noc19-ce27-lec12

[Music] [Music] in the previous lectures we learnt the responses of a single degree of freedom system under free vibrations and under harmonic and periodic forces from now onwards we will learn the responses of single degree of freedom structures under any arbitrary excitation so as a first step let us understand the response to unit impulse so what is an impulse if a large force acts on a body for a very short time that is called impulsive force and the time integral will be finite that means the force integrated over time will be the constant it will be a finite value and the quantity of this impulse is equal to force times time so what is the unit impulse if the value of this impulse is equal to 1 we can call this unit impulse so let's understand it in detail so here we have a force acting for a very small duration say at time is equal to tow so here this force the magnitude of the force is equal to 1 by Epsilon and it is acting for a time Epsilon so what will be the value of the impulse it would be 1 by epsilon times epsilon that is 1 so this is a unit impulse function so if this force is acting for a large duration that means the value of the force will be less so the product should be 1 so if epsilon is large then force will be smaller so what if this epsilon tends to 0 so if epsilon tends to 0 for this product to be equal to 1 the force should be close to infinity so as epsilon tends to 0 the magnitude of the force becomes infinite but even then the magnitude of the impulse will be same as 1 because it is force times the duration when the duration is small forceful peak but again the magnitude of the product will be equal to 1 so this is still a unit impulse force so such a force is known as unit impulse so how will we represent this type of forces in our analysis so mathematically we can represent this impulse force using Dirac Delta function so what is directed at a function so this function Delta at t minus 2 is defined us plus infinity if t is equal to 2 and it is equal to 0 otherwise so this function represents this force so when T is equal to tow that means at this position the value becomes infinity and at all other locations the value is 0 so this Dirac Delta function can be used to mathematically represent this impulsive forces now let's see how the unit impulse is initiating a vibration in a single degree freedom system so let's go back to our Newton's second law the Newton's second law states that rate of change of momentum of a body is equal to the applied force so when the forces applied to a body there will be a change of moment so momentum is mass times velocity so there will be a rate of change of momentum so D by DT of momentum will be equal to the applied force this is Newton's second law so and if we have a constant mass if the mass is constant with respect to time we can write this as mass times acceleration so the force will be equal to mass times acceleration so we can integrate this force this equation over time and we would get the Ellis's will become integral t1 to t2 P DT is equal to mass times if you integrate acceleration that will be equal to velocity so that would be the change in the velocity so we would get this impulse is equal to change in momentum so what does this mean so if we apply the unit impulse at time tow that impulse will impart a velocity to the mass and that velocity will be equal to 1 by him so for the impulse the change in velocity will be created so that velocity will be equal to 1 by him so what would be the displacement so the displacement before the impulse and up to the impulse will be equal to zero so the single degree of freedom system is at rest initially so a unit impulse acting on the single degree of freedom system causes free vibration of the system due to this initial velocity one by M so the impulse acting on that body is giving it an initial velocity and that initial velocity will cause that system to vibrate and the free vibration it is free vibration because after the impulse there's no force acting on the system so the system is the free vibration so now let's calculate the free vibration response so we have to derive that the free vibration response of a damped single degree of freedom system is given by this and it depends upon the initial displacement and the initial velocity and in this expression Sita as the damping ratio and Omega D is the natural frequency times square root of 1 minus Zeta square and this is known as the natural frequency of the damped system so we had derived this earlier so we will use this to find the response to unit impulse so the unit impulse response function this the response due to unit impulse force that say time equal to tow so we just learned that when we apply a unit impulse to the single degree of freedom system it is equivalent to an initial velocity given to the system and the Denisha velocity is equal to 1 by n and that initial displacement is equal to 0 the impulse is acting on the body at time is equal to 2 so we have X dot o and X toe known to us so for any damped system using this expression we can write the unit impulse response function that is represented as function H so H T minus 2 that is the response to unity bus then a unit impulse is acting at tow we can substitute the value of this displacement and velocity in this expression so the displacement is 0 so this term will vanish and we have this term with X naught dot replaced by our X dot tow so the expression for the unit impulse response for the dam system this this so we can just rearrange the terms and we would get 1 by M Omega D e to the power minus Zeta Omega M t minus 2 sine Omega D t minus 2 so this is the response of a dam system under unit impulse so for undamped system this unit impulse response function is equal to 1 by M Omega L sine Omega t minus 2 so we don't have this term and Omega D will be equal to Omega and then it's a undamped system this is the plot for both these responses so as you can see we have giving an initial velocity to the system through this unit impulse so this initial velocity that is the slope of this response curves at t is equal to tow will be equal to 1 by M that is equal to the initial velocity now let's find the response to arbitrary force so we will use the unit impulse response function to calculate the response to arbitrary force let's see how it is done arbitrarily varying force can be represented as a sequence of infinitely short impulses so we have this force and it is having an arbitrary shape its varying arbitrarily with time so this type of force we can split it into multiple short impulses so we can represent this force as the sequence of short impulses so just like one impulse is acting immediately after the other so we can represent it like this and we can calculate the response for each of this short impulses so the response of a system to this one impulse at say time - will be equal to the impulse will be equal to the force times the duration of thimbles so that would be the force at that time - that is p2 multiplied by the infinite simmer duration that is d2 so this will be this short impulse at time toe and the response of the system due to that impulse will be the impulse times the unit impulse response so this HT - toe is the response to a unit impulse so the response of an impulse is equal to P tow dito will be that times the unit impulse function so this is possible since we are dealing with linear system so if the system was nonlinear this will not work so a response of a linear system to any arbitrary force is equal to the sum of the responses of all small small impulses so the response due to this arbitrary force can be calculated by integrating this expression over time so the response XT will be equal to integral 0 to t P tau H t minus tau D tau so this P tau and D tau will give the value of thimbles and H is the puritan bus response so we will get the response due to this force so if we have 0 initial conditions that is at time is equal to 0 if our displacement and velocities let's say then responses will be for dam systems we can find the responses this where this P tau is the expression for this arbitrary force similarly we can find the response for an undamped system by ignoring this damping terms so that would be 1 by M Omega L integrals over time P tau sine Omega n t minus tau D tau so we can use these integrals to find the response to a nar 'but rarey force and these integrals are known as Duhamel integrals now we will use these Duhamel integrals to find the response of single degree freedom systems so first let's find the response due to a step force so in step force constant force P note this suddenly applied so at T is equal to 0 this force P note is suddenly applied and after that the forces constant so it's a constant force but before its application the force was zero so this was a sudden jump it was suddenly applied force so let's see how the response of single degree of freedom system for this kind of a step force so will this be equal to the static response because in the static case we have a constant force right so what is the difference here so in the static analysis we assume that the load is applied on the structure very gradually so that it's dynamic nature can be ignored but here the value of the force is constant but it is suddenly applied so we will get some dynamic effects and we will see how much that effect is so we can find this response using three Hamel integral so the response of an undamped system is given by this so UT is equal to one by my integral over time beta sine Omega NT data so we know that this forcing function is equal to a constant V node so we can substitute and carry out this integral so if you integrate this will be P naught by M Omega M in the grill sign so we would get P naught by K 1 minus cos Omega NT so this will be the displacement response of a single degree of freedom system under this force over this is for a undamped system so if we plot this expression the responses like this so this is the response divided by P naught by K K so we have seen that this P naught by K is the static response so if this was a static force P naught then the response would have been P naught by K so in this normalized response curve we can see that this single degree of freedom system is oscillating about the normalized response equal to 1 so that means it is oscillating about the static deformation position it is not oscillating about 0 that is it is not oscillating about its original equilibrium position but it shifts by P naught by K that is the static deformation and it oscillates about that position and the maximum amplitude as we can see from this curve is 2 P naught by K so the normalized response as to the maximum response is 2 so the amplitude is 2 P naught by K that is twice the static displacement so that means if we suddenly apply a constant load P naught response amplitude will be twice as much as the static load P not so because of this sudden application there will be dynamic effects induced and that will be twice the static displacement now let's see the response for a damped single degree of freedom system so we have this damping term in the Duhamel integral so if you carry out this integral we would get this expression we are not going to the derivation of this you can do it by yourself so if you plot this we would get a decaying response so this is because of the effect due to damping this does not go on continuously as in the case of under damped systems but this decays after some times the response converges to the static response that is P naught by K so that means after some time this dynamic effects die out and the response is equal to static response now let's see the response to a ramp force that means a gradually increasing force so this force can be represented as P note T by T R so it is linearly increasing and after some time it is reaching the value P naught so let's see how the response to this force looks like so again for an N damped SDRF system we can use Duhamel integrals and substitute the value of VT here so it would be integral 1 by M Omega M integral P naught T by T R sine Omega n t minus 2 detour same as the previous expression we just have to replace the forcing function so if we carry out this integral we would get the response as P naught by K that is the static response t by TR minus sine Omega NT by Omega L th so let's or this so this is the plot for this expression so this blue curve shows total response beauty so as we can see as the forces increasing the response also increases but it has some dynamic effects so it is oscillating and that oscillation is above the static response so this dotted red line indicates the static response that is P naught by K T by T R so as the force is increasing the static response will also increase and our dynamic response will be an oscillation about this static response so now let's see what is the response to a step force with some finite rise time so this is the combination of the last two cases we have a step force but instead of suddenly increasing the force we have a wrap in between so it is gradually increasing and then the forces constant throughout the time so depending upon the duration of this ramp the effects will change so let's calculate the response so in this case the forcing function is equal to P naught T by T R and that is when the timeless less than TR so after T are the forcing function changes then it is equal to P naught so let's find the response when time is less than TR that's T less than T are we here in this zone that means the ramp forces acting so the response in this zone will be equal to what we saw earlier so the response will be this P naught by d by dr minus sine Omega NT divided by Omega NT R so this will be valid when T is less than T R so let's plot this we have seen this earlier so this will be oscillating about the static response as the force is increasing the response will also gradually increase so when T is less than T R this is the response now we have to find the response Minh T is greater than T R so what will be the response well this time is more than T are so will that be equal to the step force so we had derived the response of this step force initially so will it be exactly equal to the step force response or it is something else when time is more than TR we have this force that is a constant force similar to a step force so will the response be same as the step force so this is the step force we have seen earlier so if we have a step force at time is equal to TR the response will be this we just derived it earlier so when when the step starts at TR the response will be this but in our case here the response of this ramp step function will be this response plus the effect due to this ramp force so because of this Ram force in this zone our system is already vibrating so at t is equal to TR it is not at rest it will have some initial conditions that is the response due to this ramp force at time is equal so the effect of those initial conditions will also affect the response after dr so our response of this system of the single degree of freedom system under this force when T greater than T R will be this response plus free vibration of the system resulting from displacement and velocity at T is equal to TR so we need to calculate the response due to this ramp function at T is equal to TR velocity and the displacement and using that we have to calculate that free vibration component and add to this step force response so our total response would be this is the response due to this step force alone and this is the the first two terms are the free vibration response due to the condition at T is equal to th and that is because of the ramp force now we can substitute the value of velocity and displacement at T is equal to TR and get the expression for the response when T created that clear so if you substitute these values we would get this expression using this we can calculate the response meant T is greater than TR so now let's see the total response of the system so when the rise time is 3.5 times TN the response looks like this so this is an undamped system responds so this period between this two peaks will be equal to the natural period depending upon the value of the rise time the nature of the vibration changes so let's see how it is changing so this plot is for a rise time equal to three point five times T so this is a large rise time so here as you can see during the ramp portion the system is oscillating but that amplitude of oscillation is very less and it is oscillating about the static response as we have discussed earlier so after T is equal to TR this is oscillating about this static response P naught by K now let's see the response for other values of rise time so this is when TR is equal to 3 times natural period so TR is slightly lesser than what we have seen earlier so here this is the response so during the ramp portion the system is oscillating but when T is equal to TR here the velocity of this responses 0 how will we know from this curve because at this point this curve has a flat slope the slope of this curve at this point is 0 so that means the velocity at T is equal to DRS 0 so when the velocity is 0 in that case this no oscillation beyond the rise time so the force is constant here and the response is also constant so this is equivalent to a static response and it is also the value is also equal to the static response so this there is no oscillation around this static response this stays as it is in the static response case so now let's see when te R is equal to 2 point 5 when it's 2.5 the amplitude of the response and this constant force face is increasing and there's also oscillation so here when T is equal to TR the velocity is non zero we can understand it from this slope so this has a positive slope so that means the velocity is not 0 so when the velocity is not zero we are getting some oscillation at this range and the amplitude of the oscillation is higher compared to three point five case so here T is the rise time is three point five TL here the amplitude was low but here the amplitude is slightly high and again the period between two peaks will be equal to the natural period because this is a undamped system now let's see for TR is equal to 2 TN here again we get a condition here that the velocity is zero and when the velocity at this position is zero then there is no oscillation the response is equal to this V naught by K so this is again equivalent to static response it does not oscillate so when the dr is further reduced when it is 1.5 times the natural period the amplitude of this vibration increases this will oscillate only if the velocity is non zero so here it is known zero so it is postulated but the amplitude will be higher than that of TR is equal to two point five over TR is equal to three point five the rise time if you reduce it further here there is no velocity so there's no solution if you further reduce it there will be oscillation but the amplitude is red increased considerably now so the amplitude is 1.6 times the static response that is V naught by K so when the rise time reduces the amplitude of the vibration increases now let's see 4 TR is equal to point two times the natural period so in this case the rise time is very no so this response is very close to the step force so in the step force when the forces suddenly applied so here in the case of step force amplitude was twice that of the static response here when the rise time is point two times DL that is when we have a small rise time the amplitude is very close to two so from this we can understand that if a constant force is applied suddenly then it can cause some dynamic effects and those effects can make the response of the system twice as much as the static response and if the force is applied gradually the response will be close to the static response so if we apply the load gradually the dynamic effects will be much less now let's conclude our observations regarding the response to step force with finite rise time so in this case the system oscillates at the natural period TN about the static response position and if the velocity is zero at the end of the run force then the system does not vibrate during the constant force phase so after the Ram force it will be a constant response it won't vibrate for smaller values of TR by TN the responses similar to that due to the suddenly applied step force so when the rise time is less the response will be equal to the suddenly applied force when TR by TN is large then the dynamic displacement oscillates about the static solution and the dynamics effects are small so if you have a large rise time then the dynamic effects are small but the structure will oscillate about its static response position now let's see the deformation response factor the deformation does factories the maximum amplitude divided by the static response so as we can see this is plotted for different values of rise time that is T R by T n so as we have seen earlier when the rise time is 0 this is equal to the suddenly applied force and this deformation response factor is 2 that means the response will be twice that of the static response as the this ratio is increasing and as the rise time is increasing this Rd is reducing and when the rise time is larger then the value of Rd is very close to 1 that means the response will be very close to the static response so depending upon the value of the rise time to the natural period the amplitude changes in the expression for this deformation response factor that is Rd can be derived from the expression for the displacement by maximizing the displacement so you can find out the maximum value of the displacement in terms of TR by TN and that is this and this is how we plot this curve now let's look at some examples so in the first example we have the single degree of freedom system and damp system and it is subjected to two impulses so the force is consisting of two impulses each of magnitude I has shown in Figure so at T is equal to zero we have one impulse in the positive direction and which has a magnitude I then after a duration TD this another impulse - I and we need to plot the displacement response of the system for various values of D D by TN and for each case show the response to individual impulses and the combined response so we need to find the response due to each of these impulses and then the combined response also plot X naught by I by M Omega n so this is like deformation response factor and indicate separately the maximal occurring at T less than or equal to TD and T greater than equal to TD that means if the maximum response in this face and this face so let's solve this the response due to this impulse functions can be calculated by using unit impulse response so we can multiply this magnitude of this impulse by the unit impulse response and we get the response corresponding to these impulses so the first response that is the response due to this first impulse this the magnitude multiplied by the unit impulse response of the undamped system so this is given for the damped system and since ours is an undamped system we can ignore this damping terms so for the first response will be I times 1 by M Omega M sine Omega M T and similarly we can calculate for the second impulse and that response would be this magnitude that is minus I multiplied by the unit impulse response at T is equal to TD so that response will be minus I times 1 by M Omega M these terms are similar and sine Omega n t minus TD so that is to locate this position and the second response is valid only when T is greater than TD so if T is less than TD this force is not acting so the effect of this will come only when T is greater than TD so now let's find the response due to both the impulses as we have seen here when T is less than TD only therefore due to this impulses there so when time is less than TD the response will be equal to x1 and when time is greater than TD the total response will be x1 plus x2 so we can add this I know mega empty then we can add this X 2 that is minus sine Omega n t minus T data so this is the total response when T is greater than TD so we can simplify this equation and we can get this expression we can reduce this to this expression now we can plot this responses the individual responses and the total responses for various values of T D by T n so first this result is for TD by TN is equal to 1 by 8 so the first impulse response is this so the amplitude of this first impulse response is X 1 by I by M Omega n is equal to 1 that means the amplitude of X 1 is equal to I by M Omega n so that is equal to the static response and the second impulse response is zero till t is equal to TD because the impulse starts only when TD is equal to time is equal to TD so til TD the responses zero so then after that we have a harmonic vibration with amplitude equal to I by M Omega n so this ratio is 1 so this is also equal to the static amplitude so if we add these two we will get this response so when these two get added up the amplitude reduces so that this amplitude is only point 8 this ratio so that means the amplitude of this total response this less than the this equivalent static response that is I by him Omega n and this is when this ratio that is TD by DN is equal to 1 by 4 so in this case the first impulse response a similar to the previous one the second one the second impulse response is little bit more delayed compared to the first one because here TD is more so this 0 portion is little longer than the previous one so when we add these two the amplitude is now close to 1.5 both the first and the second impulse responses they're having amplitude 1 but when we add together because of this phase difference the total amplitude will be either less or more than the individual Android amplitudes so that will depend upon the phase difference so when T D by TN is equal to half this phase difference between these two responses this still more so this response is same as the previous one and this one is a little bit more delayed response so if we can add these together we will get this and in this case both the amplitudes and up and we get a combined amplitude is equal to 2 that means the response of this function XT the maximum response of this function XT is equal to twice I by M Omega so let's see this for one more ratio so when TD by TN is 1 the delay between these two responses is equal to TN so once I show this this second is maple starts after one cycle of this first impulse so if we add these together for T greater than TD these responses the second impulse and the first impulse get cancelled out so then we get zero oscillations the responses 0 beyond TD so when time is less than TD then the total response is equal to the first impulse response but after TD the response is zero because the response is due to both these cancel out each other now let's move on to the second part of the problem and where we have to find out the maximum response so first let's find out the some response during time is less than TD so when time is less than TD only this first impulse response this existing so as you can see here this oscillate s-- with natural period TN and as you can see here it starts with zero displacement so the peak the first peak appears when T is equal to TN by 4 that is 1/4 of the natural period so in TL the cycle completes so the first weeks appears at TN by 4 so the number of peaks in during this time depends upon the value of this TD by TN ratio if TD is much less then we won't be getting this peak so if TD is less than this TN by 4 then we will not be getting this peak but if the value of TD is more than TN by 4 then we will be having this week so depending upon the value of this TD by TN we can find out the maximum response so the first peak occurs at TN by 4 so at the peak the response the maximum responses like this X naught by I by M Omega M is equal to 1 so that means X naught is equal to I by M Omega so this is the maximum response when time is less than T D and this D D is longer than this TN by 4 otherwise the speed won't happen so thus TD must be longer than TN by 4 for at least one peak to develop during this duration so if TD is shorter than TN by 4 there won't be any peak Deb I think in this time zone but the response will increase from 0 to utd and the value of that maximum displacement can be calculated using the first impulse response and we can calculate this as this so the maximum deformation during time between 0 and TD is X naught by I by M Omega M is equal to sine 2pi t d by d n if TD by TN s less than or equal to 1 by 4 and this ratio is equal to 1 if this ratio is more than 1 by 4 so now we need to find the maximum response when T is higher than TD so for that we can add the combined response due to the first impulse and the second impulse and maximize that expression and then find the maximum value of this response so if you maximize the expression we can find the Maxima at X naught is equal to this so that is X naught by I by M Omega n is equal to 2 modulus of sine PI TD by TM so this X T when T greater than TD has a maximum value of I by M Omega n multiplied by this so that is the maximum value when T is muda TD so let's plot these so the maximum response during T less than TD is given by these two equations this plotted here so if T D by T n is less than 1 by 4 the response increases from 0 to 1 and when it is more than one by four the amplitude is 1 and maximum response when T is greater than TD this follows this curve and that will be according to this expression so now we can find the maximum response for all the values of T that is when T is less than TD and when this 20 is greater than D D so if we combine these two plots we get the overall maximum response and that would be this so this is like these two plots plotted together so the maximum overall response will be the envelope of these two curves so that is the maximum of these two values so if we plot the envelope of these two curves we would get this so the overall maximum response of this single degree of freedom system this undamped single degree of freedom system will change like this for different values of T D by T n this is a deformation response factor and this is also known as response spectra so that means this shows the maximum response of this system for all the values of T D by T so this gives the response spectra so in the next example the properties of an elevated water tank is given the elevated water tank shown below B's 100 point 0 3 Kip's when it is full of water the tower has a lateral stiffness of 8.2 Kip's per inch treating the water travel as a single degree of freedom system estimate the maximum lateral displacement due to each of the two dynamic forces shown without any exact dynamic analysis instead use your understanding of how the maximum respond depends on the ratio of the rise time of the applied force to the natural vibration period of the system neglect damping so we can treat this elevated water tank as an undamped single degree of freedom system the mass is given the stiffness is also given we need to find out the maximum lateral displacement so two forcing functions are given so we don't have to do any exact solution but depending upon the rise time we need to find an approximate solution based on our understanding of how the maximum response depends on the ratio of the rise time to the natural period so let's find this out so the system properties are given mass is given the weight is given so the mass can be calculated as weight by gravitational acceleration so we can calculate the mass the stiffness is given so we can find the natural frequency and the natural period so we can calculate the natural period here as 2 pi square root of M by K so we get the natural period is 1 point 1 2 seconds now looking at this different forcing function we have two cases in both the cases the maximum amplitude of the force is same that is 50 Kip's but the rise time changes so in the first one the rise time is 2 less that is point two and in the second one the rise time is very large that is 4 and the natural period is only one point 1 2 seconds so to solve this we can find out the ratio of the rise time and the natural period so in the first case it is point 1 7 9 so that means compared to the natural period this rice timers very low so since the rise time of the force is relatively short and the structure will see this excitation as a suddenly applied force so since this nice time is too short so this force will be like a suddenly applied force so we learned early that when a suddenly applied step force is acting on a system the maximum amplitude is approximately equal to twice that of the static displacement so we know the static response as P naught by K so the maximum response will be twice that so we can calculate we know P naught we know the value of K so we can calculate the maximum value of the response as two times P naught by K that would be equal to twelve point two changes so this is not an exact solution but this is a approximate solution so for the second case the ratio of the rise time and the natural period is four by one point one two that is three point five seven this is very large the rise time of the force is relatively low and it will affect the structure like a static force so since the rise time is more the dynamic effect due to the application of the load is less so this is more or less like a static force so the maximum response will be approximately equal to the static response that is P naught by K so that would be equal to six point one so we can calculate the approximate value of the maximum response using this relationship [Music] [Music] [Music]