Hello, in this lecture, we are going to continue our study of beams to look at what happens inside these beams in terms of internal force, of stress and of deformation. We will finally establish together a table to directly determine the maximum deflection, particularly as a function of the various effects which act on a beam. Here, we find again something that we just did in the previous video, so I am not going to get into the details, but here we have the effective depth z which is equal to half of the depth of our rectangular beam whose width is t. You remember, we had subjected this beam to two loads of QRd, with a support reaction of QRd on the left and on the right. And we had developped the formula: QRd is equal to h over 2a times N. That is enough for us, we had developped other formulae, it allows me to say that N is equal to QRd times 2a over h. I am now going to introduce a symbol which I will use a lot thereafter. This symbol here means "proportional to". For example, the length a, I can admit for my beams, is proportional to the span. Since I have N which is equal to QRd times 2a, I can also say that N is proportional to Q, I am going to forget Rd, times l divided by h. We are going proceed, in the remainder of this video, only with proportionality relationships, which will enable us to avoid to solve certain equations which are quite complex, while still giving us a very interesting perspective on the relative importance of the various parameters in this configuration. So, here we look at the stresses and the deformations. I still have, I have written it here, z which is equal to h over 2. We have the stress sigma which is equal to N over A, that is proportional. N, we know it, is proportional to Q·L over h. And now, we still divide by A and then for A, we know that the cross-section is equal to t times h, so we can also say that it is proportional to Q times L over t times h square. And then the strain epsilon is equal to sigma, the stress, divided by the modulus of elasticity E, and is also proportional to Q times L over E, modulus of elasticity, t·h squared. We now move to bending. We have already observed when I did this demonstration in the previous video, that we are dealing with a curvature. The curvature is this continuous changing in the orientation of these little lines still perpendicular to the axis, but which change their orientation. I think that the word "curve" is quite clear in the common language. We generally rather look at the radius of curvature than at the curvature. The radius of curvature is, actually, the distance between the neutral axis of our cross-section and the center of the circle which would be tangent to our shape. What we can notice here, is that we had here an epsilon on either side. We are going to say that we had, in the upper half of the cross-section, a negative epsilon, in the lower part, a positive epsilon, one half on each side. On the other side, thre is here another small element, same thing, epsilon over 2, and then we have here the depth, in this case, that is really the total depth of our cross-section. If now we take a cross-section which has a double depth, we will still have the same epsilon over 2 on the left and on the right for the compression and for the tension. And what we can immediately notice is that the radius of curvature... we are not going to be able to enter, to touch the centre of the circle here because the radius of curvature - I should have drawn it till here - here, the radius of curvature is much larger than on the left here. So, what we can notice here is that now there is a proportionality between r and one over epsilon. r is proportional to one over epsilon. That is to say that if I increase epsilon, the radius of curvature is going to decrease. If I decrease epsilon, the radius of curvature is going to increase. What we can see on the right part is that here I have doubled the depth, the radius of curvature doubled, so, actually the radius of curvature is proportional to h over epsilon. Here, it concerns the radius of curvature r. Now, we are going to look at the curvature psi. Psi is defined as one over r, the inveres of the radius of curvature, so psi is naturally proportional to epsilon over h. since r is proportional to h over epsilon. We now look at the slope, here, the slope of our beam. The maximum slope in this case, can be found near the supports. The slope, if you remember what you have seen, for those who followed advanced mathematics courses, the slope is the integral of the curvature over the length. In practice, this means that the slope phi is proportional to psi - I draw your attention on the fact that they are not the same greek letters. The letter phi from which comes our f, we draw it without raising the pencil, the letter psi, we make a kind of trident. - so phi is proportional to psi times the length, that is to say, it is proportional to Q·L squares over E·t·h cubed. The maximum deflection w. Here, the maximum deflection, in this case, is located in the middle. That is often near the middle. The maximum deflection is the integral of the slope over the length, again. So the maximum deflection w is proportional to phi times L, that is to say proportional to Q·L cubed over E·t·h cubed. The stiffness is the property that measures the ability of a beam not to deform too much. That is defined by the load we apply divided by the maximum deflection. This value is itself proportional to E·t·h cubed over L cubed. In a structure, the maximum allowable deflection is usually defined as the a relative value, that is to say that codes give us w as a fraction of the span should be approximately equal to one over 300. Maybe this value can be one over 250 or one over 350, this will depend from the various national codes, this is not very important at the level of this course. If, however, we have fragile elements, that means elements, for example walls with tiles or highly glazed elements, we will rather have an admissible value for the maximum deflection of one 500th of the span. And we know that the maximum relative deflection is proportional to Q·L squared over E·t·h cubed. And the stiffness Q over w. When we have concentrated loads and that there are not too many, one or two, we will get something which is going to be proportional to Q times L over h. The stress is going to be proportional to Q times L over t·h squared. The strain epsilon is going to be proportional to Q times l over E t h square. The curvature psi is proportional to Q times L over E·t·h cubed. The slope phi is proportional to Q·L square divided by E·t·h cubed. The maximum deflection w is proportional to Q·L cubed over E·t·h cubed. The maximum relative deflection w over L is proportional to Q·L squared divided by E·t·h cubed. The stiffness Q over w is proportional to E·t·h cubed over l cubed. Finally, without getting into the details of how these values have been obtained, I can give you the most important values for an uniformly distributed load. The maximum deflection is proportional to lowercase q, that is to say the load in kiloNewtons per meter, the uniformly distributed load, times L to the 4th, divided by E·t·h cubed. The stiffness is proportional to E·t·h cubed over L to the 4th. Let's now come back to this diagram, you remember that we had developped it together before. There is always compression in the upper part of these cross-sections. Here, we have a cross-section which is composed of 2 superimposed beams, that is slightly different, there is a bit of compression in the middle but actually there are two separated beams. And then tension on the bottom. We have seen before that this had consequences in terms of strength. The formula was Q rd is equal to h squared divided by 4 times t over A times fd. And then we have seen that if we give a factor one times to this beam, this one, twice wider, had a factor 2 times, this one, twice deeper, had a factor 4 times, and then this one which was composed of two superimposed beams, only had a factor 2 times. If we now look at the stiffness, that is to say Q over w which is proportional to E·t·h cubed, and not squared anymore, divided by L cubed, so we have, if we give here a factor one to this first beam, when we are going to double the width, we are going to double t, so it is simply going to be twice larger. However, when we are going to double the depth, we are going to have 2 cubed, so this beam with a double depth will have 8 times the stiffness of the initial beam, and then when we place two beams on top of each other, we will again have only a factor 2 as advantage. So, what we can see, is that both in terms of strength and stiffness, it is favorable to place the material over the depth of the beams, into one unique element. We have also seen that the shape of the cross-section has a quite decisive importance on the stiffness of the beam.
