When we want to proceed with the dimensioning of steel beams, we are going to proceed in a way which is similar to what we have seen with concrete before. It is important - we are going to see why - to use Newtons and millimeters as units, that is a bit different from what we usually have but that is not very serious, and then we will look a bit at the dimensions of these steel shapes. We are going to start with exactly the same condition, we are going to predimension a steel beam with large flanges, which has a span of 10 meters, a depth of 1.32 meters, we do not know the width of the flanges. Here you go, we have our steel beam. The first thing we are going to do is to put the screen to scale. This time we do not have a distance of ten or or one hundred but a distance of ten thousand between the supports. We are now going to insert the forces, the forces have a value of one hundred kiloNewtons every meter, that is to say one hundred thousand Newtons. You can see that the applet always automatically puts the forces to scale, se we cannot see when we insert forces if they are ten, one hundred, or one thousand times too large. So, we have to be careful, I insert here the left support and then with control, the right support. I am now going to define the material properties. We have said that it has a strength of 235 Newtons per square millimeter, and then I am going to give it a width of 300 millimeters, at first, I activate the resolution pushing control and we can see something very thin which appears and if I make it go down into the cross-section, here I can see that I would only need, for the width of 300 millimeters which I have indicated, a steel thickness of approximately 13.5 millimeters. We are going to be able to do something a bit more efficient, probably decreasing the width: instead of a width of 300 millimeters, I put 100 millimeters. And then I restart the resolution, we can indeed see a slightly larger thickness, and I can move the supports in such a way that the internal forces remain inside the cross-section. Here you go, we can see that I only need now an upper flange with a width of 100 millimeters and a thickness of approximately 41.5 millimeters. So I could order a steel beam asking for a 41 or 42 millimeters thick upper flange and probably approximately the same thing for the lower flange. That would absolutely be possible. However, for steel shapes, cross-sections are standardized. Here, I show you a part of the Swiss table but that is available almost everywhere in the world for the double T cross-sections. There are many more, in some countries, there are many more than in Switzerland, so it depends on the countries but what we can very clearly notice is that there are nominal depths which vary between one hundred and one thousand millimeters so up to one meter. The profile I had before - 1.32 meters - that is not standard, that is not going to be possible. And then we have three range of weigths: HEA, HEB and HEM, if we take one of 320, you can see that the HEA weighs 97.6 kilos per meter, the HEB 127 kilos and the HEM 245 kilos so a very big difference of weight for these different shapes. We will see that it also changes their efficiency. So, for the pre-dimensioning, we are going to have at once a variation of the depth and of the thickness and then we are going to use a special grid for the free dimensioning of beams. If you start the i-Cremona applet, in the folder containing the common pictures, you will find at the beginning, this picture which is called "Abaque poutre". I have activated it here, I am now going to put the screen to scale, I have a maximum distance of 15 000 between this point here, for example, and this point there, so I put it to scale. Then, I am going to insert my forces, my forces have a value of 100 000 Newtons, I stay here in Newtons and in millimeters, which I always place in the middle of the interval, that is why we have these intervals, here, so I have 6, 7, 8, 9 and 10 forces. The supports are going to be located at the level of zero and of 10 000. I place them a bit higher, as you can see, to already take into account the position of the center of gravity of the lower flange and afterwards, I define the properties of my material. For example, here, a grade 355 steel with a width of 300 millimeters, which, you will see, is quite standard for what we want to do. I activate the resolution pushing control, and the depth, well, I am going to see, I could for example, take here a cross-section which is, let's say, 700 millimeters deep, so I am going to have here a depth of approximately 700 and here, it is said to me that with 700, I need to have a width of the upper flange... of both flanges, of about 17 millimeters. We are at the level of 700 millimeters, here and we want a thickness, the thickness can be read on T and we get roughly 17. So here, it would work with the HEA 700, but it has 27, so the depth is probably too large. I come back to my applet and I am going to decrease this depth to, let's say, 600. For 600 millimeters, I would need a width of 300 and a thickness of about 20 millimeters according to the applet. I am now at 600 millimeters, I need a width of 300, we can see that the width 2c is 300, that is constant over a large portion so it is no coincidence that I have chosen this value And then, we have a thickness of 25 and, what has been asked to me was a thickness of 20, so that is still too large. We are going to move to a solution with a depth of 500 millimeters. With 500 millimeters, I would need a thickness of approximately 25 millimeters. 500 millimeters here, I would need a thickness of 25 which does not work with the HEA, so I can have a look at the HEB, this would be a good solution. This is the HEB 550 and it weighs 187 kilos per meter, compared to the HEA 600 which weighs 178 kilos per meter. So, if we have a bit a space, we can easily use the HEA which makes us save some material. The question is, what happens - could we only have, for example, a beam depth of 400 millimeters? I come back here, I activate the funicular polygon and I try to go to 400 millimeters, that is here and we start looking at the thickness, which is significant, I need a thickness of approximately 31 millimeters. At the level of 400: here, I have 19 millimeters, it is not going to fit, but however, here, it would work with a HEM 400 which however weighs 256 kilos so much it is heavier than the previous value. We could maybe choose a smaller HEM but we can see that the weight remains almost constant. What is interesting - I am maybe going to take off the first solution I have put because it was far too conservative. We can have the orange solution with a HEA which will be a HEA 600, or the pink solution with a HEB 500, or else finally the blue solution with a HEM 400. So, the process of dimensioning of a steel double T cross-section consists in the choice of a cross-section according to its efficiency and its weight. Finally, we can also pre-dimension a composite steel-concrete beam, like I showed you in the lecture about the superimposed beams, so we have a steel beam in the lower part, a system of connecting dowels and then a concrete slab in the upper part. Look, this is schematically drawn here, obviously, what we do not know is the width of this concrete slab and also its thickness. Here we have the applet which has already been put to scale, in meters by the way since we are going to work with concrete and in which I have already inserted the forces. So, we are now going to define these dimensions, I am going to consider a strength for the concrete of 20 megaNewtons per square meter and a width of 0.8 meters like before, like for the T cross-section and I am going to activate the resolution with the funicular polygon and here once again, like for the T cross-section, what we can see is that we only need, for a width of 0.8 meters, about 60 millimeters of compression so actually this slab here could be thinner, it corresponds well, by the way, to the principle of composite construction in which we try to have relatively thin concrete slabs. The part in tension, that is interesting, the tension is not only supported by the lower part in this case, we can absolutely consider that the center of gravity of the lower part of the cross-section is where the internal forces are going to be supported so the dimensioning is a bit more complicated in this case here, we are not going to get into the details but these 1.25 megaNewtons can be carried by a considerable amount of steel which is placed in the lower part of the cross-section. In this video, we have seen how to proceed to the pre-dimensioning of various types of beams using the i-Cremona applet. We have seen that it is very important to use coherent working units, good units being megaNewton and meter for the concrete structures, Newton and millimeter for the steel structures. we have applied this dimensioning method to various types of cross-sections, first to rectangular cross-sections for a concrete beam and then to T-shaped cross-sections then to double T shapes for the steel cross-sections as well as for a composite steel-concrete beam. We have seen that the predimensioning can be done quite quickly, I draw your attention on the fact that the final dimensioning will have to take into account other parameters which are not dealt with within the framework of this introductory course.
