AALBORG UNIVERSITY ESBJERG, DENMARK -...

Lecture in Nonlinear FEM

on

the Building- and Civil Engineering sectors 8.th. semester

for

the Building- and Civil Engineering, B8k, andMechanical Engineering, B8m

AALBORG UNIVERSITY ESBJERG, DENMARK

*****************

Theme:Design of marine constructions.

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http://www.aue.auc.dk/education/m/b8km.html

http://www.aue.auc.dk/education/m/b8km.html

http://www.aue.auc.dk/

Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark

Outline: Updated: 16. februar 2004

1. Introduction Notes2. Geometrical nonlinearity - strain measures Cook 17.1, 17.93. Geometrical nonlinearity - appl. in buckling analysis Cook 17.104. Stress stiffness Cook 18.1-18.45. Buckling Cook 18.5-18.66. Material nonlinearity - introduction Cook 17.3-17.47. Material nonlinearity - solution methods Cook 17.6, 17.28. Contact nonlinearity Cook 17.89. Nonlinear dynamic problems Cook 11.1-11.510. Nonlinear dynamic problems Cook 11.11-11.18

Literature:

Noter→ A. Kristensen: http://www.aue.auc.dk/education/m/m-tools.html

Cook→ Cook, R. D. 2002: Concepts and applications of finite element anal-ysis.John Wiley & Sons

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http://www.aue.auc.dk/education/m/m-tools.html


3. Geometrical nonlinearity - appl. in buckling analysis

Programme:

Last time 4

Solution of nonlinear equations 5

The Newton-Raphson procedure 6

The Modified Newton-Raphson procedure 8

The BFGS procedure 10

Example 1: Torsional spring on a clamped beam 12

Example 2: Torsional spring on a clamped column 15

Example 3: Simple rod 17

Examples

Assignments

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Last timeThe nonlinear equations of equilibrium in the residual formulation is givenby:

r(d, f ) = p(d)− f = 0 (1)

The load is applied in load-steps n = 1,2, . . ..

The tangent-stiffnessmatrix [KT] is given as

[KT] = [K0]+ [KL({d})]+ [Kσ({σ})] (2)

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Solution of nonlinear equationsGeometric nonlinearity behaviour relates to that during loading changes ingeometry (deformations, displacements) will have an effect on the load-displacement response.

Geometric nonlinear FEA can be categorized into three groups:

1. large displacements/large rotations/large strains

2. large displacements/large rotations/small strains

3. large displacements/small rotations/small strains

The Newton-Raphson solution procedure is described generally (1D), butthere exist various solution schemes, which improve the classical Newton-Raphson procedure, e.g. the procedure outlined by Broyden-Fletcher-Goldfarb-Shanno (BFGS).

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The Newton-Raphson procedureThe Newton-Raphson procedure for a typical increment:

load-step n = 1,2, . . .

f n = f n−1+∆ f n

dn0 = dn−1

iterations i = 0,1, . . .

compute rni = p(dn

i )− f n

stop iterations when ||rni ||< εstop|| f n||

compute KT(dni )δdn

i (factorization with time)

solve the equation of equilibrium δdni = −KT(dn

i )−1rn

i (forward-backward substitu-tion)

update the displacement dni+1 = dn

i +δdni

for number of load-step

where εstop= 0,01−0,001.

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The Newton-Raphson procedureNAFEMS p. 248: Figure 5.4.

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The Modified Newton-Raphson procedureThe Modified Newton-Raphson procedure for a typical increment:

load-step n = 1,2, . . .

f n = f n−1+∆ f n

dn0 = dn−1




compute rni = p(dn

i )− f n



i )−1rn



i +δdni



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The Modified Newton-Raphson procedureNAFEMS p. 248: Figure 5.5.

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The BFGS procedureThe BFGS procedure is a Quasi-Newton procedure, which is based on se-cant methods, where KT(dn

i )−1 is updated. NAFEMS p. 252: Figure 5.6.

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The BFGS procedureThe BFGS Quasi-Newton procedure for a typical increment:

load-step n = 1,2, . . .

f n = f n−1+∆ f n

dn0 = dn−1




compute rni = p(dn

i )− f n


if i > 0, update KT(dni )−1 by BFGS inverse update scheme


i )−1rn



i +δdni



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Example 1: Torsional spring on a clamped beamExample on large displacements/large rotations/small strains NAFEMS p.17: Figure 1.13.

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Example 1: Torsional spring on a clamped beamMoment equilibrium about the hinge (in the deformed state):

PLcosθ = M P =Kθ

Lcosθwhere K is the torsional stiffness of the spring and M = Kθ. If the angle θ issmall, then cosθ→ 1 and the linear equation of equilibrium is obtained:

P =KθL

NAFEMS p. 18: Figure 1.14.

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Example 1: Torsional spring on a clamped beam

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Example 2: Torsional spring on a clamped columnExample on large displacements/large rotations/small strains NAFEMS p.46: Figure 2.5.

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Example 2: Torsional spring on a clamped columnMoment equilibrium about the hinge (in the deformed state):

PLsinθ = M P =Kθ

LsinθThis equation of equilibrium is interesting as it has two solutions

θ = 0 for arbitrary P andPLK

=θ

sinθP =

KθL

These are shown on figure 2.5. The intersection point between the twocurves of equilibrium is termed the bifurcation point.

Linearization of the term: θ→ 0 so sinθ→ θ and the equation of equilibriumbecomes:

(K−PL)θ = 0

Equilibrium is obtained for each value of P if θ = 0, but if P = K/L then thecolumn is in equilibrium for each value of θ.

This equation of equilibrium represents, in the simplest form, a linearizedbuckling problem (instability problem, Euler buckling). The critical force Pc =K/L is termed the elastic critical buckling load.

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Example 3: Simple rodExample on large displacements/small rotations/small strains. This simpleexample provide a good illustration of geometric nonlinear analysis, espe-cially a discussion of strain measures, discussion of stress, equilibrium andthe notation of the tangent- stiffness - NAFEMS p. 48: Figure 2.6.

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Example 3: Simple rodThere is an initial tensile force N0 and the assumption small strain imply thatthe cross-sectional area A of the rod is assumed constant. The equation ofequilibrium is derived for the deformed configuration:

Nsinα−P = Nu+h

L−P = 0 (3)

Introducing the constitutive conditions, i.e. it is assumed a linear elasticmaterial with elasticity modulus E, provide that:

(N−N0) = EAε

where ε is the strain.

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Example 3: Simple rodEngineering strain εE is the traditional definition used in theory based oninfinitesimal strains:

• Engineering stress:

σE =PA0

= E · εE

where A0 is the cross-sectional at beginning of the test.

• Engineering strain:

εE =L−L0

L0

where L is the current length and L0 is the length on the tensile rod atthe beginning of the test.

• Poisson’s ratio:ν =−εr

εa

where εr is the strain in radial direction and εa is the strain in the axialdirection.

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Example 3: Simple rod

• True stress:σ =

PA

where A is the current cross-sectional area of the tensile rod.

• True strain:

ε = ln

(LL0

)

• Stress-strain curve for true strain can be expressed by:

σ = K · εn

where K is the coefficient and n is deformation hardening exponent.

Since nonlinear problems often encounter large displacements and smallstrains it is imperative to apply a strain definition which yield zero strain forarbitrary rigid body movements. It is reasonable to employ L2 instead of Las basis for the strain definition.

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Example 3: Simple rodAn advantage to employ L2 is that this strain definition is relatively easy togeneralize in 2D and 3D continuum. Applying the engineering strain defini-tion and multiplying by L+L0 provide

εE =(L−L0)(L+L0)

L0(L+L0)=

(L2−L20)

L0L+L20

=(L2−L2

0)

L20

(LL0

+1

)

=(L2−L2

0)

L20

(L0

L0− L0

L0+

LL0

+1

) =(L2−L2

0)L2

0(2+ εE)

For small strains is εE≈ 0 and can be removed from the numerator. Therebythe Green-Lagrangian strain definition εG is obtained:

εG =L2−L2

0

2L20

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Example 3: Simple rodFor this example the engineering strain definition is used:

εE =L−L0

L0

where

L0 =√

a2+h2 = a

√1+

(ha

)2

and

L =√

a2+(h+u)2 = a

√1+

(h+u

a

2)

The expression for the strain εE becomes complex. This problem was avoidedby George Green by introducing another strain measure termed Green’sstrain:

εG =L2−L2

0

2L20

This yield:

εG =(

hL0

)(uL0

)+

12

(uL0

)2

(4)

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Example 3: Simple rodThe relation between the two main strain measures is:

εG− εE

εE=

12

εE ⇔ εG = εE +12

ε2E

It is assumed that N0 = 0 to obtain a simplified expression and the stress isdetermined by

σ =NA

= EεE

AsεG

εE= 1+

12

εE this expression can be written

σ =NA

= EεE = E

(12

(LL0

+1

))−1

εG (5)

For small strains it is given that L/L0≈ 1. This is illustrated by NAFEMS p.72: Figure 2.22.

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Example 3: Simple rod

The figure illustrates, the 2. Piola-Kirchhoff spænding [S] which is similar tothe rotated Cauchy stress [σ′].

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Example 3: Simple rodGenerally it is given:

• The Cauchy stress σ′ (the current force per unit deformed area) is workconsistent with the engineering strain εE

• The 2. Piola-Kirchhoff stress S(transformed forces per unit un-deformedarea) is work consistent with Green-Lagrangian strain εG

In this example with small strains the difference between εE and εG canneglected and thereby the constitutive equation

(N−N0) = EAεG (6)

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Example 3: Simple rodSubstitution of the constitutive equation 6 in the equation of equilibrium3 and inserting the Green-Lagrangian strain 4 under the assumption thatL/L0≈ 1 due to small strains yields

P = N

(u+h

L0

)=

EA

L30

(hu+

12

u2

)

︸︷︷︸a)

(h+u)︸︷︷︸b)

+N0

(u+h

L0

)(7)

It is shown that these two types of nonlinearities is included in the equationof equilibrium:

a): a nonlinear strain definition is applied

b): equilibrium is derived for the deformed state

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Example 3: Simple rodNotice the equation of equilibrium 7 is derived for the deformed situation,now referring to the original configuration, i.e. the variables A, h and L0 -NAFEMS p. 50: Figure 2.7.

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Example 3: Simple rodThe equation of equilibrium 1 is rewritten to

r(d, f ) = p(d)− f = 0 ⇒ r = N

(u+h

L0

)−P = 0 ⇒ r = p(u)−P = 0

The incremental equation of equilibrium has the form

KTδu =−r

as the indices is removed. It can be seen that

r = p(u)−P = N

(u+h

L0

)−P

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Example 3: Simple rodThe tangent-stiffness matrix KT can now be determined by

KT =dp(u)

du=

ddu

(u+h

L0N(u)

)

=ddu

(u+h

L0

)N+

(u+h

L0

)dNdu

=ddu

(u+h

L0

)N+

(u+h

L0

)ddu

(EAεG+N0)

=ddu

(u+h

L0

)N+

(u+h

L0

)ddu

(EA

(hu

L20

+u2

2L20

))

=NL0

+(

u+hL0

)EAL0

(u+h

L0

)

where dN/duhas been determined from the constitutive equation 6 and theexpression for the Green-Lagrangian strain, i.e. equation 4.

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Example 3: Simple rodThe expression for the tangent-stiffness can be rewritten providing:

KT = K0+KL +Kσ

where K0 =EAL0

(hL0

)2

is the linear stiffness, KL =EAL0

[2

uh

+(u

h

)2](

hL0

)2

is

the initial shear stiffness and Kσ =NL0

is the initial stress stiffness.

K0 is known from static analysis (small displacements).

KL reflects the effects on the stiffness due to changes in the displace-ment.

Kσ is the effect on the stiffness due to, e.g. membrane forces.

K0 and Kσ is used for linear stability analysis. Now the incremental equationof equilibrium can be solved by:

KTδu =−r where r = N

(u+h

L0

)−P

This example should illustrate the effects, which should be included in ageometric nonlinear analysis.

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