Maxwell model - 立命館大学

Inelastic Deformation

Shinichi Hirai

Dept. Robotics, Ritsumeikan Univ.

Shinichi Hirai (Dept. Robotics, Ritsumeikan Univ.) Inelastic Deformation 1 / 32

Agenda

1 One-dimensional Inelastic Deformation

2 Multi-dimensional Inelastic Deformation

3 Finite Element Method in Inelastic Deformation


Elastic/viscoplastic/rheologicaldeformation

elastic

rheological

plastic


Maxwell model

E c

E : Young’s modulusc : viscous modulusεela: strain at elastic elementεvis: strain at viscous elementε: strainσ: stress

ε = εela + εvis

σ = Eεela, σ = c εvis

stress-strain relationship in Maxwell model:

σ +E

cσ = E ε


Maxwell modelordinary differential equation of the first order:

σ +E

cσ = E ε

stress at time t:

σ(t) =

∫ t

0

Ee−Ec (t−t ′)ε(t ′) dt ′

In general,

σ(t) =

∫ t

0

r(t − t ′) ε(t ′) dt ′

Function r(t − t ′): relaxation function


Three-element model

E

c1

c2

E : Young’s modulusc1, c2: viscous moduliεvoigt: strain at Voigt elementεvis: strain at viscous elementε: strainσ: stress

ε = εvoigt + εvis

σ = Eεvoigt + c1εvoigt, σ = c2ε

vis

stress-strain relationship in three-element model:

σ +E

c1 + c2σ =

c1c2

c1 + c2ε +

Ec2

c1 + c2ε


Three-element modelordinary differential equation of the first order:

σ +E

c1 + c2σ =

c1c2

c1 + c2ε +

Ec2

c1 + c2ε

stress at time t:

σ(t) =

∫ t

0

r(t − t ′) ε(t ′) dt ′

where

r(t − t ′) =Ec2

c1 + c2e− E

c1+c2(t−t ′)

(1 +

c1

E

d

dt

)


Isotropic deformation modelselastic deformationspecified by a constant E :

σ = Eε

2D isotropic elastic deformationspecified by two constant λ and µ (Lame’s constants):

σ = (λIλ + µIµ)ε

Iλ =

1 1 01 1 00 0 0

, Iµ =

2 0 00 2 00 0 1


1

Isotropic deformation modelsviscoelastic deformationspecified by an operator E + c d/dt:

σ =

(E + c

d

dt

)ε

2D isotropic viscoelastic deformationspecified by two operators λ and µ:


λ = λela + λvis d

dt, µ = µela + µvis d

dt


Isotropic deformation modelsviscoplastic deformationspecified by a convolution with a relaxation function:

σ(t) =

∫ t

0

r(t − t ′) ε(t ′) dt ′

2D isotropic viscoplastic deformationspecified by two relaxation functions:

σ(t) =

∫ t

0

R(t − t ′) ε(t ′) dt ′

R(t − t ′) = rλ(t − t ′)Iλ + rµ(t − t ′)Iµ


Isotropic deformation modelsviscoplastic deformationa relaxation function:

r(t − t ′) = E exp

{−E

c(t − t ′)

}

2D isotropic viscoplastic deformationtwo relaxation functions:

rλ(t − t ′) = λela exp

{−λela

λvis(t − t ′)

}

rµ(t − t ′) = µela exp

{−µela

µvis(t − t ′)

}


Isotropic deformation modelsrheological deformationspecified by a convolution with a relaxation function:

σ(t) =

∫ t

0

r(t − t ′) ε(t ′) dt ′

2D isotropic rheological deformationspecified by two relaxation functions:

σ(t) =

∫ t

0

R(t − t ′) ε(t ′) dt ′

R(t − t ′) = r rheoλ (t − t ′)Iλ + r rheo

µ (t − t ′)IµShinichi Hirai (Dept. Robotics, Ritsumeikan Univ.) Inelastic Deformation 12 / 32

Isotropic deformation modelsrheological deformationa relaxation function:

r(t − t ′) =Ec2

c1 + c2e− E

c1+c2(t−t ′)

(1 +

c1

E

d

dt

)

2D isotropic rheological deformationtwo relaxation functions:

r rheoλ (t − t ′) =

λelaλvis2

λvis1 + λvis

2

exp

{− λela

λvis1 + λvis

2

(t − t ′)

}(1 +

λvis1

λela

d

dt

)

r rheoµ (t − t ′) =

µelaµvis2

µvis1 + µvis

2

exp

{− µela

µvis1 + µvis

2

(t − t ′)

}(1 +

µvis1

µela

d

dt

)


Nodal elastic forcesstress-strain relationship


a set of elastic forces applied to nodal points:

elastic force = −(λJλ + µJµ)uN

from stress-strain relationship to nodal force setreplacing Iλ by Jλ, Iµ by Jµ, and ε by uN in thestress-strain relationship yields the elastic force set


Nodal viscoelastic forcesstress-strain relationship

σ = (λelaIλ + µelaIµ)ε + (λvisIλ + µvisIµ)ε

replacing Iλ by Jλ, Iµ by Jµ, and ε by uN in thestress-strain relationship yields a viscoelastic force set

⇓a set of viscoelastic forces applied to nodal points:

viscoelastic force = − Jλ(λelauN + λvisuN)

− Jµ(µelauN + µvisuN)


Nodal viscoplastic forcesstress-strain relationship

σ(t) = Iλ

∫ t

0

rλ(t − t ′) ε(t ′)dt ′ + Iµ

∫ t

0

rµ(t − t ′) ε(t ′)dt ′

replacing Iλ by Jλ, Iµ by Jµ, and ε by uN in thestress-strain relationship yields a viscoplastic force set

⇓a set of viscoplastic forces applied to nodal points

viscoplastic force = − Jλ

∫ t

0

rλ(t − t ′) uN(t ′) dt ′

− Jµ

∫ t

0

rµ(t − t ′) uN(t ′) dt ′


2

Nodal viscoplastic forcesintroduce

fλ =

∫ t

0

λela exp

{−λela

λvis(t − t ′)

}uN(t ′) dt ′

fµ =

∫ t

0

µela exp

{−µela

µvis(t − t ′)

}uN(t ′) dt ′

Vectors fλ and fµ have dimension of force/length

viscoplastic force = −Jλfλ − Jµfµ

fλ = −λela

λvisfλ + λelauN = −λela

λvisfλ + λelavN

fµ = −µela

µvisfµ + µelauN = −µela

µvisfµ + µelavN


Nodal viscoplastic forces

ordinary differential equation of the first order:

σ +E

cσ = E ε

⇓

fλ = −λela

λvisfλ + λelauN = −λela

λvisfλ + λelavN

fµ = −µela

µvisfµ + µelauN = −µela

µvisfµ + µelavN


Nodal rheological forcesstress-strain relationship

σ(t) = Iλ

∫ t

0

r rheoλ (t−t ′)ε(t ′)dt ′+Iµ

∫ t

0

r rheoµ (t−t ′)ε(t ′)dt ′

replacing Iλ by Jλ, Iµ by Jµ, and ε by uN

⇓

rheological force = − Jλ

∫ t

0

r rheoλ (t − t ′) uN(t ′) dt ′

− Jµ

∫ t

0

r rheoµ (t − t ′) uN(t ′) dt ′


Nodal rheological forcesintroduce

fλ =

∫ t

0

λelaλvis2

λvis1 + λvis

2

e− λela

λvis1

+λvis2

(t−t ′)(uN +

λvis1

λelauN

)(t ′) dt ′

fµ =

∫ t

0

µelaµvis2

µvis1 + µvis

2

e− µela

µvis1

+µvis2

(t−t ′)(uN +

µvis1

µelauN

)(t ′) dt ′

Vectors fλ and fµ have dimension of force/length

rheological force = −Jλfλ − Jµfµ

fλ = − λela

λvis1 + λvis

2

fλ +λelaλvis

2

λvis1 + λvis

2

vN +λvis

1 λvis2

λvis1 + λvis

2

vN

fµ = − µela

µvis1 + µvis

2

fµ +µelaµvis

2

µvis1 + µvis

2

vN +µvis

1 µvis2

µvis1 + µvis

2

vN


Nodal rheological forces

ordinary differential equation of the first order:

σ +E

c1 + c2σ =

c1c2

c1 + c2ε +

Ec2

c1 + c2ε

⇓

fλ = − λela

λvis1 + λvis

2

fλ +λelaλvis

2

λvis1 + λvis

2

vN +λvis

1 λvis2

λvis1 + λvis

2

vN

fµ = − µela

µvis1 + µvis

2

fµ +µelaµvis

2

µvis1 + µvis

2

vN +µvis

1 µvis2

µvis1 + µvis

2

vN


FE formulation of viscoplastic deformation

a set of equations of elastic deformation:

−KuN + fext + AλA − MuN = 0

elastic force

⇓a set of equations of viscoplastic deformation:

−Jλfλ − Jµfµ + fext + AλA − MuN = 0

viscoplastic force


FE formulation of viscoplastic deformation

uN = vN[M −A

−AT

] [vN

λA

]=

[−Jλfλ − Jµfµ + fext

AT(2αvN + α2uN)

]

fλ = −λela

λvisfλ + λelavN

fµ = −µela

µvisfµ + µelavN


FE formulation of rheological deformation

a set of equations of elastic deformation:

−KuN + fext + AλA − MuN = 0

elastic force

⇓a set of equations of rheological deformation:

−Jλfλ − Jµfµ + fext + AλA − MuN = 0

rheological force


3

FE formulation of rheological deformation

uN = vN[M −A

−AT

] [vN

λA

]=

[−Jλfλ − Jµfµ + fext

AT(2αvN + α2uN)

]

fλ = − λela

λvis1 + λvis

2

fλ +λelaλvis

2

λvis1 + λvis

2

vN +λvis

1 λvis2

λvis1 + λvis

2

vN

fµ = − µela

µvis1 + µvis

2

fµ +µelaµvis

2

µvis1 + µvis

2

vN +µvis

1 µvis2

µvis1 + µvis

2

vN


Example (Sample Program)

simulation movie


Example (Sample Program)

simulation movie


Summary

one-dimensional inelastic deformationMaxwell model for viscoplastic deformation

three-element model for rheological deformation

2D/3D inelastic deformationisotropic deformation models

formulating nodel force sets

finite element formulation


Simulating Inelastic DeformationReport #7 due date : Jan. 29 (Fri)Simulate the deformation of a rectangular inelasticobject shown in the figure.

P2P3 is fixed to the floor.P1 and P4 may slide on the floor.[ 0, tpush ] push P14P15 downward[ tpush, thold ] keep P14P15

[ thold , tend ] release P14P15

Use appropriate values of geometrical and physicalparameters of the object.


Simulating Inelastic Deformation

P5

P3

P6

P2P1

P7

P4

P9 P10 P11

P13 P14 P15

P8

P12

P16


AppendixLet us solve the following ordinary differential equation:

x + ax = u(t)

Assuming x(0) = 0, Laplace transform of the aboveequation yields

sX − aX = U

Thus, we have

X (s) =1

s − aU ,

implying that x(t) is the convolution of eat and u(t).Consequently,

x(t) =

∫ t

0

ea(t−τ)u(τ) dτ


AppendixDifferentiating

x(t) =

∫ t

0

ea(t−τ)u(τ) dτ = eat

∫ t

0

e−aτu(τ) dτ

with respect to t, we have

x = aeat

∫ t

0

e−aτu(τ) dτ + eat · e−atu(t)

= a

∫ t

0

ea(t−τ)u(τ) dτ + u(t)

= ax + u(t),

which coincides with the ordinary differential equation.Shinichi Hirai (Dept. Robotics, Ritsumeikan Univ.) Inelastic Deformation 32 / 32

4

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Maxwell model - 立命館大学