 In the previous lecture, we discussed about the inelastic seismic response of 2D and 3D multistory frames. For both, we can have two situations. One is a strong column weak beam, the other is a strong beam weak column. When we consider the case of strong beam weak column, we generally take a shear frame model so that hinges can form only in the columns. In the other case, when the columns are strong and beams are weak, then the hinges forms in the beams. So for both types of the cases, we described how to perform the analysis. When the beam is weak and the columns are strong, then even for the case of three-dimensional frame, the analysis is not complex because we do not have to take into consideration the bidirectional interaction because the hinges form in the beams and the beams undergo only one-directional bending. Therefore, one can calculate easily the bending moment at the end sections of the beam and check whether that moment is equal to mp or not. The same case is with the 2D frame for the case when the hinge is forming in the beam. Every problem that is encountered in solving the cases where the beam is weak and column is strong and the hinges are forming on the beams, in that case one has to find out the rotations for finding out the bending moment in the beams. That rotation is to be calculated using the condensation relationship. There we obtain a relationship between the theta and delta. After finding out the delta for an incremental time interval of delta t, then we also find out the incremental rotation that takes place at the cross section. Then add this incremental rotation with the rotation at the previous time step, find out the total value of the rotation, also the total value of the deflections, sway deflection, find out then the bending moment at the cross section. In doing so, we have brought in a factor called alpha 1 and alpha 2 which are the ratio between the beam rigidity to the column rigidity. So, when the system is in the elastic state, the ratio Eib by Eic can be easily calculated and one can form the stiffness matrix. However, when the system goes into the plastic state or in other words when the plastic hinges have formed in the beams, then one has to be cautious in finding out the values of alpha 1 and alpha 2. The formulation that we discussed was with respect to a general condition that is the behavior, the material behavior or the force deformation behavior is not idealized as elastic perfectly plastic or bilinear model, but by a non-linear hysteretic system in which the stiffness changes at every point. So, in that case in the beginning of the solution, when thus we draw a tangent to the initial point in the curve and that gives the initial stiffness or tangent stiffness of the system and then we can calculate Eib and Eic based on that and find out the values of alpha 1. But subsequently as we proceed with the integration, then over a time of delta t, it is difficult to find out the values of Eib by Eic. In that case, the way it is to be calculated was explained in the previous slide that is we bring in the concept of R and this R can be calculated iteratively and using the value of R, one can find out the values of alpha 1 and alpha 2 to be used for generating the stiffness matrix at the time t plus delta t. So, with that background, the entire scheme was illustrated with the help of a very simple example that is example 6.5 where we had got 6 rotational degrees of freedom and 3 translation degrees of freedom. These rotational degrees of freedom are condensed out. We get a stiffness matrix with respect to the degrees of freedom delta. So, with the help of that delta, we carry out our analysis and for every incremental step we calculate delta theta and add it to the previous value of theta to get the value of theta at the current time step. So, for this problem at 1.36 second, the values of the displacements, acceleration, velocity or the yield rotation and the maximum permissible rotation they are given in this table and from that one can see that at joint 1 and joint 2 in the beam, the moment is equal to the MP value that is here the adjoint 1 and 2 here and for this beam adjoint 1 and joint 2, we have the values of 50 knp. So, these two sections are yielding and therefore, this beam does not contribute to the overall stiffness of the structure. So, therefore, we said that to 0 in obtaining the stiffness matrix of the system and that is how we calculated the total stiffness matrix of the system which is shown here and from that a stiffness matrix we calculated the condensed stiffness matrix which is a 3 by 3 stiffness matrix. With that stiffness matrix K delta, we now calculate the incremental displacement for the next increment of the delta T. So, this was the scheme and for the calculation and therefore, we see that at every instant of time t we have to check in the beam members whether there is a plastic moment is there at a particular cross section and if there is a plastic moment coming at a particular cross section then we assume for the next analysis a simple hinge at that particular cross section and carry out our solution with a modified stiffness matrix KT and how to take care of the case when the bending moment at a particular time exceeds the value of mp then how to rectify that that was also discussed. The same concept will be subsequently discussed in connection with the push over analysis that is the subject of our discussion today. Now push over analysis is one of the very popular method of non-linear analysis or rather static equivalent non-linear analysis that is performed for earthquake loading. In fact, the way the response spectrum method of analysis is a very good equivalent in static load analysis for earthquake forces. In the same fashion the push over analysis is a very good non-linear static analysis for the inelastic dynamic analysis for earthquake forces. The push over analysis is carried out for many purposes the chief among them is the push over analysis for finding out the behavior of the structure at the inelastic stage during the earthquake that is the inelastic analysis that we carry out for the earthquake forces that inelastic dynamic analysis is replaced by a equivalent and static non-linear analysis that I mentioned before. Second thing is that for finding out the performance criteria of the structures in the inelastic range that also is obtained with the help of a push over analysis or equivalent static push over analysis that helps designers to understand the different states inelastic states of the system when the structure goes into the inelastic zone during earthquake and one can specify certain criteria like immediate occupancy criteria or failure criteria etcetera and therefore, attaching some kind of performance level of the structure in the inelastic state. This is also known as the performance based analysis and design and it is routinely carried out for most of the multi degree of freedom systems specially the multi-story building frames in order to find out or assess the performance level of the structure in the inelastic state. The push over analysis is incorporated in most of the standard softwares that are available these days like SAP 2000 or the ANSI or ABACA for all these softwares we have push over analysis and these push over analysis requires some kind of data in the beginning and we will describe what are the kind of data that is needed for performing the push over analysis. The push over analysis provides a load deflection curve, a single load deflection curve and that single load deflection curve gives us an information about the state of the system after yielding, but the system is then idealized as if as a single degree of freedom system. So, in an overall sense the behavior of the structure in the inelastic range is understood using the push over analysis. So, the main thing that the push over analysis provides is a single load deflection curve from 0 loading to the ultimate failure state. Load is representative of equivalent static load taken as a mode of the structure and total load is conveniently taken as the base shear. So, the important thing over here is that the loads that is acting in the frame that load is considered to be distributed in any fashion or in any reasonable fashion. However, one prefers for the earthquake analysis that the load should be in accordance with the mode shape or in other words according to the shape of the first mode the loads are distributed. Now, once we do that then the sum of the load gives us the base shear and typically the by load we mean the base shear in this case the deflection may be represented by any deflection however it can be conveniently taken as the top deflection of the structure. So, the push over analysis finally provides us a load deflection curve where load is the base shear and the deflection is the top displacement of the frame. Now, the push over analysis can be force or displacement control depending upon whether we increment the force or increment the displacement. So, what we do is that either we gradually increase the displacement in the structure and look into the behaviour of the structure or we increment the force and then we study the behaviour of the structure. For both incremental non-linear static analysis is performed and for that what we require is the K matrix which is a transient K matrix that means this K matrix goes on changing with deflection, but over an incremental displacement we assume that the stiffness or the transient stiffness matrix that does not change. So, with the help of that we perform a linear analysis within the small incremental displacement and find out the values of incremental responses. The matrix at the beginning of each increment is obtained to find out the response for or the over that increment that is if the increment is at a deflection level of delta 1 then for the response at delta 1 plus delta we use the stiffness matrix that is developed at a deflection stage of delta 1. However, this may have to be modified in certain cases those modifications will describe little later. The displacement control push over analysis is preferred in most of the cases because the analysis can be carried out up to a desired displacement level. So, one can stop the push over analysis at any displacement required displacement stage and saying that we are interested to find out the behaviour of the structure up to this displacement level. So, from that point of view the displacement control push over analysis is preferred. The analysis can be carried out to any desired level and this desired level obviously depends upon as I told you the displacement level or the force level that we wish to impose on to the structure as the final displacement or final load. Following input data are required in addition to the fundamental mode shape. So, as I told you that the loads are distributed along the height of the building according to the mode shape of the structure. However, it is not necessary that one should consider the mode shape or the first mode shape of the structure one can assume any reasonable distribution also. The other information that is required is that one has to assume a collapse mechanism for this structure and this collapse mechanism is somewhat difficult to assess for a multistory frame structure. However, one can assume any kind of collapse mechanism in the sense that in order to stop the analysis we can say that we will perform the analysis up to this particular state when certain numbers of plastic hinges have occurred into the structure and that we consider as the collapse state. The structure may not have actually collapsed. In many cases the entire analysis is performed unless we find that there is a singularity in the matrix or in other words the solution cannot be carried out further. At that stage we have gone down and we will say the structure has collapsed and these kind of collapses which are not again a desirable kind of collapses that may occur even before the complete collapse. So, this collapse mechanism that we are defining to stop or calculation is an important thing or sometimes a premature or collapse state can come into picture because of the singularity of the matrix and at that stage one has to stop and the mechanism that we get we say that is the failure mechanism. Next important information that is to be provided is the moment rotation relationship of yielding cross section. Now this moment rotation relationship of yielding sections that are to be obtained from the cross sectional properties of the beams and the columns including the reinforcement. By knowing the percentage of reinforcement in the beams and columns one can find out a moment rotation curve. Once we get the moment rotation curve for the cross sections then those moment rotation curves are provided as an input for the analysis. So, the whatever be the number of cross sections that we consider where the plastic hinges are assumed to form for those cross sections we provide the moment rotation relationship. Next input that is necessary is the limiting displacement and now the limiting displacement may be provided so that either we cut off our analysis before the complete collapse. In that case we can give a limiting displacement which is much smaller displacement compared to the displacement that takes place at the time of complete collapse. So, in order to trace the complete load deformation or load displacement curve that is up to the collapse state we assume some kind of displacement which is generally a large displacement and give it as an input to the structure so that the solution automatically stops before all the actual displacement that we have provided is achieved. Next important information that is required is the rotational capacity of plastic hinge that is also very important in many cases we say that if the rotation at a plastic hinge exceeds certain value then we say that there is a rotational failure at that plastic hinge. Now this kind of failure is generally achieved for many a time and once we say that a particular plastic hinge has exceeded the value of the permissible rotation then we marked that particular cross section and say that that particular cross section has completely failed although the structure has not collapsed completely and we identify in the process of our calculation the plastic hinges which have exceeded their rotational capacity and the plastic hinges which have not crossed their permissible limit. Displacement control pushover analysis is carried out in the following steps. We choose a suitable displacement interval that is delta delta 1 for the top story of the frame corresponding to this delta delta 1 we find out the delta delta 1 at different levels of the frame that is the for the rth level the value of delta delta 1 r is equal to delta delta 1 multiplied by the mode shape coefficient for that particular rth floor. So that is how one can get the values of the displacements at different floor levels once we have assumed some value of incremental displacement at the top story of the frame. This gives a vector of displacement the index 1 denotes that this is the first increment of our displacement. Once we obtain the displacement vector then our incremental displacement vector then we multiply it with the stiffness matrix K. Now these stiffness matrix in the beginning or for the first increment is an elastic stiffness matrix K. However as the calculation proceeds this K becomes a transient stiffness matrix and it goes on changing depending upon the displacement level or depending upon the plastications that take place in the frame at different cross sections. At any time that is say at the nth increment of the load the total base shear that can be calculated by adding up the total load or total load will be is equal to summation of the delta p over the n increments. Similarly the base shear can be calculated by summing up the base shear for each increment of loading up to the nth increment. The deflection delta 1n that again is equal to the displacement increments that is given to the top floor up to the nth increment. So that is how one can get delta 1n and delta vn or vn and this plot of v vn versus delta 1n that is continuously obtained as we go on incrementing the displacement. At the end of each increment moments are checked at all potential locations of the plastic hinge and that part is little complicated in the sense that first one has to find out the value of theta n at the cross sections from the condensation relationship and as I described before we first find out the delta theta at all the sections where the yielding may take place from the incremental displacement. Once we get the incremental displacement those incremental rotations those incremental rotations are added to the previous rotation to find out the current value of the rotations. Knowing the value of rotation and the steady displacement one can find out the bending moment at the desired cross sections of the structure. If it is found that at any particular cross section the moment value is equal to the mp value then for subsequent increment what we do is that we assume an ordinary hinge at that particular cross section and find out the total stiffness matrix of the structure that is for that element we assume a hinge ordinary hinge at the section where the plasticification has taken place. Also during the calculation we go on calculating the rotations at the hinges or rotations at the plastic hinges. So even if we assume an ordinary hinge at the time of the analysis for the cases where the section has undergone on yielding and there is a plastic moment or the moment is equal to plastic moment after the analysis is performed we find out the rotation incremental rotation also for the ordinary hinge. Now once we get that rotations incremental rotation then we add this to the previous rotation to find the final rotation. So therefore we can find out the rotations at the hinges or the plastic hinges and we keep a record of that in order to check whether the permissible rotation capacity is exceeded in a plastic hinge or not. If we find that there is a case where the rotational capacities in sufficient number of hinges have taken place then it may so happen that a rotational failure may precede the actual collapse mechanism. The VV versus delta 1 is traced up to the desired displacement level or the collapse state. Now during this calculation procedure the kind of iterations that are involved or that let me explain. First iteration is that as we go on incrementing the displacements at a particular level then say we take delta 1 and from this delta 1 we construct the delta delta 1 vector with that delta delta 1 vector we get the delta p vector then check for the Mi that is the moment at the cross sections the desired cross sections where the plastic hinges are likely to form and check whether the moment there is greater than MP or not. The moment being just equal to MP is a rare case therefore in most of the cases we find that the moment is may be greater than a value of MP. If it is less than value of MP then of course that particular cross section is a state in the elastic state. Now if the this is not greater than MP that means it is elastic or if it just is equal to MP then to handle that is very simple we simply put a ordinary hinge at that particular plastic hinge and obtain a stiffness matrix and carry out the solution as before that means we go for the next increment. If it is greater then what we do is that for those cross sections we set Mi to be is equal to MP that obviously introduce some kind of imbalance into the overall equilibrium equation. So what we do is that we can have two alternative two options one is that after setting those values of the moment 2 is equal to MP then we calculate a revised value of the stiffness matrix that means for those cross sections where the moment has exceeded the value of MP at those cross sections we introduce a ordinary hinge and calculate a revised stiffness matrix and after that we find out an average stiffness matrix that is the stiffness matrix before the increment that plus these modified or the updated stiffness matrix these we add together and divide it by 2 to obtain an average stiffness and with that average stiffness we now calculate the delta P that is the load increment for the value of delta delta 1 that is the incremental displacement that we have considered. So therefore at the end with this calculation we get a different values of delta P vector and after the solution that means once we have this solution it is automatically assumed that the cross sections where we have set Mi to be is equal to MP at those cross sections yielding have taken place and for next iteration or next displacement increment we use these values of KT. So in this way one can perform an iteration or an iteration may be required whenever the bending moment is exceeding the value of MP at any particular cross section. The other way is to proportion as I told you before in the plus in elastic analysis that is the value of the MPs that is obtained after the solution those out of that we take the value of MP which is the largest one and this is done for the case when more than one sections have moments greater than the value of MP. So there we take the greatest value of the MP and then we proportion the incremental displacement in such a way that the bending moment at that particular cross section just become equal to MP. In that case what will happen for that particular cross section Mi will be simply is equal to MP for other cross sections obviously the values will not be equal to the value of MP in most of the hinges the or plastic cross sections the values would become less than the value of MP because we have proportioned the value of delta delta I. However these proportioning may not always lead to all sections having a value of M less than MP. However with little bit of you know expertise one can proportion the value of delta delta I or the incremental displacement in such a way so that we get a plastication only at only at one cross section. In that case the incremental displacement becomes non-uniform and but that does not affect the solution and one can continue in this particular fashion that is we can go for the next increment of displacement with a plastic hinge forming at one cross section only at a time. These kind of calculation may require little more time to trace the load deflection behaviour it is better that one goes for an average stiffness technique to take care of this kind of situation. The problem of a frame is solved here using the push over analysis this is the moment rotation curve for a particular beam and we can see that we are specifying it will not be Q Y and Q C it will be theta Y and theta C theta Y indicates the yield rotation and correspondingly we define a yield moment and theta C is the rotational capacity or the maximum rotation that we allow at the plastic hinge. The properties of the frame is given over here we see that we have grouped the cross sections in C 1, C 2 and B 1, B 2, C 1, C 2 are the two groups of columns and B 1 and B 2 are the two groups of the beams. So from the ground first four and second four level we have one kind of cross section and then their M Y value is given and the corresponding yield rotation is given and the last column shows the rotational capacity for that particular cross section. Second group is a third, fourth, fifth and sixth story levels there we have reduced B and D and therefore the yield moment is also reduced the rotation permissible rotation is given that is the yield rotation is given and the maximum rotation that is allowed that is also provided in the same way the B 1 and B 2 that these two groups of the beams for that the dimensions and the other required values are given over here. One can see that the again there is a change in cross section for B 1 and B 2 level. The solution provides us a result like the one which is shown over here. The different displacements and the corresponding base shear that are shown in the tabular form over here and the plastic hinges that has taken place at a different levels of displacement that has been noted down that is 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5 and so on and when we come to the last value then we see that there is there are sufficient number of sections where plastic hinges have taken place. If we see look into the base shear we can see that the base shear is increasing up to 346 that is the last but one last but one row not last but one row above or two this is the value which is has the maximum value of the base shear and after that the base shear drops down to 307 and then it slightly increases 308. The base shear versus displacement plot is shown over here and these goes on increasing and these value is about 346 and then there is a drop in the displacement or the base shear. This is the distribution of the lateral forces that is considered along the height of the building and this kind of mode shape is assumed for the analysis. So, one can get a plot like this from the push over analysis and if we go up to the state of collapse the plastic hinges that were traced and shown in the table they are shown over here and one can see that the 17 plastic hinges have occurred and at that stage the solution had to be stopped or the calculation had to be stopped. The reason for this is that if we look at this particular joint we find that the plastic hinges are forming at all the 4 cross sections surrounding this joint as a result of that there is a rotational failure that has taken place at this particular joint and because of this rotational failure the stiffness matrix became ill conditioned and therefore the calculation had to be stopped. However these also corresponded to a situation where there is a sudden drop of the base shear and one can say that this is almost a state of failure. Not necessarily that one may have to stop the solution or the calculation procedure at such a premature stage unless we have this kind of situation. If this kind of situation does not take place then perhaps more number of plastic hinges can form into the structure and one can have a calculation. Calculations may be continued further or the incremental displacements can be provided and the solution procedure can be carried out till a desired collapse state or collapse mechanism is formed by way of the plastic hinges forming at different cross sections and the plastic hinges thus will not provide a situation of premature failure like this. So, it depends upon the problem to problem one can pursue or continue the incremental displacement up to the state of a desired collapse mechanism or the collapse mechanism that they have assumed and one can get that particular collapse mechanism or achieve that collapse mechanism by way of incrementing the displacement. It may so happen that one may not go or may not be able to go up to that state before that a premature collapse state can occur. Now, with the help of the push-over analysis some important things extracted as I told you in the beginning that is first thing is that one can see the different states of the system after the yielding that is how the plastic hinges are forming at different cross sections and what are the nature of those plastic hinges where at the plastic hinges the rotational capacity has exceeded or not and the plastic hinges where still some rotation can take place all those informations can be achieved during the process of the calculation and from that one can define some state of the system after yielding or the performance level of the structure at different states. That is very important in so far as the performance based analysis is concerned. However, one of the important thing that we try to obtain in the case of by doing a push-over analysis is that to find out an overall ductility of the system in that case the basic philosophy with which the push-over analysis is carried out is that we convert the entire system into a single degree of freedom system and try to draw the load versus deflection curve. The load is given by VB and the displacement is given by delta. So, one can find out a performance point like this that means this is a load deflection curve and one can have a bispectrum drawn from the response spectrum, the displacement response spectrum and acceleration response spectrum and from that one can construct a bispectrum by eliminating the time period scale and that bispectrum can be obtained with an assumed value of the increased damping and a tentative equivalence stiffness for the entire system considered as a single degree freedom system that is when it is vibrating only in a first mode. Now, with that one can get a performance point but this performance point can be updated by successive iterations that is once we get a particular performance point then one can find out a devised value of the equivalent stiffness and the equivalent damping can be obtained from the loop hysteresis loop or giving these value of maximum value of displacement and get a revised value of equivalent stiffness and damping and find out and a bispectrum curve. So, that way we can go on doing this particular iteration till we find that the performance point that is achieved in two successive iterations are the same and from there one can obtain the value of the ductility required ductility that is the maximum displacement that we obtain divided by the elastic displacement that is when the structure first gets into the in elastic state or the first plastic hinge is formed that is taken as the yield displacement and delta becomes the final displacement ratio between the two gives a overall ductility for the system.