Total lagrangian formulation of a bar element and path ... · Total lagrangian formulation of a bar...

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:

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

[∇∆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

ui =N∑

J=1NJuJ

i , ∆ui =N∑

J=1NJ∆uJ

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

︸︷︷︸

∂·∂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

∂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

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

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

BTL DBLdV ∆un +��

δ∆unT∫V

BTNLSiBNLdV ∆un =

��δ∆unT

NT FdV +��δ∆unT

NT TdA−��δ∆unT

BTL SidV

u = ui + ∆u⇒ δu = δ∆u

BTL DBLdV ∆un +

BTNLSiBNLdV ∆un =

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

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 =

BTNLSiBNLdV

−Ri = fext − finti

fext =∫V

NT FdV +∫A

NT TdA, finti =

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:

1− ξ2

1 + ξ

]Jacobian: J = L

Shape function derivative:

N,1 = N∂X1

= J−1∂N∂ξ

= 1L[−1 1

Bar element

By considering displacements along both directions:

1− ξ2 0 1 + ξ

0 1− ξ2 0 1 + ξ

︸︷︷︸

∂u1∂X1∂u2∂X1

−1 0 1 0

0 −1 0 1

︸︷︷︸

Bar element

Displacement derivative:

∂ui∂X1

= N,1uni = 1

L[−1 1

= u2i − u1

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

(∂u1∂X1

(∂u2∂X1

=( l cos θ

L − 1)

( l cos θL − 1

( l sin θL

= l2 − L2

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

[−1 0 1 0

]︸︷︷︸

∂∆u1∂X1

+( l cos θ

L − 1)

︸︷︷︸∂u1∂X1

[−1 0 1 0

]︸︷︷︸

∂∆u1∂X1

+( l sin θ

)︸︷︷︸

∂u2∂X1

[0 −1 0 1

]]︸︷︷︸

∂∆u2∂X1

∆u11

∆u12

∆u21

∆u22

Bar element

e11 = lL2

[− cos θ − sin θ cos θ sin θ

]︸︷︷︸

∆u11

∆u12

∆u21

∆u22

︸︷︷︸

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

−1 0 1 0

0 −1 0 1

︸︷︷︸

BNL=N,1

∆u11

∆u12

∆u21

∆u22

︸︷︷︸

Bar element

If a linear stress strain relation is employed then:

S11 = E E11 = E l2 − L2

where E is Young’s modulus

∂S11∂E11

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

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

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

−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

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

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)

]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

g(ui , λi)+ ∂g

(ui , λi)∂u ∆u + ∂g

(ui , λi)∂λ

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

−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

λ− λi

∆λ = λ− λi , ∆u = KT−1(λfext − fint

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

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)

KT −fext(ui − u0)T

(λi − λ0)

[ ∆u∆λ

−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

(λ0 − λ0)

[ ∆u∆λ

−g i

[KT −fext0 0

] [∆u∆λ

−g i

[KT −fext0 0

] [∆u∆λ

−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

−g i

In the above, superscript 1 refers to the predictor step

Illustration of arc-length methods:

Crisfield: Riks:

Total lagrangian formulation of a bar element and path ... · Total lagrangian formulation of a bar...

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