ENGINEERING BREAKDOWN

New post about FEA! In this occasion, I bring you some theoretical background for two types of elements which are very useful for modelling certain structures: bars and beams.

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Let’s start from the beginning. A bar is basically an element which can resist only axial loading. Therefore each of its nodes has one degree of freedom, i.e. a displacement along the longitudinal axis of the element. On the other hand, each node of a beam presents not only one but three degrees of freedom: displacement in the longitudinal axis, displacement in the transverse direction and rotation.

As you should already know, Finite Element Analysis requires the a stiffness matrix (K) so, in order to illustrate this, in this post I will show you how the K matrix can be derived for bar elements. Note that the process for obtaining the stiffness matrix of a beam element is similar but a bit more tedious.

Consider a bar with two nodes and one axial force acting on each node. Hence, it will result in two displacements as shown in the figure.

If we now consider just a part of the element, on one side we’ll have node 1 (where force 1 is acting) and another internal force on the other end, due to equilibrium. That internal force can be expressed as the stress multiplied by the cross sectional area and, in a similar way, the stress can be written as a function of the Young’s modulus and strain. Finally, the strain can be determined by differentiating the displacement with respect to coordinate “x”. Nothing too complicated I hope! Then, applying equilibrium, it is easy to derive the following equation:

Integrating the previous equation, we obtain the following:

Now we have to determine the value of the constant. In order to do so, we just need to apply the boundary condition from node 1. Hence:

Then, using the second boundary condition, the expression for the first force is obtained:

Finally, applying equilibrium to the overall component, the second force is easily determined:

Rearranging both expressions in matrix notation, we obtain the usual expression for bar elements in Finite Element Analysis: 

After saying all of this, I should probably mention that in Abaqus, bar and beam elements are created using a simple line. Then in the “Property” module, the corresponding section should be defined: “truss” for bars and “beam” for the other case.. In addition, if beam elements are used, it is extremely important to check their orientations. That can be easily done in the module which was mentioned above. Finally, Abaqus also gives the option to visualise the beam profile. For instance, if you are modelling an I-beam, you can see the complete section modifying the “View” options on the upper toolbar. Although users can also plot the results on the beam profile, it is not recommended, since it can lead to a wrong interpretation of the results.

To conclude with this post, I would like to state that depending on the analysis, we need to decide which type of element would be the most appropriate for our particular case. For instance, if you are simulating the response of a reinforced concrete beam, the reinforcements are usually modelled as truss elements, since their mission is to resist tensile loads. On the other hand, if you have a complex assembly and you want to include a fastener, what most FE analysts do is to use a beam element and apply rigid connections between the parts.

I hope you enjoyed the post and if you still want more, I will be back soon with a practical example for modelling a simple Cantilever beam! Cheers!