 In the previous lecture we started discussing about problems of structural stability analysis, we will continue with that topic, and this lecture also will be by and large introductory in nature, so we will begin by quickly recalling what we discussed in the previous lecture, so one of the problem we considered was that of a beam column that is a beam element which carries both transfers and axial loads, so in the absence of axial loads we have, we know how to analyze the structure, the questions that we are now considering is what is the effect of presence of axial loads on the transfers response of the beam, so we showed that response can be generically written as response with P equal to 0 multiplied by a modification function, and this modification is interesting because this function is not well behaved for some values of P, specifically it becomes unbounded that would mean the response becomes unbounded, so it is this possibility of structural behavior that we are interested in investigating. So we considered the problem of a beam, single span beam carrying a concentrated load at a distance C, and we analyzed this problem and we showed that for a case of a single concentrated load at the mid-span, the mid-span deflection was delta naught is a mid-span deflection when P is 0, and that is modified by a function called chi of U which is 3 tan U – U divided by U cube, and similarly the slope at the support is slope when P equal to 0 multiplied by a modification function, and similarly maximum bending moment is modified by this function, the U is a load parameter is a function of P, and this function chi of U, epsilon of U, and xi of U are known as stability functions, and their behavior as a function of U is shown here, what is interesting is in the, as U approaches value of pi by 2 this magnification factor becomes unbounded, so presence of an axial load can have dramatic effects in the neighborhood of this critical values. In this region when load is relatively small you can see that the value is very close to 1, and we could afford to ignore the effect of axial loads on transverse response, but if you are somewhere in this region the response becomes significantly amplified. Now another important feature that we noticed was the principle of superposition is no longer valid in its traditional form, it needs to be modified, for example if there is a beam carrying loads W1 and W2 and axial load P, if we analyze this beam problem under the action of W1 alone, W2 alone and P alone and add the responses we will not get the response to this system of loading, what on the other hand a particular form of superposition is still possible where we will analyze the response under W1 and P, W2 and P, and if we add these two responses we will get response for this system. Using this modified version of principle of superposition and treating this as a building block, the solution to this problem as a building block we solved a few problems, specifically we solved the problem of a beam loaded by two end couples, and we derived the flexibility matrix for this system which relates the end rotations to the applied moments, and this is the flexibility matrix, and in the limit of P going to 0 this reverts back to the traditional flexibility matrix, and in the P in the neighborhood of P critical these elements of this matrix become unbounded thereby indicating critical behavior that should be of engineering interest. We also considered what would happen if axial load is applied eccentrically, so in real applications one cannot apply axial loads truly axially, and one possible way of depicting that is the load is applied within eccentricity E, we analyze this problem and we showed that the maximum bending moment in this case is given by M0 into secant U, if a moment is applied without the presence of axial load M0 will be the moment, but due to the application of simultaneous presence of axial load this is a modification factor, and the bending stress due to this we have computed, and this is commonly used in design of steel structures. Now we started discussing briefly about indeterminate structures, suppose we have a property cantilever carrying UDL and an axial load P, so what we did was we released this fixity condition and the problem became statically determinate, we have solved this problem, and we considered another problem where to this system we added a unknown moment M, now this moment M was selected such that the rotation at B is 0, when we add these two solutions the net rotation at B must be equal to 0. Now in today's lecture we will continue discussing behavior of indeterminate structures and also we will discuss issues related to imperfections and ideal situations, and we will consider a few typical situation arising in structural stability analysis through a few examples so that we become familiar with the issues that we need to eventually address using computational tools. The material that I am going to present in today's lecture are covered in the books by Timoshenko and Gehre and Simetsis and Hodges, and some of these examples have been taken from these references. So we will now consider, early in the previous example we considered a proper cantilever beam, now we will consider a beam which is fixed at the two ends carrying a UDL, so what we will do is again we will release the two fixity conditions and replace it by and conditions are shown here, and then we will solve another problem where the beam is loaded by two moments MA and MB. Now because of this load Q naught there will be an end rotation here and because of these moments there will be another end rotation here, will this MA and MB are unknowns they need to be selected such that the rotations that are caused here and rotations that are caused here when added become 0, because in the built up structure there is no rotations at the end. The idea of splitting the problem into these two situations is we have already solved these two problems and we are using principle of superposition in the modified way that we enunciated, so we select MA and MB such that net rotations at A and B are 0, so by symmetry you can, we can reduce that this MA must be equal to MB because loading is symmetric, boundary condition is symmetric, so this will be M naught, so the condition when translated into the solution that we already derived becomes this QL Q by 24EI chi of U plus this must be equal to 0 and based on that we get M naught, so once M naught is determined I know the complete solution to this problem and this problem, this problem is already solved, once M naught is determined this problem also gets solved and we can add the responses to construct response of the beam for this situation. Now I leave it as an exercise for you to verify that the maximum bending moment in this case at X equal to L by 2 is given by this, this is the maximum bending moment in the absence of axial load P and this is a modification factor. Now this is a special case where the load is now a concentrated load, we can again we release the end loads and solve these 2 problems and again select MA and BMB so that theta A is 0 and theta B is 0, here there is no advantage of symmetry, so we have to find out MA and MB separately, so we get the 2 equations which we should be using to find MA and MB. Now this approach can be generalized to obtain generalized 3-moment equations, similarly we can get slope deflection equations, you can do moment distribution whatever you traditionally doing we could do this, so this is a very classical set of tools, the thing that we are trying to do is we are just investigating how all these well-known results can get modified because of now our additional consideration of presence of axial loads. Now I will now go on considering few situations as I said so which sensitizes us to the kind of issues that are of relevance in the dealing with stability problems. Now let us consider a beam which is supported as shown here, an idealization for this situation is we will replace these 2 members by equivalent rotary springs as shown here, and how do we evaluate these rotary springs? We will consider these 2 members separately and find out what moment I should apply here so as to produce unit rotation, so K theta 1 is a force required to cause unit rotation at EI, that is why this K theta 1, similarly K theta 2 will be the moment required to produce unit rotation at B and that is K theta 2, that K theta 1 and K theta 2 will be function of properties of this member, once this is done the problem is now how do I solve this problem? So we consider this problem where there are no end springs we have solved this problem, now again the solution to this problem now needs to be handled as follows, we select MA and MB here again MA and MB are unknowns, instead of solving this problem we are solving this, instead of solving this problem we solve this problem first and the rotation here won't be 0 but it will be MA by K theta 1, so K theta 1 is already known, so that must be equal to this and MB similarly is given by this equation and by solving these 2 equations we will be able to find MA and MB and subsequently we will get solution to the original problem as shown here. Now in the previous lecture we considered the effect of eccentricly applied loads, that is this axial load was not applied truly axially but at the support there was slight eccentricity now why must eccentricity be limited only at the point of application of the load at the supports, there can be imperfections which is distributed throughout the beam, suppose if we consider that situation, suppose the beam has an initial imperfection which is Y naught of X and then we apply axial loads P, so the equation for equilibrium of this system will be EIY double prime plus the bending moment at this section will be P into initial imperfection plus the deflection due to YM, so this is given by this, so here again we must understand the reason why we are writing this equation in this form is that we are looking for equilibrium positions in the neighborhood of 0 displacement position and this equation is valid only under the assumption that there exists this neighboring equilibrium state, so if it exists this will be the equilibrium position, so that point must be borne in mind. So now we can deal with this equation, so this is the equation, now PY naught can be taken to the right hand side, so this behaves like a distributed load now, and there is an axial load all, so they are not independent they are related by the same multiplication factor P, that means initial imperfection here creates the same effect as that of an externally applied load, now to proceed further what we can do is we can expand Y naught of X in Fourier series involving sine functions and this if we decide to retain the first term in this approximate for sake of discussion A1 sine pi X by L, so we are considering an imperfection which can be reasonably be approximated by a half sine wave, now I substitute it here, so I again divide by EI and call P by EI as lambda square, so I get Y double prime plus lambda square Y as this, prime is differentiation with respect to X, so this, now we can this solution to this equation will have a complementary function and a particular integral, now this is a boundary value problem, so the boundary conditions are at X equal to 0 and X equal to L, so at X equal to 0 Y is 0 and at X equal to L Y is 0, so if I impose these two conditions I will get Y of X to be in this form, okay. Now if I now consider the total displacement at the cross section it will be Y of X plus Y naught of X and that I can write in this form, now that means final deflection is the initial imperfection multiplied by a modification factor, so this is the initial imperfection because of presence of axial loads and initial imperfection gets amplified through this factor, mind you there is no transverse load in this problem, it's only the axial load on a imperfect beam, so the modification factor here is given by this, now lambda square we know is square root P by EI and we can substitute that and a critical condition will occur when lambda square goes to pi square by L square, so if this goes to 0 then this modification factor becomes unbounded and that this condition will be satisfied when P is P critical which is the Euler buckling load for the beam, so if P equal to P critical slightest imperfection would get dramatically modified, that means whatever the slight deformation of the beam that already exists as a imperfection because of axial loads it can get dramatically modified provided the axial load is in the neighborhood of this critical load, okay. So the axial loads reaching the critical value again turns out to be of engineering importance, so we can now simplify this expression we have the total displacement, now maximum bending moment will be P into this at X equal to L by 2, so if I put that and substitute for lambda square in terms of P critical and P I will get the maximum bending moment will be PA into this. Now the axial stress if you find due to axial load P there is a P by A plus this is a contribution due to initial imperfection and this is again like a second formula that we have encountered earlier this is the Perry Robertson formula which again is commonly used in metal structure design. Now we can make a remark here that if you examine the nature of this equation it is Y double prime plus lambda square Y is minus A lambda square sine pi X by L and these are the boundary conditions and these are the solutions. Now as the response grows that means if you are in the neighborhood of P equal to P critical the nonlinearity steps in, so material nonlinearity can come in and so on and so forth and consequently the amplitudes may be limited by a possible nonlinear behavior of the system, although a linear theory depicts that amplitude becomes unbounded moment other nonlinearities are included because response amplitudes have become large then the amplitudes may not really tend to infinity. Now we can compare to gain insight into the behavior we can compare this situation with problem of resonance in single-degree freedom systems, see the governing equation here is Y double prime plus lambda square Y excited by a sinusoidal function, so suppose if I now consider a single-degree freedom system undamped driven harmonically by harmonic load the equation will be X double dot plus omega square X is P cos lambda T, so let's assume that the system starts from rest, the main mathematically although the field equation has the same form here the important difference is this is a boundary value problem whereas this is an initial value problem, okay, accepting for that the structure of the differential equation is exactly the same, so the nature of complementary function and particular integral everything will be same but once you start imposing the boundary conditions here or the initial conditions here the solution would differ, so here X of T we know we have got it in this form, so at resonance a critical condition is reached as lambda tends to go to omega, just as P becomes P critical here, so in that case the solution now we know it becomes a sine lambda T which is modulated by a linearly varying function of time. Here as lambda goes to omega and T tends to infinity, X of T tends to infinity, so this is the response, this is how the time envelope grows and this is how, so this shoots to infinity as T tends to infinity. Now in this case what limits this amplitude even in a linear system what limits the amplitude is presence of damping, okay, and of course if response becomes large here also nonlinearity can come in but even when nonlinearity is not included in the response the response here doesn't become unbounded, because damping presence of damping ensures that at resonance in steady state the response amplitude is limited, so this difference must be I mean if you analyze the difference you will get further insight into how this structure is behaving. Now we have considered so far three types of problems, we considered actually loaded single span beams in a way presence of a transverse load or this load being applied eccentrically or there being a presence of imperfection or a combination of all of them in a way depict situations which are far from ideal, an ideal situation will be there is a straight beam carrying no lateral loads, transverse loads, and it is loaded axially by truly axial loads, there is no initial imperfection, there is no transverse load, there is no eccentricity in applying P, that is the ideal situation. The departure from ideal as we have seen can create amplification of the response especially when P is in the neighborhood of certain critical values. Now why this happens? So to understand this we can ask the question how about the study of the ideal situation itself, does it tell us anything about why this system is behaving in this manner, so to initiate that discussion we can consider the ideal situation where there are no transverse loads, P is applied truly axially and beam axis is straight with no imperfections, so we can do a thought experiment, we consider beam loaded as shown here satisfying these specific requirements as felt out, what we will do is we will increment P in steps of, P I will start with 0, delta P to delta P so on and so forth. Now at every increment in load, at 0 of course this transverse displacement is 0, so what I am plotting here is a load deflection diagram here where load is plotted on Y axis and delta is on the X axis. Now when P equal to 0 of course response is 0 we start here, now P is increased to delta P, then what we do is we pluck the beam and allow it to vibrate, and because the beam will be damped so there are two possibilities the beam may return to its original position or it may not return to original position, it so happens that it will return to the original position if P is small, okay we will find explanations for all this, this is some kind of a preview, as we go on doing this for 2P, 3P etc. we will soon reach a stage where when you pluck the beam it oscillates but it doesn't return to its original position but takes some other position, so the final position that it takes depends on in which direction you have plucked and by how much you have plucked, so the load deflection diagram here has two branches, if you call this as a branch this as a one there are three branches, okay, compare this with situation where in the ideal linear a transversely loaded beam and if you are loading this by P and measuring delta this will be a straight line, okay there is a single branch, it will go to infinity in a linear model, so I mean that itself that may not be realistic but as far as mathematical model is concerned that is what is implied in assuming that system is linear, now there is always a single branch here, but here and there for a given P is only one delta, so unique single branch for a load deflection diagram will be seen for this type of system, whereas here for P equal to P critical there is no unique response, the uniqueness is lost, okay, so what does all this mean? So what we are doing is we are applying load P in increments as 0 delta P to delta B at each value of P we pluck the beam from its equilibrium position and see if it returns to its original position, up to P equal to P critical the beam returns to its original position there would be a single load deflection path, at P equal to P critical the beam does not return to its original position and occupies a neighboring equilibrium state, we say that the load deflection path has now bifurcated into two new paths, that is as a consequence the load deflection path ceases to be unique, there seems to be three paths possible, we don't know there may be more this analysis we have to carry out the analysis to verify, how to characterize this mathematical, I mean of course these observations are made based on from hindsight gained by doing a mathematical analysis, so we will see now how all these statements are actually made, what should be P such that adjoining equilibrium position becomes possible, is a question we will ask, okay, so we will assume that under the action of axial load P the beam can assume an adjoining equilibrium state, so this is adjoining equilibrium state and the deflection at a point X is Y of X, transverse deflection, so that means we are starting with an a priori assumption that an adjoining equilibrium position is indeed possible, under that assumption we can write now the equation EI D square Y by DX square equal to PY, minus PY where the PY is a bending moment at cross section X, and Y is at X equal to 0 is 0, at X equal to L is again 0, now the question we are asking is for Y equal to 0 is a solution for any value of P, are there any values of P for which non-trivial Y of X is possible, that means it satisfies this equation and the boundary conditions but Y is not 0, so again this is an eigenvalue problem but now it is associated with the differential operator, second order linear differential operator, so the question is for what values of P does this equation admit a non-trivial solution, so we can solve this equation I divide by EI and call P by EI as lambda square and this solution is A cos lambda X plus B sin lambda X, at X equal to 0 Y is 0, so A would become 0, at X equal to L B sin lambda L is 0, now either B can be 0 or sin lambda L can be 0, if B is 0 Y of X is 0 which is a trivial solution that's not what we are looking for, therefore for non-trivial solution sin lambda L must be equal to 0 or lambda L is N pi, so where N runs from 1 to infinity, so at least we have answered one question now, whenever P is such that lambda L is N pi and adjoining equilibrium position is possible and there are infinity of them, and the loads appear as related to the eigenvalues of this problem and the deflected shapes are given by the eigenfunctions associated with these eigenvalues and in this case the eigenfunctions are sin N pi X by L, there exist infinitely many values of P given by P N equal to N square pi square EI by L, N equal to 1 to infinity at which adjoining equilibrium states become possible, the lowest value of P for which such adjoining equilibrium states are possible is called the critical load and it is called the Euler's buckling load, so now equipped with this analysis we can redraw the load deflection diagram now, so here if your P is less than P critical our load deflection path will trace this branch, and as P becomes P B is equal to P 1 as you pluck the beam it will occupy different positions, and if you go on increasing the load there are other branches that are possible, so it turns out that there are infinity of P criticals possible and the load deflection diagram will have infinite number of branches, but of course from engineering point of view we are interested in the lowest value of P at which the structure may, structure would lose its stability, and we would keep the load P we will design the structure in such a way that the load P, critical value of load P is below this, the P critical is placed sufficiently high so that the axial loads experienced by the structure would not come close to the critical values, the buckling mode shapes are sinusoidal functions this is the first mode, this is second mode, and this is third mode and so on and so forth, now how about other boundary conditions, so suppose if you have a beam with two ends clamped, now a clarification can be made at this point when we write problems here, when I write P like this for a fixed beam it must be understood that what is being prevented here is the rotation about this axis, and the displacement about this action, the displacement in this direction is permitted, so this convention of showing this end as fixed by using this notation it should be understood that the beam is permitted to move in the U direction, this direction, otherwise load P will not have any effect on the beam, so this issue must be understood in many of the books this is how it is represented, but unless the axial point at which the load acts the beam is permitted to move in the direction of the load all these discussions would not be relevant, okay, now for indeterminate structures like this we will start with the fourth order equation relating EI D4Y by DX4 is equal to the applied loads, and in this we will include the effect of axial loads that is then the governing equation will be P D4Y by DX4 plus P D square Y DX square is equal to 0, so sometime later in the class this lecture we will derive this equation using Hamilton's principle, but I am assuming that you are familiar with this equation, so in this revision where I am assuming that you know this, so again I divide by EI I get this equation, and the solution this is a fourth order equation there will be four integration constant A cos lambda X plus B sine lambda X plus C X plus D is the complementary function. Now the boundary conditions are in terms of Y and Y prime, so Y prime is given by this, and there are four constants and four boundary conditions, moment we impose those four conditions we get four equations, and those four equations can be written in this form, so for non-trivial solution the determinant of this inverse of this matrix must not exist, and for that the determinant of this matrix must be equal to 0, and this is a characteristic equation which governs lambda, which lambda is the axial load parameter, so those values of P which satisfy this equation permit the existence of a adjoining equilibrium state. So if you simplify this there are two solutions that become possible, sine lambda L equal to 0 or tan lambda L by 2 equal to lambda L by 2, this leads to one infinity of solutions, this leads to another infinity of solutions, we will see that this leads to anti-symmetric modes and this leads to symmetric modes, and the lowest value of critical load is given by this branch, and PN is given here and P critical in this case turns out to be 39.478 by L square into EI, so the buckling mode shape is given by this corresponding to this branch and the first one is depicted here. The second branch tan lambda L by 2 equal to lambda L by 2 if 1 1 solves the first root is given by 4.493 and the P critical value is 80.763 EI by L square as you can see this is higher than this, therefore this is the critical value, this is the first root of this branch tan lambda L by 2 equal to lambda L by 2, and Pi is the first root of this, and this is the first critical load from this branch and this is the first critical load from this branch, among these two this is the lowest therefore this is the critical load for the beam. So using similar logic now we can consider other boundary conditions like a cantilever beam, a property cantilever, and I have given some broad hints on how this can be analyzed, you have to write the fourth order equation there will be 4 constants, 4 boundary conditions use them write the characteristic equation solve, and you will get the characteristic equation in this case as cos lambda L equal to 0, here tan lambda L equal to lambda L, and the critical loads for this case is given by this, and for this case it is given by this. This boundary condition here at the fixed end Y and Y prime are 0, at this top end it is the boundary condition is given by, now one of the earlier lectures we have derived this you can refer back, and in the towards the end of this lecture also we are going to see something about this. Now the buckling mode shapes have like mode shapes in vibration, they here also they have orthogonality properties that can be verified by considering a pair of eigen solution, nth eigenfunction and kth eigenfunction, the eigenvalue and the eigenfunction, the nth eigenfunction will satisfy this equation, kth eigenfunction will satisfy this PN and PK are the nth and kth eigenvalues. So what we do, we multiply the first equation by, the second equation by phi N of X and integrate from 0 to L, I get this equation, and that these equations, these integrals I will integrate by parts, and doing it twice we can show that this is something that we have done several times, so we can see that after we do this we get this equation reduces to this. Now this terms inside these braces will be 0 for classical boundary conditions like clamped simply supported free and on vertical rollers, this will be 0, this we have seen earlier in context of vibration, so we need not have to get into the details again. So from this equation we have got now this equation, similarly by multiplying the first equation by phi K and integrating with respect to X and carrying out the integration by parts twice I get a similar equation where the subscripts N and K are reversed, so we get this, if I subtract these two equations I get PK minus PN into this equal to 0, and if I assume that PK is not equal to PN this integral 0 to L phi N prime X, phi K prime F X DX equal to 0 for N not equal to K, and consequently using any one of these equations we can show that EI phi N double prime of X and phi K double prime of X DX is 0 for N not equal to K. So if you recall in vibration problems the mode shapes that we obtained were orthogonal with respect to M and EI, here again we get similar orthogonality relations. Now what I will do now is I will consider a series of 4 or 5 problems again aimed at gaining familiarity with how to deal with problems of stability analysis, many of these problems can serve as benchmarks when we develop finite element method to handle these problems. So what I am doing here we will provide exact solutions within the scope of the beam theory that we are using. Suppose I have now a property simple supported beam in which a load is applied at the mid span, okay. Now again I assume that a neighboring equilibrium state is possible and I want to know at what value of P such equilibrium state is possible. So now if I consider segment AB I can write the equation EIY double prime is minus P delta X by L, the reactions are found here, you take moments about this here RA into L is P into delta, so RA will be this and RC will be by considering equilibrium in the horizontal direction I get this. So using that I will be able to write the equation for segment AB and I have Y is AX square plus B, this is the solution and for segment BC I get a similar equation but now when I am in segment BC when I write bending moment I should include effect of P also, so that will delta minus Y will be the arm for taking moments and that PY will come here and consequently the complementary function here will be having sine and cosine terms. So we can analyze this problem and we will be able to get the solution for X in the segment 1 and segment 2 and the boundary conditions now there are 4 constants of integration, the boundary conditions will be at X equal to 0 Y is 0, X equal to L Y is 0, and Y is delta at X equal to L, I am assuming the load is applied at this load is being applied at the mid span, so at Y equal to delta at X equal to L by 2, and then there is a continuity of dy by dx at this X equal to L by 2, so I have these 4 conditions and I can solve for that and I will get the critical value to be given by this. The next example is an example of a say cantilever beam which is loaded by its own weight, imagine this is a tree in a forest and how tall the tree can be before it buckles, is that type of question if we ask, this is a problem that is discussed in the book by Timoshenko and Gere, these are the formulation is from this resource, so what we do is we consider again the existence of a neighboring equilibrium position and write the equation for bending moment at a point X, the bending moment is contributed by the weight into the corresponding arms, so for this infinitesimal, this thing is a d psi, so if M is a mass per unit length, M into d psi is the mass and into gravity is the force and bending moment at this section will be Y of psi minus Y of X, that is the arm that this is the MG into Y of psi minus Y of X into d psi is the bending moment at this section due to this infinitesimal element, and if I integrate from X to L I will get the bending moment due to weight of the beam above this point, so that is what we have written. Now we can simplify this, we differentiate this equation with respect to X and carry out a simplification we get this equation and here you can see there is a X here, so it has a variable coefficient, so one of the way to do it is to make these substitutions and we can show that this equation becomes this and solution can be obtained using power series expansions and we can show that the critical length K is given by this, okay. Now another example this again to exercise our mind, suppose we consider a beam and if we assume that at the end there is a rigid element, so the deflection of the beam will be such that this will be a straight line, this won't bend, it is not flexible, so it is again loaded by P, I want to know if this is a configuration of the structure what should be the P at which a neighboring equilibrium position becomes possible, so as I said this is a rigid element and this length is A and this remaining length of flexible part is L-A, so how do we write this? Again the equation is EI Y double prime plus PY equal to 0 and this is Y double prime plus lambda square Y equal to 0 and this is a complementary function A cos lambda X plus B sine lambda X, it X equal to 0, Y is 0, so A becomes 0, now the interesting thing is what happens here, so at X equal to L-A the deflection will be A theta, okay, this deflection is A theta and Y prime of X is given by B lambda cos lambda X, so continuity of slope at X equal to L-A I know the slope is theta because a rigid element I get this equation, so we can treat now B and theta as unknowns and we have two equations and we can write this equation in this form, so for non-trivial solutions the determinant of this must vanish and this is a characteristic equation, so once we solve this for different values of A we will be able to find the critical value of P at which the structure buckles or a neighboring equilibrium position becomes possible. Another problem suppose there is a beam and I want to add a spring here in this example at the mid-span with a view to increase its axial load carrying capacity, so how far I can increase the critical load value by selecting this K, that is the question, so how do we tackle this problem? First I will ignore the presence of this spring and solve the problem I get this deflection delta say in the absence of that, and I will now select I will apply a force K delta here so that this deflection becomes 0, so the governing equation will be EIY double prime plus PY equal to K delta by 2 into X, and again dividing by P and using the notation lambda square we get this equation, this has a complementary function and a particular integral, at X equal to 0, Y is 0 therefore A is 0, now Y prime of X you can obtain by differentiating this, and at the mid-span since structure is symmetric and spring is also placed symmetrically this must be equal to 0 and this helps us to find the value of B, therefore this is Y of X. Now at X equal to L by 2 this is my delta, now that would mean this if you take on the other side delta cannot be 0 therefore this must be equal, the term here must be equal to 0 and that leads to this characteristic equation, so we can solve this and find out the critical value of load U for a given value of K. Now if you really solve this we can see that if you rewrite this equation, the characteristic equation I can write at tan U by U equal to 1 minus 4P by KL, if K goes to 0 then U goes to pi by 2 and that is our Eulerian buckling load this is this, now as K tends to infinity then this is tan U equal to U and U becomes 4.493 and for K tending to infinity the critical value will be given by this, that would mean by inserting the spring you cannot increase the carrying capacity beyond this, if you don't have the spring your carrying capacity will be this, by inserting the spring no matter what you do you need not have to select a spring you know you cannot exceed this carrying capacity, okay, so this is the highest value to which the load carrying capacity can be increased. Another example, imagine there is a single span beam and in the end there is a rigid disc and this disc is loaded by load P, okay, and this disc can rotate along with the beam, so in the deformed geometry the disc could have become like this and the beam deflection will be something like this, that means if an adjoining equilibrium position is possible it may be like this. Now we will find the reactions now, see if this is angle is phi and this is A, so this load will be acting at L minus this distance A sine phi, A by 2 sine phi and this load will be acting at L plus A by 2 sine phi and consequently the reaction, you can take moments and find out the reactions, these reactions are functions of the angle through which the disc has rotated, which is the angle through which the right hand support, right hand the beam has rotated at the right support. So this is the equation, now bending moment, the equilibrium equation is given by this, once we find the reaction we can write the equation for the equilibrium and by dividing by EI etc. I get this equation. Now again the complementary function is this and this is a particular integral and we impose a condition at X equal to 0 and X equal to L, Y must be equal to 0 and we get this value of A and if we now approximate sine phi by phi I get Y of X to be given by this, so now dy by dx, see phi is an unknown here still we have to find that, so I will find dy by dx and evaluate it at X equal to L, moment I do that and I know it is minus phi, so I get an characteristic equation, phi cannot be 0 therefore the terms inside the parenthesis must be 0 and from this I get the load, critical value of the load parameter 3 by AL, therefore P critical is given by this. Now we will quickly review the equation of motion for actually loaded beam as shown here using Hamilton's principle, so we consider for sake of discussion, for purpose of discussion a single span beam which is supported on two transfer springs K1 and K3, K1, K2, K3, K4, K1 and K3 are translation springs, K2 and K4 are rotational springs, so suppose this beam is deflected like this and an element DX would have become element DS and using the geometry here, geometry of this triangle we know that DS square root DX square plus dy square and we can reorganize this term to pull out, take the square root, take DX outside and I get this and by expanding this and retaining only the first term I get this equation, so this is the length of this, what was DX has become DS, now the total length of this curved beam profile is integration of DS, so it is 0 to L, this integral DS which is nothing but integral of this quantity from 0 to L, so this I can write it as L plus half integral 0 to L dy by DX whole square DX. Now SS bar minus L is the change in the length that is delta, so that is given by this equation, why we need this now because we want to find out the work done by this load on the beam, so we need to know how much the beam has moved, so this is very important unless P does work on the beam its effect will not be felt, so if you go back to the convention that we use in notations this is very important, so the kinetic energy is half MOI dot square this, strain energy has now contribution due to bending which is given by this and there is strain energy stored in the springs that is given here and there is work done by axial loads which is P into delta which is given by this, so now therefore Lagrangian will be this and this is the additional term due to the strain energy in the springs, so we can now use the variational argument this we have discussed in first or the second lecture of this course so that can be used and we if you do that we will get the governing equation in this form and if you use the convention of prime for a special derivative and dot for a time derivative we get this equation. And the set of four boundary conditions you can show that they are given by this set of four equations, so these needs to be augmented with the two initial conditions or initial displacement and velocity, to this of course we can add the effect of any external excitations or time dependent boundary conditions as in case of earthquakes of course damping terms and so on and so forth, so this would be the governing equation would then be the generalization of static beam column for a dynamic situation, okay, so in our analysis subsequently we will consider some of this as we go along. Now what we have discussed till now although I discussed this governing equation of dynamical systems we have basically focused on elastic stability of statically loaded systems, so before we embark on finite element formulation what we will do is we will consider questions of stability of dynamical systems, so we will quickly review what are fixed points of a non-linear vibrating system, what is meant by stability of those fixed points that we will have bearing on some of the discussions to follow, so in the next lecture we will consider non-linear dynamical systems and investigate the fixed points and their stability and in doing so we will introduce a notion of bifurcation and see how the ideas can be developed further, so with this we will close this lecture.