Skip to main content

piecewise - Solving a discontinuous differential-algebraic equation system for plasticity behaviour


I need to solve a discontinuous equation which is typical in theory of plasticity. For a simple case I get the following equation system (reformulated for numerical implementation):


$$\begin{align*} s(t) &= \frac{\sigma(t)}{C_1} + s_{ep}\\ s_{ep}'(t) &= \begin{cases}\frac{C_1}{C_1+C_2}s'(t) & \text{for } |\sigma(t)-C_2 s_{ep}(t)| \ge \sigma_{gr} \land \sigma(t)s'(t)>0\\0 & \text{otherwise}\end{cases} \end{align*}$$


with "zero" initial conditions. I'd like to get the solution for $s(t)$ for given parameters $C_1$, $C_2$, $\sigma_{gr}$ and a known function $\sigma(t)$. I assumed: $\sigma(t) = 40000\sin(0.02t)$, $C_1=80000$, $C_2 = 20000$, $\sigma_{gr} = 15000$. This should give a hysteresis loop on a plane $\sigma(t)-s(t)$.


So in Mathematica I tried to use automatic discontinuity handling by defining the second equation using a Piecewise function:



σ[t_] := 40000*Sin[0.02*t];
eq1 = s[t] == σ[t]/C1 + sep[t];
eq2 = sep'[t] == Piecewise[{{C1/(C1 + C2)*s'[t], (σ[t]*s'[t] > 0) && ((σ[t] - C2*sep[t] >=  σgr) || (σ[t] - C2*sep[t]<=-σgr))}}, 0];
eqSys := {eq1, eq2, s[0] == 0, sep[0] == 0};
ndsolve=NDSolve[eqSys, {s[t], sep[t]}, {t, 0, 1000}]
disp[t_] := Evaluate[s[t] /. ndsolve];
sTab = Table[disp[t][[1]], {t, 0, 1000, 1}];
σTab = Table[σ[t], {t, 0, 1000, 1}];
ListPlot[Transpose[{sTab, σTab}], PlotRange -> All, GridLines -> Automatic]


Unfortunately I get:



NDSolve::tddisc: NDSolve cannot do a discontinuity replacement for event surfaces that depend only on time. >>



and the results are incomplete or the algorithm crashes. I also tried using WhenEvent with "DiscontinuitySignature" but with no success. This approach gives good results only for a linear monotonic function of $\sigma$, e.g. $\sigma(t) = 50t$.


I wrote a module to solve this using a simple first order Runge-Kutta so I obtained the solution but this is only a simple model. I'm sure Mathematica can solve this with its build-in methods. That would really save me a lot of work writing my own procedures.



Answer



Edit:
Using a helper function fh will result in no messages and no need to set extra options.


σ[t_] := 40000 Sin[0.02 t]

C1 = 80000;
C2 = 20000;
σgr = 15000;

fh[t_?NumericQ, x_, y_] := Piecewise[{{C1/(C1 + C2)*y, 
  (σ[t]*y > 0) && ((σ[t] - C2*x >= σgr) || (σ[t] - C2*x <= -σgr))}}, 0]

sol = NDSolve[{s[t] == σ[t]/C1 + sep[t], sep'[t] == fh[t, sep[t], s'[t]], 
        s[0] == 0, sep[0] == 0}, {s[t], sep[t]}, {t, 0, 1000}];


s[t_] = s[t] /. sol // First;
ParametricPlot[{s[t], σ[t]}, {t, 0, 10^3}, PlotRange -> All, 
  AspectRatio -> Full, GridLines -> Automatic]


plot



Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more