Simple Rod Example: Obtaining Nodal Displacements Using Implicit Methods

A simple example of a rod, constrained at one end and loaded at the other end is used to introduce the basic concepts of FEA.

The objective of the analysis is to find the displacement of the free end of the rod, the stress in the rod, and the reaction force at the constrained end of the rod.

The first step of any finite element simulation is to discretize the actual geometry of the structure using a collection of finite elements. Each finite element represents a discrete portion of the physical structure. The finite elements are joined by shared nodes. The collection of nodes and finite elements is called the mesh. The number of elements per unit of length, area, or volume is referred to as the mesh density. In a stress analysis, the displacements of the nodes are the fundamental variables for which a solution is required. Once the nodal displacements are known, the stresses and strains in each finite element can be determined easily.

Consider a simple example of a rod, constrained at one end and loaded at the other end. The rod is modeled with two truss elements, which can carry axial loads only.

Axially-loaded rod

The discretized model with the node and element labels and the free-body diagrams for each node in the model are shown below.

Discretized model of rod

Free-body diagram for each node

In general each node will carry an external load applied to the model, P, and internal loads, l, caused by stresses in the elements attached to that node. For a model to be in static equilibrium, the net force acting on each node must be zero; that is, the internal and external loads at each node must balance each other. For node a this equilibrium equation can be obtained as follows.

Assuming that the change in length of the rod is small, the strain in element 1 is given by

ε11=ub-uaL,
where u a and u b are the displacements at nodes a and b , respectively, and L is the original length of the element.

Assuming that the material is elastic, the stress in the rod is given by the strain multiplied by the Young's modulus Ε.

σ11=Ε11ϵ11

The axial force acting on the end node is equivalent to the stress in the rod multiplied by its cross-sectional area, A. Thus, a relationship between internal force, material properties, and displacements is obtained.

Ia1=σ11A=Eε11A=EAL(ub-ua).

Equilibrium at node b must take into account the internal forces acting from both elements joined at that node. The internal force from element 1 is now acting in the opposite direction and so becomes negative. The resulting equation is

Pb-EAL(ub-ua)+EAL(uc-ub)=0.

For node c the equilibrium equation is

Pc-EAL(uc-ub)=0.

For implicit methods, the equilibrium equations need to be solved simultaneously to obtain the displacements of all the nodes. This requirement is best achieved by matrix techniques; therefore, we write the internal and external force contributions as matrices. If the properties and dimensions of the two elements are the same, the equilibrium equations can be simplified as

{PaPbPc}-(EAL)[110121011]{uaubuc}=0.

In general, it might be that the element stiffnesses, the E A / L terms, are different from element to element; therefore, write the element stiffnesses as K 1 and K 2 for the two elements in the model. We are interested in obtaining the solution to the equilibrium equation in which the externally applied forces, P , are in equilibrium with the internally generated forces, I . When discussing this equation with reference to convergence and nonlinearity, we write it as

{ P } - { I } = 0.

For the complete two-element, three-node structure we, therefore, modify the signs and rewrite the equilibrium equation as

{PaPbPc}-[K1-K10-K1(K1+K2)-K20-K2K2]{uaubuc}=0.

In an implicit method, this system of equations can then be solved to obtain values for the three unknown variables: u b , uc, and Pa ( ua is specified in the problem as 0.0).

Once the displacements are known, we can go back and use them to calculate the stresses in the truss elements. Implicit finite element methods require that a system of equations is solved at the end of each solution increment.

Note: In contrast to implicit methods, an explicit method, does not require the solving of a simultaneous system of equations or the calculation of a global stiffness matrix. Instead, the solution is advanced kinematically from one increment to the next.