AALBORG UNIVERSITY ESBJERG, DENMARK -...
Transcript of 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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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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
The Newton-Raphson procedureNAFEMS p. 248: Figure 5.4.
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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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 KT(dni )δdn
i (factorization with time)
iterations i = 0,1, . . .
compute rni = p(dn
i )− f n
stop iterations when ||rni ||< εstop|| f n||
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
The Modified Newton-Raphson procedureNAFEMS p. 248: Figure 5.5.
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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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 KT(dni )δdn
i (factorization with time)
iterations i = 0,1, . . .
compute rni = p(dn
i )− f n
stop iterations when ||rni ||< εstop|| f n||
if i > 0, update KT(dni )−1 by BFGS inverse update scheme
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
Example 1: Torsional spring on a clamped beam
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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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Nonlinear FEA, Dep. of Computational Mechanics, AAU Esbjerg, Denmark
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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