Review of the FEM solution for the torsion problem of a rectangular cross section.

Nasser Abbasi

Introduction

This is a review of the FEM solution to the torsion problem of a rectangular cross section beam. First a description of the problem is given, then a description of the FEM method is shown, followed by a simple numerical worked example.

The problem

The problem is to solve the Poisson 2D problem for rectangular cross section. This equation is the mathematical model for a beam under torsion as described in the following diagram.


problem_review.png

Since we are using a triangle elements for the FEM mesh, the cross section mesh is the preferred mesh to use as shown in this diagram.


problem_review_1.png

The big picture

Before going into the details of the FEM solution it might be useful to look at the big picture.


problem_review_2.png

The following diagram shows more description of the methods.


problem_review_3.png

Mathematical derivation

Derivation of the symmetric weak form of the 2D Poisson equation

The following diagram shows the steps to obtain the symmetric weak for of the 2D Poisson PDE


problem_review_4.png

Converting the symmetric weak form equation from the global Cartesian coordinates system to natural coordinates system

Converting the integral equations from the global Cartesian coordinates system to what is called the natural coordinates system (the local coordinates system) is a standard step used in FEM.

"A local coordinates system that relies on the element geometry for its definition and whose coordinates range between zero and unity within the element is known a natural coordinates system. Such system have the property that one particular coordinate has unit value at one node of the element and zero value at the other nodes: its variation between nodes is linear" Note_1

Integration of shape functions when they are written in the natural coordinates are simplified since the origin is now located on the element. These are the main reasons for changing from global coordinates to the natural coordinates. For simple geometries, one can avoid having to do this coordinates transformation, but in general and in practice it is the standard procedure to do.

I found that most of the technical and mathematical difficulties involved are in this step. So more details will be spend on this.

The global coordinates of the element is shown in this diagram


global_coordinates.png

Given an equation or expression where the independent variables in the equation are $x,y$ (the global Cartesian coordinates system) and we wish to express this same equation using the independent variables $\zeta,\eta$, then we perform coordinates transformations.

Given that MATH and MATH , we first find the differentials of the old coordinates system (i.e. $dx,dy$) in terms of the differentials of the new coordinates system ($d\zeta,d\eta$)

The matrix that represents this mapping between the differentials in the old coordinates system and the new coordinates system us called the Jacobian (some books call the determinant of this matrix as the Jacobian). It is important to note that this mapping is between the differentials of the independent variables in the two coordinates system, and not between the variables themselves.

Hence we write MATH

$J$ is also written as MATH

The main use for the Jacobian is in change of variables from one coordinates system to another, and also in performing area and volume integrals.


problem_review_4_1.png

Hence, converting an integral from the global coordinates to the natural coordinates can be done as follows

MATH

When the natural coordinates are area coordinates (which is the case here), we should modify the above to become

MATH

The area coordinates MATH are illustrated in this diagram


area_coordinates.png

It is important to realize that the shape functions $N_{1},N_{2},N_{3}$ used will be the same as the area coordinates.

Let us now start from the symmetric weak form equation, with the goal to convert it to the natural coordinates (see previous diagram for the derivation of this equation)

MATH

Since the first integral above is carried along the boundaries of the whole domain itself (not along the boundaries of the invidual elements themselves) and since we set the value of the test function $v$ to be zero at the boundaries of the domain, the first part of the above integral is zero. Hence the above integral become

MATH

In the following derivations, everything is done on an element $j$, hence all the $u,v,$ and element nodes coordinates $x_{1},y_{1}$, etc.. should have a superscript $j$ on, as in MATH. To make things easier to read, I will not put the superscript $j$ but will add it back at the end.

Consider first the second integral from (1):

MATH

Which can be rewritten as$\ $MATH

Consider the first integral from above

MATH

The above is written with reference to the global coordinates system. However, We want our trial and test functions to be defined in the natural coordinates system (where things are simpler). So we need a way to transformthe above integral (2A) to the natural coordinates system.

Assume we have the mapping MATH and MATH (we will see how to obtain this mapping below). This mapping tells us how the global coordinates themselves change as a function of the natural coordinates. Now we can use differentiation chain rule to see how the trial and test functions themselves change relative the global coordinates.

MATH

Similarly for the test function

MATH

To make things more clear, we rewrite the above using matrix notation. For the trial function

MATH

and similarly for the test function

MATH

From (3) and (4), we see the following inverse transformations

MATH

and

MATH

Now transpose the column vector in (6) to be a row vector because that is how it is laid out in the integral (2A), ( and remember to change the order when transposing a product)

MATH

Now we are ready to convert the integral $I_{2}$ in eq (2A) to the natural coordinates system (these are area coordinates, notice the integral limits and the order of integration)

MATH

Where we used the standard relationship that MATH

Remember to put $d\eta$ first before $d\zeta$ since the inner limit is on $\zeta.$

Now that we have everything in the natural area coordinates system, we can do the integration. One small point left, which is to determine the differentials involved in (7).

For this we now need to decide on the actual form of the trial and test functions and on the mapping between the global and the natural coordinates system. The following explains this part, we will come back to the above integral once we have obtained the differentials MATH and determined the Jacobian.

The following diagram shows the linear transformation we will use. This is a standard transformation where the natural coordinates are called the area coordinates described more below.


natural_coordinates.png

We see from the above diagram that

MATH

From the above we obtain the following differentials

MATH

Now, we consider the trial and test functions. based on the above transformation shown in eq (8), We see that the linear trial and test functions can also be written in similar transformation

MATH

Again, immediately, we obtain the following differentials from the above expressions

MATH

Hence the Jacobian can now be evaluated (see eq(3) for reference)

MATH

And its inverse is

MATH

And

MATH

Now that we have all the differentials needed, we can now go back to the integral in eq (7) and compute it:

MATH

Now we can evaluate $K^{j}$.

MATH

The integrand is

MATH

Where

MATH

Hence MATH

But the integrand is a constant, hence we take it out of the integral

MATH

Now we evaluate MATH

MATH

Hence

MATH

But from (10) we see that MATH

and the area of a triangle with corners at MATH is given by

MATH

Hence we get MATH

Therefore we can replace MATH by $2A^{j}$ everywhere. Rewrite the local stiffness matrix in tems of the local element area:

MATH

Now that we have $K^{j}$ we plug it back into eq (11)MATH

And now that we completed this integral we go back to eq (2) :

MATH

We need to work on the second integral above MATH and transform it to the natural coordinates.

MATH

Where

MATH

MATH

MATH

$\ $Hence

MATH

But MATH

Hence the above becomes

MATH

Now we have the integral in eq (2) completed. We now have our local equations completed. Here it is. We next need to assemble them.

MATH

Where MATH

Note on the shape functions

Looking at the trial function in eq (9), repeated here

MATH

Hence we see that

MATH

And since we are looking for a trial function to be of the form MATH we see from the above that the 3 shape or basis functions are the following

MATH

And since we are using the Galerking method, where the test function uses the same basis functions as the trial function, we can write the test function as

MATH

Assembly of the global stiffness matrix

The global stiffness matrix $K$ is always square and symmetric and positive definite. (At least for structural analysis). Recall a positive definite matrix $K$ is one such that for any nonzero vector $x$ we always have $x^{\ast}Ax>0$ where $x^{\ast}$ is the conjugate of $x.$ Properties of positive definite matrix is that all its eigenvalues are positive, and it has positive determinant, and hence a positive definite matrix is always invertible.

In addition, the global stiffness matrix is banded. This means that all non-zero elements are found along bands close to the main diagonal of the matrix. Within the band itself, some values can be zero.

The width of the band is a function of the numbering of the nodes used. Different node numbering can result in smaller band width. We want to have as small a band width as possible to take advantage of some numerical methods that can utilize banded matrices.

Band width can be reduced if we keep the node numbering in each element as close as possible to each others.

Now that we have found the local stiffness matrix $K^{j}$ for element $j$ we can assemble the global stiffness matrix as shown in this diagram. The direct stiffness construction method is used. This is explained in the following diagram


assemble.png

Assembly of the global load vector

This follows in similar fashion as above. The 3 elements Load vector $\left\{ f\right\} $ for element $j$ is added to the entries of the global load vector $\left\{ F\right\} $ using the node numbering mapping.

Modification of the final global stiffness matrix and load vectors and final solution

Now we have the following equation

MATH

Where $K$ is the assembled global stiffness matrix and $\left\{ F\right\} $ is the assembled global load vector. Before we solve for $\left\{ u\right\} $, which is the stress function at all the nodes, we must modify $K$ and $F$ to take care of the given boundary conditions. I attach below 2 pages from a book which gives a good explanation and small example on this point.

Now that we have the modified $K^{\ast}$ and $F^{\ast}$ you can solve for $\left\{ u\right\} $ using your favorite linear equations solver.


fix1.jpg


fix2.jpg

Conclusion

The following are the main steps in solving the torsion problem described in this report.


summary.png

References

  1. Methods of computer modeling in Engineering & the sciences. Volume 1. Satya N. Atluri. Tech Science Press

  2. Lecture notes, MAE 207. Spring and Fall 2006. UCI. Instructor: Professor Atluri SN.

  3. Applied finite element analysis. Larry Segerlind.

  4. The finite element method for engineers. Kenneth Huebner.

  5. Mathematical methods in the physical sciences. 2nd edition. Mary Boas.

  6. Fellow students reports and code from MAE 207 projects: Roy Culver , Paul Nylandres, Q Wang. see other related reports on my MAE 207 class web page 

http://12000.org/my_courses/UCI_COURSES/CREDIT_COURSES/spring_2006/spring_MAE_207/index.htm