) and (11.5)? If evaluated at the reference configuration aligned with the X axis, c m = 1, s m = e m = γ m =, and we get EA EA GA 1 2 GA GA 1 2 GA 1 2 K M = GA EI GA 1 2 GA EI GA EA EA GA 1 2 GA GA 1 2 GA 1 2 GA EI GA 1 2 GA EI GA (11.51) This is the well known linear stiffness of a shear-locked C beam. As noted in the discussion of , this element does not perform as well as a C 1 beam when the beam is thin because too much strain energy is taken by shear in the antisymmetric bending mode. The following substitution 11 17

18 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION device, introduced by MacNeal, 5 removes that deficiency in a simple way. The shear rigidity GA is formally replaced by 12EI / 2, and magically (11.51) becomes ˆK M = EA EA 12EI 6EI EI 6EI 3 2 6EI 2 EA 12EI 3 6EI 2 4EI 6EI 6EI 2 2 2EI EA 12EI 3 6EI 2 2EI 6EI 2 4EI. (11.52) This is the well known linear stiffness matrix of the C 1 (Hermitian) beam based on the Bernoulli- Euler model. That substitution device is called the residual bending flexibility (RBF) correction. 6 Its effect is to get rid of the spurious shear energy due to the linear kinematic assumptions. If the RBF is formally applied to the nonlinear material stiffness one gets ˆK M = K a M + ˆK Mb, where K a M is the same as in (11.5) (because the axial stiffness if not affected by the substitution), whereas K b M and Ks M merge into ˆK b M = EI 3 12sm 2 12c m s m 6a 1 s m 12sm 2 12c m s m 6a 1 s m 12c m s m 12cm 2 6c m a 1 12c m s m 12cm 2 6c m a 1 6a 1 s m 6c m a 1 a 2 2 6a 1 s m 6c m a 1 a sm 2 12c m s m 6a 1 s m 12sm 2 12c m s m 6a 1 s m 12c m s m 12cm 2 6c m a 1 12c m s m 12cm 2 6c m a 1 6a 1 s m 6c m a 1 a 3 2 6a 1 s m 6c m a 1 a 2 2 in which a 1 = 1 + e m, a 2 = 4 + 6e m + 3e 2 m and a 3 = 2 + 6e m + 3e 2 m. Remark MacNeal actually proposed the more refined substitution (11.53) replace 1 GA by 1 GA s EI (11.54) where GA s is the actual shear rigidity; that is, A s is the shear-reduced cross section studied in Mechanics of Materials. The result of (11.54) is the C 1 Hermitian beam corrected by shear deformations computed from equilibrium considerations. (This beam is derived in Chapter 13 of IFEM Notes [263.) If the shear deformation is negligible, the right hand side of (11.54) is approximately 2 /(12EI ), which leads to the substitution used above.. 5 See reference in Notes and Biblioghraphy. 6 RBF can be rigurously justified through the use of a mixed variational principle, or through a flexibility calculation

19 11.7 TANGENT STIFFNESS MATRIX Geometric Stiffness Matrix The geometric stiffness K G comes from the variation of B while the stress resultants in z are kept fixed. To get a closed form expression it is convenient to pass to indicial notation, reverting to matrix notation later upon index contraction. et the entries of K G, B, u and z be denoted as K Gij, B ki, u j and z k, where indices i, j and k range over 1 6, 1 6, and 1 3, respectively. Call A j = B/ u j, j = 1,...6. Then using the summation convention, K Gij δu j = δb T z dx = B ki δu j z k dx = A j ki u z k dxδu j, (11.55) j whence K Gij = z k A j ki d X, (11.56) Note that in carrying out the derivatives in (11.56) by hand one must use the chain rule because B is a function of e, γ and θ, which in turn are functions of the node displacements u j. To implement this scheme we differentiate B with respect to each node displacement in turn, to obtain: A 1 = B = 1 u X1 A 2 = B = 1 u Y 1 A 3 = B θ 1 = N 1 A 4 = B = 1 u X2 A 5 = B = 1 u Y 2 A 6 = B θ 2 = N 2 [ N1 sin ω N 2 sin ω N 1 cos ω N 2 cos ω [ N1 cos ω N 2 cos ω N 1 sin ω N 2 sin ω [ sin ω cos ω N1 (1 + e) sin ω cos ω N 2 (1 + e) cos ω sin ω N 1 γ cos ω sin ω N 2 γ [ N1 sin ω N 2 sin ω N 1 cos ω N 2 cos ω [ N1 cos ω N 2 cos ω N 1 sin ω N 2 sin ω [ sin ω cos ω N1 (1 + e) sin ω cos ω N 2 (1 + e) cos ω sin ω N 1 γ cos ω sin ω N 2 γ,,,,,. (11.57) To restore matrix notation it is convenient to define W Nij = A j 1i, W Vij = A j 2i, W Mij = A j 3i, (11.58) as the entries of three 6 6 weighting matrices W N, W V and W M that isolate the effect of the stress resultants z 1 = N, z 2 = V and z 3 = M. The first, second and third row of each A j becomes 11 19

20 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION Node Displacements u Eqs (1.5), (1.8) Element displacement field w = [u, u, θ Eq. (1.9) X Y T Generalized strains h = [ e, γ, κ Displacement gradients w' = [u', u', θ' Eq. (1.15) T X Y T Eq. (1.29) Stress resultants z = [ N,V, M T vary U: Strain energy U δu = Eq. (1.34) _ T T T z B dx δu = p δu vary p: Internal forces p _ T T δp = (B δz + δb z) dx δu = (K + K ) M G Eq. (1.4) Tangent stiffness matrix K = K M + K G Figure Roadmap for the derivation of the T plane beam element. the j th column of W N, W V and W M, respectively. The end result is W N = 1 W V = 1 N 1 sin ω N 2 sin ω N 1 cos ω N 2 cos ω N 1 sin ω N 1 cos ω N1 2(1 + e) N 1 sin ω N 1 cos ω N 1 N 2 (1 + e) N 1 sin ω N 2 sin ω N 1 cos ω N 2 cos ω N 2 sin ω N 2 cos ω N 1 N 2 (1 + e) N 2 sin ω N 2 cos ω N2 2(1 + e) N 1 cos ω N 2 cos ω N 1 sin ω N 2 sin ω N 1 cos ω N 1 sin ω N1 2γ N 1 cos ω N 1 sin ω N 1 N 2 γ N 1 cos ω N 2 cos ω N 1 sin ω N 2 sin ω N 2 cos ω N 2 sin ω N 1 N 2 γ N 2 cos ω N 2 sin ω N2 2γ 11 2 (11.59) (11.6)

21 11. Notes and Bibliography and W M =. Notice that the matrices must be symmetric, since K G derives from a potential. Then K G = (W N N + W V V ) d X = K GN + K GV. (11.61) Again the length integral should be done with the one-point Gauss rule at ξ =. Denoting again quantities evaluated at ξ = byanm subscript, one obtains the closed form s m s m c m c m K G = N m 2 + V m 2 s m c m 1 2 (1 + e m ) s m c m 1 2 (1 + e m ) s m s m c m c m s m c m 1 2 (1 + e m ) s m c m 1 2 (1 + e m ) c m c m s m s m c m s m 1 2 γ m c m s m 1 2 γ m c m c m s m s m c m s m 1 2 γ m c m s m 1 2 γ m in which N m and V m are N and V evaluated at the midpoint. Result of running the fragment of Commentary on Element Performance. (11.62) The material stiffness of the present element works fairly well once MacNeal s RBF device is done. On the other hand, simple buckling test problems, as in Exercise 11.3, show that the geometric stiffness is not so good as that of the C 1 Hermitian beam element. 7 Unfortunately a simple substitution device such as RBF cannot be used to improve K G, and the problem should be viewed as open. An intrinsic limitation of the present element is the restriction to small axial strains. This was done to facilitate close form derivation. The restriction is adequate for many structural problems, particularly in Aerospace (example: deployment). However, it means that the element cannot model correctly problems like the snap-through and bifurcation of the arch example used in Chapter 8, in which large axial strains prior to collapse necessarily occur Derivation Summary Figure 11.9 is a roadmap that summarizes the key steps in the derivation of the internal force and tangent stiffness matrix for the C plane beam element. Notes and Bibliography For detailed justification the curious reader may consult advanced FEM books such as Hughes [399. The RBF correction was introduced by R. H. MacNeal in [473. The derivation of the Timoshenko plane beam element for linear FEM is presented in Chapter 13 of the IFEM Notes [ In the sense that one must use more elements to get equivalent accuracy

22 Chapter 11: THE T PANE BEAM EEMENT: FORMUATION EXERCISE 11.1 Homework Exercises for Chapter 1 The T Plane Beam Element: Formulation and e(2) XY given in (11.27) simulta- [A+C:2 Show that the displacement field that generates the measures e (1) XY neously as G strains is [ [ x = y X + u X Y sin θ u Y + Y cos θ + 2X sin θ. (E11.1) Show that if this displacement field is selected, all G strains exactly vanish for an arbitrary RBM. EXERCISE 11.2 [A:2 Obtain the linearized strain e XX associated with the field (E11.1) using the polar decomposition of its deformation gradient tensor. EXERCISE 11.3 [A+C:3 (Research level) Rederive p and K matrix for the displacement field (E11.1), using the exact G strain e XX