Part A Plasticity - ETH Z

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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 - 27 September, 2018

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

Evauation of: x2 = x1 + ∆x1 = x1 −(df

f (x1)

f (x2)

f (x3)

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 (x + ∆x)f2 (x + ∆x)

...fn (x + ∆x)

f1 (x)f2 (x)

...fn (x)

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

......

. . ....

∂fn∂x1

∂fn∂x2

· · · ∂fn∂xn

∆x2...

x2...xn

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

......

. . ....

∂fn∂x1

∂fn∂x2

· · · ∂fn∂xn

f1 (x)f2 (x)

...fn (x)

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 = Hk : horizontal elastic vel.

]v e = 0 : vertical elastic vel.

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

u = ue + up

up = λm

] 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)

], Ke =

[k 00 0

] Ke : elastic stiffness matrix

H : horizontal force rate[Ns

]V : vertical force rate

]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)

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 =

], r =

]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 Kem

Lumped Plasticity Model

[I Kem

[Ke − KemnT Ke

nT KemKem

nT KemnT Ke

nT Kem−1

nT Kem

The inverse of square block matrix A =

[A11 A12

A21 A22

]reads,

A−1 =

[A−1

11 + A−111 A12B−1A21A−1

11 −A−111 A12B−1

−B−1A21A−111 B−1

]where,

B = A22 − A21A−111 A12

The Matrix Cookbook

Lumped Plasticity Model: Tangent Stiffness

Instantaneous tangent stiffness of the spring-slider system:

Ke − KemnT Ke

nT Kem

[k 00 0

] [1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

Ke − KemnT Ke

nT Kem

[k 00 0

] [1 tanϕ

tanψ tanψtanϕ

] [k 00 0

Ke − KemnT Ke

nT Kem

[k 00 0

[k2 00 0

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:

Ke − KemnT Ke

nT Kem

[1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

Lumped Plasticity Model: Plastic Multiplier

Instantaneous plastic multiplier of the spring-slider system:

Ke − KemnT Ke

nT Kem

]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

(Ke − KemnTKe

∆r =

(Ke − KemnTKe

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 Newton-Raphson algorithm is used to build a return mappingalgorithm:

{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

]where:

re = uj + Ke∆uj+1 is the elastic predictor.

r1j+1 = rj and ∆λ1

j+1 = 0 is the initialization.

k is the iteration index.

{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:

{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

nT DemKem

nT KemnT

nT Kem−1

nT Kem

] [εkrεkf

]where:

Linear Jacobian → convergence in one iteration !!!

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

rk+1j+1

∆λk+1j+1

[rkj+1

∆λkj+1

[∂r∂εr

∂r∂εf

∂∆λ∂εr

∂∆λ∂εf

][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

Kj+1 =∂rj+1

∂uj+1=

∂rj+1

∂∆uj+1= −

∂rj+1

∂εr

∂εr∂∆uj+1

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

constant∂uj = ∂uj+1

Consistent Tangent Stiffness: Spring-Slider System

For the spring-slider system, the consistent tangent stiffness matrixreads:

εr = rj+1 − re + Kem∆λj+1 = rj+1 −

(rj + Ke

(∆uj+1 −m∆λj+1

))[∂r∂εr

∂r∂εf

∂∆λ∂εr

∂∆λ∂εf

[I− KemnT

nT KemKem

nT KemnT

nT Kem−1

nT Kem

Kj+1 =∂rj+1

∂uj+1=

∂rj+1

∂∆uj+1= −

∂rj+1

∂εr

∂εr∂∆uj+1

(I− KemnT

∂ (∆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

∆λj+1

[∂εr∂r

∂εr∂∆λ

∂εf∂r

∂εf∂∆λ

]−1 [εr

]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) (Force Control)

So far we focused on calculating the force response of a singleelement given a displacement trial ...

... but we want to solve the static response of a model subjected toan external load history.

Therefore, we need to solve the following balance equation,

r (uj)− fj ,ext = 0

where j is the analysis step index and,

uj : displacement vector

r (uj) : restoring force vector

fj ,ext : imposed load vector

Nonlinear Static Analysis (r,u) (Force Control)

1: for j = 1 to J do2: uj ← uj−1

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

i ri,j6: Kj ← Kj + ZT

i Ki,jZi

7: end for8: res← rj − fj,ext9: repeat

10: jac← Kj

11: uj ← uj − jac−1res12: for i = 1 to I do13: [ri,j ,Ki,j ]← element (Ziuj)14: rj ← rj + ZT

i ri,j15: Kj ← Kj + ZT

i Ki,jZi

16: end for17: res← rj − fj,ext18: until ‖res‖ >= Tol19: end for

Nonlinear Static Analysis (r,u) (Displacement Control)

In case the the restoring force is bounded, balance may not besatisfied for high loads ...

... imposing a displacement history is more convenient in this case.

Therefore, we need to solve both balance and compatibilityequations, {

r (uj) + LTλj = 0

Luj − uj ,ext = 0

where j is the analysis step index and,

uj : displacement vector

r (uj) : restoring force vector

uj ,ext : imposed displacementvector

λj : imposed load vector

L : collocation matrix

Nonlinear Static Analysis (r,u) (Displacement Control)

1: for j = 1 to J do

2: uj ← uj−1

3: λj ← λj−1

4: for i = 1 to I do5:

[ri,j ,Ki,j

]← element

)6: rj ← rj + ZT

i ri,j

7: Kj ← Kj + ZTi Ki,jZi

8: end for

9: res←[

rj + LTλjLuj − uj,ext

]10: repeat

11: jac←[

]12: ∆x← −jac−1res

13: uj ← uj + ∆x (1)

14: λj ← λj + ∆x (2)

15: for i = 1 to I do16:

[ri,j ,Ki,j

]← element

)17: rj ← rj + ZT

i ri,j

18: Kj ← Kj + ZTi Ki,jZi

19: end for

20: res←[

rj + LTλjLuj − uj,ext

]21: until ‖res‖ >= Tol

22: end for

Part A Plasticity - ETH Z

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