TimoshenkoBea - Copia

Elemento Viga de Timoshenko

99 9.3 X -ALIGNED REFERENCE CONFIGURATION

is used to simplify the kinematics, but then most of the shear is removed to restore the correct stiffness.4 As aresult, the name C0 element is more appropriate than Timoshenko element because capturing the actualshear deformation is not the main objective.

Remark 9.2. The two-node C1 beam element is used primarily in linear structural mechanics. (It is in factthe beam model used in the Introduction to FEM course.) 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.

9.3. X-Aligned Reference Configuration

9.3.1. Element Description

We consider a two-node, straight, prismatic C0 plane beam element moving in the (X, Y ) plane, asdepicted in Figure 9.7(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 following section, once invariant strain measured are obtained.The reference element length is L0. The cross section area A0 and second moment of inertia I0with respect to the neutral axis5 are defined by the area integrals

A0 =A0d A,

A0Y d A = 0, I0 =

A0Y 2 d A, (9.1)

In the current configuration those quantities become A, I and L , respectively, but only L is frequentlyused in the TL formulation. The material remains linearly elastic with elastic modulus E relatingthe stress and strain measures defined below.As in the previous Chapter the identification superscript (e) will be omitted to reduce clutter untilit is necessary to distinguish elements within structural assemblies.The element has the six degrees of freedom depicted in Figure 9.4. These degrees of freedom andthe associated node forces are collected in the node displacement and node force vectors

u =

uX1uY11uX2uY22

, f =

fX1fY1f1fX2fY2f2

. (9.2)

The loads acting on the nodes will be assumed to be conservative.

4 The FEM analysis of plates and shells is also rife with such paradoxes.5 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.

99

2

Euler-Bernoulli C1:

95 9.2 BEAM MODELS

motion

current configuration

reference configuration

Current cross section

Referencecross section

X

X

, x

Y, y

uX (X)

uY (X)

Z (X) (X)

Figure 9.2. Definition of beam kinematics in terms of the three displacement functionsuX (X), uY (X) and (X). The figure actually depicts the EB model kinematics.In the Timoshenko model, (X) is not constrained by normality (see next figure).

rotation occurs about a neutral axis that passes through the centroid of the cross section.TimoshenkoModel. This model corrects the classical beam theory withfirst-order shear deformationeffects. In this theory cross sections remain plane and rotate about the same neutral axis as the EBmodel, but do not remain normal to the deformed longitudinal axis. The deviation from normalityis produced by a transverse shear that is assumed to be constant over the cross section.Both the EB and Timoshenko models rest on the assumptions of small deformations and linear-elastic isotropic material behavior. In addition both models neglect changes 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.

9.2.3. Finite Element Models

To carry out the geometrically nonlinear finite element analysis of a framework structure, beammembers are idealized as the assembly of one or more finite elements, as illustrated in Figure 9.3.The most common elements used in practice have two end nodes. The i th node has three degrees offreedom: two node displacements uXi and uYi , and one nodal rotation i , positive counterclockwisein radians, about the Z axis. See Figure 9.4.The cross section rotation from the reference to the current configuration is called in both models.In the BE model this is the same as the rotation of the longitudinal axis. In the Timoshenko

95


Chapter 9: THE TL TIMOSHENKO PLANE BEAM ELEMENT 96

motion

current configuration

reference configuration finite element idealizationof reference configuration

finite element idealizationof current configuration

Figure 9.3. Idealization of a geometrically nonlinear beam member (as taken,for example, from a plane framework structure likethe one in Figure 9.1) as an assembly of finite elements.

uX1

uY1

uX2

uY2

1

2

uX1

uY1

uX2

uY2

1

2

1 2 1 2

(b) C (Timoshenko) model

X, x

Y, y

(a) C (BE) model1 0

Figure 9.4. Two-node beam elements have six DOFs, regardless of the model used.

model, the difference = is used as measure of mean shear distortion.2 These angles areillustrated in Figure 9.5.Either the EB or the Timoshenko model may be used as the basis for the 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 EB model is used for ordinary beams. But here a twist appearbecause of finite element considerations. This twist is one that has caused significant confusionamong FEM users over the past 25 years.

2 It is instead of because of sign convention, to make eXY positive.

96







A0 =A0d A,

A0Y d A = 0, I0 =

A0Y 2 d A, (9.1)


u =

uX1uY11uX2uY22

, f =

fX1fY1f1fX2fY2f2

. (9.2)



99

3

Euler-Bernoulli C1:








A0 =A0d A,

A0Y d A = 0, I0 =

A0Y 2 d A, (9.1)


u =

uX1uY11uX2uY22

, f =

fX1fY1f1fX2fY2f2

. (9.2)



99

4

Euler-Bernoulli C1:


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

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

(a) (b)

C1

element with same DOFsC1

C0

Figure 9.6. Sketch of the kinematics of two-node beam finite element models based on(a) Euler-Bernoulli beam theory, and (b) Timoshenko beam theory. Thesemodels are called C1 and C0 beams, respectively, in the FEM literature.

What would be the first reaction of an experienced but old-fashioned (i.e, never heard aboutFEM) structural engineer on looking at Figure 9.6? The engineer would pronounce the C0 elementunsuitable for practical use. And indeed the kinematics looks strange. The shear distortion impliedby the drawing appears to grossly violate the basic assumptions of beam behavior. Furthermore, ahuge amount of shear energy would be require to keep the element straight as depicted.The engineer would be both right and wrong. If the two-node element of Figure 9.6(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 devices.The most common are:1. Selective integration for the shear energy.2. Residual energy balancing.These devices will be used without explanation in some of the derivations of this Chapter. Fordetailed justification the curious reader may consult advanced FEM books such as Hughes.3In this Chapter the C0 model will be used to illustrated the TL formulation of a two-node, geomet-rically nonlinear beam element.Remark 9.1. As a result of the application of the aforementioned devices the beam element behaves like a BEbeam although the underlying model is Timoshenkos. This represents a curious paradox: shear deformation

3 T. J. R. Hughes, The Finite Element Method, Prentice-Hall, 1987.

98








A0 =A0d A,

A0Y d A = 0, I0 =

A0Y 2 d A, (9.1)


u =

uX1uY11uX2uY22

, f =

fX1fY1f1fX2fY2f2

. (9.2)



99

5

Timoshenko C0: 97 9.2 BEAM MODELS

normal to deformed beam axis

90

normal to referencebeam axis X

// X (X = X)

ds

direction of deformedcross section

_ =

_

Note: in practice


6

Timoshenko C0: Cinemtica II

915 9.4 ARBITRARY REFERENCE CONFIGURATION

C0

C

= +

uX1(X , Y )1 1

1(x ,y )1 1

2(x ,y )2 2

2(X , Y )2 2uY 1 uX2

uY 21

2

X, x

Y, y

// X// X_

// Y_

// Y_

X_

Y_

Figure 9.8. Plane beam element with arbitrarily oriented reference configuration.

in which = +, and where use is made of (9.24) in a key step. These derivatives were checkedwith Mathematica. Collecting all of them into matrix B:

B = 1L0

[ cos sin L0N1 cos sin L0N2sin cos L0N1 (1+ e) sin cos L0N2 (1+ e)

0 0 1 0 0 1

]. (9.26)

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

9.4.2. Constitutive EquationsBecause 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 XsXY

]=[

s1s2

]=[

s01 + Ee1s02 + Ge2

]=[

s01s02

]+[

E 00 G

] [e1e2

]= s0 + Ee, (9.27)

where E is the modulus of elasticity and G is the shear modulus. We introduce the prestressresultants

N 0 =

A0s01 d A, V 0 =

A0

s02 d A, M0 =

A0Y s01 d A, (9.28)

915

Elemento Viga de Timoshenko Chapter 9: THE TL TIMOSHENKO PLANE BEAM ELEMENT 910

1 2

=

P(x,y)

C

X

XP

YC

YC

P (X,Y)o

C(x ,y )

C (X,0)C (X)

ooY

u

YPu

XCu XC

yy

xCx

u

CC

1

2

1 2

(X)

XY

C(X+u ,u )

X

X Y

u (X) = uu (X) = u

1

2

C0

C

(a) (b)

X, x

Y, y

oL

_L

Figure 9.7. Lagrangian kinematics of the C0 beam element with X -aligned referenceconfiguration: (a) plane beam moving as a two-dimensional body; (b) reductionof motion description to one dimension measured by coordinate X .

9.3.2. MotionThe kinematic assumptions of the Timoshenko model element have been outlined in 9.2.2. Ba-sically they state that cross sections remain plane upon deformation but not necessarily normal tothe deformed longitudinal axis. In addition, changes in cross section geometry are neglected.To analyze the Lagrangian kinematics of the element shown in Figure 9.6(a), we study the motion ofa particle originally located at P0(X, Y ). The particle moves to P(x, y) in the current configuration.The projections of P0 and P along the cross sections at C0 and C upon the neutral axis are calledC0(X, 0) and C(xC , yC), respectively. We shall assume that the beam cross section dimensions donot change, and that the shear distortion

Vector de fuerzas internas

917 9.6 THE STIFFNESS MATRIX

9.5. The Internal ForceThe 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

U =L0

(N e + V + M ) d X = L0zT h d X =

L0zTB d X u. (9.33)

Here h and B are defined in equations (9.16) and (9.26) whereas z collects the stress resultants inC as defined in (9.28) through (9.30). Because U = p T u, we get

p =L0BT z d X . (9.34)

This expression may be evaluated by a one point Gauss integration rule with the sample point at = 0 (beam midpoint). Let m = (1 + 2)/2, m = m + , cm = cosm , sm = sinm ,em = L cos(m )/L0 1, m = L sin( m)/L0, and

Bm = B|=0 =1L0

cm sm 12 L0m cm sm 12 L0msm cm 12 L0(1+ em) sm cm 12 L0(1+ em)0 0 1 0 0 1

(9.35)

where subscript m stands for beam midpoint. Then

p = L0BTmz =cm sm 12 L0m cm sm 12 L0msm cm 12 L0(1+ em) sm cm 12 L0(1+ em)0 0 1 0 0 1

T [ N

VM

](9.36)

9.6. The Stiffness MatrixThe first variation of the internal force vector (9.34) defines the tangent stiffness matrix

p =L0

(BT z+ BT z) d X = (KM +KG) u = K u. (9.37)This is again the sum of the material stifness KM and the geometric stiffness KG .9.6.1. The Material Stiffness MatrixThe material stiffness comes from the variation z of the stress resultants while keeping B fixed.This is easily obtained by noting that

z =[NVM

]=[ E A0 0 0

0 GA0 00 0 E I0

][e

]= S h, (9.38)

where S is the diagonal constitutive matrix with entries E A0, GA0 and E I0. Because h = B u,the term BT z becomes BTSB u = KM u whence the material matrix is

KM =L0BTSB d X . (9.39)

917

8

Timoshenko C0:

Matriz de rigidez tangente

Rigidez Material:


This integral is evaluated by the one-point Gauss rule at = 0. Denoting again by Bm the matrix(9.35), we find

KM =L0BTmSBm d X = KaM +KbM +KsM (9.40)

where KaM , KbM and KsM are due to axial (bar), bending, and shear stiffness, respectively:

KaM =E A0L0

c2m cmsm cmmL0/2 c2m cmsm cmmL0/2cmsm s2m mL0sm/2 cmsm s2m mL0sm/2

cmmL0/2 mL0sm/2 2mL20/4 cmmL0/2 mL0sm/2 2mL20/4c2m cmsm cmmL0/2 c2m cmsm cmmL0/2cmsm s2m mL0sm/2 cmsm s2m mL0sm/2

cmmL0/2 mL0sm/2 2mL20/4 cmmL0/2 mL0sm/2 2mL20/4

(9.41)

KbM =E I0L0

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

(9.42)

KsM =GA0L0

s2m cmsm a1L0sm/2 s2m cmsm a1L0sm/2cmsm c2m cma1L0/2 cmsm c2m cma1L0/2

a1L0sm/2 cma1L0/2 a21L20/4 a1L0sm/2 cma1L0/2 a21L20/4s2m cmsm a1L0sm/2 s2m cmsm a1L0sm/2cmsm c2m cma1L0/2 cmsm c2m cma1L0/2

a1L0sm/2 cma1L0/2 a21L20/4 a1L0sm/2 cma1L0/2 a21L20/4

(9.43)in which a1 = 1+ em .9.6.2. Eliminating Shear Locking by RBFHow good is the nonlinear material stiffness (9.42)-(9.43)? If evaluated at the reference configura-tion aligned with the X axis, cm = 1, sm = em = m = 0, and we get

KM =

E A0L0 0 0 E A0L0 0 0

0 GA0L012GA0 0 GA0L0

12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0

E A0L0 0 0E A0L0 0 0

0 GA0L0 12GA0 0 GA0L0

12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0

(9.44)This is the well known linear stiffness of the C0 beam. As noted in the discussion of Section9.2.4, this element does not perform as well as the C1 beam when the beam is thin because too

918





KaM =E A0L0




(9.41)

KbM =E I0L0

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

(9.42)

KsM =GA0L0





KM =

E A0L0 0 0 E A0L0 0 0


12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0

E A0L0 0 0E A0L0 0 0

0 GA0L0 12GA0 0 GA0L0

12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0


918





KaM =E A0L0




(9.41)

KbM =E I0L0

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

(9.42)

KsM =GA0L0





KM =

E A0L0 0 0 E A0L0 0 0


12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0

E A0L0 0 0E A0L0 0 0

0 GA0L0 12GA0 0 GA0L0

12GA0

0 12GA0 E I0L0 +14GA0L0 0 12GA0 E I0L0 +

14GA0L0


918

9

Matriz de rigidez tangente

Rigidez Geomtrica:

921 9.7 A COMMENTARY ON THE ELEMENT PERFORMANCE

the j th column ofWN ,WV andWM , respectively. The end result is

WN = 1L0

0 0 N1 sin 0 0 N2 sin0 0 N1 cos 0 0 N2 cos

N1 sin N1 cos N 21 L0(1+ e) N1 sin N1 cos N1N2L0(1+ e)0 0 N1 sin 0 0 N2 sin0 0 N1 cos 0 0 N2 cos

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

(9.52)

WV = 1L0

0 0 N1 cos 0 0 N2 cos0 0 N1 sin 0 0 N2 sin

N1 cos N1 sin N 21 L0 N1 cos N1 sin N1N2L00 0 N1 cos 0 0 N2 cos0 0 N1 sin 0 0 N2 sin

N2 cos N2 sin N1N2L0 N2 cos N2 sin N 22 L0

(9.53)andWM = 0. Notice that the matrices must be symmetric, sinceKG derives from a potential. Then

KG =L0

(WN N +WV V ) d X = KGN +KGV . (9.54)

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

KG = Nm2

0 0 sm 0 0 sm0 0 cm 0 0 cmsm cm 12 L0(1+ em) sm cm 12 L0(1+ em)0 0 sm 0 0 sm0 0 cm 0 0 cmsm cm 12 L0(1+ em) sm cm 12 L0(1+ em)

+ Vm2

0 0 cm 0 0 cm0 0 sm 0 0 smcm sm 12 L0m cm sm 12 L0m0 0 cm 0 0 cm0 0 sm 0 0 smcm sm 12 L0m cm sm 12 L0m

.

(9.55)

in which Nm and Vm are N and V evaluated at the midpoint.9.7. A Commentary on the Element PerformanceThe material stiffness of the present element works fairly well once MacNeals RBF device is done.On the other hand, simple buckling test problems, as in Exercise 9.3, show that the geometricstiffness is not so good as that of the C1 Hermitian beam element.10 Unfortunately a simple

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

921

10

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