Total lagrangian formulation of a bar element and path ... · Total lagrangian formulation of a bar element and path following methods Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati,

Total lagrangian formulation of a bar element andpath following methods

Prof. Dr. Eleni ChatziDr. Giuseppe Abbiati, Dr. Konstantinos Agathos

Lecture 5 - 26 October, 2017

Institute of Structural Engineering, ETH Zurich

October 26, 2017

Institute of Structural Engineering Method of Finite Elements II 1

Outline

1 Introduction

2 Weak form

3 Finite element formulation

4 Total Lagrangian formulation of a bar element

5 Path-following Methods


Learnig goals

Understanding how geometrically nonlinear finite elements areformulated

Gaining a basic understanding of path following solutionmethods

Gaining a basic understanding of how the above areimplemented

Gaining a basic understanding of how the above can be used incommercial software (ABAQUS)


Significance of the lecture

Applications:

Structures undergoing large displacements and/orrotations such as cables, arches and shells

Materials such as elastomers and biological/soft tissue

Modeling of plastically deforming materials


Weak form

We recall:

Weak form (TL):∫V

S : δEdV =∫V

F · δudV +∫A

T · δudA

Linearized weak form:

∫V

e·D·δedV +∫V

Si : δηdV =∫V

F·δudV +∫A

T·δudA−∫V

Si : δedV

where:

e = 12

[∇∆u +∇∆uT +∇

(ui)T ∇∆u +∇∆uT∇ui

]η = 1

2

[∇∆uT∇∆u

]


Finite element formulationCoordinates and displacements are interpolated from thecorresponding nodal quantities using FE interpolation:

Xi =N∑

J=1NJX J

i , xi =N∑

J=1NJxJ

i

ui =N∑

J=1NJuJ

i , ∆ui =N∑

J=1NJ∆uJ

i

or in matrix form:

X = NXn, x = Nxn

ui = Nuni , ∆u = N∆un

Index J and superscript n refer to nodal quantities, NJ are the FEshape functions.


Finite element formulation

Typically isoparametric elements are used:

NJ = NJ (ξ1, ξ2, ξ3)

where ξ1, ξ2, ξ3 are the isoparametric coordinates. Then:

∂·∂ξ1

∂·∂ξ2

∂·∂ξ3

=

∂X1∂ξ1

∂X2∂ξ1

∂X3∂ξ1

∂X1∂ξ2

∂X2∂ξ2

∂X3∂ξ2

∂X1∂ξ3

∂X2∂ξ3

∂X3∂ξ3

︸︷︷︸

J

∂·∂X1∂·∂X2∂·∂X3

⇒

∂·∂X1∂·∂X2∂·∂X3

= J−1

∂·∂ξ1

∂·∂ξ2

∂·∂ξ3

The same transformation can be used for the current configuration.Institute of Structural Engineering Method of Finite Elements II 7


Utilizing the above, the quantities ∇u,∇∆u can be computed.

For example:

∂u1∂X1

=N∑

J=1

∂NJ∂X1

uJi

∂NJ∂X1

= J−111∂NJ∂ξ1

+ J−112∂NJ∂ξ2

+ J−113∂NJ∂ξ3

where J−1ij is the element ij of matrix J−1

ij .



Utilizing the above, the linear and non-linear part of the strainincrement can be written in terms of the nodal displacementincrements:

e = BL∆un ⇒ δe = BLδ∆un

η = 12∆unT BT

NLBNL∆un ⇒ δη = δ∆unT BTNLBNL∆un



By taking into account that:

u = ui + ∆u⇒ δu = δ∆u

Substituting the above in the weak form we obtain:

δ∆unT∫V

BTL DBLdV ∆un + δ∆unT

∫V

BTNLSiBNLdV ∆un =

δ∆unT∫V

NT FdV + δ∆unT∫A

NT TdA− δ∆unT∫V

BTL SidV




u = ui + ∆u⇒ δu = δ∆u


��δ∆unT

∫V

BTL DBLdV ∆un +��

δ∆unT∫V

BTNLSiBNLdV ∆un =

��δ∆unT

∫V

NT FdV +��δ∆unT

∫A

NT TdA−��δ∆unT

∫V

BTL SidV




u = ui + ∆u⇒ δu = δ∆u


∫V

BTL DBLdV ∆un +

∫V

BTNLSiBNLdV ∆un =

∫V

NT FdV +∫A

NT TdA−∫V

BTL SidV


Finite element formulationBy taking into account that:

u = ui + ∆u⇒ δu = δ∆uSubstituting the above in the weak form we obtain:

KT︷︸︸︷∫V

BTL DBLdV

︸︷︷︸KL

+∫V

BTNLSiBNLdV

︸︷︷︸KNL

∆un =

−Ri︷︸︸︷∫V

NT FdV +∫A

NT TdA

︸︷︷︸fext

−∫V

BTL SidV

︸︷︷︸fint



The tangent stiffness matrix, residual and force vectors are thenwritten as:

KT = KL + KNL

KL =∫V

BTL DBLdV , KNL =

∫V

BTNLSiBNLdV

−Ri = fext − finti

fext =∫V

NT FdV +∫A

NT TdA, finti =

∫V

BTL SidV



Substituting the above, the linearized equilibrium equation can beobtained in terms of the nodal unknowns:

KT∆un = −Ri

KT∆un = fext − finti


Bar element

Next the tangent stiffness matrix and force vector are developed forthe two noded bar element.

It is convenient to write:

u21 = u1

1 + l cos θ − L

u22 = u1

2 + l sin θ


Bar element

FE shape function for a two noded bar:

N =[

1− ξ2

1 + ξ

2

]Jacobian: J = L

2

Shape function derivative:

N,1 = N∂X1

= J−1∂N∂ξ

= 1L[−1 1

]


Bar element

By considering displacements along both directions:

u1

u2

=

1− ξ2 0 1 + ξ

2 0

0 1− ξ2 0 1 + ξ

2

︸︷︷︸

N

·

u11

u12

u21

u22

and

∂u1∂X1∂u2∂X1

= 1L

−1 0 1 0

0 −1 0 1

︸︷︷︸

N,1

·

u11

u12

u21

u22


Bar element

Displacement derivative:

∂ui∂X1

= N,1uni = 1

L[−1 1

]·

u1i

u2i

= u2i − u1

iL

For i = 1:

∂u1∂X1

= u21 − u1

1L = u1

1 + l cos θ − L− u11

L = l cos θ − LL = l cos θ

L − 1

For i = 2:

∂u2∂X1

= u22 − u1

2L = u1

2 + l sin θ − u12

L = l sin θL


Bar element

By taking into account that ∂ (·)∂X2

= 0 the Green-Lagrange strain atthe previous step (only component 11 is of interest) assumes theform:

E11 = ∂u1∂X1

+ 12

(∂u1∂X1

)2+ 1

2

(∂u2∂X1

)2

=( l cos θ

L − 1)

+ 12

( l cos θL − 1

)2+ 1

2

( l sin θL

)2

= l2 − L2

2L2


Bar element

By taking into account that ∂ (·)∂X2

= 0 the linear part of the strainincrement (only component 11 is of interest) assumes the form:

e11 = ∂∆u1∂X1

+ ∂u1∂X1

∂∆u1∂X1

+ ∂u2∂X1

∂∆u2∂X1


Bar element

e11 =[1

L

[−1 0 1 0

]︸︷︷︸

∂∆u1∂X1

+

+( l cos θ

L − 1)

︸︷︷︸∂u1∂X1

1L

[−1 0 1 0

]︸︷︷︸

∂∆u1∂X1

+

+( l sin θ

L

)︸︷︷︸

∂u2∂X1

1L

[0 −1 0 1

]]︸︷︷︸

∂∆u2∂X1

∆u11

∆u12

∆u21

∆u22


Bar element

e11 = lL2

[− cos θ − sin θ cos θ sin θ

]︸︷︷︸

BL

∆u11

∆u12

∆u21

∆u22

︸︷︷︸

∆un


Bar element

The nonlinear part of the strain increment is:

η = 12

[(∂∆u1∂X1

)2+(∂∆u2∂X1

)2]=[∂∆u1∂X1

∂∆u2∂X1

]·

∂∆u1∂X1∂∆u2∂X1

∂∆u1∂X1∂∆u2∂X1

= 1L

−1 0 1 0

0 −1 0 1

︸︷︷︸

BNL=N,1

·

∆u11

∆u12

∆u21

∆u22

︸︷︷︸

∆un


Bar element

If a linear stress strain relation is employed then:

S11 = E E11 = E l2 − L2

2L2

where E is Young’s modulus

Then:

∂S11∂E11

= E


Bar element

Linear part of the stiffness matrix:

KL =∫V

BTL EBLdV

Employing that V = AL, where A the cross section of the bar in thereference configuration, we obtain:

KL = EA l2

L3

cos2θ cosθsinθ −cos2θ −cosθsinθ

sin2θ −sinθcosθ −sin2θcos2θ sinθcosθ

Symm sin2θ


Bar element

Similarly the nonlinear part of the tangent stiffness matrix is:

KNL =∫V

BTNLS11BNLdV

By carrying out the integration we obtain:

KNL = EAl2 − L2

2L3

1 0 −1 00 1 0 −1−1 0 1 00 −1 0 1


Bar element

Finally the force vector can be obtained as:

f =∫V

BTL S11dV

Performing the integration:

fint = EA l2 − L2

2L2

−cosθ−sinθcosθsinθ


The Newton-Raphson method

We recall the linearized equilibrium equation:

KT∆un = −Ri = fext − finti

Typically external loads are scaled by a factor λ:

KT∆un = −Ri = λfext − finti



For a given value of λ, the above corresponds to a Newton-Raphsoniteration where:

R = Ri + ∂Ri

∂un ∆un = finti − λfext + KT∆un = 0

The iteration is repeated until

|R| <= tol

where tol is the tolerance.



Typically:

Different (increasing) values are given to the load factor

Displacements are obtained using the Newton-Raphson method

The response of the structure is obtained for progressivelyincreasing loading

The load-displacement curve of the structure can be obtained



The above procedure is called load control and can be illustrated asfollows:

Increment load

Apply load increment

Solve withNewton-Raphson

Repeat for nextincrement




Increment load







Increment load







Increment load







Increment load







Increment load






In many cases the solution path is not monotonic

Phenomena such as snap-through and snap-back can be present

The Newton-Raphson method will either fail or not provide thefull path in those cases

Such phenomena are often of interest in practice e.g.when theovercritical behavior of a structure needs to be known


Example

Solution path involving snap-through:


Example



Example



Example



Path-following methods

In this category of methods:

It is attempted to follow the full solution path includingsnap-back and snap-through phenomena

The load factor is included as an additional unknown in thenonlinear system of equations

An additional equation (constraint) is added in the system

The choice of this constraint leads to different methods



After adding ∆λ as an unknown and the constraint as an equationthe new system of equations is obtained:[

R (u, λ)g (u, λ)

]=[

00

]where:

u is the vector of nodal unknownsλ is the load factorR is the residual of the Newton-Raphson methodg is the additional constraint



The above system can be solved by employing the Newton-Raphsonmethod. The linearization of the system yields: R

(ui, λi

)+∂R

(ui, λi

)∂u ∆u + ∂R

(ui , λi)∂λ

∆λ

g(ui , λi)+ ∂g

(ui , λi)∂u ∆u + ∂g

(ui , λi)∂λ

∆λ

=[

00

]

where:

∆u is the nodal displacement increment∆λ is the load factor increment



The above system can be solved by employing the Newton-Raphsonmethod. The linearization of the system yields:[

KT −fexthT s

] [∆u∆λ

]=[

Ri

−g i

]

where:

∆u is the nodal displacement increment∆λ is the load factor increment

h Is the gradient of g with respect to u: h = ∂g∂u

s is the derivative of g with respect to λ: s = ∂g∂λ



The coefficient matrix in this system is not symmetric.

To exploit the symmetry and sparsity of KT the followingpartitioning procedure is employed:

∆uI = KT−1fext, ∆uII = KT

−1Ri

∆λ = −g i + hT ∆uII

s + hT ∆uI , ∆u = ∆λ∆uI + ∆uII



The choice of the additional constraint g is crucial

Different alternatives exist

In some cases problem specific constraints are needed

Some widely used alternatives are demonstrated in the following


Path-following methods: Load control

Setting g = λ− λ results in load control:

g = λ− λ ⇒ h = ∂g∂u = 0, s = ∂g

∂λ= 1

g i = λi − λ

[KT −fext0 1

] [∆u∆λ

]=[

Ri

λ− λi

]

∆λ = λ− λi , ∆u = KT−1(λfext − fint

i)


Path-following methods: Displacement control

Setting g = T · u− u results in displacement control:

g = T · u− u ⇒ h = ∂g∂u = TT , s = ∂g

∂λ= 0

g i = T · ui − u

[KT −fextT 0

] [∆u∆λ

]=[

Ri

u − T · ui

]

In the above T =[

0 0 . . . 1 . . . 0 0]

where the entry 1corresponds to a selected dof



Displacement control in a pathinvolving snap-through

Displacement control in a pathinvolving snap-back










Path-following methods: Arc-length

Setting g =√

(u− u0)T · (u− u0) + (λ− λ0)2 −∆s results in arclength control (Crisfield 1981):

g i =√

(ui − u0)T · (ui − u0) + (λi − λ0)2 −∆s

h = ∂g∂u =

(ui − u0)

g , s = ∂g∂λ

=(λi − λ0)

g

KT −fext(ui − u0)T

g

(λi − λ0)

g

[ ∆u∆λ

]=[

Ri

−g i

]

In the above, superscript 0 refers to the beginning of the step



We observe that for i = 0 the system becomes:

KT −fext(u0 − u0)T

g

(λ0 − λ0)

g

[ ∆u∆λ

]=[

Ri

−g i

]




[KT −fext0 0

] [∆u∆λ

]=[

Ri

−g i

]




[KT −fext0 0

] [∆u∆λ

]=[

Ri

−g i

]

The coefficient matrix is singular!

To obtain the load factor at the first iteration a predictor step isintroduced.



For the predictor step the system is solved for the external loads:

∆up = KT−1fext

Then the increment of the load factor is computed as:

∆λp = ± ∆s‖∆up‖



The sign of the increment is determined by the current stiffnessparameter:

κ = fextT ∆u

∆uT ∆u



Setting g = (∆up)T (u− u1)+ ∆λp (λ− λ1) results in analternative method for arc length control (Riks 1972):

g i = (∆up)T(ui − u1

)+ ∆λp

(λi − λ1

)

h = ∂g∂u = ∆up, s = ∂g

∂λ= ∆λp

[KT −fext

(∆up)T ∆λp

] [∆u∆λ

]=[

Ri

−g i

]

In the above, superscript 1 refers to the predictor step



Illustration of arc-length methods:

Crisfield: Riks:


Documents

Total lagrangian formulation of a bar element and path ... · Total lagrangian formulation of a bar element and path following methods Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati,