FEM for elastic-plastic problems
Jerzy Pamin
e-mail: [email protected]
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