The Finite Element Method for the Analysis of Non-Linear ......Non-Linear and Dynamic Systems: Computational Plasticity Part I Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr. Konstantinos

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 4 - Part A - 28 October, 2020

Institute of Structural Engineering Method of Finite Elements II 1

Learning Goals

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

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

To understand how to use the Return Mapping and N-Rmethods in this problem.

References:

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


https://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structureshttps://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structureshttps://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structures

Material Nonlinearity

Almost all materials are inherently non-linear.Meaning that:

F 6= KU

A typical Force-Displacement curve for such response looks like this:


Material Nonlinearity

Important to note: Not all Nonlinear behavior is Inelastic (e.g.hyperelastic materials)

Source: Dirk Mohr - Lecture 4 (ETHZ)


https://ethz.ch/content/dam/ethz/special-interest/mavt/virtual-manufacturing/ivp-dam/Studium/Vorlesungsunterlagen/Dynamic%20Behavior%20of%20Materials%20and%20Structures/Downloads/Lecture4/Lecture%204%20(Dirk%20Mohr,%20ETH,%20Integration%20Algorithms%20in%20Plasticity).pdf

Lumped Plasticity: a Spring-Slider System

Before we visit the continuum mechanics equations, let us firstexamine a simple (lumped) plasticity model.

if the applied force H is smaller than adhesion, sliding isprevented

if the applied force H is higher than adhesion, sliding starts

The deformation u is decomposed in an elastic and a plastic part as:

u = ue + up





Let us slightly refine the model, to better reflect an actual system byexamining a block sliding on a rough surface (De Borst, Crisfield, etal. 2012):

The horizontal displacement u of point A is initially caused bydeformation (elongation) of the spring (ue), since for low forcelevels, the adhesion and the friction between the block and the floorprevent sliding of the block.When the maximum shear force - exerted by adhesion and friction -is exhausted, sliding (up) occurs.



The problem is therefore decomposed as follows:

u = ue + up → u̇ = u̇e + u̇p

ue is elastic since, upon removal of the force, the deformation in thespring also disappears.

However, the displacements of the block - once triggered - do notdisappear (permanent). Deformations that cannot be recoveredduring unloading are termed inelastic or plastic.



The surface between the floor and the sliding block is not perfectlysmooth ⇒horizontal sliding (up) will also generate a vertical displacement ofthe block, vp.

These horizontal and vertical displacement components can beassembled in corresponding vectors:

up =

[up

vp

]ue =

[ue

v e

]

For the present example, the vertical elastic force component v e hasno physical meaning and is simply equal to zero.



The elastic displacement ue is linked to the horizontal force H viaHooke’s law:

H = kue

where k is the spring constant.If H = 0⇒ ue = 0

For the plastic displacement up, this is not the case.When slipping initiates it continues to grow under constant force.



We will assume that during slipping the rate of the inelasticdeformation u̇p can be determined.For the moment, let us assume that the ratio between the horizontal“plastic” velocity u̇p and the vertical “plastic” velocity v̇p can bedefined from experiments and is defined by the dilatancy angle ψ:

tan(ψ) =v̇p

u̇p

This turns out to be a function of the total accumulated plasticdisplacement/slip (equivalent plastic strain), which is linked to theso-called “hardening law”.For now, let us ignore this.



The overall displacement vector u is therefore decomposed as follows:

u = ue + up → u̇ = u̇e + u̇p

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.

u̇ = u̇e + u̇p

u̇e =

[u̇e

v̇ e

] u̇e = Ḣk : horizontal elastic vel. [ms ]v̇ e = 0 : vertical elastic vel.

[ms

]



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

u̇ = u̇e + u̇p

u̇p = λ̇m

m =

[1

tanψ

] λ̇ : plastic multiplier [m]tanψ = v̇pu̇p : ratio between plastic vert.and horiz. velocities [d .l .]

The value of the plastic multiplier λ̇ can be determined from therequirement that during plastic flow, the stresses remain bounded.



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

ṙ = Ke u̇e = Ke (u̇− u̇p)with,

ṙ =

[Ḣ

V̇

], Ke =

[k 00 0

] Ke : elastic stiffness matrixḢ : horizontal force rate

[Ns

]V̇ : vertical force rate

[Ns

]Institute of Structural Engineering Method of Finite Elements II 15


But when does sliding initiate?

We mentioned that slip occurs when the maximum shear force -exerted by adhesion and friction - is exhausted.

However, a formal criterion needs to be provided that sets theborderline between purely “elastic” displacements and slip (sliding ofthe block), i.e., when “plastic” displacements occur.

For our present system we only have two force components, namely,H and V . The simplest assumption is that sliding starts when theCoulomb friction, defined by the friction angle, φ augmented withsome adhesion, c , is fulfilled.



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 elongationf (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the blockf (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→ u̇p = 0→ u̇e = u̇

ṙ = Ke u̇



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


f (H,V , ϕ, c) = H + Vtanϕ− c = 0→ u̇p 6= 0→ u̇e = u̇− u̇p{ṙ = Ke (u̇− u̇p)ḟ = 0



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


ḟ (H,V , ϕ, c) = Ḣ + V̇ tanϕ = nT ṙ = 0

with n =

[1

tanϕ

], ṙ =

[Ḣ

V̇

]Institute of Structural Engineering Method of Finite Elements II 20

Lumped Plasticity Model

As long as we stay on the yielding surface, both of the followingconditions must be fulfilled:

{ṙ = Ke (u̇− u̇p)ḟ = 0

→

{ṙ = Ke

(u̇− λ̇m

)ḟ = 0

→

{ṙ = Ke

(u̇− λ̇m

)nT ṙ = 0

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

[I Kem

nT 0

] [ṙ

λ̇

]=

[Ke u̇

0

]


Lumped Plasticity Model

[I Kem

nT 0

] [ṙ

λ̇

]=

[Ke u̇

0

]↓[

ṙ

λ̇

]=

[Ke − KemnT Ke

nT KemKem

nT KemnT Ke

nT Kem−1

nT Kem

][u̇0

]

The inverse of square block matrix A =

[A11 A12A21 A22

]reads,

A−1 =

[A−111 + A

−111 A12B

−1A21A−111 −A

−111 A12B

−1

−B−1A21A−111 B−1]

where,

B = A22 − A21A−111 A12The Matrix Cookbook


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

Lumped Plasticity Model: Tangent Stiffness

Instantaneous tangent stiffness of the spring-slider system:

ṙ =(

Ke − KemnT KenT Kem

)u̇

λ̇ =(

nT Ke

nT Kem

)u̇

[Ḣ

V̇

]=

[k 00 0]−

[k 00 0

] [1

tanψ

] [1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

] [1

tanψ

][u̇v̇

]




ṙ =(


)u̇

λ̇ =(

nT Ke

nT Kem

)u̇

[Ḣ

V̇

]=

[k 00 0]−

[k 00 0

] [1 tanϕ

tanψ tanψtanϕ

] [k 00 0

]k

[u̇v̇]




ṙ =(


)u̇

λ̇ =(

nT Ke

nT Kem

)u̇

[Ḣ

V̇

]=

[k 00 0]−

[k2 00 0

]k

[u̇v̇]

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:

ṙ =(


)u̇

λ̇ =(

nT Ke

nT Kem

)u̇

λ̇ =

[1 tanϕ

] [k 00 0

][1 tanϕ

] [k 00 0

] [1

tanψ

][u̇v̇

]


Lumped Plasticity Model: Plastic Multiplier

Instantaneous plastic multiplier of the spring-slider system:

ṙ =(


)u̇

λ̇ =(

nT Ke

nT Kem

)u̇

λ̇ =

([k 0

]k

)[u̇v̇

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


Integration of the Force-Displacement Response

Discussion

The rate equations listed in the previous slide are in generalnon-symmetric, since in general φ 6= ψ, which further implies thatm 6= n.

As a consequence, the matrix formed by the outer product mnT willbe non-symmetric, thus rendering the tangential stiffness matrix thatsets the incremental relation between ṙ and u̇ also non-symmetric.

The example we used is a special (simplified) case, but is used hereas an intro to continuum plasticity, which follows in the next lectureand where the off-diagonal terms do not vanish.



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



Elastic domain:

f (r) < 0

↓ṙ = Ke u̇

↓∆r = Ke∆u

Plastic domain (yielding surface):

f (r) = 0

↓

ṙ =

(Ke − K

emnTKe

nTKem

)u̇

↓

∆r =

(Ke − K

emnTKe

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

{̂rj+1, ∆λ̂j+1} :

{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = 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 ∆λ1j+1 = 0 is the initialization.

k is the iteration index.




{̂rj+1, ∆λ̂j+1} :


↓[rk+1j+1

∆λk+1j+1

]=

[rkj+1

∆λkj+1

]−[

I KemnT 0

]−1 [εkrεkf

]where:







{̂rj+1, ∆λ̂j+1} :


↓[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 !!!Institute of Structural Engineering Method of Finite Elements II 29

Consistent Tangent Stiffness

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

{̂rj+1, ∆λ̂j+1} :


↓[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, ∆λ̂j+1} :


↓

Kj+1 =∂rj+1∂uj+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

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

εr = r̂j+1 − re + Kem∆λ̂j+1 = r̂j+1 −

(r̂j + K

e(

∆uj+1 −m∆λ̂j+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− K

emnT

nTKem

)Ke

with,

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

constant∂uj = ∂uj+1


Return Mapping Algorithm: Code Template

1: ∆uj+1 ← uj+1 − uj2: re ← rj + Ke∆uj+13: if f (re) ≥ 0 then4: rj+1 ← re5: ∆λj+1 ← 06: εr ← rj+1 − re + Kem∆λj+17: εf ← f (rj+1)8: repeat

9:

[rj+1

∆λj+1

]←

[rj+1

∆λj+1

]−

[∂εr∂r

∂εr∂∆λ

∂εf∂r

∂εf∂∆λ

]−1 [εrεf

]10: εr ← rj+1 − re + Kem∆λj+111: εf ← f (rj+1)12: until ‖ε‖ >= Tol13: Kj+1 ← − ∂r∂εr

∂εr∂uj+1

14: else if f (re) < 0 then15: rj+1 ← re16: Kj+1 ← Ke17: 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−13: for i = 1 to I do4: [ri,j ,Ki,j ]← element (Ziuj)5: rj ← rj + ZTi ri,j6: Kj ← Kj + ZTi Ki,jZi7: end for8: res← rj − fj,ext9: repeat

10: jac← Kj11: uj ← uj − jac−1res12: for i = 1 to I do13: [ri,j ,Ki,j ]← element (Ziuj)14: rj ← rj + ZTi ri,j15: Kj ← Kj + ZTi Ki,jZi16: 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 do2: uj ← uj−13: λj ← λj−14: for i = 1 to I do5:

[ri,j ,Ki,j

]← element

(Ziuj

)6: rj ← rj + ZTi ri,j7: Kj ← Kj + ZTi Ki,jZi8: end for9: res←

[rj + L

TλjLuj − uj,ext

]10: repeat11: jac←

[Kj L

T

L 0

]12: ∆x← −jac−1res13: uj ← uj + ∆x (1)14: λj ← λj + ∆x (2)15: for i = 1 to I do16:

[ri,j ,Ki,j

]← element

(Ziuj

)17: rj ← rj + ZTi ri,j18: Kj ← Kj + ZTi Ki,jZi19: end for20: res←

[rj + L

TλjLuj − uj,ext

]21: until ‖res‖ >= Tol22: end for


Documents

The Finite Element Method for the Analysis of Non-Linear ......Non-Linear and Dynamic Systems: Computational Plasticity Part I Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr. Konstantinos