2002-cc kr-elastoviscoplastic fe analysis in matlab

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J. Numer. Math., Vol. 10, No. 3, pp. 157–192 (2002)

c VSP 2002

Elastoviscoplastic Finite Element analysisin 100 lines of Matlab

C. Carstensen and R. Klose †

Received 3 July, 2002

Abstract — This paper provides a short Matlab implementation with documentation of the P1 finiteelement method for the numerical solution of viscoplastic and elastoplastic evolution problems in 2Dand 3D for von-Mises yield functions and Prandtl-Reuß flow rules. The material behaviour includesperfect plasticity as well as isotropic and kinematic hardening with or without a viscoplastic penali-sation in a dual model, i.e. with displacements and the stresses as the main variables. The numericalrealisation, however, eliminates the internal variables and becomes displacement-oriented in the end.Any adaption from the given three time-depending examples to more complex applications can easilybe performed because of the shortness of the program and the given documentation. In the numerical2D and 3D examples an efficient error estimator is realized to monitor the stress error.

Keywords: finite element method, viscoplasticity, elastoplasticity, Matlab

1. INTRODUCTION

Elastoplastic time-evolution problems usually require universal, complex, commer-

cial computer codes running on workstations or even super computers [11,15]. The

argument of keeping commercial secrets hidden inside a black-box, leaves the typ-

ical user without any idea what exactly is going on behind the program’s user-

friendly surface. The difference between a mathematical and a numerical model

is fuzzy. Often, users do not care about nasty details such as quadrature rules orthe exact material laws realised. But sometimes it does matter whether a regu-

larised or penalised discrete model is solved, how the termination of an iterative

process is steered, and what post-processing led to both, equilibrium and admissi-

bility of the approximate stress field. In particular, if the solution process is part

of guaranteed error control, it is quite important to know if discrete equilibrium is

fulfilled exactly or not. Finally, the use of more than one black-box in a chain is

likely to be less efficient. In summary, to aim efficiency (e.g. by well-adapting au-

tomatic mesh-refinements) and reliability we have to prevent, at least confess, all Institute for Applied Mathematics and Numerical Analysis, Vienna University of Technology,

Wiedner Hauptstraße 8-10, A-1040 Vienna, Austria

†Mathematisches Seminar, Christian-Albrechts-Universitat zu Kiel, Ludewig-Meyn-Str. 4, D-24098 Kiel, Germany



158 C. Carstensen and R. Klose

the numerical crimes. Those range from modelling errors (e.g. data errors, coarse

geometrical resolution, wrong mathematical model, etc.) to variational crimes (e.g.

material laws fulfilled at discrete points only or just in some approximate sense).

This work aims a clear presentation of clean algorithms to compute a few model

examples. Their implementation in Matlab (as one possible higher scientific com-puting tool) is the content of this paper. The provided software may serve as both,

reference material for future numerical examples and as scientificly proved public

domain source for modifications in other applications for research and educational

purposes.

This paper continues our efforts on the Laplace and Navier–Lame problem [2,3]

towards more complicated material behaviour to advertise and to open the field of

computational plasticity to a larger community of applied mathematicians and engi-

neers. Within a small strain framework, the rheological models of this work includeperfect plasticity, elastoplastic evolution with kinematic or isotropic hardening in

the setting of von-Mises yield conditions and a Prandl-Reuß flow rule [11,15]. All

variants are available for a viscoplastic regularisation with a Yosida-regularisation,

known as viscoplasticity due to Perzyna.

The time discretisation yields a minimisation problem under severe constraints

for each time step. The discrete problem is solved with a Newton-Raphson scheme

and Matlab’s standard direct solver. The main principle of the presented method

is, that, in each time step, the stress tensor is expressed explicitly and applied in

Cauchy’s first law of motion.

We expect that the reader is familiar with Matlab as well as with linear elasticity

and otherwise refer, e.g. to [3,13].

The paper is organised as follows. In Section 2 we introduce the viscoplastic

and elastoplastic problem and their mathematical model. In Section 3 we perform a

time discretisation and calculate the explicit expression of the stress tensor for the

different models of hardening. In Section 4 the discretisation in space together with

the data representation of the triangulation, the Dirichlet and Neumann boundary is

considered. Furthermore the iterative solution of the discrete problem is explained.Section 5 concerns the calculation of the stiffness matrix. Section 6 discusses post

processing routines, one for previewing the numerical solution and one for an a

posteriori error estimation. Numerical examples in Section 7 illustrate the usage

of the provided tools for four benchmark examples in two and three space dimen-

sions.

The programs are written for Matlab 6.1 but adaption for previous versions ispossible. For a numerical calculation it is necessary to run the program

with (user-specified) files

,

,

, , as well as the subroutines , , and . Their meaning andusage is explained throughout the paper and examples are provided in the later sec-tions. The graphical representation is performed with the function and theerror is estimated in

. All files with the code and the data for allexamples can be downloaded from

.



Elastoviscoplastic FE analysis 159

2. RHEOLOGICAL MODEL

2.1. Material models

The material body occupies a bounded Lipschitz domainΩ

in

d

and deforms anytime t , 0 t T . Given a displacement field u u t H 1

Ω

d (i.e., the functionalmatrix Du exists almost everywhere in Ω, is measurable, and satisfies

Ω Du

2 d x

∞) the total strain reads

ε

u

Du

DuT

2

ε

u jk

u j

k

uk

j

2

j

k

1

d

In the engineering literature, ε u

is known as the linear Green strain tensor. Thedefault model for small-strain plasticity involves an additive split

ε ε ε

e

σ σ σ

p

ξξξ

of the total strain in an elastic part, e

σ σ σ

1σ σ σ , and an irreversible part p

ξξξ

.The plastic strain p

ξξξ

depends on further internal variables ξξξ . The stress field

σ σ σ

σ σ σ t L2

Ω;

d d sym

of the Cauchy stress tensor σ σ σ x t

is linked to the elas-

tic, reversible strain. The linear relation is provided by

1 , the compliance tensor,which is the inverse of the fourth order elasticity tensor

. In the numerical exam-ples, a two-parameter model (λ and µ are the Lame constants) in linear isotropic,homogeneous elasticity,

e

λ tr e I

2µ e, is employed.

The evolution law for the plastic strain p is more involved and requires theconcept of admissible stresses, a yield function, and an associated flow rule.The kinematic variables p and ξξξ form the generalised strain P

p

ξξξ

. The cor-responding generalised stress reads ΣΣΣ

σ σ σ α α α , where α α α

m

msym describes internal

stresses.We denote by K the set of admissible stresses, which is a closed, convex set,

containing 0, and is defined by

K

ΣΣΣ : Φ

ΣΣΣ

0

The yield function Φ describes admissible stresses by Φ ΣΣΣ 0 and models eitherperfect plasticity or hardening. The dissipation functional ϕ

ΣΣΣ

, defined by

ϕ σ σ σ α α α

1

2ν inf

σ σ σ τ τ τ α α α β β β

2 : τ τ τ β β β

d d sym

m msym

Φ τ τ τ β β β

0

(2.1)

approaches the indicator function of the set K, namely

ϕ

ΣΣΣ

0

Φ

ΣΣΣ

0

∞

Φ

ΣΣΣ

0(2.2)

as ν 0. The time derivative of P is given by the flow rule

P ∂ϕ ΣΣΣ σ σ σ α α α

d d sym

m msym :

τ τ τ β β β

d d sym

m msym

ϕ

σ σ σ

α α α

σ σ σ :

τ τ τ σ σ σ

α α α :

β β β α α α

ϕ

τ τ τ

β β β

(2.3)




Here, A : B ∑m j

k 1 A jk B jk denotes the scalar products of (symmetric) matrices

A B

m m . For ν

0 the flow rule results in

P

p

ξξξ

1

ν

σ σ σ τ τ τ

α α α β β β

(2.4)

for

τ τ τ

β β β

uniquely determined as the projection of

σ σ σ

α α α

onto K , i.e.,

τ τ τ

β β β K

with

τ τ τ

β β β

:

Π

σ σ σ

α α α

uniquely determined by

dist

σ σ σ

α α α K

σ σ σ τ τ τ

α α α β β β

σ σ σ

α α α Π

σ σ σ

α α α

inf σ σ σ τ τ τ α α α β β β

: τ τ τ β β β

d

d sym

m

msym

Φ τ τ τ β β β

0

(2.5)

The following examples involve the von-Mises yield function in various modi-

fications.

Example 2.1 (Viscoplasticity). In the case of perfect viscoplasticity, there is nohardening and the internal variables ξξξ are absent. The yield function is given by

Φ

σ σ σ

dev

σ σ σ σ y

(2.6)

With (2.6) in (2.4) and s

:

max 0

s s

we obtain

p

1

ν

1 σ y

dev

σ σ σ

dev

σ σ σ

(2.7)

Example 2.2 (Viscoplasticity with isotropic hardening). Isotropic hardeningis characterised by the scalar hardening parameter α

0 and the constant H

0,the modulus of hardening. The yield function is

Φ

σ

α

dev

σ σ σ σ y

1 H α

(2.8)

With (2.8) in (2.4) we obtain

p

ξ

1

ν

1

1 H 2σ 2 y

1

1

α H

σ y dev σ

dev

σ σ σ

H σ y dev σ σ σ

(2.9)

In our model the plastic part of the free energy is given by ψ p

12 H 1ξ

2. The internalforce is defined by α

∂ψ p

∂ξ , which means α H 1ξ , where H 1 is a positive

hardening parameter.

Example 2.3 (Viscoplasticity with linear kinematic hardening). In kinematichardening the yield function is given by

Φ

σ σ σ

α α α

dev

σ σ σ

dev

α α α

σ y

(2.10)

With (2.10) in (2.4) we obtain

p

ξξξ

1

2ν

1

σ y

dev

σ α

dev

σ σ σ α α α

dev

σ σ σ α α α

(2.11)




For linear kinematic hardening, the plastic part of the free energy has the form ψ p

12

k 1 ξξξ

2. The internal force is defined by α α α ∂ψ p

∂ξξξ , which means α α α

k 1ξξξ ,with a positive parameter k 1.

Remark 2.1 (Elastoplasticity for ν 0). In the present situation the positivenumber ν may be seen as a viscose penalty which leads to equalities (such as (2.7),(2.9), (2.11)). The elastoplastic limit for ν

0 leads in (2.3) to a variational in-

equality [11,15]. This model will be included subsequently as one may considerν

0 in the formulae below.

2.2. Equilibrium equations

The stress field σ σ σ L2

Ω;

d

d sym and the volume force f

L2

Ω;

d are related by

the local quasi-static balance of forces,

divσ σ σ f

0

(2.12)

On some closed subset Γ D of the boundary with positive surface measure, we as-sume Dirichlet conditions while we have Neumann boundary conditions on the(possible empty) part Γ N . The two components of the displacement u need not sat-isfy simultaneously Dirichlet or Neumann conditions, i.e., Γ D and Γ N need not bedisjoint. With M

L∞

Γ D

d d , w H 1

Ω

d , and a surface force g L2

Γ N

wehave

Mu

w on Γ D

σ σ σ n

g on Γ N

(2.13)

Let Π denote the projection onto the set of admissible stresses K. The plastic prob-lem is then determined by the weak formulation: Seek u

H 1

Ω

d that satisfiesMu w on Γ D that, for all v H 1 D

Ω

d : v H 1

Ω

d : Mv

0 on Γ D ,

Ωσ σ σ

u

: ε ε ε

v

d x

Ωf

v d x

Γ N

g v ds

ε ε ε

u

1σ σ σ

ξξξ

α α α

1

ν

σ σ σ Πσ σ σ

α α α

Πα α α

a.e. in Ω

(2.14)

3. TIME DISCRETISATION AND ANALYTIC EXPRESSION OF THESTRESS TENSOR

3.1. Time discretisation scheme

A generalised midpoint rule serves as a time-discretisation. In each time stepthere is a spatial problem with given variables

u

t

σ

t

α

t

at time t 0 de-noted as u0 σ 0 α 0 and unknowns at time t 1 t 0 k denoted as u1 σ 1 α 1).Time derivatives are replaced by backward difference quotients. The time discreteproblem reads: Seek uϑ

H 1

Ω

d that satisfies Muϑ

w on Γ D that, for all




v H 1 D

Ω

d : v H 1

Ω

d : Mv

0 on Γ D ,

Ωσ σ σ uϑ : ε ε ε v d x

Ωf ϑ

v d x

Γ N

gϑ v ds

1

ϑ k

ε ε ε

uϑ u0

1

σ σ σ ϑ σ σ σ 0

ξξξ

α α α t ϑ

ξξξ

α α α t 0

1

ν

σ σ σ ϑ Πσ σ σ ϑ

α α α ϑ Πα α α ϑ

(3.1)

where σ σ σ ϑ

1 ϑ

σ σ σ 0

ϑσ σ σ 1 with 1

2

ϑ

1. We define

A :

ε ε ε

uϑ u0

ϑ k

1 σ σ σ 0ϑ k

(3.2)

With this notation in the flow law, we have

A

1 σ σ σ ϑ ϑ k

1

ν

id Π

σ σ σ ϑ

(3.3)

To obtain an expression of

1σ σ σ ϑ in terms of trσ σ σ ϑ and devσ σ σ ϑ we note that σ σ σ ϑ

λ tr eϑ I

2µ eϑ . There are constants α , β , γ and δ such that

σ σ σ ϑ

γ tr eϑ I

δ dev eϑ

eϑ

α trσ σ σ ϑ I

β devσ σ σ ϑ

(3.4)

Inserting of (3.4a) in (3.4b) we have

eϑ

d αγ tr eϑ I

βδ dev eϑ

α

1

d 2γ

β

1

δ

(3.5)

According to (3.4a) the stress tensor σ σ σ ϑ reads

σ σ σ ϑ

γ tr eϑ I

δ

eϑ 1

d tr eϑ I

γ

δ

d

tr eϑ I

δ e

λ tr eϑ I

2µ eϑ

(3.6)This shows δ

2µ and γ

λ

2µ

d . Inserting this in (3.4a) we deduce

1σ σ σ ϑ

1

d 2λ

2d µ trσ σ σ ϑ I

1

2µ devσ σ σ ϑ (3.7)

3.2. Analytic expression of the stress tensor

In this subsection we derive explicit expressions for the stress tensor σ σ σ ϑ for thedifferent cases of hardening. With the connection σ σ σ ϑ

σ σ σ ϑ

A

uϑ

between σ σ σ ϑ and uϑ the elastoplastic problem is determined by a nonlinear variational problem:

Seek uϑ H 1

Ω

d that satisfies Muϑ

w on Γ D and, for all v H 1 D

Ω

d :

v

H 1

Ω

d

: Mv

0 on Γ D

,

Ωσ σ σ ε ε ε uϑ

u0

1σ σ σ 0 : ε ε ε v d x

Ωf ϑ

v d x

Γ N

gϑ v ds (3.8)

We solve (3.3) for σ σ σ ϑ in the flow laws of Examples 2.1–2.3.




For viscoplasticity with isotropic hardening these constants are

C 1 :

λ

2µ

d

C 2 :

βν

1

H 2σ 2 y

ϑ k

1

β H 1 H 2σ 2 y

C 3 :

ϑ k σ y

1

α 0 H C 4 :

H 1 H 2ϑ k σ 2 y

ν

1 H 2σ 2 y

(3.18)

and in the limit ν 0 for plasticity with isotropic hardening these constants read

C 1 :

λ

2µ

d

C 2 :

ϑ k

1

β H 1 H 2σ 2 y

C 3 :

ϑ k σ y

1

α 0 H

C 4 :

H 1 H 2ϑ k σ 2 y

(3.19)

The plastic phase occurs for

devϑ k A

β

1

α 0 H

σ y

(3.20)

In the elastic phase the stress tensor reads

σ σ σ ϑ C 1 tr ϑ k A I

2µ devϑ k A

(3.21)

Proof. The discretised version of the flow law is

A

1 σ σ σ ϑ ϑ k

1

ν

1

1

H 2σ 2 y

1

1

α ϑ H

σ y

devσ σ σ ϑ

devσ σ σ ϑ (3.22)

H 1

1 α ϑ

α 0ϑ k

1ν

11

H 2σ 2 y

1

1

α ϑ H

σ y

devσ σ σ ϑ

H σ y devσ σ σ ϑ

(3.23)

We consider the plastic phase 1

1

α ϑ H

σ y devσ σ σ ϑ

0. With

1σ σ σ ϑ

α trσ ϑ I

β devσ σ σ ϑ in (3.22) we infer

ϑ k dev A

β

ϑ k

ν

1

1 H 2σ 2 y

1

1 α ϑ H σ y

devσ σ σ ϑ

devσ σ σ ϑ (3.24)

and so

devϑ k A

β

ϑ k

ν

1

1

H 2σ 2 y

1

1

α ϑ H

σ y

devσ σ σ ϑ

devσ σ σ ϑ

(3.25)

The second component of the flow rule, (3.23), can be solved for α ϑ . Inserting of α ϑ in (3.25), solving for

devσ σ σ ϑ

and inserting the resulting expression in (3.24)yields an expression for devσ σ σ ϑ . We obtain

devσ σ σ ϑ

C 3

C 2

devϑ k A

C 4

C 2

devϑ k A

(3.26)

From the first component of the flow rule we conclude trσ σ σ ϑ

C 1d trϑ k A. With

devσ σ σ ϑ 1 α ϑ H σ y the expression for devσ σ σ ϑ in the elastic results from (3.22).

The plastic phase occurs for devσ σ σ ϑ

1

α ϑ H

σ y. With (3.26) this is equiv-alent to 1

β

1

α 0 H

σ y devϑ k A

0.




Theorem 3.3 (Viscoplasticity and plasticity with kinematic hardening). For viscoplasticity and plasticity with kinematic hardening there exit constants C 1 , C 2 ,

and C 3 such that the stress tensor reads

σ σ σ ϑ

C 1 tr

ϑ k A

I

C 2

C 3

dev

ϑ k A

βα α α 0

dev

ϑ k A

βα α α 0

devα α α 0

(3.27)For viscoplasticity with kinematic hardening these constants are

C 1 : λ

2µ d C 2 :

ϑ kk 1

2ν

ϑ k

ϑ kk 1

2µ

ν

µ C 3 :

ϑ k σ yϑ k

ϑ kk 1

2µ

ν

µ (3.28)

and in the limit ν

0 for plasticity with kinematic hardening these constants read

C 1 : λ 2µ d C 2 : ϑ kk 1ϑ k

ϑ kk 1

2µ

C 3 : ϑ k σ yϑ k

ϑ kk 1

2µ

(3.29)

The plastic phase occurs for

dev

ϑ k A βα α α 0

βσ y

(3.30)

In the elastic phase the stress tensor reads

σ σ σ ϑ

C 1 tr

ϑ k A

I

2µ devϑ k A

(3.31)

Proof. The discretised version of the flow law reads

A

1 σ σ σ ϑ ϑ k

1

2ν

1 σ y

dev

σ σ σ ϑ α α α ϑ

dev

σ σ σ ϑ α ϑ α ϑ α ϑ

1

k 1

α α α ϑ α α α 0ϑ k

(3.32)We consider the plastic phase 1

σ y dev


0. With

1σ σ σ ϑ α trσ ϑ I

β devσ σ σ ϑ ,α α α ϑ α α α 0 k 1 ϑ k A

α trσ ϑ I

β devσ σ σ ϑ

(3.33)

This yields in the flow rule that

dev

ϑ k A βσ σ σ ϑ

ϑ k

2ν

1

σ y

dev

σ σ σ ϑ α α α 0 k 1ϑ k A k 1βσ σ σ ϑ

dev σ σ σ ϑ

α α α 0 k 1ϑ k A k 1βσ σ σ ϑ

The absolute value on both sides results in an equation for dev

σ σ σ ϑ α α α 0

k 1ϑ k A

k 1βσ σ σ ϑ . We solve for this modulus and substitute it in the above identity. This

shows

dev


ϑ k

2ν

1

σ y2ν ϑ k

dev ϑ k A βσ σ σ ϑ σ y

dev


k 1ϑ k A

k 1βσ σ σ ϑ

(3.34)




Multiplying both sides of (3.34) with

2ν ϑ k dev

ϑ k A βσ σ σ ϑ σ y togetherwith some basic transformations shows

2ν

ϑ k

k 1

dev ϑ k A

βσ σ σ ϑ

σ y

dev

ϑ k A βσ σ σ ϑ dev

ϑ k A βσ σ σ ϑ dev


(3.35)

On both sides of (3.35) we build the absolute value and obtain

dev


dev


σ yk 1

2ν

ϑ k (3.36)

which can be inserted in (3.35),

dev σ σ σ ϑ

α α α 0

dev ϑ k A

βσ σ σ ϑ

dev


σ yk 1

2ν

ϑ k dev


(3.37)

Adding the term dev


β devα α α 0 on both sides of (3.37) yields with somebasic transformations

dev


dev

ϑ k A βα α α 0

β dev


dev σ σ σ ϑ

α α α 0

σ y

k 1 2ν ϑ k

dev

σ σ σ ϑ

α α α 0

(3.38)

Because of 2ν ϑ k k 1

0 and with

dev σ σ σ ϑ

α α α 0

σ y

0 from (3.36) we cancalculate the absolute value of dev


as

dev σ σ σ ϑ

α α α 0

ϑ kk 1

2ν dev

ϑ k A βα α α 0

ϑ k σ yϑ k

βϑ kk 1

2βν

(3.39)

We insert (3.39) in (3.38) to obtain

devσ σ σ ϑ C 2 C 3 dev ϑ k A βα α α 0 dev ϑ k A βα α α 0 devα α α 0 (3.40)

From the first component of the flow rule we have trσ σ σ ϑ C 1d tr ϑ k A . The expres-sion for devσ σ σ ϑ in the elastic phase is obtained from (3.32) with 1

σ y

dev

σ σ σ ϑ

α α α ϑ

0.The plastic phase occurs for 1

σ y dev


0. With (3.36), (3.37)and (3.39) this is equivalent to

dev

ϑ k A βα α α 0

βσ y.

4. NUMERICAL SOLUTION OF THE TIME-DISCRETE PROBLEM

This section is devoted to the framework of the spatial discretisation of one timestep(3.1) with data structures in Subsection 4.1 and the iteration in Subsection 4.2. Theheart of the matter is the assembling of the tangential local stiffness matrix explainedin Section 5.




0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Γ

Γ

Γ

Γ

D

N

N

D

N

Figure 1. Plot of initial triangulation for the example of Subsection 7.1.

4.1. Discretisation in space

Suppose the domain Ω has a polygonal boundary Γ , we can cover Ω by a regulartriangulation

, i.e., Ω

T

T , where the elements of

are triangles for d

2and tetrahedrons for d

3. Regular triangulation in the sense of Ciarlet [10] meansthat the nodes

of the mesh lie on the vertices of the elements, the elements of thetriangulation do not overlap, no node lies on an edge of an element, and each edge

E Γ of an element T

belongs either to Γ N or to Γ D.

Matlab supports reading data from files given in ASCII format by

files.Figure 1 shows the initial mesh of a two dimensional example and Figure 2 showsthe corresponding data files. The file

contains the coordinatesof each node. The two real numbers per row are the x- and y-coordinates of eachnode. The file

contains for each element the node numbers of thevertices, numbered anti-clockwise.

The files

and

contain the two node numberswhich bound the corresponding edge on the boundary.

In the discrete version of (2.14), H 1

Ω

and H 1 D

Ω

are replaced by finite di-mensional subspaces

and D

V

: V

0 on Γ D . The discrete problem




1.0000 0.0000

2.0000 0.0000

2.0000 1.00002.0000 2.0000

1.0000 2.0000

0.0000 2.0000

0.0000 1.0000

0.5000 0.8660

0.8660 0.5000

1.3989 0.3748

1.0241 1.02410.3748 1.3989

1 2 102 3 10

3 11 103 4 114 5 115 12 115 6 126 7 127 8 128 11 128 9 11

9 10 119 1 10

1 26 7

2 3

3 44 55 67 88 99 1

Figure 2. Data files , , , and forthe initial triangulation in the example of Subsection 7.1. The files , , and are listed inSection 7.

reads: Seek Uϑ with

Ωσ σ σ ε ε ε Uϑ U0

1σ σ σ 0 : ε ε ε V d x

Ωf ϑ V d x

Γ gϑ V ds V (4.1)

Let

denote a triangulation of Ω and

the set of all nodes in

, then let

ϕ ϕ ϕ 1

ϕ ϕ ϕ dN

ϕ 1e1

ϕ 1ed

ϕ N e1

ϕ N ed

be the nodal basis of the finite dimensional space

, where N is the number of nodes of the mesh and ϕ z is the scalar hat function of node z in the triangulation

,i.e., ϕ z z

1 and ϕ z y

0 for all y

with y z. Then (4.1) is equivalent to

F p :

Ωσ σ σ ε ε ε Uϑ U0 C 1σ σ σ 0

: ε ε ε ϕ ϕ ϕ p

d x

Ωf ϑ ϕ ϕ ϕ p d x

Γ gϑ ϕ ϕ ϕ p ds

0

(4.2)for p

1 dN . F p can be decomposed into a sum of a part Q p which depends on

uϑ and a part P p which is independent of uϑ . Thus we have F p :

Q p P p with

Q p :

Ωσ σ σ

ε ε ε

Uϑ U0

C 1σ σ σ 0

: ε ε ε

ϕ ϕ ϕ p

d x (4.3)

P p :

Ωf ϑ ϕ ϕ ϕ p d x

Γ gϑ ϕ ϕ ϕ p ds

(4.4)

4.2. Iterative solution

This subsection describes the iterative solution of (4.2) by a Newton–Raphsonscheme which is realised in the Matlab program

, listed at the end of this




section. One step of the Newton iteration reads

DF

Uk ϑ

Uk 1ϑ

DF

Uk ϑ

Uk ϑ

F

Uk ϑ

(4.5)

The discrete displacement vector Uk ϑ is expressed in the nodal basis as Uk ϑ

∑dN 1 U k

ϑ

ϕ ϕ ϕ

, U k

ϑ 1 U k ϑ dN

are the components of Uk ϑ . The matrix DF is defined

as DF U k

ϑ

1 U k ϑ

dN pq :

∂ F p U k ϑ

1 U k ϑ

dN

∂ U k ϑ

q

(4.6)

At the end of this section the Matlab program

, a finite element programfor the two dimensional case is listed. In the following line 1

shows the function call with input

,

,

and

together with the vectors

and

, which are the start-vector uϑ of theNewton iteration and the displacement u0 at time t 0. The last three variables in theparameter list are the times t 0, t 1, and the variable ϑ . Lines 2–7,

show the assembling of the stiffness matrix DF and the stiffness vector Q. Line 2initialises DF , Q, and P, line 3 defines a vector which contains the numbers of degrees of freedom for each element. Lines 4–7 perform the assembling of DF andQ with a call of subroutine

. The element with number j contributes thedegrees of freedom

. The local stiffness matrix and the local stiffnessvector of element number j have to be added at the positions

L

L

of the global

stiffness matrix and L of the global stiffness vector.In the following lines 8–22




the vector F

Q P is determined. P contains the work of the volume forces f

and the surface forces g at time t ϑ . The values of f are provided by the subroutine

and evaluated in the centre of gravity xS yS

of T , to approximate the integral

T f ϕ ϕ ϕ j d x in (4.2),

T f

ϕ ϕ ϕ r d x

1

3 T f k xS yS k :

mod r

1

2

1

(4.7)

The integral

E g ϕ ϕ ϕ k ds in (4.2) that involves the Neumann conditions is approxi-

mated with the value of g in the centre

x M

y M

of the edge E with length E

,

E g ϕ ϕ ϕ r ds

12

E gk x M y M k : mod r 1 2 1 (4.8)

The following lines 24–33

take into account Dirichlet conditions whereas gliding boundary conditions are al-lowed as well. Gliding boundary conditions such as those along symmetry axes, areimplemented different from Dirichlet conditions (for all components). The displace-ment is fixed merely in one specified direction and possibly free in others. A generalapproach to this type of condition reads, for each node on the Dirichlet boundary,

m1...

md

U 1...

U d

w1...

wd

(4.9)

where m j

1

d , w j

, and U 1 U d

are the degrees of freedom for theconsidered node on the Dirichlet boundary. With the normal vector n along Γ andthe canonical basis e j, j

1 d , in

d , gliding and classical Dirichlet conditionsand the lack of Dirichlet-type conditions on certain parts of Γ can all be included by

n0...0

U 1...

U d

0

eT 1...

eT d

U 1...

U d

uD

0...0

U 1...

U d

0 (4.10)




respectively. Line 24 determines all nodes on the Dirichlet boundary and stores themin the variable

. Let there be n Dirichlet nodes. Line 26 creates a matrix

of size 2n

2 and a vector

of size 2n

1. The rows 2 j

1

1 up to 2 j for

j

1

n of

and

contain for each Dirichlet node the matrix

m1

md

T and

the vector w1 w2 . For the discrete problem, the generalised Dirichlet conditionslead to the linear system

BU

w (4.11)

where each row of B

2n 2 N contains the value of some m j at a node on the

boundary and w

2n contains the corresponding values w j at this node. The matrix

is calculated in lines 27–31 from the entries of

. A row m j

0 is not included in B and W , which is realised in line 33.

Between lines 34 –45

the Newton iteration takes place. Coupling the set of boundary conditions throughLagrange parameters with (4.5) leads to the extended system

DF Uk ϑ BT

B 0

Uk 1ϑ

λ λ λ

b

w

(4.12)

with b : DF U k

ϑ

Uk ϑ

f U k

ϑ

(4.13)

This system of equations is solved in line 37 by the Matlab solver. Matlab makes useof the properties of a sparse, symmetric matrix for solving the system of equations

efficiently. Line 37 determines the parts Uk 1ϑ and λ of the solution. Line 38 to 43

assemble the local stiffness matrix and stiffness vector. Line 32 calculates the set

of free nodes. There are m free nodes z1 zm

. The iteration process onlycontinues in line 34, if the relative residual F U k

ϑ

z1 U k

ϑ

zm

is greater than a given

tolerance and if the maximum number of iterations is less than 100.

4.3. 2D Finite Element program




4.4. 3D Finite Element program




5. TANGENTIAL LINEAR SYSTEM OF EQUATIONS

The tangential stiffness matrix DF in (4.6) as well as the vector F in (4.2) are ex-pressed as a sum over all elements T in

. Each part of the sum defines the localstiffness matrix M and the vector of the right-hand side R for the correspondingelement. Let k 1 through k K be the numbers of the nodes of an element T . Thereare dK basis functions with support on T , namely, ϕ ϕ ϕ π

T 1

ϕ k 1 e1

ϕ ϕ ϕ π

T

d

ϕ k 1 ed ϕ ϕ ϕ π

T

2K

1

ϕ k K e1 ϕ ϕ ϕ π

T

dK

ϕ k K ed , where e j is the j-th unit vector.The function ϕ k j is the local scalar hat function of node k j. The function π maps theindices 1 dK of the local numeration with respect to T to the global numeration.In the sequel we want to compute the local stiffness matrix M and a local vector

R for element T . For simplicity we write ϕ ϕ ϕ 1

ϕ ϕ ϕ dK instead of ϕ ϕ ϕ π

T 1

ϕ ϕ ϕ π

T

dK

.




The local vector R is defined as

Rr :

T σ σ σ

ε ε ε

Uϑ U0

1σ σ σ 0

: ε ε ε

ϕ ϕ ϕ r

d x

r

1

dK

(5.1)

The local stiffness matrix for element T is defined, for r s 1 dK , by

M rs

∂ Rr

∂ U k ϑ s

∂

∂ U k ϑ s

T σ

ε

dK

∑ 1

U k ϑ

ϕ ϕ ϕ

U0

1σ 0

: ε

ϕ ϕ ϕ r

d x

(5.2)

5.1. Notation for triangular elements

The Matlab implementation of M and R can be done in a compact way by defininga matrix

by

:

ε 11

ϕ j

ε 12

ϕ j

ε 21

ϕ j

ε 22

ϕ 1

6 j

1

1

2

1

2 Dϕ j 11 Dϕ j 12 Dϕ j 21 Dϕ j 22

6 j 1

Dϕ j 11

Dϕ j 21

Dϕ j 12

Dϕ j 22

6 j

1

(5.3)

with

∂ϕ k 1∂ x

∂ϕ k 1∂ y 0 0

0 0 ∂ϕ k 1

∂ x

∂ϕ k 1∂ y

∂ϕ k 2∂ x

∂ϕ k 2∂ y 0 0

0 0 ∂ϕ k 2

∂ x

∂ϕ k 2∂ y

∂ϕ k 3

∂ x

∂ϕ k 3

∂ y 0 0

0 0

∂ϕ k 3

∂ x

∂ϕ k 3

∂ y

∂ϕ k 1∂ x 0

∂ϕ k 1∂ y 0

0 ∂ϕ k 1

∂ x 0 ∂ϕ k 1

∂ y∂ϕ k 2∂ x 0

∂ϕ k 2∂ y 0

0 ∂ϕ k 2

∂ x 0 ∂ϕ k 2

∂ y∂ϕ k 3

∂ x 0 ∂ϕ k 3

∂ y 0

0

∂ϕ k 3

∂ x 0

∂ϕ k 3

∂ y

(5.4)

With the coordinates x1 y1

, x2 y2

, x3 y3

of the vertices of the element T , theentries of

and

are stored in the matrix

,

∂ϕ k 1

∂ x

∂ϕ k 1

∂ y∂ϕ k 2

∂ x

∂ϕ k 2

∂ y∂ϕ k 3

∂ x

∂ϕ k 3

∂ y

1 1 1 x1 x2 x3

y1 y2 y3

1

0 01 00 1

(5.5)

With

from (5.5) the following Matlab lines calculate the matrices

and

.




5.3. Local stiffness matrix M and local vector R

The subsequent algorithms perform a fall differentiation between the elastic and theplastic phase. In the plastic phase the local stiffness matrix M and the local vector

R for element T are obtained by evaluating (5.2) and (5.1) for the different cases of hardening. In the elastic phase M and R are well known, c.f. [3]. The Matlab rou-tine

calculates the local stiffness matrix M according to (5.2) and the firstsummand of (5.1). With M and R a step of the assembling procedure is performedin

.

Example 5.1 (Perfect viscoplasticity for 2D and 3D). The Matlab programcalculating the local stiffness matrix and the vector R depending on the element T

is listed at the end of this example for the 2D and 3D case, together with a specialimplementation of the deviator and trace function, called and in the 2Dcase and

and

in the 3D case. The 2D program is described below. In line 2global variables are transfered to the routine,

C 1 : λ

2µ d C 2 : ν βν ϑ k C 3 :

ϑ k σ y βν ϑ k C 4 :

2µ

In lines 4–7 the matrix is calculated as described in the previous section.In line 7 we define the matrix v :

ε

Uϑ U0

1σ σ σ 0 and store it row wise. Inline 8 the area of element T is determined. In line 9 a fall differentiation is per-formed. If

dev

v σ y

2µ

0 the plastic phase occurs, otherwise the elasticphase. Depending on the result of this distinction, a constant C 5 and a vector C 6 aredefined by

C 5 :

C 2 C 3 dev

v if

dev

v σ y

2µ

0

2µ else(5.10)

C 6 :

C 3 dev

v

3

ε

ϕ j

:

v

6 j

1 if dev

v σ y

2µ

0

0 0 0 0 0 0

T else.

(5.11)In lines 14 and 15 an update of the local stiffness matrix and the first summand of the vector R is performed. The components of the local stiffness matrix read

M jk T

C 1 tr

ε

ϕ j

tr

ε

ϕ k

C 5 dev

ε

ϕ j

: ε

ϕ k

C 6 j dev

v

: ε

ϕ k

(5.12)The components of the first summand of the local vector R read

R j T

C 1 tr

v

tr

ε

ϕ j

C 5 dev

v

: ε

ϕ j

(5.13)

The Matlab routines

,

and

for the 2D case are listed below.




The Matlab routines

,

and

for the 3D case are listed below.

Example 5.2 (Isotropic hardening in 2D). The Matlab program calculatingthe local stiffness matrix and the local vector R is listed at the end of this exam-ple. In line 2 global variables C 1 up to C 6 are transfered to the routine,

C 1 : λ µ C 2 :

ν

2µ 1 H 2σ 2 y ϑ k

1

1

2µ H 1 H 2σ 2 y

C 3 :

ϑ k σ y

1

α 0 H

C 4 :

H 1 H 2ϑ k σ 2 y

ν

1

H 2σ 2 y

C 5 :

2µ C 6 :

1

2µ

1

α 0 H

σ y

Lines 3–8 are analog to the last example and perform the calculation of

,




define the vector v and determine the element area T . In line 9 the fall differentiationis performed. If

dev

v

C 6

0 the plastic phase occurs, otherwise the elasticphase. Depending on the result of this distinction, constants C 7 and a vector C 8 aredefined by

C 7 :

C 3

C 2 dev

v

C 4

C 2 if dev

v

C 6

0

C 5 else(5.14)

C 8 j :

C 3 C 2 dev

v

3 ε ϕ j

: v

if dev

v C 6

0

0 0 0 0 0 0

T else.(5.15)

In lines 14 and 15 an update of the local stiffness matrix and the first summand of the local vector R is performed. The components of the local stiffness matrix read

M jk

T

C 1 tr

ε ϕ j

tr ε ϕ k C 7 dev

ε ϕ j

: ε ϕ k

C 8 j dev

v

: ε ϕ k

(5.16)The components of the first summand of the vector R read

R j T

C 1 tr

v

tr

ε

ϕ j

C 7 dev

v

: ε

ϕ j

(5.17)

The Matlab routine

is listed below.

Example 5.3 (Kinematic hardening in 2D). The Matlab program calculatingthe local stiffness matrix and the local vector R is listed at the end of this example.In line 2 global variables C 1 up to C 4 are transfered to the routine,

C 1 : λ µ C 2 :

ϑ kk 1

2ν

ϑ k

ϑ kk 1

2µ

ν

µ

C 3 :

ϑ k σ yϑ k

ϑ kk 1

2µ

ν

µ C 4 :

2µ

Lines 3–8 are analog to the last example. The strain tensor

is calcu-lated, and matrices v and v1 are defined, with v :

ε ε ε

Uϑ U0

1σ σ σ 0, v1 :




1

2µ α α α 0. In line 9 the fall differentiation is performed. If dev

v σ y

2µ

0the plastic phase occurs, otherwise the elastic phase. Depending on the result of thisdistinction, a constant C 5, a vector C 6 and a Kronecker delta d are defined by

C 5 :

C 3

dev v

C 2 if dev v

σ y 2µ 0C 4 else

(5.18)

C 6 j :

C 3 dev

v

3

ε

ϕ j

: v

if dev

v σ y

2µ

0

0 0 0 0 0 0

T else

(5.19)

d :

1 if dev

v σ y

2µ

0

0 else.(5.20)

In lines 14 and 15 an update of the local stiffness matrix and the first summand of the local vector R is performed. The components of the local stiffness matrix read

M jk T

C 1 tr

ε

ϕ j

tr

ε

ϕ k

C 5 dev

ε

ϕ j

: ε

ϕ k

C 6 j dev

v

: ε

ϕ k

(5.21)The components of the first summand of the vector R read

R j T

C 1 tr

v

tr

ε

ϕ j

C 5 dev

v

: ε

ϕ j

d dev

α 0

: ε

ϕ j

(5.22)

The Matlab routine

is listed below.

6. POSTPROCESSING

6.1. Displaying the solution

Our two dimensional problems model the plain stress condition. In that case, thecomplete stress tensor σ

3 3sym has the form

σ

σ 11 σ 12 0σ 12 σ 22 0

0 0 0




with σ 33

12λ µ λ σ 11 σ 22

. It then follows for devσ 2, where dev A :

A

13

tr AId 3 3 and A

:

∑n j

k

1 A2 jk

1 2is the Frobenius norm, that

devσ 2

µ 2

6 µ λ 2

12

σ 11 σ 22

2

2 σ 212 σ 11σ 22

The function

, listed at the end of this subsection, calculates the variable

in lines 2–9, which stores the nodal values of σ h . Here σ h is the stress calculatedby the finite element method and σ h is a smoother approximation of the discretestress σ h, which may be used for error control [8]. Line 10 determines the shearenergy density

devσ h

2

4µ

and stores it in the variable

. In line 13 the Matlabroutine

is used to draw the deformed mesh with a magnification of the

displacement, which is fixed in line 11. Grey tones are attached to the displayedmeshes which are proportional to the elastic shear energy density

devσ h

2

4µ .

The Matlab routine

is used to draw triangulations for equal types of elements. Every row of the matrix

determines one polygon where the x-, y- and z-coordinate of each cornerof this polygon is given by the corresponding entry in

,

and

. For a 2D-problem,we use for the z-coordinate the same value (zero) in all mesh points. The values

together with the options

determine the grey tone of thearea. The grey tone of each node is determined by

, that of the rest of the area islinearly interpolated.

6.2. A posteriori error indication

The averaged stress field σ σ σ h, allows an a posteriori error estimation by comparing itto the discrete (discontinuous) stress σ σ σ h. For one time step, the computable quantity

ηh σ σ σ h

σ σ σ h L2 Ω

(6.1)




defines the error estimate [8] (for pure Dirichlet conditions – Neumann boundaryconditions require modifications [9])

σ σ σ

σ σ σ h L2

Ω

C ηh

(6.2)

Here,σ σ σ h may be any piecewise affine and globally continuous stress approximation.It is emphasized that the error bound (6.2) holds only if the time-discretization erroris neglected. It is generally believed that the accumulation error (in time) cannot becaptured by averaging (in space). However the quantity ηh may monitor the localspatial approximation error and a large size of ηh indicates a larger spatial error andmotivates refinements.

Theorem 6.1 (Efficeny). Assume that the stress field σ σ σ at time t is smooth, i.e.

σ σ σ H 1 Ω; d

d sym , then ηh is an efficient error estimator up to higher order terms inthe sense that

ηh

4 σ σ σ σ σ σ h L2

Ω

h o t

(6.3)

where generically h o t O h2Dσ L2

Ω

σ σ σ

σ σ σ h L2

Ω

.

Proof. The triangle inequality provides, for all τ τ τ h which are globally continousand

-piecewise affine (like σ σ σ h), that

σ σ σ h

τ τ τ h

L2 Ω

σ σ σ

σ σ σ h

L2 Ω

σ σ σ

τ τ τ h

L2 Ω

σ σ σ

σ σ σ h

L2 Ω

h

o

t

(6.4)

for the nodal interpolation τ τ τ h I σ σ σ of σ σ σ [5]. One can prove that (with τ τ τ h globalycontinous and

-piecewise affine but arbitrary otherwise)

η M :

minτ τ τ h

σ σ σ h

τ τ τ h L2

Ω

σ σ σ h

σ σ σ h L2

Ω

4η M

This is shown for ab arbitrary constant in [6,14,16] and for the constant 4 and quitegeneral boundary conditions in [7]. The main argument is a (local) inverse estimate,

caused by equivalence of norms on finite dimensional vector spaces.The combination of the two inequalities leads to

ηh σ σ σ

σ σ σ h L2

Ω

4η M

4 σ σ σ

σ σ σ h L2

Ω

4 σ σ σ

τ τ τ h L2

Ω

for any τ τ τ h, in particular for τ τ τ h I σ σ σ .

It is emphasized once more that the reliability (6.2) is problematic and holdsonly for one time step and for large hardening and large viscosity [8].

The Matlab routine at the end of this subsection calculates at time t t 1.Lines 3–4 initialise

ω z

, the patch of the hat function at node z, which is stored in and σ σ σ h, which is stored in . Line 5 determines weights and barycen-tric coordinates by a two dimensional quadrature rule. In our algorithm we haveσ σ σ h

z

ω zσ σ σ h d x

ω z

. Lines 6–10 calculate the area of the elements (line 7), the




patch size ω z

(line 8) and

ω zσ σ σ h d x (line 9) by a loop over all elements. In line 11,

σ σ σ h is obtained by dividing

ω zσ σ σ h d x through the patch size. Lines 12–16 compute

, where for σ σ σ h four interpolation points are used on each element T .

7. NUMERICAL EXAMPLES

7.1. On a plate with circular hole

A two dimensional squared plate with a hole, Ω 2 2

2 B 0 1 is submitted

to time dependent surface forces g

t

600t

n at the top ( y

2) and the bottom( y

2), where n denotes the outer normal to ∂ Ω. The rest of the boundary is

traction free. As the problem is symmetric only a quarter of Ω is discretised. The

boundary conditions are specified in files

,

, and

:




0.1

0.2

0.3

0.4

0.5

0.6

−0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

Γ

Γ

Γ N

Γ D

N

N

Γ D

Figure 3. Deformed mesh for membrane with hole in the example of Subsection 7.1 calculated with3474 degrees of freedom. The grey tone visualises devσ h

2 4µ as described in Subsection 6.1.

This numerical example from [17] models perfect viscoplasticity with Young’smodulus E

206900, Poisson’s ratio ν

0

29, yield stress σ y

450 and vanishinginitial data for the stress tensor σ 0.

The solution is calculated with 3474 degrees of freedom in the time intervalfrom t

0 to t

0

4 by the implicit Euler method in 8 uniform steps of lengthk 1 20. The material remains elastic between t 0 and t 0 2. Figure 3 showsthe deformed mesh at t

0

2, Table 1 shows the number of iterations in Newton’s

method and the estimated error for each time step.Table 1.

Number of iterations and indicated error for membrane with

hole in the example of Subsection 7.1 for various time steps.

step iterations η

σ h

σ h 2

σ h 2

1 1 0.04322 1 0.04323 1 0.04324 5 0.0436

5 5 0.04626 5 0.05057 6 0.05658 6 0.0642




7.2. Numerical examples for Cooks’ membrane problem

In the second numerical example we solve the viscoplastic problem with isotropichardening for a two-dimensional model of a tapered panel of plexi glass described

by Ω

conv

0

0

,

48

44

,

48

60

,

0

44

. The panel is clamped at one end( x

0) and subjected to a shearing load g

0 t

on the opposite end ( x

48) withvanishing volume force f . The boundary conditions are specified in files

,

, and

:

For plexi glass we set E

2900 and ν

0

4. This example is often referred to

as the Cooks’ membrane problem and constitutes a standard test for bending domi-nated response. We calculate the solution with 8450 degrees of freedom in the timeinterval [0,0.1] with vanishing initial data for the displacement u0, the stress tensorσ 0 and the hardening parameter α 0. The time step in the implicit Euler method isk

0

01. The plastic parameters are H

1000, H 1

1 and σ y

0

1. The prob-lem becomes plastic at t

0

02. Figure 4 shows the deformed mesh at t

0

1 for2178 degrees of freedom, while Table 2 shows the number of iterations in Newton’smethod and the estimated error for each time step.

Table 2.

Number of iterations and indicated error for Cook’s mem-

brane problem in the example of Subsection 7.2 for various

time steps.

step iterations η

σ h

σ h 2

σ h 2

1 1 0.06742 7 0.06743 11 0.06754 12 0.06755 13 0.0680

6 14 0.06847 14 0.06958 14 0.07069 14 0.0723

10 14 0.0737




0.5

1

1.5

2

2.5

3

x 10−5

−10 0 10 20 30 40 50−10

0

10

20

30

40

50

60

70

Γ

Γ

Γ Γ N N

N

D

Figure 4. Deformed mesh for Cooks’ membrane problem in the example of Subsection 7.2 calculated

with 2178 degrees of freedom. The grey tone visualises

devσ

h

2

4µ

as described in Subsection 6.1.

7.3. Numerical example for axissymmetric ring

The third numerical example is taken from [4]. We solve a viscoplastic model withkinematic hardening. The geometry is a two-dimensional section of a long tubewith inner radius 1 and outer radius 2 (Fig. 6). We have no volume force, f

0 buttime depending surface forces g1

r

ϕ

t

ter and g2

r

ϕ

t

t

4er . The systemis required to keep centered at the origin and rotation is prohibited. The boundaryconditions are specified in files

,

, and

:




The exact solution is

u r ϕ t ur r t er

σ r ϕ t σ r r t er er σ ϕ eϕ eϕ

p r ϕ t Pr r t er er eϕ eϕ

(7.1)

with er

cosϕ

sinϕ

, eϕ sinϕ

cosϕ

and with a

µ

λ ,

2µ

2µ

λ ,

ur

r

t

t

2µ r

1

3

I

R

t

r

4a

µ r

r

R

t

t

2µ r

1

3 I R t

4r

4a

µ r

I r r r R t

(7.2)

σ r r t

t

r 2

8

3a I R t

1

4

1

r 2

r R t

t

r 2

8

3 a

I

R

t

1

1

r 2

2a

I

r

r

R

t

(7.3)

σ ϕ r t

∂ r σ r

∂ r (7.4)

Pr r t

0 r R t

σ y

2

a

k 1

1 R2

r 2

r

R

t

(7.5)

I r

σ y

2

a

k 1

ln r

1

2 R2

r 2 R2

(7.6)

With α

4a

3 a k 1

, the radius of the plastic boundary R t

is the positiveroot of

f R 2α ln R α

1 R2

α

2

σ yt

(7.7)




Table 3.

Number of iterations, indicated error, and exact error for axissymmetric ring in

the example of Subsection 7.3. This numerical example validates our claim that

η can be an accurate error guess.

step iterations e

σ

σ h 2

σ 2 η

σ h

σ

h 2

σ h 2 e

η1 1 0.0512 0.0510 1.00

. . . . . . . . . . . . . . .

14 1 0.0512 0.0510 1.0015 3 0.0521 0.0515 1.0116 4 0.0533 0.0529 1.0117 4 0.0549 0.0543 1.0118 4 0.0565 0.0560 1.0119 4 0.0584 0.0579 1.01

The material parameters are E

70000 for Young’s modulus, ν

0

33 for Pois-son’s ratio and the yield stress is σ y

0

2.The solution is calculated in the time interval [0,0.2] with 3202 degrees of free-

dom and vanishing initial data for the displacement u0, the stress tensor σ 0 and thehardening parameter α 0. The time step in the implicit Euler method is k

0

01. Thehardening parameter is k 1

1. The material becomes plastic at t

0

16. Table 3shows the error and the estimated error for the different time steps together withthe number of iterations in Newton’s method. Figure 5 shows the deformed mesh at

t

0

18, displacements magnified by the factor 10.

7.4. Numerical example in three dimensions

As a three-dimensional example we solve a perfect viscoplastic problem on the cube

0

1

3

0

0

5

3. The face with x

1 is Dirichlet boundary, the rest is Neumannboundary. The volume force f is zero while the surface force is g tn, if the normalvector n points in y-direction. The boundary conditions are specified in files

,

, and

:




0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

x 10−7

−0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

Figure 5. Deformed mesh for axissymmetric ring in the example of Subsection 7.3 calculated with3202 degrees of freedom. The grey tone visualises devσ h

2 4µ as described in Subsection 6.1.

2gf=0

Ω

g1

1 2

E

70000ν

0

33

σ y

0

2

k 1

1

g1 ter

g2 t

4er

Figure 6. Axissymmetric ring in the example of Subsection 7.3 with initial triangulation of a quarterof the ring endowed with symmetrie boundary conditions.




−0.5

0

0.5

1

−0.20

0.20.4

0.60.8

11.2

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

xy

z

Figure 7. Deformed mesh for 3D example in Subsection 7.4. The grey tone visualises the squaremean of the eigenvalues of the stress tensor.

The solution is calculated in the time interval [0,1] with stepsize 0.05 and van-ishing initial data for the displacement u0 and the stress tensor σ 0 and for materialparamters E

206900, ν

0

29, and yield stress σ y

450. Up to t

0

2 s thematerial is entirely elastic. The deformed mesh for 10359 degrees of freedom att

0

15 s is shown in Fig. 7.

Acknowledgments. It is our pleasure to thank Dr. Jochen Alberty and AndreGeilenkothen for discussions. The second author (R. K.) thankfully acknowledgesthe partial support by the German research foundation at the Graduiertenkolleg 357’Effiziente Algorithmen und Mehrskalenmethoden’. Parts of this work was done atthe Institute for Applied Mathematics and Numerical Analysis, Vienna Universityof Technology whose hospitability is greatfully acknowledged.

REFERENCES

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11. W. Han and B. D. Reddy, Plasticity: Mathematical Theory and Numerical Analysis. Springer,New York, 1999.

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13. D. Redfern and C. Campbell, The Matlab5 Handbook . Springer, 1998.

14. R. Rodriguez, Some remarks on Zienkiewicz-Zhu estimator. Int. J. Numer. Meth. in PDE , 10,625 – 635.

15. J. C. Simo and T. J. R. Hughes, Computational Inelasticity. Springer, 2000.

16. R. Verfurth, A Review of A Posteriori Error Estimation and Adaptive Mesh-Refinement Tech-niques. Wily-Teubner, 1996.

17. P. Wriggers et al., Benchmark Perforated Tension Strip. In: Communication in Talk at ENU- MATH Conference in Heidelberg, 1997.

8. APPENDIX

8.1. Main program in Example 1




8.2. Calculation of the stress in Example 1

8.3. Main program in Example 4

8.4. Calculation of the stress in Example 4

2002-cc kr-elastoviscoplastic fe analysis in matlab

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