Engineering analysis + design software

User Area > Advice

Isoparametric Finite Element Analysis Formulation

The isoparametric formulation of finite elements is recognised as being most effective in practical analyses, having the following advantages

  • It readily allows the coordinates and displacements at any point within an element to be found using an assumed displacement and coordinate variation

  • Different element shapes and numbers of nodes are easily generated

  • Material and geometric properties may vary over an element and are properly accounted for by use of appropriate values at the element Gauss Points

The isoparametric formulation uses the following two relationships

         {u} = [N}{d}

Where:

{u} are the displacements at any point within an element

{d} are the displacements at the nodes of an element

This relates the displacements at any point within an element to the nodal displacements according to an assumed displacement variation ([N])

         {x} = [N}{X}

Where:

{x} are the coordinates of any point within an element

{X} are the coordinates of the nodes of an element

This relates the coordinates at any point within an element to the nodal coordinates according to an assumed displacement variation ([N])

The matrix [N] is known as the shape function or interpolation matrix and is a function of the natural coordinate system used. It is used in equations (1) and (2) to determine the variation of both the coordinates and the displacements within an element. That is, for a geometrically well-defined element shape, the shape function matrix establishes a unique relationship between the displacements and coordinates at any point within an element with those at the nodal positions.

As an example, consider how the stiffness matrix of a 3-noded bar element is generated with the isoparametric formulation. This element has an axial degree of freedom at each node only and a natural coordinate system based on x is used as follows

 

By the use of a quadratically varying shape function assumption the element can exactly model a quadratically varying displacement (and coordinate) field. Such a variation is given as

 

If the strains are calculated as

 

From the relationship {u} = [N]{d}, we have

 

Where [B] defines the “strain-displacement” matrix as

 

However it has been previously seen that [N] is specified in terms of x rather than x. Therefore, using the chain rule, this may be written as

 

Giving

 

The term  is worthy of closer inspection since it turns out to be of significance in the finite element method. Using {x} = [N]{X}, we have

  If x2 is (reasonably) assumed the midpoint, then x2 = x1+L/2 and x3 = x1+L, so that

Reducing to

 

Mathematically, dx/dx represents a “scale factor” between the two coordinate systems and is commonly termed the Jacobian. This is an important association and is commonly denoted as [J], so that, for this 3-noded bar example

 

In this case [J] is a simple value for the length of the bar, although in general, it will be a matrix with a dependency on the natural coordinate system.

The strain-displacement matrix defined above is then given by

 

And the strain as

 

Indicating that, although a quadratic shape function will simulate a quadratic displacement distribution across an element, the strain (and, therefore, stress) will actually vary linearly. Similarly a linear shape function will lead to constant strain values within an element. For more information on shape functions click here.

The element stiffness matrix is defined as

 

Where [D] is called the “modulus” or “material stiffness” matrix and, in the case of the bar element, is defined as

Where E is the elastic modulus for the material. Normally the stiffness matrix would be evaluated using numerical integration. For now the solution is obtained analytically as follows

Since

 

When, when expanded, gives

After multiplication of the separate matrix terms, we have

 

Integrating with respect to x gives

 

And evaluating over the natural coordinate limits of 1, yields

Which, after simplification, gives the element stiffness matrix as

To check that this stiffness matrix produces the expected structural response, consider the 3-noded bar (diagram at beginning of this page) of cross sectional area (A) =1, elastic modulus (E) = 100 and length (L) = 10, axially supported at node 1 and subjected to a concentrated axial load of 5 at node 3. If the displacements at nodes 1, 2 and 3 are d1, d2 and d3 respectively, the element displacement vector becomes

 

Where d1 is set to zero as a result of the applied support condition. The element force vector specifies the applied load at node 3 together with the reaction force (R1) at the supported node 1 as

 

Given the static equilibrium equation

This gives

Extracting the three equations as

 

Solving for d2, d3 and R1 gives

 

As expected.


Finite Element Theory Contents

Local Coordinate System

Shape Functions

Numerical Integration

The Jacobian Matrix


innovative | flexible | trusted

LUSAS is a trademark and trading name of Finite Element Analysis Ltd. Copyright 1982 - 2022. Last modified: November 29, 2022 . Privacy policy. 
Any modelling, design and analysis capabilities described are dependent upon the LUSAS software product, version and option in use.