Question: Find displacement field ξ(x,t) and stress field σ(x,t) complying with the system of equations

Transcription

1 4 An example of solution to a poblem of an elasto-plastic stuctue Small petubations : ε = 2 ( ξ + T ξ, Equilibium : div ( σ + F = 0, Constitutive laws: f (σ 0, ε = S : σ + ε p, ε p = λ f σ (σ, λ f (σ = 0. Bounday Conditions: T = σ.n = T d su S T, ξ = ξ d su S ξ. Loading path : F(x,t,T d (x,t,ξ d (x,t σ(x,t, ξ(x,t. Question: Find displacement field ξ(x,t and stess field σ(x,t complying with the system of equations

2 6 Tosion of an elastoplastic cylindical ba (These Slides ae adapted fom a classoom by P. Suquet Elastic pefectly plastic mateial, homogeneous, isotopic and obeying to von Mises citeion. z=h e_ z α Bounday Conditions: at z = 0 : ξ = ξ θ = 0, T z = 0, α z h at z = h : ξ = 0, ξ θ = α, T z = 0, at = R : T = 0. z=0 Zeo body foces, initial state = natual state without stess

3 7 Génealized loading paamete : P e = T.ξ ds = Qq, q = α tosion angle, Q z=h = M = 2π R 0 σ θz 2 d tosion couple. Elastic Solution: ξ kinematically admissible (K.A., σ statically admissible (S.A., and linked by the elastic law ξ = ξ θ e θ, σ = τ( ( e z e θ + e θ e z, τ max τ ξ θ = αz h, µα τ( = h, M = µi h α, M I = πr4 2. µi h α R

4 8 Fist plastification Monotonous loading α(t. Small α : elastic egime τ( = µα h. τ τ max von Mises Citeion : σ eq σ 0, σ = τ( ( e z e θ + e θ e z, R σ eq = ( 3 /2 2 s i js i j σ eq = 3τ, von Mises : τ( k, k = σ 0 3. The citeion is fist eached fo: = R, α = α 0 = kh µr, M = M 0 = πkr3 2, k = σ 0 3. Fo α α 0, M M 0 all the ba is in the elastic egime

5 9 Elasto-plastic egime Constuction of a solution, stating fom physically intuitive assumptions. H : Stess state = simple shea in the plane (e θ,e z : σ = τ( ( e z e θ + e θ e z, σ S.A. τ(, τ = k a R. H2 : Plastic Zone = a R pogessing fom the exteio to the cente of the ba. plastic zone τ a k R elastic coe a R Shea τ in elastic zone?

6 0 Elastic Zone Equations to be fulfiled in elastic zone: σ = C : ε, divσ = 0, Bounday conditions has not change at z = 0 and z = h, σ.e =a = σ.e =a +. Foces applied by the plastic zone on the elastic zone? σ =a +.e = k ( e z e θ + e θ e z.e = 0. ==> Tosion of an elastic ba whose exteio adius is a. τ( = µα h. Matching of elastic zone / plastic zone τ(a = k, a = kh αµ = α 0 α R. k τ zone plastique a R a 0 as α +. Elastic eseve. elastic coe

7 Plastic defomation plastic zone τ a k R elastic coe a R Nomality ule ε p = 3 2 λ s σ eq, s = τ( ( e z e θ + e θ e z, ε p = ε p ( e θ e z + e z e θ, ε p = 0 0 a, = 3 2 λ τ σ eq = 3 2 λ( a, ε = ε el + ε p = ṡ 2µ + ε p = ε θz ( eθ e z + e z e θ, ε θz = α 2h 0 a, ε θz = 3 2 λ( a R.

8 2 Constuction of a K.A. displacement field Gometical and kinematical compatibility: by seaching ξ = ξ θ e θ ε θz = 2 ξ θ z h 0 2 ε θz dz = ξ θ (h, = α, λ = 0 0 a, λ = α 3h a R. λ is then positive tas long as α is positive which coesponds to a loading condition. By integation ξ θ = αz/h. Conclusion: a solution (σ,ξ to the poblem has been built.

9 3 Relation (Q,q, Couple-Twist angle. R M = 2π τ( 0 2 d ( a µα 3 = 2π d + 0 h ( 4 = M 0 3 ( α α R a k 2 d Remak : when α +, M M u = 4 3 M 0. M u M 0 M α 0 α The cuve (q,q = (α,m looks like a hadening cuve, even the mateial is pefectly plastic

10 3 Fomula in cylindical coodinates Vectos U = u. e + v. e θ + w. e z. scala function f(, θ, z. stess field (M. Difféents opeatos in cylindical fame : gad( u = u, (u,θ v u,z v, (v,θ + u v,z w, w,θ w,z div( u = (u, + v,θ + w,z ot u = w,θ v,z u,z w, (v, u,θ u 2 v u = 2,θ v + 2 u 2,θ w gad(f = f, f,θ f,z u 2 v 2 f = (f,, + 2f,θθ + f,zz div = σ, + σ θ,θ + σ z,z + σ σ θθ σ θ, + σ θθ,θ + σ θz,z + 2σ θ σ z, + σ θz,θ + σ zz,z + σz (