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F O U N D A T I O N S O F C I V I L A N D E N V I R O N M E N T A L E N G I N E E R I N G

No. 6 2005

Publishing House of Poznan University of Technology, Pozna2005ISSN 1642-9303

Andrius GRIGUSEVICIUS, Stanislovas KALANTAVilnius Gediminas Technical University

OPTIMIZATION OF ELASTIC-PLASTIC BEAM

STRUCTURES WITH HARDENING USING FINITE

ELEMENT METHOD

Elastic-plastic beam structures, subjected by distributed loads, optimizationproblems are considered in this paper. Mixed method is suggested to form static andgeometrical equations by setting the finite element interpolation functions of internalforces and displacements. That allows forming optimization problems with restrictedmiddle cross-sections of beams. General expressions of static and geometrical equationsare presented for a beam subjected by a distributed load. The optimization mathematicalmodels of elastic-plastic beam structures with linear hardening are formulated as

quadratic programming problems. Results of numerical example are presented andbriefly discussed.

Key words: elastic-plastic beam structures, optimization, linear hardening,quadratic programming, distributed loading.

1. INTRODUCTION

Systems subjected only by concentrated loads are considered in manyworks ([1-12]) intended for beam structures optimization. Optimizationalgorithms of beam systems subjected by a distributed load are still notsufficiently developed. Of course, a distributed load can be changed into asystem of some concentrated loads. However, in this case the size of the problemincreases considerably. Therefore, additional difficulties emerge, since successof a solution of a non-linear optimization problem also depends on the size ofthe problem. Trying to avoid these difficulties second order equilibrium finite

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Optimization of elastic-plastic beam structures with hardening using finite element method 33

principal of minimum complementary the energy geometrical equations arebeing formed

,kTkk uA (2.3)

or using physical equations,

,0 kTkkkk uASD (2.4)

where ku - displacement vector corresponded by the element static equations

(2.2), k - nodal strains vector. Element flexibility matrix

kl

kk

T

kk dxxx0

HdHD (2.5)

and distortion (initial or plastic strains) vector

kl

k

T

kk dxxx0

00 H . (2.6)

Here kd - flexibility matrix of infinitesimal element; kl - finite element length.

b)

a)

Qk2k1Q

Nkk2M

kNMk1

1 2

21

k4P

k5Pk6P

k1P

k2P

P

y

x

kl

k3

Fig. 1. Finite element under bending and compression or tension with 6 degrees of

freedom: a) generalized forces; b) nodal internal forces For instance, static equations of a simple element under bending andcompression or tension (Fig. 1), described by linear internal functions are

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34 Andrius Griguseviius, Stanislovas Kalanta

k

k

k

kk

kk

k

k

k

k

k

k

k

k

k

k

k

k

k

N

M

M

ll

ll

M

N

Q

M

N

Q

P

P

P

P

P

P

2

1

2

2

1

1

6

5

4

3

2

1

11

111

1

11

P (2.7)

Geometrical equations i.e. strain expressions of this element are derived

using equation (2.3):

34111

kkk

k

k uuul

, 64121

kkk

k

k uuul

,

52 kkk uu ,

(2.8)

where the directions of displacements kiu correspond to the directions of

generalized forces kiP . Such a method of creation of element equations is called a static one.However, an alternative method of forming element equations is possible, whichwe call a mixed method.

We will create an alternative mixed finite element by setting interpolationfunction of internal forces (2.1) and displacements interpolation function

kku

kx

ky

kx

ky

kx

ky

k xu

u

x

x

xu

xux uH

H

Hu . (2.9)

We are formulating the equations of this element with reference to the principleof virtual work:

kk l

kTT

k

l

k

T

kkTk dxxxdxxx

00

uSSS A ,

where operator

2

2

2

2

dx

ddx

d

TA .

Using functions (2.1) and (2.9) we get such an equation

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kl

kuTT

kTkk

Tk

k

dxxx uHHSS

0

A .

From here the geometrical equations are

kkkl

kuTT

kk

k

dxxx uBuHH

0

A , (2.10)

where matrix

k

l

kuTT

kk dxxx0

HHB A . (2.11)

Then from virtual works equality

kTk

T

kk

T

kk

T

k SBuSPu (2.12)

arise static equations

kTkk SBP . (2.13)

Thus static and geometrical equations of the element can be formedtwofold making directly static equations and indirectly geometrical equations

or vice versa.If while forming (2.2) and (2.13) equations the same generalized forces

kP are considered, then

kTkkk SBSA .

Furthermore, when internal forces vectors kS of equilibrium and mixed elementare equivalent then

Tkk BA .

Consequently considering different generalized forces kP or vectors kS is

possible to get alternative element expressions of static and geometricalequations.

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3. STATIC AND GEOMETRICAL EQUATIONS OF THE

ELEMENT SUBJECTED BY DISTRIBUTED LOAD

Often several elements of a beam system are subjected by uniformdistributed loads of intensity yp and sometimes xp , too. When formulating

optimization problems a second order equilibrium element, one degree offreedom of which corresponds to integral displacement (Fig. 2), is usually usedto model such beams. Distribution of internal forces in such an element isdescribed by these interpolation functions [14]:

,24231 322

22

2

12

2k

kk

k

kk

k

kk

k Ml

x

l

xM

l

x

l

xM

l

x

l

xxM

311 kk

k

k

k Nl

xN

l

xxN

.

(3.1)

31

k4P

k5Pk6P

k1P

k2PPk3

y

x

kl

Qk3k1Q

Nk1k3M

k3NMk1

1 3

21k3u

uk2

uk1

uk6uk5

uk4

=pkx2

k7P

k8P

uk7

uk8 3

2

=pkya)

b)

c)

Fig. 2. Finite element with integral displacements 7ku and 8ku : a) generalized

forces; b) internal forces of nodes; c) generalized displacements

The stress state of this element is defined by a vector of internal forces Tkkkkkk NNMMM 31321 ,,,,S . Internal forces functions (3.1) are not

compatible with static equations of beam under bending and compression ortension

yp

dx

xMd

2

2

,

xpdx

xdN

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a priori. Therefore, to perform these conditions, additional generalized forces

7kP and 8kP are introduced and two additional equations are formed

32122

2

7 24

kkk

k

kk MMM

ldx

xMdP ,

318

1kk

k

kk NN

ldx

xdNP ,

where 7kP and 8kP are the intensities of the forces distributed along the element

length.Static equations of this element are

3

1

3

2

1

222

3

3

3

1

1

1

8

7

6

5

4

3

2

1

11

484

11

3411

1

143

k

k

k

k

k

kk

kkk

kkk

kkk

kx

ky

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

N

N

M

M

M

ll

lll

lll

lll

pp

M

N

Q

M

N

Q

PP

P

P

P

P

P

P

P (3.2)

These equations correspond to geometrical equations (2.4), which contain thedisplacements

kl

kyk dxxuu0

7 , kl

kxk dxxuu0

8

and the element flexibility matrix derived from (2.5) formula is

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k

k

k

k

k

k

k

k

k

kk

A

I

A

I

A

I

A

IEI

l

55,2

5,25215,0181

5,012

15D . (3.3)

where E - elasticity modulus of material, kA and kI - element cross-section

area and moment of inertia.Further aim is to create equations for the element subjected by a distributeload (Fig. 3) when 7kP and 8kP are concentrated generalized forces and 7ku ,

8ku - second node displacements.

31

k4P

k5Pk6P

k1P

k2PPk3

21k3u

uk2

uk1

uk6uk5

uk4

2

k7P

k8P

uk7

uk8

3

a)

b)

Fig. 3. Element subjected by a distribute load with linear displacements of middle node:a) generalized forces, b) nodes displacements

The vertical displacements of this element are described by the fourth degreepolynomial [16]

74

4

3

3

2

2

63

4

2

32

44

4

3

3

2

2

33

4

2

32

14

4

3

3

2

2

163216

238145

254818111

k

kkk

k

kkk

k

kkk

k

kkk

k

kkk

ky

ul

x

l

x

l

x

ul

x

l

x

l

xu

l

x

l

x

l

x

ul

x

l

x

l

xxu

l

x

l

x

l

xxu

, (3.4)

and displacements xukx - by second degree polynomial

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822

52

2

22

2

4223

1 kkk

k

kk

k

kk

kx ul

x

l

xu

l

x

l

xu

l

x

l

xxu

. (3.5)

Expressing (2.11) formula we derive such a matrix of geometrical equations:

105,125,2105,25,12

3214

311

1

642

322

32

321

114

31

15

1

kkk

kkk

kkk

k

lll

lll

lll

B (3.6)

We can see that it is not equal to the transpose matrix of static equations (3.2).Therefore static and geometrical equations of finite elements shown in Fig. 2 andFig. 3 are different. The reason for this are different degrees of freedom of 7ku

and 8ku . Element shown in Fig. 3 is recommended to apply as it does notinvolve integral displacements.

External nodal forces kiF of the considered element, equivalent to thelinear distributed load, are derived from formulas

,8,5,2,

;7,6,4,3,1,

0

0

idxpxHF

idxpxHF

k

k

l

kxkiki

l

kykiki

(3.7)

where xHki - shape functions of the polynomials (3.4) and (3.5). The valuesof these forces are shown in Fig. 4.

31

k4F

k5F

k6F

k1F

k2F

Fk3

2

k7F

k8F

=p lkx k

6

=p lky k60

2

= p lky k307

=2p lkx k

3

= p lky k3016

= p lky k

60

2

-

=p lkx k

6

= p lky k307

Fig. 4. External nodal forces of the element

If the beam is subjected only by a linear distributed load ( 0kxp ), then itis possible to model it by the element shown in Fig. 2 or Fig. 3. However in thiscase kkk NNN 31 and distribution of the displacements xukx in theelement is linear. Therefore with the view of the decrease of the number ofequations and variables it is preferable to use the element with 7 degrees of

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freedom for such beams. This element is derived from the element describedabove, eliminating the displacement 8ku . The stress state of such an element is

described by a vector Tkkkkk NMMM ,321 ,,S and external forces052 kk FF . The matrix of geometrical equations is derived from matrix (3.6)

by only changing the forth row to this one 0,0,15,0,0,15,0 and eliminatingthe last row and the last column.

4. EQUATIONS OF FREE ORIENTED ELEMENT

In second and third chapters static and geometrical equations ofconsidered finite elements are presented in local coordinate xy system. However

beam constructions with beam elements which local coordinate xy axis is notparallel to global all system coordinate ''yx axis are often occur in engineeringpractice. Static and geometrical equations of all construction are being createdwith reference to finite elements equations and are described by global forces Pand displacements u . Therefore the equations of all elements must be expressed

by global forces and displacements.Relation between generalized forces of free oriented element (Fig. 5) in

local and global coordinate systems is described by equation

'kkk PTP , (4.1)

where transformation matrix

cossin sincos

1cossinsincos

1cossinsincos

kT (4.2)

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Pk5'

Pk5Pk6

P'

k4

Pk4

P'k8

Pk8P'k7

Pk7

P'k1Pk1

Pk2

P'k2

Pk3

'x

y '

x

y

Fig. 5. Local and global forces of the finite element

Analogically displacements

'kkk uTu (4.3)

By using formula (4.1) we derive static equation of the element in theglobal coordinate system:

kkkk SAPT '

orkkk SAP

'' , (4.4)

where

kTkkkk ATATA

1' . (4.5)

Analogically geometrical equation

'''kkkkkkkk uBuTBuB

or

0uBSD ''0 kkkkk , (4.6)

where

kkk TBB ' . (4.7)

External nodal forces are interdependent by the equation

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kTkk FTF

' .

5. ELEMENT STRENGTH CONDITIONS

Strength conditions of a beam construction with linear hardening materialwill be analyzed accepting that axial and shear forces work is negligible incomparison with bending moments work and can be ignored. These strengthconditions must be secured in all above-mentioned structure design cross-sections (i.e. finite elements nodes):

,00 ipii MMM ,00 ipii MMM ....,,2,1 ni (5.1)

here iM0 - limit moment of the i-th cross-section; n - the number of design

cross-sections; piM - plastic moment of the i-th cross-section. Plastic moments

vector of all structure can be expressed through plastic strains vector:

pp HM (5.2)

here H - hardening matrix.

MM

Mpk

0k

MiM0

Mpi

i

li( h )

Mi

l

M0

MpjjM

( l )

m1(t)

m2

mj

j

( e )

l

jM x

( s)( r )

ll

Fig. 6. Internal forces distribution in the beam

Hardening matrix for a certain type of finite element can be derived

similarly like an elastic stiffness matrix 1 DK . Plastic deformations can occurin some parts of the finite element (Fig 6). Additional design points (h,r,s,e,l,t) need to be introduced to bind these parts. Flexibility growth matrix for any

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plastic zone can be derived using finite element method technique. For instanceflexibility increment matrix of plastic zone ihis:

;9518209155 12102440123020

25

30242

143001

11

210

440

231

323232

323232

333

222

0

2

2

22

22

22

1

0

1

iiiiiii

iiiiiii

iiii

l

ii

ii

iil

Tp

T

ihpih

I

l

lll

lll

dxxx

x

x

ll

ll

ll

Idx

I

ii

BkkBd

(5.3)

here I - cross-section inertia moment; - hardening ratio; l - length of thefinite element; B - matrix of polynomial multipliers expressed through beamnodes i, j, k coordinates; ihB - matrix of polynomial multipliers expressed

through beam nodes i,r,hcoordinates; pk , k- polynomial coordinates vectors

for plastic zone ih and finite element k respectively. Second and third rows ofthis matrix are irrelevant for finite element flexibility increment matrix sincethese rows correspond additional design points r and h. Analogously deriving

matrices for others plastic zones and eliminating additional design points rowswe get such a flexibility increment matrix of k-th finite element:

333

32

31

322

311

32

31

333

52

2525

25

30jjjj

mmmmmmmm

iiii

pk

I

l

d . (5.4)

Second row of this matrix is derived adding corresponding members of plasticzoneslmand mtflexibility increment matrices.

Hardening matrix of all construction is derived inversing flexibilityincrement matrices:

1 pkdiagdH . (5.5)

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6. CONSTRUCTION DISCRETE MODEL EQUATIONS

The equations of a whole construction are being formed from thecorresponding equations of the finite elements sk ,...,2,1 , conjugated to entirediscrete model.

Strength conditions of the construction from linear hardening material aredescribed by inequality

0SSHS re0 (6.1)

or0SHS 0 (6.2)

where TnSSS ,...,, 21S - internal forces vector; - quasi-diagonal matrixdiagonally filled with the blocks of matrices k , sk ,...,2,1 ; H - hardening

matrix created considering to material hardening in the plastic sections; eS -

elastic internal forces vector; rS - residual internal forces vector. Plasticmultipliers vector is related with plastic strains vector by flow rule

T

p , 0 ,

0SHS 0T .(6.3)

Construction discrete model, same as a finite element, static andgeometrical equations can be formulated in two ways directly formulatingstatic equations and indirectly geometrical equations or vice versa.

Lets say that the displacements and forces of all elements are described

by vectors Tsuuu ,...,, 21u , T

sPPP ,...,, 21P , T

sFFF ,...,, 21F . Thedisplacements of the construction discrete model nodes are described by vector

Tmuuu ,...,, 21u and forces acting in the direction of these displacements are

described by vector TmFFF ,...,, 210F , where m discrete model degree of

freedom.Direct formulation of geometrical equations. Geometrical equationsof separate unconnected elastic-plastic finite elements sk ,...,2,1 are described

by system

0uBDS T , (6.4)

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where kdiag DD , kdiag BB - quasi-diagonal matrices diagonally filledwith the blocks of matrices kD and kA . Here is considered, that distortions

Tp0 .

Separate elements are being connected to the whole discrete model bydisplacement compatibility equations

Cuu , (6.5)

which describe the relation between finite elements nodal displacements u andnodal displacements of discrete model. Depending on them we derive

geometrical equations of discrete model

0BuDS T , (6.6)

where

CBB . (6.7)

Unknowns of elastic-plastic constructions in analysis and optimizationproblems are frequently considered as the residual internal forces rS and

displacements ru as opposed to the real ones. Since re uuu and elasticstrains meet with the equation

0BuDS ee , (6.8)

therefore, from the equation (6.6) can be derived geometrical equation, whichrelates residual strains and displacements of the construction

0BuDS rT

r . (6.9)

The principle of virtual displacements will be used for static equationcreation. With reference to equality of virtual works

FuS TT or FuSBu TTT (6.10)

we derive

FSB T . (6.11)

There are static equations of discrete model. Statically possible residual internalforces rS have to meet with the equation

0SB rT . (6.12)

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Direct formulation of static equations. Static equations of discretemodel can be formulated directly, using finite elements static equations (2.2) anddisplacements compatibility equations (6.5).

Static equations of all elements (2.2) compounds such a system of staticequations

SAP , (6.13)

where kdiag AA - quasi-diagonal static equations matrix, diagonal blocksof which are matrices of separate blocks kA . These equations describe only

forces equilibrium inside every element but do not describe forces equilibriumbetween elements. Discrete model static equations of every separate element andall system we derive with reference to the principal of virtual displacements.

Virtual works of internal and external forces of construction are correlatedwith equation

0FuFuPu TTT

or FuPCu TTT . (6.14)

From here

,FPC T (6.15)

where

.0FFCF T (6.16)

Considering to (6.13) we derive such discrete model static equation:

FAS , (6.17)

where

ACA T . (6.18)

From equation (6.17) follows

0AS r , (6.19)

which meets with residual internal forces of the construction.Thus static and geometrical equations of a finite elements system can be

formulated using one of above described methods. Equation systems created onthe ground of these methods are equivalent though expressions of these systemscan be different. Applying either method we will derive identical equationsystems only when the internal forces vectors kS and displacements vectors ku

will be the same. Then we will derive TBA .

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7. MATHEMATICAL MODELS OF OPTIMIZATION

PROBLEMS

Optimization problem of constructions of elastic-plastic linear hardeningmaterial is being formulated as follows: for given construction design schemaand loading find the internal forces, displacements and optimal beam limitinternal forces satisfying given strength and stiffness conditions.

Designing beam constructions usually the same cross-sections are selectedfor some beam groups. Lets say that the limit internal forces of such beamgroups are described by the optimized parameters vector 0M . Relation between

vectors 0S and 0M is described by the equation

00 GMS , (7.1)

where G - construction configuration matrix. Stiffness conditions are describedby displacement restrictions

uLu , (7.2)

and optimal criterion is minimal value of function

00 MS T

f , (7.3)

here - weight coefficient vector which members are proportional to the totallength of the same cross-section beams; u - maximal values of restricteddisplacements.

Restraints system of the considered optimization problem is made fromthree condition groups: equations (static and geometrical) that describeconstruction stress-strain state and strength and stiffness restrictions. Thus,considering to above used notation, construction optimization is expressed bysuch a model: find

0min MT , (7.4)

when

0SB rT

or 0AS r ,0BuDS r

Tr ,

0SSHGM re0 ,

0SSHGM reT

0 , 0 ,

uuuL re , 0M 0 .

(7.5)

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Unknowns in this problem are vectors of residual internal forces rS and

displacements ru as well as limit internal forces and plastic multipliers vectors

0M and . This is the problem of nonlinear and non-convex programming. Asshown in [14], for the ideal plastic case such a problem can be recomposed intoequivalent quadratic programming problem, which for the positively definitequadratic matrix has only one solution. Applying the technique proposed in [14]we derive such a model: find

reTT

SSHGMM 00min , (7.6)

when

0SB rT or 0AS r ,

0BuDS rT

r , 0 ,

0SSHGM re0 ,

uuuL re , 0M 0 .

(7.7)

Using the solution of static and geometrical equations

UDBBDBu TTTr 111 , (7.8)

DDBBDBBDS

TTTTr

11111 (7.9)

we can eliminate residual internal forces and displacements from this model.Then we derive such a problem: find

SHGMM eTT

00min , (7.10)

when 0SHGM e0 ,

uUuL e , 0M 0 , 0 .

(7.11)

8. NUMERICAL EXAMPLE

Described method is realized for three-span beam (Fig. 7). Beam material

elastic modulus kPa10205 6E , yield limit kPa2400000 . A list ofdisplacement restrictions is written in Table 1. The beam cross-section is I-type(IPE). The dependences among beam cross-section inertia moment I, area Aand limit bending moment 0M according to [17] are:

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3300

bAaM , 11

bAaI , (8.1)

here 1a , 3a , 1b , 3b - cross-section shape coefficients (according to [17] for IPE

cross-section 7918716,01a , 8294723,03a , 319975,21b , 660459,13b ).

2m2m 6m 3m 2m15m

20 kN 10 kN/m 40 kNm

M M M01 02 03

Fig. 7. Beam design schema

Optimization of this beam is performed for two cases: beam material is ideal elastic-plastic ( 0 );

beam material is elastic plastic with linear hardening ( 8101,6 ). Iterative solving of the mathematical model (7.6 - 7.7) was used for this

problem. Cross-section inertia moments and relative plastic zones lengths wasfreely chosen in the first iteration. Obtained results (Table 1) shows, that thedisplacements their limit meanings reached in the first and second spans for bothcases. Objective function (OF) meaning in the second case is smaller by 11,3,and this shows the expediency of hardening evaluation aiming to use lessmaterial resources.

Table 1. Data and results of numerical example

System Displacementrestrictions, cm

Obtaineddisplacements, cm

Obtained limitbending

moments, kNmIdeal elastic-plastic

0 ;78337OF ,

I span 041 ,ur

II span 062 ,ur

III span 053 ,ur

I span 041 ,ur

II span 062 ,ur

III span 3313 ,ur

331301 ,M

562502 ,M

222603 ,M

Elastic-plastic withhardening

81016 , ;

54299OF ,

I span 041 ,ur

II span 062 ,ur

III span 053 ,ur

I span 041 ,ur

II span 062 ,ur

III span 4643 ,ur

661001 ,M

532802 ,M

141703 ,M

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9. CONCLUSIONS

1. Optimization algorithms of beam structures subjected by distributed loadsare improved in this paper. Mixed method is suggested to form static andgeometrical equations by setting the finite element interpolation functions ofinternal forces and displacements.

2. General expressions of static, geometrical equations and strength conditionsof a beam element subjected by a distributed load are formed using mixedmethod. That allows forming optimization problems with restricted middlecross-sections of the beams. Formulating the equations by static method

there is no such a possibility.3. Formed general finite element equations allow to unify the equationsformation of all construction. It is shown, that using static and mixedmethods we can form a lot of alternative equations systems for the oneconstruction. These equations systems are equivalent but their numericalexpressions will be equal only when internal forces vectors kS and

displacements vectors ku will be the same for the both methods.4. The optimization mathematical model with non-linear conditions are

modified to quadratic programming problem and applied for optimization ofstructures with linear hardening and subjected by distributed loads.

REFERENCES

1. yras A.A.: Mathematical Models for the Analysis and Optimization ofElastoplastic Structures, New York: John Wiley, 1983, 121 p.

2. Borkowski A.: Analysis of Skeletal Structural Systems in the Elastic andElastic-plastic Ranges, Warshawa: PWN Elsevier, 1988, 200 p.

3. Atkoinas J.: Design of elastoplastic systems under repeated loading,Vilnius: Science and Encyclopedia Publishers, 1994, 148 p. (in Russian).

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A. Grigusevicius, S. Kalanta

OPTYMALIZACJA PRTOWYCH ELASTYCZNO-PLASTYCZNYCHKONSTRUKCJI WYPRODUKOWANYCH Z WZMACNIAJCEGO SIMATERIAU METODELEMENTW SKOCZONYCH

S t r e s z c z e n i e

W rozprawach optymalizacji prtowych elastyczno-plastycznych konstrukcjirozpatruje sinajczciej systemy obciane siami skupionymi. Praca jest powiconaudoskonaleniu algorytmw optymalizacji prtowych konstrukcji obcionych siami

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cigymi. Rwnania statyczne oraz geometryczne elementu skoczonego i caejkonstrukcji proponuje sitworzymetodkombinowan, zadajc funkcje interpolacyjneprzemieszcze oraz si. To sprzyja sformuowaniu zadania optymalizacji, w ktrymmona ograniczy ugicia rodkowego przekroju w elementach zginanych.Przedstawiono oglne formy rwna statycznych oraz geometrycznych dla prtaobcianego cigym obcieniem. Zadania optymalizacji prtowych elastyczno-plastycznych konstrukcji wyprodukowanych z liniowo wzmacniajcego simateriau sprzedstawione jak zadania kwadratowego programowania. Przedstawione s rezultatyanalizy obliczeniowej.

Received, 26.08.2003.

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