Hi, I am trying to caculate for a given stress the corresponding strain in the Von Mises plasticity example. I know that this results in a non-linear equation.
Not notice I have the following equations:
σᵗ = C ⊡ (ϵ - ϵᵖ) J₂ = 0.5 * dev(σᵗ) ⊡ dev(σᵗ) # second invariant σᵗₑ = sqrt(3.0*J₂) # effective trial-stress (von Mises stress) σʸ = σ₀ + H * k φᵗ = σᵗₑ - σʸ # Trial-value of the yield surface h = H + 3*G μ = φᵗ / h # plastic multiplier c1 = 1 - 3G * μ / σᵗₑ s = c1 * dev(σᵗ) # updated deviatoric stress σ = s + vol(σᵗ) # updated stress
The only thing which I have not given is \epsilon. Now I reordered and substitute everything:
σ = (1 - 3G*((sqrt(1.5*dev(C ⊡ (ϵ - ϵᵖ)) ⊡ dev(C ⊡ (ϵ - ϵᵖ))) - (σ₀ + H * k) ) / (H + 3*G) )/ (sqrt(1.5* dev(C ⊡ (ϵ - ϵᵖ)) ⊡ dev(C ⊡ (ϵ - ϵᵖ))))) * dev(C ⊡ (ϵ - ϵᵖ)) + vol(C ⊡ (ϵ - ϵᵖ))
Is there a “nice” way to solve this equation somehow?