Table 2

Stress-updating procedure – small strain

(a) Initialise
k=0, αdn+1(0)=αdn, σn+1(0)=σn+1trial
(b) Check the yield condition and evaluate residuals
fn+1(k)=f(σn+1(k),αdn+1(k))rn+1(k)={σn+1(k)(1αdn+1)σn+1trialαdn+1(k)αdn}+Δt{γ˙(1αdn+1(k))Deσfn+1(k)α˙d}If rn+1(k)<ϵr and fn+1(k)<ϵf, thenset ()n+1=()n+1(k) and exitElse
(c) Compute the algorithmic modulus
An+1(k)=[An+1σσAn+1σαdAn+1αdσAn+1αdαd]
where
An+1σσ=I+[Δγσσ2fn+1(k)+σfn+1(k){σΔγn+1(k)}T](1αdn+1(k))DeAn+1σαd=[Δγσαd2fn+1(k)+σfn+1(k)αdΔγn+1(k)](1αdn+1(k))De+σn+1trialΔγn+1(k)σfn+1(k)An+1αdσ={σΔαdn+1(k)}TAn+1αdαd=1αdΔαdn+1(k)
(d) Evaluate increment of stresses and internal variable
{Δσn+1(k+1)Δαdn+1(k+1)}=(An+1(k))1:rn+1(k)
(e) Update stresses and internal variable
{σn+1(k+1)αdn+1(k+1)}={σn+1(k)αdn+1(k)}+{Δσn+1(k+1)Δαdn+1(k+1)}Set k=k+1 and go to (b)

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