When designers and engineers are selecting proper materials to build something they have to pay special attention to various aspects, like their strength, durability, flexibility, and if they are able to withstand the different substances they will be exposed to.
A very important characteristic of materials is how much they will deform when being pulled, crushed or bent in the building or device they are a part of. These stresses exert forces on the bonds between the atoms that compose a solid object, changing the distances between them or even changing their positions. These changes translate into a macroscopic deformation, which can be temporary or permanent.
If a pulling force applied to the atomic bonds of a material is high enough, it can cause these bonds to break, which ends up in a macroscopic crack or even a fracture. Just before breaking, some materials elongate to a final length. This value determines the percent elongation of the material, which provides very valuable information for engineers to ensure the quality and durability of their designs. Let’s discover how to calculate this parameter.
How to calculate percent elongation
- Determine the initial length of the specimen of interest, l0.
- Perform a tensile test until the specimen breaks.
- Put both remaining pieces together and determine the specimen’s final length at failure, lf.
- Calculate the percent elongation using the following equation:
What is the percent elongation
Materials are made up of atoms. These bond to each other in different ways depending on the type of elements involved. The nanoscopic bond type can determine some of the properties of the material at the macroscopic level. Let’s see a few examples.
In metals, atoms set their uppermost electrons —called valence electrons— free, which can easily diffuse throughout the entire volume. This is why metals are usually very good electric conductors: once an electric potential is applied to a metal piece, free electrons can easily react to it by diffusing around.
On the other hand, some materials called ceramics tend to be electric isolators. In this case, almost all of the atoms’ valence electrons are bound to one of the elements composing the material. In consequence, electrons are not free to flow when an electric potential is applied. Since heat is also partly transmitted through the transport of excited electrons from the hot to the cold side of the material, ceramics are also bad thermal conductors.
In these two examples we can identify two specific types of bonds: the metallic bond and the ionic bond. Another important type of bond is the covalent bond, found in many organic materials. These three types of bonds are called the primary or chemical bonds atoms form in solids.
Each of these types of bonds possesses unique characteristics which, as shown before, can determine a material’s electric, thermal, optical and also mechanical properties. This last type of property refers to the behaviour a material exhibits when subjected to different types of forces.
Now, it is very useful to predict how a material will respond to forces, for example, when using it in the framework of a building, in a bridge, in furniture, or in an infinite amount of applications. Let’s discover some important types of mechanical properties of materials.
Imagine you pull an elongated metal piece from both ends. When you do so, the atoms on which the force is applied transmit it to their neighbouring atoms across the bond joining them together. These then transmit the force further until the entire sample is stretched. Finally, the force you apply on both ends is distributed among the bonds inside the sample and the resistance you feel is due to them withstanding it.
How much a material can withstand a pulling force depends directly on the type of bonds it forms. So, as we mentioned before, the nanoscale structure influences the macroscopic behaviour. In general, the stronger the atomic bonds are in a material, the more it will resist an external force applied on it.
Let’s go back to our metal sample. If you pull harder and harder you will probably manage to elongate it, causing it to deform. A sample’s deformation, whether temporary or permanent, is measured through its strain, which is defined as:
Where l0 is the sample’s initial length, and li the instantaneous length you measure when applying the force, as the following image shows. As you might have probably realized, strain is a unitless value that indicates how big the deformation is with respect to the original length of the sample.
If you think about it, stretching a thin piece of metal to a strain of 0,01 (1% longer than it originally was) is probably easier than stretching a thicker one to the same amount of strain. This is, of course, because there are more atoms bonded together in the latter, so the effect per bond is less than in the first case.
This is why forces acting on solid bodies are usually defined in terms of their cross sectional area. This is the area you measure when performing a transverse cut to the metal sample, like the following image shows.
The force acting upon a certain area is called a stress, which is defined as:
Where F is the applied force and A the cross sectional area it acts upon. Stress has units of N/m2, which are also called Pascal (Pa). Now let’s do a thought experiment to determine the relation between the stress a sample is subjected to and the strain it exhibits.
A tensile test is a type of characterization method performed on an elongated sample with a special geometry defined in a technical standard. In this test, the sample is fixed by both ends to a special machine which can both apply a pulling or tensile force and measure the resulting strain at the same time. The result of this type of test is a stress-strain curve, which looks like this for many materials:
This curve gives us very valuable information about the material’s mechanical properties. For example, for low values of stress, its relation to strain is linear, so that =E. This part of the curve is called the elastic region, since the material can stretch and then recover its original size after removing the stress. The proportionality constant E is therefore called the modulus of elasticity, or Young’s modulus. The higher its value, the steeper the linear region to the left of the curve, and the stiffer the material is.
In general, the higher the stress the sample is subjected to, the higher its strain will be. After a certain point, the groups of atoms that compose the material will accommodate at a microscopic level, and align with the applied force. This leads to unrecoverable deformations, since atoms have moved too far to go back to their initial positions. This happens in the right region of the stress-strain curve, and since deformations are permanent, it is called the plastic region.
In the plastic region, atoms will eventually align completely with the applied force, which means it will finally act entirely on the bonds between them. You can think of this as a bunch of intertwined strings whose ends are strapped together. If you pull from both ends, the strings will first align with the force before it starts acting parallel to each of them. See the following image.
This alignment process causes the material to deform less and less with increasing stress, due to the atomic bonds being very stiff. This leads to what is called work hardening or strain hardening of the sample. On the stress-strain curve this can be seen as a significant reduction of its slope. After this stage, the maximum force the material will withstand depends solely on how strong the atomic bonds are, which consequently depends on the type of bond in place (metallic, ionic, covalent, etc.).
Eventually, the material’s atomic bonds will not be able to withstand the applied force anymore and will break. This is what we call failure. Different materials can resist different amounts of stress before breaking. The highest point of the curve, called the ultimate tensile strength (UTS) indicates the maximum stress the material can withstand before failure, and it is a critical parameter for engineers and designers. Imagine you want to build a bridge. You certainly need to know how much weight your design will be able to resist if made of aluminum, steel, concrete, or any other construction material.
Just as materials possess a maximum strength, they also exhibit a maximum strain when finally breaking. Materials that exhibit high amounts of strain when subjected to a force are called ductile, and those that do not deform significantly are called brittle. If you want to experiment with the stress-strain curve, go ahead and check this simulator built by MechaniCalc, where you can input a material’s Young’s modulus (elastic modulus), yield strength and ultimate strength and generate the plot.
If we let our metal sample reach failure, and then take the two remaining pieces and join them together, we will be able to measure the final length it reached before breaking, as the following image shows. The corresponding strain, calculated with equation 2, is called the strain at failure or final strain, lf.
Since strain is expressed as a proportion of the original length, and is not an absolute value, it is useful to compare different materials. If we multiply it by 100%, we obtain the so-called percent elongation, which indicates how much a material will deform, as a percentage of its original dimension, before breaking. It is therefore a percent variation of the sample’s length.
Percent elongation formula
As mentioned before, percent elongation is a type of percent variation and, as such, refers to the proportion of the increase of a tensile test sample’s length to its initial length. In this case, we define an initial value (initial specimen length) and a final value (length of specimen at failure). Since we want to determine the variation from the initial to the final value, we must measure it with respect to the former. Percent elongation is therefore calculated as:
Where, l0 is the sample’s initial length, and lf its final length just before breaking. In order to apply this equation, the length of the test probe needs to be measured precisely according to a technical standard. Usually a caliper is used for this task.
Exercise 1: calculate the percent elongation of aluminum 6063-T5 alloy according to the following stress-strain curve.
Image edited from Wang, et al. (2016)
Answer: 7,2%
If a hanging bridge’s vertical supports are made of 2,3 m bars of aluminum 6063-T5 alloy, and its length is measured every week for safety, how long will the pieces be when needing replacement if the maximum allowed strain is 70% of the strain at failure?
Answer: 2,4 m
Sources:
Wang, Y.Q. & Zhongxing, Wang & Yin, F.X. & Yang, L. & Shi, Y.J. & Yin, J.. (2016). Experimental study and finite element analysis on the local buckling behavior of aluminium alloy beams under concentrated loads. Thin-Walled Structures. 105. 44-56. 10.1016/j.tws.2016.04.003.
