the tl plane beam element: formulation§11.2 beammodels §11.1.introduction in this chapter the...

11The TL Plane

Beam Element:Formulation

11–1

Chapter 11: THE TL PLANE BEAM ELEMENT: FORMULATION

TABLE OF CONTENTS

Page§11.1 Introduction . . . . . . . . . . . . . . . . . . . . . 11–3

§11.2 Beam Models . . . . . . . . . . . . . . . . . . . . 11–3§11.2.1 Basic Concepts and Terminology . . . . . . . . . . . 11–3§11.2.2 Beam Mathematical Models . . . . . . . . . . . . 11–4§11.2.3 Finite Element Models . . . . . . . . . . . . . . 11–5§11.2.4 Shear Locking . . . . . . . . . . . . . . . . 11–7

§11.3 X -Aligned Reference Configuration . . . . . . . . . . . . . 11–8§11.3.1 Element Description . . . . . . . . . . . . . . 11–8§11.3.2 Element Motion . . . . . . . . . . . . . . . . 11–9§11.3.3 Displacement Interpolation . . . . . . . . . . . . 11–11§11.3.4 Strain-Displacement Relations . . . . . . . . . . . . 11–11§11.3.5 *Consistent Linearization . . . . . . . . . . . . . 11–12§11.3.6 *Rigid Body Motion Check . . . . . . . . . . . . . 11–13

§11.4 Arbitrary Reference Configuration . . . . . . . . . . . . 11–13§11.4.1 Strain-Displacement Matrix . . . . . . . . . . . . 11–14§11.4.2 Constitutive Equations . . . . . . . . . . . . . . 11–15

§11.5 Strain Energy . . . . . . . . . . . . . . . . . . . . . 11–15

§11.6 Internal Force Vector . . . . . . . . . . . . . . . . . 11–16

§11.7 Tangent Stiffness Matrix . . . . . . . . . . . . . . . . . 11–16§11.7.1 Material Stiffness Matrix . . . . . . . . . . . . . 11–16§11.7.2 Eliminating Shear Locking by RBF . . . . . . . . . . 11–17§11.7.3 Geometric Stiffness Matrix . . . . . . . . . . . . 11–19

§11.8 Commentary on Element Performance . . . . . . . . . . . . 11–22

§11.9 Derivation Summary . . . . . . . . . . . . . . . . . . 11–22

§11. Notes and Bibliography . . . . . . . . . . . . . . . . . 11–22

§11. Exercises . . . . . . . . . . . . . . . . . . . . . . 11–23

11–2

§11.2 BEAM MODELS

§11.1. Introduction

In this Chapter the finite element equations of a geometrically nonlinear, two-node Timoshenkoplane beam element are formulated using the Total Lagrangian (TL) kinematic description. Thederivation is more typical of the general case. It is still short, however, of the enormous complexityinvolved, for instance, in the FEM analysis of nonlinear three-dimensional beams or shells. In factthe latter are still doctoral thesis topics.

In the formulation of the bar element in Chapter 9, advantage was taken of the direct expression ofthe axial strain in terms of reference and current element lengths. That shortcut bypasses the useof displacement gradients, and allows the reference configuration to be arbitrarily oriented. Thesimplification works equally well for straight bars in three-dimensional space.

A more systematic but lengthier procedure is required with more complicated elements. The pro-cedure requires going through the displacement gradients to construct a strain measure. Sometimesthis measure is too complex and must be simplified while retaining physical correctness. Thenwork-conjugate stresses are introduced and paired with strains to form the strain energy functionof the element. Repeated differentiations with respect to node displacements yield the expressionsof the internal force vector and tangent stiffness matrix. Finally, a transformation to the globalcoordinate system may be required.

In addition to giving a better picture of the general procedure just outlined, the beam elementillustrates the treatment of rotational degrees of freedom.

§11.2. Beam Models

§11.2.1. 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 Mechanicsof Materials courses. Such a basic knowledge is assumed here. The following material recapitulatesdefinitions 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 itssupports. By “rod-like” it is meant that one of the dimensions is considerably larger than the othertwo. This one is called the longitudinal dimension and defines the longitudinal direction or axialdirection. Directions normal to the longitudinal direction are called transverse. The intersectionof planes normal to the longitudinal direction with the beam are called cross sections, just as forbar elements. The beam longitudinal axis is directed along the longitudinal direction and passesthrough the centroid of the cross sections.1.

Beams may be used as isolated structures. But they can also be combined to form frameworkstructures. This is actually the most common form of high-rise building construction. Individualbeam components of a framework are called members, which are connected at joints. Frameworkscan be distinguished from trusses by the fact that their joints are sufficiently rigid to transmit bendingmoments between members.

1 If the beam is built of several materials, as in the case of reinforced concrete, the longitudinal axis passes through thecentroid of a modified cross section. The modified-area technique is explained in elementary courses of Mechanics ofMaterials

11–3


motion

current configuration

reference configuration

Figure 11.1. 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 spatialbeams, which are used in three-dimensional frameworks and are subject to forces applied alongarbitrary directions.

A plane beam resists primarily loading applied in one plane and has a cross section that is symmetricwith respect to that plane. Plane frameworks, such as the one illustrated in Figure 11.1, areassemblies of plane beams that share that symmetry. Those structures can be analyzed with two-dimensional idealizations.

A beam is straight if the longitudinal direction is a straight line. A beam is prismatic if its crosssection is uniform. Only straight, prismatic, plane beams will be considered in this Chapter.

§11.2.2. Beam Mathematical Models

Beams are actually three-dimensional solids. One-dimensional mathematical models of planebeams are constructed on the basis of beam theories. All such theories involve some form ofapproximation that describes the behavior of the cross sections in terms of quantities evaluated onthe longitudinal axis. More precisely, the element kinematics of a plane beam is completely definedby the two models described below if the following functions are given:

• Axial displacement u X (X)

• Transverse displacement uY (X) (also called lateral displacement)

• Cross section rotation θZ (X) ≡ θ(X): angle by which cross section rotates.2

Here X denotes the longitudinal coordinate in the reference configuration. See Figure 11.2. In thesequel 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 beamtheory, and is the one covered in elementary treatments of Mechanics of Materials. It accounts forbending moment effects on stresses and deformations. Transverse shear forces are recovered fromequilibrium but their effect on beam deformations is ignored (more precisely: the strain energydue 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

§11.2 BEAM MODELS

motion


reference configuration

Current cross section

Referencecross section

X, x

X

u (X)Y

u (X)X

Zθ (X) ≡ θ(X)

Y, y

Figure 11.2. Definition of beam kinematics in terms of the three displacement functions: u X (X),uY (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 11.3.

motion


reference configuration finite element idealizationof reference configuration

finite element idealizationof current configuration

Figure 11.3. Idealization of a geometrically nonlinear beam member (as taken, for example, from aplane 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 neutralaxis 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, butdo not remain normal to the deformed longitudinal axis. The deviation from normality is producedby 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 linear-elastic isotropic material behavior. In addition both models neglect any change in dimensions of thecross sections as the beam deforms. Either theory can account for geometrically nonlinear behaviordue to large displacements and rotations, as long as the other assumptions hold.

11–5


uX1 X1

Y1

X2

Y2

Y1

X2

Y2

u

u

u

θ1

θ2

u

u

u

u

θ1

θ2

1 2 1 2

(b) C (Timoshenko) model

X, x

Y, y

(a) C (BE) model1 0

Figure 11.4. Two-node beam elements have six DOF, regardless of which model is used.

§11.2.3. Finite Element Models

To carry out the geometrically nonlinear finite element analysis of a framework structure, beammembers are idealized as an assembly of finite elements, as illustrated in Figure 11.3. Beamelements used in practice have usually two end nodes. The i th node has three DOF: two nodedisplacements u Xi and uY i , and one nodal rotation θi , positive counterclockwise in radians, aboutthe Z axis. See Figure 11.4.

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 Timoshenkomodel, the shear distortion angle is γ = θ − ψ , as shown in Figure 11.5. The mean shear strainhas the opposite sign: γ = −γ = ψ − θ , so as to make the shear strain eXY positive, as per theusual conventions of structural mechanics.

Either the BE or the Timoshenko model may be used as the basis for the TL beam element formu-lation. Superficially it appears that one should select the latter only when shear effects are to beconsidered, 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 confusionamong 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 ofshear deformation, finite elements based on this model are in fact simpler to construct! Here arethe two key reasons:

(i) Separate kinematic assumptions on the variation of cross-section rotations are possible, asmade evident by Figure 11.5. Mathematically: θ(X) may be assumed independently of u X (X)

and uY (X). As a consequence, two-node Timoshenko elements may use linear variations inboth displacement and rotations. On the other hand, a two-node BE model requires a cubicpolynomial for uY (X) because the rotation θ(X) is not independent.

(ii) The linear transverse displacement variation matches that commonly assumed for the axialdeformation (bar-like behavior). The transverse and axial displacement assumptions are thensaid to be consistent.

11–6

§11.2 BEAM MODELS

normal to deformed beam axis

Angles are positiveas shown. Note that γ = θ − ψ

whereas γ = −γ = ψ − θ

90

normal to referencebeam axis X

// X (X = X)

ds

ψdirection of deformedcross section _

_ _γ

γψ

θ

Current nodeline (CNL)

Figure 11.5. Illustrates total and BE section rotations θ and ψ , respectively, in the plane beam Timoshenkomodel. The mean shear distortion angle is γ = θ − ψ , but we take γ = −γ = θ − ψ as shear strain measure, tomatch usual sign conventions of structural mechanics. For elastic deformations of engineering materials |γ | << 1.Typical values for |γ | would be O(10−3) 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 illustratedby the elements contrasted in Figure 11.6. Although as pictured there both elements have sixdegrees of freedom, the internal kinematics of the Timoshenko model is far simpler.

§11.2.4. Shear Locking

In the FEM literature, a BE-based model such as the one shown in Figure 11.4(a) is called a C1

beam because this is the kind of mathematical continuity achieved in the longitudinal directionwhen a beam member is divided into several elements (cf. Figure 11.3). On the other hand, theTimoshenko-based element pictured in Figure 11.4(b) is called a C0 beam because both transversedisplacements, as well as the rotation angle θ , preserve only C0 continuity.

What would be the first reaction of an experienced but old-fashioned (i.e, “never heard aboutFEM”) structural engineer on looking at Figure 11.6? The engineer would pronounce the C0

element unsuitable for practical use. And indeed the kinematics looks way wrong. The sheardistortion grossly violates the basic assumptions of beam behavior. And indeed, a huge amount ofshear 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) wereconstructed with actual shear properties and exact integration, an overstiff model results. Thisphenomenom is well known in the FEM literature and receives the name of shear locking. To avoidlocking while retaining the element simplicity it is necessary to use certain computational devicesthat 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. Forexplanatory literature references see Notes and Bibliography and Appendix S. In this Chapter theC0 model will be used to illustrated the TL formulation of a two-node, geometrically nonlinearbeam element.

11–7


2-node (cubic) elementfor Euler-Bernoulli beam model:plane sections remain plane andnormal to deformed longitudinal axis

2-node linear displacement & rotations element for Timoshenko beam model:plane sections remain plane but notnormal to deformed longitudinal axis

C1

element with same DOFsC1

C0

(a) (b)

1 2 1 2

Figure 11.6. Contrasting kinematics of 2-node beam FEM models based on (a) BE beam theory, and (b)Timoshenko beam theory. These are called C1 and C0 beam elements, respectively, in the FEM literature.

Remark 11.1. As a result of the application of the aforementioned devices the beam element behaves like a BEbeam although the underlying model is Timoshenko’s. This represents a curious paradox: shear deformationis used to simplify the kinematics, but then most of the shear will be removed to restore the correct physicalbehavior.3 As a result, the name “C0 element” is more appropriate than “Timoshenko element” becausecapturing the actual shear deformation is not the main objective.

Remark 11.2. The two-node C1 beam element is used primarily in linear structural mechanics. It is in factthe beam model used in the IFEM course [?, Chapter 12].) This is because some of the easier-constructionadvantages cited for the C0 element are less noticeable, while no artificial devices to eliminate locking areneeded. The C1 element is also called the Hermitian beam element because the shape functions are cubicpolynomials specified by Hermite interpolation formulas.

§11.3. X-Aligned Reference Configuration

§11.3.1. Element Description

We consider a two-node, straight, prismatic C0 plane beam element moving in the (X, Y ) plane, asdepicted in Figure 11.7(a). For simplicity in the following derivation the X axis system is initiallyaligned with the longitudinal direction in the reference configuration, with origin at node 1. Thisassumption is relaxed in the next section, once invariant strain measured are obtained.

The reference element length is L0. The cross section area A0 and second moment of inertia Izz0

with respect to the neutral axis4 are defined by the area integrals

A0 =∫

A0

d A,

∫A0

Y d A = 0, I0 =∫

A0

Y 2 d A, (11.1)

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 = 0.

11–8

§11.3 X -ALIGNED REFERENCE CONFIGURATION

1 2

P(x,y)

P (X,Y)0 C (X)0

XPuXCu

x ≡ xx

C(x ,y )C C

C (X,0)0

1

2

θ(X)

C(X+u ,u )X Y

XCXu (X) = u

C0

C

(a) (b)

X, xX, x

Y, y Y, y

0L1(−L /2,0 )0

1(x ,y )1 1

2(x ,y )2 2

2(L /2,0 )0

ψψ

ψSection rotation isθ=ψ+γ=ψ−γ

_

Py ≡y

P0Y ≡Y

P0 C0X≡X ≡X

Cy CP

X

reduction to 1D

//X//Xψ

L

θ1 γ1

θ2 γ2

YYC

u (X) = uCurrent node

line (CNL)

Figure 11.7. Lagrangian kinematics of TL C0 plane beam element with X -aligned reference configuration:(a) beam moving as 2D body; (b) reduction of motion description to 1D as measured by material coordinate X .

In the current configuration those quantities become A, Izz and L , respectively, but only the currentlength L is frequently used in the TL formulation. The material remains linearly elastic, withYoung’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 clutteruntil it is necessary to distinguish elements within structural assemblies.

The element has the six degrees of freedom depicted in Figure 11.4. These degrees of freedom andthe associated node forces are collected in the node displacement and node force 6-vectors

u = [ u X1 uY 1 θ1 u X2 uY 2 θ2 ]T , f = [ fX1 fY 1 fθ1 fX2 fY 2 fθ2 ]T .

(11.2)

The external loads applied to the nodes will be assumed to be conservative.

§11.3.2. Element Motion

The kinematic assumptions of the Timoshenko model element were outlined in §11.2.2. Basicallythey state that cross sections remain plane upon deformation, but not necessarily normal to thedeformed longitudinal axis. In addition, changes in cross section geometry are neglected.

To work out the Lagrangian kinematics of the element shown in Figure 11.6(a), we study the motionof a generic particle located at P0(X, Y ) in the reference configuration, which moves to P(x, y) inthe current configuration. The projections of P0 and P on the neutral axis, along the cross sectionsat C0 and C are called C0(X, 0) and C(xC , yC), respectively. We denote by u X0 = u(X, 0) anduY 0 = u( X, 0) the displacements of the neutral axis points.

To formulate the exact reference-to-current mapping it is convenient to decompose the motion intothe three stages illustrated in Figure 11.8. The two intermediate configurations are denoted byC and C with particle coordinates x, y and x, y, respectively. The mapping from C0 to C is

11–9


(3) plane rigid rotation by angle ψ (positive CCW)

(1) translation and stretching along X0

(2) rigid translation along Y

~

^

Rotation center CR

Figure 11.8. Decomposition of TL C0 plane beam motion into 3 stages.

x = X + u X0 − Y tan γ, y = Y . From C to C is x = x, y = y + uY 0 = Y + uY 0. FromC to C is x = R cos(ψ + α), y = R sin(ψ + α), in which R =

√(u X0 − Y tan γ )2 + Y 2 and

tan α = Y/(u X0 − tan γ ).

The exact kinematics is[xy

]=

[xC − Y (sin ψ+ sin γ cos ψ)

yC + Y (cos ψ− sin γ sin ψ)

]=

[xC − Y [sin θ + (1− cos γ ) sin ψ]yC + Y [cos θ + (1− cos γ ) cos ψ]

]. (11.3)

in which ψ + γ has been replaced by θ . But xC = X + u XC and yC = uY C . From now we denoteu XC and uY C simply as u X and uY , respectively. Assume for simplicity that γ is constant over theelement. Expand (11.3) in Taylor series in γ up to O(γ 4):[

xy

]=

[u X + X + 1

2 Y γ 3 cos θ − Y (1 + 12 γ 2 + 7

24 γ 4) sin θ

uY + Y (1 + 12 γ 2 − 7

24 γ 4) cos θ + 12 Y γ 3 sin θ

](11.4)

If we assume the shear distortion is very small, setting γ → 0 in (11.4) gives the simplifiedLagrangian description of the motion

[xy

]=

[X + u X − Y sin θ

uY + Y cos θ

], (11.5)

in which u X , uY and θ are functions of X only. This form, which is correct up to O(γ 2), can bedirectly obtained from the last of (11.3) on replacing 1− cos γ by 0. This concludes the reductionto 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 Xderivative:

w =[ u X (X)

uY (X)

θ(X)

], w′ = dw

d X=

[ du X/d XduY /d Xdθ/d X

]=

[ u′X

u′Y

θ ′

], (11.6)

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 followingdifferential relations

1 + u′X = s ′ cos ψ, u′

Y = s ′ sin ψ, s ′ = ds

d X=

√(1 + u′

X )2 + (u′Y )2, (11.7)

11–10

§11.3 X -ALIGNED REFERENCE CONFIGURATION

in which ds is the differential arclength in the current configuration; see Figure 11.5. If u′X and u′

Yare constant over the element,

s ′ = L/L0, 1 + u′X = L cos ψ/L0, u′

Y = L sin ψ/L0. (11.8)

Remark 11.3. Replacing 1− cos γ by zero in (11.3) is equivalent to saying that Y (1− cos γ ) can be neglectedin 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) hasthe virtue of being purely kinematic (i.e., no material properties appear) and of leading to (exactly) eY Y = 0.

§11.3.3. Displacement Interpolation

For a 2-node C0 element it is natural to express the displacements and rotation functions as linearin the node displacements:

w =[ u X (X)

uY (X)

θ(X)

]= 1

2

[ 1 − ξ 0 0 1 + ξ 0 00 1 − ξ 0 0 1 + ξ 00 0 1 − ξ 0 0 1 + ξ

]

u X1

uY 1

θ1

u X2

uY 2

θ2

= N u, (11.9)

in which ξ = 2X/L0 is the isoparametric coordinate that varies from ξ = −1 at node 1 to ξ = 1at node 2, and N as usual denotes the shape function matrix. Differentiating this expression withrespect to X yields the displacement gradient interpolation:

w′ =[ u′

Xu′

Yθ ′

]= 1

L0

[ −1 0 0 1 0 00 −1 0 0 1 00 0 −1 0 0 1

]

u X1

uY 1

θ1

u X2

uY 2

θ2

= N′ u. (11.10)

§11.3.4. Strain-Displacement Relations

The deformation gradient matrix of the motion (11.5) is

F =[ ∂x

∂ X∂x∂Y

∂y∂ X

∂y∂Y

]=

[1 + u′

X − Y κ cos θ − sin θ

u′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 displacementgradient matrix is

G = F − I =[

u′X − Yκ cos θ − sin θ

u′Y − Yκ sin θ cos θ − 1

], (11.12)

From the definition (8.12), the plane portion of the Green-Lagrange (GL) strain tensor follows as

e =[

eX X eXY

eY X eY Y

]= 1

2 (FT F − I) = 12 (G + GT ) + 1

2 GT G. (11.13)

11–11


This expands to the component form

e = 12

[2(u′

X−Yκ cos θ) + (u′X−Yκ cos θ)2 + (u′

Y −Yκ sin θ)2 −(1 + u′X ) sin θ + u′

Y cos θ

−(1 + u′X ) sin θ + u′

Y cos θ 0

](11.14)

It is seen that the only nonzero strains are the axial strain eX X and the shear strain eXY +eY X = 2eXY ,whereas eY Y vanishes. Through the consistent-linearization techniques described in §11.3.5 below,it can be shown that under the small-strain assumptions made precise therein, the axial strain eX X

can be replaced by the simpler form

eX X = (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 arrangedin the strain vector

e =

e1

e2

=

[eX X

2eXY

]=

[(1 + u′

X ) cos θ + u′Y sin θ − Y θ ′ − 1

−(1 + u′X ) sin θ + u′

Y cos θ

]=

[e − Yκ

γ

].

(11.16)

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 thefollowing generalized strain vector:

h = [ e γ κ ]T . (11.18)

Because of the assumed linear variation in X of u X (X), uY (X) and θ(X), e and γ only depend onθ whereas κ is constant over the element. Making use of (11.8) one can express e and γ in thegeometrically invariant form

1 + e = s ′ cos(θ − ψ) = L cos γ

L0, γ = −s ′ sin(θ − ψ) = L sin γ

L0(11.19)

In theory one could further reduce e to L/L0 and γ to L γ /L0, but those “simplifications” wouldactually complicate the strain variations taken in the following Section.

§11.3.5. *Consistent Linearization

The derivation of the consistent linearization (11.15) is based on the polar decomposition analysis of thedeformation gradient. Introduce the matrix

Ω(α) =[

cos α − sin α

sin α cos α

], (11.20)

which represents a two-dimensional rotation (about Z ) through an angle α. Since Ω is an orthogonal matrix,ΩT = Ω−1. The deformation gradient (11.11) can be written

F = Ω(ψ)

[s ′ 00 0

]+ Ω(θ)

[−Y θ ′ 00 1

]. (11.21)

11–12

§11.4 ARBITRARY REFERENCE CONFIGURATION

where s ′ is defined in (11.7) Premultiplying both sides of (11.21) by Ω(−θ) gives the modified deformationgradient

F = Ω(−θ)F =[

s ′ cos(θ − ψ) − Y θ ′ 0−s ′ sin(θ − ψ) 1

](11.22)

Now the GL strain tensor 2e = FT F − I does not change if F is premultiplied by the orthogonal matrix Ω

because FT ΩT ΩF = FT F. Consequently 2e = FT

F − I. But if the strains remain small, as it is assumed inthe Timoshenko model, the following are small quantities:

(i) s ′ − 1 = (L/L0) − 1 because the axial strains are small;

(ii) Y θ ′ = Yκ 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 + L + higher order terms, where L is a first-order linearization in the small quantities s ′ − 1, Y θ ′

and γ = ψ − θ . It follows that

e = 12 (L + LT ) + higher order terms. (11.23)

Carrying out this linearization one finds that eXY and eY Y do not change, but that eX X simplifies to (11.15). Itcan also be shown that 2eXY = γ

.= γ within the order of approximation of (11.23).

§11.3.6. *Rigid Body Motion Check

It is instructive to check whether the foregoing displacements and strain fields can correctly represent rigidbody motions (RBM). An arbitrary RBD of the element can be represented by a rigid translation uX R, uY R,along X and Y , resectively, and a rotation of angle θR about the translated position. Denoting the currentcoordinates by xR, yR the motion is given by[

xR

yR

]=

[cos θR sin θR

− sin θR cos θR

] [X + u X R

Y + uY R

]. (11.24)

The displacements are u X = xR − X R and uY = yR − Y . The X derivatives are

u′X = cos θR − 1, u′

Y = − sin θ, θ ′R = κ R = 0. (11.25)

This motion is inserted into the following strain measures:

e(1)

X X = (u′X−Yκ cos θ) + 1

2 (u′X−Yκ cos θ)2 + 1

2 (u′Y −Yκ sin θ)2,

e(2)

X X = (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)

X X and e(1)

XY are those in the GL strain tensor (11.13). Normal strain e(2)

X X isthe result of the linearization (11.15), while the shear strain e(2)

X X includes a correction not derivable from theGL formulas. Inserting the RBM described by (11.24) and (11.25) into the shear and normal strain measures(11.27) one obtains

e(1)

X X = 0, e(2)

X X = 2 sin2 θR, e(1)

XY = − 12 sin(2θR), e(2)

XY = 0. (11.27)

Conclusion: the strain measures e = e(2)

X X and γ = e(1)

XY used in the strain vector (11.18) do not capture exactlyRBM as the rotation angle θR gets large. For that to be achieved, it would be necessary to pick e = e(1)

X X andγ = e(2)

XY . There is no displacement field, however, that produces both of these as components of the GL straintensor. See also Exercise 10.1.

11–13


C0

C

ϕ

φ = ψ+ϕ

ϕ

ψ

uX1

1(X ,Y )1 1

1(x ,y )1 1

2(x ,y )2 2

2(X ,Y )2 2

uY1uX2

uY21θ

2θ

X, x

Y, y

γ// X

γ// X

γ// X_γ// Y

_

γ// Y_

γX_

γY_

M0 N 0V0

MN

V

C0

C

(a) (b)

Figure 11.9. Plane beam element with arbitrarily oriented reference configuration:(a) kinematics, (b) internal stress resultants.

§11.4. Arbitrary Reference Configuration

In the general case the reference configuration C0 of the element is not aligned with X . Thelongitudinal axis X forms an angle ϕ with X , as illustrated in Figure 11.9(a). The six degreesof freedom of the element are indicated in that Figure. Note that the section rotation angle θ ismeasured from the direction Y , normal to X , and no longer from Y as in Figure 11.6.

Given the node coordinates (X1, Y1) and (X2, Y2), the reference angle ϕ is determined by

cos ϕ = X21/L0, sin ϕ = Y21/L0, X21 = X2 − X1, Y21 = Y2 − Y1, L20 = X2

21 + Y 221. (11.28)

The angle φ = ψ +ϕ formed by the current longitudinal axis with X (see Figure 11.5) is determinedby

cos φ = cos(ψ + ϕ) = x21

Lsin φ = sin(ψ + ϕ) = y21

L, (11.29)

in which x21 = x2 − x1, y21 = y2 − y1 and L may be obtained from the motion as

x21 = X21 + u X2 − u X1, y21 = Y21 + uY 2 − uY 1, L2 = x221 + y2

21. (11.30)

Solving the foregoing trigonometric relations for ψ gives

cos ψ = X21x21 + Y21 y21

L L0= X21(X21 + u X2 − u X1) + Y21(Y21 + uY 2 − uY 1)

L L0,

sin ψ = X21 y21 − Y21x21

L L0= X21(Y21 + uY 2 − uY 1) − Y21(X21 + u X2 − u X1)

L L0.

(11.31)

It follows that L sin ψ and L cos ψ are exactly linear in the translational node displacements. Thisproperty simplifies considerably the calculations that follow.

11–14

§11.5 STRAIN ENERGY

§11.4.1. Strain-Displacement Matrix

For the generalized strains it is convenient to use the invariant form (11.19), which does not dependon ϕ. The variations δe, δγ and δκ with respect to nodal displacement variations are required in theformation 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

∂u X1= ∂[L cos(θ − ψ)/L0 − 1]

∂u X1= ∂[L(cos θ cos ψ + sin θ sin ψ)/L0 − 1]

∂u X1

= −X21 cos θ + Y21 sin θ

L20

= − cos ϕ cos θ + sin ϕ sin θ

L0= −cos ω

L0,

(11.32)

in which ω = θ + ϕ. Use was made of (11.31) in a key step. These derivatives were checked withMathematica. Collecting all of them into matrix B:

B = 1

L0

[ − cos ω − sin ω L0N1γ cos ω sin ω L0N2γ

sin ω − cos ω −L0N1 (1 + e) − sin ω cos ω −L0N2 (1 + e)0 0 −1 0 0 1

]. (11.33)

Here N1 = (1 − ξ)/2 and N2 = (1 + ξ)/2 are abbreviations for the element shape functions(caligraphic symbols are used to lessen the chance of clash against axial force symbols).

§11.4.2. Constitutive Equations

Because the beam material is assumed to be homogeneous and isotropic, the only nonzero PK2stresses are the axial stress sX X and the shear stress sXY . These are collected in a stress vector srelated to the GL strains by the linear elastic relations

s =[

sX X

sXY

]=

s1

s2

=

[s0

1 + Ee1

s02 + Ge2

]=

s0

1s0

2

+

[E 00 G

] e1

e2

= s0 + Ee, (11.34)

in which E is the modulus of elasticity and G is the shear modulus. We introduce the initial stress(prestress) resultants

N 0 =∫

A0

s01 d A, V 0 =

∫A0

s02 d A, M0 =

∫A0

−Y s01 d A. (11.35)

These define the axial forces, transverse shear forces and bending moments, respectively, in thereference configuration. We also define the stress resultants

N = N 0 + E A0 e, V = V 0 + G A0 γ, M = M0 + E I0 κ. (11.36)

These represent axial forces, tranverse shear forces and bending moments in the current config-uration, respectively, defined in terms of PK2 stresses. See Figure 11.9(b) for signs. These arecollected in the stress-resultant vector

z = [ N V M ]T . (11.37)

11–15


§11.5. Strain Energy

As in the case of the TL bar element, the total potential energy = U − P is separable becauseP = λqT u. The strain (internal) energy is given by

U =∫

V0

[(s0)T e + 1

2 eT Ee]

dV =∫

A0

∫L0

[(s0

1 e1 + s02 e2) + 1

2 (Ee21 + Ge2

2)]

d A d X . (11.38)

Carrying out the area integrals while making use of (11.34) through (11.37), U can be written asthe sum of three length integrals:

U =∫

L0

(N 0e + 12 E A0 e2) d X +

∫L0

(V 0γ + 12 G A0γ

2) d X +∫

L0

(M0κ + 12 E I0 κ2) d X ,

(11.39)

The three terms in (11.39) define the energy stored through bar-like axial deformations, sheardistortion and pure bending, respectively.

§11.6. Internal Force Vector

The internal force vector can be obtained by taking the first variation of the internal energy withrespect to the node displacements. This can be compactly expressed as

δU =∫

L0

(N δe + V δγ + M δκ

)d X =

∫L0

zT δh d X =∫

L0

zT B d X δu. (11.40)

Here h and B are defined in equations (11.18) and (11.33), respectively, whereas z collects the stressresultants in C as defined in (11.35) through (11.37). Because δU = p T δu, we get

p =∫

L0

BT z d X . (11.41)

This expression may be evaluated by a one point Gauss integration rule with the sample point atξ = 0 (beam midpoint). Let θm = (θ1 + θ2)/2, ωm = θm + ϕ, cm = cos ωm , sm = sin ωm ,em = L cos(θm − ψ)/L0 − 1, γm = L sin(ψ − θm)/L0, and

Bm = B|ξ=0 = 1

L0

−cm −sm − 1

2 L0γm cm sm − 12 L0γm

sm −cm12 L0(1 + em) sm −cm

12 L0(1 + em)

0 0 −1 0 0 1

(11.42)

where subscript m stands for “at beam midpoint.” Then

p = L0BTmz =

−cm −sm

12 L0γm cm sm

12 L0γm

sm −cm − 12 L0(1 + em) −sm cm − 1

2 L0(1 + em)

0 0 −1 0 0 1

T NVM

(11.43)

11–16

§11.7 TANGENT STIFFNESS MATRIX

§11.7. Tangent Stiffness Matrix

The first variation of the internal force vector (11.41) defines the tangent stiffness matrix

δp =∫

L0

(BT δz + δBT z

)d X = (KM + KG) δu = K δu. (11.44)

This is again the sum of the material stifness KM and the geometric stiffness KG .

§11.7.1. 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

δz =

δNδVδM

=

[ E A0 0 00 G A0 00 0 E I0

] [δeδγ

δκ

]= S δh, (11.45)

where S is the diagonal constitutive matrix with entries E A0, G A0 and E I0. Because δh = B δu,the term BT δz becomes BT SB δu = KM δu, whence the material matrix is

KM =∫

L0

BT SB d X . (11.46)

This integral is evaluated by the one-point Gauss rule at ξ = 0. Denoting again by Bm the matrix(11.43), we find

KM =∫

L0

BTmSBm d X = Ka

M + KbM + Ks

M (11.47)

where KaM , Kb

M and KsM are due to axial (bar), bending, and shear stiffness, respectively:

KaM = E A0

L0

c2m cmsm −cmγm L0/2 −c2

m −cmsm −cmγm L0/2cmsm s2

m −γm L0sm/2 −cmsm −s2m −γm L0sm/2

−cmγm L0/2 −γm L0sm/2 γ 2m L2

0/4 cmγm L0/2 γm L0sm/2 γ 2m L2

0/4−c2

m −cmsm cmγm L0/2 c2m cmsm cmγm L0/2

−cmsm −s2m γm L0sm/2 cmsm s2

m γm L0sm/2−cmγm L0/2 −γm L0sm/2 γ 2

m L20/4 cmγm L0/2 γm L0sm/2 γ 2

m L20/4

(11.48)

KbM = E I0

L0

0 0 0 0 0 00 0 0 0 0 00 0 1 0 0 −10 0 0 0 0 00 0 0 0 0 00 0 −1 0 0 1

(11.49)

KsM = G A0

L0

s2m −cmsm −a1L0sm/2 −s2

m cmsm −a1L0sm/2−cmsm c2

m cma1L0/2 cmsm −c2m cma1L0/2

−a1L0sm/2 cma1L0/2 a21 L2

0/4 a1L0sm/2 −cma1L0/2 a21 L2

0/4−s2

m cmsm a1L0sm/2 s2m −cmsm a1L0sm/2

cmsm −c2m −cma1L0/2 −cmsm c2

m −cma1L0/2−a1L0sm/2 cma1L0/2 a2

1 L20/4 a1L0sm/2 −cma1L0/2 a2

1 L20/4

(11.50)

in which a1 = 1 + em .

11–17


§11.7.2. Eliminating Shear Locking by RBF

How good is the nonlinear material stiffness given by (11.49) and (11.50)? If evaluated at thereference configuration aligned with the X axis, cm = 1, sm = em = γm = 0, and we get

KM =

E A0L0

0 0 − E A0L0

0 0

0 G A0L0

12 G A0 0 −G A0

L0

12 G A0

0 12 G A0

E I0L0

+ 14 G A0L0 0 − 1

2 G A0 − E I0L0

+ 14 G A0L0

− E A0L0

0 0 E A0L0

0 0

0 −G A0L0

− 12 G A0 0 G A0

L0− 1

2 G A0

0 12 G A0 − E I0

L0+ 1

4 G A0L0 0 − 12 G A0

E I0L0

+ 14 G A0L0

(11.51)

This is the well known linear stiffness of a shear-locked C0 beam. As noted in the discussion of§11.2.4, this element does not perform as well as a C1 beam when the beam is thin because toomuch strain energy is taken by shear in the antisymmetric bending mode. The following substitutiondevice, introduced by MacNeal,5 removes that deficiency in a simple way. The shear rigidity G A0

is formally replaced by 12E I0/L20, and magically (11.51) becomes

KM =

E A0L0

0 0 − E A0L0

0 0

0 12E I0

L30

6E I0

L20

0 −12E I0

L30

6E I0

L20

0 6E I0

L20

4E I0L0

0 −6E I0

L20

2E I0L0

− E A0L0

0 0 E A0L0

0 0

0 −12E I0

L30

−6E I0

L20

0 12E I0

L30

−6E I0

L20

0 6E I0

L20

2E I0L0

0 −6E I0

L20

4E I0L0

. (11.52)

This is the well known linear stiffness matrix of the C1 (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 theRBF is formally applied to the nonlinear material stiffness one gets KM = Ka

M + KMb, where KaM

is the same as in (11.50) (because the axial stiffness if not affected by the substitution), whereas

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.

11–18


KbM and Ks

M merge into

Kb

M = E I

L30

12s2m −12cmsm 6a1L0sm −12s2

m 12cmsm 6a1L0sm

−12cmsm 12c2m −6cma1L0 12cmsm −12c2

m −6cma1L0

6a1L0sm −6cma1L0 a2L20 −6a1L0sm 6cma1L0 a3L2

0−12s2

m 12cmsm −6a1L0sm 12s2m −12cmsm −6a1L0sm

12cmsm −12c2m 6cma1L0 −12cmsm 12c2

m 6cma1L0

6a1L0sm −6cma1L0 a3L20 −6a1L0sm 6cma1L0 a2L2

0

(11.53)

in which a1 = 1 + em , a2 = 4 + 6em + 3e2m and a3 = 2 + 6em + 3e2

m .

Remark 11.4. MacNeal actually proposed the more refined substitution

replace1

G A0by

1

G As+ L2

0

12E I0(11.54)

where G As is the actual shear rigidity; that is, As is the shear-reduced cross section studied in Mechanicsof Materials. The result of (11.54) is the C1 Hermitian beam corrected by shear deformations computedfrom equilibrium considerations. (This beam is derived in Chapter 13 of IFEM Notes [273].) If the sheardeformation is negligible, the right hand side of (11.54) is approximately L2

0/(12E I0), which leads to thesubstitution used above.

.

§11.7.3. Geometric Stiffness Matrix

The geometric stiffness KG comes from the variation of B while the stress resultants in z are keptfixed. To get a closed form expression it is convenient to pass to indicial notation, reverting tomatrix notation later upon “index contraction.” Let the entries of KG , B, u and z be denoted asKGi j , Bki , u j and zk , where indices i , j and k range over 1–6, 1–6, and 1–3, respectively. CallA j = ∂B/∂u j , j = 1, . . . 6. Then using the summation convention,

KGi jδu j =∫

L0

δBT z d X =∫

L0

∂ Bki

∂u jδu j zk d X =

∫L0

A jki zk d X δu j , (11.55)

whence

KGi j =∫

L0

zk A jki d X , (11.56)

Note that in carrying out the derivatives in (11.56) by hand one must use the chain rule because Bis a function of e, γ and θ , which in turn are functions of the node displacements u j . To implementthis scheme we differentiate B with respect to each node displacement in turn, to obtain:

11–19


A1 = ∂B∂u X1

= 1

L0

[0 0 N1 sin ω 0 0 N2 sin ω

0 0 N1 cos ω 0 0 N2 cos ω

0 0 0 0 0 0

],

A2 = ∂B∂uY 1

= 1

L0

[0 0 −N1 cos ω 0 0 −N2 cos ω

0 0 N1 sin ω 0 0 N2 sin ω

0 0 0 0 0 0

],

A3 = ∂B∂θ1

= N1

L0

[sin ω − cos ω −N1 L0(1 + e) − sin ω cos ω −N2 L0(1 + e)cos ω sin ω −N1 L0γ − cos ω − sin ω −N2 L0γ

0 0 0 0 0 0

],

A4 = ∂B∂u X2

= 1

L0

[0 0 −N1 sin ω 0 0 −N2 sin ω

0 0 −N1 cos ω 0 0 −N2 cos ω

0 0 0 0 0 0

],

A5 = ∂B∂uY 2

= 1

L0

[0 0 N1 cos ω 0 0 N2 cos ω

0 0 −N1 sin ω 0 0 −N2 sin ω

0 0 0 0 0 0

],

A6 = ∂B∂θ2

= N2

L0

[sin ω − cos ω −N1 L0(1 + e) − sin ω cos ω −N2 L0(1 + e)cos ω sin ω −N1 L0γ − cos ω − sin ω −N2 L0γ

0 0 0 0 0 0

].

(11.57)

To restore matrix notation it is convenient to define

WNi j = A j1i , WV i j = A j

2i , WMi j = A j3i , (11.58)

as the entries of three 6 × 6 “weighting matrices” WN , WV and WM that isolate the effect of thestress resultants z1 = N , z2 = V and z3 = M . The first, second and third row of each A j becomesthe j th column of WN , WV and WM , respectively. The end result is

WN = 1

L0


0 0 −N1 cos ω 0 0 −N2 cos ω

N1 sin ω −N1 cos ω −N 21 L0(1 + e) −N1 sin ω N1 cos ω −N1N2L0(1 + e)

0 0 −N1 sin ω 0 0 −N2 sin ω


N2 sin ω −N2 cos ω −N1N2L0(1 + e) −N2 sin ω N2 cos ω −N 22 L0(1 + e)

(11.59)

WV = 1

L0



N1 cos ω N1 sin ω −N 21 L0γ −N1 cos ω −N1 sin ω −N1N2L0γ

0 0 −N1 cos ω 0 0 −N2 cos ω

0 0 −N1 sin ω 0 0 −N2 sin ω

N2 cos ω N2 sin ω −N1N2L0γ −N2 cos ω −N2 sin ω −N 22 L0γ

(11.60)

and WM = 0. Notice that the matrices must be symmetric, since KG derives from a potential. Then

KG =∫

L0

(WN N + WV V ) d X = KG N + KGV . (11.61)

11–20


X

X

Y

Y

Eqs (10.5), (10.8)

Eq. (10.9)

Node Displacements u

Element displacement field w = [u , u , θ]

Displacement gradients w' = [u' , u' , θ' ]

Generalized strains h = [ e , γ, κ ]

Stress resultants z = [ N ,V, M ]

Strain energy U

Eq. (10.15)

Eq. (10.29)

T

T

T

T

Tangent stiffness matrix K = K + K M G

Internal forces p

Eq. (10.34)

Eq. (10.40)

TT TδU = ∫ z B dX δu = p δuL 0vary U:

_

T Tδp = ∫ (B δz + δB z) dX δu = (K + K )L0

M Gvary p:_

Figure 11.10. Roadmap for the derivation of the TL plane beam element.

Again the length integral should be done with the one-point Gauss rule at ξ = 0. Denoting againquantities evaluated at ξ = 0 by an m subscript, one obtains the closed form

KG = Nm

2

0 0 sm 0 0 sm

0 0 −cm 0 0 −cm

sm −cm − 12 L0(1 + em) −sm cm − 1

2 L0(1 + em)

0 0 −sm 0 0 −sm

0 0 cm 0 0 cm

sm −cm − 12 L0(1 + em) −sm cm − 1

2 L0(1 + em)

+ Vm

2

0 0 cm 0 0 cm

0 0 sm 0 0 sm

cm sm − 12 L0γm −cm −sm − 1

2 L0γm

0 0 −cm 0 0 −cm

0 0 −sm 0 0 −sm

cm sm − 12 L0γm −cm −sm − 1

2 L0γm

.

(11.62)

in which Nm and Vm are N and V evaluated at the midpoint.

11–21


Result of running the fragment of

§11.8. Commentary on Element Performance

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 stiff-ness is not so good as that of the C1 Hermitian beam element.7 Unfortunately a simple substitutiondevice such as RBF cannot be used to improve KG , and the problem should be viewed as open.

An intrinsic limitation of the present element is the restriction to small axial strains. This wasdone 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 modelcorrectly problems like the snap-through and bifurcation of the arch example used in Chapter 8, inwhich large axial strains prior to collapse necessarily occur.

§11.9. Derivation Summary

Figure 11.10 is a roadmap that summarizes the key steps in the derivation of the internal force andtangent stiffness matrix for the C0 plane beam element.

Notes and Bibliography

For detailed justification the curious reader may consult advanced FEM books such as Hughes’ [424]. TheRBF correction was introduced by R. H. MacNeal in [504]. The derivation of the Timoshenko plane beamelement for linear FEM is presented in Chapter 13 of the IFEM Notes [273].

7 In the sense that one must use more elements to get equivalent accuracy.

11–22

Exercises

Homework Exercises for Chapter 10

The TL Plane Beam Element: Formulation

EXERCISE 11.1

[A+C:20] Show that the displacement field that generates the measures e(1)

XY and e(2)

XY given in (11.27) simulta-neously as GL strains is [

xy

]=

[X + u X − Y sin θ

uY + Y cos θ + 2X sin θ

]. (E11.1)

Show that if this displacement field is selected, all GL strains exactly vanish for an arbitrary RBM.

EXERCISE 11.2

[A:20] Obtain the linearized strain eX X associated with the field (E11.1) using the polar decomposition of itsdeformation gradient tensor.

EXERCISE 11.3

[A+C:30] (Research level) Rederive p and K matrix for the displacement field (E11.1), using the exact GLstrain eX X .

11–23

the tl plane beam element: formulation§11.2 beammodels §11.1.introduction in this chapter the...

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