F8. Prototype models for plastic response; integration of plastic flow; structural analys of beam problem.

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1 F8. Prototype models for plastic response; integration of plastic flow; structural analys of beam problem. Prototype model: Perfect plasticity Consider model beavior: Figure 1 stablis reological model: spring combined wit slider! Figure 2 Free energy formulated in internal plastic strain ε p fl κ =ε p ψ@ε, ε p ; θd = 1 2 Hε εp αθl 2 σ= ψ ε = Hε εp αθl, K = ψ ε p =σ Inelastic deformation in slider spontaneously as stress exceeds yield stress fl rate independent beavior Plastic micro stress K same as continuum stress σ, cf. Maxwell model for linear visco-elasticity! stablis reduced dissipation

2 2 Notes_F8.nb = K ε p 0OK proper formulation of ε p =? stablis evolution of plastic deformation in deformation in slider: Introduce: yield function: φ@σd =» σ» σ y, σ y = yield stress Yield criterion: φ@σd =» σ» σ y = 0 Figure 3 Postulate evolution of deformation in slider as il ε p = 0ifφ<0 Helastic loadingl iil ε p =λ sign@σd wit λ 0 if φ=0 Hplastic loadingl Reconsider dissipation Figure 4 il = K ε p def = 0OK! iil = K ε p =σsign@σd λ =» σ» λ 0 OK since λ 0 Plastic multiplier λ 0 φ =»σ» σ = sign@σd=gradient of yield function fl ε p =λ φ wen φ@σd = 0 Plastic evolution ε p =λ φ controlled by loading (or Kun-Tucker) conditions φ 0, λ 0,φλ = 0 ü Consequences of loading conditions - Continuum tangent relation Assume α θ=0 fl σ=hε ε p L σ = Hε ε p L Distinguis 1L φ@σd < 0 σ = ε = σ e 2L φ@σd = 0 σ = Hε ε p L = σ e Conclusion: lastic loading HL λ sign@σd Plastic loading HPL

3 Notes_F8.nb 3 σ = σ e λ sign@σd φ 0, λ 0,φλ = 0 = Consider tangent stiffness beavior! Assume φ@σd = 0, and increase loading so tat φ@σ +σ dtd =φ@σd dt dt 0 Beavior controlled by linearized loading conditions, i.e. σ = σ e λ sign@σd = φ 0, λ 0,φ λ = 0 In tis case, we ave φ = sign@σd σ = sign@σd Hσ e σ = σ e λ sign@σd =...= 0 HPL Plastic loading (P) for φ@σd = 0 occurs if λ > 0 for ε > 0 if sign@σd = 1 λ > 0 for ε < 0 if sign@σd = 1 Stress state remains on te yield surface! λ sign@σdl = 0 λ = sign@σd σ e = sign@σd ε Prototype model: Isotropic ardening plasticity Consider model beavior Figure 5 stablis reological model: spring combined wit slider+spring!

4 4 Notes_F8.nb Figure 6 Free energy formulated in internal plastic strain ε p and internal ardening variable κ fl ψ@ε, ε p, κ; θd = 1 2 Hε εp αθl H κ2 σ= ψ ε = Hε εp αθl Reduced dissipation wit =σ p ε p + K κ =σε p + K κ 0OK proper formulation of ε p =? σ p =σ= ψ ε p = Hε εp αθl, K = ψ κ = H κ Inelastic deformation in slider spontaneously as stress exceeds yield stress fl rate independent beavior Plastic micro-stress K σ! xtra contribution K κ in dissipation due to micro stress K in slider! Caracterization fl H > 0 Hardening material beavior H = 0 Perfectly plastic material beavior H < 0 softening material beavior Hassociated wit fracturel Formulate evolution of plastic deformation in slider: Introduce: yield function: φ@σ, KD :=» σ» K σ y wit σ y = initial yield stress Postulate evolution ε p for lastic loading () and plastic loading (P): HL ε p = 0ifφ@σ, KD < 0 HPL ε p =λ φ =λ sign@σd κ =λ φ K = λ = wit λ 0 for φ=0 stablis dissipation il =σ p ε p + K κ =σε p H κ κ def = 0OK!

5 Notes_F8.nb 5 iil = Hσ sign@σd KL λ = H»σ» KL λ =σ y λ 0 OK since λ 0 Plastic multiplier λ 0 φ =»σ» σ = sign@σd=gradient of yield function fl ε p =λ φ wen φ@σd = 0 Plastic evolution ε p =λ φ φ 0, λ 0,φλ = 0 controlled by loading (or Kun-Tucker) conditions Prototype model: Non-linear creep (Norton's creep law) Solution procedures - Structural analysis à Numerical integration flow rules - Backward uler Metod ü Isotropic ardening plasticity model Consider time interval I td subdivided into a finite number of timesteps mtim, i.e I = 80, t 1,..., t n, t n+1 < n=1,mtim Apply B finite differences to rate formulated state equation ε σ = +ε p B ε = σ + εp ε = σ σ n + ε p σ=σ n + ε ε p =σ e ε p wit σ =σ σ n and te elastic "trial" stress σ e :=σ n + ε. lastic loading () fl ε p := 0 fl σ =σ e if φ@σ e,k n D 0. Plastic loading (P) fl Apply B finite differences to elastic-plastic "flow rule": ε p φ =λ =λ sign@σd B ε p = λsign@σd κ φ =λ K = λ B κ = λ K = K n H κ = K n + H λ xplicit expression for integrated stress: σ =σ e λ sign@σd fl sign@σd = sign@σ e D» σ» =» σ e» λ φ@»σ», KD =» σ e» λ HK n + H λl σ y =» σ e» λ H λ σ y K n = 0 λ =» σe» σ y K n + H fl Integrated stress = φ el + H wit φ el = φ@»σ e», K n D σ=σ e ε p =σ e λ sign@σ e D =σ e σ=σ e» σ e» σ y K n + H» σ e» σ y K n + H σ e» σ e» =σe H σy+kn»σ e» + H sign@σ e D (1) (2)

6 6 Notes_F8.nb Tangent relationsip dσ dε =? wit dσ dε = d»σ e» dε + H sign@σe D = d»σ e» dε + H sign@σe D = + H = H + H (3) d» σ e» dε = sign@σ e D (4) ü Linear creep law ü Non-linear Norton creep law à Analysis of beam cross section - solution strategy for isotropic ardening plasticity ü General Consider beam cross section Figure 9 ü Problem Apply bending moment about y-axis and normal force along x-axis of beam. Compute (for given bending moment and axial force) uniaxial stress distribution along cross section eigt ü Analysis Kinematics: Assume uler-beroulli teory fl Plane beam sections remain plane after deformation:

7 Notes_F8.nb 7 Figure 10 Described via 1) curvature κ =κ@xd = w and 2) axial elongation ξ =ξ@xd = u Local uniaxial strain of "beam fiber" at given value of x along considered beam: ε@zd =κz +ξ=h1, zl J ξ κ N Constitutive law: Specify te relation σ = f Hε@z, tdl =?; Assume elasticity σ=ε@zd Stress resultants (section forces): Specify te definition of te bending moment and te axial force! fl N = σ A A M = σ z A A Out-of-balance force stablis out-of-balance force vector defined as g@ pd = i N Ÿ A σ A y κ j k M Ÿ A σ z A z := 0 wit p = J ξ N { Note! From kinematics we ave uniaxial fiber strain ε@zd = H1, zl J ξ ξ N ε=h1, zl J κ κ N Consider as generic case: isotropic ardening plasticity σ=σ e» σ e» σ y K n sign@σ + H e D σ e relates to current time step t = n t + t ; In context of beam cross section we obtain: σ e =σ n + ε wit ε@zd := H ξ + κ z L ü Main solution procedure - Non-linear solution strategy 1) For eac loadstep at time t = n t + t wit 0 t = 0

8 8 Notes_F8.nb 2) stablis p i D 3) Solve for g@ pd := 0; consider Newton procedure D i ξ i+1 = g i wit D = g p 4) Update improvement of solution = tangent stiffness matrix p i+1 = p i + ξ i+1 5) Ceck convergence, Compute error e =» g i» e tol go to step 2 e > tol got step 6 6) Set incremental soltion p = p i,σ = n σ; goto step 1. 7) nd of procedure ü Spatial integration - out-of-balance force Problem: determine generalized stresses fl consider "layerwise" numerical integration: Subdivide integration interval I D ( = is te cross section eigt) into finite segments 8 z k = ê Nint< k=1,nint Carry out integration as: -Normal force, integration interval subdivided into layers Nint N = σ b z = k=1 σ k b k z k wit σ k = Jσ e -Bending moment, integration interval subdivided into layers» σ e» σ y K n + H sign@σ e DN k Nint M = σ zb z = k=1 σ k b k z k z k ü Tangent stiffness matrix Consider linearization: we ave dg = D d p wit D = g p dg = J dn dm N = J 1 N dσ b x = 9dσ = z D = a J 1 z z z 2 N b x were (sow as omework) a = ε = tangent stiffness moduli =...= H ε d ε = a H1, zl J d ξ d κ N= + H

9 Notes_F8.nb 9 ü xample: lastic-plastic response of symmetric beam cross section Consider te symmetric beam cross section in te figure below. Figure 11 Te material is assumed to be isotropiic elastic-plastic wit linear ardening, wereby te integrated stress (according to te discussion above) becomes σ=σ e ε p =σ e λ sign@σ e D = (7) 9 λ = φ el + H =» σ =σe e» σ y K n sign@σ + H e D In te present case we assume total deformation teory wic means tat te (total) strain is applied in one single step, i.e. ε ε, λ λ = σ= ε» σe» σ y + H + H σ = ε = 1 + H H» ε» σ yl sign@εd + H Te absolute value of te uniaxial stress is tus given by:» ε»» σ» = 9 +H HH» ε» +σ yl H» ε» σ yl H» ε» σ yl = HH» ε» +σ yl + H HL HPL In tis case te strain of te cross section is given by ε@zd =κz sign@εd sign@zd for κ>0 Te elastic range of te cross section is given by σ y =ε y@z y D =κz y =κ el 2 z y 2 I κ el κ Hence, te integrated stress of te cross section is given by wit σ=» σ» sign@εd M (8) (9) (10) (11) σ=9 κ z +H HH κ z +σ z y Te bending moment is defined by»z» L HL HPL wit σ=» σ» sign@εd (12)

10 10 Notes_F8.nb 2 z y M = 2 σ zb z = 2 ba σ z z Tis is evaluated as: z y 2 ba κ z 2 z H z y 2 σ z z = z y 2 JH κ z 2 +σ y z 2» z» N z (13) M@κD = b 3 12 κ if κ κ el = 2 σ y M@κD = H + H wic can be plotted as: b 3 12 J 1 + m H I κ el κ M2 + 3 σ y H 1 κ J1 I κ el κ M2 NN κ if κ>κ el M_@κD Responses wit H 20 and H κ Figure 12

a) Give an example of a case when an (s,s) policy is not the same as an (R,Q) policy. (2p)

roblem a) Give an example of a case wen an (s,s) policy is not te same as an (R,) policy. (p) b) Consider exponential smooting wit te smooting constant α and moving average over N periods. Ten, tese two