Hello. In this video, we will see how to proceed with the pre-dimensioning of a beam using the i-Cremona applet. You will see that it requires a more intense use of the applet than what we have done so far. In particular, we will have to be careful with units. We will have to specify the dimensions of the cross-section, and the properties of materials, at least, of one material per beam, and then we will see how we can choose the dimensions of a cross-section on the basis of this pre-dimensioning. So far, and in the rest of the course, we did not have to worry about units. That means that if you prefer to work in Newtons and millimeters, you can, but you can also work in Newtons and centimeters, in kiloNewtons and meters, and so forth. For the applet, this is always possible, but to avoid confusions, I recommend you 2 sets of units: either a set with Newtons and millimeters, so the spans, if we have a 10 meters span, we are actually going to have 10 000 millimeters, if we have a load of 100 kiloNewtons, that is 100 000 Newtons, a distributed load of 100 kiloNewtons per meter, it becomes 100 Newtons per millimeter, a strength of concrete of 20 Newtons per square millimeters, or here, that is 500 Newtons per square millimeters, that does not change. So this, that is the setup for steel or concrete beams, if you wish to work in millimeters in concrete, there is no problem. However, if you use other units, well, conversions will be required since the loads, for example, if you have them in kiloNewtons and the distances in millimeters, so you should have the strength in kiloNewtons per square millimeter, and this is a little problem. Another solution is to use a unit which we have not seen that much: the megaNewton, one megaNewton is equal to 1000 kiloNewtons, that is a million Newtons, and the meter, that is a good unit for the concrete beams with the applet. So we will have a span of 10 meters, a load of 0.1 megaNewtons per meter, that is equal to 100 kiloNewtons, a concentrated load of 100 kiloNewtons becomes a load of 0.1 megaNewtons, a load of 100 kiloNewtons per meter becomes a load of 0.1 megaNewtons per meter, and then the good news is that a Newton per square millimeter is equal to a megaNewton per square meter, because there is one million on each side, so we still have the same strength, which we know, of 20 megaNewtons per square meter or for steel, of 500 megaNewtons per square meter. Well, here is typically the kind of structures with which we want to work. That is a beam, we are going to give it dimensions, a span of 10 meters, we then have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 loads, placed at intervals of one meter, but you can see that we start at half a meter from the support, in such a way that each time, each load is equal and represents a width of 1 meter. So here, we are going to say a load of 100 kiloNewtons, so 100 kiloNewtons per meter, we are going to look for the width, and also the effective depth, we do not know compression and tension forces either, we are going to look for them. What I decided is that the total depth of this beam will be equal to 0.132 times the span, which means that this depth here is 1.32 meters. So the ratio between the span and the depth is equal to 1.32. We now push the button to start the applet, we are here in this applet, we are going to work in megaNewtons and in meters. The first thing to do is to define the geometry of the screen, or the scale of the screen, clicking on this button here, which we have not used yet. On the bottom, the distance between 2 points appears, that is a standard value of 100, we want a value of 10, so we put 10. Afterwards, where do we have these 10 meters? We have them between here and there. Thus, I first click here, then I draw out the mouse keeping this clear blue line till the end of the 10 meters here. I release it, we do not see anything, but the screen has been scaled. If you want, you can redo it several times, if you have not done something proper, but it is not necessary to reach a very high precision, because, one more time, that is a pre-dimensioning. We now want to introduce the loads, in megaNewtons, by default it would be 100 kiloNewtons, but what we want, is 0.1 megaNewtons. And I insert these ten forces. What matters is that the line of action is correct. We can put them a bit higher, a bit lower. Now, I insert the left support here, but I place it at the level of the reinforcement bar, we will see later why this is important. And pressing control, I do the same thing with the right support at the level of the reinforcement. I can now start the funicular polygon, and then I know that the entire compression must remain inside the material, so the funicular polygon is going to pass somewhere around here with approximately 1.2 megaNewtons of compression. Here I will have tension. You now understand why I put the supports at this level here, because at this level here, the red tensile force exactly superimposes with the position I have indicated with the reinforcement bar. Otherwise, that would not have been the case, the consequence would be that the effective depth would be larger than in reality. That is good but not sufficient. We are going to deactivate this and we are going to use another button here, that we have never used, which enables us to define the geometry and the materials of our cross-section, the strength, we have said that it is concrete with 20 megaNewtons per square meter, then afterwards, we need to provide the width of the beam. We are going to start with a width of 0.3 meters. We now want to activate the funicular polygon for it to take into account this geometry. We are going to deactivate this red force, just for aesthetical reasons, we are going to activate the geometry of the funicular polygon pushing at the same time the control button. Doing this, you can see that now the funicular polygon is shown with a thickness. Here, it is said for example that we have 1.23 megaNewtons and the dimensions of the compressed zone are 0.3 meters, this is the defined width, times 0.21. Actually, we must remain inside the cross-section, but we can go higher, so we move this line here till it exactly reaches the top of the concrete cross-section. We can see that the compression decreased to 1.05 and that 0.3 times 0.17 is the dimension of the compressed zone somewhere here on the top of our cross-section. So the variable that we have, these 30 centimeters, we can see that it is largely enough actually. We have a very good z, but we do not use the concrete which is in the middle, so we are going to try another width, a bit more extreme. I am going to put 0.1 meters. So with 0.1 meters, we have to reactivate it, and we can see that here, we get a very large thickness. We are going to try, we will roughly manage to put it inside the cross-section, then here, what we can is is that we have a thickness of 0.1 times 0.65, that means a width of 0.65, that is more than half of the 1.32 meters of the depth of our cross-section, and this is not allowed by the codes. This value should be at least under 0.5, and even a bit smaller, that would not be bad. Moreover, we can see that the internal force is quite larger, 1.31 megaNewtons, which we also have in tension here, so that would require quite a lot of reinforcement, so we are going to choose an intermediate solution, with a width of 0.15 meters. So here, we come back... And here, we are clearly under the value of 0.5. We have an internal force which have decreased to approximately 1.15 megaNewtons, and thus, that would be the solution. Here, we know that we must have a tensile internal force of 1.16 megaNewtons. We know that here, the compression is equal to 1.15 megaNewtons, and the tension has approximately the same value. 1.15 megaNewtons, that is 1150 kiloNewtons, so we can calculate the effective cross-section that we need saying: that is 1150 times 10 power 3, to obtain Newtons, divided by the strength of 500. 1150 times 10 power 3 divided by 500, which gives us 2300 square millimters of steel. How can we proceed to the dimensioning? Well, we must now look at the reinforcement cross-sections which are available in Switzerland. Actually, that is probably the same kind of diameters everywhere, here we have even diameters until 22 millimeters, every 2, then after, it jumps to 26, 30, 34, 40. The diameters are almost all easily available until 26 millimeters. Beyond, this is a bit more difficult to get. And then if we have the number of bars, and I remind you that we are looking for a solution to obtain 2300 square millimeters, looking at this table, we find for example here 2280, that is a bit smaller, or else here, with 2510 square millimeters, that is 8 20 millimeters diameter bars. So here, the reinforcement would be composed of 8 20 millimeters diameter bars. That is here, 8 diamater 20. We do not have much space since we have decided that the width of the beam would be 150 millimeters. So probably, what we are going to do, is that we are going to place 4 below and 4 above, the lever arm will be slightly reduced as a consequence, and then we will need a bit more reinforcement. But since 8 diameter 20 provides us a bit of strength reserve, that is likely to be sufficient. Here I give you again the process to pre-dimension a beam. That is particular for a rectangular concrete beam, but it is going to work for a lot of other beams. So the first thing, which is very important, is to define the scale of the screen. Afterwards, we are going to specify the forces, being careful to the units. Then we will introduce the supports. We are going to activate the resolution of the funicular polygon using the control button on the keyboard until we get a satisfactory combination. A very elegant solution is to have a beam whose the cross-section is optimized, very wide at the top, narrower at the bottom, but sufficiently wide however to be able to place the reinforcement. You have seen that before, we were a bit cramped with the reinforcement, if we put a small foot like this one, it gives us more space to place the reinforcement, and then here, on the top, we will have a larger width. Here in the applet, I have already defined the scale of the screen, the loads, etc. What I am going to do is to define the material. Well, that is still concrete with 20 megaNewtons per square meter, and then its width, let's say the spacing between these beams is 0.8 meters, so the width here on the top is 0.8 meters, I define this, and then I activate the funicular polygon pushing control and I can see that the effect of this larger width is very significant since I obtain at once a smaller internal force, only of approximately one megaNewton, because the effective depth is larger, and on the other hand, this compressed zone is very small since it is only 60 millimters deep, 0.06 meters, that is 60 millimeters. So I could make this zone here very small. Actually, for constructive reasons, we are going to make it thicker, maybe around 15 centimeters wide, but it would be possible to make it smaller. What we can see is that this shape enables to be doubly efficient. Concrete is going to be very little compressed and the internal force in steel is going to decrease. Here, to fill this table, I am going to let you do it yourself, but we would very likely have slightly less reinforcement with this solution and a solution which is absolutely feasible with reinforced concrete.
