Theory of Finite Element Analysis

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:

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 Fi = kij*uj, where Fi is the force in direction i, uj is the displacement in direction j, and kij is the "stiffness" coefficient relating Fi to uj. In our particular example of a horizontal truss element, we have the following system of equations:

F1 = k11u1 + k12u2 + k13u3 + k14u4

F2 = k21u2 + k22u2 + k23u3 + k24u4

F3 = k31u1 + k32u2 + k33u3 + k34u4

F4 = k41u1 + k42u2 + k43u3 + k44u4

Alternatively, in matrix form:

The matrix kij 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 Fi is equal to kj1. 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 Fi's equal our ki1'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: ki2 = ki4 = 0. Finally, a displacement u3 = 1 will result in forces just opposite to those from u1 = 1, so ki3 = -ki1. 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:

x - direction: R2 - F2(element3) - F2(element2) = 0

y - direction: R1 - F1(element3) - F1(element2) = 0

We want to solve for R1 and R2. Obtaining the nodal forces F2(element3), F2(element2), F1(element3), and F1(element2) from our previously determined matrices we get:

R1 = AE/L ( 3u1/2 - u2/2 - u3 - u5/2 + u6/2 )

R2 = AE/L ( -u1/2 + u2/2 + u5/2 - u6/2 ).

Similarly, from equilibrium of nodes 2 and 3 we obtain:

R3 = AE/L ( -u1 + u3 )

R4 = AE/L ( u4 - u6 )

R5 = AE/L ( -u1/2 + u2/2 + u5/2 - u6/2 )

R6 = AE/L ( u1/2 - u2/2 - u4 - u5/2 +3u6/2 )

We can now combine all of our external forces into one matrix:

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:

u1 = - PL/AE

u2 = -4PL/AE

u6 = -PL/AE

This application of FEA to a simple three-member truss shows in general how the method works. Most applications to engineering problems, however, are much more complex. Such analyses require large numbers of elements and nodes in order to accurately represent the physical system being studied. These analyses inevitably require the application of a computer.

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.

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