Advanced Computational Mechanics - Appunti

Elastoplasticity3.1 Question 10

Elastoplasticity in 1D (linear isotropic/kinematic hardening): formulation of the governing equations (in rates) for the time-continuous problem.

Elastoplasticity is a rate-independent phenomenon which is irreversible. When we enter elastoplasticity, we can always apply the principle of virtual displacements:

"Z #L LZ XX T TT TA B σ (ε) dx = δu N q dxδue e0 0ee e e| {z } | {z }F FIe EeT TeTe

Imposing δu = δU C we get:

X XT Te T Te −δU = δU =⇒ R (U ) = F F (U ) = 0C F C FIe Ee E Ie e

The nonlinear system is solved using Newton-Raphson algorithm.

∂F ∂F ∂u ∂FX XTe TeI Ie e Ie= C = C C e∂U ∂u ∂U ∂ue ee eL L Z Z∂F dσ ∂ε dσ E A −110T TIe − −= A B dx = A B B dx = [L α (u u ) ]2 1 e −1 12∂u dε ∂u dε L0 0e e

Advanced computational mechanics 163

distinguish strain in two parts: one elastic and the other plastic ε = ε + ε .

Governing equations for the rate problem:

• e pε̇ = ε̇ + ε̇ additive decomposition of strain
• eσ̇ = E ε̇ elastic law
• p|σ| − | | →φ (σ, λ) = σ (λ) yield function, where λ is the hardening parameter λ̇ = ε̇ λ =yt pR | | ≥ε̇ dt 00
• σ = σ + Hλ linear hardening law, with H hardening modulusy 0
• pε̇ = λ̇ signσ plastic flow rule
• ≤ ≥φ 0, λ̇ 0, φ λ̇ = 0 loading/unloading conditions

Figure 3.1: Linear isotropic hardening behaviour

Determination of the tangent modulus:

∆σ ∆σ ∆σ ∆σ ∆σ ∆σ EHe pE = E = H = =⇒ ∆ε = ∆ε + ∆ε =⇒ = + =⇒ E =t te p∆ε ∆ε ∆ε

E E H E + HtAdvanced computational mechanics 173. Elastoplasticity 3.2. Question 11Determination of the plastic multiplier using the consistency condition:|σ| −φ (σ, λ) = (σ + Hλ)0∂φ ∂φ p− − −φ̇ = σ̇ + λ̇ = signσ σ̇ H λ̇ = signσ E (ε̇ ε̇ ) H λ̇ =∂σ ∂λ− − −= signσ E ε̇ signσ E λ̇signσ H λ̇ = signσ E ε̇ λ̇ (E + H)Enforcing consistency condition φ̇ = 0 we get: E signσ ε̇λ̇ = E + HGoverning equations for the rate problem (linear kinematic hardening):• e pε̇ = ε̇ + ε̇ additive decomposition of strain• eσ̇ = E ε̇ elastic law• |σ − −φ (σ, α) = α| σ yield function, where α is the back stress0•

&palpha;˙ = H ε˙ evolution of the back stress• ∂φp plastic flow ruleε˙ = λ˙ ∂σ• ≤ ≥φ 0, λ˙ 0, φ λ˙ = 0 loading/unloading conditions∂φ ∂φ − −φ˙ = σ˙ + α˙ = sign (σ α) ( σ˙ α˙)∂σ ∂αImposing consistency condition we get: p p p−σ˙ = α˙ =⇒ E (ε˙ ε˙ ) = H ε˙ =⇒ ε˙ (E + H) = E ε˙ =⇒ E− −=⇒ λ˙ sign (σ α) (E + H) = E ε˙ =⇒ λ˙ = sign (σ α) ε˙E + H3.2 Question 11Elastoplasticity in 1D (linear isotropic hardening): formulation of the (strain-driven) incre-mental problem and solution by the return mapping algorithm.In order to face elastoplasticity we need to discretize time in pseudo-time, meaning that we define thetime increment Δt between time instants t

and t . At time t we know all the quantities of ourn n+1 npninterest ε , ε , σ and λ . The evolution process in the step is regarded as strain driven since ann nnincrement of total strain ∆ε is given. The time-continuous constitutive equations are enforced at t ,n+1with time derivatives expressed as difference quotients using values at t and t :n n+1∂(·) ∆(·)=∂t ∆tt n+1Governing equations for the finite-step problem:pn+1 pn+1en+1 |σ | −− φ = (σ + Hλ )ε = ε + ε σ = E ε εn+1 n+1 n+1 n+1 n+1 0 n+1p ≥ ≤∆ε = ∆λ signσ ∆λ 0 φ 0 φ ∆λ = 0n+1 n+1 n+13.2.1 Return mapping algorithmWe initially assume the step is elastic and define the trial elastic quantities:p,trtr p tr tr tr|σ | −σ := σ + E∆ε ε := ε λ = λ φ := (σ +

Hλ ) = 0 =⇒ ∆λ E [σ + H (λ + ∆λ)] = 0n+1 n+1 0 n+1 0 nn+1

We conclude that: tr tr|σ | − (σ + Hλ ) φ0 nn+1 n+1∆λ = =E + H E + H

The other quantities at time step t are:n+1 ∆λ E pn+1tr tr tr pn tr− − σ = εσ = σ E∆λ signσ = 1 ε + ∆λ signσ λ = λ + ∆λn+1 n+1 nn+1 n+1 n+1 n+1tr ||σ n+13.3 Question 12Elastoplasticity in 3D (von Mises yield criterion with linear isotropic hardening, associa-tive flow rule): formulation of the governing equations (in rates) for the time-continuousproblem.

We define Lamè constants as: νE Eλ = µ = G =−(1 + ν) (1 2ν) 2 (1 + ν)

The stress state in a point can be written as:σ = 2µε + λε δ (3.1)ij ij kk ijMultiplying the equation by δ we get:ij σ 2µ + 3λkkσ δ =

2µε δ + λε δ δ =⇒ σ = 2µε + 3λε =⇒ = σ = ε = Kε (3.2)ij ij ij ij kk ij ij kk kk kk m kk kk3

Where σ is the mean stress and K is the bulk modulus. Multiplying equation 3.2 by δ and thenm ijsubtracting it from equation 3.1 we get:

σkk− −σ δ = 2µε + λε δ Kε δij ij ij kk ij kk ij3

Defining the deviatoric stress and strain tensors: εσkk kk−− δ e = ε δs = σij ij ij ij ij ij3 3

We can rewrite: 2µ −s = 2µe + + λ K ε δ = 2µeij ij kk ij ij3| {z }=0

The elastic energy density is:  1 1 σ ε 1 ε σ σ ε kk kk kk kk kk hh·W = σ ε = s + δ e + δ = s e + s δ + e δ + δ δ = ij ij ij ij ij ij ij ij ij

ij ij ij ij ij2 2 3 3 2 3 3 9 | {z } | {z } | {z }s =0 e =0 =3hh hh 1 σ ε 1 s σ ε 1 s s σ ε J σ ε kk hh ij kk hh ij ij kk hh 2 kk hh= s e + = s + = + = + = W + Wij ij ij dev vol2 3 2 2µ 3 2µ 2 6 2µ 6

Advanced computational mechanics 193. Elastoplasticity 3.3. Question 12√ −Where J is the second invariant of stress deviator tensor. Von Mises yield criterion reads J c = 02 2√ −3J σ (σ = uniaxial yield stress). In fact, in case of uniaxial stress σ = 0 exceptor, alternatively, 2 0 0 ijfor σ :11 2 1 1 1 4 1 1 12 222 233 2 2−s = σ s = s = σ =⇒ J = s + s + s = σ + + = σ11 11 22 33 11 2 11 11 113 3 2 2 9 9 9 3At yielding σ = σ :11 0 1 p2 −J = σ =⇒ 3J σ = 02 2 0033.3.1 von Mises associative model with linear isotropic hardeningpij• eijε = ε + ε additive decomposition of

strainsij• ehkσ = D ε elastic lawij ijhkq• 32 −φ = s s (σ + Hλ) yield functionij ij 0spij• ∂φ 3 √ ij=ε̇ λ̇ = λ̇ plastic flow rule∂σ 2 3 s sij hk hk2• ≤ ≥φ 0, λ̇ 0, φλ̇ = 0 loading/unloading condition r39 s s 2ij ijpij pij p p2 2λ̇ =ε̇ ε̇ = λ̇ =⇒ λ̇ = ε̇ ε̇ij ij2∥s∥4 2 3vMMatrix notation:    λ + 2µ λ λ µ 0 0 D 011 λ λ + 2µ λ 0 µ 0= Dε D = D = =D&sig