A simple FEM formulation for large deﬂection 2D frame ... simple FEM... · A simple FEM...

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Comput. Methods Appl. Mech. Engrg. xxx (2004) xxx–xxx

www.elsevier.com/locate/cma

A simple FEM formulation for large deflection 2Dframe analysis based on position description

H.B. Coda *, M. Greco

Structural Engineering Department, S~ao Carlos School of Engineering, University of S~ao Paulo, Av. Trabalhador Sao Carlense 400,

S~ao Carlos, S~ao Paulo 13566-960, Brazil

Received 8 January 2003; received in revised form 17 October 2003; accepted 2 January 2004

Abstract

This study presents a simple formulation to treat large deflections by the finite element method. The proposed

formulation does not use the displacement concept. It considers position as the main unknown variable of the problem.

The strain determination is performed directly from the proposed position concept. A non-dimensional space is created

and relative curvature and fibers length are calculated for both reference and deformed configurations and used to

directly compute the strain energy at general points. This procedure is very simple and presents good results, as shown

in the example section. The simplicity of the resulting formulation may be considered as its main attribute. Two-

dimensional frame problems are analyzed by the proposed formulation.

� 2004 Elsevier B.V. All rights reserved.

Keywords: FEM; 2D frame analysis; Position description

1. Introduction

The analysis of structures that exhibit large deflections is of great importance in the current engineering.

The increasing search for economy and optimal material application leads to the conception of very flexiblestructures. As a consequence, the equilibrium analysis in the non-deformed position is no more acceptable

for most of applications. In this sense, a lot of studies have been developed in this engineering field. For

example, the analytical solution of slender bars and their simple composition have been studied by various

authors over the last years [1–12]. This approach is quite complicate and not general, as the superposition

of effects is not valid for non-linear applications.

In order to create automatic, general and reliable tools for the analysis of largely deflected structures,

various researchers have presented important contributions regarding finite element procedures [13–30,32–

40]. These researches are very important to the development of the human knowledge of the subject,clarifying and opening the understanding of the current researchers.

* Corresponding author. Tel./fax: +55-16-2739482.

E-mail address: [email protected] (H.B. Coda).

0045-7825/$ - see front matter � 2004 Elsevier B.V. All rights reserved.

doi:10.1016/j.cma.2004.01.005

mail to: [email protected]

2 H.B. Coda, M. Greco / Comput. Methods Appl. Mech. Engrg. xxx (2004) xxx–xxx

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One should explicitly mention the notable study by Simo et al. [40] due to the completeness of itsmathematical considerations and range of applications. It seems, from general consulted references, that no

other research presents such a high scientific standard on the non-linear frame analysis subject.

In the present study a simple engineering language is used to present a geometrical non-linear formu-

lation based on position description. The position description presented here makes use of an intermediate

non-dimensional space that allows defining a non-linear ‘‘engineering’’ strain measure calculated from a

relative fiber length for different positions of the body analyzed.

The principle of minimum potential energy is applied, considering a simple linear hyper-elastic consti-

tutive relation. Euler–Bernoulli kinematics is adopted due to its simplicity. Various examples are shown atthe end of the paper, comparing the numerical results achieved with the analytical and other numerical

solutions found in the literature.

2. The simple positional formulation

The principle of minimum potential energy can be written using position considerations (not displace-

ments). Considering only the conservative elastic problem results in

P ¼ Ue � P ; ð1Þ

where P is the total potential energy, Ue is the strain energy and P is the potential energy of the applied

forces.

Adopting linear constitutive relation for hyper-elastic materials, the strain energy can be written for the

reference volume V0 as

Ue ¼ZV0

ue dV0 ¼ZV0

1

2reee dV0; ð2Þ

where re is defined here as the ‘‘engineering stress’’, i.e., the energy conjugate of the proposed ‘‘non-linear

engineering strain’’ ee. The strain energy is assumed to be zero in a reference position, called non-deformed.

The potential energy of applied forces is written as

P ¼X

FX ; ð3Þ

where X is the set of positions independent of each other, which may be occupied by a point of the body. It

is interesting to note that the potential energy of the applied forces may not be zero in the referenceconfiguration. The total potential energy is written as (Fig. 1)

P ¼ZV0

1

2reeedV0 �

XFX : ð4Þ

In order to perform numerically the integral indicated in Eq. (4), it is necessary to map the geometry of

the studied body (the accepted geometric approximation) and to know its relation with the strain mea-

surement adopted. Fig. 2 gives the central line general geometry of a curve over a plane (2D frame bar).More detailed information about the kinematics of the considered continuum is given in Fig. 3. The

curve of Fig. 2 can be parameterized as a function of a non-dimensional variable n (varying from 0 to 1). In

this study a linear approximation for position in x direction and a cubic approximation for y direction are

imposed. When it is done addition care should be taken when the curve is not a one valued function of x.Adopting approximation previously described and the nodal parameters shown in Fig. 2, one writes

x ¼ X1 þ lxn; ð5Þ

Ue

eU0F

Fx0

x

x

Fig. 1. Total potential energy written for a body in two different positions.

y

y2

y1

x2xx1

2θ

1θ ξ0 1

Fig. 2. Curve in a 2D space.

(0,1)

ξ

η(1,1)

(1,-1)(0,-1)

h / 2

h / 2 0fds0

s0

TR

(x ,y )R

(x ,y )R0

R

R0

f

RN

s

h / 2

h / 2(x ,y )P P

N

(x ,y )P0P0 T

ds

z( )η

Fig. 3. Auxiliary non-dimensional space and simple mapping.

H.B. Coda, M. Greco / Comput. Methods Appl. Mech. Engrg. xxx (2004) xxx–xxx 3

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where

lx ¼ ðX2 � X1Þ: ð6Þ

Relating n to y by a cubic approximation, one writes

y ¼ cn3 þ dn2 þ enþ f ð7Þ

with

c ¼ ½tanðH2Þ þ tanðH1Þ�lx � 2ly ; ð8aÞ

d ¼ 3ly � ½tanðH2Þ þ 2 tanðH1Þ�lx; ð8bÞ

e ¼ tanðH1Þlx; ð8cÞ

f ¼ Y1; ð8dÞ

where ly ¼ Y2 � Y1.In this section our attention is concentrated on the simple definition of our strain measurement leaving

the proof of its objectivity to the following section. As Euler–Bernoulli hypothesis is adopted, only the

longitudinal strain is considered. Imagine a fiber parallel to the central line with an initial length defined by

ds0. After deformation, its length becomes ds and the following non-linear engineering strain is defined [41]:

ee ¼ds� ds0

ds0: ð9Þ

At this point the novelty is the calculation of the proposed strain measure that can be performed by

relative length calculations referred to the non-dimensional space represented here by variable �n� (see Fig.2), i.e.

ee ¼ds� ds0

ds0¼ ds=dn� ds0=dn

ds0=dn: ð10Þ

The values ds0=dn and ds=dn can be considered as ‘‘auxiliary’’ stretches calculated regarding the non-

dimensional space. In this study the physical reference configuration R0 is a straight frame bar.

In the initial configuration for the central line passing trough the mass center of the bar one has

ds0dn

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidx0dn

� �2

þ dy0dn

� �2s

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðl0xÞ

2 þ ðl0yÞ2

q¼ l0 ð11Þ

or

ds0 ¼ l0dt; ð12Þwhere l0 is the initial length of the finite element. A general configuration, for any instant, is described by

the approximation defined in Fig. 2. For this case the central line auxiliary stretch in computed as

dsdn

��central

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidxdt

� �2

þ dydt

� �2s

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðlxÞ2 þ ð3ct2 þ 2dt þ eÞ2

qð13Þ

or

dscentral ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðlxÞ2 þ ð3ct2 þ 2dt þ eÞ2

qdn: ð14Þ


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Applying Eq. (10), for central line one calculates

ecentral ¼1

l0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðlxÞ2 þ ð3ct2 þ 2dt þ eÞ2

q� 1: ð15Þ

Following usual engineering procedures, for Euler–Bernoulli (originally straight) frame element, the

strain at a fiber which is ‘‘z’’ distant from the central line can be written as

ee ¼ ecentral þ1

rz; ð16Þ

where 1=r is the exact curvature of the central line depicted in Fig. 2. This curvature is given as a function of

n only, as one can see in [31]:

1

r¼ dx

dnd2y

dn2

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðdy=dnÞ2 þ ðdx=dnÞ2

q� �3

: ð17Þ

Replacing the known approximations, i.e. Eqs. (5)–(8), Eq. (17) becomes

1

r¼ lxð6ct þ 2dÞ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffil2x þ ð3ct2 þ 2dt þ eÞ2

q� �3,

: ð18Þ

Substituting Eq. (16) into Eq. (2) results in

UeðespecÞ ¼E2

ecentral

�þ 1

rz�2

¼ 1

2ðecentralÞ2

þ 2ecentral1

rzþ 1

rz

� �2!: ð19Þ

In order to calculate the strain energy it is necessary to integrate the specific strain energy (UeðespecÞ) overthe initial volume of the analyzed body, as the proposed strain measure is, by nature, a Lagrangian variable.

3. Objectivity of the proposed strain measure

Before describing the numerical method based on Eqs. (4) and (19) to solve geometrical non-linear

problems, the objectivity of the proposed strain measure should be considered in order to allow theapplication of the resulting formulation to general large deflection analysis as emphasized in important

papers as [40–42].

The general position of points for the analysed continuum is described by a simple mapping from a

rectangular auxiliary space, see Fig. 3.

From Fig. 3 one writes the mapping, respecting Euler–Bernoulli kinematics, by the following relations:

XP ¼ XP0ðnÞ � zðgÞ dYP0dn

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidYP0dn

� �2

þ dXP0dn

� �2s ; ð20Þ

YP ¼ YP0ðtÞ þ zðgÞ dXP0

dn1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

dYP0dn

� �2

þ dXP0dn

� �2s ; ð21Þ

where P is a general point of the continuum and P0 is a point of the central line of the element.This mapping represents a deformation from the auxiliary non-dimensional space to any real position of

the body, therefore the usual non-linear continuum mechanics concepts may be applied. The deformation

gradient of this mapping is given by


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A ¼

dXP

dndXP

dgdYPdn

dYPdg

2664

3775: ð22Þ

One can calculate the stretches (kn and kg) following the reference directions ‘‘n’’ and ‘‘g’’ defined here by

the following unit vector Mt ¼ ½1; 0�T and Mg ¼ ½0; 1�T as follows [41]:

kn ¼ kðMnÞ ¼ A1

0

� ��; ð23Þ

kg ¼ kðMgÞ ¼ A0

1

� �� ¼ 1: ð24Þ

It is important to note that for this formulation the third direction is considered non-deformable andtherefore a principal direction, i.e., k3 ¼ 1. The angle between vectorsMn andMg, after deformation, is easily

calculated, resulting:

cos h ¼ 1 0½ � AT A� 0

1

� �¼ 0: ð25Þ

It means that no distortion occurs following the prescribed kinematics, therefore the calculated stretches are

the principal ones, i.e., kn ¼ k1 and kg ¼ k2. This reasoning naturally results in

J ¼ k1k2k3 ¼ k1: ð26ÞAs J is objective [41], the stretch

k1 ¼dsdt

ð27Þ

is also objective.The proposed strain measure, expression (10), is composed of stretch (27) and (for this particular case)

constant value ds0=dt, therefore one concludes that it is objective.

4. The numerical method

The technique proposed is similar to any finite element method except the determination of strain, where

the auxiliary space is originally used. It is necessary to integrate the specific strain energy, Eq. (19), over thebar volume. Integrating it over the cross-section area results in

Ue ¼EA2

ðecentralÞ2 þEI2

1

r

� �2

: ð28Þ

Now one integrates the strain energy per unit of length, Eq. (28), along the original length of the bar, i.e..

Ue ¼Z 1

0

EA2

ðecentralÞ2 þEI2

1

r

� �2

l0 dn ¼ l0

Z 1

0

ue dn: ð29Þ

As the strain energy is written as a function of nodal parameters it is necessary to differentiate the Total

Potential Energy, expression (4), regarding nodal parameters, to obtain the equilibrium statement. In order

to do it one should reorganize Eq. (4) as follows:


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P ¼ l0

Z 1

0

ue dn� Fx1X1 � Fy1Y1 �M1H1 � Fx2X2 � Fy2Y2 �M2H2; ð30Þ

where ðX1; Y1;H1;X2; Y2;H2Þ are nodal positions and ðFx1; Fy1;M1; Fx2; Fy2;M2Þ are their conjugate forces. Asthere is no singularity in the strain energy integral one can derive Eq. (30) regarding nodal positions. The

next equation shows this procedure for position X1.

oPoX1

¼ l0

Z 1

0

oueoX1

dn� Fx1 ¼ 0: ð31Þ

The numerical strategy is to develop derivatives inside integrals and integrate them numerically

regarding the non-dimensional variable n. As it can be noted the numerical integral result is not linear

regarding nodal positions. Therefore, one writes the above system of equations in the following generic

way:

g1ðX1; Y1;H1;X2; Y2;H2Þ ¼ f1ðX1; Y1;H1;X2; Y2;H2Þ � Fx1 ¼ 0;

g2ðX1; Y1;H1;X2; Y2;H2Þ ¼ f2ðX1; Y1;H1;X2; Y2;H2Þ � Fy1 ¼ 0;

g3ðX1; Y1;H1;X2; Y2;H2Þ ¼ f3ðX1; Y1;H1;X2; Y2;H2Þ �M1 ¼ 0;

g4ðX1; Y1;H1;X2; Y2;H2Þ ¼ f4ðX1; Y1;H1;X2; Y2;H2Þ � Fx2 ¼ 0;

g5ðX1; Y1;H1;X2; Y2;H2Þ ¼ f5ðX1; Y1;H1;X2; Y2;H2Þ � Fy2 ¼ 0;

g6ðX1; Y1;H1;X2; Y2;H2Þ ¼ f6ðX1; Y1;H1;X2; Y2;H2Þ �M2 ¼ 0;

ð32Þ

or in a compact notation:

oPopi

¼ giðpjÞ ¼ fiðpjÞ � Fi ¼ 0; ð33Þ

where pi is a generalized parameter and indices are related to nodal positions by ð1; 2; 3; 4; 5; 6Þ ¼ðX1; Y1;H1;X2; Y2;H2Þ.

In a vector representation one has

gðp; F Þ ¼ 0 ð34Þor

f ðpÞ � F ¼ 0: ð35ÞIt is important to observe that in this study the applied forces are not dependent on space. Space dependent

forces are easily implemented if desired. The vector function gðpÞ is non-linear regarding nodal parameters(p and F ). To solve Eq. (34) one can use the Newton–Raphson procedure, i.e.

gðpÞ ¼ 0 ffi gðp0Þ þ rgðp0ÞDp ð36Þor

Dp ¼ �½rgðp0Þ��1gðp0Þ; ð37Þwhere p is any position and p0 is the initial position.

At this point all usual words of non-linear analysis could be introduced, but the reader is invited to

understand the procedure as a simple non-linear system solver. One can calculate the Hessian (of strain

energy) matrix rgðp0Þ from expressions (30) and (33), as

rgðp0Þ ¼ gi;kðp0Þ ¼ fi;kðp0Þ � Fi;‘; ð38Þ


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where i ¼ 1–6; k ¼ 1–6 for parametric positions and ‘ ¼ 7–12 for forces. It is easy to achieve the following

representation:

rgðp0Þ ¼ l0

Z 1

0

ue;ik dtjp0 � di‘: ð39Þ

In order to solve Eq. (37) one needs to calculate gðp0Þ, i.e.

gðp0Þ ¼ l0

Z 1

0

ue;i dtjp0 � Fi: ð40Þ

The iterative (Newton–Raphson) process is summarized as follows:

(1) Assume p0 as the initial configuration (non-deformed). Calculate gðp0Þ following Eq. (40).

(2) For this p0 calculate the Hessian matrix per unity of length, ue;ikjp0 . Integrate this value as indicated in

(39) and the result is the gradient of g at p0.(3) Solve the system of Eq. (37) and determine Dp(4) Update position p0 ¼ p0 þ Dp. Return to step 1 until Dp is sufficiently small.

Theoretically the process is not incremental. However dividing the total loading (or prescribed position)

in cumulative steps helps to start the iterative procedure at a position nearer to the final desired result,

reducing the number of iterations. The incremental procedure is summarized as follows:

(a) p0 Initial position(b) p0 ¼ p0 þ Df , where Df is an increment of load or position stored in a single vector

(c) f1; 2; 3; 4g iterations

(d) Return to item (b)

The algebraic steps necessary to implement the proposed kinematics and approximations are given in the

electronic appendix hosted at the Internet address: http://www.set.eesc.sc.usp.br/docentes/coda/grupo/

mgreco/Appendix.pdf.

5. Numerical examples

5.1. Pure bending of an Euler beam

This example is an Euler beam initially horizontal, clamped at one end and subjected to an applied

moment at the other. The adopted parameters are: L ¼ 500 cm, A ¼ 20 cm2, E ¼ 2� 106 N/cm2 and

I ¼ 2000 cm4. Six load steps and one hundred finite elements were used to run this problem until an entire

lap of the geometry. The solution at each load step is obtained in seven iterations. A tolerance of 10�6 for

absolute position residuals is adopted. A mesh of ten finite elements is also used to run the problem; more

iterations are needed to reach the same accuracy achieved for one hundred finite elements.In Fig. 4 the horizontal and vertical positions of the free node are compared with the analytical solution.

It is important to note that intermediate values have been used in order to show the total path behaviour.

In Fig. 5 the deformed shapes during �loading� are depicted for 10 finite elements. Again intermediate

positions are shown in order to illustrate the example.

In Fig. 6 results for 10 and 100 finite elements are compared with the analytical solution for bending

moment/rotation relation.

http://www.set.eesc.sc.usp.br/docentes/coda/grupo/mgreco/Appendix.pdf

http://www.set.eesc.sc.usp.br/docentes/coda/grupo/mgreco/Appendix.pdf

0

1

2

3

4

5

6

7

-200 -100 0 100 200 300 400 500 600

Analytic vertical Analytic horizontal Numerical vertical Numerical horizontal

Position [cm]

θ [ra

d]

Fig. 4. Horizontal and vertical positions for analytical and numerical cases.

-150 -100 -50 0 50 100 150 200 250 300 350 400 450 5000

50

100

150

200

250

300

350

400 θθ

θθθθθθ

=0,62831853 rad =1,25663706 rad =1,88495559 rad =2,51327412 rad =3,14159265 rad =3,76991118 rad =5,02654825 rad =6,28318531 rad

Y [c

m]

X [cm]

Fig. 5. Deformed shapes for a different prescribed rotation (h).


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It must be noted that Simo et al. [40] solved this example using Reissner kinematics and quadratic

approximation for displacement and rotation. They achieved the exact solution for each load step with only

two iterations and ten finite elements.

5.2. Euler elastica

A clamped bar, Fig. 7, with length L ¼ 2 m subjected to the action of an increasing compressive load

from 0 to 37,100 kN is analyzed. The physical properties are: I ¼ 2:425� 10�5 m4, A ¼ 0:0175 m2 and

E ¼ 210� 109 Pa. The Euler critical load is Pcr ¼ 3141:5 kN. Ten finite elements are used to solve this

problem.

In Fig. 8 a position curve of the loaded point is shown for initial eccentricities of L=1000, L=100 and

L=10, measured at the top of the column. Only four load steps are necessary to run this problem and a

tolerance of 10�6 for absolute position residuals is adopted. The initial shape is considered parabolic. In

Fig. 8 the critical reference value is used to show the accuracy of results. In Fig. 9 various deformed shapes,for eccentricity L=100, show the pre-critical, critical and post-critical behaviors of the column.

0

5.000.000

10.000.000

15.000.000

20.000.000

25.000.000

30.000.000

35.000.000

40.000.000

45.000.000

50.000.000

55.000.000

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

Analytic Numerical (10 FE) Numerical (100 FE)

θ [rad]

M [N

.cm

]

Fig. 6. Relation between the applied bending moment and the rotation at the loaded end.

P

2m

Fig. 7. Analyzed column.


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5.3. Arch subjected to vertical load

This example shows the behavior of a parabolic arch (for example a footbridge) subjected to a central

load (position control). The analysis starts from the initial position and keeps showing the post-critical

behavior of the arch. The adopted geometry and physical properties are: Lx ¼ 500 cm, Ly ¼ 30 cm, A ¼ 20cm2, E ¼ 2000 N/cm2 and I ¼ 200 cm4. Shapes for important steps are given in Figs. 10–13. In Fig. 14 the

position of the central point versus the actual loading is depicted and compared with results obtained using

ADINA�. To run this example 100 finite elements are used without considering symmetry and a step of 0.1

N is considered to show the total path. The program is capable of running this example using a 1 N step

together with the position control.

It is interesting to comment that from steps 1 to 429 the shape pattern is the one depicted in Fig. 11 and

at step 400 the two �hills� become the half height of the �valley�, when referred to zero, and the sign of the

force changes. From steps 418 to 471 almost no change in load is noted and �hills� turn into �valleys�. This isthe critical region where the structure assumes indifferent positions. After step 574 the position pattern

depicted in Fig. 13 is maintained.

Fig. 14 shows ADINA� results until position �10, where indifferent equilibrium is reached.

0.0 0.5 1.0 1.5 2.0 2.5 3.00

4.000

8.000

12.000

16.000

20.000

24.000

28.000

32.000

36.000

40.000

1st order Eccentricity=0.1% L Eccentricity=1.0% L Eccentricity=10.0% L

P [k

N]

X [m]

Fig. 8. Lateral position for the loaded point.

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6

LOAD LEVEL P=100 kN P=2900 kN P=3100 kN P=3300 kN P=3900 kN P=5400 kN P=13100 kN P=37100 kN

X [m]

Y [m

]

Fig. 9. Deformed shapes for eccentricity L=100.


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5.4. Two bar truss

This example is usual to test the large strain behavior of trusses. It consists in applying position control

for a simple truss structure, Fig. 15. Only one finite element has been used due to the symmetry of the

problem. In Fig. 16 a position versus load curve is depicted. The proposed formulation, using non-linear

Fig. 10. Initial position.

0 50 100 150 200 250 300 350 400 450 500-40-20

02040

Y [c

m]

X [cm]

Fig. 11. Step 429 inside the critical zone, last step in the pre-critical shape.

0 50 100 150 200 250 300 350 400 450 500-40-20

02040

Y [c

m]

X [cm]

Fig. 12. Step 430 inside the critical zone, first step in the post-critical shape.

0 50 100 150 200 250 300 350 400 450 500-40-20

02040

Y [c

m]

X [cm]

Fig. 13. Step 610, the shape is almost the initial one in the reverse position.


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engineering strain measure, achieves an exact solution for one iteration. Another strain measure is also used

in this example. It is called here the logarithm strain measure and is given by

e ¼ ds� ds0ds

¼ ds=dn� ds0=dnds=dn

:

The adopted physical properties are h ¼ a ¼ 1, I ¼ 10; 000, E ¼ 1 and A ¼ 0:5.

-40 -30 -20 -10 0 10 20 30-90.0

-80.0

-70.0

-60.0

-50.0

-40.0

-30.0

-20.0

-10.0

0.0

10.0

ADINA Positional formulation

Forc

e [N

]

Y [cm]

Fig. 14. Position of the central point versus applied force (position control).

Fig. 15. Two bars truss.

1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 -2.5 -3.0-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2 Engineering strain measure Logarithmic strain measure

Forc

e

X

Fig. 16. Position versus reaction.


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As expected the logarithm strain measure provided a stiffened behavior in compression and a softenedbehavior in tension. In this case the same linear relation between strain and stress is adopted for both strain

measures.

5.5. Pined fixed diamond frame

Jenkins et al. [11] solved this example analytically, using elliptic integrals. Mattiasson [2] performed the

numerical evaluation of the elliptic integrals providing a valuable test for finite element formulations of

large deflections. The following properties are adopted to run the problem: L ¼ 1, E ¼ 1, I ¼ 1 andA ¼ 1000. Fig. 17 shows the diamond frames and the measured ‘‘displacements’’. Displacements are cal-

culated here by the difference between positions to allow comparisons with Ref. [2]. Symmetry is adopted.

Twenty finite elements were used to run this problem. In Figs. 18 and 19 the achieved displacements are

depicted together with the analytical solution calculated by Mattiasson [2]. The maximum achieved errors

Fig. 17. Diamond frame, static scheme for tension ðP > 0Þ and compression ðP < 0Þ situations.

0

1

2

3

4

5

6

7

8

9

10

-0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3

Analytic Numeric - Tension Numeric - Compression

UX/L

PL2 /E

I

Fig. 18. Horizontal displacement versus load for diamond frame.

0

1

2

3

4

5

6

7

8

9

10

-1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4

Analytic Numerical - Tension Numerical - Compression

UY/L

PL2 /E

I

Fig. 19. Vertical displacement versus load for diamond frame.

-1.5 -1.2 -0.9 -0.6 -0.3 0.0 0.3 0.6 0.9 1.2 1.5-0.30

0.00

0.30

0.60

0.90

1.20

1.50

1.80 LOAD LEVEL

P=0 P=1.5 P=10

Y

X

Fig. 20. Compression diamond frame; deformed shapes.

-1.5 -1.2 -0.9 -0.6 -0.3 0.0 0.3 0.6 0.9 1.2 1.5-0.30

0.00

0.30

0.60

0.90

1.20

1.50

1.80 LOAD LEVEL

P=0 P=2 P=10

Y

X

Fig. 21. Tension diamond frame; deformed shape.


ARTICLE IN PRESS


ARTICLE IN PRESS

are 1% in tension and 0.3% in compression. The adopted tolerance is Tol ¼ 10�5 in position. In Figs. 20 and21 some deformed shapes are depicted.

6. Conclusions

In this paper a consistent and simple formulation is proposed to solve geometrically non-linear plane

frame problems. It is based on position description and calculates strains from relative lengths and cur-

vatures using an intermediate non-dimensional space. From this procedure an engineering non-linear strainmeasure is presented. In order to show the didactic possibilities of the technique, a simple engineering

language is used. The formulation developed can be used for practical applications and presents good

convergence and accuracy.

Acknowledgements

To Dr. S.H.S. N�obrega for running ADINA� example and to FAPESP (Fundac�~ao de Amparo �aPesquisa do Estado de S~ao Paulo) for the financial support.

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