Nonlinear finite elements/Homework 2/Solutions

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Problem 1: Classification of PDEs[edit | edit source]


Solution[edit | edit source]

Part 1[edit | edit source]

What physical processes do the two PDEs model?

Equation (1) models the one dimensional heat problem where represents temperature at at time and represents the thermal diffusivity.

Equation (2) models the transverse vibrations of a beam. The variable represents the displacement of the point on the beam corresponding to position at time . is where is Young's modulus, is area moment of inertia, is mass density, and is area of cross section.

Part 2[edit | edit source]

Write out the PDEs in elementary partial differential notation

Part 3[edit | edit source]

Determine whether the PDEs are elliptic, hyperbolic, or parabolic. Show how you arrived at your classification.

Equation (1) is parabolic for all values of .

We can see why by writing it out in the form given in the notes on Partial differential equations.

Hence, we have , , , , , , and . Therefore,

Equation (2) is hyperbolic for positive values of , parabolic when is zero, and elliptic when is less than zero.

Problem 2: Models of Physical Problems[edit | edit source]


Solution[edit | edit source]

  • Heat transfer: :
  • Flow through porous medium: :
  • Flow through pipe: :
  • Flow of viscous fluids: :
  • Elastic cables: :
  • Elastic bars: :
  • Torsion of bars: :
  • Electrostatics: :

Problem 4: Least Squares Method[edit | edit source]


The residual over an element is


In the least squares approach, the residual is minimized in the following sense

Solution[edit | edit source]

Part 1[edit | edit source]

What is the weighting function used in the least squares approach?

We have,

Hence the weighting functions are of the form

Part 2[edit | edit source]

Develop a least-squares finite element model for the problem.

To make the typing of the following easier, I have gotten rid of the subscript and the superscript . Please note that these are implicit in what follows.

We start with the approximate solution

The residual is

The derivative of the residual is

For simplicity, let us assume that is constant, and . Then, we have


Now, the coefficients are independent. Hence, their derivatives are

Hence, the weighting functions are

The product of these two quantities is


These equations can be written as



Part 3[edit | edit source]

  1. Discuss some functions that can be used to approximate

For the least squares method to be a true "finite element" type of method we have to use finite element shape functions. For instance, if each element has two nodes we could choose

Another possiblity is set set of polynomials such as

Problem 5: Heat Transefer[edit | edit source]


where is the temperature, is the thermal conductivity, and is the heat generation. The Boundary conditions are

Solution[edit | edit source]

Part 1[edit | edit source]

Formulate the finite element model for this problem over one element

First step is to write the weighted-residual statement

Second step is to trade differentiation from to , using integration by parts. We obtain

Rewriting (3) in using the primary variable and secondary variable as described in the text book to obtain the weak form

From (4) we have the following

Part 2[edit | edit source]

Use ANSYS to solve the problem of heat conduction in an insulated rod. Compare the solution with the exact solution. Try at least three refinements of the mesh to confirm that your solution converges.

The following inputs are used to solve for the solution of this problem:

  pi = 4*atan(1) !compute value for pi
  ET,1,LINK34!convection element
  ET,2,LINK32!conduction element
  R,1,1!AREA = 1
  MP,HF,1,25 !convectivity
  emax = 2
  length = 0.1
  *do,nnumber,1,emax+1 !begin loop to creat nodes
     n,emax+2,length,0!extra node for the convetion element
  type,2 !switch to conduction element
  TYPE,1 !switch to convection element
  e,emax+1,emax+2!creat zero length element



ANSYS gives the following results

Node Temperature
1 50.000
2 27.590
3 5.1793
Node Heat Flux
1 -4.4821
4 4.4821

Part 3[edit | edit source]

Write you own code to solve the problem in part 2.

The following Maple code can be used to solve the problem.

> restart;
> with(linalg):
> L:=0.1:kxx:=0.01:beta:=25:T0:=50:Tinf:=5:
> # Number of elements
> element:=10:
> # Division length
> ndiv:=L/element:
> # Define node location
> for i from 1 to element+1 do
>   x[i]:=ndiv*(i-1);
> od:
> # Define shape function
> for j from 1 to element do
>   N[j][1]:=(x[j+1]-x)/ndiv;
>   N[j][2]:=(x-x[j])/ndiv; 
> od:
> # Create element stiffness matrix
> for k from 1 to element do
>   for i from 1 to 2 do
>     for j from 1 to 2 do
>       Kde[k][i,j]:=int(kxx*diff(N[k][i],x)*diff(N[k][j],x),x=x[k]..x[k+1]); 
>     od;
>   od;
> od;
> Kde[0][2,2]:=0:Kde[element+1][1,1]:=0:
> # Create global matrix for conduction Kdg and convection Khg
> Kdg:=Matrix(1..element+1,1..element+1,shape=symmetric):
> Khg:=Matrix(1..element+1,1..element+1,shape=symmetric):
> for k from 1 to element+1 do
>   Kdg[k,k]:=Kde[k-1][2,2]+Kde[k][1,1];
>   if (k <= element) then
>     Kdg[k,k+1]:=Kde[k][1,2];
>   end if;
> od:
> Khg[element+1,element+1]:=beta:
> # Create global matrix Kg = Kdg + Khg
> Kg:=matadd(Kdg,Khg):
> # Create T and F vectors
> Tvect:=vector(element+1):Fvect:=vector(element+1):
> for i from 1 to element do
>   Tvect[i+1]:=T[i+1]:
>   Fvect[i]:=f[i]:
> od:
> Tvect[1]:=T0:
> Fvect[element+1]:=Tinf*beta:
> # Solve the system Ku=f
> Ku:=multiply(Kg,Tvect):
> # Create new K matrix without first row and column from Kg
> newK:=Matrix(1..element,1..element):
> for i from 1 to element do
>   for j from 1 to element do
>     newK[i,j]:=coeff(Ku[i+1],Tvect[j+1]);
>   od;
> od;
> # Create new f matrix without f1
> newf:=vector(element,0):
> newf[1]:=-Ku[2]+coeff(Ku[2],T[2])*T[2]+coeff(Ku[2],T[3])*T[3]:
> newf[element]:=Tinf*beta:
> # Solve the system newK*T=newf
> solution:=multiply(inverse(newK),newf);
> exact := 50-448.2071713*x;
> with(plots):
> # Warning, the name changecoords has been redefined
> data := [[ x[n+1], solution[n]] <math>n=1..element]:
> FE:=plot(data, x=0..L, style=point,symbol=circle,legend="FE"):
> ELAS:=plot(exact,x=0..L,legend="exact",color=black):
> display({FE,ELAS},title="T(x) vs x",labels=["x","T(x)"]);