so good morning to you we are now on to the second lecture in this course on advanced structural analysis we will be continuing reviewing basic structural analysis so these are the seven modules we are on module one and in this module one we basically have three parts the introduction we covered yesterday today we look at statically determined structures and in the next class we will review work in energy methods as i mentioned earlier the textbook that we will be consistently referring to is structural analysis which is authored by me that book has five parts we finish with part one in the session yesterday today hopefully we will try to cover most of part two in a quick manner so let us come back to the big picture we are dealing with a structure we are actually dealing with a model of a structure in this course in structural analysis our structure will be a skeletal structure we are removing the spatial elements from the structure the structure is made up of elements and the elements could be space frame elements plane frame elements grid or beam elements truss elements of course the most generic is the space frame element all other elements are special cases of that one element and the elements are interconnected with joints those joints are modeled usually as rigid joints or pin joints we have discussed these you can also have semi rigid joints and at the boundaries of the structure you have supports and it is important to identify them correctly and there are different types the most generic is elastic and when you restrain movements you can make it fixed or guided fixed or hinged or roller on this structure we apply loads and the loads are of two types direct actions which you indicate with those arrow marks or you can have indirect loads and there are three types as we discussed you have support settlements you could have construction errors you could have environmental changes and we are interested in finding the response of the structure both the force response and to some extent the displacement response so if you look at the structure in its entirety you have two fields you have the force field some of the items in the force field come from the loads which you know but many of the items you do not know and they come from the response they include the support reactions and the internal forces and you also have the displacement field which is related to the force field most of the displacements are unknown so they are in the response side but sometimes displacements come as a kind of indirect loading so they come to the load side now most structures are indeterminate they should be stable definitely and the indeterminacy could be either viewed for the purpose of analysis as static indeterminacy or kinematic indeterminacy you have a choice in your solution path of indeterminate structures you could use the force method of analysis or the displacement method of analysis if you were to do things on your own manually without the help of a computer then you would obviously choose that path which involves less effort so you have to look at the degree of indeterminacy which is the degree of static indeterminacy versus the degree of kinematic indeterminacy and of course you have to decide on what level you want to do this analysis so ah should do you need to do a non-linear analysis which is little complicated as you know or is it enough to do a linear analysis do you need to do a dynamic analysis or is it enough to do a static analysis do you need to do a probabilistic or stochastic analysis where you have uncertainties you have uncertainties in the loads for sure you also have uncertainties in the material properties in the structure and therefore you will have uncertainties in the response or shall we assume that everything is deterministic well in this course in structural analysis we take the the simplest form of analysis that is we assume that the analysis is static its linear and its deterministic that is a starting point and its good enough in practice we should also know in advance courses how to do nonlinear analysis we will get a taste of it in seventh module of this course how to do dynamic analysis often we convert dynamic loads to equivalent static loads or we apply factors called impact factors or dynamic amplification factors to taker of the amplification caused by the dynamism and we take care of the uncertainties through load factors in design so if you are really skilled at analysis you should reach to a level where you can do non-linear dynamic probabilistic analysis but we are beginners and so we should first learn how to do linear static deterministic analysis so this is the scope that is covered in analysis of determinants structures you know we all begin with finding support reactions which we know well enough we should know how to analyze beams both find out the bending moment shear forces as well as finding deflections in slopes we should know how to analyze trusses analyze funicular systems like cables and ideal arches analyze plane frames and learn how to draw influence lines we have done all this so here we'll just do a quick review a quick overview overview of some of these topics now if you look closely at the response of the structure the force response and the displacement response you will find that the correct response the force response must satisfy equilibrium which means it must be statically admissible and the displacement response must be compatible it that preserves the integrity of the structure the elements are interconnected the way they are supposed to be if the joints are rigid then the angle between the connecting elements do not change so your correct response must be both statically and kinematically admissible and there's a third requirement that is the force displacement relationships should be satisfied which in turn depends on the constitutive relationships in the material the stress strain relationships and so on and you can satisfy equilibrium in a variety of ways the way that we normally refer to is direct equilibrium you are familiar with this this comes from newton's law the resultant force on that structure is supposed to be zero and if you interpret that in terms of scalar quantities in cartesian coordinates you end up with six degrees of six equations of equilibrium sigma fx equals 0 f y equal to 0 i have z equals 0 sigma m x equals 0 sigma m y equals 0 and sigma m z equal to 0 this is called direct equilibrium very straight forward elementary if we are dealing with the planar structure and most of our example structures are planar which means all the elements in the structure lie in one plane and the loads also act on that plane then we can simplify and eliminate some of these equations which of these equations would you eliminate let's assume that the structure is in the x y plane so the z uh f z doesn't need to be applied what else can you get rid of m x and m y because you are taking moments about the z axis which is in the plane of x y so if you are dealing with a planar structure for every free body or for the overall free body your number of equations reduced from 6 to 3. we will also look at simple ways of calculating deflections in beams and you're familiar with the direct integration method the moment area method and the conjugate b method we will look at in a subsequent class the principle of virtual work which can also be used to find the force response and the displacement response and we can also use the energy theorems to do exactly the same but that's something we'll reserve for later right now let us do the first thing that we normally learn in structural analysis so here is the example of a two span continuous beam as you can see that there are four reactions possible three vertical reactions and one horizontal that horizontal reaction is clearly equal to zero if you apply sigma f x equal to zero but you have three vertical reactions and you have only two equations of equilibrium left right you can mix and match the equation but you have only two independent equations and if you apply sigma fy equal to zero it would turn out that va plus vb plus vc will add up to the applied load 800 kilo newton and if you take moments about the point b for example you can show that va minus vc is 400 that's all that you get you can write down more equations but they will be linearly dependent you cannot get so this is not enough to solve the problem you can imagine having multiple solutions for example i have shown you here four possible solutions and each could be correct because they all these solutions satisfy those two equations you have three unknowns two equations so you can have multiple solution sets so which of these is correct let us say i have i am sure each one of you in this room about 70 75 students here can come up with your own unique combination of these solution sets how do i know which one is correct or closest to the truth is there a simple way to find out can i ask all of you to do some extra effort and then from that i will be able to judge which is closest to the correct solution sir what is the additional quantity that i can ask you all to calculate compatibility is a big thing displacement reflection energy energy if i ask you all to calculate the strain energy in the beam see once you have these forces you have the bending moments once you have bending moment distribution you can calculate strain energy which is an internal energy internal elastic energy so you will all have different values of strain energy which you think is closest to the correct solution the one which is least so you know there are powerful ways of doing this but the other thing to do as some of you suggested is to check whether you satisfy compatibility or not now very clearly the first solution for example 400 400 0 will not satisfy compatibility because certainly vc cannot be 0 if that contact is established between the support and the b the only way that could turn out to be a correct solution is if there is no contact right which means you will probably get a deflected shape ah which looks like which looks like can you see the first deflection shape the dotted line that means there is no contact it is like there are supports only at a and b and it will the beam will definitely lift off from the support c then you could get this solution there is nothing wrong with that but it does not it is not kinematically admissible so the correct solution which you will learn to do is one in which the beam comes back to the support c and you will find that there is an uplift at that location and you have a negative force which means a force acting essentially downward and we can prove these numbers later right so the correct solution must not only satisfy equilibrium but it must also satisfy compatibility you will find that if your structure is statically determinate you do not need to explicitly satisfy compatibility it gets automatically satisfied it is enough to satisfy equilibrium and you can prove this will come to the proof when we study matrix methods ah subsequently at this point i want you to realize that when we draw diagrams like this we tacitly assume that a b and c are support locations that is these are reactions but this diagram does not reveal anything and it is necessary to know how to distinguish between a free body diagram and a loading diagram this is only a free body diagram if you want a loading diagram then this would be the correct picture to draw you do not show the support reactions you show the supports the big difference between this diagram and the previous diagram is that this diagram has with it an associated deflection diagram that means whatever deflected shape you draw you must show that the deflection is 0 at a b and c and that there is a slope compatibility at b right so this is an important distinction to draw because for the same free body diagram i can have multiple load loading diagrams for example this is the correct solution for this problem but look at this picture here i put an elastic spring at the middle and if that spring stiffness is 0 it is a simply supported beam and for that free body diagram this is perfectly acceptable or you could have something like this you do not have any supports at a and c those springs are very flexible and you still satisfy that free body diagram so this is an important distinction to make note of so we start with support reactions we are we begin with a simply supported beam a simply supported beam is one in which you you want stability so you need to have three constraints in a planar structure so essentially you have two vertical reactions and one horizontal reaction and if there is no horizontal load applied to that system that horizontal reaction is zero so essentially you have two vertical reactions and and you could have any loading what i have marked here w is the resultant load now the structure that you put on top of the supports could be a beam could be a truss it could be a frame could be an arch could be anything could be a boulder like we have shown here but the reactions do not change and the reactions are statically determinate and they are very easy to determine if you mark the left side as a and the right is b then if you take moments about a you can find the reaction at b it is very straightforward if you want to find the reaction at a take moments about b and intuitively you know these results are correct because the load w will be shared by v a and v b in some proportion its in proportion to to the distances so for v a the distance b by l the ratio b by l is the fraction of the load w and for reaction at b it's a by n so this is a intuitive simple way of understanding ah simply supported beam reaction let us take a slightly more complicated problem this is not simply supported here you have two hinge supports which means you have the possibility of a horizontal reaction coming in this would be normally statically indeterminate or over rigid but for the fact that there is an internal hinge provided at the crown and so what is this kind of arch called they did this deliberately why did they do this deliberately to avoid indeterminate why was traditionally people afraid of dealing with indeterminacy well they didn't know how to analyze indeterminate structures for for many centuries but even after they did they still avoided it in fact if you take our own country in india most of the bridges even today are simply supported they are not continuous they are just rigid why is that yeah because you have possibility of indirect loading and unless you are absolutely sure that there is no relative dis support settlement which is possible if you are on very hard strata or you are on piles or you are not you are able to handle temperature and shrinkage in other associated problems then you can confidently go ahead with this because we have the tools to analyze these structures so that hinge at b gives you an additional equation what is that equation that you get momentum b is zero moment about any point on that screen is zero so don't say b after b zero so the bending moment inside the structure in the arch at b is zero there is no moment transfer possible so to really take use of that condition it is called the equation of condition well in any case the vertical reactions are very easy to compute because that that 32 kilo newton load in the middle will get shared equally and that 10 into 2 that is 20 kilo newton load will get shared in the proportion vb by lva by l so vertical reactions are determinant the horizontal reaction is what we wish to now find out so you take a free body you you can take either the left or the right here taking the right free body is more convenient because you do not have any loads to be shown in that portion and then you invoke the condition that the bending moment at b is zero it is called an equation of condition and you can easily solve and get the horizontal reaction okay so this is the solution for a typical problem like this but there is nothing unique about that arch we found that the solutions were va equal to 34 kilo newton v c equal to 18 kilo newton and the horizontal reactions uh 22.5 for this structure but take a look at this structure this is a three inch portal frame um with the same height overall in the same span and the same loading located at the same locations so you will find that these solutions that you have calculated for the arch also hold good for this frame also hold good for this structure where a b is inclined as long as the loads are at the same place and they are the same and the distances overall are the same you get the same reaction so that is an important point to note when you have a statically determinate system the type of structure is not important it could be an arch it could be a frame it could be a beam it could be a cable okay now we finish support reactions by direct equilibrium we go to the next topic which is beams and we will touch on the essential point so first let's look at the basic kinematic variables what are the kinematic variables which define the displacement in a beam deflection slope slope curvature curve very good so if you take a curved shape of that cantilever beam as shown we will assume deflection delta x to be positive if pointing upwards it really does not matter but for consistently consistency let's assume upward deflection is positive then the slope of this is called slope or its also called rotation it is a derivative of that deflection and the derivative of the slope turns out to be approximately equal to the curvature the curvature is actually 1 divided by the radius of curvature at that point and you know that for small deformations it turns out to be approximately equal to delta double dash okay so this is something we begin with and then you have static variables what are the static variables for a beam you have the load intensity which is usually given to you shear force and bending bending moment right perfect so again these are the sign conventions we will assume q of x is the load intensity usually measured in kilo newton per meter and i want you to understand that when the intensity changes along the span you can't afford to talk about meters so it is actually more correct to use the language newton per millimeter because it can change every millimeter but it is equivalent to kilo newton per meter ok so that is important to note so that is q of x the derivative so how is it related what is the relationship what are the basic differential equations that relate these three quantities shear force is the derivative of okay so you can prove this by taking a small element of the beam and just applying direct equilibrium and you can prove that the load intensity is equal to the derivative of the shear force the shear force is the derivative of the bending moment and what else can you prove bending moment is related to the kinematic variables how either well if you invoke yeah that's right if you invoke simple bending theory the euler bernoulli principle m by i is equal to e by r 1 by r is curvature so you can prove that bending moment is nothing but e i into delta double dash of 5 whichever you wish to say so that is the relationship between static variables and kinematic variables and actually you can express all the static variables in terms of deflection e i is called the flexural rigidity it is a product of e is the elastic modulus assuming homogeneous elastic material i is the second moment of area it's an area property so the you can show that the shear force is e i into delta triple dash and therefore q is e i into the fourth derivative of the deflection so you can if someone asks you what is the basic differential equation for a beam you can either choose to explain it this way e i into fourth derivative of this deflection is the load intensity or you could say e i into curvature is bending moment both are correct right so let us quickly assimilate how this these concepts can be applied to drawing bending moment and shear force and deflection diagrams i have shown here two cantilever beams with a linearly varying load you can write mathematical equations the total load is w and you can write q of x i put a negative sign because in the sign convention i have shown you q is positive when pointing upwards but if it is a gravity load it will point downwards so i put a minus sign there now i am going to draw two shapes of the shear force diagram which is related to the intensity diagram this is one shape and this is another shape let us say the top shape is a and the low one is b which of these two shapes is correct for that first cantilever b a is correct why is a correct right so you have to look at the slopes you have to look at that relationship q is equal to ds by dx so let me use this pointer sorry right so you will find that the slope of the shear force diagram here the slope is zero its tangential here must be the load intensity which is zero whereas in the lower case there is a definite slope so its so actually this diagram belongs belongs there for the other b so you can see where the slope is 0 where the load intensity is 0 and where the slope is maximum there actually the load intensity is max so it is a very easy way of understanding the relationship and you can write equations if you wish what about the bending moment diagram well it takes this shape the slope of the bending moment diagram is the shear force and the slope of the bending moment diagram is similar to the shape of the curvature diagram all you need to do is to divide the bending moment diagram by e i and you've got the curvature diagram you integrate the curvature diagram twice you get the deflection diagram after you apply the boundary condition so here if you have a linearly varying load intensity your shear force will vary quadratically your bending moment will vary with a cubic variation therefore your curvature will also have a cubic variation therefore your slope or rotation will have a fourth order variation and your deflection will have a fifth order variation and actually you can get these quantities by just integrating but there is an easy method of finding deflections and that is a conjugate b method which we'll look at shortly now normally you are given problems where you are given the load intensity and you have to derive everything you have to derive the shear force bending moment deflection and so on but it's good to tease yourself with a different kind of problem where you are given a bending moment diagram for example this diagram which is a simple diagram a triangular shape and you are asked to predict the loading diagram what load could have cost this so what load could have cost this simply supported beam with a concentrated node that is the standard answer everybody will give because from the bending moment diagram you can pull out the shear force diagram from the shear force diagram you can pull out the free body diagram which is not the same as loading diagram and from the free body diagram you can pull out how many loading diagrams any number so the simply supported is one case which is the easiest to understand most familiar to us so this is what you said okay can can we have a cantilever beam for example yes why not you can have a cantilever beam here but this is not complete you must also have this so ah the free body diagram shows you three arrow marks p p by two p by two it is your choice which of them you want to treat as loads in which you want to treat as reactions because the free body does not change so you can have multiple loading diagrams having the same free body diagram and its interesting to take this one step further and see how does the deflected shape change for example this would be the deflected shape for the simply supported beam it will be symmetric what would it be for the cantilever you're right so you will find the deflected shape for the cantilever will be something like that and these two deflected shapes are related why are they related you're dealing with a kinematic variable so you must talk kinematic language don't talk about bending moments in free bodies they have the same curvatures at every point in that beam the radius of curvature is exactly the same so all you need to do is to have a rigid body rotation of one diagram the curvature does not change but the deflections change and that is what you do you rotate this first diagram anti-clockwise by theta and you will get the other diagram so it is very important to understand these relationships so that is 2 theta at the free end of the cantilever let us talk about some other interesting practical issues where a lack of proper understanding of basic structural analysis can lead to to mistakes now i am going to report to you some common mistakes that many practicing engineers make so take a look at a footing like this it is called a isolated footing okay you have a slab which you can treat like a beam and let us say there are three columns and ah let's say it's a symmetric system where you have three identical column nodes p p p and the spans are equal l and l and you have a pressure acting from below right that's the soil pressure and that soil pressure must be within this you know safe bearing capacity and the settlements must be acceptably low so longitudinally this is called the rigid footing assumption you assume that the soil pressures are uniform and so the total load is three p and if the width of the footing is b and the total length is l then the area total area in contact with the soil is 2 l into b and that is your soil pressure and if you want to treat it like a beam then in kilo newton per meter you can multiply that soil pressure by b again and you get get a uniformly distributed load acting upward right so this is called the rigid foundation assumption the assumption is the foundation is so rigid that you can ignore the deformations in it it will move together as a whole now the analysis is sometimes done by treating this as a two span continuous beam okay and i will show you how it is an inverted beam so this is a model done do you find anything wrong with this this is done by some engineers that is finally you have to design that footing you have to find the bending moments and shear forces in that footing so you have this load acting from below you analyze it as an indeterminate structure two span continuous beam if you do that i will show you the results you will get a shear force diagram like that and a bending moment diagram right i have drawn the bending moment diagram on the tension side as our normal convention so it will look like that clear don't worry about those values they are 100 accurate you can check them out is there anything wrong with this actually they are not rigid sorry actually they are not residents you're right actually the rigid foundation assumption is a big assumption but it's very commonly made in practice now let's accept that assumption for that assumption is this result correct sitting on the ground you're assuming a uniform pressure the deformations well they this is how people do in practice so you have to give me a clear reason why this wrong okay let me prove to you that it is wrong why is it wrong because if you work backwards and find the reactions at a b and c you will get something interesting you will get 9 p by 16 15 p by 8 and 9 p by 16 but you began by saying its p p p so there itself ah you can see that there is an inconsistency right so it is wrong it is wrong and let us do something more interesting let us say the central column actually carried very heavy loads and you had practically very little loads on the two extremities right so let us take an extreme situation where instead of p p p the loads are 0 3 p and 0. if you were to do this kind of modeling your shear force diagram and bending moment diagram will not change as long as the loading is symmetric agreed and that's something radically wrong so what's the correct solution how do we handle this you'll be surprised i've asked this question of practicing engineers in companies and they have not been able to answer i want to i am happy with the rigid foundation assumption i just need the correct bending moment diagram and shear force diagram and obviously it must depend on the applied loads you give me any combination of loads p p p or 0 3 p 0 i should get 2 different bending moment diagram how do i do that well the answer is surprisingly easy the answer is so the answer is you do not need to have jumped into the assumption of a two span continuous beam because the moment you assumed rigid foundation assumption you got a free body which is complete by itself right can't you draw a shear force in bending moment diagram from the free body so this is the confusion that is there in many engineers you mix up things you mix up free body with loading diagram once you've got a free body diagram that is all that you need you have got the statics complete you got the shape of the shear force diagram you got the shape of the minimum diagram and its easy to draw this is consistent and if it was 0 3 p 0 you got a completely different shear force and many more diagrams so the answer is so simple okay so once you make the assumption everything falls into play do not complicate things which are assumed to be simple but you are right if you want to do an exact solution then you should put springs and you know there are assumptions the soil has a modulus of subgrade reaction which you which soil test can do and you will find the actual behavior is going to be this is the behavior assumed that everything goes down by the same amount which is not true the actual behavior will be something like that this is sometimes referred to as the beam on elastic foundation analysis lets not worry about this for the present let me come to another important topic related to the beam you can draw the bending moment diagram if someone gave you the loading diagram now here is a situation where i have where i have a cantilever beam whose dimensions are given 1 meter long cross section is 75 mm wide and 100 mm deep the elastic modulus is also given to you and let us imagine the beam is right here in front of us and i let's say i pull down the free end and i release it and it will start vibrating and i leave the hall go for a cup of tea one of you takes a picture of it using a digital camera and first you've got a picture like that let's say you got a picture like that right now does the beam have bending moments yes or no why because it's bent because you remember our discussions in the last class because i have asked this question very recently in other engineering college and they couldn't give the answer because it is bending because a straight beam has changed its curvature there has to be bending moments how do you compute those bending moments from the deflection how do you get it from the deflection derivative of deflection so you can digitize a diagram and get for different location let's say you have 10 locations on that beam you can get the deflections and then you know that there is a relationship between curvature and bending moment so second derivative of deflection so you can use any suitable numerical technique like a central difference formulation get the slopes from this picture from the slopes get the curvature as a second derivative multiply that with e i you got the bending moment diagram so please remember and this is the displacement approach when you begin learning structural analysis you learn the force approach that means given some load you can find the bending moment and shear force but you can also use a deflection approach the displacement approach from the deflected shape you can extract the bending moment diagram which will look like this so the other question is can't you use the other method also surely some loading is responsible for this bending moment how do you find those loads well those loads are imaginary loads they were given a name by newton what are those forces called they call inertial forces they are caused by the accelerations of the the masses in the beam and they probably they take a shape which is similar to the ah deflected shape if the beam is executing simple harmonic motion then you you know by definition simple harmonic motion is one in which the acceleration is proportional to the diff deflection and so it would be that but actually it may not be executing simple harmonic motion so you have multiple modes and so on okay so how do you calculate deflection in beams very quickly let us say you were given this problem you were given q naught you surely can write an expression for bending moment let us say you have done it correctly i have given you the answer there from the bending moment you get curvature by simply dividing that expression by e i from the curvature how do you get rotation just integrate it integrate it once more you get deflection but you end up with some constants so you'll have two constants how do you solve for those constants one at a time yes apply boundary conditions in this case kinematic boundary conditions so at x equal to zero the deflection is zero so c delta is zero at x equal to l the deflection is also zero so you get the second constant c theta plug it back into both the equations and you've got the two equations but engineers should are more comfortable by looking at figures one picture can speak a thousand words because the old chinese saying so you should draw graphs so if you draw the graph of of curvature which is like the bending moment diagram it will look like that the maximum curvature is at the mid span no not at mid span how do you know where the maximum curvature is well where the moment is maximum where is the moment maximum where the shear force is zero so you can locate that point and find that curvature this is the shape of your rotation or slope and the slopes are negative on the left's region and positive in the right region and this is the shape of your deflection diagram it turns out to be negative because our definition of delta x is pointing upwards positive and you can locate the maximum deflection by finding where the derivative of delta in other words theta is 0 and you can write these expressions okay so we've done this before but practicing structural engineers are not do they want to be economical with their investment on time so they want to find short fast ways of doing one of the most powerful methods is the conjugate beam method so you have the kinematic variables and you have the static variables and you see that there's a definite relationship in each of these streams if you go travel upwards you if you integrate you get from one the other if you integrate curvature you get slope if you integrate slope you get deflection if you integrate load intensity you get shear force if you integrate shear force you get bending moment so there is a kind of an analogy between these two streams and that analogy was put to good use in the conjugate beam method so if you have the same beam in which you apply the curvature diagram as a load intensity diagram then the shear force in that beam would naturally correspond to the slope or the rotation and the bending moment in that imaginary beam would correspond to deflection this is a simple concept brilliant solution to finding deflection so here goes the statement if we visualize an equivalent beam called the conjugate beam with appropriate boundary conditions with the distributed loading having an intensity q of x equal to the curvature phi of x acting on the conjugate beam then at any location x the shear force s of x will be equal to the slope theta of x in the real beam and the bending moment m of x will be equal to the deflection delta of x so you can use that argument to prove the the boundary conditions that you need to apply on the conjugate beam at the fixed end where the given conditions are the deflection and slope are zero it would imply that the bending moment and shear force in the conjugate beam should be zero and they will always be zero when you have a free end you do not have the possibility of a concentrated load at the free end because the curvature diagram will be always continuous it will never have singularities so the fixed end becomes free the free end becomes fixed the hinged or roller will remain as it is simply supported the intermediate support becomes an internal hinge because you have a discontinuity in slope there and the internal hinge will be replaced by an intermediate support in the conjugate b very simple demonstration take a cantilever beam you want to find the deflection and the slope at the free end well first so you want to find delta max and theta max first draw the bending moment diagram then draw the conjugate beam the free end becomes fixed and the fixed end becomes free on that conjugate beam put the curvature diagram which means bending moment divided by e i and then you find you just analyze the beam find the support reaction at b you will find that the slope at b is given by the shear force of the vertical reaction at b and the bending moment is given by gives the ah deflection at b so you reflection is p l cube by three its a standard formula and slope is p l square by two e i very easy little more complicated problem here i i'll quickly rush through it it's a little complicated because you've got a propped cantilever with an internal hinge and you are asked to find the slopes at b c and d and the deflections at b and d so it's difficult but the easiest way you can solve such a difficult problem is the conjugate beam method so let us just go through the steps what do you first need to do analyze the beam draw the bending moment diagram you can also try drawing the deflected shape it will look like that draw the free body diagram it's easy to analyze once you separate out you know the parent and the child you can draw the shear force diagram you've got the bending moment diagram okay let's just accept it you've got the bending moment diagram then what do you do that is for the conjugate that divided by ei is the loading diagram on your conjugate beam what are the boundary conditions in your conjunctivity support is fixed b you have an intermediate support c internally see you leave in place the intermediate support and a d fix it perfect so this is the given beam and this is the conjugate beam okay conjugate beam the boundary condition that has shown now instead of putting on that beam that big complicated diagram it is good to separate out the curved parts from the straight line parts just for simplicity in calculation so if you do that and you do superposition you can actually find out the centroids of the resultant forces and analyze it so look carefully at the values that you need to compute because at b you will get two values of slope because the slope on the left side and the right side are not going to be equal but that's because you have a discontinuity in the shear force diagram in the conjugate beam because you have a support there so you can calculate all these very accurately let's just accept this this is something you should have learned in basic structural analysis just cover the concept here okay now accidentally i think we are running out of time conjugate beam method also gives a method for solving some type of indeterminate problems where the indeterminacy is not more than two for example take this fixed beam okay how do you find the fixed end moments in this beam well you can visualize this as a simply supported beam with the original loading plus the hogging fixed end moments right and if you draw the conjugate beam for this you will find the fixed fixed beam becomes free free and you can superimpose the sagging bending moment diagram called the free bending moment diagram divided by e i and the hogging bending moment diagram and if you divide by e i you got those two diagrams and all you need to do is to equate the resultant forces and apply the moment equilibrium which means the centroid should match and you can easily establish simple formulas for fixed end moments right so we will continue with this in the next class thank you you