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I would like to do some finite element calculations in axisymmetric cylindrical coordinates. I wish to calculate stress in terms of {r,z} coordinates. The radial stress, circumferential stress, axial stress and shear stress are $$\left\{\sigma _r,\sigma _{\theta },\sigma _z,\tau \right\}$$

The equilibrium equations are

$$ \frac{\partial \sigma _r}{\partial r}+\frac{\sigma _r-\sigma _{\theta }}{r}+\frac{\partial \tau }{\partial z}=0 $$

$$ \frac{\partial \sigma _z}{\partial z}+\frac{\partial \tau }{\partial r}+\frac{\tau }{r}=0 $$ $$ \frac{1}{r}\left(\frac{\partial \sigma _{\theta }}{\partial r}\right)=0 $$

The stress strain relationships are

$$ \sigma _r=\frac{Y}{(\nu +1) (2 \nu -1)}\left((\nu -1) \epsilon _r-\nu \left(\epsilon _{\theta }+\epsilon _z\right)\right) $$

$$ \sigma _{\theta }=\frac{Y}{(\nu +1) (2 \nu -1)}\left((\nu -1) \epsilon _{\theta }-\nu \left(\epsilon _r+\epsilon _z\right)\right) $$

$$ \sigma _z=\frac{Y}{(\nu +1) (2 \nu -1)}\left((\nu -1) \epsilon _z-\nu \left(\epsilon _r+\epsilon _{\theta }\right)\right) $$ $$ \tau =\frac{Y}{2 (1+\nu )}\left(\gamma _{\text{rz}}\right) $$

Finally the strain displacement relationships are $$ \epsilon _r=\frac{\partial u}{\partial r} $$ $$ \epsilon _z=\frac{\partial w}{\partial z} $$ $$ \epsilon _{\theta }=\frac{u}{r} $$ $$ \gamma _{\text{rz}}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r} $$

Where {u,w} are the displacements in the radial and axial directions. Somehow we have to get these equations into the mathematica form that is required for differential equations.

As a starting point these equation in the Wolfram language are

(* Equilibrium Equations *)
 eqn1 = {
       D[σr[r, z], r] + (σr[r, z] - σθ[r, z])/
        r + D[τ[r, z], z],
       D[σz[r, z], z] + D[τ[r, z], r] + τ[r, z]/r,
       1/r D[σθ[r, z], r]
       };
(* Stress Strain *)
eqn2 = {
   σr[r, z] -> 
    Y/((ν + 1) (2 ν - 
        1)) ((ν - 
          1) ϵr - ν (ϵθ + ϵz)),
   σθ[r, z] -> 
    Y/((ν + 1) (2 ν - 
        1)) ((ν - 
          1) ϵθ - ν (ϵr + ϵz)),
   σz[r, z] -> 
    Y/((ν + 1) (2 ν - 
        1)) ((ν - 
          1) ϵz - ν (ϵr + ϵθ)),
   τ[r, z] -> Y/(2 (1 + ν)) γrz
   };

(* Strain Displacement *)
eqn3 = {
   ϵr -> D[u[r, z], r],
   ϵz -> D[w[r, z], z],
   ϵθ -> u[r, z]/r,
   γrz -> D[u[r, z], z] + D[w[r, z], r]

   };

I have tried putting Inactive around the derivatives and getting the equations into some sort of matrix form but can't make it work out.

Here is a mesh as a minimum working example

r0 = 1;
r1 = 6;
r2 = 8;
z1 = 4;
z2 = 5;
mesh = ToElementMesh[
   RegionUnion[Rectangle[{r0, 0}, {r2, z2}], 
    Rectangle[{r0, z2}, {r1, z1 + z2}]]
   ];
mesh["Wireframe"]

Mathematica graphics

If we can make a stress operator, say, aso then the formulation for NDSolve would be

{uif, wif} = 
 NDSolve[aso == {NeumannValue[1, r == r2 && 0 <= z <= z2], 0}, {u, 
   w}, {r, z} ∈ mesh]

A plain stress operator is given under Coupled PDEs in Help but I have not been able to put it in that form. Can someone see how to do it? Thanks.

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  • $\begingroup$ Not familiar with linear elasticity, but if I understand the related wiki page correctly, the stress formulation is equivalent to the system above? $\endgroup$ – xzczd May 23 at 6:16
  • $\begingroup$ @xzczd Thanks for the cross reference to the wiki page. I think for the finite element formulation we need a solution in terms of displacements. Consequently, we need to eliminate stress and strain, expressing them in terms of displacements. As the problem is linear I think this is equivalent to matrix multiplication . However, as the matrices involve derivatives we need to maintain their operator status so it becomes more complicated. $\endgroup$ – Hugh May 23 at 6:29
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I think I have finally got a method for getting the coefficients of the differential equation. (user21 has been useful as ever).

The starting point is to consider the two equations that are the basis for solving differential equations. From the tutorial: Solving Partial Differential Equations with Finite Elements

Mathematica graphics

I think there are some typos here . The key point is that in the second set of equations all the terms should be positive while in the system of partial differential equations the corresponding terms should be negative. I also think that the Mathematica graphics should be on the right. I am going to call the second equation part1 with coefficients c, α and γ and then the partial differential equation and Neumann equation become

Mathematica graphics

Here part2 are the terms needed to complete the differential equation. The steps I am going to use to derive the equations are as follows. Let me know if there is a simpler method.

  1. Make a symbolic version of all the needed coefficients.
  2. Starting from the second equations make a symbolic version of each term .
  3. Equate these terms to those in the actual equation and make replacement rules for the coefficients. This gives us part1.
  4. Take the divergence of -part1 and then subtract it from the differential equation to get part2.
  5. Equate symbolic terms for part2 with the actual terms and thus get replacement rules for these coefficients.
  6. Generate the coefficients.
  7. Solve using the coefficients and InitializePDECoefficients( because I don't know how to make an Inactive version of the differential equation). That can be another question.

Examination of the equations (and some experimentation) shows that we only need c, α, β and a coefficients. The full list of the names of these coefficients are given under InitializePDECoefficients as follows

Mathematica graphics

Here are the equations for the stress and the partial differential equations.

ClearAll[σr,σz,τrz,τzr,σθ]
stress={σr[r,z],τrz[r,z],τzr[r,z],σz[r,z]};
(* ddisp={(u^(1,0))[r,z],(u^(0,1))[r,z],(w^(1,0))[r,z],(w^(0,1))[r,z]}; *)
ddisp = {Derivative[1,0][u][r,z], Derivative[0,1][u][r,z], Derivative[1,0][w][r,z], Derivative[0,1][w][r,z]};
disp={u[r,z],w[r,z]};

(*  stress in terms of strain *)
eqn1={
σr[r,z]==Y/((ν+1) (2 ν-1)) ((ν-1) ϵr-ν (ϵθ+ϵz)),
τrz[r,z]==Y/(2 (1+ν)) γrz,
τzr[r,z]==Y/(2 (1+ν)) γzr,
σz[r,z]==Y/((ν+1) (2 ν-1)) ((ν-1) ϵz-ν (ϵr+ϵθ)),
σθ[r,z]==Y/((ν+1)(2ν-1)) ((ν-1)ϵθ-ν(ϵr+ϵz))
};

(* strain in terms of displacement *)
eqn2={
ϵr-> D[u[r,z],r],
ϵz-> D[w[r,z],z],
ϵθ->u[r,z]/r,
γrz-> D[u[r,z],z]+D[w[r,z],r],
γzr-> D[u[r,z],z]+D[w[r,z],r]
};
(* Get the equations for stress in terms of displacements *)
sol1=First@Solve[eqn1/.eqn2,Join[stress,{σθ[r,z]}]]//Simplify;
(* differential equation to be solved *)
pde={
D[σr[r,z]/.sol1,r]+D[τrz[r,z]/.sol1,z]+(σr[r,z]/r)-σθ[r,z] /r,
D[σz[r,z]/.sol1,z]+D[τzr[r,z]/.sol1,r]+(τzr[r,z]/r)
}/.sol1//Simplify;

Now we have the first step and form the symbolic version of the coefficients and the equations.

n = 2;
space = {r, z};
vars = {u[r, z], w[r, z]};
ccoffs = Array[c, {n, n, n, n}]; (* DiffusionCoefficients *)
αcoffs = Array[α, {n, n, n}];  (* ConservativeConvectionCoefficients *)
βcoffs = Array[β, {n, n, n}];  (* ConvectionCoefficients *)
acoffs = Array[a, {n, n}];   (* ReactionCoefficients *)


cT = Table[Sum[ccoffs[[i, j]].Grad[vars[[j]], space], {j, n}], {i, n}];
αT = Table[Sum[αcoffs[[i, j]] vars[[j]], {j, n}], {i, n}];
βT = Table[Sum[βcoffs[[i, j]].Grad[vars[[j]], space], {j, n}], {i, n}];
aT = Table[Sum[acoffs[[i, j]] vars[[j]], {j, n}], {i, n}];

Next are steps 2 and 3 and a check to see that we have formed part1 correctly

part1 = {{σr[r, z], τrz[r, z]}, {τzr[r, 
       z], σz[r, z]}} /. sol1 // Simplify;

{vec1, mat1} = CoefficientArrays[part1, ddisp] // Normal // Simplify;

{vec1S, mat1S} = CoefficientArrays[cT, ddisp] // Normal // Simplify;

sub1 = Thread[Flatten[mat1S] -> Flatten[mat1]];

{vec2, mat2} = CoefficientArrays[part1, disp] // Normal // Simplify;

{vec2S, mat2S} = 
  CoefficientArrays[αT, disp] // Normal // Simplify;

sub2 = Thread[Flatten[mat2S] -> Flatten[mat2]];

(* Check we have modeled part1 correctly *)
check1 = cT + αT /. Join[sub1, sub2] // Simplify;

part1 == check1 // Simplify

True

The next step is to find part2. The pde is equal to zero so we can multiply by any value and not change its meaning. Here to fit with the negative value of part1 the pde is multiplied by -1. We get part2 and again match coefficients between symbolic and actual values.

part2 = -pde - Div[-part1, space] // Simplify;

{vec3, mat3} = CoefficientArrays[part2, ddisp] // Normal // Simplify;

{vec3S, mat3S} = CoefficientArrays[βT, ddisp] // Normal // Simplify;

sub3 = Thread[Flatten[mat3S] -> Flatten[mat3]];

{vec4, mat4} = CoefficientArrays[part2, disp] // Normal // Simplify;

{vec4S, mat4S} = CoefficientArrays[aT, disp] // Normal // Simplify;

sub4 = Thread[Flatten[mat4S] -> Flatten[mat4]];

(* Check that part2 is correctly modelled*)
check2 = βT + aT /. Join[sub3, sub4];

part2 == check2 // Simplify

True

Thus the coefficients are

coffs = {ccoffs, αcoffs, βcoffs, acoffs} /. 
  Join[sub1, sub2, sub3, sub4]

Mathematica graphics

Now for an example. First a module to solve the equation.

Needs["NDSolve`FEM`"];
ClearAll[mySolve];
mySolve[{u_, w_}, {r_, z_}, mesh_, ubc_, 
  wbc_, {cCoffs_, αCoffs_, βCoffs_, aCoffs_}] := Module[
  {df, ccc, cv, rc, vd, sd, methodData, initCoeffs, initBCs, 
   discretePDE, load, stiffness, damping, mass, split, discreteBCs},
  df = "DiffusionCoefficients" -> -cCoffs;
  ccc = "ConservativeConvectionCoefficients" -> -αCoffs;
  cv = "ConvectionCoefficients" -> βCoffs;
  rc = "ReactionCoefficients" -> aCoffs;

  (*variable data*)
  vd = NDSolve`VariableData[{"DependentVariables", 
      "Space"} -> {{u, w}, {r, z}}];

  (*Solution data*)
  sd = NDSolve`SolutionData["Space" -> ToNumericalRegion[mesh]];

  (*Initialize the finite element data with the variable and solution \
data.*)
  methodData = InitializePDEMethodData[vd, sd];

  (*Initialize the partial differential equation coefficients.*)
  initCoeffs = InitializePDECoefficients[vd, sd, df, ccc, cv, rc];

  (*Initialize the boundary conditions.*)
  initBCs = InitializeBoundaryConditions[vd, sd, {ubc, wbc}];

  (*Compute the discretized partial differential equation.*)
  discretePDE = DiscretizePDE[initCoeffs, methodData, sd];
  {load, stiffness, damping, mass} = discretePDE["SystemMatrices"];

  (*Compute how the incidents are split between the two dependent \
variables.*)
  {i1, i2, i3} = methodData["IncidentOffsets"];
  split = {i1 + 1 ;; i2, i2 + 1 ;; i3};

  (*Discretize the initialized boundary conditions.*)
  discreteBCs = DiscretizeBoundaryConditions[initBCs, methodData, sd];

  (*Deploy the boundary conditions in place.*)
  DeployBoundaryConditions[{load, stiffness}, discreteBCs];

  {load, stiffness, split}]

Edit

I have changed the answer from here on so that we can have an example which compares a 2D and 3D case. This is at the request of Alex Trounev who would like to make further tests.

The 2D case first

r0 = 1;
r1 = 6;
r2 = 8;
z1 = 5;
z2 = 9;
peram = {Y -> 10^3, ν -> 33/100};
mesh = ToElementMesh[
   RegionUnion[Rectangle[{r0, 0}, {r2, z1}], 
    Rectangle[{r0, z1}, {r1, z2}]], MaxCellMeasure -> 0.05];
mesh["Wireframe"]

Mathematica graphics

stress = 20;(* stress applied to structure *)

{load, stiffness, split} = mySolve[{u, w}, {r, z}, mesh,
   {DirichletCondition[u[r, z] == 0, z == 0]},
   {DirichletCondition[w[r, z] == 0, z == 0],
    NeumannValue[stress, z == z2]},
   coffs /. peram
   ];
solution = LinearSolve[stiffness, load];
uif = ElementMeshInterpolation[{mesh}, solution[[split[[1]]]]];
wif = ElementMeshInterpolation[{mesh}, solution[[split[[2]]]]];

ClearAll[σz];
σz[r_, z_] := Evaluate[-((Y (ν u[r, z] - r (-1 + ν) 
\!\(\*SuperscriptBox[\(w\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, z] + r ν 
\!\(\*SuperscriptBox[\(u\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, z]))/(r (1 + ν) (-1 + 2 ν))) /. 
    Join[{u -> uif, w -> wif}, peram]];
p11 = Plot[wif[r, z2], {r, r0, r1}, PlotRange -> All, 
  PlotStyle -> {Orange}]
p12 = Plot[σz[r, z2], {r, r0, r1}, 
  PlotRange -> {All, {18, 22}}, PlotRange -> All, 
  PlotStyle -> {Orange}]
p13 = Plot[σz[r, z1], {r, r0, r1}, PlotRange -> {All, All}, 
  PlotRange -> All, PlotStyle -> {Orange}]
p14 = Plot[uif[r2, z], {z, 0, z1}, PlotRange -> All, 
  PlotStyle -> {Orange}]

The above will produce plots which we can compare to the 3D case. To make the 3D case we need the mesh as follows

r0 = 1;
r1 = 6;
r2 = 8;
z1 = 5;
z2 = 9;
peram = {Y -> 10^3, ν -> 33/100};
reg1 = BoundaryDiscretizeRegion[Cylinder[{{0, 0, 0}, {0, 0, z1}}, r2]];
reg2 = BoundaryDiscretizeRegion[
   Cylinder[{{0, 0, z1}, {0, 0, z2}}, r1]];
reg3 = RegionUnion[reg2, reg1];
reg4 = BoundaryDiscretizeRegion[Cylinder[{{0, 0, 0}, {0, 0, z2}}, r0]];
reg = RegionDifference[reg3, reg4]

mesh2 = ToElementMesh[reg, MaxCellMeasure -> 0.1];
mesh2["Wireframe"]

Mathematica graphics

The stress operator is the standard 3D operator

ClearAll[stressOperator]; 
stressOperator[
  Y_, ν_] := {Inactive[
     Div][{{0, 0, -((Y*ν)/((1 - 2*ν)*(1 + ν)))}, {0, 0, 
       0}, {-Y/(2*(1 + ν)), 0, 0}}.Inactive[Grad][
      w[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, -((Y*ν)/((1 - 2*ν)*(1 + ν))), 
       0}, {-Y/(2*(1 + ν)), 0, 0}, {0, 0, 0}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-((Y*(1 - ν))/((1 - 2*ν)*(1 + ν))), 0, 
       0}, {0, -Y/(2*(1 + ν)), 0}, {0, 
       0, -Y/(2*(1 + ν))}}.Inactive[Grad][
      u[x, y, z], {x, y, z}], {x, y, z}], 
  Inactive[Div][{{0, 0, 0}, {0, 
       0, -((Y*ν)/((1 - 
              2*ν)*(1 + ν)))}, {0, -Y/(2*(1 + ν)), 
       0}}.Inactive[Grad][w[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, -Y/(2*(1 + ν)), 
       0}, {-((Y*ν)/((1 - 2*ν)*(1 + ν))), 0, 0}, {0, 0, 
       0}}.Inactive[Grad][u[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-Y/(2*(1 + ν)), 0, 
       0}, {0, -((Y*(1 - ν))/((1 - 2*ν)*(1 + ν))), 0}, {0,
        0, -Y/(2*(1 + ν))}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}], 
  Inactive[Div][{{0, 0, 0}, {0, 
       0, -Y/(2*(1 + ν))}, {0, -((Y*ν)/((1 - 
              2*ν)*(1 + ν))), 0}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, 0, -Y/(2*(1 + ν))}, {0, 0, 
       0}, {-((Y*ν)/((1 - 2*ν)*(1 + ν))), 0, 0}}.Inactive[
       Grad][u[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-Y/(2*(1 + ν)), 0, 0}, {0, -Y/(2*(1 + ν)), 0}, {0,
        0, -((Y*(1 - ν))/((1 - 2*ν)*(1 + ν)))}}.Inactive[
       Grad][w[x, y, z], {x, y, z}], {x, y, z}]}

stress = 20;
{uif2, vif2, wif2} = NDSolveValue[{
     stressOperator[Y, ν] == {
       0,
       0,
       NeumannValue[stress , z == z2]},

     DirichletCondition[u[x, y, z] == 0, z == 0],
     DirichletCondition[v[x, y, z] == 0, z == 0],
     DirichletCondition[w[x, y, z] == 0, z == 0]
     } /. peram, {u, v, w}, {x, y, z} ∈ mesh2];

ClearAll[σz1];
σz1[x_,y_,z_]:=Evaluate[(Y ((-1+ν) (w^(0,0,1))[x,y,z]-ν ((u^(1,0,0))[x,y,z]+(v^(0,1,0))[x,y,z])))/((1+ν) (-1+2 ν))/.Join[{u-> uif2,v-> vif2,w-> wif2},peram]];
p21=Plot[Evaluate[Table[wif2[r Cos[θ],r Sin[θ],z2],{θ,0,(19 2π)/20,(2π)/20}]],{r,r0,r1},PlotStyle->{Gray},PlotLegends->LineLegend[{Gray,Orange},{3D,2D}]]
p22=Plot[Evaluate[Table[σz1[r Cos[θ],r Sin[θ],z2],{θ,0,(19 2π)/20,(2π)/20}]],{r,r0,r1},PlotStyle->{Gray},PlotRange->{All,{19.5,20.5}},PlotLegends->LineLegend[{Gray,Orange},{3D,2D}]]
p23=Plot[Evaluate[Table[σz1[r Cos[θ],r Sin[θ],z1],{θ,0,(19 2π)/20,(2π)/20}]],{r,0,r1},PlotStyle->{Gray},PlotRange->All,PlotLegends->LineLegend[{Gray,Orange},{3D,2D}]]
p24=Plot[Evaluate[Table[uif2[r2 Cos[θ],r2 Sin[θ],z]Cos[θ]+vif2[r2 Cos[θ],r2 Sin[θ],z]Sin[θ],{θ,0,(19 2π)/20,(2π)/20}]],{z,0,z2},PlotStyle->{Gray},PlotRange->All,PlotLegends->LineLegend[{Gray,Orange},{3D,2D}]]
Show[p21,p11]
Show[p22,p12]
Show[p23,p13]
Show[p24,p14]

The plots when combined in Show are as follows

Mathematica graphics

Mathematica graphics

Mathematica graphics

Mathematica graphics

I have plotted the 3D case along a number of lines and all in gray. There is some spread in the 3D values. The 2D plots overlay the 3D ones. The third plots shows some discrepancy at the edge of the region when r = 6. However, this is a stress along a radial line and I don't think it is easy to get the stress directly at the point where the radial line exits the body.

These plots do seem to agree. There is an issue about are the grids compatible for there mesh density? I don't know how do this. Please make comparisons using other code.

The bit I don't know how to do is to put the coefficients into an equation with appropriate Inactivates. Any ideas?

Edit user21: I deleted my answer; to the best of my knowledge this answer is correct. I am going to show the inactive operator for a top level NDSolveValue call. To do these I usually derive the coefficients, just like you did. The I use 'NDSolve`ProcessEquations' to get the state object and extract the FEM coefficients from that until the inactive form and the coefficients match. I started with

op = {
  Inactive[Div][-coffs[[1, 1, 1]].Inactive[Grad][u[r, z], {r, z}], {r,
     z}],
  Inactive[Div][-coffs[[1, 2, 1]].Inactive[Grad][u[r, z], {r, z}], {r,
     z}]
  }

Evaluated the code below to check that the coefficients match and continued to add terms until I arrived at:

op = {Inactive[
      Div][-coffs[[1, 1, 1]].Inactive[Grad][u[r, z], {r, z}] + 
      Inactive[Times][-coffs[[2, 1, 1]], u[r, z]], {r, z}] + 
    Inactive[Div][-coffs[[1, 1, 2]].Inactive[Grad][w[r, z], {r, z}] + 
      Inactive[Times][-coffs[[2, 1, 2]], w[r, z]], {r, z}] + 
    coffs[[3, 1, 1]].Inactive[Grad][u[r, z], {r, z}] + 
    coffs[[3, 1, 2]].Inactive[Grad][w[r, z], {r, z}] + 
    coffs[[4, 1, 1]]*u[r, z] + coffs[[4, 1, 2]]*w[r, z],
   Inactive[Div][-coffs[[1, 2, 1]].Inactive[Grad][u[r, z], {r, z}] + 
      Inactive[Times][-coffs[[2, 2, 1]], u[r, z]], {r, z}] + 
    Inactive[Div][-coffs[[1, 2, 2]].Inactive[Grad][w[r, z], {r, z}] + 
      Inactive[Times][-coffs[[2, 2, 2]], w[r, z]], {r, z}] + 
    coffs[[3, 2, 1]].Inactive[Grad][u[r, z], {r, z}] + 
    coffs[[3, 2, 2]].Inactive[Grad][w[r, z], {r, z}] + 
    coffs[[4, 2, 1]]*u[r, z] + coffs[[4, 2, 2]]*w[r, z]
   };

This is the proposed inactive form of the operator and I check that it matches the input coefficients by extracting them from the state object.

{state} = 
  NDSolve`ProcessEquations[{op == {0, 
       NeumannValue[stress, z == z2]} /. peram, 
    DirichletCondition[{u[r, z] == 0, w[r, z] == 0}, z == 0]}, {u, 
    w}, {r, z} \[Element] mesh];
pdecd = state["FiniteElementData"]["PDECoefficientData"];

Setup:

numCoff = coffs /. peram;

Check:

numCoff[[1]] // MatrixForm
pdecd["DiffusionCoefficients"] // MatrixForm

numCoff[[2]] // MatrixForm
pdecd["ConservativeConvectionCoefficients"] // MatrixForm

numCoff[[3]] // MatrixForm
pdecd["ConvectionCoefficients"] // MatrixForm

numCoff[[4]] // MatrixForm
pdecd["ReactionCoefficients"] // MatrixForm

The sign differece in the first two is OK, since that is added when given to InitializePDECoefficients. The structural difference in the third coefficient is also nothing to worry about. It's just the internal representation that differs from the input form.

We can then use:

{uifND, wifND} = 
  NDSolveValue[{op == {0, NeumannValue[stress, z == z2]} /. peram, 
    DirichletCondition[{u[r, z] == 0, w[r, z] == 0}, z == 0]}, {u, 
    w}, {r, z} \[Element] mesh];

And get:

Show[{mesh["Wireframe"["MeshElement" -> "BoundaryElements"]], 
  ElementMeshDeformation[mesh, {uifND, wifND}, "ScalingFactor" -> 10][
   "Wireframe"[
    "ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}]

This is the same result one gets when using the low level FEM functions as above. A final note, the operator op above as terms resulting in zero when looked at closely, but I wanted to show how such an inactive version of the operator would look in a more general scenario.

enter image description here

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  • $\begingroup$ @user21 I think I have got my coefficients. Does this look like a proper derivation to you? I don't get as many 'r's as you. This may be because the system is modified to avoid the singularity at the origin. I need to experiment more to examine that. I have all the coefficient types you mentioned. This can probably be generalised but that is future work. $\endgroup$ – Hugh Jun 24 at 17:56
  • $\begingroup$ Concerning the typos: Yes, in the mentioned equation the signs of the coefficients should read -alpha12 and -alpha22. That's already corrected and will be updated with the next version. A correct version is in the ref page of InitializePDECoefficients $\endgroup$ – user21 Jun 25 at 4:39
  • $\begingroup$ The gamma terms have opposite signs as shown in the link given above. It looks as if this is not relevant for your case anyways, so that's not an issue. $\endgroup$ – user21 Jun 25 at 4:41
  • 1
    $\begingroup$ @user21 definition of coffs added. Thanks for looking at the code. $\endgroup$ – Hugh Jun 25 at 11:22
  • 1
    $\begingroup$ @user21 I added a cross check with COMSOL as an answer and the results look quite similar. $\endgroup$ – Tim Laska Jun 26 at 10:52
5
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To derive the system of equations we use the code

(*Equilibrium Equations*)
eqn1 = {D[σr, r] + (σr - σθ)/r + 
    D[τ, z], D[σz, z] + D[τ, r] + τ/r, 
   1/r D[σθ, r]};
(*Stress Strain*)
eqn2 = {σr -> 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵr - ν (ϵθ + ϵz)), \
σθ -> 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵθ - ν (ϵr + ϵz)), \
σz -> 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵz - ν (ϵr + ϵθ)), \
τ -> Y/(2 (1 + ν)) γrz};
(*Strain Displacement*)
eqn3 = {ϵr -> D[u[r, z], r], ϵz -> 
    D[w[r, z], z], ϵθ -> u[r, z]/r, γrz -> 
    D[u[r, z], z] + D[w[r, z], r]};


eq = {σr = 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵr - ν (ϵθ + ϵz)), \
σθ = 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵθ - ν (ϵr + ϵz)), \
σz = 
    Y/((ν + 1) (2 ν - 1)) ((ν - 
          1) ϵz - ν (ϵr + ϵθ)), \
τ = Y/(2 (1 + ν)) γrz} /. eqn3;

sys = {D[eq[[1]], r] + (eq[[1]] - eq[[2]])/r + D[eq[[4]], z], 
  D[eq[[3]], z] + D[eq[[4]], r] + eq[[4]]/r, 1/r D[eq[[2]], r]};

Here we get three equations for two unknowns. The third equation can be integrated independently of the first two. It gives expression to $\sigma _\theta$. Then we can build a solution using FEM. To solve a specific problem, one needs to know the boundary conditions. Here is an example of deformation under compression:

r0 = 1;
r1 = 6;
r2 = 8;
z1 = 4;
z2 = 5; Y = 10^3; \[Nu] = 1/3;
<< NDSolve`FEM`
mesh = ToElementMesh[
   RegionUnion[Rectangle[{r0, 0}, {r2, z2}], 
    Rectangle[{r0, z2}, {r1, z1 + z2}]]];
mesh["Wireframe"]
{ufun, wfun} = 
 NDSolveValue[{sys[[1]] == NeumannValue[-10, z == z1 + z2], 
   sys[[2]] == 0, 
   DirichletCondition[{u[r, z] == 0, w[r, z] == 0}, z == 0]}, {u, 
   w}, {r, z} \[Element] mesh]

mesh = ufun["ElementMesh"];
Show[{
  mesh["Wireframe"[ "MeshElement" -> "BoundaryElements"]],
  ElementMeshDeformation[mesh, {ufun, wfun}][
   "Wireframe"[
    "ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}]

fig1

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  • $\begingroup$ Thank you for your answer. I will study this (may take a day or two) you seem to have assembled the equations without developing a stress operator -which is novel. Also, can you put in a NeumannValue i.e. a stress with your formulation? $\endgroup$ – Hugh May 24 at 10:21
  • $\begingroup$ @Hugh Yes, we can get NeumannValue (stress) using this code. I didn't add anything, just used your equations. $\endgroup$ – Alex Trounev May 24 at 11:08
3
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Cross Check

The following is a quick cross check in COMSOL as requested by @user21 in @Hugh's answer.

I meshed the model at physics based settings of Fine and Extremely Fine corresponding to 1,285 and 17,295 triangles, respectively. Hugh's model was 1716 triangles on my machine. I only reproduced the top plot, but you can see that there is good agreement with the lower resolution COMSOL model.

COMSOL Hugh Comparison

An animation of the stress evolution shows singularities at the right angle bend and in the lower right corner where the model is pinned.

Pinned animation

Applying a roller constraint to the bottom edge eliminates the singularity in the lower right corner, but you will need to add a fillet to remove the other.

Roller Constraint

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  • $\begingroup$ Thanks Tim, much appreciated! $\endgroup$ – user21 Jun 26 at 11:15
  • $\begingroup$ Well done Tim. Just to confirm did you do a 3D or 2D simulation? $\endgroup$ – Hugh Jun 26 at 11:54
  • $\begingroup$ @Hugh This was a 2D axisymmetric simulation. I could do a 3D if desired, but not until tomorrow. $\endgroup$ – Tim Laska Jun 26 at 12:00

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