solution of nonlinear dynamics response - part ii

Topic 14

Sections 9.3.1, 9.3.2, 9.3.3, 9.5.3, 8.2.4

9.6,9.7,9.8,9.11

Contents:

Mode superposition analysis in nonlinear dynamics

Substructuring in nonlinear dynamics, a schematicexample of a building on a flexible foundation

Study of analyses to demonstrate characteristics ofprocedures for nonlinear dynamic solutions

Example analysis: Wave propagation in a rod

Example analysis: Dynamic response of a three degree offreedom system using the central difference method

Example analysis: Ten-story tapered tower subjected toblast loading

Example analysis: Simple pendulum undergoing largedisplacements

Example analysis: Pipe whip solution

Example analysis: Control rod drive housing with lowersupport

Example analysis: Spherical cap under uniform pressureloading

Example analysis: Solution of fluid-structure interactionproblem

Solution ofNonlinearDynamicResponse-Part II

Textbook:

Examples:

14-2 Nonlinear Dynamic Response - Part II

References: The use of the nonlinear dynamic analysis techniques is described withexample solutions in

Bathe, K. J., "Finite Element Formulation, Modeling and Solution ofNonlinear Dynamic Problems," Chapter in Numerical Methodsjor Partial Differential Equations, (Parter, S. V., ed.), Academic Press, 1979.

Bathe, K. J., and S. Gracewski, "On Nonlinear Dynamic Analysis UsingSubstructuring and Mode Superposition," Computers & Structures, 13,699-707,1981.

Ishizaki, T., and K. J. Bathe, "On Finite Element Large Displacementand Elastic-Plastic Dynamic Analysis of Shell Structures," Computers& Structures, 12, 309-318, 1980.

TH E S OLV\T ION 0 F

TH E ']) 'YNAM Ie EQUlLJ B-

~IUH ElluATIONS CAN

'5 ACHIVE~ US1I'Jb

'DIR=,-T "I.NTE{;RATJON

Me-PtobS,

_ E,>PL\C tT lNTE. 5R.

S~B'STR lACTu'R 11\15

w~ I)lSCUSS i~ESE l~H'

NIQuES ~'R'EFL'I I~ \t\\S

LEe.n,fi O~

1>E N b\.A LtA M

\

EXS P\?E wlil?

'RES"PONS,c SOUA110N

Topic Fourteen 14-3

. A""AL't~IS OF CR.b

1-1011\'>11\1&

SOL-I.tTION OF "RE::C;?ONS~

OF S'PHE~llAL CAP

At-IAL'ISI1 i"E. lEST)

THE 1:>E:TAIL.'S, Of'

nlESE "?'R~~LEl'-\

S,bLuTICNc:. "~E

61\1E III 1 N THE"f~'PE'RS SE.E

I

STUb,! oLAlbE

Markerboard14-1


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Mode superposition:

The modes of vibration change due tothe nonlinearities, however we canemploy the modes at a particular timeas basis vectors (generalizeddisplacements) to express theresponse.

This method is effective when, innonlinear analysis,

- the response lies in only a fewvibration modes (displacementpatterns)

- the system has only localnonlinearities

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The governing equations in implicit timeintegration are (assuming no dampingmatrix)M HLltO(k) + TK ~U(k) = HLltR _ HLltF(k-1)- - -

Let now T = 0, hence the method ofsolution corresponds to the initial stressmethod.

Using

The modal transformation gives

H.:1tX(k) + 0 2 ax(k) = T (H.:1tR _ H.:1tF(k-1)

Topic Fourteen 14-5

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where

n2 = [W~"W~] = ~r ... ~s]

.equations cannot be solvedindividually over the timespanCoupling!

Typical problem:

~~========:::::::D!I~

Pipe whip: Elastic-plastic pipeElastic-plastic stop

Nonlinearities in pipe and stop. Butthe displacements are reasonably wellcontained in a few modes of thelinear (initial) system.

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1~ Nonlinear Dynamic Response - Part II

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Substructuring

Procedure is used with implicit timeintegration. All linear degrees offreedom can be condensed out priorto the incremental solution.

Used for local nonlinearities:Contact problemsNonlinear support problems

p-0

_0

oSlip

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Example:

Ten storybuilding Finite element

model

-"master" node

- substructureinternal node

Substructuremodel

/t AK

master dot

substructureinternal dot

master dot

substructureinternal dot

master dot

Topic Fourteen 14-7

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Here

tA

( 4) tK = K + Llt2 M + Knonlinear~ toil mass~1I nonlinear stiffnessI matrix effectsall linearelement contributions

A tK + K nonlinear- -

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After condensing out all substructureinternal degrees of freedom, we obtaina smaller system of equations:

~entries from condensed

1_.........7rUbstruclures

master dof

Major steps in solution:

Prior to step-by-step solution,establish Bfor all mass and constantstiffness contributions. Staticallycondense out internal substructuredegrees of freedom to obtain Be.We note that

t A A t.!Sc = .!SC + .!Snonlinear

condensed i. all nonlinear effects7 A 4

from K= K+ .lt2 MalllinearJ "-total mass matrix

element contributions

For each time step solution (and eachequilibrium iteration):

- Update condensed matrix, Ke, fornonlinearities.

- Establish complete load vector for alldegrees of freedom and condense outsubstructure internal degrees of freedom.

- Solve for master dof displacements,velocities, accelerations and calculate allsubstructure dof disp., veL, ace.

The substructure internal nodal disp., veL,ace. are needed to calculate the completeload vector (corresponding to all dof).

Solution procedure for each time step(and iteration):

tu t+4tU-, -,tU' t+4t(j-, t+4tRA _ t+4tRA _ t+41U _ -,.. _ _c _c t+4tU"tu

Topic Fourteen 14-9

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substructuredegrees offreedomcondensedout

usingcondensedeffectivestiffnessmatrix tKe


Example: Wave propagation in a rodTransparency

14-13

R

Uniform, freely floating rod

/

R

1000 N+------

L = 1.0 mA = 0.01 m'Z.p = 1000 kg/m3

E = 2.0 x 109 Pa

time

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Consider the compressive force at apoint at the center of the rod:

-.B-1' .5 'I' .5 'I

t* = time for stress waveto travel throughthe rod1000 N

Compressiveforce

IA

The exact solution for the force atpoint A is shown below.

1....-_--+-__\--_-+-_--+__ time1/2 t* t* % t* 2 t*

We now use a finite element mesh often 2-node truss elements to obtainthe compressive force at point A.

All elements uniformly spaced

Topic Fourteen 14-11

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R

Central difference method:

The critical time step for this problem is

Llt = L Ic = t* ( 1 )cr e number of elements

Llt > Lltcr will produce an unstablesolution

We need to use the inital conditionsas follows:

aMOO~=OR

~0 _ HUj -

mjj

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exact

/

1500

Using a time step equal to atcr we obtainthe correct result: For this special

case the exactsolution is obtained

Finite elements

~100QCompressiveforce (N) 500

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t*

-500

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Using a time step equal to ! atcr , thesolution is stable, but highlyinaccurate.

Finite elements1500

1000Compressiveforce (N) 500..

-500.

time

1500

1000Compressiveforce (N) 500

Now consider the use of thetrapezoidal rule:

A stable solution is obtained withany choice of at.

Either a consistent or lumpedmass matrix may be used. Weemploy a lumped mass matrix inthis analysis.

Trapezoidal rule, dt = dterlcDM' .i.nitialconditions computed using MOU = OR.- The solution is inaccurate.

Finite element solution,/"10 element mesh

AA/AA exact solution

A A/A


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-500

t* 2t* timeAA

14-14 Nonlinear Dynamic Response - Part 11

1500

1000Compressiveforce (N)

500

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Trapezoidal rule, dt = dtcrlcDM' zeroinitial conditions.

- Almost same solution is obtained.Finite element solution,

/ 10 element mesh

(!) (!) (!) exact solution(!) '" ~ f

(!)(!)(!

(!)(!)

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t*

-500

Trapezoidal rule, dt = 2dtcrlcDM- The solution is stable, althoughinaccurate.

(!)2t* time(!)

1500Finite element solution,/10 element mesh

.t> at = 2atcrlcDM1000

Compressiveforce (N)

500

o ....t>

-500

t*

.t>exact solution~

I!:>

2t* time


Trapezoidal rule, at = ! atcrlCDMTransparency

14-23

~

~

~


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Trapezoidal rule, at = atcrlCDMFinite element solution, 100 element mesh

1500 /

O,+-----"'-L-----t------'"t-++t-tt1Ht----time

-50

Now consider a two-dimensional modelof the rod: tz

1 0 element 5 L- y. m./>/ .

For this mesh, atcr =P t*/(10 elements)because the element width is less thanthe element length.

If At = t*/(10 elements) is used, the solutiondiverges

-In element 5,

IT 1>- (1000 N)zz 0.01 m2

at t = 1.9 t*

Example: Dynamic response of threedegree-of-freedom systemusing central difference method


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kL = 1 Ibf/ftm = 1 slug

X1 = X2 = X3 = 0

*1 = 0.555 ft/sec*2 = 1.000 ft/sec*3 = 1.247 ft/sec

FomeL{:0.95

Displacement Ix2 - x3 1(~tcrit)linear = 1.11 sec(~terit)nonlinear = 0.14 sec


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Results: Response of right mass

3

D' 2ISp.

(ft) 1

X1 0 HJ--:-':.-----i~:---___='::n--:--~-t(sec)-1

-2

-3.: .:1t = 0.05 sec0: .:1t = 0.15 sec

Response of center mass:Transparency

14-30

.:1 =0.05 sec.

.:1 = 0.15 sec..:O'

2Disp.(ft) 1

X2 0 1--If....L-- ~-----J~--....I...--t(sec)

-1

-2

Response of left mass:

Disp.(ft)

X320'

~\,\:>,

.... 0--

t(sec)


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.: d=0.05 sec.0: d=0.15 sec.

Force (Ibf) in center truss:

TIME dt=0.05 dt=0.159.0 -0.666 -0.700

12.0 -0.804 -0.87715.0 0.504 0.50318.0 0.648 -0.10021.0 -0.132 -0.05924.0 -0.922 0.550

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Example: 10 story tapered tower

3.2 m--1~

Pressureinducedbyblast

Applied load (blast):

32 m

Girder properties:E=2.07x 1011 Pav=0.3A=0.01 m2

As =0.009 m2

1=8.33x 10-5 m4

p = 7800 kg/m3

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Forceperunitlength(N/m)

1000

o+---+-----+---+--~---o 50 100 150 200

time (milliseconds)

Purpose of analysis:

Determine displacements,velocities at top of tower.

Determine moments at base oftower.

We use the trapezoidal rule and alumped mass matrix in the followinganalysis.

We must make two decisions:

Choose mesh (specifically thenumber of elements employed).

Choose time step ~t.

These two choices are closely related:

The mesh and time step to be useddepend on the loading applied.

Topic Fourteen 1421

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Some observations:

The choice of mesh determinesthe highest natural frequency (andcorresponding mode shape) that isaccurately represented in the finiteelement analysis.

The choice of time step determines the highest frequency ofthe finite element mesh in whichthe response is accurately integrated during the time integration.

Hence, it is most effective tochoose the mesh and time stepsuch that the highest frequencyaccurately "integrated" is equal tothe highest frequency accuratelyrepresented by the mesh.

The applied loading can be represented as a Fourier series whichdisplays the important frequenciesto be accurately represented bythe mesh.

Force perunit length(N/m)

Consider the Fourier representation ofthe load function:

f(t) = ~o +I (ancos(2'ITfnt) +bnsin(2'ITfntn=1

Including terms up to

case 1: fn= 17 Hzcase 2: fn= 30 Hz

The loading function is represented asshown next.

Fourier approximation including termsup to 17 Hz:

/APPlied load

/FOUrier approximation

o 100 200time (milliseconds)

Thpic Fourteen 14-23

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Fourier approximation including termsup to 30 Hz:

2000Force perunit length(N/m)

1000

100 200time (milliseconds)

We choose a 30 element mesh,a 60 element mesh and a 120element mesh. All elements are2-node Hermitian beam elements.

30 elements 60 elements 120 elements

Thpic Fourteen 14-25

Determine "accurate" natural frequen- .cies represented by 30 element mesh:

From eigenvalue solutions of the 30and 60 element meshes, we find

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rate

curate

mode natural frequencies (Hz)number 30 element mesh 60 element mesh

1 1.914 1.9142 4.815 4.828 accu3 8.416 8.480

14 12.38 12.585 16.79 17.276 21.45 22.47

17 26.18 28.088 30.56 29.80

inac

Calculate time step:

Tco = 117 Hz = .059 sec

~t= 21 Tco = .003 sec A smaller time step would accurately

"integrate" frequencies, which are notaccurately represented by the mesh.

A larger time time step would notaccurately "integrate" all frequencieswhich are accurately represented bythe mesh.

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Determine "accurate" natural frequencies represented by 60 element mesh:

From eigenvalue solutions of the 60and 120 element meshes, we find

rate

ate

mode natural frequencies (Hz)number 60 element mesh 120 element mesh

5 17.27 17.286 22.47 22.49 accur7 28.08 28.148 29.80 29.759 32.73 33.85

10 33.73 35.0611 36.30 38.96 inaccu

Calculate time step:Transparency

14-46 Teo = 31 Hz= .033 sec

Llt == 2~ Teo = .0017 sec

The meshes chosen correspond tothe Fourier approximations discussedearlier:

30 element mesh _.---" Fourier approximationincluding terms upto 17 Hz.

60 element mesh _.-_a Fourier approximationincluding terms upto 30 Hz.

Pictorially, at time 200 milliseconds,we have (note that the displacementsare amplified for visibility):


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30 elements 60 elements

Pictorially, at time 400 milliseconds,we have (note that the displacementsare amplified for visibility):

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30 elements 60 elements


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Consider the moment reaction at thebase of the tower:

40

20

M(KN-m)

500250

time (milliseconds)

Ol-+----------if---+-------+--

- : 30 elements/VVV: 60 elements

-20

-40

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Consider the horizontal displacementat the top of the tower:

~6IJ .lementsr u

.06 11.04u30 .i.1s(m) .02

0250 500

-.02 time (milliseconds)

-.04

-.06

Consider the horizontal velocity at thetop of the tower:

'Ibpic Fourteen 1429

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.6

V(m/s)

-.2

-.4

-.6

Comments:

IV

11

The high-frequency oscillationobserved in the moment reactionfrom the 60 element mesh isprobably inaccurate. We note thatthe frequency of the oscillation isabout 110Hz (this can be seendirectly from the graph).

The obtained solutions for thehorizontal displacement at the topof the tower are virtually identical.

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Example: Simple pendulum undergoinglarge displacements

~ length = 304.43 cm

tip/"mass = 10 kg

g = 980 cm/sec2

1Initial conditions:

0 = 9000 = 0

One truss element with tip concentratedmass is employed.

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Calculation of dynamic response:

The trapezoidal rule is used tointegrate the time response.

Full Newton iterations are used toreestablish equilibrium during everytime step.

Convergence tolerance:ETOL= 10- 7

(a tight tolerance)


Choose Llt = 0.1 sec. The followingresponse is obtained: Transparency

14-55

6 time (sec)

last obtained solutionSolution procedurefailed during next

'i~ep90

-90

-45

e 45(degrees)

o+---!'r--+---P--+----.:::I---+---

The strain in the truss is plotted:

An instability is observed.Transparency

14-56

time(sec)

2

5x10~5

strain

-10x10- 5


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The instability is unchanged whenwe tighten our convergence tolerances.

The instability is also observedwhen the BFGS algorithm isemployed.

Recall that the trapezoidal rule isunconditionally stable only in linearanalysis.

Choose at = 0.025 sec, using theoriginal tolerance and the full Newtonalgorithm (without line searches).

The analysis runs to completion.

pFinite element solution

6 4(degrees)

0time

-45 (sec)

-9


The strain in the truss is stable: Transparency14-59

strain

3x10- 6

2x 10-6

finite element solution,At=.025 sec

1

4o+----!IL..-----+l~---lL-_fL_->''----

8 12 time (sec)

1 X 10-6

It is important that equilibrium be accuratelysatisfied at the end of each time step: Transparency

1460Finite element solution, at = .025 sec.,equilibrium iterations used asdescribed above.

f90

-45

-90

6 45(degrees) time (sec)

O-l---lf--~"'I=-b-==+-~+=-+=I=:---12"\Finite element solution,at = .025 sec., noequilibrium iterationsused.


5x10- 5

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Although the solution obtained withoutequilibrium iterations is highlyinaccurate, the solution is stable:

Finite element solution, At=O.025 sec.,

10 x 10- 5 no equilibrium iterations used.

strain f Finite element solution, 8t=O.025 sec.,equilibrium iterations used asdescribed above.

4 8 12 time (sec)

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Example: Pipe whip analysis:

360 P=6.57x 105 Ib

~ 2775=~OxJ30

~~-z-diameternot drawn restraint 5.75to scale

all dimensions in inches

Determine the transient responsewhen a step load P is suddenlyapplied.

Finite element model:

Six Hermitian beam elements

~>---".-------


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The analysis is performed using

Mode superposition (2 modes)

Direct time integration

We use, for each analysis,

Trapezoidal rule

- Consistent mass matrix

A convergence tolerance ofETOL = 10-7 is employed.

Eigenvalue solution:

Mode 1, natural frequency = 8.5 Hz

Mode 2, natural frequency = 53 Hz

Choice of time step:

We want to accurately integrate thefirst two modes:

at == 2~ Tco = 2~ ((freqUency ~f mode 2))

=.001 sec

Note: This estimate is based solely on a linearanalysis (Le, before the pipe hits therestraint and while the pipe is still elastic).

Determine the tip displacement:

'Ibpic Fourteen 14-37

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time Transparency(milliseconds) 14-68

0 2 4 6 8 100

~~

tip -2(!)~

disp. (!) Gap(in) (!)

-4 ~ (!)~

~~ (!)

C!l - mode superposition ~-6 ~ - direct integration


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Determine the moment at the built-inend of the beam:

time(milliseconds)

0 2 4 6 8 10Moment 0(Ib-in) (!) ~ (!)

-1 X 107 (!) ~~

-2X 107 (!) ~ (!)~

-3x 107 (!) (!)(!) (!)~~

-4 X 107(!) - mode superposition~ - direct integration

r~ii 12, 5,n2.1 I

r-- ....~EI, constant


Slide14-1

l

l

l

L 38 9 In

d 0.1 in

-x

lit'~H~

WAnalysis of CAD housing with lower support

TIME ISECONlSTIP

DEFLECTION(INCHES) 0,.--_~~__~=--_~O.o-,3!..-_~:!..-_~~_~~_~0.07

-0.02

-0.04

- PETERSON AND BATHE-0.06

o DIRECT INTEGRATION

-0.08 '" MODE 5UPERPOS ITION12 MODES)

-0.10

CRD housing tip deflection

Slide14-2


Slide14-3

p

wR = 22.27 in.h = 0.41 in.e= 26.67"

E = 1.05 X 107 Ib/in2

v = 0.3cry = 2.4 X 104 Ib/in2

ET = 2.1 X 105 Ib/in2

p = 9.8 X 10-2 Ib/in3

Slide14-4

Ten a-node axisymmetric els.Newmark inte (8 = 0.55, (X = 0.276)2 x 2 Gauss integration ~L-consistent mass 600Ib/in~~t = 10f-Lsec, T.L. 0 TIME

Spherical cap nodes under uniform pressure loading

TIME - msec

o 02 04 0.6 0.8 1.0~---r--.-------,---,------,

DEFLECTIONW.-Inches

002

0.04

006

0.08

Dynamic elastic-plastic response of a spherical cap.p deformation independent

TIME - msec

o 02 04 06 08 1-0~--.-------,--.-----.------,


Slide14-5

DEFLECTIONWo- inches

002

004

006

Newmark integration(1= D.5, llC- 0.25)

008

Response of the cap using consistent and lumpedmass idealization

TIME - msec

..,....-_0.:,..::.2__0::.,..4_--=0.,...6::....---=.0...:8:...--.:.,1.0

DEFLECTION

We-IncheS

002

004

006

008

Nagarajan& Popov "

Consistent massNewmark integration

r:5"=O~)

~'.'\.....

Slide14-6

Effect of numbers of Gauss integration points on thecap response predicted

1442 Nonlinear Dynamic Response - Part II

Slide14-7

BLIND FLANGEJ ~ULSE GUN 3-Inch FLEXIBLE NI 200 PIPE~ \m P: r~~p,B ~

1._-----3---'nc-"~:I-GI~D-P-IP-E ~_"*,,.,J JNICKEL 200

E= 30'10' PSIET= 73.7 110

4 PSI 030P = B.3I.10- 4 SLUG;FTITo = 12.B'103 PSI IN

WATERK = 32 liO' PSIP = 9.36 10-\~FT

Slide148

Analysis of fluid-structure interaction problem(pipe test)

2500

2000

1500PRESSURE

IPSI) 000I

500

1.0 20TIME (MSEC)

3.0 4.0

PRESSURE PULSE INPUT

MESH MODEL FOR NI PIPE

t[[ttr'----.If..-~.'""~

I RIGID PIPE Iyt,_.----"'-'~_

Finite element model


Slide14-9

~ilDI~d

f.lI'lR'l,I(~Tdl

- ~Ol~d[lP(RIIII(IO'Al

-ilD,,.AOP(HllIlIIIAl

Slide14-10

I.~ 1.0 2.5llllflllSE(1

Pili II I I! IN f HOM NICOl l PIP~ IP 31

1.0 ,.5 l.~ lP :.~ 4.0 0l-,....,.-4,....,r-tt-~..,."....,._"II[IIIIS[(1

PI T: AT . _~ ,.. '1010 IOI(_E l PIP!

'-dOIU[lPfRIM(Nlo\l

o w w w w ~ w w WIIf1l(llIIS(CI

PillAr ~IN INIONICKHPlP( !pel

-ADINA[IPfHIIII[NTdl

o I.C U J.C 4.C b,() '.0 atHlf :t,l5[:

Contents:

Textbook:

Reference:

Topic 15

Use of ElasticConstitutiveRelations in TotalLagrangianFormulation

Basic considerations in modeling material response

Linear and nonlinear elasticity

Isotropic and orthotropic materials

One-dimensional example, large strain conditions

The case of large displacement/small strai~ analysis,discussion of effectiveness using the total Lagrangianformulation

Hyperelastic material model (Mooney-Rivlin) for analysisof rubber-type materials

Example analysis: Solution of a rubber tensile testspecimen

Example analysis: Solution of a rubber sheet with a hole

6.4, 6.4.1

The solution of the rubber sheet with a hole is given in

Bathe, K. J., E. Ramm, and E. 1. Wilson, "Finite Element Formulationsfor Large Deformation Dynamic Analysis," International Journal forNumerical Methods in Engineering, 9, 353-386, 1975.

USE OF CONSTITUTIVERELATIONS

We developed quite general kinematicrelations and finite elementdiscretizations, applicable to small orlarge deformations.

To use these finite elementformulations, appropriate constitutiverelations must be employed.

Schematically

K = ( BT C B dV, F = ( BT T dVJv \ Jv .J.constitutive relations enter here

For analysis, it is convenient to use theclassifications regarding the magnitudeof deformations introduced earlier:

Infinitesimally small displacements

Large displacements / large rotations,but small strains

Large displacements / large rotations,and large strains

The applicability of material descriptionsgenerally falls also into thesecategories.

Topic Fifteen 15-3

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15-4 Elastic Constitutive Relations in T.L.F.

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Recall:

Materially-nonlinear-only (M.N.O.)analysis assumes (models only)infinitesimally small displacements.

The total Lagrangian (T.L.) andupdated Lagrangian (U.L.)formulations can be employed foranalysis of infinitesimally smalldisplacements, of large displacementsand of large strains (considering theanalysis of 2-D and 3-D solids).

~ All kinematic nonlinearities arefully included.

We may use various material descriptions:

Material Model

Elastic

HyperelasticHypoelasticElastic-plastic

CreepViscoplastic

Examples

Almost all materials, for smallenough stressesRubberConcreteMetals, soils, rocks under highstressesMetals at high temperaturesPolymers, metals

ELASTIC MATERIAL BEHAVIOR:

In linear, infinitesimal displacement,small strain analysis, we are used toemploying

Topic Fifteen 15-5

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stress +dO"

Linear elastic stress-strainrelationship

\1 = EtedO" = E de

strain

For 1-0 nonlinear analysis we can use

stress Nonlinear elasticstress-strainrelationship

slope Ie l


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We can generalize the elastic materialbehavior using:

te tc tQUiJ. = 0 ijrs oErs

doSt = oCijrs doErsThis material description is frequentlyemployed with

the usual constant material moduliused in infinitesimal displacementanalysis

rubber-type materials

Use of constant material moduli, for anisotropic material:

JCijrs = oCiys = A. 8t 8rs + fJ.(8 ir 8js + 8is 8y)

Lame constants:

A.= Ev E(1 + v)(1 - 2v) , fJ. = 2(1 + v)

Kronecker delta:

., = {a; i ~ j-81t 1; i = j-

Thpic Fifteen 15-7

Examples:

2-D plane stress analysis:

Transparency15-9

1 v o

o

1 - v

1v

o 0

/ , (,2 + Ie )corresponds to 0812 = J.L OE12 0e-21

EOc = 2- 1 - v

2-D axisymmetric analysis:

v 0 v1 - v 1 - v

v 1 0 vE(1 - v) 1 - v 1 - v

Q = (1 + v)(1 - 2v)0 0

1 - 2v02(1 - v)

v v 01 - v 1 - v

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local coordinate -1system a-b~e =

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For an orthotropic material, we alsouse the usual constant material moduli:Example: 2-D plane stress analysis

1 Vab 0Ea - Eb

1 0Eb

a sym.

(1 + v)(1 - 2v)

f

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Sample analysis: One-dimensionalproblem:

Material constants E, v

A

E (1 - v)

LConstitutive relation: dS11 = EdE 11


Material constants E, vIn tension:

A

~E (1 - v)

t~ (1 + v)(1 - 2v)

0L JConstitutive relation: 6S11 = E6E 11


Material constants E, vIn tension:

In compression: A

IT E(1-v)ILl t~ (1 + v)(1 - 2v)

I-----=-0L---.:...--- ft - t

Constitutive relation: 0811 = E oE 11

Topic Fifteen 15-9

Transparency15-13

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Transparency15-15

Transparency15-16

We establish the force-displacementrelationship:

t t 1 (t )2oEn = OU1,1 + 2 OU1,1tL _ L

-----oc-

Using tL = L + t~, riSn = EriE 11, wefind

LThis is not a realistic materialdescription for large strains.

The usual isotropic and orthotropicmaterial relationships (constant E, v,Ea , etc.) are mostly employed inlarge displacement / large rotation, butsmall strain analysis.

Recall that the components of the2nd Piola-Kirchhoff stress tensor andof the Green-Lagrange strain tensorare invariant under a rigid bodymotion (rotation) of the material.

- Hence only the actual strainingincreases the components of theGreen-Lagrange strain tensor and,through the material relationship, thecomponents of the 2nd PiolaKirchhoff stress tensor.

- The effect of rotating the material isincluded in the T.L. formulation,

tiF = f. tisl tis dVV'T~

includes invariant under arotation rigid body rotation

Thpic Fifteen 15-11

Transparency15-17

Transparency15-18


Transparency15-19

Pictorially:

01::22

~

r

1 ..-

..... "01::211 f j -61::1101::12.

Deformation to state 1(small strain situation)

Rigid rotation fromstate 1 to state 2

Transparency15-20

For small strains,1 1 1 1OE11 , OE22 , OE12 = OE21 1,1S 1C 1o i} = 0 ijrs oE rs ,a function of E, v

6Si} . 17i}

Also, since state 2 is reached by arigid body rotation,

2 1 2S 18OEi} = oEy. , 0 i} = 0 i}l

27 = R 17 RT"--"'

rotation matrix

Applications:

Large displacement / large rotation butsmall strain analysis of beams, platesand shells. These can frequently bemodeled using 2-D or 3-D elements.Actual beam and shell elements willbe discussed later.

Linearized buckling analysis ofstructures.

Topic Fifteen 15-13

Transparency15-21

Frame analysis:

Axisymmetricshell:

2-D/ plane stress

elements

2-D/ axisymmetric

elements

Transparency15-22


doSi} = oCyrs doErs--s---- Ii ciw

t taOCi}aOCrs

Transparency15-23

Transparency15-24

General shell:

3-D /continuumelements

Hyperelastic material model:formulation of rubber-type materials

t ariwOSi} = -t-

aoEij- t t---S- oC iys OCrs

where

ciw = strain energy density function (perunit original volume)

Rubber is assumed to be an isotropicmaterial, hence

dw = function of (11 , 12 , 13)where the Ii'S are the invariants of theCauchy-Green deformation tensor (withcomponents dC~):

11 = Jc;;

12 = ~ (I~ - JCij JC~)

b = det (dc)

Example: Mooney-Rivlin material law

Jw = C1 (11 - 3) + C2 (12 - 3)'-' '-'

material constants

withb = 1-s-incompressibility constraint

Note, in general, the displacementbased finite element formulationspresented above should be extended toinclude the incompressibility constrainteffectively. A special case, however, isthe analysis of plane stress problems.

Topic Fifteen 15-15

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Transparency15-27

Special case of Mooney-Rivlin law:plane stress analysis

time 0 time t

Transparency15-28

For this (two-dimensional) problem,

JC11 JC12 0

Jc = JC21 JC22 0o 0 JC33

Since the rubber is assumed to beincompressible, we set det (JC) to 1 bychoosing

tc 1033= t t t t(OC11 OC22 - OC12 OC21 )

We can now evaluate 11, 12 :

11 = dC11 + dC22 + Cc tc 1 tc tC)o 11 0 22 - 0 12 0 21

1 tc tc dC11 + dC222=0 110 22+(tC tc tc tC)o 11 0 22 - 0 12 0 21

1 (tC)2 1 (tc 2- 2 0 12 - 2 0 21 )

The 2nd Piola-Kirchhoff stresses are

t aJw aJw ( remember JOSii. = ateo = 2 atc JCiL = 2 JEiL + 8r

I' oc.. y. 0 Y. ir ir

= 2 :JCij- [C, (I, - 3) + Co (i:, - 3)]

- 2 C al1 + 2 C al2- 1 ----r--C 2 ----r--Cao y. ao y.

Topic Fifteen 15-17

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Performing the indicated differentiationsgives

Transparency15-32

This is the stress-strain relationship.

We can also evaluate the tangentconstitutive tensor oC ijrs using

_ a2 JwOc.I/rs - at e .. ate

OVI} Ovrs

a2I1 a2h= 4 C1 t t + 4 C2 t t~C~~Cffi ~C~~Cffi

etc. For the Mooney-Rivlin law

Example: Analysis of a tensile testspecimen:

I 30.5

Mooney-Rivlin constants:

C1 = .234 N/mm2

C2 = .117 N/mm2

thickness = 1 mm

J..371

'Thpic Fifteen 15-19

Transparency15-33

All dimensions in millimeters

Finite element mesh: Fourteen a-nodeelements Transparency

15-34

Gauge~length I

Constraineddisplacements

"""'t---i

Lt1 R2'2

1520 Elastic Constitutive Relations in T.L.F.

Results: Force -deflection curvesTransparency

15-355

4

Applied 3load(N) 2

1

Gaugeresponse--........

Totalresponse~

10 20 30

Transparency15-36

Extension (mm)

Final deformed mesh (force = 4 N):

b t +d to j I; ;\~

/Deformed

r--- 20 in ----1 P = 90 Ib/in2

Topic Fifteen 15-21

Slide15-1

T20in

1p =1.251l10J~2

In 4

b = I in (THICKNESS)

lOin

Analysis of rubber sheet with hole

~ ....... ........ ... __ 150 Ib

_3001b

10 in -------1Finite element mesh

Slide15-2

1522 Elastic Constitutive Relations in T.L.F.

150 r------,r-----,,..--,,....-----,----,............--.....,.--:7--,

1210s642O~__L.__--'........_--'........_--'__---L__.......

o

,,,,/AJ,,

100 t---1-7-'--I+--I7"--;----+---;II

~ ] TL. SOLUTION,P ~ b AND 'DING

50 I------+-~--+----+-~"" t-2P--= B P w

Slide15-3

W[in)

Static load-deflection curve for rubber sheet with hole

Slide15-4

[in]10

c

o 6 10 12 14 16 18 20 22 ~n]

Deformed configuration drawn to scale ofrnhh~r ~h~~t with hole (static analysis)

Topic Fifteen 15-23

Slide15-5

I(sec]0.20/

~~-_///0.12

B

0.080.04

P (Ib)

"C,~/-10M

At = 0.0015 sec /--if/ \

/ \/ \

w[in)'0

Displacements versus time for rubbersheet with hole, T.L. solution

Contents:

Textbook:

Example:

Topic 16

Use of ElasticConstitutiveRelations inUpdatedLagrangianFormulation

Use of updated Lagrangian (U.L.) formulation

Detailed comparison of expressions used in totalLagrangian (T.L.) and V.L. formulations; strains,stresses, and constitutive relations

Study of conditions to obtain in a general incrementalanalysis the same results as in the T.L. formulation, andvice versa

The special case of elasticity

The Almansi strain tensor

One-dimensional example involving large strains

Analysis of large displacement/small strain problems

Example analysis: Large displacement solution of frameusing updated and total Lagrangian formulations

6.4, 6.4.1

6.19

SO FAR THE USE OFTHE T.L. FORMULATION

WAS IMPLIED

Now suppose that we wish to use theU.L. formulation in the analysis. Weask

Is it possible to obtain, using the U.L.formulation, identically the samenumerical results (for each iteration)as are obtained using the T.L.formulation?

In other words, the situation is

Program 1

Topic Sixteen 163

Transparency16-1

Transparency16-2

Only T.L. formulationis implemented

- Constitutive relations are

JSij- = function of displacements

doSij- = oCijrs doErs

Information obtained from physicallaboratory experiments.

P

~~

Program 1 results

16-4 Elastic Constitutive Relations in u'L.F.

Transparency16-3

Program 2 Question:1-------------1

Only U.L. formulation How can we obtainis implemented with program 2

- Constitutive relations are identically the samet'Tij. = ~ CD results as aredtSij. = ~ obtained from

'--- ----' program 1?

Transparency16-4

To answer, we consider the linearizedequations of motion:

~v OCijrS oers 80ei~dV +~v JSijo80'T)i~OdV]

~T.L.

_ t+~t t 0- m- 0v OSi~ 80ei~ dV

Terms used in the formulations:

T.L. U.L.formulation formulation Transformation

tfc dV 1tdV dV = oP tdVv tv P

t tOeij, OT)i} teij' tTl i}

Oei} = OXr,i OXs,} terst tOT)i} = OXr,i OXs,} tT)rs

OOeij, OOT)i} Oteij, OtT) i}OOei} = JXr,i JXS ,} OtersOOT)i} = JXr,i JXS ,} OtT)rs

Derivation of these kinematicrelationships:

A fundamental property of JCi} is that

JCi} dOXi dOx} = ~ (CdS)2 - (OdS)2)

Similarly,

t+dJci} dOXi dOx} = ~ ((t+dtds)2 - (OdS)2)

and

tCrs dXr dtxs = ~ ((t+dtds)2 - CdS)2)

Topic Sixteen 16-5

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Transparency16-7

timeD

Transparency16-8

Fiber dO~ of length ods moves tobecome dt~ of length tds.

Hence, by subtraction, we obtain

OCi}doXi dOx} = tCrs dtxr dtxs

Since this relationship holds forarbitrary material fibers, we have

t tOCi} = OXr,i OXs,pCrs

Now we see thatt t t t

oe~ + o'T)~ = OXr,i OXs,} ters + OXr,i OXS ,} t'T)rs

Since the factors 6Xr,i 6Xs,~ do notcontain the incremental displacementsUi, we have

ttl" .oe~ = OXr,i oXs,pers ~ Inear In Ui

t t d t" "oTJ~ = OXr,i oXsJ tTJrs ~ qua ra IC In Ui

In addition, we have

oOei} = 6Xr,i 6Xs,} oters

OOTJi} = 6Xr,i 6Xs,} OtTJrs

These follow because the variation istaken on the confi~uration t+Llt andhence the factors OXr,i 6Xs,} are taken asconstant during the variation.

Topic Sixteen 16-7

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16-8 Elastic Constitutive Relations in U.L.F.

Transparency16-11

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We also have

T.L. U.L.Transformation

formulation formulation

0

JSi} t ts POt 01'i} o i} = t t Xi,m 1'mn t Xj,nP

0

oCijrs tC~rsC Po 0 Coo

o its = t t Xi,a t Xj,b t abpq t Xr,p t Xs,qP

(To be derived below)

Consider the tangent constitutivetensors oC~rs and tCiys :

Recall that

dOSi} = oCijrS doErs

~d'ff '1dtS" = tG dtE I erentlaIi Irs rs increments.s--

Now we note thato

dOSii = ~ PXi a PXi b dtSabp , "

doErs = JXp,r Jxq,s dtEpq

Hence

(~ PXi,. ?Xj.b d,Sab) =DC". (6xp,r D'xq,S d'E~... I I T

doSik doErs

Solving for dtSab gives

d,S.b = (~6x.'i 6xb.j. DCijrs 6xp,r Jx."s) litEpq\ '

tCabpq

And we therefore observe that thetangent material relationship to be usedis

t

C _Pt t C t tt abpq - 0-=- OXa,i OXb.J- o iJ"s oXp,r oXq,sP

Topic Sixteen 16-9

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Transparency16-14

16-10 Elastic Constitutive Relations in D.L.F.

Transparency16-15

Transparency16-16

Now compare each of the integrals appearing inthe T.L. and U.L. equations of motion:

1) fav dSij.8oeij. dV =ftvtTij.8tet tdV ?, ,

True, as we verify by substituting theestablished transformations:

Lv (~ PXi,m ~mn PX~n ) (dXr,i dxs.;. 8ters) dV. IS . 30eiL

o t =r tTmn 8ters (~Xi,m dXr,i) (~Xj.,n dxs,j.) ~p dV~v p

3m 3ns -1-dV-

=r tTmn 8temn tdVJtv

2) fov dSt8011t dV =JvtTt8tl1t tdV ?I I


fav (:PPXi,m tTmn PX~n) (dXr,i dxs,j. 8tl1rs) dV. P " ,. .

riSt 30TJij-

=r tTmn 8tl1rs (PXi,m dXr,i)(?X~,n dxs,~) ~P dVJov P----=---~~

3m 3ns IdV

= r tTmn 8tl1mn tdVJtv

3) fov oCijrs oers 80eij dV = t tC~rs ters &e~ tdV ?I I


fa (0 )Po 0 0 00v tp t Xi,a t Xj,b tCabpq t xr,p t Xs,q X

\ I I

OCijrs

(dXk,r dxe.s tekf) (dXm,i dxn.j- 8temn) dV\ T I \ J

Provided the establishedtransformations are used, the threeintegrals are identical. Therefore theresulting finite element discretizationswill also be identical.

(JKL + JKNd dU = t+~tR - JF

(~KL + ~KNd dU = t+~tR - ~F

Topic Sixteen 16-11

Transparency16-17

Transparency16-18

dKL = asL

dKNL = asNL

dE = tE

The same holds foreach equilibrium iteration.


Transparency16-19

Transparency16-20

Hence, to summarize once more,program 2 gives the same results asprogram 1, provided

CD ~ The Cauchy stresses arecalculated from

tt _Pt ts tTiJ- - 0:::- OXi,m 0 mn oXj,n

P ~ The tangent stress-strain law is

calculated fromt

C _Pt t C t tt ifs - up OXi,a OXj,b 0 abpq oXr,p oXs,q

Conversely, assume that the materialrelationships for program 2 are given,hence, from laboratory experimentalinformation, tTij- and tCijrs for the U.L.formulation are given.Then we can show that, provided theappropriate transformations

ots - POt 0o it - -t tXi,m Tmn tXj,n

Po

C - Po 0 Cooo ijrs - -t tXi,a tXj,b t abpq tXr,p tXs,qP

are used in program 1 with the T.L.formulation, again the same numericalresults are generated.

Hence the choice of formulation (T.L.vs. U.L.) is based solely on thenumerical effectiveness of the methods:

The ~BL matrix (U.L. formulation)contains less entries than the tiBLmatrix (T.L. formulation).

The matrix product BT C B is lessexpensive using the U.L. formulation.

If the stress-strain law is available interms of tis, then the T.L. formulationwill be in general most effective.

- Mooney-Rivlin material law-Inelastic analysis allowing for large

displacements / large rotations, butsmall strains

Topic Sixteen 16-13

Transparency16-21

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Transparency16-23

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THE SPECIAL CASEOF ELASTICITY

Consider that the components JCi1S aregiven:

ts tc ta i} = a ijrs aE rsFrom the above discussion, to obtainthe same numerical results with theU.L. formulation, we would employ

tt _ P t (tc t) tTi} - op aXi,m a mnrs aE rs aXj.n

t

C Pt t C t tt ilS = 0 aXi,a aXj,b a abpq aXr,p aXs,qP

We see that in the above equation, theCauchy stresses are related to theGreen-Lagrange strains by atransformation acting only on the mand n components of JCmnrs .

However, we can write the total stressstrain law using a tensor, ~C~rs, byintroducing another strain measure,namely the Almansi strain tensor,

tTi} = ~C~s !E~~/AlmanSi strain tensor

ttca _ P t t tc . t ttits - ~ aXi,a aXj.b a abpq aXr,p aXs,q

P

Definitions of the Almansi strain tensor:

~Ea = ! (I - ~XT ~X) / a:uk- 2 - - - aXJ-ta 1(t t t t)tE = - tU" + tU' . - tUk' tUk .I'" 2 I.... ~.I .I.~

A symmetric strain tensor, ~E~ = ~E;

The components of ~E a are notinvariant under a rigid body rotationof the material.

Hence, ~Ea is not a very useful strainmeasure, but we wanted to introduceit here briefly.

Topic Sixteen 16-15

Transparency16-25

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Transparency16-27

Example: Uniaxial strain

t t~ 1 (t~)2of 11 = 0L +"2 0L

strain

1.0

-1.0=--_~~

Green-Lagrange

Engineering

~:.--_- Almansita

1.0 L

Transparency16-28

It turns out that the use of ~Clrs withthe Almansi strain tensor is effectivewhen the U.L. formulation is used witha linear isotropic material law for largedisplacement / large rotation butsmall strain analysis.

'Ibpic Sixteen 16-17

In this case, ~C~s may be taken as

t . atCijrS = A8ij- 8rs + f.L(8ir 8js + 8is 8jr)

j

= tCiys constantsPractically the same response iscalculated using the T.L. formulationwith

JCijrS = A 8it 8rs + f.L(8ir 8js + 8is 8jr)\

Transparency16-29

= oC iys constants

MALLET ET AL

TL au L SOLUTIONS

r-~---{

Ii

TWELVE &-NODE ELEMENTSFOR HALF OF ARCH

Slide161

E =10 X 108 Ib./in.2v= 0.2

04030.201

20

30

40

LOAD P [Ib]

VERTICAL DISPLACEMENT AT APEX Wo [in]

Load-deflection curve for a shallowarch under concentrated load

16-18 Elastic Constitutive Relations in V.L.F.

Transparency16-30

The reason that practically the sameresponse is calculated is that therequired transformations to obtainexactly the same response reduce tomere rotations:

Namely, in the transformations from~C~s to JCabPq, and in the relationbetween oCijrs and tCiys ,

o-.e ..!- 1 [JXi.~ = JX = JR Jutp -, r

..!- tR-0_

Transparency16-31

However, when using constant materialmoduli (E, v) for large strain analysis,with

totally different results are obtained.

Topic Sixteen 16-19

E= -:-:--_E--'-:-(1---,---:-.--:v)__=__:_(1 + v) (1 - 2v)

Transparency16-32

A

Lt - tBefore, we used 0811 = E oE11.

. t - t aNow, we consider 'T11 = E tE 11

Consider the 1-D problem alreadysolved earlier:

Material constants E, v

Here, we haveTransparency

16-33

t tp1'11 = ~

A

U tL 0L tAt E- t asing = + '-1, 'T11 = tE11,obtain the force-displacementrelationship.

we

1620 Elastic Constitutive Relations in u'L.F.

Transparency16-34

~tp = E2A (1 - C~r)-t-----jL-----+-- t~

Transparency16-35

Example: Corner under tip load

L

h A

L = 10.0 m}!! = _1h = 0.2 m L 50b = 1.0 mE = 207000 MPav = 0.3

Finite element mesh: 51 two-dimensionala-node elements

25 elements-I

Topic Sixteen 16-21

Transparency16-36

JRf'+t""-----"""1JI-'""'1J

not drawnto scale

25elements

:~I I I I I

All elements areplane strainelements.

Transparency16-37

Consider a nonlinear elastic analysis.For what loads will the T.L. and U.L.formulations give similar results?


Transparency16-38 For large displacement/large

rotation, but small strain conditions, the T.L. and U.L. formulations will give similar results.

- For large displacement/largerotation and large strain conditions, the T.L. and U.L. formulations will give different results,because different constitutiverelations are assumed.

Beam elements,.-:T.L. formulation

ol&-===----+-----+---_+_-o 5 10 15Vertical displacement of tip (m)

Results: Force-deflection curve Over the range of loads shown, the T.L.

and U.L. formulations give practicallyidentical results

The force-deflection curve obtained withtwo 4-node isoparametric beamelements is also shown.

62-D elements,T.L. and U.L.formulations~4

Force(MN)2

Transparency16-39

Deformed configuration for a load of 5 MN(2-D elements are used):

Undeformed 1Rr--------IIIIII Deformed, load = 1 MN

Deformed, load=5 MN

Numerically, for a load of 5 MN, we have,using the 2-D elements,

T.L. formulation U.L. formulationvertical tip

15.289 m 15.282 mdisplacement

The displacements and rotations are large.However, the strains are small - they canbe estimated using strength of materialsformulas:

Cbase = M~~2) where M (5 MN)(7.5 m)

3%

Topic Sixteen 16-23

Transparency16-40

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Contents:

Textbook:

Example:

References:

Topic 17

Modeling ofElasta-Plastic andCreepResponse Part I

Basic considerations in modeling inelastic response

A schematic review of laboratory test results, effects ofstress level, temperature, strain rate

One-dimensional stress-strain laws for elasto-plasticity,creep, and viscoplasticity

Isotropic and kinematic hardening in plasticity

General equations of multiaxial plasticity based on ayield condition, flow rule, and hardening rule

Example of von Mises yield condition and isotropichardening, evaluation of stress-strain law for generalanalysis

Use of plastic work, effective stress, effective plasticstrain

Integration of stresses with subincrementation

Example analysis: Plane strain punch problem

Example analysis: Elasto-plastic response up to ultimateload of a plate with a hole

Computer-plotted animation: Plate with a hole

Section 6.4.2

6.20

The plasticity computations are discussed in

Bathe, K. J., M. D. Snyder, A. P. Cimento, and W. D. Rolph III, "On SomeCurrent Procedures and Difficulties in Finite Element Analysis of Elastic-Plastic Response," Computers & Structures, 12, 607-624, 1980.

17-2 Elasto-Plastic and Creep Response - Part I

References:(continued)

Snyder, M. D., and K. J. Bathe, "A Solution Procedure for Thermo-Elastic-Plastic and Creep Problems," Nuclear Engineering and Design, 64,49-80, 1981.

The plane strain punch problem is also considered in

Sussman, T., and K. J. Bathe, "Finite Elements Based on Mixed Interpolation for Incompressible Elastic and Inelastic Analysis," Computers& Structures, to appear.

. WE "D/SC.L{ SS"E D INTHE "PRE\llO\A ANAl'lSIS

- WE GENE~AU2E OuR

HObEUN6 CONSlbERA.lloNS

10 2-1) ANb3.-1)

STRES,S 5\1v..A.TI\lNS

Markerboard17-1


Transparency17-1

Transparency17-2

MODELING OF INELASTICRESPONSE:

ELASTO-PLASTICITY, CREEPAND VISCOPLASTICITY

The total stress is not uniquelyrelated to the current total strain.Hence, to calculate the responsehistory, stress increments must beevaluated for each time (load) stepand added to the previous totalstress.

The differential stress increment isobtained as - assuming infinitesimallysmall displacement conditions-

dO'ij- = C~s (ders - de~~)

where

C~s = components of the elasticitytensor

ders = total differential strain increment

de~~ = inelastic differential strainincrement

The inelastic response may occurrapidly or slowly in time, depending onthe problem of nature considered.Modeling: In ~Iasticity, the model assumes that

de~s occurs instantaneously with theload application.

In creep, the model assumes thatde~~ occurs as a function of time.

The actual response in nature can bemodeled using plasticity and creeptogether, or alternatively using aviscoplastic material model.

- In the following discussion weassume small strain conditions,hence

we have either a materiallynonlinear-only analysis

or a large displacement/largerotation but small strainanalysis

Topic Seventeen 17-5

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Transparency17-5

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As pointed out earlier, for the largedisplacement solution we would usethe total Lagrangian formulation andin the evaluation of the stress-strainlaws simply use

- Green-Lagrange strain componentfor the engineering strain components

and

- 2nd Piola-Kirchhoff stress components for the engineering stresscomponents

Consider a brief summary of someobservations regarding materialresponse measured in the laboratory

We only consider schematically whatapproximate response is observed; no detailsare given.

Note that, regarding the notation, no time, t,superscript is used on the stress and strainvariables describing the material behavior.

MATERIAL BEHAVIOR,"INSTANTANEOUS"

RESPONSE


Transparency17-7

Tensile Test:

Crosssec:tionalarea Ao-z...

Assume small strain conditions

behavior in compressionis the same as in tension

Hence

e - eoe = --::---=-eop

(J' = -Ao

engineeringstress, (J

assumed

II

II

II

II

~ ", ', ,.",;'......._-----

fracture

x ultimatestrain

5'

ngineeringtest strain, e

Constant temperature

Transparency17-8


Transparency17-9

Transparency17-10

Effect of strain rate:(J'

de. .dt Increasing

e

Effect of temperature(J'

temperature is increasing

e


MATERIAL BEHAVIOR, TIMEDEPENDENT RESPONSE Transparency

17-11

Now, at constant stress, inelasticstrains develop.

Important effect for materials whentemperatures are high

Typical creep curveTransparency

17-12

Engineering strain, efracture

Tertiaryrange

Secondaryrange

a = constanttemperature = constant

Instantaneousstrain {(elastic and Primaryelasto-plastic) range

--:..+----=--+------'--~--+------l-

time


Transparency17-13

Effect of stress level on creep strain

ex

temperature = constant

x ---.s---::fracture

I

time

Effect of temperature on creep strainTransparency

17-14

e-S" fracture

x ""x

u=constant

time

MODELING OF RESPONSE

Consider a one-dimensional situation:tp

~tu h= A

~ jt.---tp 'e ~ ~L L

We assume that the load is increasedmonotonically to its final value, P*.

We assume that the time is "long" sothat inertia effects are negligible(static analysis).


Transparency17-15

Load plasticityeffects

P*:..-....r_--,pt-redominate

.creep effects

fTransparency

17-16

time-dependent inelastic strainsare accumulated - modeled ascreep strains

'-"time inte~al t* (small)without time-dependentinelastic strains

time


strain

tOe

trr = E teEtelN = teP

1

rry- - - - -

Plasticity, uniaxial, bilinearmaterial model stressTransparency

17-17

Transparency17-18

Creep, power law material model:eC = ao

'Ibpic Seventeen 17-13

Viscoplasticity:

Time-dependent response is modeledusing a fluidity parameter 'Y:

e =


Transparency17-21 PLASTICITY

Transparency17-22

So far we considered only loadingconditions.

Before we discuss more generalmultiaxial plasticity relations, considerunloading and cyclic loadingassuming uniaxial stress conditions.

Consider that the load increases intension, causes plastic deformation,reverses elastically, and again causesplastic deformation in compression.

load

elastic

plastic

time

Bilinear material assumption, isotropichardening (T ET

(Ty

e

plastic strain (T~e~

plastic strainI---+-l e~

Bilinear material assumption, kinematichardening (J ET

(Ty

plastic strain e~

e


Transparency17-23

Transparency17-24


Transparency17-25

Transparency17-26

MULTIAXIAL PLASTICITY

To describe the plastic behavior inmultiaxial stress conditions, we use

A yield condition

A flow rule

A hardening rule

In the following, we consider isothermal(constant temperature) conditions.

These conditions are expressed using astress function tF.

Two widely used stress functions are the

von Mises function

Drucker-Prager function


von Mises

tF =~ ts ts.. _ tK2 I} II'Transparency

17-27

Drucker-Prager

t ~a._II ; t(j =3

1 t t- s .. s ..2 If If

We use both matrix notation and indexnotation: Transparency

17-28

dCT11dCT22dCT33dCT12dCT23dCT31

dCT =

matrix notationnote that both de~2and de~1 are added

de~1de~2

de~3

de~2 + de~1de~3 + de~2de~3 + de~1


Transparency[ de~, de~2 de~3 ]17-29

d P- de~1 de~2 de~3eij -de~1 de~2 de~3

indexnotation

[ dcrl1 d

2) Flow rule (associated rule):

P t atFdeii. = X. -at

I ITi}

where tx. is a positive scalar. 1-D equivalent:

d P 2 t\ te11 = 3 I\. IT

d P 1 t\ te22 = - 3 I\. IT

d P 1 t\ te33 = - 3 I\. IT

3) Stress-strain relationship:

dIT = CE (d~ - d~P)

1-D equivalent:

dIT = E (de11 - de~1)

Thpic Seventeen 17-19

Transparency17-31

Transparency17-32


Transparency17-33

Transparency17-34

Our goal is to determine CEP such that

dO" = CEP de- ~-

\instantaneous elastic-plastic stress-strain matrix

General derivation of CEP :

Define

Using matrix notation,


Transparency17-35

tgT = [tq11 : tq : tq :: 22: 33:

't. t Ii P22: P33 i

tIt It]P12 i P23 i P31

We now determine tx. in terms of de:Using tF = 0 during plastic deformations,

t atF atF pd F = -t- d(Jii. + --:::t=Pat deii.a(Ji} I ei} I

= 'gT do- - 'ET~tx.tg

=0

Transparency17-36

17-22 Elasto-Plastic and Creep RespoDse - Part I

Transparency17-37

Transparency17-38

Also

tgTdO" = tgT (QE (d~ - d!t))L . t

The flow rule assumption may bewritten as

HencetgT dO" = tgT (CE (d~ _ tA tg))= tA tQT tg-

Solving the boxed equation for tA gives

Hence we can determine the plasticstrain increment from the total strainincrement: I ..

tota strain Increment

p tgTCEd~~tde = ( ) 9~ - tQTtg + tgT CEtg

plastic strainincrement

Example: Von Mises yield condition,isotropic hardening

Two equivalent equations:


Transparency17-39

Transparency17-40

t ~O"Y=T

(trr t )2 (t t)2 (t t)2

~~\+t-t~/?~principal stresses

tF _ 1 t t t. t _ 1 t 2- 2: sij.~ K , K - 3 0"Y~5 t

d . t " (Tmm ~eVlatonc s resses: Sij- = (Tij- - -3- Uij


Transparency17-41 Yield surface

for plane stressEnd view ofyield surface

Transparency17-42

We now compute the derivatives of theyield function.

First consider tpy.:

t _ atF _ a (1 t t 1 t 2py. - - atef! - - atef! 2 sy. SiJ- - 3 O'y)

II' t(J"it fixed II'

(t(J"ij- fixed implies tSij- is fixed)

What is the relationship between tcryand the plastic strains?

We answer this question using theconcept of "plastic work".

The plastic work (per unit volume) isthe amount of energy that isunrecoverable when the material isunloaded.

This energy has been used increating the plastic deformationswithin the material.

Pictorially: 10 example


Transparency17-43

stress

~et

Shaded area equalsplastic work t wp:

'----------strain

Transparency17-44

rtePIjo In general, 'Wp =Jo Tay.de~


Transparency17-45

Transparency17-46

Consider 1-0 test results: the currentyield stress may be written in terms ofthe plastic work.

Shaded area =:

Wp = ~ (~T - i) (t(1~ - (1~)'---------- strain

We can now evaluate tPij- - whichcorresponds to a generalization of the1-D test results to multiaxial conditions.

t _ 2 t (d\JY atWp)/at(1Y

Pij- - 3 (]'Y dtWp ater ate~\ t /.

(effective stress)

(increment iiieffective

t plastic strain)result is obtained

Alternatively, we could have used thatdtWp = tcr dtet

where

and then the sameusing

t .. _ 2 t (d\ry ateP

)Pit - 3 (J'y dteP ate~

Next consider tq~:


Transparency17-47

Transparency17-48

17-28 ElastoPlastic and Creep Response - Part 1

We can now evaluate gEP:

symmetric

Transparency17-49

cEP = _E_- 1+v

de11 de22 2 de122

'I I I

1 - V (') V 'I I I Q. 1 , I-- - ~ 511 I -- - ~ 511 522 . - P 511 512 .1-2v 11-2v I I I- - - - - - - - - - - - - - - - -:2: -1- I" - - - - - - I .......

I 1 - V (') I I Q. ' , II 1 _ 2v - ~ 522 I I - P 522 512 IL 1_ I" - - - - - - I -

I . I - ~ '533 1512 I ...I I I

-'1-----2 'I - -~('512) I .I 2 I-------1-

I I

Transparency17-50

3 1 ( 1 )where 13 = - f22 O"y 1 + g E ET 1 + v3 E - ET E

Evaluation of the stresses at time t+ ~t:

The stress integrationmust be performed ateach Gauss integrationpoint.

We can approximate the evaluation ofthis integral using the Euler forwardmethod.

Without subincrementation:

J:t+~le I+~I I- e e

Ie CEP d~ . CEP ,d~"/ - - -- t

With n subincrements:

den ~t

t+~--s-n

+ ...


Transparency17-51

Transparency17-52

den

t+(n-1)Lh


Transparency17-53

Pictorially:

tsubincrements

t+8t

Transparency17-54

Summary of the procedure used tocalculate the total stresses at timet+dt.Given:

STRAIN = Total strains at time t+dtSIG = Total stresses at time tEPS = Total strains at time t

(a) Calculate the strain incrementDELEPS:

DELEPS = STRAIN - EPS

(b) Calculate the stress incrementDELSIG, assuming elastic behavior:

DELSIG = CE * DELEPS(c) Calculate TAU, assuming elastic

behavior:TAU = SIG + DELSIG

(d) With TAU as the state of stress,calculate the value of the yieldfunction F.

(e) If F(TAU) < 0, the strain incrementis elastic. In this case, TAU iscorrect; we return.

(f) If the previous state of stress wasplastic, set RATIO to zero and goto (g). Otherwise, there is atransition from elastic to plastic andRATIO (the portion of incrementalstrain taken elastically) has to bedetermined. RATIO is determinedfrom

F (SIG + RATIO * DELSIG) = 0since F = 0 signals the initiation ofyielding.


Transparency17-55

Transparency17-56

17-32 ElastoPlastic and Creep Response - Part I

Transparency17-57 (9) Redefine TAU as the stress at start

of yieldTAU = SIG + RATIO * DELSIG

and calculate the elastic-plasticstrain increment

DEPS = (1 - RATIO) * DELEPS(h) Divide DEPS into subincrements

DDEPS and calculateTAU ~ TAU + CEP * DDEPS

for all elastic-plastic strainsubincrements.


1........--- 2b ---I.~I

p

Slide17-1

PLASTIC ZONE

Plane strain punch problem

Slide17-2

P/l Ii.I

z 10- .....I

lOl- A ,,"~.

Finite element model of punch problem


Slide17-3

STRESS, 0' lplll

'00

"200 Ib.'CW.. h \. "... "".. T,.

- - - THEORTICAL

-- - -----0I 0 2.0 30DISTANCE FROM CENTERLINE, b (inl

Solution of Boussinesq problem-2 pt. integration

Slide17-4

STRESS, 0' lpoi)

200

'00

VIZ

er".. T,Z

- - - TH(ORETlCAl

-----------10 2.0 30DISTANCE FROM CENTERLINE, b (in)

Solution of Boussinesq, problem-3 pt. Integration

APPLIEDLOAD,P/2kb

p

~ 2 pi In,a S"pI inl


Slide17-5

0.05 QIQ 0.'5DISPLACEMENT, ,,/b

0.20

Load-displacement curves for punch problem


Transparency17-58

Transparency17-59

Limit load calculations:

p

rTTTTl

o

Plate is elasto-plastic.

Elasto-plastic analysis:

Material properties (steel)a

740 -I--

a(MPa)

~Er = 2070 MPa, isotropic hardening

'-E=207000 MPa, v=0.3

e This is an idealization, probably

inaccurate for large strain conditions(e > 2%).

TIME = 0LOAD = 0.0 MPA

TIME - 41LOAD - 512.5 MPA

TIME - 62LOAD - 660.0 MPA

=====..:=r~--.,.,......~"-----....----....~-----,.._-...-.~....,....,11...,!!!!....~....

\


ComputerAnimationPlate with hole

Contents:

Textbook:

References:

Topic 18

Modeling ofElasto-Plastic andCreepResponse Part II

Strain formulas to model creep strains

Assumption of creep strain hardening for varying stresssituations

Creep in multiaxial stress conditions, use of effectivestress and effective creep strain

Explicit and implicit integration of stress

Selection of size of time step in stress integration

Thermo-plasticity and creep, temperature-dependency ofmaterial constants

Example analysis: Numerical uniaxial creep results

Example analysis: Collapse analysis of a column withoffset load

Example analysis: Analysis of cylinder subjected to heattreatment

Section 6.4.2

The computations in thermo-elasto-plastic-creep analysis are describedin

Snyder, M. D., and K. J. Bathe, "A Solution Procedure for Thermo-Elastic-Plastic and Creep Problems," Nuclear Engineering and Design, 64,49-80, 1981.

Cesar, F., and K. J. Bathe, "A Finite Element Analysis of QuenchingProcesses," in Numerical Methods jor Non-Linear Problems, (Taylor,C., et al. eds.), Pineridge Press, 1984.

18-2 Elasto-Plastic and Creep Response - Part II


The effective-stress-function algorithm is presented in

Bathe, K. J., M. Kojic, and R. Slavkovic, "On Large Strain Elasto-Plasticand Creep Analysis," in Finite Element Methods for Nonlinear Problems (Bergan, P. G., K. J. Bathe, and W. Wunderlich, eds.), SpringerVerlag, 1986.

The cylinder subjected to heat treatment is considered in

Rammerstorfer, F. G., D. F. Fischer, W. Mitter, K. J. Bathe, and M. D.Snyder, "On Thermo-Elastic-Plastic Analysis of Heat-Treatment Processes Including Creep and Phase Changes," Computers & Structures,13, 771-779, 1981.

CREEPWe considered already uniaxialconstant stress conditions. A typicalcreep law used is the power creep laweO = ao (181 t~.

time

Aside: other possible choices for the creeplaw are

eC = ao exp(a1 cr) [1 - exp(-a2 c~:r4 t)]+ as t exp(ae cr)

eC

= (ao (cr)a1) (ta2 + a3 t84 + as ta6) exp (te +-2~3.16)~'-'

temperature. in degrees C

We will not discuss these choices further.

Topic Eighteen 183

Transparency18-1

Transparency18-2


Transparency18-3

The creep strain formula eC = aO (Tal ta2

cannot be directly applied to varying stresssituations because the stress history doesnot enter directly into the formula.

-----(12Ccreep strain not affected

_-.J~---(11 by stresshistory priorto t1

time

"time

decrease in the creepstrain is unrealistic

f----.......-------(11

~---,...--------(12

Example:

J I:eC

Transparency18-4

The assumption of strain hardening:

The material creep behavior dependsonly on the current stress level andthe accumulated total creep strain.

To establish the ensuing creep strain,we solve for the "effective time" usingthe creep law:

_ totally unrelatedteC = aQ \:ra1~~ to the physical

time

(solve for t)

The effective time is now used in thecreep strain rate formula:

Now the creep strain rate dependson the current stress level and onthe accumulated total creep strain.

Topic Eighteen 18-5

Transparency18-5

Transparency18-6


Transparency18-7

Transparency18-8

Pictorially: Decrease in stress

to'-1-__------------

4---1------------- a'+-------------- time

0'2

material/response

~__~;::::::::.. a'

~~------------time

Increase in stress

to'-+--------------~------........'-------- a'

+---------------- time0'2

material--.-:::::::.~---=--..:::>

response

-4:..-------------- time

Reverse in stress (cyclic conditions)

-+----+--------time


Transparency18-11 We define

,....---.....t- 3 t t(J = 2 Si~ sy. (effective stress)

t-Ce = (effective strain)

Transparency18-12

and use these in the uniaxial creeplaw:

eC = ao cr81 {82

The assumption that the creep strainrates are proportional to the currentdeviatoric stresses gives

te' C t ts (' M' I t" 'ty)i~ = ry i~ as In von Ises p as ICI

try is evaluated in terms of the effectivestress and effective creep strain rate:

3 t.:.Ct ery = 2 tcr

Using matrix notation,

d~c = Coy) (D to") dt" ;/deviatoricstresses

For 3-D analysis,

Topic Eighteen 18-9

Transparency18-13

D=

! -~ -~! -~

!

symmetric1

11

In creep problems, the timeintegration is difficult due to the highexponent on the stress.

Solution instability arises if the Eulerforward integration is used and thetime step ~t is too large.- Rule of thumb:

A -c -< _1 (t-E)L.l~ - 10 ~

Alternatively, we can use implicitintegration, using the a-method:

HaatO" = (1 - a) to" + a HatO"

Transparency18-14

1810 Elasto-Plastic and Creep Response - Part II

Transparency18-15

Iteration algorithm:t+~ta(l-1) = ta +_(k) _

cE [~(1-1 L ~t t+a~t'Y~k-=-1{) (0 t+a~t!!~k-=-1{)]

we iterate ateach integrationpoint

r

xxx

x

x

k = iteration counter at each integration point

s

Transparency18-16 a>V2 gives a stable integration

algorithm. We use largely a = 1.0.

In practice, a form of NewtonRaphson iteration to accelerateconvergence of the iteration canbe used.

Choice of time step dt is nowgoverned by need to convergein the iteration and accuracyconsiderations.

Subincrementation can be employed.

Relatively large time steps can beused with the effective-stressfunction algorithm.

THERMO-PLASTICITY-CREEPstress

O'y3-+-__Plasticity: O'y2-+-_-+-

0'y 1---l---74L...-----r--Increasingtemperature

strain

Topic Eighteen 18-11

Transparency18-17

Transparency18-18

Creep:

creepstrain

Increasingtern erature

time


Transparency18-19

Transparency18-20

Now we evaluate the stresses usingt+~te

I+AIQ: = IQ: + r, -TCEd(~ _ ~p _ ~c _ ~TH)J,~ .-5~

thermalstrains

Using the ex-method,

I+AIQ: = t+AICE{ [~ _ ~p _ ~c _ ~TH]

+[I~ _ I~P _ t~C _ I~TH]}

where

e = I+Ale - Ie- --

and~p = Llt (t+a~t~) (D t+a~ta)

~c = Llt (t+a~t'Y) (D t+a~ta)

e;H = (t+~ta t+~t8 - ta t8) Oi.j-where

ta = coefficient of thermal expansion attime t

t8 = temperature at time t

The final iterative equation is

- Lit C+Cldt~~k~1{) (0 t+Cldt(J~k~1{)

- Lit (t+Cldt)'~k~1{) (0 t+Cldt(J~k~1{)

_~TH]and subincrementation may also beused.

Numerical uniaxial creep results:


Transparency18-21

5m

Area =1.0 m2

Uniaxial stress 0"

Creep law:eC = ao (0")a1 ta2

stress in MPat in hrE = 207000 MPav = 0.3

Transparency18-22


Transparency18-23

The results are obtained using twosolution algorithms:

ex = 0, (no subincrementation) ex = 1, effective-stress-function

procedure

In all cases, the MNO formulation isemployed. Full Newton iterationswithout line searches are used with

ETOL=0.001RTOL=0.01RNORM = 1.0 MN

Transparency18-24

1) Constant load of 100 MPaeC = 4.1 x 10~11 (a)3.15 to.8

0.1 ~t = 10 hra=1

(J' = 100 MPa

displacement(m) 0.05

1000500

time (hr)

0+-------+------+----o


Transparency18-25

at = 10 hr


~t = 10 hr(1=1

0.1disp.(m)0.0

3) Stress reversal from 100 MPa to-100 MPa

eC = 4.1 x 10-11 (cr)3.15 to.8

(J' = 100 MPa

Transparency18-27

500 1000 time (hr)

-0.05

Transparency18-28

4) Constant load of 100 MPaeC = 4.1 x 10-11 (cr)3.15 t.4

.01

disp.(m)

.005

~t = 10 hr(1=1 (J' = 100 MPa

1000 time (hr)500

0+- --+ -+-__

a


Transparency18-29

(T = 100 MPa

5) Stress increase from 100 MPa to200 MPa.06 eC = 4.1 x 10-11 (cr)3.15 t.4

8t=10hrex = 1

.04disp.(m)

.02

1000 time (hr)500

0,+-- --+ +- __

o

(T = 100 MPa

6) Stress reversal from 100 MPa to-100 MPa

eC = 4.1 x 10-11 (cr)3.15 t.4.01 8t = 10 hr

disp. ex = 1(m)

.005

Transparency18-30

O'+- -+~;;;:::__----+-------1000 time (hr)

(T = -100 MPa-.005


Transparency18-31

Consider the use of a = 0 for the"stress increase from 100 MPa to 200MPa" problem solved earlier (case #5):

.06

.04disp.(m)

.02

~t = 10 hr

1000 time (hr)500

0,+- -+ +-__o

Transparency18-32

Using dt = 50 hr, both algorithmsconverge, although the solution becomesless accurate for a - O.

.06

.04disp.(m)

.02

~t = 50 hr

(!)(!)

(!)


Using dt = 100 hr, (X = a does notconverge at t = 600 hr. (X = 1 still givesgood results.

.06

Transparency18-33

~t=100hr.04

disp.

(m) a = 1, ~t = 10 hr.02 ~

a=1--1000 time (hr)500

Oe--------t-~---__+_-o

E=2x106 KPav=O.Oplane stressthickness = 1.0 m

Example: Column with offset load

R

-11-0.75Transparency

18-34

10 m

Euler buckling load = 41 00 KN


Transparency18-35

Transparency18-36

Goal: Determine the collapse responsefor different material assumptions:

Elastic

Elasto-plastic

Creep

The total Lagrangian formulation isemployed for all analyses.

Solution procedure:

The full Newton method withoutline searches is employed with

ETOL=0.001

RTOL=0.01

RNORM = 1000 KN

Mesh used: Ten 8-node quadrilateralelements

~ 3 x 3 Gauss integration~ used for all elements

Elastic response: We assume that thematerial law is approximated by

ts tc to ij = 0 ijrs OCrswhere the components JCijrS areconstants determined by E and v (aspreviously described).

Topic Eighteen 1821

Transparency18-37

Transparency18-38

5000

Appliedforce(KN) 2500

Eulerbuckling ioad

tPPlied forcenLateraljJr displacementO+-----t----+----+---a 2 4 6

Lateral displacement of top (m)


Transparency18-39

Transparency18-40

Elasto-plastic response: Here we use

ET = 0cry = 3000 K Pa (von Mises yieldcriterion)

and

JI+At E

t+LltS = cis + o-CEP doE0_ - tE 0- -

0-

whereoCEP is the incremental elastoplastic constitutive matrix.

Plastic buckling is observed.

Elastic2000

Appliedforce(KN)

1000Elasto-plastic

O+---t-----t---+---+---+---o .1 .2 .3 .4 .5

Lateral displacement of top (m)

Creep response:

Creep law: eC = 10-16(cr)3 t (t inhours)No plasticity effects are included.

We apply a constant load of 2000K N and determine the time history ofthe column.

For the purposes of this problem, thecolumn is considered to havecollapsed when a lateral displacementof 2 meters is reached. Thiscorresponds to a total strain of about2 percent at the base of the column.

We investigate the effect of differenttime integration procedures on theobtained solution:

Vary At (At = .5, 1, 2, 5 hr.)

Vary ex (ex=O, 0.5, 1)

'Ibpic Eighteen 18-23

Transparency18-41

Transparency18-42


Transparency18-43

Collapse times: The table below liststhe first time (in hours) for which thelateral displacement of the columnexceeds 2 meters.

a=O a=.5 a=1dt=.5 100.0 100.0 98.5dt= 1 101 101 98dt=2 102 102 96dt=5 105 105 90

Pictorially, using dt =0.5 hr., ex =0.5,we haveTransparency

18-44Time= 1 hr

(negligible creepeffects)

Time=50 hr(some creep

effects)

IfIIIfIII

Time= 100 hr(collapse)


1.0

0.5

Transparency18-45

Choose at = 0.5 hr.- AI1 solution points are connected

with straight lines.2.5

a=1 a=Ocollapse '- ~ a = .5

2.0+----~--------IJ~--

Lateraldisp. 1.5(m)

20 40 60 80 100' 120time (hr)

Effect of a: Choose at=5 hr.- All solution points are connected

with straight lines.2.5

collapse2.0,-+--__----:. ---->~-~

Lateral 1.5disp.(m) 1.0

0.5

Transparency18-46

O+---+---+---+----f--t---......o 20 40 60 80 100 120

time (hr)


Transparency18-47

We conclude for this problem:

As the time step is reduced, thecollapse times given by ex = 0,ex = .5, ex = 1 become closer. Forat = .5, the difference in collapsetimes is less than 2 hours.

For a reasonable choice of timestep, solution instability is not aproblem.

r

I=-"T

II

zf-rl~2Ra

I I IS: : S!IL..............--JIL-IL.....L..I..L.IL..&II~I..LIu..111'"'r',~

Ra = 25 mm ~

Analysis of a cylinder subjected to heat treatment


Slide18-1

1400 70

/\ k SlidecWs/kgDC W/mDC 18-2

/\c

800

400 20

o 200 400 600 ODC 900

Temperature-dependence of the specific heat,~, and the heat conduction coefficient, k.

18-28 ElastoPlastic and Creep Response - Part II

Slide18-3

1.2

0.4

----------------, .40

v

.32

Temperature-dependence of the Young's modulus, E,Poisson's ratio, II, and hardening modulus, ET

600 ,..-----------------..

Slide18-4

200

o

Temperature-dependence of the material yield stress

Thpic Eighteen 18-29

Slide18-5

600 0 c 900o

-2

-6

-4

a 1O-5oc-/r------- --.

2

Temperature-dependence of the instantaneous coeffiCientof thermal expansion (including volume change due to

phase transformation), a

,.,..-------------------,

1SII+---+------t---+---"~_+---\--I

_+---+------t---""''''


Slide18-7

1000 r-------------------------~

e c r----------=-==~_600

10 3D to 60 txI 100 t.oc JOt)

temperatureIn element "

2 J. 6810

surfacet~mperaturl

1 J. 6 8 1

IranSformar-;;;: --------mr~rval--------====~~=--

600

100

Surface and core temperature; comparison betweenmeasured and calculated results

Slide18-8

1000..----------------------,

500

Nmm2

-500Urr

-1000-'------------------~

Measured residual stress field

1000.,-------------------,


Slide18-9

500

o

-500

\\

\

\ 1Ra0.8 " 1.0

\\

\\

\

-1000.L-------------------'

Calculated residual stress field

Contents:

Textbook:

References:

Topic 19

Beam, Plate, andShell ElementsPart I

Brief review of major formulation approaches

The degeneration of a three-dimensional continuum tobeam and shell behavior

Basic kinematic and static assumptions used

Formulation of isoparametric (degenerate) general shellelements of variable thickness for large displacementsand rotations

Geometry and displacement interpolations

The nodal director vectors

Use of five or six nodal point degrees of freedom,theoretical considerations and practical use

The stress-strain law in shell analysis, transformationsused at shell element integration points

Shell transition elements, modeling of transition zonesbetween solids and shells, shell intersections

Sections 6.3.4, 6.3.5

The (degenerate) isoparametric shell and beam elements, including thetransition elements, are presented and evaluated in

Bathe, K. J., and S. Bolourchi, "A Geometric and Material NonlinearPlate and Shell Element," Computers & Structures, 11, 23-48, 1980.

Bathe, K. J., and L. W. Ho, "Some Results in the Analysis of Thin ShellStructures," in Nonlinear Finite Element Analysis in StructuralMechanics, (Wunderlich, W., et al., eds.), Springer-Verlag, 1981.

Bathe, K. J., E. Dvorkin, and L. W. Ho, "Our Discrete Kirchhoff and Isoparametric Shell Elements for Nonlinear Analysis-An Assessment,"Computers & Structures, 16, 89-98, 1983.

19-2 Beam, Plate and Shell Elements - Part I


The triangular flat plate/shell element is presented and also studied in

Bathe, K. J., and L. W. Ho, "A Simple and Effective Element for Analysis of General Shell Structures," Computers & Structures, 13, 673681, 1981.

STRUCTURAL ELEMENTS

Beams

Plates

Shells

We note that in geometrically nonlinearanalysis, a plate (initially "flat shell")develops shell action, and is analyzedas a shell.

Various solution approaches have been proposed:

Use of general beam and shelltheories that include the desirednonlinearities.

- With the governing differentialequations known, variationalformulations can be derived anddiscretized using finite elementprocedures.

- Elegant approach, but difficultiesarise in finite element formulations: Lack of generality Large number of nodal degrees

of freedom

Topic Nineteen 19-3

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Use of simple elements, but a largenumber of elements can modelcomplex beam and shell structures.

- An example is the use of 3-nodetriangular flat plate/membraneelements to model complex shells.

- Coupling between membrane andbending action is only introducedat the element nodes.

- Membrane action is not very wellmodeled.

bendingf membraneartificial

.ISs stiffness

I

\~3 / degree of freedom with\/_~ artificial stiffness~'/..z

L5}~--xl

X1

Stiffness matrix inlocal coordinatesystem (Xi).

Example:Transparency

19-4

Isoparametric (degenerate) beam andshell elements.

- These are derived from the 3-Dcontinuum mechanics equationsthat we discussed earlier, but thebasic assumptions of beam andshell behavior are imposed.

- The resulting elements can beused to model quite general beamand shell structures.

We will discuss this approach in somedetail.

Basic approach:

Use the total and updated Lagrangianformulations developed earlier.

Topic Nineteen 19-5

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We recall, for the T.L. formulation,

f HAtS.. ~HAtE .. 0dV _ HAt(jJ}Jov 0 II'U 0 II' - ;:ILLinearization ~

'v OCiirs oers 80ei~ dV +f,v JSi~8o"ly. dV

= HAtm - f,v JSi~8oey.dV

Also, for the U.L. formulation,

JVHA~Sij.8HA~Eij. tdV = HAtm

Linearization ~

Jv tCifs ters 8tei} tdV +Jv~i~8t'T\ij. tdV

= H At9R - f",t'Tij. 8tei}tdV

Impose on these equations the basicassumptions of beam and shell -action:

1) Material particles originally on astraight line normal to the midsurface of the beam (or shell)remain on that straight linethroughout the response history.

For beams, "plane sections initiallynormal to the mid-surface remainplane sections during the responsehistory".

The effect of transverse sheardeformations is included, andhence the lines initially normal tothe mid-surface do not remainnormal to the mid-surface duringthe deformations.

Topic Nineteen 19-7

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not 90 in general

time t

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2) The stress in the direction "normal"to the beam (or shell) mid-surface iszero throughout the response history.

Note that here the stress along thematerial fiber that is initially normalto the mid-surface is considered;because of shear deformations, thismaterial fiber does not remainexactly normal to the mid-surface.

3) The thickness of the beam (or shell)remains constant (we assume smallstrain conditions but allow for largedisplacements and rotations).

Topic Nineteen 19-9

FORMULATION OFISOPARAMETRIC

(DEGENERATE) SHELLELEMENTS

To incorporate the geometricassumptions of "straight lines normal tothe mid-surface remain straight", and of"the shell thickness remains constant"we use the appropriate geometric anddisplacement interpolations.

To incorporate the condition of "zerostress normal to the mid-surface" weuse the appropriate stress-strain law.

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rX2

tv~ = director vector at node k

ak = shell thickness at node k(measured into direction of tv~)

Shell element geometryExam~: 9-node element


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Element geometry definition:

Input mid-surface nodal pointcoordinates.

Input all nodal director vectors at time O. Input thicknesses at nodes.

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r---X2

- material particle(OXi)

Isoparametric coordinate system(r, s, t):

- The coordinates rand s aremeasured in the mid-surfacedefined by the nodal pointcoordinates (as for a curvedmembrane element).

- The coordinate t is measured inthe direction of the director vectorat every point in the shell.

s

Topic Nineteen 19-11

Interpolation of geometry at time 0:Transparency

19-17N N

k~' hk Xr ,+ ~ k~' a~ hk V~imid-surface effect of shell

only thicknessmaterialparticlewith isoparametriccoordinates (r, s, t)

hk = 2-D interpolation functions (asfor 2-D plane stress, planestrain and axisymmetric elements)

X~ = nodal point coordinatesV~i = components of V~

Similarly, at time t, 0t-coordinate

t ~ h t k


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To obtain the displacements of anymaterial particle,

t t 0Ui = Xi - Xi

Hen

solution of nonlinear dynamics response - part ii

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