Tag Archives: Cantilever

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.

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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.

Consider the structure from Fig.1 and try to analyse it with simple structural mechanics. Firstly, we need to determine the reaction forces and moment at the fixed end caused by load P. Applying equilibrium as shown in Fig.2, the following results are found:

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Figure 2. Equilibrium applied to the Cantilever beam

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Then, applying equilibrium to a portion “x” of the beam from the fixed end (see Fig. 3), the stress distribution can be easily calculated by applying, once again, equilibrium:

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Figure 3. Equilibrium applied to a section of the Cantilever beam

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While the axial stress is null and the shear stress is constant along the length of the beam, the bending moment depends on the position. This means that the maximum moment that the structure experiences is located precisely at the fixed end, rather than at the loaded point! Therefore, the fixed end is the region which suffers the most, resulting in designers adding extra material or other types of reinforcements in that area, in opposition to what other people mights have thought! If you don’t believe this, you can always hold a small stick in one had and apply a vertical displacement on the free end with your other hand: the stick will bend and break at the end that you’re holding!

My point with this post is that this simple question can be asked to you in an interview, and you should not jump into the first “logical” response that comes into your mind. Just spend some time thinking about the problem and you’ll see how basic structural mechanics can help you! And you thought you would never use all these basic theories you were taught at university…

C is for “Cantilever beam”

As I promised a few weeks ago, I’m back with a tutorial! In this occasion, a cantilever beam will be modelled in Abaqus/Standard. What is more, the importance of defining a good mesh (not only the element size matters!) will be illustrated with several examples.


So, first things first. A cantilever beam is a structure which has one of its ends fully constrained. This means that all degrees of freedom are restricted. An example is presented in the following figure:

beam

A similar case as the one presented above will be modelled using the Finite Element package Abaqus/Standard. The structure can be created using different types of elements: beam, bar, solid or shell. I have decided to model it using shell elements, since it will allow me to show the influence of the element size and type in a very simple way.

In order to create the component we need to be in the Part module. We should select the 3D deformable shell (planar) option. Then we will be able to create the geometry. In this case a simple rectangle of 20mm x 100mm will be enough. Please bear in mind that Finite Element codes are unitless and all the parameters need to be defined in a consistent system of units. I have decided to use mm, GPa and kN.

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