Finite Element Analysis (FEA) is a numerical method which provides solutions to problems that would otherwise be difficult to obtain. In terms of fracture, FEA most often involves the determination of stress intensity factors. FEA, however, has applications in a much broader range of areas; for example, fluid flow and heat transfer. While this range is growing, one thing will remain the same: the theory of how the method works.

The most efficient method of learning is by example. Therefore, I would like to present to you a simple FEA problem: the case of a three-member truss. The method of solution to this problem should demonstrate the basic concepts of FEA which are present in any analysis.

Before introducing specific quantities for our example, let's first take a look at our structure:

The overall objective of our analysis will be to determine the displacements of the truss members given the load P.

The first thing we must do is choose our elements. For our situation this is easy: each truss member should be one element. Further division would accomplish nothing, since each truss member can only support axial loads.

Let us now examine a single truss member:

(more info. on nodes)

Nodes are located at each end of the bar, each of which can have displacements in the x and y directions. The displacements are denoted u₁, u₂, u₃, and u₄. Corresponding forces due to these displacements are F₁, F₂, F₃, and F₄. The bar has a uniform cross-sectional area A and Young's Modulus E. The general relationship between force and displacement is F_i = k_ij*u_j, where F_i is the force in direction i, u_j is the displacement in direction j, and k_ij is the "stiffness" coefficient relating F_i to u_j. In our particular example of a horizontal truss element, we have the following system of equations:

The matrix k_ij is called the " stiffness matrix." It is the matrix which defines the geometric and material properties of the bar. Stiffness matrices are a fundamental part of FEA. These matrices always define inherent properties of the system being studied. For the system at hand, we need to determine the stiffness matrix. The way we will go about doing this may seem a little strange at first, but try to follow the reasoning as it does make sense. Let's begin by assuming u₁ = 1 and u₂ = u₃ = u₄ = 0. Then our matrix takes the form:

Each force F_i is equal to k_j1. Now, recall from mechanics of materials that the displacement of a rod is given by u = FL/AE. With displacement u₁ = 1, force 1 is F₁ = AE/L. To maintain equilibrium, we must also have a force F₃ = -AE/L:

Since our F_i's equal our k_i1's, we have:

It important to remember that our element can support only axial loads. Therefore, displacements u₂ and u₄ can not give rise to stresses in the bar since these displacements are perpendicular to the axis of the bar. Thus, the stiffness coefficients of these displacements must be zero: k_i2 = k_i4 = 0. Finally, a displacement u₃ = 1 will result in forces just opposite to those from u₁ = 1, so k_i3 = -k_i1. Our stiffness matrix is:

It must be emphasized that the stiffness matrix just derived is only valid for bars parallel to the x-axis. Through a similar derivation it can be shown that the stiffness matrix for any bar oriented at an angle "theta" to the x-axis is:

where c = cos"theta" and s = sin"theta". Note that when "theta" = 0, this stiffness matrix reduces to the one we derived for a horizontal bar.

Now knowing the stiffness matrix for any axially loaded bar, we can apply it to a real situation with specific quantities. Consider the following truss:

The displacements and external forces are:

Note the symbols we are using: R is an external force on the truss; F is an internal force resulting from the stresses imposed on the structure during a displacement. Knowing the orientations of each element, we can set up matrices for them. Using "theta" = 90 degrees for element 1, "theta" = 135 degrees for element 2, and "theta" = 0 degrees for element 3 we obtain the following matrices:

Element 1:

Element 2:

Element 3:

We can now generate a set of equilibrium equations for each node. Consider the following figure:

The nodal forces (resulting from element displacements) must be equal and opposite to the externally applied forces. Note that we have all forces drawn in positive x and y directions. Thus, for equilibrium at node 1:

Now recall what we are trying to do here: given a load P, we want to solve for the displacements at each node. Observing that node 2 is pinned and that node 3 is on a roller, the displacements u₃, u₄, and u₅ must equal 0. These values are quite important because without them we wouldn't be able to solve the problem. As a matter of fact, values such as these are always needed in finite element analyses; they are known as " boundary conditions." Next, we must state the reactions which are known from our particular loading. We can see from the truss that R₁ = 0, R₂ = -P, and R₆ = 0. Entering the known displacements and reactions into our matrix we get:

This matrix reduces to:

We can now finish our problem by solving this matrix for u₁, u₂, and u₆:

For Virginia Tech engineering students interested in FEA, one undergraduate course is available: ESM 4734 - An Introduction to the Finite Element Method. In this course, students study the theory and application of FEA to problems in various fields of engineering and applied sciences. The pre-requisite for the course is: ESM 2074 - Computational Methods.

References

Finite Element Analysis on Microcomputers, Nicholas M. Baran, McGraw-Hill Book Company, 1988.

Finite Element Primer, Bruce Irons and Nigel Shrive, John Wiley & Sons, Inc., 1983.

Table of Contents

Submitted by Jason Midkiff

Virginia Tech Materials Science and Engineering

http://www.eng.vt.edu/eng/materials/classes/MSE2094_NoteBook/97ClassProj/num/midkiff/theory.html

Last updated: 4/29/97