 So welcome back, most common encountered cases of dynamic loading on a structure is the ground excitation due to an earthquake and as an instructional engineer, you will have to design structure to resist the earthquake forces generated by the ground shaking. So in today's class, we are going to learn how to set up the equation of motion of any structure subject to ground excitation and investigate further into that. Then we are going to look into a different type of loading which would demonstrate the difference between or which will demonstrate the effect of dynamic loading. So we are going to talk about how to apply a pulse load. So if a load P0 is suddenly applied, then we will see what would be the dynamic effect on the structure. And then we are going to solve some problems in which we are going to simplify our structure to a spring mass damper system and then solve for its response. So we are going to start our lecture today and in today's lecture, we are just going to be reviewing some of the concepts that we did in the last class and then we are going to be solving some problems here. So let us start with our equation of motion that we did last class. In the last class, we said that if we have let us say equation of motion as Fd over Fi plus Fd plus Fs equal to the applied force. This was our equation of motion and we considered two representation. The first representation was basically a frame representation. This representation is better suited for people with the structural engineering background and what we assume that the whole mass of the frame is concentrated at a level and then under the action of an external load Pt, it actually deforms by deformation let us say u and if it is a damped system then there was also damper connected in between. So this was the first representation and the same representation could also be understood using another representation we discussed which was basically a spring mass damper representation. The spring mass representation damper is used more frequently in mechanical vibration subject or in physics. So in this spring mass representation, again I have the same mass m here, the stiffness k damping coefficient c and under the action of this applied load here it undergoes our displacement u. So these two representation we considered and if we had let us say a linear system. So for a linear system we said that my spring force or the stiffness force would be simply k times u and for linear damper it would be simply c times velocity so that I can write my equation of motion as m u double dot plus c u dot plus k u and that should be equal to p of t and this is a second order differential equation. So this was in general when there is a force p of t that was being applied on the structure now we also saw that if we had an earthquake excitation in which there was a ground so for an earthquake excitation the same equation can be adapted to be written in a different form. So for an earthquake excitation with ground acceleration represented by u g t and displacement basically represented by u g t here and the acceleration u g double dot t. So we saw that the total displacement of the this frame that we had there would be a rigid component due to so rigid motion of the body due to ground displacement so this is u g of t and then there would be a further deformation due to the response of the structure so in this case it would be something like this because initially I had a mass here which finally shifted here. So this remember that this is u g of t here and then the displacement of the structure I have represented as representing as u of t. So the total displacement of the structure can be written as ground displacement plus the relative displacement of the structure or the single degree of freedom representation of the structure. Okay and same would be the relationship if you double differentiate it you would again get u g of double of t and then u double dot t. So this is the representation now similarly if you consider the free body diagram of this remember that there is only f i f d and f s here there is no external for as such that we had at the top of the structure p of t okay and we said that for a inertial force is always due to total acceleration however the damping and the stiffness force is due to the deformation of the structure. So the rigid motion which is basically u g of t which is the rigid offset it is not going to produce any stiffness force in the structure because it does not produce any deformation in the structure okay similarly for the damping the damping is contributed by the velocity in the structure with respect to its base and for that representation you could only have velocity with the structure is deforming and undergoing let us say a velocity u dot okay. So that is why we could write it as m of u g of t plus the relative acceleration and f d I can write it as c of u of t and then f s s a of u of t and that should be equal to 0 and I was able to write this equation as reformulated and write it as okay u t I will bring the u g double dot t on the right hand side then write it as this term here which is basically the same as the equation that we discussed earlier this equation okay or the equation here except there is a difference that now we have a p effective here which is represented by the mass of the structure times the ground acceleration and this is basically the earthquake force that acts on the structure so earthquake force on a structure is acts opposite to the direction of ground acceleration and it is mass times mass of the structure times the ground acceleration okay so just keep that in mind okay so the next step that comes that if I have a second order differential equation how do we solve it and so if I have this equation of motion here as again consider let me write that again how do I solve this so depending upon whether it is a free vibration okay free vibration so free for free vibration your p t is equal to 0 or whether it is an undamped system so for an undamped system your c is equal to 0 or a damped system okay c is not equal to 0 I could obtain different solution of the value basically the expression for u of t okay and that we are going to do in the subsequent chapter when we are going to study free vibration and the response of undamped and the damped system subject to periodic and non periodic loadings okay but what I want to show you here today is that to a simple example how a load that is applied suddenly or you can call it like you know dynamically how does it differ from the load that is applied statically and how would actually the response would differ okay so for example let me take an example here in this example I am basically applying a step force in which the load p naught is applied okay but it is applied suddenly okay so I have this time variation of the load and then I need to find out what would be the response of the system so I am considering basically an undamped system so that my c is equal to 0 just for the sake of demonstration and I want to find out what is my basically the peak displacement now if you remember from your undergraduate classes before if you were told that there is a force p naught in a spring that has a spring constant as k the deformation it was set and I am going to represent it as u static it could be simply calculated as p naught by k correct now let us see if I apply this step force suddenly and then if I have available the time variation of the PT okay then I am going to solve the equation of motion and then I am going to see whether my dynamic response is any different from this value and how much actually is that difference okay so in this case I have given the force PT which is p naught for all time greater than equal to 0 okay and my because it is an undamped system my equation of motion becomes Mu equal dot sku is equal to PT which is here is p naught so my goal is to solve this equation okay and we are going to come to like you know in subsequent chapters how do we solve this differential equation I just want to give an overview that how do we solve the second order differential equation okay so for a second order differential equation okay the total solution ut is represented by some of a particular solution and then some of complementary solution which is also called homogeneous solution okay so a particular solution is any solution that satisfies this equation okay so any solution that satisfies the equation your equation of motion can be used as a particular solution so for in this case I can directly look at that if I substitute up equal to p naught by k then the double differential of upt is actually 0 and then I get p naught equal to p naught so this could be used as a particular solution okay and then I have to find out complementary or homogeneous solution which is basically finding out the solution of this equation here k of uc of t equal to 0 so we set the right hand side equal to 0 and that is why we call it a homogeneous solution okay and then we try to find out the value of or the expression for uc of t okay so in this case for this kind of differential equation I am going to assume my uc of t as e to the power lambda t okay and when I substitute in the above equation what do I get as m e to the power or let me say m lambda square because it is a double differential e to the power lambda t okay plus k e to the power lambda t okay so it should be m lambda square plus k equal to 0 okay now I know that this term cannot be equal to 0 it does not give me any feasible solution so what I am left with is this expression here if I quit that to 0 I get lambda as plus minus under root minus k by m which can be further written as plus minus i to represent the under root minus 1 and this one I will write it as k by m again I will write as this minus i omega n omega n is basically the natural frequency of the system that you are considering we are going to come back to that in a later chapter just right now you like you know it interpret as any other constant through which I am representing this quantity under root k by m okay so I have two roots here lambda 1 which is i omega n and the lambda 2 which is minus i omega n okay so if I have two roots the solution of this complementary this equation right here okay can be written as a linear combination of both roots okay so I would write that as remember that I had assumed this has to be the solution so linear combination of the two solution a e to the power lambda 1 t and b e to the power lambda 2 t okay which would nothing but e a to the power i omega n t plus b e to the power minus i omega n t okay and I also remember from my knowledge of complex number that I can write e i to the power i x equal to cos x plus i sin x and if I replace i i x by minus x I can write this e to the power minus x s again cos x minus i sin x okay so I will do that here okay and then rearrange the terms to obtain a plus b cos omega n t plus i a minus b sin omega n t okay so this and this these two are unknown coefficients that I need to find out I am going to represent them with two unknown coefficients here and I can write it as at c cos omega n t plus d sin omega n t okay so this is the complementary or the homogeneous solution so the total solution of the differential equation I can write it as u particular plus u complementary okay and this would be p naught by k plus c cos omega n t plus d sin omega n t now you see that we consider a second order differential equation and now we have two unknown constants so we need two conditions to actually find out the specific solution to our differential equation okay and where those two conditions are going to come from well it comes from the initial conditions for a dynamic problem okay so the initial conditions in a equation of motion would be your initial velocity and initial displace sorry initial displacement and initial velocity okay and in this case because it is given that initially like you know they were at rest suddenly a force is applied so I can assume you can assume that u 0 and u dot 0 both to be equal to 0 okay for my case okay so you can substitute that in this equation here to obtain the value of the constant c and d okay and you will see that okay you will get the final solution as p naught by k 1 minus cos omega n t right so this is the solution that I have okay due to the step force that is being applied to my single degree of freedom system so let me just write it again here or redraw it again so I have p t here and then I have t here and this is p naught now remember had it been a static problem then my u static would have been just p naught by k so if this is my u of t n so this is the response this is the applied force and this is the response for a static problem I would have this displacement as p naught by k but I can see that my dynamic displacement actually is not p naught by k but it actually fluctuates due to time variation of cos of omega n t and if you try to plot that it would look like something like at t equal to 0 your displacement is 0 okay and at t equal to let us say pi what you will find your displacement is actually minus this term would be minus 1 so that we will get 2 p naught by k so if I draw this time displacement this value here is 2 p naught by k which is the maximum dynamic displacement okay which would occur when cos omega n t is minus 1 so I can write u max is equal to 2 p naught by k okay which is 2 times the u static that I had determined okay so you can see that if I had applied the load p naught statically or for a static problem I would get a displacement which is equal to p naught by k however if I apply the same load suddenly like a step force then I am getting a peak displacement which is 2 times the static displacement okay so this is the effect of the structural dynamics that is coming into the play here okay and this too is as you would study later it is called dynamic amplification of the response and there are different factors to define it okay so this example was just to demonstrate that if a load p naught is applied statically or if it is applied dynamically then your response are actually or the peak responses are actually different and for this specific case it was 2 times the static displacement okay so I hope this problem will give you a better idea of what the dynamic nature of problem could do to the response of a single degree of freedom system okay alright if that is clear to you what we are going to do now we are going to do or solve some examples related to these type of equation of motion okay so and then see how do we tackle different problems okay so in example one here what do I have here is actually a slab okay slab that is supported and four columns okay so these are the slabs that are supported on four columns okay and the dimension of these this slab is b in this direction and b along this direction okay and it is given that these four columns with respect to these axis is x and y have stiffnesses kx the lateral stiffnesses kx and the lateral stiffness ky in the two perpendicular directions so each of these columns would have these stiffnesses into direction okay and what you need to do for a small angular deformation of the slab in plane angular deformation so let us say slab is actually rotating okay like this in plane you need to set up the equation of motion for this slab here alright so I would like you to pause this and then try to solve this problem alright let us discuss the solution of this problem okay so let us consider the top view of this slab that I have here okay slab let me just draw it like this okay so this is the slab that I have and if I rotate it if it has an angular rotation let me again draw the deflected position of this slab which would be I just redraw it here okay would be something like something like this okay sorry for the bad drawing but let me I mean figuratively like it does the job okay so what is happening here it is rotating by theta here okay so that at all points it is rotating by theta with respect to its original position okay now remember as it rotates okay at any corner the horizontal displacement is the same displacement that I have here which is represented by the half of this length which is d by 2 okay times the angle so this displacement horizontal displacement is d by 2 theta similarly the vertical displacement at this point is nothing but the half of this length which is b by 2 times this angular displacement which is b by 2 times theta and same would be the displacement at all the four corners okay and like you know if you want you can do the same thing you can derive the same thing by considering the angular displacement by plotting a line and then doing like this okay this is just a simple way of considering that whatever the horizontal displacement at this point okay at this point here would be the same horizontal displacement at this point the vertical displacement of this point would be the vertical displacement of the slab at horizontal position okay now when it tries to rotate it like that what will happen yes there the column would apply a force in this direction which would be kx times the horizontal displacement which is d by 2 theta and in this direction which would be okay ky times b by 2 times theta and same would be the case at other corners as well for example here it would apply in this direction okay and in this direction okay and at this point it is going to apply in this direction and in this direction and at this point it is going to apply in this direction and in this direction and the one thing that is common that at each corner each pair of these forces are actually creating clockwise moment about this point the center of this slab. So if I want to set up the equation of motion, I need to sum up the moment due to all these forces that are being applied by the column on the slab at all of its corners. And then there is another force or the moment basically the inertial moment due to rotation of this slab which would be opposite to the direction of rotation and it would be i times theta double dot. So I can simply write it as i times theta double dot and then take the moment of these column forces that are being applied on the slab. So it would be first force would be kx times d by 2 times theta and the moment of this force about the center would be again d by 2 plus ky b by 2 times theta and the moment about center would be multiplying would be obtained by multiplying with the lever on b by 2. And this times 4 as you can see it is producing the same moment at all the four corners. So I will multiply it with 4 to get the final equation. So I can write it as final equation of motion as kx d square plus ky b square times theta equal to 0. So this is my final equation of motion for this given problem. Once that is clear let us discuss another example. Example 2 here in this example what do I have? I actually have a rigid bar here. So let me draw a rigid bar. It is connected to a pin connection, pin support at one third of its length and there are three springs each of stiffness k connected to this rigid bar at length l by 3 l by 3 and l by 3 from each other. And what I need to do again is to set up the equation of motion for an angular rotation of this bar ok. And the mass of the bar is m and as I said the total length is l ok. So take some time and solve this problem ok. Let us discuss the solution to this problem ok. First let me say let me draw the initial position and the defined position for an angular rotation. So let us say it rotates in clockwise direction by an angle theta. So let us say the rotation is in this direction by angle theta ok. Due to this rotation the springs at the leftmost end would be stretched by a deformation and I can find that out as l by 3 times theta and this second spring is going to compressed by l by 3 times theta and the third spring would be 2 l by 3 times theta ok. So in terms of forces this spring is going to apply downward force and these two springs are going to apply upward forces like this ok. And if you consider the free body diagram due to clockwise rotation there would be an inertial movement which would be applied anticlockwise ok. And that would be the I of this or the moment of inertia of this rod about the point of rotation times theta double dot ok. So if I take an moment of all these forces and equate it to 0 about the point of rotation O which is the pin support ok. Let us see what do I get? I get I theta double dot plus the moment of this force all these three forces spring forces are in the same direction the counter clockwise direction ok. So it would be k l by 3 which is the spring force times the lever arm about the point of rotation is l by 3 ok. For the second spring k l by 3 times theta times l by 3 and then the right most spring would be k 2 l by 3 times theta into 2 l by 3 and that should be equal to 0 ok. Now the I of this bar about its about the point of rotation would be I of this bar about its center of mass plus the mass of the bar times the distance of the point of rotation from the center of mass which I can calculate as this ok. And this I can write as m l square by 12 and this as m l square by 36 which eventually gives me m l square by 9 ok. So I will substitute that here m l square by 9 plus this I would get as 6 k l square by 9 times theta equal to 0 there is another theta double dot term here ok. So the final equation of motion would be m l square or I can just divide that and I can write it as theta double dot plus 6 k divided by m times theta equal to 0. So this is the final equation of motion for this system that is shown here ok alright. Let us go ahead and do example 3 ok. So in example 3 what do I have it is not a very difficult problem but I just want to demonstrate you the concept of rotation and twisting ok. So in this case what do I have I have a thin shaft here ok which is rigidly connected to a disc ok. The polar moment of inertia of this shaft ok can be calculated from its diameter d the radius of this disc is r ok the mass of this disc is m ok the shear modulus of this shaft is actually g ok let me see do you need anything else yeah you need the length of this shaft. Now can you imagine if I try to provide twist to this disc ok what is going to happen remember that disc in its own plane is very rigid compared to the twisting stiffness of the shaft. So all the twisting is going to happen in the shaft and if you remember if I have 2 spring in series let us say k1 and k2 ok where let us say k1 is less less than k2. So k equivalent is actually decided by the whatever the most flexible spring is in this case it would be approximately k1 ok and that you can find out from the expression 1 by k1 plus 1 by k2 right if k1 is very very small than k2 this quantity would be very very large to this quantity so that you can neglect it so that when you invert it k equivalent becomes equal to k1 ok. So in this case disc own stiffness torsional stiffness is much higher than the shaft stiffness ok and that is why I am just going to consider the whole resistance is coming from the shaft and if you remember from your solid mechanics class if I have a shaft of length l ok and then shear modulus g and polar moment of inertia j ok and if I apply a twisting moment on this shaft ok the relationship between this twisting moment and the twist in this shaft so if I project it here let us say this is deforming by theta this I can write it as jg by l times theta which is the basically the rotational stiffness or the twisting stiffness of this shaft right here ok. So if you consider the free body diagram of this disc and then assume that it is rotating by theta remember the same would be the rotation in the shaft ok so shaft is going to apply and equal an opposite moment which I have here which would be m of the shaft let us say it is ms ok and because I am rotating it by theta what will happen I need to again include in the free body diagram a rotational inertia ok which would be whatever the mass moment of inertia of this disc times theta double dot ok so if I take moment of all the forces about center I can write it as i theta double dot plus ms that should be equal to 0 and I know that my ms is this ok so I write it as mr square by 2 ok plus jg by l times theta that should be equal to 0 and if you remember the polar moment of inertia of a shaft is given by the formula pi d4 by 32 ok so that you can use here alright so this was our third example ok let us go through another example which is example over here ok and in this example what do I have I have two concentric discs which are rigidly connected to each other ok so I am going to draw the figure here and there is a string which is connected to this internal disc and there is a mass hanging from this disc ok and there is a spring connected to the outer disc and then the stiffness of this spring is k the internal radius is r and the external radius is 2r the mass here is m ok so you need to set up the equation of motion for this system ok so pause the video here and then take some time to attempt this problem and see what do you get as final answer ok alright let us discuss the problem ok now for this setup that I have here let us consider the motion from the initial equilibrium position remember it is not the undeformed position but the equilibrium position that I am considering alright and if you remember gravity load need not to be considered if I am considering the motion from the equilibrium position and if initially the gravity load was being balanced by a spring ok deformation in the spring which is the case here so I do not need to consider that gravity load in the equation of motion or the free body diagram ok so if I consider the free body diagram of the mass here ok alright what are the forces acting on this mass mass the mass is going down so there would be inertial force which is acting upwards which would be m u double dot and then there is a tension in this string here alright and I am saying that this is coming down by u now can I say if this string comes down by u ok the circumferential rotation of this inner disc would also be u so that the angular rotation of the inner disc ok so I am saying that this is coming here by u ok so that this theta that I get ok would be how much ok this theta would be u divided by the inner radius of the inner disc ok which is r ok and same would be the rotation of the outer disc as well however the circumferential rotation distance would be actually 2r times theta which if I substitute the value of theta it would be 2u and the same would be the extension in this spring here ok so the force in the spring would be actually k times 2u ok which is the deformation in the string and that would be the force that would be applied on this outer disc by this spring so if I consider free body diagram of the whole disc arrangement ok I would have force 2ku acting here I have force t which is acting here ok remember that this do not have any mass themselves ok so if I take the moment of all the forces about the center of the disc I get t times r is equal to 2ku times 2r so that gives me t as 4ku so now I know that what is my force in the string I can write down the equation of motion of this mass ok which would be nothing but mu double dot plus t and that should be equal to 0 which I can write it as mu double dot plus 4u and that should be equal to 0 alright ok now let us see our example 5 an example 5 I again have a spring mass system except in this case ok I do not have a block but I have a rotational disc here ok and I need to find out what is the equation of motion if there is rolling without slipping so if that is the condition and these constants are given to you ok the radius is basically r here you need to find the equation of motion of this system here alright ok so take some time and then try to attempt this problem so that we are going to come back to again this later and then solve it alright.