We’re going to introduce in this topic the two key variables required to define the fundamental behaviour of materials. These variables are stress and strain.

In many engineering applications it is required to assess how a structure deforms or behaves under a load. Making such an assessment requires some knowledge about the material constituting the structure itself. For example you couldn’t sit on the chair made of rubber. You need a stronger material like hard plastic or timber. So when it comes to characterising a material there are two questions we need to answer. One is, how much does a material deform under a load? The second questions is, how much load does it take for the material to break? And to answer these questions we need to define stress and strains. Let’s see why with an experiment.

# Strain Experiment

We take a short rubber band and we’re going to place this rubber band in intention. We start by measuring its initial length, which comes to 54 millimetres and we place weights on this hanger to stretch the rubber band. It now stretched to 62 millimetres. Let’s write these results on the board. So we have a short rubber band. We have an initial length of 54 millimetres. A final length of 62 millimetres. That give us an elongation that I will call delta L of 8 millimetres. Not let’s repeat the experiment with a longer rubber band.

So, again we place the hanger on the band and measure the initial length. This comes to 102 millimetres. We place the weight on the hanger and stretch the rubber band and measure the final length again and it comes at about 117 millimetres. Let’s write the results on the board. So we have the long rubber band. The initial length is 102 millimetres and the final length is 1117 millimetres. That gives us an elongation delta L of 15 millimetres. What that shows is if you use the elongation or changing length to characterise the materials behaviour you don’t have the same number. However it is the same material. It’s the same rubber. The material has the same width, the same thickness but we have different length and we have different elongations. So we can’t use elongation to characterise the material’s behaviour. However let’s see what proportion of the initial length, L not this delta L represents. If we divide the change in length by the initial length for the shorter of the band we get 8 millimetres over 54 and that’s about 0.15 or 15%. For the long rubber band we have delta L over L not equal 15/102 and that’s about 0.15 or 15%. We now see that the numbers do not match. We now see characterised the material’s behaviour, its capacity to stretch under a load regardless of the geometry of the specimen we’ve tested in this case the initial length. These parameter delta L over L not I not is noted epsilon usually in mechanics and this is a strain. Epsilon is a length over a length and hence, as no unit.

# Stress Experiment

Now let’s perform another experiment regarding stresses. In this experiment, we’re going to pull on two polystyrene specimens of different cross sections and we’re going to bring them to failure. What we’re going to use for this experiment as well is a dynamometer which is graduated in new tons that will give us an indication of the load it took to actually break the specimens. So let’s start with these the specimen having the smallest cross section. So there is a system inside the dynamometer that records the maximum load we’ve reached. I will now pull on it and break it. Here we go. Let’s see now what load it took to reach failure. It’s about 42 new tons as you can see there. let’s write the results on the board. So we have the small specimen and the force to failure is 42 new tons. Now let’s repeat the experiment with a larger specimen.

We use a dynamometer again and similarly I’ll pull until we’ve reached failure. Here we go. We now have reached about 85 new tons. Let’s write the result on the board. Once again, we see that the numbers do not match. If we use the force as an indication of resistance of the material we cannot basically tell how strong the polystyrene is without considering the cross section of the specimen. To some extent, these results aren’t surprising because we have different cross section. There is more material to break on the specimen than on this one. So it’s not a surprise to have a larger number here because it’s actually all this material that contributes to the resistance.

Let’s do a simple calculation and divide this force by the cross sectional area of the specimen. For the small specimen F over A is 42 new tons divided by 0.01, which is 10 millimetre thickness times 0.015 which is 15 millimetres across and that gives us about 280,000 new ton per metre squared. These numbers are expressed in metres. Now let’s do the same calculation for the large specimen. We now have F over A equal to 85 divided by 0.01 which is, again, the thickness of the specimen times 0.03, which is the width of the specimen and that gives us about 280,000 new ton per square metres. So we now see that the two numbers do match. We’ve managed to quantify the resistance to the polystyrene regardless of the geometry of the specimen in this case the cross sectional area. So for a one dimensional tensile test, the force divided by the cross sectional area F over A I called the stress noted sigma and a stress, as you’ve seen here as a unit of newton over an area metre square or also called pascal and noted Pa. Stress and strain a critical in engineering because of they help quantifying a material’s behaviour regardless of the geometry of the specimen tested. So initial lens, for example, diameter or cross sectional area.

In these experiments, we’ve used only simple one dimensional testing elongation or breakage, but in engineering in many cases, the stress distribution is far more complex and it’s often a 3D situation. In that case defining the stress and strain is far more complex than we’ve done here, but this is not covered in this course. I invite you to refer to the additional resources if you want more information on this topic.