Tag Archives: Structural Design

Classic question about the use of a Cantilever beam for designing vehicle structures

Good afternoon everyone! I know it’s been a while since the last post but I’ve been (and still am) very busy with all kind of simulations, tests and writing papers and my doctoral thesis. Hopefully, I’ll manage to write some more articles during the summer! I recently had a conversation with some senior engineers from a F1 team regarding Cantilever beams and some erroneous assumptions which are commonly made, so I wanted to discuss it with you! Hope you enjoy this brief post!

A few weeks ago, I had the chance to speak with three top F1 designers and we had a chat about a certain question regarding the use of the Cantilever beam as a tool to design some vehicle structural components. First of all, let’s remind what this type of configuration is. A Cantilever beam is a structure which is fully constrained at one end, having a vertical load applied at the other end of the beam to study the effect of bending, as illustrated in Fig. 1.

Figure 1. Schematic of a Cantilever beam

This type of structure is very useful when designing certain components, since they can be simplified to this well-known beam, reducing the number of variables and being able to define simpler design targets. The thing is that usually, in reality, the components usually have some part of its length reinforced (e.g. thicker walls), so two questions arise: why is this non-homogeneous beam common and where should that reinforcement be placed?We agreed that a lot of people answer very quickly that it should be placed at the free end of the beam, i.e. where the load is applied. According to these people, the reason for this is pretty obvious, since that end will suffer the greatest deflection (I will write another post soon where I derive this and discuss some ways to calculate it by hand!). Hence, if that region was reinforced, the deflection would be smaller and the structure would be better in terms of bending performance. But, is this true? Let’s have a thought.

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How to solve a Finite Element problem using hand calculations

Basically, when we want to determine the forces and displacements in a certain structure using Finite Element Analysis (FEA), what we are doing is creating a system of equations that relates the stiffness of the elements to the displacements and forces in each node. When we run a simulation, we do not see all the calculations. For that reason, today I want to illustrate a simple case that can be easily solved by hand applying that methodology.

Before getting started, just think of a spring. Everyone has come across the Hooke’s law at a certain point during school. It states that the force in the spring is proportional to a constant “k” multiplied by the variation in length of the spring. FEA follows the same principle, but in this case the “k” constant is the stiffness matrix and the variation in length is a vector of displacements and rotations, depending on the case.

Let’s study a simple static case. Our structure consists of two bar elements connected at a common node, where a load “P” is applied. The other two nodes have both horizontal and vertical displacements constrained (see the boundary conditions). For this particular case, the reactions in nodes 1 and 3 and the displacements of node 2 are requested. I have solved the problem by hand following a few steps that, based on my experience, can be generalised for more complex problems. Pretty much, the┬ásummary of the methodology is: Continue reading