[granzer]

I have been trying to study FEM for past couple of days and this question has been bugging me for a while now.

I have read that the order of the element is the order of the polynomial used to approximate/represent the field variable in that element. If we consider a one-dimensional, 2 degrees of freedom element (with 2 nodes) the polynomial would be a linear polynomial and if this element happens to have an interior node (so totally 3 nodes, and 3 DOF), then the polynomial would be a quadratic polynomial.

I have also read that the degree of the polynomial we choose also depends on the degrees of freedom of the element. So, if we consider a 4 DOF element, then the polynomial would be cubic polynomial. But even though we use a cubic polynomial the element is still a linear element.

Here is my confusion: Would a quadratic polynomial still be able to represent a one dimensional, 3 noded elements (i.e. with 1 interior node) with 2 DOF at each node (so a 6 DOF element)? I am asking this because usually a 1 dimension 3 noded element would be called a quadratic element...but the textbooks always assume a single degree of freedom at each node. Would this element (with 3 nodes) still be called a 'quadratic-element' if the number of degrees of freedom per node is increased to 2?

[FMDenaro]

Originally Posted by granzer

I have been trying to study FEM for past couple of days and this question has been bugging me for a while now.

I have read that the order of the element is the order of the polynomial used to approximate/represent the field variable in that element. If we consider a one-dimensional, 2 degrees of freedom element (with 2 nodes) the polynomial would be a linear polynomial and if this element happens to have an interior node (so totally 3 nodes, and 3 DOF), then the polynomial would be a quadratic polynomial.

I have also read that the degree of the polynomial we choose also depends on the degrees of freedom of the element. So, if we consider a 4 DOF element, then the polynomial would be cubic polynomial. But even though we use a cubic polynomial the element is still a linear element.

Here is my confusion: Would a quadratic polynomial still be able to represent a one dimensional, 3 noded elements (i.e. with 1 interior node) with 2 DOF at each node (so a 6 DOF element)? I am asking this because usually a 1 dimension 3 noded element would be called a quadratic element...but the textbooks always assume a single degree of freedom at each node. Would this element (with 3 nodes) still be called a 'quadratic-element' if the number of degrees of freedom per node is increased to 2?

The degree g of a lagrangian polynomial in 1D is equal to g=n-1 where n is the number of nodes where the values are prescribed. Given therefore a shape function of a degree g you can use it for any variable you have.

[granzer]

Originally Posted by FMDenaro

The degree g of a lagrangian polynomial in 1D is equal to g=n-1 where n is the number of nodes where the values are prescribed. Given therefore a shape function of a degree g you can use it for any variable you have.

@FMDenaro This may be a very basic question. I am referring to the FEM textbook by David V Hutton. In this when the author talks about higher order 1D element, say with 4 nodes, I know that is cubic polynomial function is needed and this element is called a 3rd order or a cubic element as we are using a cubic element.

Now in the same book when the author is talking about a 1D (2 noded) beam element which has 2 DOF at each node (so totally 4 DOF) again a cubic polynomial is used. Here it is not explicitly stated that this is a cubic element. But since a cubic polynomial is used shouldn't this also be called a cubic element?

[Eifoehn4]

Dear granzer,

do you understand the difference between Langrange- and Hermite interpolation?

This is the answer to your question.

[FMDenaro]

Originally Posted by granzer

@FMDenaro This may be a very basic question. I am referring to the FEM textbook by David V Hutton. In this when the author talks about higher order 1D element, say with 4 nodes, I know that is cubic polynomial function is needed and this element is called a 3rd order or a cubic element as we are using a cubic element.

Now in the same book when the author is talking about a 1D (2 noded) beam element which has 2 DOF at each node (so totally 4 DOF) again a cubic polynomial is used. Here it is not explicitly stated that this is a cubic element. But since a cubic polynomial is used shouldn't this also be called a cubic element?

An element with 2 nodes define g=1 for the esplicit lagrangian polynomial but g=3 for an implicit Hermite (or Padè) interpolation. This is the typical case of the cubic spline. Of course, the resolution requires to solve an implicit equation

[granzer]

Originally Posted by Eifoehn4

Dear granzer,

do you understand the difference between Langrange- and Hermite interpolation?

This is the answer to your question.

Originally Posted by FMDenaro

An element with 2 nodes define g=1 for the esplicit lagrangian polynomial but g=3 for an implicit Hermite (or Padè) interpolation. This is the typical case of the cubic spline. Of course, the resolution requires to solve an implicit equation

After studying about Hermite interpolation I know that it not only ensures continuity of the given DOF (let's take x-displacement) but also the continuity of its derivative till a given order of the derivative. So applying this understanding to my question I know come to understand why this is important as the rotation DOF is the derivative of the displacement DOF.

Now knowing this I would like to refine my previous question with which I think should clear this up for me.

Let's take 2 cases. Each case comparing a 1D 2-noded element and a 1D 3-noded element (ie with 1 interior node):

Case 1: Each node has 2 DoF for the 2 elements. Let them be x-displacement and rotation. In this case, I know we use hermite shape function. So the approximation function for x-displacement will be given as

u(x) = sum(ui * Ni) + sum(u'i * Mi)

Where u(x) is the approx function for displacement, ui is the nodal displacement, Ni shape function for nodal displacement, u'i is the derivative of nodal displacement at the node, and Mi shape function for derivative of displacement.
a) What about the value for rotation DOF? Will, there be an approximation function for the rotation DOF as well?
b) Taking the 2 noded element, I know that this element has a total of 4 DOF. For this will we be using a cubic polynomial? Also, what will the order of the element be? (Is it a cubic element even though it has just 2 nodes?)
c) Taking the 3 noded element, I know that this element has a total of 6 DOF. For this will we be using a 5th order polynomial? Also, what will the order of the element be? ( Is it a 5th order element or will it be a quadratic element as it has 3 nodes?

Case 2: Lets again consider that each node has 2 DOF. But this time let the DoF be x-displacement and temperature (or x-displacement and y-displacement).
a) In this case would the approximation function for x-displacement be :

u(x) = sum(ui * Ni) + sum(Ti * Mi)
where Ti is the nodal temperature and Mi is the shape function for temprerature DoF.
b) What about Temperature? Would there be approximation function for temperature DoF too? (some thing like T(x) = sum(Ti * Si) + sum(ui * Qi) ??
c) What will the order of the element be for the 2 noded element and the 3 noded element?

Please help me out with this. Thank you