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<title>An introduction to vibration</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


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The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

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<item n="speechevent">Lecture</item>

<item n="acaddept">Engineering</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG</item>

<item n="module">unknown</item>





<u who="nm0855"> <kinesic desc="overhead projector is on showing transparency" iterated="n"/> at the end of the last lecture <pause dur="1.1"/> i'd really set out for you the the basic <pause dur="0.2"/> ideas of multimode vibration and promised to revise that with <pause dur="0.2"/> an example <pause dur="0.4"/> # <pause dur="0.4"/> <trunc>a</trunc> and that's what this is <pause dur="1.9"/> this is <pause dur="0.2"/> in the <pause dur="0.2"/> handout that you had at the start of this section of the course it's one of the the two <pause dur="0.2"/> worked example sheets in there <pause dur="0.7"/> # <pause dur="0.3"/> just so that you get all the detail right this is a case where we do have subscripts that are easy to get wrong so so let's see what's going on <pause dur="0.5"/> so there shouldn't be any need to do more than just have a quick look through this <pause dur="1.8"/> and it it's more or less the same as we set up in the simple example last week <pause dur="0.3"/> except that there are now three masses <pause dur="1.9"/> four springs <pause dur="0.2"/> two rigid walls <pause dur="1.5"/> # <pause dur="1.0"/> in the notes that you have there are <trunc>r</trunc> <pause dur="0.2"/> rather more words describing what goes on between this so <kinesic desc="indicates point on transparency" iterated="n"/> this is just an edited down version of the sheet that you have <pause dur="1.9"/> but note the basic principle as before <pause dur="0.4"/> is that for each mass in turn M-one M-two <pause dur="0.2"/>

et cetera <pause dur="0.6"/> we set up <pause dur="0.4"/> the Newton's equation of motion <pause dur="0.9"/> we have <pause dur="0.2"/> force <pause dur="1.6"/> generated by springs so we have terms like K-two A-two-minus-A-one <pause dur="0.3"/> K-one <pause dur="0.3"/> A-one-minus-zero <pause dur="0.2"/> in this case <pause dur="0.5"/> and that's going to equal an acceleration <pause dur="0.5"/> because it's a sinusoidal motion <pause dur="0.3"/> the acceleration becomes just omega-squared <pause dur="0.5"/> times the <pause dur="0.4"/> amplitude so we get <pause dur="1.1"/> mass times acceleration as a force M-one A-one omega-N-squared equals <pause dur="0.8"/> spring constant times the deflection <pause dur="0.3"/> spring constant times the deflection <pause dur="0.4"/> as we go along <pause dur="0.6"/> the the chain of events <pause dur="0.5"/> so that's how we set it up basically and we do that with a separate equation <pause dur="0.5"/> for each of the masses M-one <pause dur="0.4"/> M-two <pause dur="0.3"/> and <pause dur="0.3"/> M-three just reduced to et cetera <pause dur="0.7"/><vocal desc="cough" iterated="n"/> # excuse me <pause dur="0.2"/> on this slide <pause dur="2.5"/> the only thing to point out about those two again just to reinforce what i said last time <pause dur="1.0"/> whether you <pause dur="0.4"/> do it just the same as this example doesn't matter but do be systematic about what you do <pause dur="0.4"/> and something that i think works well is to start always with the highest <pause dur="0.7"/> subscripts <pause dur="0.5"/>

to the left-hand side as you go through so i recommend that <pause dur="0.4"/> given we're looking at M-one and <unclear>we'd</unclear> interested in the spring either side <pause dur="0.3"/> start with K-two <pause dur="0.3"/> write down its extension as A-two-minus-A-one <pause dur="0.6"/> then go to K-one <pause dur="0.2"/> and have A-one-minus <pause dur="0.2"/> and here <pause dur="0.4"/> it's a wall i've written in the zero <pause dur="0.3"/> formally just to stress that that's a wall that doesn't move at that end <pause dur="0.3"/> # but obviously you could just write K-one A-one <pause dur="0.2"/> if you wish <pause dur="0.4"/> but if you do it like that and notice the same thing <kinesic desc="indicates point on transparency" iterated="n"/> three is there then it goes to twos then it goes down to one <pause dur="0.5"/> if you systematize it like that <pause dur="0.2"/> then the signs come out in that nice pattern of <pause dur="0.4"/> brackets with a negative number inside <pause dur="0.2"/> minus the other one with brackets with the negative number inside <pause dur="0.3"/> and you're less likely just to make the little slips <pause dur="0.4"/> that that can upset these sort of procedures <pause dur="1.0"/> so <pause dur="0.8"/> that's each equation <pause dur="0.4"/> and normally <pause dur="0.4"/> we would <pause dur="0.2"/> for larger problems then put this into <pause dur="1.5"/> a matrix notation omega-N-squared is a constant of this so we can just

have <pause dur="0.6"/> some set of A <pause dur="0.2"/> amplitude vector A <pause dur="0.9"/> is some sort of stiffness matrix <kinesic desc="indicates point on transparency" iterated="n"/> here <pause dur="0.4"/> multiplied by the same <pause dur="0.3"/> vector <pause dur="0.2"/> A <pause dur="1.2"/> which is just done by <kinesic desc="indicates point on transparency" iterated="n"/> rearranging here this is with the Ks outside the brackets but just turn that round so you get the As <pause dur="0.2"/> outside and the Ks inside <pause dur="0.5"/> and you can get this form <pause dur="0.5"/> but notice that it's convenient <pause dur="1.1"/> to divide through in each row by M-one by M-two and so on so that appears <pause dur="0.5"/> in the denominator of each term inside there <pause dur="0.7"/> now that's <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/> very typically of how <pause dur="1.1"/> books present it <pause dur="0.5"/> you might argue that you're used to seeing K <pause dur="0.2"/> as the stiffness <pause dur="0.3"/> so using K for this stiffness divided by <trunc>math</trunc> mass <pause dur="0.3"/> might not be such a bright idea but a lot of the books do that so i've stuck with that but just notice <pause dur="0.3"/> it is a sort of stiffness matrix <pause dur="0.4"/> but it's modified by those masses <pause dur="1.3"/> if you get it to that form <pause dur="0.4"/> then by far the best approach for larger problems <pause dur="0.3"/> is then to give it to a solver <pause dur="0.3"/> and ask for the eigenvalues <pause dur="0.6"/> which correspond to the <pause dur="0.3"/> natural frequencies <pause dur="0.2"/>

squared the omega-N-squareds <pause dur="0.6"/> and <pause dur="0.3"/> the eigenvectors which give you the mode shapes <pause dur="0.9"/> if you don't fancy doing that or it's a problem you're doing by hand <pause dur="0.4"/> then <pause dur="0.4"/> you're probably better off just solving this <kinesic desc="indicates point on transparency" iterated="n"/> as a set of simultaneous equations <pause dur="0.3"/> at that level rather than worrying about the matrix formality <pause dur="0.4"/> the matrices do help on big problems they're not essential <pause dur="1.0"/> so <pause dur="0.4"/> if you do it <pause dur="0.4"/> in normal ways you solve for omega-N <pause dur="0.2"/> and the ratio <pause dur="0.4"/> of the amplitudes <pause dur="0.3"/> to <pause dur="0.8"/> say the amplitude A-one remember that we always have <pause dur="0.3"/> one too many variables we have the natural frequency <pause dur="0.4"/> plus <pause dur="0.2"/> N <pause dur="0.2"/> amplitudes <pause dur="1.5"/> so that's how you solve for that and it's not worth working the details out on it <pause dur="1.1"/> if you use the eigenvalue eigenvector <pause dur="0.2"/> formally through something like Matlab or a similar program <pause dur="0.5"/> you get the answers directly <pause dur="0.9"/> you can of course <pause dur="0.7"/> solve the mode shapes directly as well from the simultaneous equations <pause dur="0.3"/> but once you have the resonant frequencies the omega-Ns <pause dur="0.3"/> you can in fact guess what's going on <pause dur="0.5"/> and

<kinesic desc="indicates point on transparency" iterated="n"/> that's what i've done with <trunc>th</trunc> with the arrows i've drawn under the diagram there <pause dur="0.6"/> they correspond to the three possible motions <pause dur="0.3"/> all three could be going the same way in phase <pause dur="0.6"/> # with the same amplitude <pause dur="0.5"/> oh # sorry with different amplitudes but in the same <pause dur="0.3"/> # phase relationship <pause dur="0.5"/> or we could have the two <pause dur="1.4"/> outside ones going the same way <pause dur="0.2"/> and the other one coming the other way <pause dur="0.5"/> # <pause dur="0.2"/> which is the bottom one or in between those where <pause dur="0.2"/> two of them are moving in one way and one the other <pause dur="0.6"/> # and if you think about it they're the only three truly independent motions you get the middle one swaps round the other two are moving <pause dur="0.4"/> together <pause dur="0.7"/> # <pause dur="0.3"/> now i i've written <pause dur="0.6"/> from this the top downwards in the way they are because that corresponds <pause dur="0.3"/> to the number of changes of direction if you look at that the top one has no change of direction <pause dur="0.4"/> the next one down has one change <pause dur="0.4"/> the one below that <pause dur="0.4"/> has two changes of direction <pause dur="0.9"/> because the change of direction corresponds to there being a node <pause dur="0.4"/> between <pause dur="0.7"/>

those two displacements <pause dur="0.5"/> and as again we discussed last week <pause dur="0.3"/> we know that <trunc>ev</trunc> <pause dur="0.3"/> the more nodes you have <pause dur="1.3"/> <unclear>the nodes</unclear> the higher mode shape that you have <pause dur="0.3"/> corresponds to increasingly high resonant frequencies that's the energy link between them <pause dur="0.4"/> so once you've got the omega-Ns <pause dur="0.4"/> just by inspection you can work out <pause dur="0.2"/> the general shape of the modes <pause dur="0.4"/> but you need to <trunc>so</trunc> and which <pause dur="0.2"/> goes with which omega <pause dur="0.3"/> but you then need to solve formally to get the actual ratio <pause dur="0.6"/> of these amplitudes the sketching can't tell you that <pause dur="1.8"/> so that's the basic example that's really all there is to <pause dur="0.2"/> most of # <pause dur="0.8"/> # <pause dur="0.3"/> axial vibration and again let just <kinesic desc="turns on overhead projector showing transparency" iterated="n"/> put flash that slide up no <pause dur="0.2"/> nothing to copy down there it's just a quick <pause dur="0.5"/> look at that just <trunc>s</trunc> sketched out just before i came down today <pause dur="0.3"/> and notice that if it were a torsional problem <pause dur="0.2"/> it's exactly the same we'd have shafts of some stiffness <pause dur="0.4"/> there'd be moments of inertia for the

rotors <pause dur="0.2"/> and a theta <pause dur="0.3"/> displacement for each one you just plug in the <trunc>form</trunc> the symbols into the <trunc>w</trunc> <pause dur="0.2"/> example over there <pause dur="0.9"/> if you have say a building frame a steel framed building with sort of uprights <pause dur="0.3"/> and fairly some concrete floors something mairly fairly massive there <pause dur="0.6"/> then you may have a sway system by which you know experimentally <pause dur="0.5"/> the values of these stiffnesses of these different <pause dur="0.2"/> storeys <pause dur="0.3"/> and you've just got effectively the slab of mass <pause dur="0.2"/> and again <pause dur="0.3"/> that is the same problem as that <kinesic desc="indicates transparency" iterated="n"/> this just corresponds to the spring experimentally <pause dur="0.2"/> this is the mass <pause dur="0.4"/> and so on and so forth <pause dur="0.4"/> so all these problems reduce <pause dur="0.2"/> whatever variables they're in <pause dur="0.5"/> to <pause dur="0.7"/> that piece of sort of handle turning sort of <pause dur="0.4"/> # mathematical analysis </u><gap reason="break in recording" extent="uncertain"/> <u who="nm0855" trans="pause"> <trunc>th</trunc> what i'd like to do on though <trunc>i</trunc> is not sort of work more complicated examples <pause dur="0.3"/> # <pause dur="0.6"/> but move on to <pause dur="0.3"/> the slightly different conditions <pause dur="0.7"/> # <pause dur="0.3"/> hinted at by that <kinesic desc="puts on transparency" iterated="n"/> little frame i just showed on the other <trunc>s</trunc> <pause dur="0.2"/>

projector <pause dur="0.8"/> # <pause dur="0.8"/> if we know experimentally what the effective lateral stiffness of the <pause dur="0.5"/> building is <pause dur="0.2"/> fine we can use that simple method <pause dur="1.2"/> if we don't and we have to work with beams transverse vibration <kinesic desc="demonstrates motion with arms" iterated="y" dur="3"/> so we have a beam that's vibrating in some <pause dur="0.2"/> motion of that sort <pause dur="0.8"/> then there are one or two tricks which are just a little bit different <pause dur="0.2"/> and you'll find that nearly all the textbooks do treat them <pause dur="0.2"/> as a separate case <pause dur="0.3"/> as i'm doing in <trunc>th</trunc> # # in these lectures <pause dur="1.8"/> so let's just very briefly <pause dur="0.3"/> look at beam vibration and it's not a thing i want to spend a lot of detail <pause dur="0.4"/> the the theory here <pause dur="0.4"/> again <pause dur="1.0"/> is mathematically <pause dur="0.4"/> tricky <pause dur="0.2"/> just thumping out the solutions from the basic information # # # is <pause dur="0.8"/> is a slog it's easy to make mistakes <pause dur="0.3"/> so all i'm really concerned with is that we get pick up on the the physical principles that are going on behind that <pause dur="0.5"/> that will <pause dur="0.3"/> get you to the level by which you can then look up solutions with a degree of confidence <pause dur="0.4"/> # when you need them <pause dur="0.5"/> so <pause dur="0.3"/> we'll deal with beams

and just to remind you what we mean by a beam <pause dur="0.3"/> it's something which is long and thin <pause dur="0.4"/><kinesic desc="demonstrates beam with arms" iterated="n"/> # so we've we've got it between supports there's a structure across here <pause dur="0.6"/> # whatever the depth of that beam is it's small compared to its <pause dur="0.5"/> its length <pause dur="0.4"/> and when we do that we can ignore <pause dur="0.4"/> generally speaking sheer <pause dur="0.6"/> effects and concentrate <pause dur="0.4"/> only <pause dur="0.3"/> on <pause dur="0.2"/> beam bending and then we can also use classical <pause dur="0.4"/> beam bending theory that you met last year <pause dur="0.4"/> and have met <pause dur="0.4"/> earlier in this course in the bit that <gap reason="name" extent="1 word"/> <pause dur="0.2"/> was dealing with <pause dur="1.0"/><vocal desc="clears throat" iterated="n"/><pause dur="1.4"/> okay well there's different things we might want to do let's start with the very simplest case <pause dur="0.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> oops <pause dur="2.0"/> i'm just drawing it <pause dur="0.3"/> out like that <pause dur="0.7"/> we have <pause dur="0.2"/> a light beam <pause dur="1.0"/> # always the easier case on perfect simple supports as shown there but it doesn't <pause dur="0.2"/> the end conditions we'll come back to <pause dur="0.5"/> a light beam in other words we can neglect its mass <pause dur="0.2"/> with a single point lumped mass <pause dur="0.4"/> stuck somewhere <pause dur="0.2"/> we're not i've drawn it near the middle because that makes it easy but there's nothing in

what i'm saying <pause dur="0.3"/> that means it has to be <pause dur="0.4"/> in the centre it can be anywhere along the beam <pause dur="1.8"/> and this might correspond to the case where we have a structural <pause dur="0.6"/> I-beam or something like that <pause dur="0.4"/> # <pause dur="0.7"/> which has got <pause dur="0.6"/> something maybe ten times the weight of that mounted near the centre of it perhaps a machine <pause dur="0.3"/> that could be vibrating and may cause the beam to vibrate <pause dur="1.8"/> so that's the scenario we set up <pause dur="0.3"/> and <pause dur="1.4"/> generally speaking when when you can set up an a <pause dur="0.9"/> a coordinate frame <pause dur="0.3"/> and if i just replace this mass <pause dur="0.3"/> by a force and just had <pause dur="0.2"/> said there is a point force pushing downwards <pause dur="0.3"/> on there <pause dur="0.3"/> what is the shape of the beam how does the beam deflect <pause dur="0.5"/> then <pause dur="0.4"/> as far as i'm concerned <pause dur="0.6"/> # that is no problem for you to solve you may worry about it but <pause dur="0.6"/> at this stage the assumption of this course <pause dur="0.4"/> is that you're <pause dur="0.3"/> comfortable solving problems like that <pause dur="0.5"/> so <pause dur="0.6"/> we put a force on there <pause dur="0.9"/> # how would you do it well there's two things you could do <pause dur="0.4"/> # <pause dur="0.2"/> let's deal with the formal one first <pause dur="0.3"/> that is you could

calculate the bending moments along that beam you'd sort of solve for the reaction forces calculate the bending moment <pause dur="0.4"/> and then we know that the deflection shape <pause dur="0.9"/> or the second derivative of the of the # <pause dur="0.3"/> deflection shape <pause dur="0.4"/> relates functionally <pause dur="0.2"/> directly to that bending moment <pause dur="1.5"/> and we get the shape out <pause dur="0.7"/> okay <pause dur="0.2"/> what do we do if it's vibrating <pause dur="0.4"/> well notice what we've done with everything else i've said i've <pause dur="0.6"/> got a force from the springs <pause dur="2.3"/> i'm now going to generate a force <pause dur="0.5"/> which is sort of mass times acceleration because i know the mass is bouncing around in that sense in other words i've got what i could call <pause dur="0.2"/> an inertia force it's the force M-A <pause dur="0.2"/> associated with the acceleration <pause dur="0.6"/> so all i need to do is substitute that inertia force in terms of the fixed F <pause dur="0.5"/> that you would have had on a a an earlier example <pause dur="0.3"/> <trunc>a</trunc> and we're away we do the same thing <pause dur="0.4"/> and i'm not going to actually work through an example here i'm just going to give you two lines of text <shift feature="voice" new="laugh"/>if <shift feature="voice" new="normal"/>you like <pause dur="0.4"/> that that <pause dur="0.2"/>

tells you <pause dur="0.4"/> that thing <pause dur="0.9"/> the inertia force then if we're moving transverse vibrations i # <trunc>tha</trunc> <trunc>tha</trunc> <pause dur="0.2"/><kinesic desc="demonstrates motion with arms" iterated="y" dur="3"/> again the mass is i say where my hands are joined it is going through like that <pause dur="0.8"/> what we have <pause dur="1.1"/> is <pause dur="0.7"/> an inertia force mass of the lump thing Y <pause dur="0.7"/> is its deflection <pause dur="0.8"/> # so Y-double-dot would be its acceleration so we've sort of got Y-<pause dur="0.5"/>double-dot or Y <trunc>D</trunc> <pause dur="0.2"/> # sorry <pause dur="0.2"/> M-double-dot <pause dur="0.2"/> Y-double-dot or M <pause dur="0.4"/> D-two-Y <trunc>D-ec</trunc> <pause dur="0.3"/> T-squared if you prefer to call it that <pause dur="0.6"/> # we have <pause dur="1.2"/> that form <pause dur="0.3"/> again we know that Y is going to be a sinusoidal motion that's the assumption of all our vibration systems <pause dur="0.3"/> so the Y-double-dot term can be replaced by omega-squared-Y <pause dur="0.3"/> again just plugging it in the derivative <pause dur="0.7"/> # approach we discussed a couple of lectures back <pause dur="0.6"/> so the inertia force then <pause dur="0.4"/> has <pause dur="0.6"/> that as a a peak value <pause dur="0.3"/> # it it's oscillating sinusoidally but that is <pause dur="0.2"/> perfectly adequate <pause dur="0.3"/> amount of data <pause dur="1.0"/> the bending moment as always is just this formula you should be very familiar with <pause dur="0.4"/> bending moment it's is equivalent to <pause dur="0.3"/> E-I <pause dur="0.9"/> D-two-Y D-X-squared <pause dur="0.2"/>

in other words the the second spatial derivative <pause dur="0.3"/> the curvature of the beam <pause dur="0.2"/> locally <pause dur="0.6"/> where E is Young's modulus I is the second moment of area of the cross section perfectly normal <pause dur="0.3"/> standard beam theory <pause dur="1.5"/> and <pause dur="0.2"/> all you do is <kinesic desc="indicates point on transparency" iterated="n"/> equate that <pause dur="0.2"/> to <kinesic desc="indicates point on transparency" iterated="n"/> that <pause dur="0.6"/> # you then need to integrate twice with respect to Y <pause dur="0.5"/> and you can solve for <pause dur="0.7"/> # omega-N-squared and and for Y from from that <pause dur="0.4"/> formula <pause dur="0.9"/> so that's formally how you go about doing the problem and <pause dur="0.6"/> it's <pause dur="0.3"/> it's not any more difficult that's all the theory all the theory you need is to remember that <kinesic desc="indicates point on transparency" iterated="n"/> there are these things called inertia forces <pause dur="0.3"/> and that you can plug them in to the beam bending formula <pause dur="0.5"/> that you've been using for a long while <pause dur="0.9"/> however <pause dur="0.2"/> it it is often <pause dur="0.2"/> quite sort of a tricky sort of thing to do <pause dur="0.5"/> and one of the big advantages we have <pause dur="0.5"/> is that a lot of the beam <pause dur="0.9"/> structural beam bending formulae and so on <pause dur="0.3"/> are tabulated for us <pause dur="0.3"/> so let can we find a short cut trick that will help <kinesic desc="reveals covered part of transparency" iterated="n"/> and i think that's something that's <pause dur="0.4"/> of more immediate

interest <pause dur="3.0"/><vocal desc="cough" iterated="n"/><pause dur="0.2"/> excuse me again <vocal desc="clears throat" iterated="n"/><pause dur="2.7"/> so <pause dur="0.3"/> an easier way for this problem <pause dur="2.0"/> always we are dealing with conditions which are linearly elastic and we're # let me <trunc>s</trunc> we said this <pause dur="0.3"/> at the start of this section of the course but let's repeat it again <pause dur="0.3"/> once we lose linear elasticity none of the stuff that i'm talking about is going to work properly <pause dur="0.3"/> so <pause dur="0.2"/> that that <pause dur="0.4"/> is just available to us as a <pause dur="0.3"/> a a prerequisite <pause dur="0.8"/> if it's linear elastic <pause dur="0.2"/> the beam <pause dur="0.2"/> just acts as a simple spring <kinesic desc="demonstrates motion with arms" iterated="y" dur="6"/> i've got this thing here if i push it downwards <pause dur="0.5"/> against its own elastic <pause dur="0.5"/> bending properties it'll try and push me back up again so it just acts as a spring <pause dur="1.8"/> and if i think of the spring as as <trunc>l</trunc> some value lambda newtons per metre in this case i'm <trunc>ju</trunc> i'm just pressing down <kinesic desc="indicates point on transparency" iterated="n"/> here and saying <pause dur="0.4"/> how much force do i need to deflect a certain amount <pause dur="0.2"/> i could actually <pause dur="0.2"/> calculate that from first principles <pause dur="0.3"/> using the sort of formulae <pause dur="0.9"/> little bit higher up on that

page <pause dur="0.3"/> or maybe i can do that experimentally i can just <trunc>s</trunc> put <pause dur="0.3"/> balance a weight on there and see how much it deflects <pause dur="0.4"/> to get an equivalent stiffness <pause dur="1.2"/> now we know that this is only a single mass and and a <unclear>sort of</unclear> spring system here because the the beam is light <pause dur="0.3"/> so we know from <pause dur="0.2"/> the elementary theory <pause dur="0.3"/> that the natural frequency will be just square root <pause dur="0.3"/> of <pause dur="0.5"/> stiffness over mass <pause dur="0.2"/> that is here lambda-over-M it was square root of K-over-M <pause dur="0.3"/> for the or or helical spring <pause dur="0.3"/> example we looked at earlier <pause dur="2.0"/> so all we need to do now is <pause dur="0.4"/> see whether we know what # lambda <pause dur="0.3"/> is <pause dur="0.5"/> we know what the mass is i assume # # in all of these problems <pause dur="0.8"/> # <pause dur="0.2"/> well <pause dur="0.2"/> # we could be <trunc>ex</trunc> experimental <pause dur="0.4"/><kinesic desc="changes transparency" iterated="y" dur="5"/> but we can actually get at it <pause dur="1.0"/> by some other methods so if i can just give you that bit as an extension to the bottom of that sheet <pause dur="2.8"/> what happens if we do just <pause dur="0.2"/> as i said experimentally <pause dur="0.2"/> sort of balance a mass on the beam and see how much it deflects <pause dur="0.3"/> well that gives us lambda <pause dur="0.2"/> and it gives it as from

that thing the static deflection <pause dur="1.0"/> is the force M-G now because we're looking at how much it would deflect <pause dur="0.7"/> # as a horizontal beam <pause dur="0.6"/> divided by this same stiffness it's the same thing that causes both effects <pause dur="0.6"/> so experimentally that's how i get at lambda <pause dur="0.4"/> but i can also get at it because i can i can look up <pause dur="0.6"/><kinesic desc="indicates point on transparency" iterated="n"/> that static deflection <pause dur="0.4"/> for <pause dur="1.0"/> a weight <pause dur="0.7"/> from the sort of tables that you get in the data book we know for a simply supported beam there is so much deflection at the centre and so on and so forth <pause dur="0.8"/> # <pause dur="0.2"/> so what we do actually <pause dur="0.2"/> if i can now take that one off <pause dur="2.6"/><event desc="takes off transparency" iterated="n"/> is <pause dur="1.2"/> okay let # let me just put it back <kinesic desc="puts on transparency" iterated="n"/> a second in fact <pause dur="0.3"/> # <pause dur="0.6"/> if we just look at this formula here <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> and just plug in <pause dur="0.6"/> to get rid of lambda there which is the sort of unknown and we don't really want to know explicitly <pause dur="0.5"/> if i just substitute from <kinesic desc="indicates point on transparency" iterated="n"/> this formula in there <pause dur="0.2"/> then i finish up with this rather magic looking formula <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/>

that <pause dur="1.6"/> the natural frequency is the square root of G <pause dur="0.5"/> upon <pause dur="0.3"/> the static deflection <pause dur="2.4"/> okay <pause dur="0.4"/> simple enough formula but it worries quite a lot of people and with some reason <pause dur="1.7"/> all the while i've been saying to you with this stuff that the the vibration <pause dur="0.3"/> does not depend at all <pause dur="0.7"/> on the <pause dur="1.3"/> atmosphere we're working in the gravitational constants the vibration is just the same on the moon as it is on the earth <pause dur="0.4"/> other things may change but that stays the same <pause dur="0.6"/> so <pause dur="0.5"/> why does G appear <kinesic desc="indicates point on transparency" iterated="n"/> in this formula <pause dur="1.4"/> # and the answer is very straightforwardly <pause dur="0.2"/> that it's <pause dur="0.2"/> because <pause dur="0.2"/> using it in this form is what we can use with the tables so don't worry about it is it's just an artificial step <pause dur="0.5"/> # in going through it because if we now look at a simple example <pause dur="2.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so just a very simple example of simply supported beam which is what <pause dur="0.2"/> i'd actually drawn on the previous sheet <vocal desc="cough" iterated="n"/><pause dur="2.3"/> we'll put the mass in the the middle because that's the one

that's easy to look up all the tables in the books have the the central <pause dur="0.9"/> force <pause dur="0.5"/> type condition <pause dur="0.5"/> and you will find that if you have a simply supported beam <pause dur="0.3"/> with a force <pause dur="0.8"/> W <pause dur="0.3"/> a weight W in the centre of it and the deflection at under that weight is W-L-cubed <pause dur="0.5"/> over forty-eight-E-I <pause dur="2.1"/> you can calculate it by bending moments but just look it up it's there in the data book it's there in all the textbooks <pause dur="1.0"/> now here the weight is simply just M-G <pause dur="0.5"/> and i think you can see what's happening now is that although i've <pause dur="0.2"/> i've lost apparently the mass and have <kinesic desc="indicates point on transparency" iterated="n"/> gained a G in there that i didn't like <pause dur="0.4"/> whenever we look up in the tables <pause dur="0.7"/> what the deflection's going to be <pause dur="0.3"/> ha ha the <trunc>ma</trunc> the <kinesic desc="indicates point on transparency" iterated="n"/> M and G come there that gives me the mass back <pause dur="0.5"/> and notice the G will always cancel out <pause dur="0.4"/> so that trick of the G in the formula <pause dur="0.3"/> above <pause dur="0.3"/> is purely because that's the way you will read it from the tables so <pause dur="0.2"/> so again the textbooks tend to quote the formula to you <pause dur="0.7"/> in that form <pause dur="0.8"/> but so we

just plug that <kinesic desc="indicates point on transparency" iterated="n"/> Y-S into <kinesic desc="indicates point on transparency" iterated="n"/> that formula <pause dur="0.5"/> # it's easy to call it omega-N-squared and lose the square root sign <pause dur="0.4"/> # and it just comes out <pause dur="0.3"/> of that form <pause dur="0.4"/> so for a simple <pause dur="0.2"/> transverse beam <pause dur="0.2"/> we can relate it to the normal beam bending <pause dur="0.4"/> criteria the the <pause dur="0.3"/> second moment of area <pause dur="0.4"/> <trunc>a</trunc> and so on and so forth <pause dur="0.4"/> very straightforwardly <pause dur="2.3"/> okay that's fine if we have just one mass <pause dur="0.3"/> let's <pause dur="0.3"/> consider <pause dur="0.4"/> still <pause dur="1.2"/> a light beam <pause dur="0.2"/> so we've still got no no significant mass in the beam itself <pause dur="0.4"/> but i'm now going to put more than one mass along it so i've got <kinesic desc="demonstrates motion with arms" iterated="y" dur="10"/> the beam along here <pause dur="0.3"/> lump of mass there lump of mass here <pause dur="0.4"/> and what can happen now is well they can both go sort of up and down together somehow or we could i suppose now have a motion <pause dur="0.7"/> of them oscillating like that <pause dur="0.5"/> i've got two masses <pause dur="0.2"/> two degrees of freedom <pause dur="0.3"/> two different modes of vibration so that all fits into the same sort of spring mass pattern that we're seeing before <pause dur="0.7"/> # but the question is how do we go about solving it <pause dur="1.2"/> well again formally <pause dur="0.2"/> and i'm not

going to work an example <pause dur="1.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> at this stage <pause dur="0.3"/> but formally if we had two or more masses <pause dur="0.6"/> we'd do something like that <pause dur="3.6"/> there are two points along the beam <pause dur="0.4"/> we <pause dur="0.6"/> obviously each has a mass M-one <pause dur="0.3"/> we allocate to those a deflection <pause dur="0.3"/> at the mass of Y-one Y-two <pause dur="0.2"/> et cetera <pause dur="0.8"/> # <pause dur="1.1"/> and again i think of it as a force <pause dur="1.3"/> a point force at each of those mass points <pause dur="0.2"/> and i can again do just what i did before <pause dur="0.4"/> we can work out the bending moments in this beam in terms of a static force set F-one F-two et cetera <pause dur="0.7"/> now because there are point forces in there <pause dur="0.2"/> you may want to <trunc>u</trunc> and more than one of them <pause dur="0.2"/> you will need to use one of the special notations <pause dur="0.3"/> what i've called here a singularity function <pause dur="0.3"/> you might know it as Macauley's notation it depends which <pause dur="0.5"/> textbook you've gone to but again from other courses <pause dur="0.3"/> you should be quite familiar with the idea <pause dur="0.4"/> of handling these extra forces coming in <pause dur="0.2"/> as we take the bending moment <pause dur="0.3"/> along the beam <pause dur="0.8"/> so we can establish a bending

moment <pause dur="0.2"/> through there <pause dur="0.8"/><kinesic desc="indicates point on transparency" iterated="n"/> we can <pause dur="0.8"/> know that each of the forces F-one is M-one Y-one omega-N-squared <pause dur="0.2"/> F-two of course just <pause dur="0.2"/> F <pause dur="0.2"/> will be <trunc>w</trunc> M-two <pause dur="0.2"/> Y-two <pause dur="0.2"/> omega-N-squared and so on <pause dur="0.2"/> so we just have a series of forces <pause dur="0.3"/> it's a more complicated version of the previous problem <pause dur="0.6"/> but what we actually do is we will set up <pause dur="0.6"/> # <pause dur="2.0"/><kinesic desc="indicates point on transparency" iterated="n"/> this F-one <pause dur="0.3"/> equals <pause dur="0.2"/> the deflection <pause dur="0.6"/> Y-one calculated from the bending moment <pause dur="0.3"/> equation in the same way <pause dur="0.7"/> and that will give us a set of equations <pause dur="0.3"/> which just the same as they would with the spring mass system <pause dur="1.3"/> # on the axial case <pause dur="0.2"/> will give us <pause dur="2.3"/> N unknowns where we have N masses <pause dur="0.3"/> Y-one <pause dur="0.2"/> through to Y-N <pause dur="0.3"/> plus omega-N-squared so we're in just the same situation as before <pause dur="0.4"/> we have one too many variables and we solve <pause dur="0.9"/> for omega-N-squared <pause dur="0.4"/> and the ratio of each of those deflections to <pause dur="0.2"/> say <pause dur="0.5"/> Y-one <pause dur="0.2"/> so it's exactly the same <pause dur="0.4"/> slog it out by hand or give it to a computer and ask it for the eigenvalues <pause dur="0.7"/> and eigenvectors <pause dur="1.0"/> now i'm not going to work that through any more today because it it

it's the sort of tricky sort of bits and pieces it's just a a a lot of <pause dur="0.7"/> tedious mathematics which gain nothing <pause dur="0.2"/> or give us nothing for the physical understanding of the problem <pause dur="0.5"/> there is a question very similar to this example <pause dur="0.7"/> # <pause dur="0.2"/> that sketch there <pause dur="0.3"/> on the example sheet <pause dur="0.4"/> we have a class in week nine when we'll discuss <pause dur="0.3"/> the results of those <pause dur="0.3"/> and as always of course well the the expectation is that you'll have had a go at those problems for yourself <pause dur="0.4"/> and found out how much you know about it <pause dur="0.4"/> before we discuss the solution <pause dur="0.3"/> that day <pause dur="0.4"/> so i am expecting <pause dur="0.3"/> that you will have a fair background to this <kinesic desc="indicates transparency" iterated="n"/> in the <pause dur="0.2"/> week nine Thursday examples class <pause dur="0.2"/> that i will actually go through <pause dur="0.6"/> the worked example on that <pause dur="1.2"/> now <pause dur="1.1"/> just to tell you in advance of that <pause dur="0.4"/> the <trunc>s</trunc> formal solution that we'll look at in the examples class <pause dur="0.3"/> uses this method primarily just so that we <pause dur="0.8"/> revise that method <pause dur="0.9"/> but <pause dur="1.2"/> just going off the top of the screen <kinesic desc="indicates point on transparency" iterated="n"/> the bit # # there is about <pause dur="0.2"/> that's the trick you can use <pause dur="0.2"/> assuming we have a

simple deflection system <pause dur="0.2"/> we can look it up in tables <pause dur="0.8"/> i hope at least some of you have <pause dur="0.2"/> sort of <trunc>ha</trunc> had already the thought look hang on there's just two forces here or might be three in a more complicated one <pause dur="0.6"/> but this is all linear elasticity <pause dur="1.0"/> i can look up in a table what would happen if there was only <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> that one <pause dur="0.3"/> <trunc>th</trunc> that's in the data book <pause dur="0.6"/> # <pause dur="1.0"/><kinesic desc="indicates point on transparency" iterated="n"/> that one's in the data book it's really just a <trunc>d</trunc> <pause dur="0.2"/> different numbers into the same problem <pause dur="0.3"/> i can <trunc>s</trunc> <pause dur="0.4"/> so i can look up the solution for each of those single force systems <pause dur="0.9"/> but it's all linear <pause dur="0.2"/> i can superpose those two solutions to get <pause dur="0.4"/> just add the deflections if i calculate the deflection Y-one and Y-two <pause dur="0.3"/> only for F-one <pause dur="0.5"/> and Y-one and Y-two only for F-two <pause dur="0.5"/> then <pause dur="0.2"/> for both forces <pause dur="0.4"/> the total deflection at Y-one is the sum of those <pause dur="0.8"/> two sort of <pause dur="0.4"/> partial Y-ones <pause dur="0.2"/> and similarly for Y-two <pause dur="0.3"/> so you can <pause dur="0.4"/> get at the system by just superposing <pause dur="0.3"/> the looked-up solutions # in in the same way <pause dur="0.2"/> and you

again <pause dur="0.2"/> then just equate <pause dur="0.4"/> plug <pause dur="0.3"/> these F values in <pause dur="0.3"/> and you get the same set of equations <pause dur="0.2"/> obviously <pause dur="0.4"/> the two methods work the same one is doing the formal linear theory <pause dur="0.3"/> the other is using tables to look up bits of the theory <pause dur="1.4"/><vocal desc="sigh" iterated="n"/><pause dur="0.4"/> it gets to be a bit of a balance it's worth the trick <pause dur="0.4"/> for a single system <pause dur="0.8"/> unless for some reason you need the other parameters as well then you know that makes it worth doing all the hard slog of mathematics <pause dur="0.7"/> if you've got three of these <pause dur="0.9"/><kinesic desc="indicates point on transparency" iterated="n"/> i reckon there's probably because the superposition means slogging out a lot of terms and then adding them together <pause dur="0.3"/> i would have thought if you've got three <pause dur="0.3"/> it's probably less actual algebraic slog <pause dur="0.4"/> to do it by the formal method <pause dur="0.5"/> two is about a break point <pause dur="0.8"/> it it's <pause dur="1.5"/> it's probably easier to think about what you're doing <pause dur="0.5"/> # <pause dur="0.2"/> doing it by superposition <pause dur="0.2"/> it might actually be slightly less work doing it by formal methods but they're about equal in terms of the effort <pause dur="0.3"/> it takes <pause dur="2.6"/><event desc="takes off transparency" iterated="n"/> so i'll leave that with you

and we'll pick that up at the examples class because again there is really no new theory there it is just a matter of <pause dur="0.2"/> confidence in using the basic ideas <pause dur="1.0"/> so what i'd like to do is is move on to <pause dur="0.2"/> the other case # of vibration which is <pause dur="0.4"/> where we <unclear>again to</unclear> transverse <pause dur="0.4"/> but we now <pause dur="0.5"/> cannot any longer ignore <pause dur="0.7"/> the mass <pause dur="0.6"/> of the beam itself <pause dur="0.7"/> and in fact we're going to go the whole hog and look at only one special case this year <pause dur="0.5"/> # <pause dur="1.0"/> we're going to look only at the case where it's # a uniform beam <pause dur="0.4"/> of some <pause dur="0.2"/> mass and there is no point masses on it at all so we've gone from one extreme only point masses <pause dur="0.5"/> and a light beam <pause dur="0.3"/> to a uniform beam and no other masses at all <pause dur="0.5"/> and this is the other one which is a worked example <pause dur="0.3"/><kinesic desc="puts on transparency" iterated="n"/> so you have a sheet of this again with some of the more text written in to just give you a better <pause dur="0.5"/> explanation than is on these slides <pause dur="1.7"/> # and again this time i've given it you because the final result <pause dur="1.0"/> is really rather a tricky # <pause dur="0.2"/> awkward <pause dur="0.4"/> looking formula <pause dur="0.9"/> so <pause dur="0.4"/> anyway <pause dur="0.3"/> what we have now then is a uniform beam again

it's going to be vibrating somehow like this through in this way <pause dur="0.4"/> # <pause dur="0.2"/> it's uniform density <pause dur="0.3"/> and with any uniform system <pause dur="0.3"/> our normal recourse is to looking at an elementary section <pause dur="0.3"/> and then attacking it with calculus <pause dur="0.4"/> so that's what we do here <kinesic desc="indicates point on transparency" iterated="n"/> let's take a short section <pause dur="0.2"/> D-X <pause dur="0.3"/> along the length of the beam <pause dur="1.1"/> # the beam is defined i've <pause dur="0.5"/> doesn't really matter but its length happens to be L <pause dur="0.2"/> but notice we need the mass per unit length now we're using the linear density idea that's so commonly used <pause dur="0.3"/> through structural mechanics <pause dur="3.9"/> right the transverse acceleration we're interested in the motion <kinesic desc="demonstrates motion on transparency" iterated="y" dur="1"/> that way <pause dur="0.3"/> so we're going to have the mass of this little element will just be M-D-X <pause dur="0.3"/> okay because it's a linear density <kinesic desc="indicates point on transparency" iterated="n"/> this thing small-M <pause dur="2.0"/><kinesic desc="demonstrates motion on transparency" iterated="y" dur="3"/> the vibration is going to be along this way <pause dur="0.2"/> and it's just a Y-double-dot if you like <pause dur="0.2"/> or D-two-Y D-T-squared <pause dur="0.3"/> but notice that we're going to be interested in the <kinesic desc="demonstrates beam with arms" iterated="y" dur="2"/> spatial distribution <pause dur="0.4"/> of the shape of the beam <pause dur="0.3"/> and in the <kinesic desc="demonstrates motion with arms" iterated="y" dur="2"/> time <pause dur="0.3"/> motion of it <pause dur="0.4"/> so to be strictly

correct <pause dur="0.4"/> we have to be working in partial derivatives at this stage hence the <pause dur="0.3"/> the symbols used here <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> so we have partial D-two-Y <pause dur="0.4"/> D-T-squared <pause dur="2.6"/> for the other case you're only # usually <pause dur="0.4"/> for these beam case you're only usually concerned <pause dur="0.3"/> with the displacement directly under the mass <pause dur="0.3"/> and so you tend to <pause dur="0.3"/> to go back to a a a full derivative there and not worry about the other plane <pause dur="0.2"/> here we have to be more precise <pause dur="1.8"/> okay <pause dur="0.6"/> if we've got this acceleration going on <pause dur="0.7"/><vocal desc="clears throat" iterated="n"/> force going on there <pause dur="0.7"/> now what the beam is seeing <pause dur="0.3"/> # <pause dur="0.3"/> # <pause dur="0.3"/> the acceleration <kinesic desc="indicates point on transparency" iterated="n"/> there has of course has a certain little force <pause dur="0.3"/> where does that force come from well it comes by <pause dur="0.8"/> # elasticity from the next bit of the beam along and all the way out <pause dur="0.2"/> to the supports <pause dur="0.3"/> # just as it does in all other <pause dur="0.2"/> bending or deflection problems <pause dur="0.5"/> so what i effectively have here <kinesic desc="indicates point on transparency" iterated="n"/> is because of the acceleration <pause dur="0.4"/> i have <pause dur="0.5"/> that <pause dur="0.6"/> load function <pause dur="0.8"/> now that's no more than saying that if this were a static problem that i've got a a <pause dur="0.6"/> a heavy beam <pause dur="0.2"/> hanging between

supports <pause dur="0.9"/> how is it going to sag <pause dur="0.3"/> well what i say is its density is M per unit length <pause dur="0.3"/> therefore i have <pause dur="0.2"/> a force <pause dur="0.4"/> M-G <pause dur="0.5"/> per unit length in other words a a linear force function of that sort and you're <pause dur="0.2"/> again you know <trunc>b</trunc> <pause dur="0.5"/> # in the data book rather than having capital W for a point weight you have a <trunc>s</trunc> <pause dur="0.2"/> lower case W for a distributed weight <pause dur="0.2"/> it's no more than that <pause dur="0.3"/> but it's our general acceleration of motion <pause dur="0.2"/> not just big-G it's obviously <trunc>dimen</trunc> # sorry little-G <pause dur="0.3"/> it's not dimensionally <pause dur="0.4"/> # any different <pause dur="3.5"/> so <pause dur="1.3"/> we have <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> that as a low density <pause dur="1.5"/> we know from basic beam theory <pause dur="0.6"/> that <pause dur="1.0"/> the fourth partial derivative of displacement <pause dur="2.5"/> is the load function <pause dur="0.8"/> # <pause dur="0.6"/> the second # you remember the second <pause dur="0.6"/> derivative gave us the bending moment <pause dur="0.2"/> and if you think about how you calculate a bending moment from a force <pause dur="0.2"/> there are two steps in in that as well so you get to the fourth derivative <pause dur="0.4"/> but <kinesic desc="indicates point on transparency" iterated="n"/> these again are equations that should be well known to <pause dur="0.2"/> to you from <pause dur="0.6"/> other courses <pause dur="2.1"/> so we get <kinesic desc="indicates point on transparency" iterated="n"/> this basic <pause dur="0.3"/> equation of motion

which occurs for all these types of problems <pause dur="0.6"/> # <pause dur="0.7"/> if we <pause dur="0.2"/> have <pause dur="0.2"/> uniform elastic properties <pause dur="0.8"/> for E <kinesic desc="indicates point on transparency" iterated="n"/> there if we have a uniform cross section <pause dur="0.6"/> # all the way along so that I is a constant and if we have uniform mass distribution so M is a constant <pause dur="0.4"/> then the thing is fairly straightforward <pause dur="0.3"/> in principle you could solve it <pause dur="0.2"/> with those things variable along the length <pause dur="0.4"/> but in general you will find that you by <pause dur="0.7"/> classical methods anyway you will not get a solution out under those conditions you will have non-linear <pause dur="0.4"/> differential equations <pause dur="0.7"/> you can solve them by computers <trunc>m</trunc> # finite element methods and such things <pause dur="0.8"/> so we're going to deal only with this simple case where everything's a constant <pause dur="0.6"/> it's a separable partial differential equation <pause dur="0.6"/> # <pause dur="0.2"/> i hope you can remember coming across those before <pause dur="0.5"/> # but you probably hoped you would never have to do <pause dur="0.2"/> anything very serious with them <pause dur="0.5"/> and that's fine by me <pause dur="0.3"/> # i'm quite happy to reduce all the <pause dur="0.2"/> difficulty of solving <kinesic desc="indicates point on transparency" iterated="n"/> that down to one

line <pause dur="0.5"/> it can be solved readily <pause dur="0.2"/><kinesic desc="makes quotation mark gesture" iterated="n"/> in quotation marks <pause dur="0.3"/> and what i mean by that is there's a perfectly standard solution <pause dur="0.2"/> you can look up in a textbook how to do it <pause dur="0.3"/> there's nothing difficult it's a just a routine turning the handles <pause dur="1.0"/> job <pause dur="0.3"/> to churn out the solution to that <pause dur="0.2"/> it's just a lot of hard work to do it <pause dur="0.9"/> as you will realize when you see that the actual formula and again don't try and write that down it's on the sheet <pause dur="0.6"/> # <pause dur="1.0"/> comes out to be that the displacement Y as a function of <pause dur="0.2"/> <trunc>displ</trunc> <pause dur="0.2"/> # <pause dur="0.5"/> the transverse displacement Y as a function of the distance along the beam X <pause dur="0.3"/> and time <pause dur="0.6"/> comes out to be this amazing thing here <pause dur="1.0"/> looking at the far end it's got <kinesic desc="indicates point on transparency" iterated="n"/> sine-omega-T-plus-phi <pause dur="0.4"/> type shape there which is what we always expect to see in these <pause dur="0.6"/> but whereas before we just sort of sad had sort of A-sine-omega-T for the single spring mass <pause dur="0.4"/> we now have something which has got four constants <pause dur="0.5"/> that's no surprise <kinesic desc="indicates point on transparency" iterated="n"/> we've got a fourth derivative there <pause dur="1.1"/> but it's got a sine term <pause dur="0.2"/> a

cos term <pause dur="0.3"/> the hyperbolic sine a hyperbolic cosine term there <pause dur="0.3"/> all those four <pause dur="0.2"/> terms can crop up all the things we can do <pause dur="0.3"/> with <pause dur="0.5"/> exponential functions in other words <pause dur="0.2"/> # come up in this equation <pause dur="0.2"/> and in general the solution is like that <pause dur="0.6"/> we've stuck an alpha <pause dur="0.2"/> in here <pause dur="0.7"/> just to save work because alpha comes out to be the whole of all the Ms and omega-squareds and everything else E-Is <pause dur="0.5"/> all go in there <pause dur="1.4"/> and all <pause dur="0.2"/> transverse uniform beams in vibration <pause dur="0.3"/> satisfy <pause dur="0.2"/><kinesic desc="indicates transparency" iterated="n"/> that formula <pause dur="1.3"/> we simplify it when we know the <pause dur="0.6"/> N conditions we know whether the Ns are fixed or pinned <pause dur="0.3"/> # <pause dur="0.2"/> we know whether <pause dur="0.4"/> # whether whether we hit it with a hammer in the middle to get it going various forcing functions <pause dur="0.3"/> so by using boundary and initial conditions <pause dur="0.3"/> we can simplify <pause dur="0.2"/> down for <pause dur="0.5"/> # <pause dur="0.6"/> real problems and find that not usually all these terms <pause dur="0.3"/> exist some of those Cs will go to zero <pause dur="0.5"/> but that's really what what we're dealing with on that level <pause dur="0.3"/> and i think you can see that <pause dur="0.4"/>

although <pause dur="0.9"/><kinesic desc="changes transparency" iterated="y" dur="12"/> # a sort of # <pause dur="0.7"/> an understanding of the principles of what's going on is important at this stage <pause dur="0.4"/> # <pause dur="1.2"/> it's not the sort of thing you're going to go into lightly you'll you'll do it only if you really have to to start solving those complex forms <pause dur="0.9"/> so anyway just working on this is still part of the stuff you've got in the notes <pause dur="0.8"/> let's just take a specific example which picks up on that formula on the previous page <pause dur="0.6"/> if we take a simply supported beam <kinesic desc="demonstrates beam with pen" iterated="n"/> so we've got a a beam across here it's just got a simple support at each end <pause dur="0.6"/> in other words one that can't <pause dur="0.5"/> it holds position doesn't transmit a moment <pause dur="0.5"/> and we then know <pause dur="0.6"/> that there will be zero N deflection <pause dur="0.2"/> and zero bending moments <pause dur="0.4"/> # at both ends of the beam underneath each support <pause dur="0.4"/> that's that's what <pause dur="0.2"/> defines what we mean by simply support <pause dur="0.2"/> <trunc>s</trunc> supported beam <pause dur="1.9"/> so when we plug those numbers in so you just slog

through that <pause dur="0.2"/> horrible formula we had last time work out the the appropriate <pause dur="0.5"/> put in # # <pause dur="1.2"/> X-equals-nought <pause dur="0.3"/> # set Y-equals-nought and see what the constants come out and so on and so forth <pause dur="0.3"/> then it turns out that <kinesic desc="indicates point on transparency" iterated="n"/> that one comes out to be <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> like that <pause dur="0.3"/> only the sine term only one of those four C <pause dur="0.3"/> C-one to C-four things <pause dur="0.3"/> is actually non-zero for that <pause dur="0.4"/> classic case <pause dur="0.2"/> if you have a cantilever <pause dur="0.3"/> you'll find you will finish up with two of the terms and there'll be a different two <pause dur="0.2"/> and so on and so forth but <pause dur="0.9"/> we just hammer it through in that way <pause dur="0.9"/> now the phi <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> there <pause dur="2.1"/> is the one that corresponds to just whether we pulled the beam down and let it go or whether it was sitting in the middle and we hit it or whatever just the same as the phi <pause dur="0.2"/> in the very simple spring mass <pause dur="0.4"/> so that's the one that's worked from the time initial condition <pause dur="1.7"/> # <pause dur="0.4"/><kinesic desc="indicates point on transparency" iterated="n"/> what we get in here notice is we do have <pause dur="0.4"/> a subtlety that's come up in this formula <pause dur="0.3"/> notice that it's not just sine-pi-X or something like that

inside there <pause dur="0.3"/> it's sine-N-<pause dur="1.9"/>pi-X <pause dur="0.2"/> where N is an integer <pause dur="0.7"/> and the reason for that is of course that when we're solving that and saying you know at what values does sine go to zero <pause dur="0.3"/> well it goes to zero at <pause dur="0.3"/> an angle of zero at an angle of pi at an angle of two-pi at an angle of three-pi <pause dur="0.2"/> and so on and so forth <pause dur="0.4"/> so all of those are possible solutions <pause dur="0.3"/> to how our string's vibrating or our beam is vibrating <pause dur="0.2"/> there's no difference between a string vibration and a beam vibration in the sense we're talking about them <pause dur="4.4"/> so <pause dur="0.8"/> we get <pause dur="0.8"/> a series of solutions here <pause dur="0.3"/> if we look at <pause dur="0.4"/> N-equals-one <kinesic desc="indicates point on transparency" iterated="n"/> in here <pause dur="0.4"/> we've got <pause dur="0.6"/> time varying behaviour but we've got spatially varying <pause dur="0.7"/> behaviour from that term <pause dur="0.3"/> and it's saying that <trunc>i</trunc> if N is one <pause dur="0.2"/><kinesic desc="demonstrates motion with arms" iterated="y" dur="16"/> then the thing is a half sine wave like that in other words it's zero at both ends <pause dur="0.3"/> and it's moving through in that sort of motion <pause dur="0.7"/> at any point along the beam <pause dur="0.6"/> it's going at a speed <pause dur="0.3"/> omega-N <pause dur="1.3"/> from that <pause dur="0.5"/> time sensitive <trunc>te</trunc> sinusoidal term <pause dur="0.3"/> but the distribution is like that <pause dur="0.5"/>

if i go up <pause dur="0.2"/> to N-equals-two i get something that <pause dur="0.3"/> has <pause dur="0.6"/> the classical sine wave shape a a node in the middle <pause dur="0.3"/> it's still any point <kinesic desc="indicates point on transparency" iterated="n"/> this way it's still going at exactly the same speed of oscillation <pause dur="1.0"/> for <pause dur="0.4"/> the relevant solution for omega-N <pause dur="0.5"/> and so for each of those modes <pause dur="0.2"/> here as we go up <pause dur="0.4"/> we're getting more <pause dur="1.1"/> turning points <pause dur="0.8"/> so we're getting <pause dur="0.4"/> a more difficult <trunc>e</trunc> <pause dur="0.3"/> energetic shape <pause dur="0.3"/> to drive <pause dur="0.3"/> we need a higher omega-N to cope with that <pause dur="0.3"/> so we know that we're going to get <pause dur="0.4"/> in theory an infinite number of these <kinesic desc="indicates point on transparency" iterated="n"/> now because there's an infinite number of solutions to that equation <pause dur="0.4"/> and we're having an infinite number of modes <pause dur="0.6"/> of increasing <pause dur="0.8"/> frequency <pause dur="1.2"/> now what's the practical side of that well what we know in practice <pause dur="1.7"/> is that <pause dur="1.7"/> it's more difficult to drive the high modes <kinesic desc="demonstrates motion with arms" iterated="y" dur="4"/> this one goes fairly easily <pause dur="0.5"/> # it doesn't take a lot of energy to do that to do it with a <pause dur="0.2"/> full wave vibration takes more energy to do one with lots and lots of little kinks up and down

think of just the <pause dur="0.3"/> the bending energy you have to put in to make the beam go into that shape <pause dur="0.3"/> and you can see it's going to be just harder to get it going like that <pause dur="0.5"/> so in practice <pause dur="0.4"/> beams rarely show more than with any <pause dur="0.7"/> non-negligible <trunc>am</trunc> amplitude <pause dur="0.3"/> more than one or two of these low order modes it's fairly rare <pause dur="0.6"/> to <trunc>g</trunc> get something with a very complicated pattern <pause dur="0.2"/> so we normally finish up with a a fairly simple practical problem <pause dur="0.4"/> # you can more or less guarantee for for all realistic <pause dur="0.5"/> conditions that <kinesic desc="indicates point on transparency" iterated="n"/> this <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> this might be quite big but that's going to have smaller amplitude significantly and the next one up's going to be very small <pause dur="0.5"/> and so on and so forth <pause dur="0.3"/> and you can argue that purely on the this <pause dur="0.4"/> # notion of the energy that's involved <pause dur="2.0"/> so <pause dur="0.9"/> if you need to do the formal theory <pause dur="0.4"/> you can slog it through <pause dur="0.5"/> if you have <trunc>ac</trunc> <pause dur="0.2"/> access to a textbook <pause dur="0.5"/> and you can remember that there is that horrible formula with the cos cosh sine and sinh terms in it <pause dur="0.6"/> you can look that

up in the textbook and just do the boundary conditions <pause dur="0.6"/> maybe for a lot of what you're doing it's okay just to say look these are the possible motions this is the one that worries me <pause dur="0.5"/> and we could then do <pause dur="0.3"/> a guess <pause dur="0.4"/> type solution rather than do it properly knowing which of these <kinesic desc="indicates point on transparency" iterated="n"/> mode shapes and knowing the mode shapes <pause dur="0.4"/> we can sketch out <pause dur="0.4"/> the shape not the amplitude but we can sketch out the shapes <pause dur="0.7"/> purely by a sort of physical intuition <pause dur="1.2"/> we could then think in terms of using approximate methods <pause dur="0.2"/> that can sort of home in on the one that we want <pause dur="0.4"/> and there are some very sophisticated <trunc>a</trunc> and very useful techniques <pause dur="0.3"/> that that are widely used for doing that <pause dur="0.2"/> but they going into those and doing those properly <pause dur="0.3"/> is really # # if you like a topic for the third year <pause dur="2.6"/> so <pause dur="0.6"/> i'm going to leave it the beam example by saying look that's the basic theory <pause dur="0.5"/> that's how you would get at the solution if you needed to and you ought to be aware of that <pause dur="0.4"/> but the basic behaviour is <kinesic desc="indicates point on transparency" iterated="n"/> this series of

omega-Ns and when you actually plug in real boundary conditions <pause dur="0.3"/> to the complicated formula <pause dur="1.0"/> life's fairly straightforward <pause dur="0.6"/> but # <pause dur="0.2"/> i mean there was no <pause dur="0.3"/> sense in which we expect you to memorize that formula for exam <pause dur="0.5"/> conditions or anything of that sort it <pause dur="0.2"/> it's just not the sort of thing that's sensible to do <pause dur="0.3"/> you look it up in a textbook if you want it <pause dur="2.0"/> so we do use approximate methods # <pause dur="0.5"/><event desc="takes off transparency" iterated="n"/> and # <pause dur="0.3"/> again <pause dur="0.7"/> when you look at the # <pause dur="1.5"/> example sheets if you haven't done so yet you'll find one of them <pause dur="0.4"/> has got an <trunc>asteri</trunc> one of the questions has got an asterisk against it <pause dur="0.5"/> because it's <pause dur="0.2"/> a a sort of <pause dur="0.7"/> an extra one to do if you're feeling # enthusiastic <pause dur="0.7"/> # it is a a case of looking at one of the earlier problems and looking how we might <pause dur="0.4"/> use <pause dur="0.3"/> approximate methods that is you know to use an approximate method <pause dur="0.3"/> you usually are iterating <pause dur="0.3"/> and trying to guess roughly what the right answer is <pause dur="0.6"/> to get you a reasonable starting point otherwise it's you

know if you guess badly it's a lot of work so it's worth <pause dur="0.2"/> a little bit of brain power to get a good guess <pause dur="0.6"/> # and there's one example that shows you that <pause dur="0.9"/> but what i'd like to spend the last few minutes today <pause dur="0.5"/> <trunc>i</trunc> is <pause dur="0.6"/> just hinting at one <trunc>w</trunc> how one or two of those <pause dur="0.4"/> # <pause dur="0.2"/> methods are developed <pause dur="0.4"/> # <pause dur="0.4"/> not <trunc>b</trunc> so much to <pause dur="1.8"/> show you the <trunc>en</trunc> enough to get you into next year's course <pause dur="0.4"/> but just to <pause dur="0.6"/><kinesic desc="puts on transparency" iterated="n"/> show how we can use <pause dur="0.2"/> energy methods <pause dur="0.2"/> alongside these other approaches <pause dur="0.3"/> you will notice that regularly <pause dur="0.4"/> i've given you a formal classical solution in Newton's laws <pause dur="0.4"/> and then said think about the energy <pause dur="0.3"/> and it will <pause dur="0.3"/> it will help you to see why this pattern of <trunc>sha</trunc> of modes and nodes and so on goes together <pause dur="1.0"/> well if that's the case then presumably there there could be good reasons for using energy <pause dur="0.3"/> more formally as the solution method and i and i think it's just worth seeing this <pause dur="0.6"/> now this first little bit it it's nothing very complicated <pause dur="0.3"/> it's the sort of stuff that is in the first year

texts so <pause dur="0.2"/> you know if again if you look at the reading list <pause dur="0.3"/> you'll find that the reading associated with this particular little section of the course <pause dur="0.4"/> is in things like Hibbler and Mariam as well as the the main texts <pause dur="0.5"/> for <pause dur="0.2"/> for this <pause dur="0.5"/> # particular session <pause dur="3.3"/> i think all we'll do here is just look at one example of it so let's just consider <pause dur="0.3"/> free vibration of a single spring mass system to stay low <trunc>l</trunc> let's let's not do anything very complicated 'cause it's only the <pause dur="0.2"/> the idea <pause dur="0.2"/> that you just are aware how the methods work <pause dur="1.8"/> if we consider it as a free vibration then <pause dur="1.0"/> in the sort of mechanical thermodynamic sense we can regard it as a closed system it it's just sitting there doing its own thing <pause dur="0.6"/> without any <trunc>s</trunc> significant influence from the outside world <pause dur="0.5"/> and if that is true <pause dur="0.4"/> then the mechanical energy within that system must be conserved <pause dur="4.3"/> now what mechanical energy do we have in a vibration problem <pause dur="0.2"/> again this is where we were

discussing <pause dur="0.3"/> this time last week <pause dur="0.3"/> we've got <pause dur="0.4"/> strain energy or spring <pause dur="0.2"/> potential energy if you like <pause dur="0.3"/> and we've got kinetic energy stored in the the <pause dur="0.5"/> the inertial part of it <pause dur="0.6"/> so we're saying that <pause dur="0.5"/> again let's take the simple case <pause dur="0.5"/> because it's <pause dur="0.2"/> thermodynamically isolated in its most simple case then we've <pause dur="0.2"/> ignored friction because that would generate heat <pause dur="0.5"/> out of the <pause dur="0.2"/> kinetic energy <pause dur="0.3"/> and we don't want that <pause dur="0.3"/> so we've got just these two forms of energy that we're swapping between kinetic and strain and back again to get the oscillation going <pause dur="0.7"/> that some of them must be a constant <pause dur="2.4"/> so <pause dur="1.3"/><kinesic desc="reveals covered part of transparency" iterated="n"/> just <pause dur="0.5"/> flogging that through <pause dur="0.2"/> if the motion is a sine wave <pause dur="0.7"/> as we're assuming everywhere else <pause dur="0.3"/> X equals A-sine-omega-T <pause dur="1.2"/> its maximum displacement obviously is A <pause dur="3.8"/> and when is that maximum <trunc>displacem</trunc> # <pause dur="0.9"/> going to occur <pause dur="0.3"/> well it's going to occur at the ends of the motion that is where the thing stops and is about to turn round <pause dur="0.8"/> so <pause dur="0.2"/> we'll get the maximum displacement A <pause dur="0.4"/> at the points where the

speed <pause dur="0.2"/> is zero <pause dur="1.0"/> when's it going fastest well <kinesic desc="demonstrates motion with arms" iterated="y" dur="6"/>it's obviously going fastest <unclear>you</unclear> see it's stopped here it's come through here stopped at the other end <pause dur="0.3"/> and it's clearly fastest <pause dur="0.5"/> when it's in the middle <pause dur="0.7"/> when there's no displacement <pause dur="0.5"/> so we'll get the maximum speed <pause dur="0.5"/> is A-omega we've been using A-omega-squared for the <pause dur="0.6"/> acceleration a lot so that should be <pause dur="0.5"/> totally familiar to you <pause dur="0.6"/> # <pause dur="0.5"/> the speed will be A-omega at displacement zero <pause dur="0.6"/> that statement i think you can see purely by physical intuition <pause dur="0.4"/> but if you want to be formal about it <pause dur="0.3"/> then just put <pause dur="0.6"/> differentiate <kinesic desc="indicates point on transparency" iterated="n"/> this with respect to time and have X-dot <pause dur="0.4"/> equals <trunc>A-omega-si</trunc> cos-omega-T <pause dur="0.5"/> and you know do do the formal maxima and minima <pause dur="0.2"/> and you will you can prove it like that but # but i think the physical intuition <pause dur="0.5"/> should be all you need to do that <pause dur="2.1"/> now this is useful because <pause dur="1.6"/> strain energy is to do with displacement <pause dur="0.8"/> kinetic energy is to do only with speed <pause dur="0.5"/> now it happens that <pause dur="0.5"/> # when we've got the

speed zero there's no kinetic energy and we've got maximum displacement and vice versa with the other one <pause dur="0.4"/> so the maximum strain energy <pause dur="0.4"/> and the maximum kinetic energy <pause dur="0.2"/> must <pause dur="0.2"/> actually be the same value <pause dur="0.5"/> and we can either sum them <shift feature="voice" new="laugh"/>formally <shift feature="voice" new="normal"/><pause dur="0.8"/> # for all values or we can just say look just look at the maximum strain look at the maximum speed <pause dur="1.0"/> from that we get <pause dur="0.2"/> strain energy <pause dur="0.2"/> kinetic energy and those two maximum values must be the same <pause dur="0.3"/> so we actually have two slightly different things we can do we can use it in that <kinesic desc="indicates point on transparency" iterated="n"/><pause dur="0.7"/> strict formulation <pause dur="0.4"/> or we can use it in in <kinesic desc="indicates point on transparency" iterated="n"/> this form <pause dur="0.2"/> and in given situations <pause dur="1.1"/> thinking out about it one way or the other <pause dur="0.5"/> tends to be a little easier <pause dur="3.1"/> so <pause dur="0.2"/> very simple example let's go right back to the <pause dur="0.2"/> very first example we did as a revision <pause dur="0.4"/> at at the start of this section <pause dur="1.4"/> single mass single spring K <pause dur="0.5"/> # and let's look at what the strain energy is well the maximum strain energy <pause dur="0.7"/> is when we have the thing extended A beyond its natural length <pause dur="0.4"/> and is <pause dur="0.5"/> a half-K times the displacement squared <pause dur="0.6"/> that's the straightforward spring <pause dur="0.2"/> Hooke's law <pause dur="0.6"/> energy formula <pause dur="4.0"/> the

other side we have <pause dur="0.5"/> the maximum kinetic energy which is half-M-V-squared <pause dur="0.2"/> and here <pause dur="0.3"/> is is the maximum V we want which is A-omega from the line before <pause dur="0.3"/> so we get half-M <pause dur="0.8"/> A-omega-all-squared <pause dur="1.2"/> reorganize that remembering that we're looking for the natural frequencies so putting the subscript-<pause dur="0.3"/>N back in there <pause dur="0.5"/> omega-N-squared <pause dur="0.3"/> equals K-upon-N <pause dur="0.3"/> # and of <pause dur="0.4"/> it course it would be a terrible shock <pause dur="0.3"/> if anything other than that result <pause dur="0.2"/> came out since it's one that we're so familiar with <pause dur="0.3"/> but just notice that <pause dur="0.4"/> that works you know drops it straight out in one line <pause dur="0.3"/> under those conditions <pause dur="0.4"/> and it will do equivalent things <pause dur="0.2"/> in more complicated <pause dur="0.7"/> situations <pause dur="0.9"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so <pause dur="1.0"/> here for this year <pause dur="0.6"/> i don't want to go any further along that particular little bit of a line than to do that and just notice that <pause dur="0.3"/> for that simple case <pause dur="1.0"/> if you like that method there's no reason why you shouldn't use it <pause dur="0.6"/> for some more complicated cases <pause dur="0.4"/> # # of <pause dur="0.3"/> complex motions <pause dur="0.3"/> we can set up <pause dur="1.6"/> special coordinate

systems <pause dur="0.3"/> we can use these sort of tricks <pause dur="0.3"/> including the one that says let's formally <pause dur="0.8"/> differentiate the energy equation and say if it's conserved then the the change with time must be zero <pause dur="0.4"/> which is the the sort of one just going off the screen at the top <pause dur="1.5"/> and <pause dur="0.4"/> if you do that then you get to a whole series of techniques <pause dur="0.3"/> going under the name of Lagrangian mechanics <pause dur="0.4"/> # and <pause dur="0.2"/> later on even if you're really desperate Hamiltonian mechanics <pause dur="0.2"/> but they're a whole series of very powerful <pause dur="0.2"/> analytical techniques <pause dur="0.7"/> whose underpinning <pause dur="0.3"/> is <kinesic desc="indicates point on transparency" iterated="n"/> this very simple example # and <pause dur="1.1"/> all i want to do today is say look if you come across these formally next year <pause dur="0.4"/> # <pause dur="0.2"/> you should go away with enough confidence <pause dur="0.3"/> if you choose a set of options by which you don't get involved with those courses <pause dur="0.8"/> but then later <trunc>i</trunc> early in your professional career <pause dur="0.3"/> do come across people throwing Lagrangians at you and things like like that and expecting you to understand it <pause dur="0.3"/> don't panic <pause dur="0.3"/> with a bit of luck just a

little reminder will say you know that's how you start <pause dur="0.3"/> it's not too much of a problem <pause dur="0.4"/> given that you need to do it to go and look in a textbook <pause dur="0.8"/> and just build it up from the sorts of things that we've been looking at <pause dur="3.5"/><event desc="takes off transparency" iterated="n"/> okay so i think that's all we'll do there can i show you one other slide it's in the note <pause dur="0.3"/> # in the # copies of the # <pause dur="0.4"/><kinesic desc="puts on transparency" iterated="n"/> slides in the library <pause dur="0.4"/> and i don't really want to talk about it in any detail at all <pause dur="0.3"/> it won't come up on an exam paper or anything of that sort <pause dur="0.2"/> it is purely information for you for the future <pause dur="0.3"/> depending on which route <pause dur="0.3"/> you might choose to go through <pause dur="0.6"/> and that is just to point out that alongside the energy methods <pause dur="1.1"/> that there <pause dur="0.3"/> are <pause dur="0.3"/> this trick of estimating where you are and using <pause dur="0.3"/> approximate processes <pause dur="0.3"/> # goes very closely <pause dur="0.3"/> that you can use approximations in all sorts of ways but <pause dur="0.2"/> Rayleigh <pause dur="0.4"/> # in <trunc>p</trunc> <pause dur="0.2"/> developed a whole series of techniques <pause dur="0.3"/> which <pause dur="0.2"/> use a mixture of energy <pause dur="0.3"/> and an approximate guess <pause dur="0.2"/>

at what's going on <pause dur="0.3"/> and you get this sort of pattern <pause dur="0.2"/> of behaviour <pause dur="0.6"/> # <pause dur="2.3"/> we can <pause dur="0.4"/> very often in real vibration problems <pause dur="0.4"/> if we know omega-N <pause dur="0.3"/> that's good enough we <trunc>n</trunc> <pause dur="0.3"/> we know that the initial <pause dur="0.2"/> the actual <trunc>ax</trunc> <pause dur="0.2"/> amplitude of the vibration depends just how we kick it <pause dur="0.3"/> so omega-N's <pause dur="0.2"/> usually the thing we're worried about <pause dur="0.8"/> if we <pause dur="1.5"/> can guess <pause dur="0.2"/> say for a <pause dur="0.2"/> <trunc>vo</trunc> <pause dur="0.2"/> oscillating beam or something like that the shape in which it's oscillating <pause dur="0.6"/> reasonably precisely <pause dur="1.1"/> we can estimate at its maximum deflection by <pause dur="0.2"/> standard beam theory say <pause dur="0.4"/> what the strain energy would be in that <pause dur="0.2"/> it's a formulae you can look up <pause dur="0.4"/> so providing we get just a a reasonable guess of that <pause dur="0.2"/> shape <pause dur="0.3"/> that would give us a maximum strain energy <pause dur="0.4"/> we could then <trunc>e</trunc> <pause dur="0.5"/> work out <pause dur="0.3"/> a maximum kinetic energy <pause dur="0.6"/> # as <pause dur="0.4"/> a sort of M-omega-squared sort of term <pause dur="0.4"/> and <pause dur="0.9"/> plug in those together it gives us <pause dur="0.3"/> an estimate of omega-squared <pause dur="0.3"/> and that's really what it says on here we we can guess at the shape <pause dur="0.5"/> # what Rayleigh discovered and what makes the method

powerful <pause dur="0.6"/> is that really quite crude guesses of that shape <pause dur="0.3"/> get you surprisingly close <pause dur="0.4"/> to the right answer <pause dur="0.7"/> and not only that <pause dur="0.4"/> they are always wrong in the same direction you always know that you've overestimated or underestimated depending on the problem you've set up <pause dur="0.4"/> # the conditions that work so it's a nice safe <pause dur="0.3"/> predictable <pause dur="0.2"/> approximate technique <pause dur="2.8"/> normally the way you find the shape is just to say okay <pause dur="0.3"/> as i did in that simple example we use a more sophisticated version of that <pause dur="0.7"/> here's a beam vibrating how how would it sag under gravity <pause dur="0.3"/> that's a good guess of the shape it will vibrate in for its lowest mode of behaviour <pause dur="0.7"/> so let's use that as the starting point for the system you know so so you use sort of common sense <pause dur="0.3"/> guesses intuition whatever you like to call it <pause dur="0.4"/> to deal

with that <pause dur="0.3"/> but some of you undoubtedly will later on in your courses come across a formal treatment of Rayleigh's method <pause dur="0.8"/> others of you <pause dur="0.5"/> perhaps more on the civil side <pause dur="0.3"/> may not deal with it directly don't get involved perhaps with earthquake engineering <pause dur="0.2"/> team at some stage <pause dur="0.3"/> and find other people using it <pause dur="0.3"/> at least you've heard the words to get you into the field <pause dur="0.2"/> and that's all i'm concerned with here just that you've heard the words <pause dur="0.2"/> it's not something that's going to get examined but i but i thought it was worth spending <pause dur="0.3"/> just a couple of minutes at the end of this session <pause dur="1.0"/> talking about it <pause dur="0.6"/> so that's it for today <pause dur="0.2"/> tomorrow we'll meet up again <pause dur="0.2"/> and we're back on examined material like <gap reason="inaudible" extent="1 sec"/> so it is stuff that's fair game <pause dur="0.3"/> we will look at vibration isolation as a subtopic <pause dur="0.4"/> # <trunc>wi</trunc> within the overall vibration picture