 Welcome back everyone. So, we are going to continue our discussion from the last class. Basically, in the last class we discussed how to set up the equation of motion by using two method. The first method used method of direct equilibrium in which we basically used equilibrium conditions to set up the equation of motion. And then the second method used method which is called influence coefficient method. And we saw that using both methods we can set up the equation of motion for a multi-degree of freedom system. So, we are going to do some examples today and apply both method to actual problems and see how to actually utilize those methods to set up the equation of motion for a multi-degree of freedom system. In the last lecture, we discussed that if we have a multi-degree of freedom system, how to formulate the equation of motion of a multi-degree of freedom system. Okay and we took the example of a shear type building in which we said that, okay we took an example of a three-story shear type building in which we said that the building has three degree of freedom represented by u1, u2 and u3 and it has masses basically lumped at each level and their storey stiffness is represented by k2, k1, k2 and k3. Okay and we found out the equation of motion of this three-story building subject to external load. Okay so these external loads were basically p3, p2, p1 applied at the respective stories. Okay and the equation of motion that we got was of this form, okay m1, 0, 0, 0, m2, 0 and 0, 0, m3. So, we got this mass matrix and then if there is also damper at each between each storey, okay then that can also be formulated in this equation of motion. And we can write this as c1 plus c2 and 0, sorry c1 plus c2 and then minus c2, 0, minus c2, c2 plus c3, 0, 0, minus c3 there should be c3 here, no minus c3 here and then this is c3. Okay with the velocities and then we had the stiffness terms which took the same form as the damping term. So, I can write this as k2 plus k3 minus k3, 0, minus k3, k3. Okay then u1, u2, u3 the displacement vector and this is equal to the applied load vector. So, in general the equation of motion of a multi degree of freedom system can be written as okay inertial force vector plus the damping force vector plus the internal force of the stiffness force vector equal to the applied force vector and this is basically the extension of the equation that we had used for a single degree of freedom system. And of course, in this case for the case that we have considered here can be written as mass times acceleration, sorry there is an acceleration terms here. So, acceleration vector plus damping would be damping matrix times velocity vector and then there is internal force that is stiffness matrix times the displacement vector and this is the applied force vector. So, this is the general form of the equation of motion and we also talked about that this equation of motion can be derived using two methods okay. In the first method we can simply consider the free body diagram of the system that is shown here. So, either we can cut the system at these three locations and write down the three equation of motion corresponding to each masses okay. And that is called the direct equilibrium method in which we are going to directly write down the equation of motion and then formulate these matrices and vectors here. In the second method we talked about the influence coefficient method okay and in the influence coefficient method we said that I am going to formulate stiffness matrix okay by considering the unit displacement at any degree of freedom and then finding out the forces required to have that unit displacement maintained at that particular degree of freedom and zero displacement everywhere okay. And the forces that we get basically those forces are the column vector okay column vector of the stiffness matrix and these coefficients are called the influence coefficient and we are going to repeat that for each degree of freedom to find out the whole stiffness matrix okay. So, we said that a ij is nothing but force at degree of freedom i okay due to unit displacement at degree of freedom j okay and the same concept can also be extended for the damping matrix in which the same thing okay so let us say c11, c21, cn1 and the damping matrix can be obtained assuming cij would be the force at dufi due to unit velocity at degree of freedom j okay similarly the mass matrix can also be found out like that okay and mij is basically force at dufi due to unit acceleration at dufj okay so in these cases what do we do we first consider unit displacement at any degree of freedom let us say j and zero displacement everywhere and we find out forces that need to be applied at each degree of freedom to maintain that state of deformation okay and that can be done using the static analysis okay. So, this is considered in the stiffness component of the structure this is considered in the damping component okay and this is considered in the mass component okay so the this component basically means that in this the stiffness component we only consider the bare frame in the damping component we only consider the dampers and in the mass component we only consider the masses in the system without the frame or the damper ok. And we know, we consider the representation of a multidigrafitum system can be written as a sum of 3 individual components stiffness component, damping component and the mass component. So, using this method as well we can formulate our equation of motion like this ok. So, in today's class what we are going to basically do employ both these method direct equilibrium method and the influence coefficient method to find out the equation of motion of different type of systems ok. So, let us do the first example ok. So, in first example basically what do we do? I am going to consider the same example that we did last class. So, I have a continuous bar, a rigid bar ok and the mass is distributed over the length l ok. And force vectors Pt and P theta are being applied at the center of this rigid bar ok and the degrees of freedom are u1 and u2 ok. So, basically and these springs have a stiffness is k1 and k2. So, the equation of motion need to be found out using both method first the direct equilibrium method and then the influence coefficient method ok. So, let us first do the direct equilibrium method. So, as we know a rigid bar in 2 dimension ok can be represented as 2 degrees of freedom. So, let us say initially the bar was here, but at any time t ok the degree of freedom u1 and u2 ok. So, this is u1 here and this is u2 here. So, this bar in 2 dimension can rotate or can translate ok. So, it can do both motion and to represent the motion we need 2 degree of freedom to represents its displaced position with respect to the initial equilibrium position ok. Now, this is the deformed position. So, we need to draw the free body diagram ok of this bar in a deformed position ok. So, let us do that now as I know as you know you have a spring here. So, when it is deformed by u1 force that would be applied here would be k1 u1 and then there is another force which is basically k2 u2. Now, I am going to apply 2 pseudo quantities here ok. So, because this bar can translate and can rotate I am going to apply the pseudo translational inertial force and the pseudo rotational moment ok against the direction of translation and rotation. So, it is translating in this direction ok. So, let me just so you will it will have the bar would have mass times acceleration at this point. Now, acceleration at that point can be written as acceleration at the end 1 and acceleration at the end 2 divided by 2 ok because it is at the middle point. So, this is the pseudo translational force. Now, it is rotating anticlockwise. So, a clockwise pseudo moment would be applied to it and that would be the moment about its center of mass let us call this ICM times the rotational acceleration which I can write as u2 minus u1 divided by L ok. Basically, this is the angle here ok. So, the angle theta here ok for this is basically you consider u2 minus u1 divided by L ok and the rotational acceleration would be just the double differentiation of that quantity right there ok. Remember that there are two forces as well here. So, you have Pt and P theta ok. So, utilizing that let us write down the equation of motion. Now, to write down the equation of motion what I am going to do I am going to first write down let us consider this as end A and end B summation of moment about point B equal to 0 what that would give me directly the equation let us write it down here ok. So, I will have IM ok times u2 minus u1 divided by L and this is basically clockwise then I have k1 u1 times L which is anticlockwise ok and then I have M u1 plus u2 divided by 2 times L by 2 ok and this quantity is basically anticlockwise and then I have force Pt which is creating clockwise moment. So, Pt times L by 2 and then a anticlockwise moment P theta ok and that is equal to 0 ok. So, with that let me just rearrange the terms here ok. So, we will have mass times remember IM the moment of inertia of this would be bar ML square by 12 ok. Now, I am going to substitute that in the equation of motion over there. So, let us do that here ML square by 12 times u2 minus u1 divided by L here ok and then again I have this term here u1 plus u2 acceleration divided by 2 times L and this becomes 4 here. There is this quantity here k1 u1 times L plus Pt L by 2 minus P theta ok. So, let us further simplify this one ok. So, this we can write it as M by 3 ok u1 plus M by 6 times u2 ok and this plus k1 u1 here and this is equal to Pt by 2 minus P theta by L ok. Similarly, what I am going to do here write down the equation of motion summation of M equal to 0 ok and let us write that down. So, when I do that it will have IM which is basically again IM times u2 minus u1 divided by L plus M times u1 plus u2 divided by 2 times L by 2 ok both clockwise alright about point A. Then I have k2 u2 which is again clockwise times L and then we have P theta which is anticlockwise and Pt which is also creating anticlockwise moment. So, if I take it to the right hand side of the equality I would get that as Pt plus. So, let me just write down Pt times L by 2 plus P theta here ok. So, this I can write down as again I can simplify this as M by 6 u1 plus M by 3 u2 plus k2 times u2 and that is equal to Pt by 2 plus P theta by L ok. So, equation 1 equation 2 can be combined together with the matrix form and written as M by 3 times M by 6 again M by 6 times M by 3 and then the acceleration vector here ok and then I have the force vector k1 0 and 0 k2 then the displacement vector and that is equal to the force vector which comes out to be Pt by 2 minus P theta by L let me write it again and then Pt by 2 plus P theta by L. So, there is some important point to note here if you look at it this is a distributed mass bar ok. So, the mass is distributed throughout the length and the degrees of freedom are actually defined at the ends of this bar ok just above k1 and k2. So, if you look at it ok we get the stiffness matrix as diagonal matrix. However, if you look at this mass matrix it is a non-diagonal matrix ok and the forces also you do not directly get the force Pt and P theta. So, and this is basically the implication of how you define your degrees of freedom ok because the degrees of freedom in this case were directly above k1 and k2 ok which represents the deformation in the two springs I get a diagonal stiffness matrix, but if you look at the displacement here it does not correspond to a single lumped mass ok it is a distributed mass and it does not correspond to the direction of the applied forces which are moment forces in moment which are Pt and P theta that is why again we do not get directly the diagonal mass matrix or a single force vector that comprises of directly the Pt and P theta ok. So, this is the equation that we obtained using the direct equilibrium method let us obtain the same equation ok or let us see what we obtain if you utilize the influence coefficient method ok influence coefficient method. Now, in the influence coefficient method let us start with the formulation of the stiffness matrix ok and if you remember to get the stiffness matrix first we are going to apply the unit displacement at each degree of freedom with zero displacement at other degrees of freedom and find out the columns of the stiffness vector ok. So, let us do that. So, in the first case we are considering u1 equal to 1 and u2 equal to 0 ok. So, basically the deflected shape would look like something like this. So, this is my bar here it has u1 equal to 1 and u2 equal to 0 ok. Now, to maintain the shape we would have to apply forces which would be the influence coefficient for the stiffness matrix. So, those forces would be at the degree freedom 1 the force due to unit displacement at degree of freedom 1 and force at degree of freedom 2 due to unit displacement at degree of freedom 1. So, these would give me k1 and k1 1 and k2 2 1 ok and remember in this case we have subject to these deformation ok k1 1 and k2 2 we only consider the stiffness components ok not the mass component or the applied force or anything ok. Now, subject to these displacement we know that we have two springs as well right because we still need to consider the stiffness component. So, these forces need to be applied, but with this displacement we know that there will be downward force at k1 which would be k1 times the deformation in that spring which is 1 and then here k2 times the deformation in that spring which is 0 ok. And again we can solve this ok. So, let us write down if I write down the equilibrium summation m v equal to 0 I can directly get as k11 times l is equal to k1 times 1 ok and times l this is equal to 0. So, k11 is nothing, but k1 similarly if I consider summation m equal to 0 I would get as k21 times l minus there is no moment created by the force k2 because it is 0 ok equal to 0. So, k2 equal to 0 ok. So, we have got the first column of our stiffness vector which is k10 I still do not know what this is second column to get that let us say u1 equal to 0 and u2 equal to 1. So, in the second case I will have the deformation state which is something similar to this one. So, u1 equal to 0 u2 equal to 1 I will have to apply the forces which are force at degree of freedom 1 due to unit displacement to degree of freedom 2 force at degree of freedom 2 due to unit displacement degree of freedom 2 and subject to this deformation state it would have the spring forces which are 0 at this point and k2 times 1 at this point. So, again utilizing similarly the equilibrium of equation I can get as k12 as 0 and k22 as sorry k2 ok. So, I have obtained k00 k2 and this is my stiffness matrix ok. So, although I have demonstrated for 2 degree of freedom we can extend it for any degrees of freedom ok. Now, let us come down to finding out the mass matrix. Now, for the mass matrix we are going to repeat the similar kind of process except now we are going to doing the same thing for the acceleration not the displacement. So, in the first case ok just so in the first case I am going to assume unit acceleration at point A and then 0 acceleration at the second degree of freedom system. So, in this case I have unit acceleration at this point and then 0 acceleration at this point and remember we only consider mass in this one there is no spring or anything in this system ok. So, to get the mass matrix we only consider the mass component of the system. So, unit acceleration 1, acceleration 0. So, it would be varying somewhere linearly between these two acceleration. Now, because the mass is distributed the inertial force on this bar ok if you consider x to represent the displacement from the or the position from the right most end then at any point the acceleration is basically x by L times u 1 u 1 double dot which is what. So, the acceleration u x is basically x by L ok. Now the inertial force would simply be alright whatever the mass ok x now mass x is basically m divided by L ok. So, the inertial force at any seam distance would be F i x equal to m by L m x times the acceleration at that point which would be x. So, m x by L square ok and now we are going to write down the equilibrium equation for this one. So, what will happen I have inertial forces which are distributed like this and to maintain the state of acceleration I need to apply force m 1 due to unit acceleration at 1 and then force a degree of freedom 2 due to unit acceleration at 1 ok. So, in this case if I consider summation m b equal to 0 then I can write it as m 1 1 times L is basically equal to the net effective inertial force which would be in this case of times L times m divided by. So, the net resultant force in this case if you consider for this one would be m divided by L ok. Now this force would be acting at distance which is 2 L by 3. So, that need to be mentioned or you could just simply write it as you know if you take the integration of it the total moment would be basically m by L times x by L times x 0 to L ok and that would give you m by 3 ok or that is not m L by 3 it is say it is m L by 3 and in this case if you look at it you get the same quantity here. So, m 1 1 is basically m by 3 ok. Similarly, if I consider summation m a equal to 0 then you will get m 2 1 times L is equal to the same quantity, but now from the left hand side it is at distance L by 3. So, here you get m 2 1 as m by 6. So, we have got the first column m by 3 and m by 6 we still sorry we still need to get the second column here ok. So, we are going to follow the same procedure and we are going to apply 1 equal to 0 and in this case u 2 equal to 1. So, now basically consider inertial forces would be acting opposite to the direction of acceleration and I need to apply force at degree of freedom point due to unit displacement degree of freedom 2, force at degree of freedom 2 due to unit displacement at degree of freedom 2 ok and then the force that we will have here remember again this is u 2 equal to 1 and the inertial force at any distance is same quantity m by L times the acceleration x by L ok. So, I again I can employ the same equilibrium equation in this case you can get as m 2 1 ok or not m 2 1, m 1 2 is m by 6 and m 2 2 as m by 3 ok. So, this is m by 3 m by 6 m by 6 m by 3 all right. So, mass matrix is also obtained as this m by 3 m by 6 and m by 6 and m by 3. So, we have obtained mass matrix we obtain the stiffness matrix one more quantity that need to be obtained is the force vector. Now, if you look at it ok I have this bar here in which the applied forces p t and p theta are actually not applied the degree of freedom u 1 and u 2 ok. So, I need to find out the equivalent force system for this. So, that p t and p theta can be basically decomposed or it can be rewritten so that along the degrees of freedom u 1 and u 2. Now, let us say the forces p 1 and p 2 ok these two forces are equivalent these are two systems are same systems ok. So, in order to achieve the same thing ok what we are going to do again we are going to write down the equation of motion ok for this system ok. So, let us go ahead and write down the equation of motion ok. So, in this case let me first this case let me just first write down p 1 plus p 2 is equal to p t ok because if you consider same thing ok. So, at this point the net resultant force at this point is p 1 plus p 2 the net resultant moment at this point is basically p 2 minus p 1 divided by L by 2 ok and this is the net moment in the anticlockwise direction at this point and that is equal to theta. So, we can solve this and we can find out p 1 as p t by 2 minus p theta by L and p 2 as p t by 2 plus p theta by L ok. And this simplification we only did because our degrees of freedom or the applied forces were not applied at the degrees of freedom that we had considered for this problem. So, now we can write down the equation of motion as m by 3 m by 6 m by 6 m by 3 and the acceleration vector ok, then the stiffness vector k 1 0 0 k 2 2 1 u 2 and then the force vector ok. And we can compare this to the direct equilibrium method and see we have obtained the same equation of motion ok. So, which method to employ in what kind of problem you would only learn through looking at the problem and doing or practicing more problem ok. So, sometimes one method is usually easier to apply for a specific type of problem compared to other method and there is no fixed rule as such ok. So, that you would only need to know, but remember that it does not matter which method you employ as long as you are doing it correctly in the end you should get the same answer ok. Although you might find one method to be little bit difficult than the other method for some type of problem ok alright. So, after this what we are going to do or for this problem let us say instead of considering the degree of freedom along the two spring ok, how do we consider the degree of freedom as the u t which represents the translational motion and u theta which represents the rotational motion. So, that the deformed position can again be this as u t and this as u theta ok and the rest of the parameters remain same a 1 k 2 ok. So, in this case again I have p t and p theta like that. So, in this case you can go ahead and you can find out the equation of motion ok. You would get equation of motion which is little bit different and I am not going to solve this system I leave it for you to solve the equation of motion and let me just write down the final equation of motion ok. So, we get as m 0 0 m l square by 12 times u t and u theta and then k 1 plus k 2 k 2 minus k 1 times l by 4 and k 2 minus k 1 l by 4 and then k 1 plus k 2 l square by 4 and u 1, u 2 and this is equal to p t times p theta. Now, notice an important difference compared to the last equation that we had written here. In this case we are defining the degrees of freedom u t and u theta which are along the center of mass and the center of rotation or we can say that it is it represents the degrees of freedom along the mass translational rotation and the rotational motions. That is why we again get the diagonal matrix in which we have the mass term and we have the moment of inertia about the center of mass ok. However, because now the degrees of freedom are not defined along the springs we get non-diagonal matrix for the spring and the force vector because the degrees of freedom are along the force vector now we directly get as p t and p theta ok. So, we get two different equation although they are not exactly different I will just we will come back to that ok depending upon the equation or the degrees of freedom that we have defined ok. Now, you might see a different formulation of equation of motion, but we will see in the next chapter a dynamic system is basically defined through its mode shapes and frequencies which are called modal properties. So, even if you see that these equation are somewhat ok in a different form basically they represent the same system because through some mathematical manipulation this can be transformed to this or this can be transformed to this ok and we use something a matrix that is called a transformat transformation matrix to do that and let us quickly see how do we do that if we consider a system ok equivalent system. So, let me just take example of this one remember we had a system. So, in the default position it looks like this. Now, first time what we did ok we consider u 1 and u 2 ok which are which were the displacement at these two locations to represent the deformed shape. In the second case we considered ut and u theta to represent the displaced position ok. Now, if we consider here ok can I say my u 1 is nothing, but ut minus u theta l by 2 and u 2 is ut plus u theta l by 2 ok. So, that I can write it in a vector form u 1 u theta u 1 u 2 is actually equal to 1 minus l by 2 and then 1 l plus l by 2 times ut and ut theta ok. This matrix here is called the transformation matrix we are going to represent it a and we will see that these two systems are basically equivalent and if we need to transform a system from ut to ut theta ok. Let us say in the second case ok that is so first case let us say the mass matrix is m the stiffness matrix is k and the displacement vector is u ok. In the second case let us say mass matrix is m dash stiffness matrix is k dash the displacement vector is u dash and if I write it like this so basically the equation is here u is equal to transformation matrix A times u dash ok. I can substitute this formulation so that you will look at it here and your stiffness would basically become k dash equal to transformation matrix transpose times this k times transformation matrix and m dash is transformation matrix transport times the initial mass matrix times the transformation matrix ok. So, these are actually related these two systems and they represent the same system and we can switch from one system to other system by utilizing these equations ok. And if you have taken a course in structural mechanics you would have learned about this transformation matrices ok alright. Once this is clear remember we have been writing down our equation of motion as mass matrix times u damping matrix time sorry acceleration vector times velocity vector and then stiffness matrix the under displacement vector as p. Now, this is for any load vector p. Now, if we consider earthquake ground excitation of multi degree of freedom system so earthquake loads ok. So, for these type basically earthquake loads what we are doing let us consider the three mass lumped mass representation of the three storey shear type building. So, in this case I will have certain displacement like due to the ground excitation let us call this u g t and let us represent the ground acceleration of u g and then the relative displacement u t ok. Now, at each at any degree of freedom let us say ok. So, let us say this is the ith degree or jth degree of freedom ok. The total displacement let us say this is j the total displacement ok at the jth degree of freedom would be the ground displacement plus the relative displacement ok of the jth degree of freedom alright ok. Now, if we write that the we can write also write down our velocity by differentiating it once and the acceleration by differentiating it twice and then substitute it in this equation of motion ok. Keeping in mind this acceleration is actually the total acceleration and this velocity and displacement are actually relative velocity and relative displacement ok. So, let me just write down that this is equal to ok. So, in terms of vector representation this can be written as let us say vector u I am writing down for acceleration u t is equal to u g which is basically vector of the same quantity ok u g throughout the like you know in a column and then I have u g t which I can write as u 1, u 2, u 3. So, this would be the relative displacement vector ok. So, because I have the same quantity u g t I can write this as unit vector it is not a unit matrix ok it is not a. So, just keep in mind this is a unit vector times u g double dot t ok plus u of t. Now, in this case if I substitute it I would get the final expression as this times the displacement sorry the relative acceleration vector ok u plus relative velocity and then relative displacement this is equal to minus m times 1 times basically the quantity u g t here and this is my force effective force vector for a seismic excitation of a multi degree of freedom system ok. So, I am going to write my p effective as minus m this one and then u g t ok. All right now one thing to notice here that in this case my degrees of freedom were in the same direction ok as the ground excitation. So, it might happen that my degrees of freedom might not be defined in that same direction as the ground excitation in that case this quantity that I get here 1 or the unity vector it might not be actually unity and then in that case we represent it as a influence vector ok which is denoted as l ok and which basically represents the relationship between the direction of the ground motion and the direction or the degrees of freedom ok and I will give you some examples to show that what I basically mean by that ok. So, let us say ok I have in this case we had a building ok. So, in that building my ground excitation u g t is along the degrees of freedom ok. So, that I could write this as 1 the influence vector is 1. However, in the second case let us say we have something like this where the masses are actually lumped and the degrees of freedom are defined it like this u 1 in this direction ok u 2 in this direction and then u 3 in this direction ok. So, in this case let us say this is u g. So, to get the influence vector ok let us let me first write down my total acceleration vector ok as or first let us say for any j th degree of freedom any j th degree of freedom u j t the total displacement as the relative displacement plus the ground displacement and if we write it in terms of vector it would be total displacement vector as a relative displacement plus the vector ok the ground vector u g. Now, if my degrees of freedom are not along the ground excitation then this we write it as u t plus some vector l times u g t ok not acceleration we are still considering the displacement here ok sorry let me write it again here. Now, what do we do to find out this influence vector we apply a unit ground displacement in whichever direction the ground excitation is applied. So, you apply a unit value of the ground displacement and then you look at it for that unit ground displacement what happens to the displacement along each degree of freedom ok. For example, in this case that we have here if I apply u g equal to 1 ok and we are doing it statically ok. So, this is just finding out the relationship. So, if I apply u g equal to 1 ok all of these degree of freedom will move by 1 correct. So, that is why my l becomes 1 1 1. However, in this case if I apply ground displacement equal to 1 u 1 moves by 1 u 2 moves by 1 u 3 basically is 0 because there is no vertical there is no displacement in the vertical direction due to the unit ground movement of 1. So, in this case basically my l becomes 1 1 and 0 ok. So, that my influence vector can be written as 1 1 0. Similarly, let us say for the same thing I have something like this instead of I have 3 masses here and instead of translational ground motion let us say I have a rotational ground motion. So, theta g is there. So, in this case also what do we need to do? Just apply a unit rotation of theta g equal to 1 and then see along each degree of freedom what is the displacement corresponding to this one. So, this would be if this height is let us say h 1 this would be 1 times remember if this angle is 1 this is 90 degree then this angle would also be unity. So, this would be h 1 times 1 along this degree of freedom if u 2 and u 3 are this this would be whatever the height that we consider let us say h 2 h 2 times 1 in the this direction and this would be whatever the length of this is let us say this is h 3 here ok. If this is rigid this connection here if this rotates by theta this would also rotate by theta and it would come down by the length times this angle. So, that would be h 3 times 1. So, the influence vector here would be h 1 h 2 and h 3 ok and once we get the influence vector we can find out the effective force vector as minus m times the influence vector ok times the ground acceleration ok. If it is a translational ground acceleration if it is a rotational ground acceleration then this would be l times theta g t ok. And we can write down the our equation of motion as same as m u plus c of u plus k of u and this is equal to the p effective ok. So, we have seen that how to set up the equation of motion for a general load vector and also for the seismic excitation by obtaining the influence vector ok and we learned basically two type of method to set up the equation of motion the first one was the direct equilibrium method and the second was with the influence coefficient method ok. So, with this we would like to conclude this chapter ok. In the next chapter we are going to study how to get the modal properties of a multi degree of freedom system ok.