Hello, in this video, we are going to look at bending, which is the resistance mode of beams, then we are going to look at how this mode works to obtain, for a given beam, its maximal bending strength. We are going to start here by a reminder about the effect of compression, first, then, of tension, on a rubber foam element. We have seen this together during the course "The Art of Structures I". The effect, when we have a known initial length of this element, we are going to call it L. If we subject it to a compression, - of course, there will be compression on the other side, but I don's take time to draw it- what is going to happen is that its length is going to decrease. - we have seen it in this video - So we are going to have here a delta l and here the length will be L minus this delta L. That is the effet of compression. In a similar way, if we subject it to tension, the element is goign to lengthen of a value delta L. The element now has a length L plus delta L. Delta L, we can read it here, that is the lengthening induced by the tensile internal force. Here on the right, I write the properties of a tie in tension, as a reminder. So a stress-strain diagram. We have a linear part whose the slope is equal to the modulus of elasticity of the material. Then we are going to reach a certain value of design strength fd before going into a larger lengthening, and finally failure. - that will not really interest us within the framework of this course - If we have, making a diagram, force-lengthening, - what we have observed on the rubber foam element - the shape would be very similar: the slope of this diagram is now equal to the stiffness of the cross-section which is the modulus of elasticity times the cross-section. Then the maximum strength will here be equal to A times the strength of the material. We could have done the same thing for the compression, but that will be enough for a reminder. In this video, we can see how, by first applying a concentrated load in the middle of this rubber beam, it bends, it curves. I can obtain a similar effect taking the beam in my hands, and making it rotate to curve it. This phenomenon is called bending. If we take a closer look, bending a bit less the beam, what happens, we have an element which was initially rectilinear with lines perpendicular to the axis. Then, when I have rotated it, when I have make this beam bend, these elements have remained perpendicular to the axis, but they have rotated. What has happened? The upper part of our beam shortens. Everywhere: on the left, on the right and in the middle. So what it means is that we have here, acting in the upper part, everywhere above this axis, we have compression. Conversely, if we look at the lower part, we can see that here, we have, on the left and on the right, we have a lengthening. So, in this part here, we have tension. This, it is new: we have never seen till now simultaneously tension and compression on the same element. Then, obviously, in the middle there is a part which is a bit Swiss, which is neutral, on which nothing happens. There is neither lengthening, nor shortening - I draw it in green - Here, that is what we call the neutral axis. At the level of the neutral axis, there is neither lengthening, nor shortening. On this basis, let's look at how we can obtain the maximal strength of a beam with a rectangular cross-section. For this, we are going to make act on this beam two loads QRd which correspond to the maximal design load. Obviously, both reactions are equal on the left and on the right to QRd. If that is the maximal load, what is going to happen, is that half of the cross-section, the upper half of the cross-section, is going to be subjected to only compression over the whole depth. And then, the lower half is also going to be subjected to tension over half of the depth. As a consequence, we will have, for the indicated arch-cable, we will have tension with a well defined geometrical position, like we are going to see it in one minute. Here I draw the corresponding arch, which also has a well defined position. And if we look at the cross-section, half of the cross-section is entirely in compression, the other half is entirely in tension. What does it means? If there is half of the cross-section, that means that this distance here must ncessarily be equal to h over four, on the top and on the bottom. This, it does not particularly interests us. What interests us is to know what there is between both. That is z which is thus equal to h over two. Here we have a cross-section - I should have indicated it - which has a depth h, and then a width t, we are going to use it in a few moments. These loads act at a distance a of the support, on the left and on the right and we can solve this system using for example a free-body around the left support. This free-body is subjected to the reaction force QRd. - it acts upwards - Then, to a tensile internal force - I am going to call it again N, like in the previous videos - and to a compressive internal force in the diagonal. What is the value of N? Well, I can look at it. I have half of the cross-section. So the center of gravity of N is going to be located here, in the red part. N in tension will be equal to t times h over two times the design strength of my material fd. Now, I want to link all these properties together. I can observe that we have again a similar triangle between this equilibrium of the node on the left support, and then here, this part of our arch-cable. What we can say is that: QRd divided by N is like h over two, that is to say z divided by a. So, QRd is equal to h over two times a times N. Now, we just saw what is the value of N, we can write that it is equal to h over two a times the value that we have taken here for N, times t times h over two times fd. And then finally, QRd is equal to h squared divided by four, times t divided by a, times fd. This formula which we just derived, is actually based on a distribution of the stresses over the depth of the cross-section, and thus, over the entire upper part, we have equal compressive stresses which have a value of fd, and over the entire lower part, we have equal tensiles stresses which have a value of fd. This is what we call a plastic distribution of the stresses. That is linked to the plasticity of the material. If we really go to the limit, that is what is going to happen and then we will obtain QRd is equal to h square over four times t over a, times fd. This is the formula which I would like you to remember for this part of the course. But that is true that, maybe the best observers among you told themselves "Oui but be careful, here, actually, I have a maximum epsilon, but afterwards, it is going to decrease, near the neutral axis, it is going to be very small." Indeed, that is something which would be closer to reality. At least, when we are at lower levels of load, it corresponds to what we call an elastic distribution of stresses. So, again, on the top, we have fd. Afterwards, we have smaller values which keep decreasing. And then, from the axis, from the top to the bottom, we have again values which increase till fd. The ones among you which will study in an analytical way the strength of materials, you will see both these distributions, and you will see that the formula for the strength, if we are based upon something elastic, it is very similar, with only one difference: it is going to be h square over six times t over a, times fd. We now want to see the influence of the cross-section on the strength. So I write again this formula which we just determined: QRd is equal to h square divided by four times t divided by a, times the strength of the material fd. We have various cross-sections, here a cross-section with a depth h and a width t. Here, still a depth h, but a width of two t. Here a depth of two h and a width of t, and here on the right, a depth of two h, but in two elements, and a width t. Let's now look at what happens for the distribution of stresses. We are going to have in the cross-section on the far-left, the upper part which is going to be in compression, the lower part which is going to be in tension. And then the effective depth z, between the centers of gravity of tension and compression, is equal to h over two. I am going to take this cross-section as reference, and I am thus going to say: this, it is one time the strength. The second cross-section, I put much more material, it is twice wider, but it has the same depth. So the lever arm z is still going to be equal to h over two. And if I look at the formula, I now have t which passed to two t, so I have twice more strength. In some ways, that is logical, I put twice more material, I have twice more strength. Let's now look at what happens if we put twice more material, but a bit differently: we put it on the depth, here I have compression over a depth h, and afterwards tension on the lower part. Now, the distance between both centers of gravity z is equal to h. Before, it was equal to h over two. If I look at the formula, t does not change, but h is replaced by h squared, and now, this cross-section has four times more strength. That is interesting, I have put twice more material, and I obtain four times the strength, because I have created not a wider cross-section, like in the case B, but a deeper cross-section. So it means that a deeper beam is more efficient than a wider beam. Now, it could happen that we are in situation in which we have two beams which we want to place on top of each other. So what is going to happen? Here, if I have two beams which are one on top of each other, actually, I will have compressieon in the upper part of the beam, and tension in the lower part of each of the beams. Actually, each of the beams will have an effective depth z equal to h over two. The consequence of this is that we are going to have twice the strength, while we have the same depth than in the solution C, but because we have put two beams which are not linked between each other, we do not have this advantage to be able to consider the entire depth at once. In this lecture about bending, we have seen that bending is the combination, in a same cross-section, of tension and compression. We have seen how to determine the strength of a rectangular beam, on the basis of hypothesese about the strength of the material, such as we already know them since "The Art of Structures I". And then we have finally seen that the shape of the cross-section has a certain importance, since certain combinations of shapes can bring us advantages with a constant amount of material.
