 Towards the end of the last lecture we briefly touched upon questions related to energy methods in stability analysis and we will be looking into those issues in detail in this lecture. Before that we will quickly recall what we studied in the previous lecture, so we considered dynamical systems of the form Y dot equal to A of Y, these are autonomous systems this is a vector of functions, so A is, this is also a vector of, Y is a vector N cross 1 and A is a vector of functions, this is independent, time doesn't appear explicitly in this, then we say that it is an autonomous system, we call points at which the system state is at rest that is Y dot of T equal to 0 as fixed points, there the equilibrium points are the fixed points and these points are obtained as roots of the equation A of Y equal to 0, so we saw that for a undamped single degree freedom system the origin is a fixed point, similarly for a damped single degree freedom system origin is a fixed point, for a system with negative linear elastic stiffness and a positive cubic non-linear stiffness we saw that the origin is a fixed point and also there are a pair of fixed points located as shown here, so non-linear systems have more than one fixed point, the kind of examples that we studied we observed that non-linear systems had more than one fixed point, whereas linear systems of this kind exhibited only one fixed point. Now we were interested in studying the stability of the motion in the neighborhood of fixed points, so if X star and Y star are the fixed points where F and G are 0, for example we consider a pair of first order equations X dot equal to F of XY and Y dot equal to G XY, the fixed points X star, Y star are obtained by solving this equation and if we perturb the motion in the neighborhood of these fixed points the perturbations evolve as per this equation and we showed that the question on whether these perturbations grow in time can be answered by studying eigenvalues of this matrix and we showed that if A is greater than 0 where after doing this eigenvalue analysis for this Jacobian matrix we wrote the eigenvalues as A plus minus I B and if the real part of the eigenvalue is greater than 0 then motions grow in time and the fixed point X star, Y star is unstable, on the other hand if real part is negative the fixed point is stable, A equal to 0 is a case where the motion will be, perturbations will be periodic, so the question on whether fixed point is stable or unstable remains unresolved, so depending on the nature of these eigenvalues we classified the fixed points as node, saddle, focus and center, so in a node both eigenvalues are real and are of the same sign, so depending on the sign the fixed point can be stable or unstable, in a saddle both roots are real with one root positive and other root negative and the fixed point is unstable because there is a positive real root, focus the roots are complex conjugates and this the fixed point could be stable or unstable depending on the sign of the real part of the roots, center roots are pure imaginary and a linearized stability analysis is inadequate to answer the questions on whether the fixed point is stable or unstable. There's another way of looking at fixed points, we can consider two possibilities, one is known as robust cases, here we talk about rappellars when both eigenvalues have positive real part, and we talk about attractors when both eigenvalues have negative real part, saddles are the ones where one eigenvalue is positive and other eigenvalue is negative, marginal cases when both eigenvalues are pure imaginary, there are other situations like higher order and non isolated fixed points where at least one of the eigenvalues will be 0, so we are not going to get into greater details on this, we'll later see how all these ideas are related to structural stability. Now we use the word bifurcations, this is relation to the number nature of the fixed points, as the parameters in a nonlinear dynamical system are changed, one observes that number of fixed points can change, the nature of the fixed points can change, the stability of the fixed points can change, so whenever such changes take place we say that the system has undergone bifurcation, so there is various nomenclature associated with bifurcations depending on before the bifurcation occurs what was the nature of the fixed points, and after the bifurcation occurs what is the nature of the fixed points, we saw that there is a node, focus, saddle, center, etc., so there is half bifurcation where a periodic solution is borne and the several you know classification of bifurcations, this is, this again we are not going to get into greater details on these issues. Now we are going to discuss in today's class energy methods for stability analysis, so what we will do is we will consider a system with n generalized coordinates, and focus attention on statically loaded structures, these energy methods are based on two axioms, the first axiom states that a stationary value of the total potential energy with respect to the generalized coordinates is necessary and sufficient condition for the equilibrium state of the system. The second axiom states that a complete relative minimum of the total potential energy with respect to the generalized coordinates is necessary and sufficient for the stability of an equilibrium state of the system, the first condition pertains to the equilibrium, the second condition pertains to the nature of the equilibrium whether it is stable or not, so this discussion is available in the monograph by Thompson and Hunt, so some of the material that I will be discussing are drawn from this resource. So we can make some remarks, see axioms are statements which are consistent with our physical experience of the world, these statements are deemed to be self-evident, and also in these axioms the kinetic energy and dissipation energy have not entered the statement of these axioms, I am drawing your attention to that fact. Now in the statement of these axioms there are issues like stationary value, relative minimum and etc., so we will quickly see a few results from optimization, so let F of X be a function where X lies between A and B, and let F prime of X star, F double prime of X star and F n minus 1 X star be 0, and F n of X star is not 0, where N is superscript N means I am considering nth order derivative, F is a function of a scalar variable here, if N is even, F of X star is minimum value of F of X at X equal to X star if F of X n X star is greater than 0, on the other hand F of X star is a maximum value of F of X at X equal to X star if F n of X star is less than 0, if N is odd F of X star is neither a maximum or a minimum, so we can see a few schematics here, this is a function F of X versus X, and we have here A1, B1, A2, B2, A3, these are known as local extrema or stationary points, obviously F prime of X is 0 in this depiction here, this A1, A2, A3 are known as local or relative maximum, B1 and B2 are known as local or relative minimum, in the range of X between A and B, A2 is the global maximum, similarly B1 is the global minimum, in this function again lying between these two points A and B, the local minimum and global minimum coincide, here this point F prime of X is 0, F double prime of X is 0, but F triple prime of X is not 0, so this is a point of inflection, okay, so this also is embedded in the statement of this theorem, now let us consider a function of several variables, let F of X be a function of X1, X2, X3, XN, if F of X is an extreme point minimum or maximum at X equal to X star, and F of X star exists then dou F by dou XI X star is equal to 0 for I equal to 1 to N, okay, these are standard results from calculus, then a sufficient condition for a stationary point X star to be an extreme point is that the Hessian matrix of F of X at X equal to X star is positive definite when X star is a minimum, positive negative definite when X star is a maximum, that means when you consider the Hessian matrix F of X, if it is positive definite when X star is minimum it is positive negative definite when X star is a maximum. What is Hessian matrix? Hessian matrix is the matrix of second order derivative of H, so this is given by this elements of H IJ is dou square F dou XI dou XJ, clearly since dou square F dou XI dou XJ is equal to dou square F dou XJ dou XI this Hessian matrix is symmetric, now we say that the Hessian matrix is positive definite if all the eigenvalues of H are greater than 0, or Q transpose HQ is greater than 0 for any choice of N cross 1 vector Q, or we have few requirements H11 absolute value of H11 must be greater than 0, the 2 by 2 determinant H11, H12, H21, H22 must be greater than 0 so on and so forth and determinant of H itself must be greater than 0. Now what we will do is we will try to use these exams and try to gain understanding of how these exams are applied by considering few examples, so these examples are drawn from this book by Simetsis and Hodges, the first example we consider is that of a rigid bar which is loaded actually and which is supported by a spring K, now as if a neighboring equilibrium position exist then the potential is half K A square tan square theta minus PL 1 minus cos theta, this is K A square tan square theta is the strain as is stored here, now theta is a generalized coordinate, so there is only one generalized coordinate, for equilibrium dou U by dou theta must be equal to 0, so you differentiate this with respect to theta I get K A square tan theta secant square theta minus PL sine theta must be equal to 0, so if you organize these terms we get the condition that sine theta into K A square by cos cube theta minus PL must be equal to 0, so this quantity can be 0 by either sine theta being 0 or the term inside this bracket being parenthesis being 0, so we get the condition sine theta equal to 0 or K A square divided by cos cube theta minus PL equal to 0, now this is the condition for equilibrium, now we want to examine the stability now, so we have to differentiate we have to find now the second derivative dou square U by dou theta square, so dou U by dou theta we have obtained this is given by this expression, and upon differentiating this with respect to theta we get these terms, and by reorganizing them we get the equation in this form. Now we will consider now the equilibrium position theta equal to 0 and see what happens to the sine of dou square U by dou theta square, so when theta equal to 0 we can see here cos theta is 1, cos cube theta is 1 this is 0, so I get dou square U by dou theta square is K A square minus PL, theta equal to 0 is stable if this is greater than 0 or P is less than K A square by L, so this is a condition for stability of theta equal to 0, so P critical obviously is K A square by L, now if theta is emanating from the second branch, okay, so in that case I have theta which satisfies this equation, now if you look at the expression for dou square U by dou theta square there is a term which is exactly equal to this and that is 0 and I am left with this term, therefore dou square U by dou theta square for theta which satisfy this condition is given by this. Now this quantity is always positive, you can see here K is positive, A square is positive, sine square theta is positive, cos 2 power 4 theta is positive, this would mean that any theta that satisfy this equation is always a stable equilibrium point, so now we can plot this initially till theta reaches this K A square, P reaches the value of K A square by L the only theta equal to 0 is the possible solution, and this is stable, this branch is actually the root of this equation, K A square by cos cube theta minus PL equal to 0 is this branch, this is always stable, and once K A square by L is crossed these points become unstable, so this is the load deflection diagram for the problem based on the two exams that we have developed. Another example, there are two rigid bars supported on spring here, these dots represent hinges and these bars are loaded actually through these loads P, now as the load P increases these two bars are on rollers, so they will move towards each other, so at some point this roller would have moved by this amount and this comes here, and this roller would have moved by this amount it comes here, and consequently this link moves downwards, right, so these loads are now doing work on, the work done by P on this system is P into this distance, and on the other hand because of this motion there will be energy stored in K, so what is the total strain energy, potential energy here, half K delta square minus 2P L square plus delta square, how do I get that? This is a rigid link therefore this length is L, so this deflection is delta, so L square plus delta square is, sorry this is a right angle triangle therefore delta square plus this square it must be L, so based on that I get this distance as L square minus delta square, so the distance by which the load has moved is L minus L square by delta square, and there are two such motions, so that's a factor 2 comes here. Now delta is a generalized coordinate, so I have got now U in terms of delta, for equilibrium I have dou U by dou delta equal to 0, so I differentiate I get this equal to 0, and this is the load deflection path which satisfies the equilibrium condition, U this equal to 0. Now for sake of simplification we can linearize this function, so I take L outside and I have dou square root of 1 minus delta by L whole square, and if delta by L is small I can approximate this as K delta minus 2P delta by L, which is equal to delta K minus 2P by L. Now equilibrium point therefore is delta equal to 0, now what is the second order derivative? Del square U by dou delta square is you have to differentiate with respect to delta I get K minus 2P by L, for delta equal to 0 to be stable this quantity in the parenthesis must be greater than 0, so I get the condition K minus 2P by L must be greater than 0, or P must be less than K L by 2. So according to the linearized, after linearizing the problem we have got this result that delta equal to 0 is a stable branch till the time I reach P reaches the value of K L by 2, after that delta equal to 0 is unstable, we are unable to trace the other paths because we have linearized here, so I have left this as an exercise you don't do this linearization but plot this curve, and evaluate the second order gradient at each of these points and see whether the sign is positive or negative, and then decide upon whether the point on this curve that is this equal to 0 curve is stable or not, so then you will be able to trace the complete solution, obviously this line will be part of the solution but there will be further information which is not encapsulated in this load deflection diagram. Now we have considered two examples with one generalized coordinate, now let us consider an example where we allow for two generalized coordinates, so this problem is conceptually similar to the one that we considered just now but it has now three rigid bars with two hinges and again loaded actually, so upon the application of these loads this roller move to the right, this roller move to the left, and these springs gets compressed, and this is the snapshot of the deformed configuration from which I will be able to write the expression for the total potential energy, this is energy stored in K1, I am calling this as K1 K2 numerically both are equal, so this is K1 delta 1 square plus half K2 delta 2 square P 3L minus this distance, this distance, so you can compute this distance by using simple geometrical arguments and we get U as a function of delta 1 and delta 2, so this is the total potential, and now I have for equilibrium I have to differentiate this function with respect to delta 1 and delta 2 and equate it to 0, so if I do that I get two equations, for example dou U by dou delta 1 I get this, and here again if I linearize, if I assume that this delta 1 is small in relation to L I can linearize this function and I will get dou U by dou delta 1 is this, and for equilibrium this must be equal to 0, a similar equation by differentiating this with respect to delta 2 can be done, you differentiate again assume that delta 2 by L is a small quantity linearize this function and I get K2 delta 2 minus P into this quantity equal to 0 as a condition for equilibrium, so there are now two conditions for equilibrium I can write them together, before that we can also think of, that is what we are doing here, dou U by dou delta 1 equal to 0, dou U by dou delta 2 equal to 0 leads to these pair of equations in the matrix form this is the equation. Now we can examine the stability by considering the second order gradients, so that is the Hessian matrix I have to find out and see whether Hessian matrix is positive definite or not, so I construct the second order derivatives and the Hessian matrix, and for Hessian matrix to be positive definite there are two criterion K minus 2P by L must be greater than 0 and determinant of this must be greater than 0, so I get this condition, first condition is K minus 2P by L must be greater than 0, second condition is this, now if K minus 2P by L is greater than 0 it automatically implies that K minus P by L will be greater than 0, and hence this condition in conjunction with this really translates to this condition K minus 3P by L must be greater than 0, so we get from this P critical to be K L by 3. Now in one of the early discussions on stability we considered the problem of snap through, we conceptually outlined how the solution might be, but now we will work out the solution in some detail, so for illustrating that we again consider two links, rigid links connected through a hinge here, and this is supported on a spring as shown here, and when P equal to 0 this bar occupies this triangular position, now as P is increased as shown here, as P is increased this point moves down and this roller moves to the right thereby compressing this spring, and this load will do the work through this distance and there will be strain energy stored here. Now we will write the expression for the total potential in this case, you can note down the nomenclature here, alpha is the angle that this link makes to the horizontal when P is 0, whereas theta is the angle made by the link to the horizontal when P is not 0, so theta is the generalized coordinate whereas alpha is a parameter of the problem. So we get the total potential to be given by the strain energy stored in the spring which is, what is the distance through which this point has moved? This distance is 2L cos alpha whereas this distance is 2L cos theta, so the difference between the two is the distance through which this roller has moved and that is here, 2L cos theta minus 2L cos alpha whole square half K. Similarly the work done by P is P into this vertical distance and this vertical distance here is L sine alpha whereas here it is L sine theta, so the distance through which this has moved is L sine alpha minus L sine theta, so that is this. So we can take out 2L outside and rewrite this expression and now we are ready to investigate the stability. For equilibrium we have to evaluate dou U by dou theta, dou U by dou theta upon differentiating this with respect to theta I get this. Second derivative I differentiate this with respect to theta and I get this expression, we can do a bit of simplification and we get dou U by dou theta and dou square U by dou theta square as outlined here. Now we examine the condition for equilibrium, the condition for equilibrium is dou U by dou theta equal to 0, so that means this quantity must be equal to 0. So we can rearrange the terms and if we do that I will get the equation as P divided by 4KL is sine theta minus cos alpha tan theta, so this is actually a load deflection curve which is non-linear, theta is the displacement and P is the applied load, so points P, theta form the points on load deflection curve and for equilibrium they should obey this relation. Now every point on this load deflection curve we have to evaluate dou square U by dou theta square and look at the sign of that quantity, if it is positive that corresponding point on this load deflection path is stable, otherwise it is unstable, so I will put this dou square U by dou theta square, I will evaluate this and for this to be greater than 0 I get this condition that cos alpha by cos theta minus cos square theta must be greater than or equal to 0, when this condition is also satisfied. Now graphically this curve looks as follows, this line that you are seeing here is the load deflection path given by this curve, this curve is this red line, when load is 0 the Y axis is the load, when load is 0 the deflection will be this, this is the theta which is alpha, this is actually alpha, this value will be alpha, as load is increased the theta goes on reducing and when it reaches this point B, so what I am doing is for every value of theta here I will evaluate dou square U by dou theta square and see what is the sign of this quantity, if it is positive it is stable otherwise it is unstable, so these points are stable and moment I reach this point B what happens is this branch will be completely unstable, so the loading path unstable positions cannot be occupied, so the structure actually snaps it comes here, this path is unstable, now upon increasing the load further it traces this path, now similarly if you start unloading the structure would be able to realize this path, but at this point it snaps to this, snaps upwards and comes to this point and again these points will be stable, okay, so this region is never occupied either in the forward path or in the reverse path, so this region is unstable region, these values of theta cannot be realized by the structure, so that is what this theory tells, and this as I said is a snap through phenomena where if you carry an umbrella during a rainy day, windy day, the curvature of the umbrella, the sign of that suddenly changes under certain velocities, so that is the kind of thing that we are talking about, but under static loads. Now we can make few observations here, to study this problem we need to use large deflection theory, if you linearize you won't get these details of this theory correctly, here at B there are no adjacent equilibrium positions possible, so in the neighborhood of B as load increases the system loses its stability and moves to C to a far away state, this is C, it doesn't come in the neighborhood, this behavior is called limit load buckling, okay, the buckling of spherical caps and arches display this behavior, that is curved elements. Now we can compare this with buckling due to bifurcation equilibrium, so if you recall when you studied this problem of an ideal beam carrying a truly axial load, Y equal to 0 was the equilibrium solution and we are interested to see if there is any other equilibrium position in the neighborhood of Y equal to 0, so by assuming that there exist in neighboring equilibrium position we wrote the equation, presuming that such a neighboring equilibrium exist and try to find out for what value of P that assumption is valid, so that led to this load deflection diagram as shown here, so when load is 0, P is 0, this delta is a mid-span deflection is 0 and as we gently increase value of P and at every increase suppose you pluck the beam and allow it to oscillate and see whether it returns to its original state or not, assuming that system is damped, it will return to its original path till P reaches a value P1, after that depending on the perturbations that you give it occupies a neighboring equilibrium position, okay, so that is the phenomenon of buckling due to bifurcation of equilibrium, whereas in limit load buckling what happens is as you go on increasing the load the structure deforms, it is now here the deflection is 0 for as the load increases, but here the as the load increases the deformation also increases, okay, so we started here as the load increases the deflection is changing, this is unlike this situation where with increase in P origin still continues to be stable, okay, and when it reaches this value a critical value it occupies a faraway position, in buckling due to bifurcation of equilibrium as load increases deformation deflection remains at 0, non-zero deflection becomes possible only when load reaches the critical value, till that time there is no deflection, whereas in limit load buckling the structure deforms with every increase in applied load, both load and deformation evolve as load increases, and the critical stage when structure becomes unstable and the system gets into an equilibrium far away from the original equilibrium position, so when that critical condition is reached the structure occupies a faraway equilibrium position, as the load approaches the critical value any small increase in the load results in large changes in deformation, even away from this critical condition with increases in load the deformation changes, whereas that doesn't happen here, so to analyze limit load buckling we need to perform a non-linear analysis, now suppose if we consider the same problem purely based on equilibrium considerations, suppose I find out the reactions and take moment about B and write the equilibrium equation, so that can be done, for example this reaction vertical reaction by symmetry is P by 2, and the force in the spring we have already derived it to be K into 2L cos theta minus 2L cos alpha, now moment about B if you take here F into this distance and this reaction into that will give me the required relation and I get the equilibrium path, that is a relation between P and theta, but this analysis does not address the questions on stability of states and hence cannot explain if all states are realizable or not, you will get this graph and if you plot it you will get a curve like this, but you will not know whether this path is realizable or not, okay, but also there will be a problem here if you consider this value of P there are three possible solutions, so which solution will be realized, you see that question has to be answered, so that depends on how you are approached that point, and this intermediate point cannot be realized at all, okay, so these details will not be delineated if I do simply an equilibrium analysis, so you need to perform a stability analysis as well, now let's now return to problems of continuum, so far we considered some simple examples involving rigid links and some simple mechanical models, so let us return to problems of say stability of beams, so let us consider this problem of ideal simply supported beam carrying truly axial loads, and we are considering whether there is an equilibrium position in the neighborhood possible or not, so I can write the expression for strain energy in the system as shown here, how do I now apply the axioms to investigate what value of P such positions are possible, so what we do is we use this Rayleigh-Ritz type of analysis, so this we have discussed earlier when we in the very first few classes when we discussed approximate methods for determination of natural frequency, so we are used to the language so I will not repeat all the details, so we start with a trial function, for example if I take Y of X to be Ax into L minus X, so as you can see here at X equal to 0, Y is 0, and X equal to L Y is 0, so at least the geometric boundary condition is satisfied, this A is the generalized coordinate, now if you look at the expression for total potential it has D square Y by DX square and DY by DX, so you find out Y prime of X I get A into L minus 2X, Y double prime of X minus 2A, substitute this into B and perform this integration I get V of A to be A square 2EI minus PL cube by 6, DV by DA simple you differentiate this with respect to A I get this, D square V by DA square I get this, now I will need these two quantities to infer equilibrium and stability, for equilibrium DV by DA must be equal to 0 that means A would be 0, and this equilibrium position is stable if the second order derivative of V evaluated at this point is positive, so what is DV by DA at A equal to 0, it is a constant everywhere, so I get this, and this is if it is greater than 0 A is stable, so for this to be greater than 0 P should be less than 12EI by L cube, so P critical according to this theory is 12EI by L cube, now if you recall we have computed the critical load for this case and first critical load is pi square EI by L cube, now this is 9.869 EI by L cube and this approximate analysis gets an approximation of 12EI by L cube to this number, so there is about 20% error in this analysis. Now if you recall the discussion on choice of trial functions when we discuss Rayleigh Ritz method in the context of determination of natural frequencies, we classified the trial functions as being admissible comparison and Eigen functions, there are geometric boundary condition and natural boundary conditions, and there is a field equation, the trial function which satisfy the geometric boundary conditions and which need not satisfy the natural boundary conditions and field equations are called admissible function. If the trial function satisfy geometric boundary condition as well as natural boundary conditions but not the field equation, they are called comparison functions. Eigen functions are those which satisfy the geometric boundary condition, natural boundary condition as well as the field equation, that means they are the exact solution. Now if you look at the trial function that we have used, Y equal to AX into L minus X is an admissible function, because the second derivative is a constant and bending moment would not be 0 at the supports. Now we can try other shape functions, for example Y of X could be a deflected profile of the beam under its own weight, suppose if I take the beam and apply a concentrated load, sorry distributed load which mimics its own self weight, this deflected profile will satisfy the geometric boundary condition, and obviously if the analysis is right the bending moment computed from this curve will be 0 at the 2 supports, so it will be a comparison function. So you can use that A into this shape can be taken as a trial function, that's one way of constructing comparison functions. We can also consider for example if it is sine N pi X by L, there are 2 eigenfunctions, okay, so if I use in the Rayleigh Ritz method sine pi X by L as a material function I will get the exact solution, but you would not know in real problems what would be the eigenfunction. So to illustrate this let us consider a single span beam which is made up of stepped cross sections, the middle half has a flexibility of EI and the other quarter have EI by 4, so I want to find out the, this structure is subjected to axial loads, and I am interested in knowing the critical value of this load P where the equilibrium position Y equal to 0 loses its stability. So I will write the expression for V, I had taking into account that this EI is different in different cross sections, so first I will integrate from 0 to 0.25L, and then 0.25L to 0.75L, and 0.75L to 0.25L, and I use EI by 4, EI and EI by 4 in these 3 cross sections respectively. The axial load of course is constant throughout, so I will take, there is no need to split this. So this is the expression for the total potential, now I can try 2 term solution, I will take A1 sine pi X by L plus A2 sine 3 pi X by L. Now if the beam are to be uniform, okay, sine pi X by L will be an eigenfunction, because it is a true mode shape of the, buckling mode shape of the beam. On the other hand if the, since the beam is inhomogeneous sine pi X L, pi X by L will be a comparison function. So I am considering these two comparison functions, and A1 and A2 are the generalized coordinates. So I can substitute this and perform this integration, if I do that I get this as my total potential in terms of generalized coordinates A1 and A2. Now condition for minimum, sorry equilibrium is dou V by dou N, dou A1 equal to 0, dou V by dou A2 equal to 0. So if I do that I get this matrix equations and the A1 equal to 0 and A2 equal to 0 is the equilibrium position. Is it stable? So I have to construct the Hessian matrix and evaluate it, the solution A1 equal to 0 and A2 equal to 0 and see whether it is positive. So the critical condition of course will be the determinant of H is 0, so if I do that I get the critical value for P to be given by this number. Now here of course if I retain only if the first term I can put A2 equal to 0 and the critical load corresponding to that will be this term equal to 0, which will be 0.864 pi square EI by L square, that is I am forcing A2 to be 0. So if only one term is retained I get the answer as 0.864 into this multiplier, if two terms are retained I get an answer 0.735 into this, it so happens that the exact solution for this problem is 0.65 pi square EI by L square, this would mean that retaining the second term reduces the error from 33% to about 13%, so it offers advantage. Now we can consider few more examples so that the ideas get fixed in our mind, so we consider now a cantilever beam loaded actually through load P as shown here, so this is the expression for the total potential, I will try and take a trial function AX square, so Y prime X is 2AX, Y double prime is 2A and YFX is admissible because at the free end we need conditions on bending moment and shear force that is will not be satisfied by this, so substitute I will get V to be this, so dou V by dou A equal to 0 is the condition for equilibrium and that means A equal to 0 is the equilibrium position. Now differentiate this with respect to A again dou V by dou A and look at the sign of this quantity at A equal to 0, if I do this for A equal to 0 to be stable I get the condition that P must be less than 3I by L square, so according to this single term approximation the critical load is 3I by L square, whereas the exact solution is 2.467 EI by L square, so we can improve upon this solution by taking YFX to be A into phi X where phi X is a deflected profile of the beam under its own weight, by that in this case what is meant is you have to consider a deflected profile of the beam this is phi of X, I don't mean weight in this direction of the applied load but in this direction, okay, so that might offer some improvement. Now we will consider one more example, we will consider two alternative trial functions, one in which the geometric boundary conditions are satisfied and another one in which the force boundary condition is also satisfied. So let us consider this property cantilever carrying this axial load P, the geometric boundary conditions are Y of 0 is 0, Y prime of 0 is 0 and Y of L is 0, so I will now select a trial function which satisfy these three conditions, so I will start with Y of X is A naught plus A1 X plus A2 X square plus A3 X cube and I will differentiate this and by using these three conditions I arrive at a trial function which is Y of X is A into X cube minus LX square, okay, this satisfies all the boundary conditions, geometric boundary conditions. You now substitute into the expression for total potential and if you perform the requisite calculation I get P critical as 30EI by L square, so I have skipped those steps you can fill it up. Now on the other hand if I now consider the force boundary condition also while constructing the trial functions, so I will now assume a fourth order polynomial for Y of X, so four of these constants I will evaluate by imposing the four boundary conditions and the fifth one will be treated as the generalized coordinate. If I do that the polynomial that I get eventually will be of this form, Y of X is AL to the power of 4, X by AL to the power of 4, etc. So this trial function is a comparison function, it satisfies the geometric boundary conditions as well as the force boundary condition, the force boundary condition is Y double prime L equal to 0, that is EI Y double prime L must be equal to 0 which is the bending moment. Now with this I get the potential V of A as shown here and I use the two XEMs and check the sign of the Hessian, check the condition for Hessian to be positive definite and I get P critical to be 21EI by L square. So summary is in first choice I had a trial function which satisfies geometric boundary condition but force boundary condition was not satisfied and I got a P critical of 30EI by L square. In the second choice both geometric and force boundary conditions were satisfied and I got this answer, the true answer is exact answer is 20.14EI by L square, so you can see that the extra effort that we made in achieving, in satisfying the boundary condition pays dividends here. Now we talked about Rayleigh Ritz method, we could also use the Galerkin's method, here the starting point will be the governing differential equation itself, so let us consider an inhomogeneous beam where EI is a function of X and the governing equilibrium equation is given by this, this can be obtained based on application of Hamilton's principle and you will be able to get this with along with all the appropriate boundary conditions, so that would be the starting point for applying Galerkin's method. Now I assume Y of X to be given by this series A n phi n of X, where phi n of X need to be at least admissible, so if I substitute this into this equation, the equation won't be satisfied, we will be left with a residue. Now as we have seen in method of weighted residuals, weighted sum of this residue is taken to be 0 and the weighting function in Galerkin's method is taken to be the trial function itself and I get a set of N equations, so if we multiply by phi K and integrate from 0 to L, I will get these terms and I can evaluate this term by parts and use the boundary conditions and I will be able to simplify this, I have not shown those simplification, we have done that when we discussed a Galerkin's method in the context of vibration analysis, so I will get equations in this form, so the new thing is there is one more integral J N K given by this, here also we can simplify this by integrating once. Then this is for K equal to 1 to N and in a matrix form I can write it as K A plus P into J A equal to 0 and this matrix J is known as stability matrix, we want now if you want a neighboring equilibrium position a non-trivial solution is needed for A, so the condition for non-trivial solution is the determinant of K plus P J must be equal to 0, so this is an eigenvalue problem with P as the eigenvalue and A the deflected profile is the eigenvector, so this leads to estimates of N values of P at which the system admits a neighboring equilibrium state, okay, so this is Galerkin's method, so the Galerkin method you know we can use it for few examples, for example if you consider a beam fixed at the two ends carrying axial loads as shown here, again I want to recall that this fixity convention that we are using here should be taken to mean that the translation and rotation here are 0, but the axial deformations are permitted, otherwise this load will not do any work on the beam, that is not what is meant by writing this fixity condition. Now how do we estimate, the problem here is to estimate PC that is a critical value of this P by using two trial functions Galerkin's method and these two trial functions the way they are derived is in one case I apply a UDL and find the solution, this is phi 1 of X, in the other case I apply half the beam load this way and other half this way, that means this load acts upwards, this load acts downwards, so the deflected profile will be something like this, so this is phi 2 of X, and they are explicitly given here and I leave it as an exercise that you use Galerkin's method and evaluate the P critical value. Now there can be other examples that can be used, for example we consider the cantilever beam deflecting under its own weight, so we found out the possibility of the structure buckling under its own weight, so this problem can again be tackled using Galerkin's method, this is also left as an exercise. Now so far we have considered simple structural elements, but when we come across built up structures like a two span beam or a portal frame or industrial shed carrying various types of loads, how do we proceed, how do we determine the equilibrium position and their stability, so here again if you can construct the global trial functions, the trial functions which are valid all across the structure you can still use Galerkin's type of approach, but they become increasingly unwieldy, so this situation we have encountered when we started discussing application of finite element method for vibration problems. So what was the remedy at that time, what we did was a structure like this is discretized into elements and the nodal displacement degrees of freedom, the nodal displacement values were taken as degrees of freedom and the field variables within an element were interpolated using the nodal values of the degrees of freedom at the nodes, and then we formulated the mass and stiffness matrix, we can do the same thing here instead of using global trial functions we can construct trial functions which are polynomials over pieces of these structural elements and then demand continuity of deformation etc. across these elements, so that is basically the idea of using finite element method for this type of problems. Now how these ideas are developed is what we will consider in the next lecture, so we will close this lecture at this point.