 Dus vandaag gaan we, voor sommige, het verhaal, kijken naar de moorsirkel. En active en passive soil failure, maar na dat, we beginnen met nieuwe toppen. Dus eerst gaan we verhaal de moorsirkel. De moorsirkel is een, laten we zeggen, een mathematical tool, een way of showing things concerning stresses en the relations between different stresses. And you probably got it in the first year with solid mechanics, people who did soil mechanics. They should have had it very often. But if you look at cutting processes, which we do in this course, you will often see more circles. So you have to understand what they do, what you can do with them. We will not do 3D moorsirkels or 3D surfaces. We will just do it 2D, two-dimensional, so it's not too complicated. Well, if you consider a cylinder of soil, and on top you have a vertical stress and horizontally you have a horizontal stress. And in fact the places where you have those two stresses, normal stresses, you don't have shear stresses. And that's one of the things, we call those stresses and also those directions principal stresses. So the stress, the surface where you only have a normal stress is called a principal stress. That's important because you will hear that word more often in the course. So you always have two principal stresses with two directions, in this case vertical and horizontal. They are always under 90 degrees with each other and later you will also see why. But you should remember there's always 90 degrees between them. Now in many, let's say simple applications, you consider the vertical stress always as a principal stress because it's the gravity of this planet. It's just the weight of the soil which gives a vertical stress. This is not always necessary. In soil cutting it often happens that the directions of those principal stresses are not horizontal and vertical. But it depends on, for example, the blade angle, it can also depend on the angle of internal friction. Dus de fact that you work with vertical and horizontal is mainly because normally you have gravity which forms one of the principal stresses. That also makes it easy because if gravity is one of the principal stresses, then it means the vertical stress is just the weight of the soil. So if you take a certain plane, a certain surface in the soil and you know the amount of soil on top of it and you know the weight of that amount, then that gives you the vertical stress. So the vertical stress is often easy to calculate. Now if we want to know what are the stresses on any plane, any surface under an angle alpha, we have to do some calculations. Well, so here we have vertical horizontal and this is the plane under an angle alpha. Now what I already told in the dredging introduction course, if you want to solve this, you have to consider that an equilibrium of stresses does not exist. You can never take the equilibrium of stresses, you can only take the equilibrium of forces. So you have to transform those stresses into forces and in this case that's not too complicated. We can make a triangle with the vertical stress and the horizontal stress and on the plane under an angle alpha, we have a sigma and a tau. For this tau, by the way, you do not yet know in which direction the tau is, it could be in the direction of the picture, it could also be in the opposite direction. But that does not matter because if you do it in the opposite direction, you will get a minus sign in front of the tau and at the end the result will still be correct. But one thing in general, you should have a good definition of what is a positive axis, what is a negative axis, what is positive, what is negative and in soil normally compression is positive en tensile is negative, while in mechanical engineering it's the other way around, in mechanical engineering in steel, usually tensile is positive and compression is negative, so you have to remember that. Ok, so if we look at this picture you see all those four stresses, but like I said you cannot take the equilibrium of stresses, you have to take the equilibrium of forces. Well, how do you go from stresses to forces? Very simple, you multiply the stress with the surface, with the cross section or surface that it applies on. Now in this case you could say, yeah, but we don't know the values for those surfaces, but you know the ratio of all those surfaces. So if we assume that this plane, so the plane under the angle alpha has a surface of one, just assume, and that means a length of one and perpendicular on the picture, you also assume a unit of length, so that's also a one. Then the sigma and the tau work on a surface of one, one square meter or whatever you want to call it. That means this one, the horizontal stress works on a surface of sine alpha and the vertical stress works on a surface of cosine alpha. So the ratio between those three surfaces is one sine alpha and cosine alpha. That means, and again perpendicular on the blackboard, we assume a length of one. So that means we know the ratio between the three surfaces. And you will see that if you would make those surfaces bigger, multiply it with some factor, then that factor will be in each force. And that means later in the equations you can divide by that factor and you will not see that factor anymore. So whether you just use the ratios or you use some actual surface, the result is the same. Ja, it doesn't make a difference. Ok, now to solve this and it's an exercise. And for some of you you already saw this, but some didn't. It's an exercise that you also have to do when you want to solve problems of cutting processes. Only in that case you have more angles, more parameters. So it's a little bit more complicated, but mainly because you have more terms in your equations. Ja, but the principle is the same. So what do we do? First we take a vertical equilibrium. And in the vertical equilibrium you look at the vertical stress. Well the vertical stress times cosine alpha gives you the force. So here this sigma v cosine alpha is the vertical force. Then on the other side we have the sigma times cosine alpha, which gives the vertical component. And because that surface is one, you don't have to multiply with anything else because the surface was one. So we get the sigma cosine alpha and you can also see that in this case the tau works in that direction by the sine of alpha. So that's your vertical equilibrium. For the horizontal equilibrium it works the same. You take the sigma times sine alpha and in this case the sine alpha is because of the surface. The surface has the value of sine alpha, so you multiply with the sine alpha. So this is the horizontal force externally. And then you have a component of sigma sine alpha against it, opposite. But the tau works in the same direction as the sigma h, so it gets a minus sign. You get tau cosine alpha. So this way you have two equilibrium equations. En dat means you can... Two equations with two unknowns because the sigma and the tau are the two unknowns. And that means you can solve it. If the number of unknowns and the number of equations is the same, you can always solve it. It's first year mathematics basically. Okay, what do we do? Well, to solve it, let's say first we want to have an equation for the sigma. If you want to have an equation for the sigma, it means you have to get rid of the tau. You don't want the tau in your equation anymore. Well, how to deal with that? If you multiply the top equation by cosine alpha and the bottom equation by sine alpha, then in both equations you get such a term. But here I have a plus sign and here I have a minus sign. So if I add them up, I get rid of the tau. So that's the way to do it. Well, it means I get... Here I get sigma phi cosine squared. Sigma cosine squared in the second equation. I get sigma h sine squared and sigma sine squared. If I add it up, I get the sigma phi cosine squared plus sigma h sine squared equals sigma. That's the equation we get. Now, if you would continue with the alpha in the equation, you would never find the solution. So for the sine squared and the cosine squared, you have to substitute something. That's basic mathematics. For the cosine squared, you can write 1 plus cosine 2 alpha divided by 2. For the sine squared, you can write 1 minus cosine 2 alpha divided by 2. That's just basic mathematics. You substitute that in this equation. Rewrite a little bit, reorder a little bit. And then you get the equation at the bottom. And in the equation at the bottom, you see you have two terms. You have a sigma phi plus sigma h divided by 2, which basically is the average of the two stresses. It's just the average, nothing else. And here you have the sigma phi minus sigma h divided by 2, which is the difference between the two stresses, the two principal stresses divided by 2. And later, we will see that this difference divided by 2 is the radius of the Mohr's circle. Here you see the cosine of 2 alpha, and that's also typical of the Mohr's circle, that in working with Mohr's circles, in the Mohr's circle, all the angles are represented by 2 times the real angle. We will see that when we see those circles. Okay, so this is the solution for the sigma. Then we want the solution for the tau. Well, if you want to have the tau, you have to get rid of the sigma. How to get rid of the sigma? Well, first of all, you want two terms that are equal, but with opposite sign. That means the top equation, which was the vertical equilibrium, we will multiply with sine alpha here, and the bottom equation, the horizontal equilibrium equation, we will multiply with minus cosine alpha, with the minus sign. You can also say, I multiply with the plus sign, and then I take the difference of the two equations. But if you multiply with the minus sign, you can just add up the two equations. It's easier to see which terms are equal, but with a negative sign. So here you see they are equal, but with a negative sign. So now if I add up the two equations, the sigma has gone, and I get this equation for the tau. In general, if you find an equation with sine alpha, cosine alpha, or sine x, cosine x, you know that basically it means I have the sine of two x divided by two. And that's also basic mathematics. So this equals the sine of two alpha divided by two. En if I substitute that, then again, here I find the difference between the two principal stresses divided by two, and I find the sine of two alpha. And that means I go to a system where every angle is two times the angle. Well, here we have those two equations, the sigma equation and the tau equation. And what I did with the sigma equation, I moved this term to the left-hand side of the equation. And then the reason is that by doing that, I can do an operation that gives me the final result. But basically those are still the same two equations. Well, now what I do is I square both equations. So I take the square of both equations. And add them up. Well, if I do that, then this is the square of this term. Oh, yeah, this is the square of this term. This is the square of this term. Here you see the square of the tau equation. And if I add them up, then you will see that cosine squared plus sine squared is one. So I get rid of those terms. And here I get the so-called circle equation. Why is it the circle equation? Well, basically, because in more circles, in general, we use a tau-sigma diagram where the sigma is the horizontal axis and the tau is the vertical axis. And that means here I have the horizontal coordinate minus the center of a circle. We don't know yet what it looks like, but it's the center of a circle. And this is the vertical component. So apparently in vertical direction I do not have any shift in horizontal direction. I have a shift, but in vertical direction I have not. And the right-hand side of the equation is the radius squared. So if you look up the standard circle equation, you will see that this is almost exactly a circle equation. Okay, what does it look like? Well, to start with, we have the horizontal axis, which is the sigma axis. We have the vertical axis, which is the tau axis. We have the two principal stresses, sigma phi in this point en sigma h in dat punt. En they are, like I said, they are always under 90 degrees because they are principal stresses. But since in the Mohr circle we multiply angles by 2, in the Mohr circle there's 180 degrees between the two principal stresses. And that means, and I also said principal stresses are on the surface where you don't have shear stresses. So they are always the two intersection points with the horizontal axis. So if you draw a circle, you have the two intersection points. That's the two principal stresses. And then you always have 180 degrees between them. Whether you go in one direction or the other direction, it's always 180 degrees. So you can see the two alpha here. That means the point sigma comma tau is here. So if, for example, if that angle alpha would be 30 degrees, then in the Mohr circle here I would see 60 degrees. If it would be 45 degrees, then in the Mohr circle I would be at the top of the circle because it would be at 90 degrees. So it's always 2 times the angle in reality. There's something more you can see. If we would have sent, and I will show you pictures of that later, if we would have sent, then you don't have what we call cohesion of or shear stress. And that means you always start in the origin. So we start with a sand-like material. It means you start in the origin. Then you can draw a line from the origin that is just a tangent at the circle. So we start here in the origin and the line just touches the circle. And then you will find a certain angle, this angle, phi. And this angle we call the angle of internal friction of the material. But just mathematically or geometrically, this angle here will also be exactly phi. That has nothing to do with soil mechanics. If you make such a graph, just mathematically, those angles are equal. So here you have a phi, here you have a phi. You also have the negative axis, but very often we only look at the positive tau axis, although the negative axis also exists because if you have a shear stress in one direction, there has to be a shear stress in the opposite direction on another plane, otherwise it's not an equilibrium. One important thing, if you would not have an equilibrium of forces, which is possible, it means you have some force left and that force will always result in accelerations. So if there is no equilibrium, there is force left, you will have an acceleration. So if you observe from experiments that the material is not accelerating, that you have a nice steady state process, that means no accelerations, there has to be an equilibrium of forces. If you would have an experiment where you see accelerations, then you also know there is no equilibrium. So you have to think of that. An equilibrium is not something that you always have, but if I have a nice steady state cutting process and the material flows nicely over my blade without any accelerations, there has to be an equilibrium of forces. So this is just one picture. Well, if we know that this is the angle of internal friction, then okay, so if we can do a test, and I showed you the test last week, the triaxial test, you can do a test to determine this angle. But one of the problems of soil is that soil is never the same. You know, even if I take the same soil and I try to prepare it exactly as the previous time en I do the test again in the triaxial equipment, I will find a phi that's maybe one degrees or two degrees different. Because it's very difficult with soil to reproduce it 100%. That's also a problem with experiments in the laboratory and I will show you that later. But if you take soil samples from practice, we call it situ in situ samples, and you would test for this angle of internal friction, each spot can have different sand or clay or whatever. The particle size distribution could be a bit different. The density could be a bit different. So you will never find exactly the same angle of internal friction, and that also counts for the other soil mechanical parameter. So if you think of the permeability or the angle of external friction, all those things will never be exactly the same if you do another test. One thing we noticed is that if you do those tests and as a company you want to build up a database of data about soils, then you can say, oh, I already did a project in that area and now we do it again. So we probably have the same soil. If you want soil parameters that you can still compare, you have to do the test in the same laboratory and if possible you let the same person do the test. Because the outcome of the test could depend even on the person who is doing the test on the whole preparation of the test. So it's that critical. It's not that it differs. It will always differ too much, but one degree difference in the internal friction angle for a dredging company could mean two or three percent less or more production. So that's why they like to have everything carried out in the same way. Well, so what do we do? We take a number of samples and we do tests at different pressures. Here you can see a case. Well, in fact in this picture I took care that all the circles have attention at the same line. But in reality, for example, this circle could be a bit bigger. This one a bit smaller and this one again a bit bigger. And then what you do is you try to draw the best fit line through those three circles and then that's some average angle of internal friction. So that's how you deal with that. And here you can see you have the sigma phi 1, phi 2, phi 3. So every time I increase the pressure, we call it the confining pressure, I increase the pressure and that means my whole circle is shifting to the right. But you don't always have sand. You can also have clay or loam. A typical thing of those kind of materials is they have strength. Sand does not really have strength of itself. So if I have loose sand, dry sand, I want to shear it. I hardly need any force. Maybe I have to overcome a little bit gravity, but that's it. But if I have clay, I need a certain force because the clay has strength of itself. Now this strength of clay follows basically from two parameters. One parameter is the shear strength of the clay. We want to shear it. How much force or stress do I need to make it shear? And the shear stress, we calculate it from the origin up and then you take the intersection point on the vertical axis. In dredging we often talk about cohesion. In soil mechanics I think they call the SU value shear strength. In dredging we usually talk about the cohesion, cohesive soils. This is the cohesion and that also means the origin of those lines is somewhere on the negative axis. So that's one. In fact all I did to draw this picture is move the vertical axis to the right. That's all I did. And then you get this intersection point. The second parameter, which is very important for us and I don't think in normal civil engineering it's too important, but in dredging it is, it's the tensile strength of the material. So how hard can I pull on a piece of clay before it breaks? You can imagine if you build a dike and you use clay in the dike to make it isolate for water flow. The clay is always under compression so I don't have to deal with the tensile strength. But when I'm dredging, when I'm cutting and I'm kind of bending the material you get tensile stresses in the material and where you get them you might get tensile failure. So the tensile strength is also important. Now the tensile strength, in this case you could say is this point, is the point where those lines start. I could call that the tensile strength. But we will see later that on the negative axis very often the shape of this line is not a straight line anymore but it's curved and it goes to somewhere in the middle in this picture. Also in this case you can do a number of tests and based on the number of tests you can determine both the angle of internal friction and the cohesion of the material. In this case because this intersection point the cohesion, you use it in many equations so if you just base it on one circle it's not enough and by the way if you just have one circle you don't even know whether it has cohesion or not because if I have one circle I can always draw a line from the origin with a tangent on that circle and I would find a very high internal friction angle which is not realistic but yeah, any line is possible. So I really need at least two circles to find a combination of internal friction and cohesion. But in fact the more circles I have the more accurate I am. Active soil failure. It's one of the examples so now we know the more circle and we know how it works. Active soil failure. What do we have? Well we assume we have a dam but it could also be a bulldozer blade or whatever. In the case of active soil failure it's not a bulldozer blade it's a dam but with passive soil failure it could be a bulldozer blade. So we assume the soil wants to fail by its own weight and when the soil is failing it wants to slide down so what would be the force maximum force I have to apply on the dam to keep it stable to keep it intact. And this is not just an imaginary example because if you look at for example ports where you have the dams between the water and the land there you always have to calculate how strong should the dam be to keep my K the ships arrive to keep it stable. So what do we have? Well we have a normal force here and a shear force here between the two we always have an angle phi internal friction angle in this case we just consider sand not clay this is only sand and further we assume on the dam we do not have friction it's the most simple example of you could say a cutting process. Here you have an angle beta with the horizontal in some books you will find the angle beta here depends on the book but we use the beta here and you have a height H. Ok, so I want to know what is the maximum force to keep it stable. Well, first I have to calculate the G the total gravitational force and the G in fact is the triangle of soil that wants to fail so you have to calculate the cross section of that triangle and that's I think half cotangent B cotangent is 1 divided by the tangent and you get the density of the soil times the G and the H squared because it's a surface so you get the H for one side en for the other side but with some signs and co-signs so this is the equation for the G the second equation we know is that in sand we don't have cohesion there's a relation between normal force and friction force and the relation is that friction force is normal force times the friction coefficient and in mechanical engineering we always work with the friction coefficient but in soil mechanics they work with the internal friction angle and the tangent of phi is equal to the friction coefficient so sometimes you use the angle sometimes you use the friction coefficient in fact right now I'm working on some research with pipeline transport the transport of sand and water in pipelines and if the particles are too heavy you will get a bad in the pipeline but if the pressure is high enough the bad will start sliding in the pipeline and in all the publications everybody talks about the friction coefficient and not about the friction angle while many of those people who do that kind of research are civil engineers but still they use the friction coefficient so it depends on the topic once somebody starts with the friction coefficient in all the publications you will see the friction coefficient and that's the way it works so we don't have cohesion we don't have adhesion and we have a smooth wall meaning no friction on the wall c is the cohesion in terms of shear strength so it's not the force the small c is the stress the strength the a is the adhesive strength the delta is the external friction angle between the soil and the dam we can write down the two equilibrium equations again so we have the horizontal equilibrium and the vertical equilibrium just try to rehearse it at home and here you can see we want to know the F but we also have the S and the N so basically you have three unknowns but we also have this equation as a third equation you have the F, the S and the N as unknowns the G is your only known term because it follows from the geometry so basically one, two, three equations so you can solve it well the first thing you have to do is solve the F that's at the top equation we do that in the same way by cross multiplying the equations with a sine or a cosine depending of from which term you want to get rid of so this is the solution for the F then this was the G we already derived that before and if you substitute the G in the F equation you get this equation now what's not very nice of that equation because we said we want to know where is the force at the maximum because that would be the maximum pressure of the soil against the dam when does that happen well the only unknown we can still play with is the beta is that angle we already solved the other unknown we still have the beta as an unknown but if you want to see at which beta is the F at a maximum or at a minimum whatever you want normally you take the derivative of an equation and you determine where is the derivative equal to zero and if you want a maximum then you also have to determine the double derivative and the double derivative has to be negative that's mathematics again but now the problem is here you have the beta in the denominator and in the denominator so if you start determining the derivative you get a very complicated equation well we can simplify it to this equation and on the next slide I will show you how to do that so first of all this was the F equation we need the derivative to be zero and the double derivative to be negative now that term that complicated term if you look at it this was the term okay I can say I take that term minus one plus one which just means I add zero that's all well you can always add zero to something it doesn't change then for the minus one I can write this denominator divided by the denominator and then it's still one well except if one of the terms would be zero then officially you can't do that but we do not assume that's the case so I have minus this term minus this term plus one en I can also see here I have cosine times sine here I have sine times cosine and if I see such two terms like sine alpha cosine beta plus sine beta cosine alpha basically that's what I have then I can add up the two angles and get one term well if I add up the two angles one angle is beta and the other angle is phi minus beta so if I add them up the result is phi phi minus beta plus beta is phi so the result is just the sine of phi so this term results in this term this way I also get rid of the minus sign that I had in front of the equation and the result is an equation where I only have the beta in this denominator ok we go back to this slide so that was this equation so I only have the beta in the denominator I want to know the maximum so when do I have the maximum f if this denominator is at a maximum because I have to deduct this term from one so if this denominator is at a maximum this term as a whole is the smallest which will result in the maximum f so I can just say ok so if I want to know the maximum of this denominator all I have to do is take the derivative of that just that one term and then I also find the solution for the whole equation so I take this term here and f is at a maximum if the small f is at a maximum so that's what I do well that gives me the solution for the derivative according to this cosine of 2 beta minus phi this is the double derivative and the first derivative is 0 when beta is pi divided by 4 plus half phi and that's one of the standard solutions in soil mechanics you will find these solutions in many applications well we also substitute this angle in this equation for the double derivative and we will find that the result is minus 2 which is negative which means we are at a maximum you can also make a graph of the function and you will also see it's a maximum but at the time when people started with this they didn't have computers so they didn't have excel and make a quick chart of such a function which you can do now and you will see on the chart on the graph where is the maximum but in those days they had to do everything analytically they didn't have those computers and so they use this kind of mathematics okay if you substitute this in the original f equation this is the result and we see well the first part is in fact from the g term just from the weight but between brackets we get a 1 minus sin phi divided by 1 plus sin phi and this term we also call the coefficient of active soil failure to give a feeling what are we talking about well if the phi would be 30 degrees then the sin of phi is 0.5 so you get 1 minus 0.5 divided by 1 plus 0.5 is 1 divided by 3 so that gives you a ratio of 1 divided by 3 going a little bit further we can also see that the relation between the horizontal stress and the vertical stress is also this Ka which means the horizontal stress normally is just one third of the vertical stress if I just go outside I would measure in the soil vertically I would measure the weight of the soil and horizontally I would measure maximum no not maximum something around one third of the vertical stress of the weight of the soil so that's active soil failure we will have a break and after the break we continue even kijken, doe je het? ja, ik doe het oké, wat lijkt het? de active failure dus je krijgt een meer cirkel de grootste stress is de vertical stress en de kleinste stress de horizontal stress en hier heb je de interne frictie angle in de soilmechanen zeen deze active soil failure want de soil zorgt voor actie dus als de soil zorgt voor actie is het gevoel dat het actief is en van het punt van de gevoel van de soilmechanen het betekent maar van het punt van de gevoel van normaal bedrijf je zou zeggen, als we actie nemen dat is actief en als we niet actie nemen dat is passief maar in soilmechanen is het gewoon reversed passive soil failure dat is de andere kant dus actief soil failure was passive soil failure is als de buitenlijke wereld iets doet dus je kunt zeggen dat ontdekking soilstructuur zoals dikes en rood en wat is based op actief soil failure passive soil failure is als de buitenlijke wereld iets doet dus dat is wat we doen in dredging in soilmechanen en civil engineering je wilt iets verbinden dat blijft daar voor 100 jaar en in dredging we willen het in een fractie van een seconde verbinden wat is de verschil wel, de foto is bijna dezelfde we hebben nog de g en de f en de n en de s maar nu de s is in de opposite richting waarom omdat de scherf is altijd opposite tot de richting van velociteit dus als we zo hard pushen dat we de soil verblijft we zullen de soil verblijven dus het zal verblijven in deze richting en dat betekent dat de scherf de opposite richting is verblijven en in actief soil failure wanneer de soil verblijft wil het verblijven dus de scherf is verblijven dus de foto is bijna dezelfde de scherf is ook bijna dezelfde we hebben dezelfde scherf voor de weight, voor de g de verschil is dat we een andere beta vinden maar de scherf is precies dezelfde omdat het de crosssector van een triangle is en dat is een standaard scherf we hebben ook de scherf tussen scherf en normaal scherf geen cohesie, geen adhesie we hebben ook de scherf want we willen de twee verblijven mechanismen verblijven horizontal en verticale scherf en als je de twee actiefsoil verblijven met passiefsoil verblijven je zal zien dat er een minus en hier een plus signen en in de andere scherf was het gewoon de opposite en dat is omdat de scherf in de opposite richting is wat doen we? we verblijven we hebben deze scherf voor de scherf de verschil met de vorige scherf is dat we nu een plus signen in de scherf we hebben een minus signen de g-scherf is dezelfde dus als we de g-scherf in de scherf verblijven we hebben deze scherf nu we willen een minimum weten want we willen de minimum verblijven niet een maximaal verblijven want als ik een boeldozer heb en ik ga tegen de scherf op de minimale scherf waar het kan verblijven het zal verblijven dus we bekijken voor een minimum dus weer moeten we de derailleur de dubbele derailleur moet nu groter zijn dan 0 want we bekijken voor een minimum en we moeten de scherf de scherf weer te vinden de derailleur hoe doen we dat bijna dezelfde je hebt dat term met de beta's in de nominator en de nominator je doet minus 1 plus 1 dus je hebt 0 nu voor de minus 1 je neemt de nominator both als nominator en de nominator er is nog 1 en nu we zien een cosine, sine, sine, cosine dus we kunnen het samen met 1 sine de enige probleem is hier je hebt een minus signen en de eerste term heeft een plus signen dus je kunt niet alleen de twee adden maar je kunt dat probleem met een plus signen met een minus signen voor deze beta want de sine van minus beta is minus de sine van beta dus je hebt dat gevoel hier heb je ook dat gedaan en sinds de cosine beta is de cosine van minus beta voor de cosine het is niet moeilijk wanneer je de sine van de argument veranderen dus je kunt dat gewoon doen oké, dus nu kan ik de twee angles dus ik heb een phi plus beta minus beta en de resultaat is 1 plus de sine van phi divide door de nominator dus dat is de manier om het te doen ik heb gevoel wanneer je dit weet het is heel simpel maar ik heb gevoel in andere lectieren dat meest studenten niet denken van deze manier om deze problemen te veranderen maar wanneer je het metematisch weet het is niet heel complex oké, dus we hebben deze solution en als we de solution analyseren we kijken voor een minimum dus wanneer is deze term op het minimum wanneer hier een plus signen is wanneer de nominator hier op het maximum de hele term zal op het minimum zijn ik wil de maximum van de nominator weten en dat betekent dat ik de derailleur en de dubbele derailleur moet negatief zijn want nu bekijk ik voor de maximum van dat nominator wat krijgen we dit is de eerste derailleur de tweede derailleur ik krijg 0 voor beta van phi 4 dus 45 degries en als ik die in de dubbele derailleur substituer ik de minus 2 die negatief is dus ik heb de maximum van dat term dus het betekent ik substituer de solution voor beta in mijn equatie en dit is de resultat en nu zie je dat het bijna dezelfde ziet alleen in het geval van passieve soilverfaling ik vind 1 plus sin phi divided by minus sin phi while in the previous case we had the minus here and the plus here so it's just a matter of changing the plusses and minuses that also means that if we look at just an example 30 degrees we get 1 plus 0.5 divided by 1 minus 0.5 gives you 3 and if we consider the relation between horizontal stress and vertical stress in this case it would mean the horizontal stress is 3 times the vertical stress so if you compare it with active failure it means it's 9 times as big the horizontal stress than in the case of active failure ja ja why is that because here we are looking at the derivative and double derivative of the f of the force and we need the minimum force so for the minimum force the double derivative has to be positive for a minimum it should be positive but now we simplify the equation to this equation and here we want the maximum of the denominator so now we are looking for a maximum again and that means we need double derivative ja ok this was the result and what does it look like well now you get the horizontal stress as the maximum principle stress and the vertical stress as the smallest principle stress when can you get this well already the example bulldozer pushing against the soil that could be a possibility another possibility is what we call bridge forming and bridge forming is for example you have a silo a silo normally is a huge cylinder with a cone at the bottom and materials should flow out they often use that in the ports to store whatever kind of material and if you need some of that material you open the silo at the bottom and the material should flow out but it is possible that the material in the silo starts pushing against the wall because of gravity in the beginning and it changes from active from active mode where the vertical stress is the biggest and in such a case the horizontal stress can become so big that actually there is a bridge in the silo which is stable and then if you open the silo at the bottom nothing will flow out because the bridge is stable and you need to do something to make it collapse well if you look at a hopper dredge when you empty a hopper dredge by opening the bottom doors or shelves and the hopper is full of sand you could also get bridge forming in the hopper and then you can open whatever you want but nothing will happen so in the bottom of a hopper dredge they have a lot of water jets to fluidize the sand and this way destroy any bridge that might occur because if they wouldn't do that you could get a bridge and the hopper will stay full nothing will happen so that's one of the dangers another case where you may have passive mode in nature is if you look at soils that used to have an ice cap on top of them in the ice age because during the ice age also in the north of holland we had maybe one kilometer of ice on the soil and because of that the stresses in the soil became much larger than just by soil weight well once the ice started melting those large stresses stay in the soil and you may get a passive mode of soil failure which means if you go to the north sea to certain locations you will find that the horizontal stresses are still bigger than the vertical stresses and if you are a dredging company and you don't know it means the cutting forces will probably be bigger and you need more power to dig the material and if you don't know that and they say oh we have this kind of soil and you think okay I can have a production of so and so much and then once you start dredging you find out the production is much less because the stresses in the soil were higher those things actually occur well here you can see in the moor circles active and passive so if we have a certain vertical stress because of the weight the small circle is the active case so the horizontal stress in the case of 30 degrees the horizontal stress would be 1 third and this is the passive case where the horizontal stress would be 3 times and when we are cutting soil we always have to deal with the passive case this is the same picture shift it a little bit and you get cohesion so you have an intersection with the vertical axis so you have some cohesion cutting mechanisms well I made some slides of a publication of the 60's of two Japanese who did some good research but their research was focused on yeah wouldn't say dry soil but soil above water level but if the soil is above water level it doesn't always mean it's dry sand could be dry but normal soil will not be 100% dry so in their paper they mentioned certain equipment and they wanted to be able to calculate cutting forces this is a laboratory system they had in fact in reality this looks like it's manual but in reality there's a motor driving it but what you can see is here a blade in this case under 60 degrees and they put a grid in the soil so you can actually see what is the soil doing how is it deforming and I must say it's an old publication but it's still one of the publications that gives you a nice overview of what is soil actually doing well they have some examples this is dry quartz sand now normally when we talk about sand it's quartz so for me this is just normal sand you see the blade here 60 degrees and you see how it's deforming with it's shearing here but it's not staying intact above the blade you can see many cracks and shear lines and that's what it looks like wet quartz sand it also shows the shear plane and yeah a little bit the same but it looks more compact and the reason for that is that we will see later when you have wet sand you get poor pressures because of the water and the poor pressures keep the sand in dry sand well you could say I have air but the air can flow so easily through the sand that we hardly have poor pressures plastic bentonite bentonite is a sort of artificial clay they use it a lot in oil drilling yeah you can see here you have some sort of plastic deformation but you can see shear planes and it is disintegrating then plastic loam for me plastic loam is also a sort of clay but the chemical composition is a little bit different but what do you see with plastic loam you see plastic deformation and the material stays intact you don't see all those cracks it stays one flow of material sometimes that's good sometimes that's not good it depends on the application why would it not be good well it flows nicely in one flow but it's so plastic and so adhesive that if you have a cutter dredge it will stick to everything where it can stick and then if you have a continuous flow once it sticks somewhere the following clay will also stick and before you know it the whole cutter head is blocked with clay you have to move up the leather of the cutter dredge remove everything and start all over again so for a cutting perspective we don't really like it because this kind of process will block everything plastic clay same kind of behavior plastic deformation so you have shear planes it is shearing but it is not failing in other materials it's actually failing so if you have a shear plane it cracks on the shear plane but in those kind of materials it's a continuous process this looks a little bit like steel cutting on a turning machine you get a continuous curl and you you have to put we call it breakers so on the turning machine you have a special piece of metal to make the curl break because also when you turn when you are turning steel you don't want this continuous curl because it can stick to everything not like clay but it can get into the machine and you don't want that compacted loam so compacted loam you can compare with very hard clay very dry hard clay and you see that here you get cracks and in fact why do you get cracks well you can consider this cutting process as bending the material and if you are bending material whatever material it is on the outside you will get tensile stresses and on the inside side you get compaction and if those tensile stresses at the bottom exceed the tensile strength you will get cracks and that's actually what is happening here this is a summary of what you just saw so this is what could happen if you have shear failure but plastic shear failure this is what happens in sand so they call this the shear type this is the flow type where you just have one shear plane but also plastic deformation so we call it the flow type and this is what happens if you get those tensile cracks in very hard loam and normally you can calculate in which angle the tensile crack will occur and then after a short while after the first crack occurs you get a secondary crack so this we call the primary crack then you get a secondary crack under 90 degrees because if you would put the process in the moor circle you see it's at 180 degrees with the first crack so that's why you get exactly 90 degrees for the secondary crack theoretically but that's what those guys found they also in their publication they show some moor diagrams well here you don't see the circles but those are all sigma tau diagrams so this is for dry quartz sand plastic bentonite you can see that the angle of internal friction is very small so it's almost horizontal and here they put a point which is the tensile strength so we will not look at the intersection point with the horizontal axis but you measure the tensile strength and that basically that means that those pieces of the line should not be there we should cut the lines there and draw a vertical line here this was plastic bentonite plastic loam looks a little bit the same here you have plastic clay and you can see the tensile strength is much further away and that means relatively my tensile strength is much larger so if I am bending that type of clay I will still get those tensile stresses but the tensile strength is bigger than the stresses so I will not get a crack this is the compacted loam and you can see the tensile strength is very small compared to the other stresses and that's why I will exceed this point and I get tensile cracks this is another summary where they say this is the shear type here we have the flow type so shear type is in sand it will fail on the line under the angle of phi we call it a failure criteria of failure line this is what happens in that plastic bentonite so it will fail on this line not on tensile strength plastic loam the flow type will also fail on this line but now it's not destructive and with that I mean if you really get a shear plane and it's shearing and it's like a fracture that's what I call destructive after that the two pieces of material are not connected anymore so that's actually what happens here but in the flow type you only get plastic deformation which means after cutting it's still intact and it still has its strength and then here the tear type it fails on the tensile strength so not on this line but it will fail on the tensile strength in rock we are not considering rock here but in rock usually you have a rather complicated cutting process where a number of those failure mechanisms occur at the same time and that makes it more complicated this shows you a little bit about the two principle stresses so we had the maximum and the minimum principle stress and here they show the lines of minimum principle stress and in this direction the lines of maximum principle stress and that means in this area you will get the tensile stress well that's what you can see over here this area B is the area where you have tensile stress in this area usually you have compressive stress and here they show the corresponding more circles so you can actually see what does it look like in the more circle here you see it again high tensile strength I will get the flow type sand has no tensile strength no cohesion it's the shear type and a small tensile strength gives you the tear type so it gives a nice overview of what you can expect and here they show it with the more circles in those areas A, B and C of the previous picture I already put this on blackboard so you can just look at those pictures if you like I can also put this publication on blackboard because for me it's always a nice introduction in what is actually happening in many pictures of failing mechanism well then they show what happens under different angles this is sand under 30 degrees dry sand so they start with 30 degrees and it looks a little bit like it's almost the flow mechanism but it's not you have discrete shear planes so the material is really shearing if you make the angle bigger 45 degrees you can see you get much more failure lines visible failure lines you have to consider that the fact that you don't see something doesn't mean it's not there because in sand it's always difficult to determine did it shear or not 60 degrees 75 degrees en dan they have 90 degrees here you can see very clearly the area where the material is shearing then they also have such a series for plastic loam you can see at a very small angle you get a crack here and one of the reasons is that the normal stresses are not high enough if the angle is too small 45 degrees we still see a crack here but you see the material is flowing over the blade as a continuous material 60 degrees we don't see the crack anymore so you get a nice continuous flow and in fact if you don't want continuous flow one of the things you can do is reduce your blade angle because you will get that crack sooner so you could say if I have a bigger angle the bending is more so I should get the crack but it's all a matter of the ratio between different stresses 75 degrees and I found that this picture in their publication was upside down but nobody noticed but when I made the slides I thought something is wrong so I just flipped the picture this is a 90 degree blade and this is the material ok now if we look at cutting processes and we want to go further with theory first we have to talk about some definitions so this is the material we have a blade under an angle alpha here we have the angle beta just like with passive and active soil failure the blade has a height hb you could also define a blade length it's a matter of definition but I use the height of the blade and here you have the we call it the thickness of the layer cut you could call it a height but I call it the thickness I'm cutting a certain layer the edge of the blade we call point A then where the shear plane comes to the surface we call point B and the top of the blade we call point C then you have a certain cutting velocity which is in the positive direction and we have two forces horizontal force and vertical force you can define many more forces finally you need two forces under 90 degrees those are the forces those are the horizontal and the vertical force what you can see is that the vertical force the positive direction is downwards so if we find a positive vertical force it means the force is downwards so it's pulling the blade into the soil and that's what happens in certain cases the types we saw I added one type and that's the curling type because we found that if you have a very thin layer of material with rather high adhesion but it could also happen under other circumstances then the normal force in the shear plane is not strong enough to push the material all the way over the blade the adhesive force is holding it up and it will either start curling or it will crack it will bend buckle whatever you want to call it in offshore drilling they call it bowling bowling of the chip so in fact this also happens when you are drilling at 5 or 10 kilometers for oil and there the thickness that you are cutting is like 0.2 millimeter so that explains why they have this problem because it's a very thin layer so thin layers high adhesion in some way gives you a curling type and that's also what happens in steel cutting then if the layer is thicker and it's always a combination of a thicker layer of material the adhesion is a little bit less and the force in the shear plane is big enough to push the material all the way over the blade I always compare it with cheese cutting if you have very young cheese it will stick to the blade and it will crack if you have medium cheese it will nicely flow over the blade and if you have very old cheese which I prefer by the way you will get a crack ja so this is the tear type with those tensile failures and then this is what happens in sand and like sand didn't have cohesion and tensile strength and you get all those discrete shear planes in sand by the way the distance between the shear planes we found from many tests in the laboratory it's like 1 to 2 centimeters it depends on the sand of course but the magnitude is 1 to 2 centimeters now cutting forces the first step is a generic equation which is valid for all types of material and that means in the picture we need to have all the forces that could occur it doesn't mean they will occur but they could occur so which forces do we have first of all the eye the eye stands for inertial force the inertial force is the force because the material has to accelerate and although I said there is no acceleration while there is no acceleration of this lump of material but before a particle passes the shear plane it has no velocity at all after it passes the shear plane it has a velocity related to the velocity of the blade so it must have accelerated so there is an acceleration force and this acceleration in force in fact is the same force that you get when you have the hose of a fire truck and you open it you feel that impulse force pushing you back with the same equation you can calculate this inertial force so that's the first force secondly the cohesive force if the material has a shear strength a cohesion then you just multiply that cohesion by the cross section of the shear plane and you find your cohesive force then we always have a normal force and if you have an angle of internal friction you get a shear force so that's the N1 and the S1 and if you combine the two you get the K1 so the K1 is not an extra force it's the sum of N1 and S1 it's just the factorial sum of the two so those are the force then in many soils underwater because dredging occurs underwater there could be poor pressures en if the poor pressures are not equal to the hydrostatic pressure you either get an over pressure or an under pressure we always talk about under pressures and this under pressure will also result in a force on the shear plane so that's the W the W stands for water so it's the water poor pressure force then we have the G which is the weight of this lump of soil and on the blade you get the A which is equivalent to the C but now it's the adhesion the adhesion force which is the external shear strength of the material so that's the A very weak clay usually has an adhesion which is almost equal to the cohesion it can never be bigger but it could be equal to dry clay but then we talk very weak clay we talk about let's say 1 to 20 kPa then it's relatively weak and the clay that you use to play with when you were little at home would be in that range it's weak clay and I think also the clay they use for pottery would be somewhere in that range but you can also have very hard clay like 400 kPa and that's very dry and then usually on the outside if you would touch it you would hardly feel any adhesion and in fact if I would even have harder clay then you have a transition between clay and certain types of rock rock doesn't have adhesion at all so the ratio between the adhesion and the cohesion is decreasing with increasing shear strength then you also have the normal force and the shear force under an angle delta delta is the angle of external friction so we had the internal friction here we have the external friction and this external friction can never be bigger than the internal friction usually it's about two-thirds of the internal friction so if you don't have any value you just take two-thirds en we have the poor pressures the W so those are the pressures then if you look at the forces on the blade because here we looked at the forces on the soil well here the forces are exactly the same but in opposite direction action is reaction so you should always look carefully am I looking at the forces on the blade or am I looking at the forces on the soil because there is just a plus minus difference now there is one more thing which we normally don't use but in the curling type if you would have a curling type of process you need this and that's the moments well most of the forces they are parallel either to the shear plane or to the blade and that means they will not result of the blade that was point A I think but which forces result in a moment well on the shear plane you have the N1 and the W1 we assume they are at the same point of action so they have the same radius so this will cause a moment and on the blade you have the normal force and the poor pressure force we also assume they are at the same point of action and this is the radius so basically I can create three conditions out of this I have my equilibrium of horizontal forces vertical forces and moments and with those three I can solve most problems we always have the beta which is left in the equations and for the beta we do roughly the same as with passive and active soil failure but the principle is that nature is lazy so nature wants to do this with a minimum of energy which we call the minimum energy principle well what is energy in such a cutting process energy is the horizontal force times the cutting velocity the vertical force is perpendicular to the velocity which contributes to your power to your energy which is consumed so if you say ok horizontal force times velocity is the cutting power and I want to minimize the cutting power it means I have to find the beta where the horizontal force is at the minimum and that's the principle we use to find the correct beta in this case we cannot solve it analytically you have to solve it numerically and that has to do with the nature of many of the equations which are non-linear and sometimes implicit and that just means you cannot solve it directly you need a computer to solve it in most cases what is the solution so I'm not going to give you all the equations solve it, this is the solution so if I have this horizontal and vertical equilibrium and I solve it then for the k2 which was the force on the blade I find this equation and then once I know the k2 I can calculate the horizontal and vertical force on the blade so that's the bottom two equations but if we look at the equation then we can see something we can see that each force is independent in the equation and that's very nice so there is no cross multiplying or anything each force is independently in the equation so here we have the pore pressures on the blade pore pressures on the shear plane the weight the inertial force cohesive force well the fact that they are independent in the equation is very nice because now we can say in a certain type of soil we can delete a number of forces if they are not applicable for that type of soil and what's left is the equation for that type of soil so we will start with dry sand if I have dry sand I will not have the W2 and the W1 so they are zero is the G important? yes because there are hardly any other forces so I need the weight of the soil so the G is important is the I important at high cutting velocities it is at low cutting velocities it's not if you say ok what's the difference between 1 meter per second and 10 meters per second that's a factor 100 because in the I the velocity is squared so probably at 10 meters per second it's relevant but at 1 meter per second probably I can neglect it so it depends on how fast I'm cutting do I have a C and an A cohesion and adhesion not in sand sand doesn't have cohesion and adhesion so in dry sand only the G and the I are left wet sand well we have water in the pores so we will have those pore pressures and those pore pressures can be rather big that means I can neglect the weight it doesn't play a role the inertial force at very high cutting speeds but not at normal cutting speeds so normally I will also neglect the I and the A it's still sand so I don't have cohesion and adhesion so I can neglect them 2 other things I should look at the delta and the phi sand has internal and external friction so they have a value so I have to take them into account then clay well clay doesn't have those pore pressures the weight can be neglected inertial force can be neglected but clay has cohesion and adhesion but clay in our case and I will explain next week doesn't have a phi and a delta you don't have internal and external friction at very high velocities at very low speed like 100 years clay could have internal friction we call it the phi is zero concept but I will explain that next week so we have the C and the A term and those two are zero well it makes the equations more simple then the last one is rock if we look at rock we don't have those pore pressures under normal conditions at atmospheric conditions G is neglectable I is neglectable rock has shear strength so we have to take the shear strength into account rock doesn't have adhesion on the outside but rock has internal and external friction so we have to take those into account so here you have one equation and depending on the type of soil you can delete a number of terms or variables and that way you can get an equation for each type of soil ok that's it for today and then tomorrow we will continue