fem for elastic-plastic problemsjpamin/dyd/cmce/cmce_lecture7_jp.pdf · ideal huber-mises-hencky...

FEM for elastic-plastic problems

Jerzy Pamin

e-mail: JPamin@L5.pk.edu.pl

With thanks to:

P. Mika, A. Winnicki, A. WosatkoTNO DIANA http://www.tnodiana.comFEAP http://www.ce.berkeley.edu/feap

Comp.Meth.Civ.Eng., II cycle

Lecture scope

Physical nonlinearity

Plastic flow theory

Computational plasticity

Simulation of plastic deformations

Final remarks

References[1] R. de Borst and L.J. Sluys. Computational Methods in Nonlinear SolidMechanics. Lecture notes, Delft University of Technology, 1999.

[2] G. Rakowski, Z. Kacprzyk. Metoda elementow skończonych w mechanicekostrukcji. Oficyna Wyd. PW, Warszawa, 2005.

[3] M. Jirasek and Z.P. Bazant. Inelastic Analysis of Structures. J. Wiley &Sons, Chichester, 2002.

Comp.Meth.Civ.Eng., II cycle

Incremental-iterative analysisNonlinear problem:fext applied in incrementst → t + ∆t → σt+∆t = σt + ∆σ

Equilibrium at time t + ∆t:ne∑e=1

AeT∫V eBTσt+∆t dV = ft+∆t

ext

ne∑e=1

AeT∫V eBT∆σ dV = ft+∆t

ext − ftint

where: ftint =∑nee=1A

eT∫V e B

Tσt dV

Linearization of the left-hand side at time t

∆σ = ∆σ(∆ε(∆u))

Equation set for an increment:

K∆d = ft+∆text − ftint

Comp.Meth.Civ.Eng., II cycle

Physical nonlinearity

K∆d = ft+∆text − ftint

Linearization of LHS at time t:

∆σ = ∆σ(∆ε(∆u))

∆σ =(∂σ∂ε

)t ( ∂ε∂u

)t∆u

D = ∂σ∂ε , L = ∂ε

∂u

Discretization: ∆u = N∆de

Linear geometrical relations → Matrix of discrete kinematic relationsB = LN independent of displacements

Tangent stiffness matrix

K =ne∑e=1

AeT∫V eBTDB dV Ae

Comp.Meth.Civ.Eng., II cycle

Plastic yielding of material

A

CB

displacement

force

P

A

+

-σy

σy

σy

σy

σy

σy

+

- -

+

CB

elastic material

equivalent plastic strain distribution

plastifiedelastic material

microscopic level

crystal shear dislocationlattice slip

Comp.Meth.Civ.Eng., II cycle

Plastic flow theory [1,3]

Load-carrying capacity of a material is not infinite, during deformationirreversible strains occur

Notions of plasticity theoryI Yield function f (σ) = 0

- determines the limit of elastic responseI Plastic flow rule εp = λm

- determines the rate of plastic strainλ - plastic multiplierm - direction of plastic flow(usually associated with the yield functionmT = nT = ∂f

∂σ )I Plastic hardening f (σ −α, κ) = 0

kinematic (κ = 0) or isotropic (α = 0)I Loading/unloading conditions:

f ¬ 0, λ ­ 0, λf = 0 (unloading is elastic)

Comp.Meth.Civ.Eng., II cycle

Plastic flow theory

Response is history-dependent, constitutive relations written in rates

Plastic flow whenf = 0 and f = 0(plastic consistency condition)

Additive decompositionε = εe + εp

Bijective mappingσ = Deεe

Introduce flow ruleσ = De(ε− λm)

Consistency

f = ∂f∂σ σ + ∂f

∂κ κ

Hardening modulus

h = − 1λ∂f∂κ κ

Substitute σ intonTσ − hλ = 0

Determine plastic multiplier

λ = nTDe εh+nTDem

Constitutive equation

σ =[De − D

emnTDe

h+nTDem

]ε

Tangent operator

Dep = De − DemnTDe

h+nTDem

Time integration necessary at the point level

Comp.Meth.Civ.Eng., II cycle

Huber-Mises-Hencky plasticity

Most frequently used is the Huber-Mises-Hencky (HMH) plastic flowtheory, based on a scalar measure of distortional energy Jσ2

I Yield functione.g. with isotropic hardeningf (σ, κ) =

√3Jσ2 − σ(κ) = 0

κ - plastic strain measure(κ = 1

σσTεp = λ)

I Associated flow ruleεp = λ ∂f∂σ

I Hardening rulee.g. linearσ(κ) = σy + hκh - hardening modulus

Comp.Meth.Civ.Eng., II cycle

Response: force-displacement diagrams

Ideal plasticity Hardening plasticity

Comp.Meth.Civ.Eng., II cycle

Plastic flow theory

Yield functions for metals:Coulomb-Tresca-Guest i Huber-Mises-Hencky (HMH)

Insensitive to hydrostatic pressure p = 13 Iσ1

Comp.Meth.Civ.Eng., II cycle

Plastic flow theory

Yield functions for soil:Mohra-Coulomb i Burzyński-Drucker-Prager (BDP)

Sensitive to hydrostatic pressure

Comp.Meth.Civ.Eng., II cycle

‘Yield’ functions for concrete (plane stress)

Kupfer’s experiment

Rankine ‘yield’ function: f (σ, κ) = σ1 − σ(κ) = 0

Inelastic strain measure κ = |εp1 |

Comp.Meth.Civ.Eng., II cycle

Computational plasticity

Return mapping algorithm→ backward Euler algorithm (unconditionally stable)

1) Compute elastic predictorσtr = σt +De∆ε

2) Check f (σtr , κt) > 0 ?If not then elasticcompute σ = σtrIf yes then plasticcompute plastic correctorσ = σtr −∆λDem(σ)f (σ, κ) = 0(set of 7 nonlinear equations for σ,∆λ)Determine κ = κt + ∆κ(∆λ)

σ

σtr

σt

f = 0

Iterative corrections still necessary unless radial return is performed.

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Elasticity, plane strain

Deformation, vertical stress σyy and stress invariant Jσ2

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Elasticity, mesh sensitivity of stresses

Stress σyy for coarse and fine meshes

Stress under the force goes to infinity (results depend on mesh density) -solution at odds with physics

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Ideal Huber-Mises-Hencky plasticity

Final deformation and stress σyy

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Ideal Huber-Mises-Hencky plasticity

Final strain εyy and strain invariant Jε2

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Ideal Huber-Mises-Hencky plasticity

0 0.2 0.4 0.6 0.8 1

Displacement

0

200

400

600

800

Forc

e

0 0.2 0.4 0.6 0.8 1

Displacement

0

200

400

600

800

Forc

e

This is correct!

For four-noded element load-displacement diagram exhibits artificialhardening due to so-called volumetric locking, since HMH flow theorycontains kinematic constraint - isochoric plastic behaviour which cannotbe reproduced by FEM model.

Eight-noded element does not involve locking.

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

Elasticity, plane strain, eight-noded elements

Deformation, vertical stress σyy and stress invariant Jσ2

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

HMH plasticity

Final deformation and stress σyy

Comp.Meth.Civ.Eng., II cycle

Brazilian split test

HMH plasticity

Final strain εyy and invariant Jε2

Comp.Meth.Civ.Eng., II cycle

Burzyński-Drucker-Prager plasticity

I Yield function with isotropic hardeningf (σ, κ) = q + α p − βcp(κ) = 0q =√

3J2 - deviatoric stress measurep = 1

3 I1 - hydrostatic pressureα = 6 sinϕ

3−sinϕ , β = 6 cosϕ3−sinϕ

ϕ - friction anglecp(κ) - cohesion

I Plastic potential f p = q + α pα = 6 sinψ

3−sinψψ - dilatancy angleNonassociated flow ruleεp = λm, m = ∂f p

∂σ

I Plastic strain measureκ = ηλ, η = (1 + 2

9 α2)

12

I Cohesion hardening modulush(κ) = ηβ

∂cp∂κ

q

p

HMH

BDP

ϕ

βcp

Huber-Mises-Hencky yieldfunction is retrieved forsinϕ = sinψ = 0

Comp.Meth.Civ.Eng., II cycle

Slope stability simulation

Gradient-enhanced BDP plasticity

Evolution of plastic strain measure

Comp.Meth.Civ.Eng., II cycle

Final remarks

1. In design one usually accepts calculation of stresses (internal forces)based on linear elasticity combined with limit state analysisconsidering plasticity or cracking.

2. In nonlinear computations one estimates the load multiplier for whichdamage/failure/buckling of a structure occurs. The multiplier can beinterpreted as a global safety coefficient, hence the computationsshould be based on medium values of loading and strength.

Comp.Meth.Civ.Eng., II cycle