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 summary of the methodology is: Continue reading

Things to take into account when making your own convertible car

Last February I participated in the Young Persons’ Lecture Competition, organised by IOM3. In particular, the local heat took place at the University of Surrey. I want to share with you the transcript of my presentation. I have to say that I tried to present a quite complex topic in a very simple way so that anybody without  an engineering background could follow it. Hope you enjoy it!


Abstract: What would happen if you removed the roof of your car? First of all, you would have a convertble vehicle to enjoy that one sunny day we have in England. Second and most importantly, you would probably be the bravest person on earth. Driving on a bad road or even going over a speed bump could have dramatic results. Using simple engineering concepts, logic and a shoe box you will be able to understand why that could happen ad how automotive companies overcome this issue.

Let’s start from the beginning. What is Strength of Materials? It is the science that studies the behaviour of solid objects when they are subjected to stresses and strains. So, first question: what kind of objects? Basically we can have 1D, 2D and… Exactly! 3D elements! Some examples could be a bar (1D), a shell or a plate (2D) and a hexaedron (3D). For this particular topic, I’m going to focus on 2D elements, since the body panels of a car can be considered as very thin shells assembled together.

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Graphene-based composite nanomaterial can prevent the overheating of lithium-ion batteries

Although the performance of lithium-ion batteries (LIBs) have been remarkably improved in the past decades, there is a big risk in the use of this type of battery: they can catch fire when they are subjected to abuse. Researchers from Stanford University have developed a nanocomposite material which can be included into the electrodes in order to prevent the explosion of the battery.


To perform in an efficient way, LIBs require operation conditions which are within a specific range of current density, voltage and temperature. Nevertheless, when they are subjected to abuse conditions, exothermic reactions can take place, leading to a fast increase in internal temperature and pressure. What does it mean? Well, our battery is likely to explode!

Current LIBs include external sensors to prevent overheating and overpressure but, unfortunately, temperature and pressure inside the cells can actually increase much faster than they can be detected by those external sensors. Because of that, several alternatives have been developed in order to include internal components to solve the problem. For example, ceramic coating has been proven to be an effective way to shut down the battery and improving the thermal tolerance. However, after the battery is shut down, it cannot be used again. Using solid-state electrolytes can be another option, but the overall performance of the battery is decreased due to their low ionic conductivity. Continue reading

Mechanics of composites

A while ago, I wrote a simple document for undergraduates in order to explain that composite materials can fail in different ways. This was created as a high level document which could be used to find useful references with regards to failure modes, basic failure criteria and damage propagation models. I wanted to share this with you in case you are new in this field or just if you simply want to learn some basics of composites!


A composite can be defined as a material which is composed of two or more constituents of different chemical properties, with resultant properties different to those of the individual components. They usually consist of a continuous phase (matrix) and a distributed phase (reinforcement). These reinforcements can be fibrous, particulate or lamellar and they are usually stiff and strong, so that they are responsible for providing the stiffness and the strength of the composite. On the other hand, the matrix provides shear strength, toughness and resistance to the environment.

Fibre reinforced composites are considered as the strongest and sometimes also the stiffest, due to:

  • Alignment of molecules or structural elements.
  • Very fine structures.
  • Elimination of defects.
  • Unique structures.
  • Statistical factors.

With regards to fibre reinforced composite materials, their main failure modes are:

  • Fibre failure induced by tension in fibre direction.
  • Fibre failure induced by compression in fibre direction.
  • Matrix fracture induced by tension.
  • Matrix fracture induced by compression.
  • Delamination

It is remarkable that fibre failure typically caused composite failure, whereas matrix failure may not cause the same drastic effect. Continue reading

Carbures is back on track

After a relatively long period of instabilities, the Spanish composite manufacturer Carbures is raising again. This is really good news for the Spanish industry and for all those engineers who are interested in this kind of advanced material.


It’s just been made public that in 2016 Carbures reached their historic record in terms of the production of aircraft components made of composite materials. As a matter of fact, their production has increased 16.2% with respect to 2015, manufacturing a total of 45,695 aircraft parts. Therefore, we can say that Carbures have returned to the place where they belong: being one of the top composite manufacturers for the European aerospace and defence sectors.

For those who don’t know the company, they produce structures for quite a few members of the Airbus fleet. For instance, some of the civil airplanes which use their components are: A320, A320NEO, A330, A340, A350 or even the impressive A380. In addition, they also contribute to the military sector (e.g. A400M). The parts which are manufactured by Carbures include from lids of the oil tanks of the engine to parts of the fuselage. Continue reading

D is for “Ductile Damage”

The FEA dictionary is back and it’s time for letter D! Today I will introduce you to one of the methods to introduce damage in your material models.


Although it was created based on the failure of metals, this damage model can be used to introduce the degradation of mechanical properties for other types of materials. This option is available in Abaqus/Standard and Abaqus/Explicit and it requires the definition of the ideal elastic-plastic behaviour of the material, a damage initiation criterion and a damage evolution response. Please note that if any of the requirements cited before is not defined, the material properties will not be degraded.

In Abaqus there are different options for the damage initiation criterion and basically they can be classified as follows:

  • Criteria for fracture of metals (ductile and shear).
  • Criteria for necking of sheet metal.

Continue reading

Geneva International Motor Show 2017

Last month I travelled to Geneva (Switzerland) in order to attend the most important motor show of the year and here you will find some of the best pictures that I took during the event.


IMG_1325For those of you who are not familiar with the automotive world, I should start this post by stating that during the year there are several events known as “motor shows”, where Original Equipment Manufacturers (OEMs) exhibit their new vehicles and technologies. Since not all OEMs go to every event, there are some motor shows which are flagged in every calendar due to their relevance in terms of the companies and public that will be attending. In these terms, the Geneva International Motor Show is considered as the most important event of the year, followed by Frankfurt.

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Recycling of Carbon Fibre Reinforced Polymers

The use of carbon fibre reinforced polymers (CFRP) is increasing every day. This type of material have been used in aerospace and automotive industries (amongst others) for years, but now the cost of manufacturing components made of carbon fibre is becoming more accessible for mass production and more companies are introducing CFRP parts in their products because of their good mechanical properties, energy absorption capability and low weight. However, since a large increment in the production is observed, companies need to be aware of the different recycling techniques that are currently available for these materials.



Nowadays there are different ways to recycle composite materials and some of them are more developed than others. However, the use of recycled carbon fibres (rCF) is not that common in industry, mainly because of the lack of confidence in their performance when compared to virgin carbon fibres (vCF). In addition, there is a clear disadvantage: rCFs cannot be used for the same applications as what they were originally designed for. Because of this, I want to introduce some of the recycling techniques which are currently available for composites.

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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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B is for “bar” and “beam” elements

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.


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.

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