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The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems: Computational

Plasticity Part I

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

Lecture 2 - 28 September, 2017

Institute of Structural Engineering Method of Finite Elements II 1

Learning Goals

To understand the Newton-Raphson algorithm in the mostgeneric form.

To understand a basic lumped plasticity model that consists ona spring-slider system.

To understand the algorithmic procedure of a nonlinear staticfinite element analysis.

References:

Ren de Borst, Mike A. Crisfield, Joris J. C. Remmers, Clemens V.Verhoosel, Nonlinear Finite Element Analysis of Solids andStructures, 2nd Edition, Wiley, 2012.

Example: Forming of a metal profile


The Newton-Raphson Method

Given the following nonlinear equation:

f (x) : R→ R

we want to find,

x : f (x) = 0

following an iterative procedure based on linearization,

f (xj + ∆xj) ≈ f (xj) +df

dx|xj ∆xj = 0

↓

∆xj = −(df

dx|xj)−1

f (xj)

↓xj+1 = xj + ∆xj


The Newton-Raphson Method (1-D)

Definition of: f (x)



Initial guess set by the user: x1



Evaluation of: f (x1) anddf

dx|x1



Evauation of: x2 = x1 + ∆x1 = x1 −(df

dx|x1

)−1

f (x1)




dx|x2




dx|x2

)−1

f (x2)




dx|x3




dx|x3

)−1

f (x3)




dx|x4




dx|x4

)−1

f (x4)



f (x5) ≈ 0→ Stop !!!


The Newton-Raphson Method (n-D)

Given the following nonlinear vector equation:

f (x) : Rn → Rn

we want to find,

x : f (x) = 0

following the same iterative procedure based on linearization,

f (xj + ∆xj) ≈ f (xj) +∂f

∂x|xj ∆xj = 0

↓

∆xj = −(∂f

∂x|xj)−1

f (xj)

↓xj+1 = xj + ∆xj

Partial derivatives ∂∂xj

replace derivatives ddx .


The Newton-Raphson Method (n-D)

Linearization of the vector function and expansion of theNewton-Raphson increment:

f1 (x1 + ∆x1)f2 (x2 + ∆x2)

...fn (xn + ∆xn)

n×1

=

f1 (x1)f2 (x2)

...fn (xn)

n×1

+

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

......

. . ....

∂fn∂x1

∂fn∂x2

· · · ∂fn∂xn

n×n

∆x1

∆x2...

∆xn

n×1

↓x1

x2...xn

j+1

n×1

=

x1

x2...xn

j

n×1

−

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

......

. . ....

∂fn∂x1

∂fn∂x2

· · · ∂fn∂xn

−1

jn×n

f1 (x1)f2 (x2)

...fn (xn)

j

n×1


Lumped Plasticity: a Spring-Slider System

This spring-slider system is the simplest plasticity model.

if force H is smaller than adhesion, sliding is prevented

if force H is higher than adhesion (right limit), sliding starts

u = ue + up → u = ue + up

u : total displacement of A [m]

ue : spring elongation (elasticdisplacement) [m]

up : block sliding (plasticdisplacement) [m]

k : spring stiffness[Nm

]ψ : dilatancy angle [rad ]

H : horizontal force [N]

V : vertical force [N]



A mathematical model of the spring-slider system is derived thatexpresses the relationship between displacement and force rates.

u = ue + up

ue =

[ue

v e

] ue = Hk : horizontal elastic vel.

[ms

]v e = 0 : vertical elastic vel.

[ms

]



A mathematical model of the spring-slider system is derived thatexpresses the relationship between displacement and force rates.

u = ue + up

up = λm

m =

[1

tanψ

] λ : plastic multiplier [m]

tanψ = vp

up : ration between plastic vert.and horiz. velocities [d .l .]



As analogously done for displacements, we define the force responserate of the spring-slider system.

r = Ke ue = Ke (u− up)

with,

r =

[H

V

], Ke =

[k 00 0

] Ke : elastic stiffness matrix

H : horizontal force rate[Ns

]V : vertical force rate

[Ns

]Institute of Structural Engineering Method of Finite Elements II 9


The following Coulomb yielding function f to define the borderlinebetween purely elastic spring elongation and plastic block sliding.

ϕ : friction angle, c : adhesion coefficient.

f (H,V , ϕ, c) = H + Vtanϕ− c < 0 : elastic spring elongation

f (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the block

f (H,V , ϕ, c) = H + Vtanϕ− c > 0 : physically impossible !!!


Coulomb Yield Function

No plastic strain occurs when the force state stays in the elasticdomain.


f (H,V , ϕ, c) = H + Vtanϕ− c < 0→ up = 0→ ue = u

r = Ke u



Plastic strain occurs when the force state belongs to the yieldingsurface.


f (H,V , ϕ, c) = H + Vtanϕ− c = 0→ up 6= 0→ ue = u− up{r = Ke (u− up)

f = 0



The force states can move either to the elastic domain or within theyielding surface (Prager’s consistency condition).


f (H,V , ϕ, c) = H + V tanϕ = nT r = 0

with n =

[1

tanϕ

], r =

[H

V

]Institute of Structural Engineering Method of Finite Elements II 13

Lumped Plasticity Model

As long as we stay on the yielding surface, both following conditionsmust be verified:

{r = Ke (u− up)

f = 0→

{r = Ke

(u− λm

)f = 0

→

{r = Ke

(u− λm

)nT r = 0

Since m and n are constant, the system is linear and therefore it’sconvenient to recast it in matrix form:

[I KemnT 0

] [r

λ

]=

[Ke u

0

]


Lumped Plasticity Model

[I KemnT 0

] [r

λ

]=

[Ke u

0

]↓[

r

λ

]=

[Ke − KemnTKe

nTKemKem

nTKemnTKe

nTKem−1

nTKem

][u0

]

The inverse of square block matrix A =

[A11 A12

A21 A22

]reads,

A−1 =

[A−1

11 + A−111 A12B−1A21A

−111 −A−1

11 A12B−1

−B−1A21A−111 B−1

]where,

B = A22 − A21A−111 A12

The Matrix Cookbook


https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf

Lumped Plasticity Model: Tangent Stiffness

Instantaneous tangent stiffness of the spring-slider system:

r =(Ke − KemnTKe

nTKem

)u

λ =(

nTKe

nTKem

)u

[H

V

]=

[k 00 0

]−

[k 00 0

] [1

tanψ

] [1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

] [1

tanψ

][uv

]




r =(Ke − KemnTKe

nTKem

)u

λ =(

nTKe

nTKem

)u

[H

V

]=

[k 00 0

]−

[k 00 0

] [1 tanϕ

tanψ tanψtanϕ

] [k 00 0

]k

[uv]




r =(Ke − KemnTKe

nTKem

)u

λ =(

nTKe

nTKem

)u

[H

V

]=

[k 00 0

]−

[k2 00 0

]k

[uv]

It is interesting to note that the spring-slider system has no stiffnesswhen the force state belong to the yielding surface.


Lumped Plasticity Model: Plastic Multiplier

Instantaneous plastic multiplier of the spring-slider system:

r =(Ke − KemnTKe

nTKem

)u

λ =(

nTKe

nTKem

)u

λ =

[1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

] [1

tanψ

][uv

]


Lumped Plasticity Model: Plastic Multiplier

Instantaneous plastic multiplier of the spring-slider system:

r =(Ke − KemnTKe

nTKem

)u

λ =(

nTKe

nTKem

)u

λ =

[k0

]k

[uv]

It is interesting to note that only plastic displacement incrementoccurs when the force state belong to the yielding surface.


Integration of the Force-Displacement Response

Force-displacement response of the spring-slider system.

Let’s imagine to turn this into a computer program:

1: function [rj+1] = elementForce (uj+1)2: ...3: end


Integration of the Force-Displacement Response

Elastic domain:

f (r) < 0

↓r = Ke u

↓∆r = Ke∆u

Plastic domain (yielding surface):

f (r) = 0

↓

r =

(Ke − KemnTKe

nTKem

)u

↓

∆r =

(Ke − KemnTKe

nTKem

)∆u

How to handle the case when we are moving from the elastic to theplastic domain?


Return Mapping Algorithm



Return mapping algorithm Step #1.

rj : initial restoring force (onset of load step j + 1).




rj : initial restoring force (onset of load step j + 1).re = rj + Ke∆uj+1 : elastic predictor of the restoring force (end of loadstep).




rj : initial restoring force (onset of load step j + 1).re = rj + Ke∆uj+1 : elastic predictor of the restoring force (end of loadstep).rj+1 = re −Kem∆λj+1 : exact restoring force (end of load step) thatsatisfies f (rj+1) = 0.



The return mapping algorithm recasts the previous equations in formof residuals that are minimized by using the Newton-Raphsonalgorithm:

{rj+1, ∆λj+1} :

{εr = rj+1 − re + Kem∆λj+1 = 0

εf = f (rj+1) = 0

↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−[∂εr∂r

∂εr∂∆λ

∂εf∂r

∂εf∂∆λ

]−1 [εkrεkf

]


Return Mapping Algorithm: Spring-Slider System

This is the specialization to the spring-slider where m and n areconstant and the solution is achieved in one iteration!!!

{rj+1, ∆λj+1} :

{εr = rj+1 − re + Kem∆λj+1 = 0

εf = f (rj+1) = 0

↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−[I KemnT 0

]−1 [εkrεkf

]where k is the iteration index and the process stops when ‖ε‖ < tol.




{r1j+1 = rj

∆λ1j+1 = 0

→

{ε1r = −Ke∆uj+1

ε1f = f (rj)

↓[rj+1

∆λj+1

]=

[rj0

]−

[I− KemnT

nTDemKem

nTKemnT

nTKem−1

nTKem

] [−Ke∆uj+1

f (rj)

]where k is the iteration index and the process stops when ‖ε‖ < tol.




{r1j+1 = rj

∆λ1j+1 = 0

→

{ε1r = −Ke∆uj+1

ε1f = f (rj)

↓[rj+1

∆λj+1

]=

[rj0

]+

Ke∆uj+1 −Kem(nTKe∆uj+1+f (rj))

nTKem(nTKe∆uj+1+f (rj))

nTKem

where k is the iteration index and the process stops when ‖ε‖ < tol.

f (rj) ≤ 0→ the displacement increment ∆uj+1 is partiallyconverted to plastic displacement ∆λj+1 !!!


Consistent Tangent Stiffness

The Jacobian computed for the last iteration of the Newton-Raphsonalgorithm provides the consistent tangent stiffness matrix:

{rj+1, ∆λj+1} :

{εr = rj+1 − re + Kem∆λj+1 = 0

εf = f (rj+1) = 0

↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−[∂εr∂r

∂εr∂∆λ

∂εf∂r

∂εf∂∆λ

]−1 [εkrεkf

]


Consistent Tangent Stiffness

The Jacobian computed for the last iteration of the Newton-Raphsonalgorithm provides the consistent tangent stiffness matrix:

{rj+1, ∆λj+1} :

{εr = rj+1 − re + Kem∆λj+1 = 0

εf = f (rj+1) = 0

↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−

[∂r∂εr

∂r∂εf

∂∆λ∂εr

∂∆λ∂εf

][εkrεkf

]↓

Kj+1 =∂rj+1

∂uj+1= −

∂rj+1

∂εr

∂εr∂uj+1

with,

∂ (∆uj+1) = ∂ (uj+1 − uj) = ∂uj+1 −��>

constant∂uj = ∂uj+1


Consistent Tangent Stiffness: Spring-Slider System


{rj+1, ∆λj+1} :

{εr = rj+1 − re + Kem∆λj+1 = 0

εf = f (rj+1) = 0

↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−

[I− KemnT

nTKemKem

nTKemnT

nTKem−1

nTKem

] [εkrεkf

]↓

Kj+1 =∂rj+1

∂uj+1= −

∂rj+1

∂εr

∂εr∂uj+1

= Ke − KemnTKe

nTKem

with,

∂ (∆uj+1) = ∂ (uj+1 − uj) = ∂uj+1 −��>

constant∂uj = ∂uj+1


Return Mapping Algorithm: Code Template

1: ∆uj+1 ← uj+1 − uj

2: re ← rj + Ke∆uj+1

3: if f (re) ≥ 0 then4: rj+1 ← re5: ∆λj+1 ← 06: εr ← rj+1 − re + Kem∆λj+1

7: εf ← f (rj+1)8: repeat

9:

[rj+1

∆λj+1

]←

[rj+1

∆λj+1

]−

[∂εr∂r

∂εr∂∆λ

∂εf∂r

∂εf∂∆λ

]−1 [εr

εf

]10: εr ← rj+1 − re + Kem∆λj+1

11: εf ← f (rj+1)12: until ‖ε‖ >= Tol13: Kj+1 ← − ∂r

∂εr

∂εr∂uj+1

14: else if f (re) < 0 then15: rj+1 ← re

16: Kj+1 ← Ke

17: end if


Associated vs. Non-Associated Plastic Flow

Some concluding remark:

nT = [1,tanϕ] : outward normal of the yielding surface (in thestress/force space)

mT = [1,tanψ] : direction of the plastic deformation flow (inthe strain/displacement space)

m = n : the plastic deformation flow and the normal to theyielding surface are co-linear. This is the so called associatedplasticity case that holds, for example, for metals.

m 6= n : the plastic deformation flow and the normal to theyielding surface are not co-linear. This is the so callednon-associated plasticity case that holds, for example, for soils.


Nonlinear Static Analysis (r,u)

We derived a procedure for calculating the force response of a singleelement given a displacement trial ...

... but we want to solve the static displacement response of a model,which combines several elements, subjected to an external loadhistory.

The corresponding balance equation reads,

uj : r (uj)− f (tj) = 0

where,

uj : global displacement vector

r (uj) : global restoring force vector

f (tj) : global external load vector

at time step j-th.


Nonlinear Static Analysis (r,u): Code Template

1: for j = 1 to J do2: uj ← uj−1

3: for i = 1 to I do4: ri,j ← elementForce (Ziuj)5: rj ← rj + ZT

i ri,j6: end for7: εr ← rj − f (tj)8: repeat9: for i = 1 to I do

10: Ki,j ← elementStiff (Ziuj)11: Kj ← Kj + ZT

i Ki,jZi

12: end for13: uj ← uj −K−1

j εr

14: for i = 1 to I do15: ri,j ← elementForce (Ziuj)16: rj ← rj + ZT

i ri,j17: end for18: εr ← rj − f (tj)19: until ‖εr‖ >= Tol20: end for


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