nm0855: at the end of the last lecture i'd really set out for you the the basic 
ideas of multimode vibration and promised to revise that with an example er a-, 
and that's what this is this is in the handout that you had at the start of 
this section of the course it's one of the the two worked example sheets in 
there er 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 so 
there shouldn't be any need to do more than just have a quick look through this 
and it it's more or less the same as we set up in the simple example last week 
except that there are now three masses four springs two rigid walls er in the 
notes that you have there are r-, rather more words describing what goes on 
between this so this is just an edited down version of the sheet that you have 
but note the basic principle as before is that for each mass in turn M-one M-
two 
et cetera we set up the Newton's equation of motion we have force generated by 
springs so we have terms like K-two A-two-minus-A-one K-one A-one-minus-zero in 
this case and that's going to equal an acceleration because it's a sinusoidal 
motion the acceleration becomes just omega-squared times the amplitude so we 
get mass times acceleration as a force M-one A-one omega-N-squared equals 
spring constant times the deflection spring constant times the deflection as we 
go along the the chain of events so that's how we set it up basically and we do 
that with a separate equation for each of the masses M-one M-two and M-three 
just reduced to et cetera [cough] er excuse me on this slide the only thing to 
point out about those two again just to reinforce what i said last time whether 
you do it just the same as this example doesn't matter but do be systematic 
about what you do and something that i think works well is to start always with 
the highest subscripts 
to the left-hand side as you go through so i recommend that given we're looking 
at M-one and we'd interested in the spring either side start with K-two write 
down its extension as A-two-minus-A-one then go to K-one and have A-one-minus 
and here it's a wall i've written in the zero formally just to stress that 
that's a wall that doesn't move at that end er but obviously you could just 
write K-one A-one if you wish but if you do it like that and notice the same 
thing three is there then it goes to twos then it goes down to one if you 
systematize it like that then the signs come out in that nice pattern of 
brackets with a negative number inside minus the other one with brackets with 
the negative number inside and you're less likely just to make the little slips 
that that can upset these sort of procedures so that's each equation and 
normally we would for larger problems then put this into a matrix notation 
omega-N-squared is a constant of this so we can just 
have some set of A amplitude vector A is some sort of stiffness matrix here 
multiplied by the same vector A which is just done by rearranging here this is 
with the Ks outside the brackets but just turn that round so you get the As 
outside and the Ks inside and you can get this form but notice that it's 
convenient to divide through in each row by M-one by M-two and so on so that 
appears in the denominator of each term inside there now that's very typically 
of how books present it you might argue that you're used to seeing K as the 
stiffness so using K for this stiffness divided by math-, mass might not be 
such a bright idea but a lot of the books do that so i've stuck with that but 
just notice it is a sort of stiffness matrix but it's modified by those masses 
if you get it to that form then by far the best approach for larger problems is 
then to give it to a solver and ask for the eigenvalues which correspond to the 
natural frequencies 
squared the omega-N-squareds and the eigenvectors which give you the mode 
shapes if you don't fancy doing that or it's a problem you're doing by hand 
then you're probably better off just solving this as a set of simultaneous 
equations at that level rather than worrying about the matrix formality the 
matrices do help on big problems they're not essential so if you do it in 
normal ways you solve for omega-N and the ratio of the amplitudes to say the 
amplitude A-one remember that we always have one too many variables we have the 
natural frequency plus N amplitudes so that's how you solve for that and it's 
not worth working the details out on it if you use the eigenvalue eigenvector 
formally through something like Matlab or a similar program you get the answers 
directly you can of course solve the mode shapes directly as well from the 
simultaneous equations but once you have the resonant frequencies the omega-Ns 
you can in fact guess what's going on and 
that's what i've done with th-, with the arrows i've drawn under the diagram 
there they correspond to the three possible motions all three could be going 
the same way in phase er with the same amplitude oh er sorry with different 
amplitudes but in the same er phase relationship or we could have the two 
outside ones going the same way and the other one coming the other way er which 
is the bottom one or in between those where two of them are moving in one way 
and one the other er 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 
together er now i i've written from this the top downwards in the way they are 
because that corresponds to the number of changes of direction if you look at 
that the top one has no change of direction the next one down has one change 
the one below that has two changes of direction because the change of direction 
corresponds to there being a node between 
those two displacements and as again we discussed last week we know that ev-, 
the more nodes you have the nodes the higher mode shape that you have 
corresponds to increasingly high resonant frequencies that's the energy link 
between them so once you've got the omega-Ns just by inspection you can work 
out the general shape of the modes but you need to so-, and which goes with 
which omega but you then need to solve formally to get the actual ratio of 
these amplitudes the sketching can't tell you that so that's the basic example 
that's really all there is to most of er er axial vibration and again let just 
put flash that slide up no nothing to copy down there it's just a quick look at 
that just s-, sketched out just before i came down today and notice that if it 
were a torsional problem it's exactly the same we'd have shafts of some 
stiffness there'd be moments of inertia for the 
rotors and a theta displacement for each one you just plug in the form-, the 
symbols into the w-, example over there if you have say a building frame a 
steel framed building with sort of uprights and fairly some concrete floors 
something mairly fairly massive there then you may have a sway system by which 
you know experimentally the values of these stiffnesses of these different 
storeys and you've just got effectively the slab of mass and again that is the 
same problem as that this just corresponds to the spring experimentally this is 
the mass and so on and so forth so all these problems reduce whatever variables 
they're in to that piece of sort of handle turning sort of er mathematical 
analysis 
nm0855: th-, what i'd like to do on though i-, is not sort of work more 
complicated examples er but move on to the slightly different conditions er 
hinted at by that little frame i just showed on the other s-, 
projector er if we know experimentally what the effective lateral stiffness of 
the building is fine we can use that simple method if we don't and we have to 
work with beams transverse vibration so we have a beam that's vibrating in some 
motion of that sort then there are one or two tricks which are just a little 
bit different and you'll find that nearly all the textbooks do treat them as a 
separate case as i'm doing in th-, er er in these lectures so let's just very 
briefly look at beam vibration and it's not a thing i want to spend a lot of 
detail the the theory here again is mathematically tricky just thumping out the 
solutions from the basic information er er er is is a slog it's easy to make 
mistakes so all i'm really concerned with is that we get pick up on the the 
physical principles that are going on behind that that will get you to the 
level by which you can then look up solutions with a degree of confidence er 
when you need them so we'll deal with beams 
and just to remind you what we mean by a beam it's something which is long and 
thin er so we've we've got it between supports there's a structure across here 
er whatever the depth of that beam is it's small compared to its its length and 
when we do that we can ignore generally speaking sheer effects and concentrate 
only on beam bending and then we can also use classical beam bending theory 
that you met last year and have met earlier in this course in the bit that 
namex was dealing with okay well there's different things we might want to do 
let's start with the very simplest case oops i'm just drawing it out like that 
we have a light beam er always the easier case on perfect simple supports as 
shown there but it doesn't the end conditions we'll come back to a light beam 
in other words we can neglect its mass with a single point lumped mass stuck 
somewhere we're not i've drawn it near the middle because that makes it easy 
but there's nothing in 
what i'm saying that means it has to be in the centre it can be anywhere along 
the beam and this might correspond to the case where we have a structural I-
beam or something like that er which has got something maybe ten times the 
weight of that mounted near the centre of it perhaps a machine that could be 
vibrating and may cause the beam to vibrate so that's the scenario we set up 
and generally speaking when when you can set up an a a coordinate frame and if 
i just replace this mass by a force and just had said there is a point force 
pushing downwards on there what is the shape of the beam how does the beam 
deflect then as far as i'm concerned er that is no problem for you to solve you 
may worry about it but at this stage the assumption of this course is that 
you're comfortable solving problems like that so we put a force on there er how 
would you do it well there's two things you could do er let's deal with the 
formal one first that is you could 
calculate the bending moments along that beam you'd sort of solve for the 
reaction forces calculate the bending moment and then we know that the 
deflection shape or the second derivative of the of the er deflection shape 
relates functionally directly to that bending moment and we get the shape out 
okay what do we do if it's vibrating well notice what we've done with 
everything else i've said i've got a force from the springs i'm now going to 
generate a force 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 
an inertia force it's the force M-A associated with the acceleration so all i 
need to do is substitute that inertia force in terms of the fixed F that you 
would have had on a a an earlier example a-, and we're away we do the same 
thing and i'm not going to actually work through an example here i'm just going 
to give you two lines of text if you like that that 
tells you that thing the inertia force then if we're moving transverse 
vibrations i er tha-, tha-, again the mass is i say where my hands are joined 
it is going through like that what we have is an inertia force mass of the lump 
thing Y is its deflection er so Y-double-dot would be its acceleration so we've 
sort of got Y- double-dot or Y D-, er sorry M-double-dot Y-double-dot or M D-
two-Y D-ec T-squared if you prefer to call it that er we have that form again 
we know that Y is going to be a sinusoidal motion that's the assumption of all 
our vibration systems so the Y-double-dot term can be replaced by omega-squared-
Y again just plugging it in the derivative er approach we discussed a couple of 
lectures back so the inertia force then has that as a a peak value er it it's 
oscillating sinusoidally but that is perfectly adequate amount of data the 
bending moment as always is just this formula you should be very familiar with 
bending moment it's is equivalent to E-I D-two-Y D-X-squared 
in other words the the second spatial derivative the curvature of the beam 
locally where E is Young's modulus I is the second moment of area of the cross 
section perfectly normal standard beam theory and all you do is equate that to 
that er you then need to integrate twice with respect to Y and you can solve 
for er omega-N-squared and and for Y from from that formula so that's formally 
how you go about doing the problem and it's it's not any more difficult that's 
all the theory all the theory you need is to remember that there are these 
things called inertia forces and that you can plug them in to the beam bending 
formula that you've been using for a long while however it it is often quite 
sort of a tricky sort of thing to do and one of the big advantages we have is 
that a lot of the beam structural beam bending formulae and so on are tabulated 
for us so let can we find a short cut trick that will help and i think that's 
something that's of more immediate 
interest [cough] excuse me again so an easier way for this problem always we 
are dealing with conditions which are linearly elastic and we're er let me s-, 
we said this at the start of this section of the course but let's repeat it 
again once we lose linear elasticity none of the stuff that i'm talking about 
is going to work properly so that that is just available to us as a a a 
prerequisite if it's linear elastic the beam just acts as a simple spring i've 
got this thing here if i push it downwards against its own elastic bending 
properties it'll try and push me back up again so it just acts as a spring and 
if i think of the spring as as l-, some value lambda newtons per metre in this 
case i'm ju-, i'm just pressing down here and saying how much force do i need 
to deflect a certain amount i could actually calculate that from first 
principles using the sort of formulae little bit higher up on that 
page or maybe i can do that experimentally i can just s-, put balance a weight 
on there and see how much it deflects to get an equivalent stiffness now we 
know that this is only a single mass and and a sort of spring system here 
because the the beam is light so we know from the elementary theory that the 
natural frequency will be just square root of stiffness over mass that is here 
lambda-over-M it was square root of K-over-M for the or or helical spring 
example we looked at earlier so all we need to do now is see whether we know 
what er lambda is we know what the mass is i assume er er in all of these 
problems er well er we could be ex-, experimental but we can actually get at it 
by some other methods so if i can just give you that bit as an extension to the 
bottom of that sheet what happens if we do just as i said experimentally sort 
of balance a mass on the beam and see how much it deflects well that gives us 
lambda and it gives it as from 
that thing the static deflection is the force M-G now because we're looking at 
how much it would deflect er as a horizontal beam divided by this same 
stiffness it's the same thing that causes both effects so experimentally that's 
how i get at lambda but i can also get at it because i can i can look up that 
static deflection for a weight 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 er so what we do actually if i can now take that 
one off is okay let er let me just put it back a second in fact er if we just 
look at this formula here and just plug in to get rid of lambda there which is 
the sort of unknown and we don't really want to know explicitly if i just 
substitute from this formula in there then i finish up with this rather magic 
looking formula 
that the natural frequency is the square root of G upon the static deflection 
okay simple enough formula but it worries quite a lot of people and with some 
reason all the while i've been saying to you with this stuff that the the 
vibration does not depend at all on the atmosphere we're working in the 
gravitational constants the vibration is just the same on the moon as it is on 
the earth other things may change but that stays the same so why does G appear 
in this formula er and the answer is very straightforwardly that it's because 
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 er in going through it because if we now 
look at a simple example so just a very simple example of simply supported beam 
which is what i'd actually drawn on the previous sheet [cough] 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 force 
type condition and you will find that if you have a simply supported beam with 
a force W a weight W in the centre of it and the deflection at under that 
weight is W-L-cubed over forty-eight-E-I 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 now here the weight is simply just M-G and i think you can see what's 
happening now is that although i've i've lost apparently the mass and have 
gained a G in there that i didn't like whenever we look up in the tables what 
the deflection's going to be ha ha the ma-, the M and G come there that gives 
me the mass back and notice the G will always cancel out so that trick of the G 
in the formula above is purely because that's the way you will read it from the 
tables so so again the textbooks tend to quote the formula to you in that form 
but so we 
just plug that Y-S into that formula er it's easy to call it omega-N-squared 
and lose the square root sign er and it just comes out of that form so for a 
simple transverse beam we can relate it to the normal beam bending criteria the 
the second moment of area a-, and so on and so forth very straightforwardly 
okay that's fine if we have just one mass let's consider still a light beam so 
we've still got no no significant mass in the beam itself but i'm now going to 
put more than one mass along it so i've got the beam along here lump of mass 
there lump of mass here 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 of them 
oscillating like that i've got two masses two degrees of freedom two different 
modes of vibration so that all fits into the same sort of spring mass pattern 
that we're seeing before er but the question is how do we go about solving it 
well again formally and i'm not 
going to work an example at this stage but formally if we had two or more 
masses we'd do something like that there are two points along the beam we 
obviously each has a mass M-one we allocate to those a deflection at the mass 
of Y-one Y-two et cetera er and again i think of it as a force a point force at 
each of those mass points and i can again do just what i did before we can work 
out the bending moments in this beam in terms of a static force set F-one F-two 
et cetera now because there are point forces in there you may want to u-, and 
more than one of them you will need to use one of the special notations what 
i've called here a singularity function you might know it as Macauley's 
notation it depends which textbook you've gone to but again from other courses 
you should be quite familiar with the idea of handling these extra forces 
coming in as we take the bending moment along the beam so we can establish a 
bending 
moment through there we can know that each of the forces F-one is M-one Y-one 
omega-N-squared F-two of course just F will be w-, M-two Y-two omega-N-squared 
and so on so we just have a series of forces it's a more complicated version of 
the previous problem but what we actually do is we will set up er this F-one 
equals the deflection Y-one calculated from the bending moment equation in the 
same way and that will give us a set of equations which just the same as they 
would with the spring mass system er on the axial case will give us N unknowns 
where we have N masses Y-one through to Y-N plus omega-N-squared so we're in 
just the same situation as before we have one too many variables and we solve 
for omega-N-squared and the ratio of each of those deflections to say Y-one so 
it's exactly the same slog it out by hand or give it to a computer and ask it 
for the eigenvalues and eigenvectors 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 tedious 
mathematics which gain nothing or give us nothing for the physical 
understanding of the problem there is a question very similar to this example 
er that sketch there on the example sheet we have a class in week nine when 
we'll discuss the results of those and as always of course well the the 
expectation is that you'll have had a go at those problems for yourself and 
found out how much you know about it before we discuss the solution that day so 
i am expecting that you will have a fair background to this in the week nine 
Thursday examples class that i will actually go through the worked example on 
that now just to tell you in advance of that the s-, formal solution that we'll 
look at in the examples class uses this method primarily just so that we revise 
that method but just going off the top of the screen the bit er er there is 
about that's the trick you can use assuming we have a 
simple deflection system we can look it up in tables i hope at least some of 
you have sort of ha-, had already the thought look hang on there's just two 
forces here or might be three in a more complicated one but this is all linear 
elasticity i can look up in a table what would happen if there was only that 
one th-, that's in the data book er that one's in the data book it's really 
just a d-, different numbers into the same problem i can s-, so i can look up 
the solution for each of those single force systems but it's all linear i can 
superpose those two solutions to get just add the deflections if i calculate 
the deflection Y-one and Y-two only for F-one and Y-one and Y-two only for F-
two then for both forces the total deflection at Y-one is the sum of those two 
sort of partial Y-ones and similarly for Y-two so you can get at the system by 
just superposing the looked-up solutions er in in the same way and you 
again then just equate plug these F values in and you get the same set of 
equations obviously the two methods work the same one is doing the formal 
linear theory the other is using tables to look up bits of the theory [sigh] it 
gets to be a bit of a balance it's worth the trick for a single system 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 if you've got three of these i 
reckon there's probably because the superposition means slogging out a lot of 
terms and then adding them together i would have thought if you've got three 
it's probably less actual algebraic slog to do it by the formal method two is 
about a break point it it's it's probably easier to think about what you're 
doing er doing it by superposition it might actually be slightly less work 
doing it by formal methods but they're about equal in terms of the effort it 
takes 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 confidence in using the basic ideas so 
what i'd like to do is is move on to the other case er of vibration which is 
where we again to transverse but we now cannot any longer ignore the mass of 
the beam itself and in fact we're going to go the whole hog and look at only 
one special case this year er we're going to look only at the case where it's 
er a uniform beam of some mass and there is no point masses on it at all so 
we've gone from one extreme only point masses and a light beam to a uniform 
beam and no other masses at all and this is the other one which is a worked 
example so you have a sheet of this again with some of the more text written in 
to just give you a better explanation than is on these slides er and again this 
time i've given it you because the final result is really rather a tricky er 
awkward looking formula so anyway what we have now then is a uniform beam again 
it's going to be vibrating somehow like this through in this way er it's 
uniform density and with any uniform system our normal recourse is to looking 
at an elementary section and then attacking it with calculus so that's what we 
do here let's take a short section D-X along the length of the beam er the beam 
is defined i've doesn't really matter but its length happens to be L but notice 
we need the mass per unit length now we're using the linear density idea that's 
so commonly used through structural mechanics right the transverse acceleration 
we're interested in the motion that way so we're going to have the mass of this 
little element will just be M-D-X okay because it's a linear density this thing 
small-M the vibration is going to be along this way and it's just a Y-double-
dot if you like or D-two-Y D-T-squared but notice that we're going to be 
interested in the spatial distribution of the shape of the beam and in the time 
motion of it so to be strictly 
correct we have to be working in partial derivatives at this stage hence the 
the symbols used here so we have partial D-two-Y D-T-squared for the other case 
you're only er usually for these beam case you're only usually concerned with 
the displacement directly under the mass and so you tend to to go back to a a a 
full derivative there and not worry about the other plane here we have to be 
more precise okay if we've got this acceleration going on force going on there 
now what the beam is seeing er er the acceleration there has of course has a 
certain little force where does that force come from well it comes by er 
elasticity from the next bit of the beam along and all the way out to the 
supports er just as it does in all other bending or deflection problems so what 
i effectively have here is because of the acceleration i have that load 
function now that's no more than saying that if this were a static problem that 
i've got a a a heavy beam hanging between 
supports how is it going to sag well what i say is its density is M per unit 
length therefore i have a force M-G per unit length in other words a a linear 
force function of that sort and you're again you know b-, er in the data book 
rather than having capital W for a point weight you have a s-, lower case W for 
a distributed weight it's no more than that but it's our general acceleration 
of motion not just big-G it's obviously dimen-, er sorry little-G it's not 
dimensionally er any different so we have that as a low density we know from 
basic beam theory that the fourth partial derivative of displacement is the 
load function er the second er you remember the second derivative gave us the 
bending moment and if you think about how you calculate a bending moment from a 
force there are two steps in in that as well so you get to the fourth 
derivative but these again are equations that should be well known to to you 
from other courses so we get this basic equation of motion 
which occurs for all these types of problems er if we have uniform elastic 
properties for E there if we have a uniform cross section er all the way along 
so that I is a constant and if we have uniform mass distribution so M is a 
constant then the thing is fairly straightforward in principle you could solve 
it with those things variable along the length but in general you will find 
that you by classical methods anyway you will not get a solution out under 
those conditions you will have non-linear differential equations you can solve 
them by computers m-, er finite element methods and such things so we're going 
to deal only with this simple case where everything's a constant it's a 
separable partial differential equation er i hope you can remember coming 
across those before er but you probably hoped you would never have to do 
anything very serious with them and that's fine by me er i'm quite happy to 
reduce all the difficulty of solving that down to one 
line it can be solved readily in quotation marks and what i mean by that is 
there's a perfectly standard solution you can look up in a textbook how to do 
it there's nothing difficult it's a just a routine turning the handles job to 
churn out the solution to that it's just a lot of hard work to do it 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 er comes out to be that the displacement Y as a 
function of displ-, er the transverse displacement Y as a function of the 
distance along the beam X and time comes out to be this amazing thing here 
looking at the far end it's got sine-omega-T-plus-phi type shape there which is 
what we always expect to see in these but whereas before we just sort of sad 
had sort of A-sine-omega-T for the single spring mass we now have something 
which has got four constants that's no surprise we've got a fourth derivative 
there but it's got a sine term a 
cos term the hyperbolic sine a hyperbolic cosine term there all those four 
terms can crop up all the things we can do with exponential functions in other 
words er come up in this equation and in general the solution is like that 
we've stuck an alpha in here just to save work because alpha comes out to be 
the whole of all the Ms and omega-squareds and everything else E-Is all go in 
there and all transverse uniform beams in vibration satisfy that formula we 
simplify it when we know the N conditions we know whether the Ns are fixed or 
pinned er we know whether er whether whether we hit it with a hammer in the 
middle to get it going various forcing functions so by using boundary and 
initial conditions we can simplify down for er real problems and find that not 
usually all these terms exist some of those Cs will go to zero but that's 
really what what we're dealing with on that level and i think you can see that 
although er a sort of er an understanding of the principles of what's going on 
is important at this stage er 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 so anyway just working on this is still part of the stuff 
you've got in the notes let's just take a specific example which picks up on 
that formula on the previous page if we take a simply supported beam so we've 
got a a beam across here it's just got a simple support at each end in other 
words one that can't it holds position doesn't transmit a moment and we then 
know that there will be zero N deflection and zero bending moments er at both 
ends of the beam underneath each support that's that's what defines what we 
mean by simply support s-, supported beam so when we plug those numbers in so 
you just slog 
through that horrible formula we had last time work out the the appropriate put 
in er er X-equals-nought er set Y-equals-nought and see what the constants come 
out and so on and so forth then it turns out that that one comes out to be like 
that only the sine term only one of those four C C-one to C-four things is 
actually non-zero for that classic case if you have a cantilever you'll find 
you will finish up with two of the terms and there'll be a different two and so 
on and so forth but we just hammer it through in that way now the phi there 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 in the very simple spring mass so that's the one that's worked from 
the time initial condition er what we get in here notice is we do have a 
subtlety that's come up in this formula notice that it's not just sine-pi-X or 
something like that 
inside there it's sine-N- pi-X where N is an integer 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 well it goes to zero at an angle of zero at an angle of pi at 
an angle of two-pi at an angle of three-pi and so on and so forth so all of 
those are possible solutions to how our string's vibrating or our beam is 
vibrating there's no difference between a string vibration and a beam vibration 
in the sense we're talking about them so we get a series of solutions here if 
we look at N-equals-one in here we've got time varying behaviour but we've got 
spatially varying behaviour from that term and it's saying that i-, if N is one 
then the thing is a half sine wave like that in other words it's zero at both 
ends and it's moving through in that sort of motion at any point along the beam 
it's going at a speed omega-N from that time sensitive te-, sinusoidal term but 
the distribution is like that 
if i go up to N-equals-two i get something that has the classical sine wave 
shape a a node in the middle it's still any point this way it's still going at 
exactly the same speed of oscillation for the relevant solution for omega-N and 
so for each of those modes here as we go up we're getting more turning points 
so we're getting a more difficult e-, energetic shape to drive we need a higher 
omega-N to cope with that so we know that we're going to get in theory an 
infinite number of these now because there's an infinite number of solutions to 
that equation and we're having an infinite number of modes of increasing 
frequency now what's the practical side of that well what we know in practice 
is that it's more difficult to drive the high modes this one goes fairly easily 
er it doesn't take a lot of energy to do that to do it with a full wave 
vibration takes more energy to do one with lots and lots of little kinks up and 
down 
think of just the the bending energy you have to put in to make the beam go 
into that shape and you can see it's going to be just harder to get it going 
like that so in practice beams rarely show more than with any non-negligible am-
, amplitude more than one or two of these low order modes it's fairly rare to g-
, get something with a very complicated pattern so we normally finish up with a 
a fairly simple practical problem er you can more or less guarantee for for all 
realistic conditions that this 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 
and so on and so forth and you can argue that purely on the this er notion of 
the energy that's involved so if you need to do the formal theory you can slog 
it through if you have ac-, access to a textbook and you can remember that 
there is that horrible formula with the cos cosh sine and sinh terms in it you 
can look that 
up in the textbook and just do the boundary conditions 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 and we could then do a guess type solution rather than 
do it properly knowing which of these mode shapes and knowing the mode shapes 
we can sketch out the shape not the amplitude but we can sketch out the shapes 
purely by a sort of physical intuition we could then think in terms of using 
approximate methods that can sort of home in on the one that we want and there 
are some very sophisticated a-, and very useful techniques that that are widely 
used for doing that but they going into those and doing those properly is 
really er er if you like a topic for the third year so i'm going to leave it 
the beam example by saying look that's the basic theory that's how you would 
get at the solution if you needed to and you ought to be aware of that but the 
basic behaviour is this series of 
omega-Ns and when you actually plug in real boundary conditions to the 
complicated formula life's fairly straightforward but er i mean there was no 
sense in which we expect you to memorize that formula for exam conditions or 
anything of that sort it it's just not the sort of thing that's sensible to do 
you look it up in a textbook if you want it so we do use approximate methods er 
and er again when you look at the er example sheets if you haven't done so yet 
you'll find one of them has got an asteri-, one of the questions has got an 
asterisk against it because it's a a sort of an extra one to do if you're 
feeling er enthusiastic er it is a a case of looking at one of the earlier 
problems and looking how we might use approximate methods that is you know to 
use an approximate method you usually are iterating and trying to guess roughly 
what the right answer is 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 a little bit of brain 
power to get a good guess er and there's one example that shows you that but 
what i'd like to spend the last few minutes today i-, is just hinting at one w-,
how one or two of those er methods are developed er not b-, so much to show 
you the en-, enough to get you into next year's course but just to show how we 
can use energy methods alongside these other approaches you will notice that 
regularly i've given you a formal classical solution in Newton's laws and then 
said think about the energy and it will it will help you to see why this 
pattern of sha-, of modes and nodes and so on goes together well if that's the 
case then presumably there there could be good reasons for using energy more 
formally as the solution method and i and i think it's just worth seeing this 
now this first little bit it it's nothing very complicated it's the sort of 
stuff that is in the first year 
texts so you know if again if you look at the reading list you'll find that the 
reading associated with this particular little section of the course is in 
things like Hibbler and Mariam as well as the the main texts for for this er 
particular session i think all we'll do here is just look at one example of it 
so let's just consider free vibration of a single spring mass system to stay 
low l-, let's let's not do anything very complicated 'cause it's only the the 
idea that you just are aware how the methods work if we consider it as a free 
vibration then 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 without any s-
, significant influence from the outside world and if that is true then the 
mechanical energy within that system must be conserved now what mechanical 
energy do we have in a vibration problem again this is where we were 
discussing this time last week we've got strain energy or spring potential 
energy if you like and we've got kinetic energy stored in the the the inertial 
part of it so we're saying that again let's take the simple case because it's 
thermodynamically isolated in its most simple case then we've ignored friction 
because that would generate heat out of the kinetic energy and we don't want 
that 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 that some of 
them must be a constant so just flogging that through if the motion is a sine 
wave as we're assuming everywhere else X equals A-sine-omega-T its maximum 
displacement obviously is A and when is that maximum displacem-, er going to 
occur well it's going to occur at the ends of the motion that is where the 
thing stops and is about to turn round so we'll get the maximum displacement A 
at the points where the 
speed is zero when's it going fastest well it's obviously going fastest you see 
it's stopped here it's come through here stopped at the other end and it's 
clearly fastest when it's in the middle when there's no displacement so we'll 
get the maximum speed is A-omega we've been using A-omega-squared for the 
acceleration a lot so that should be totally familiar to you er the speed will 
be A-omega at displacement zero that statement i think you can see purely by 
physical intuition but if you want to be formal about it then just put 
differentiate this with respect to time and have X-dot equals A-omega-si cos-
omega-T and you know do do the formal maxima and minima and you will you can 
prove it like that but er but i think the physical intuition should be all you 
need to do that now this is useful because strain energy is to do with 
displacement kinetic energy is to do only with speed now it happens that er 
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 so the maximum strain energy and the maximum 
kinetic energy must actually be the same value and we can either sum them 
formally er for all values or we can just say look just look at the maximum 
strain look at the maximum speed from that we get strain energy kinetic energy 
and those two maximum values must be the same so we actually have two slightly 
different things we can do we can use it in that strict formulation or we can 
use it in in this form and in given situations thinking out about it one way or 
the other tends to be a little easier so very simple example let's go right 
back to the very first example we did as a revision at at the start of this 
section single mass single spring K er and let's look at what the strain energy 
is well the maximum strain energy is when we have the thing extended A beyond 
its natural length and is a half-K times the displacement squared that's the 
straightforward spring Hooke's law energy formula the 
other side we have the maximum kinetic energy which is half-M-V-squared and 
here is is the maximum V we want which is A-omega from the line before so we 
get half-M A-omega-all-squared reorganize that remembering that we're looking 
for the natural frequencies so putting the subscript- N back in there omega-N-
squared equals K-upon-N er and of it course it would be a terrible shock if 
anything other than that result came out since it's one that we're so familiar 
with but just notice that that works you know drops it straight out in one line 
under those conditions and it will do equivalent things in more complicated 
situations so here for this year i don't want to go any further along that 
particular little bit of a line than to do that and just notice that for that 
simple case if you like that method there's no reason why you shouldn't use it 
for some more complicated cases er er of complex motions we can set up special 
coordinate 
systems we can use these sort of tricks including the one that says let's 
formally differentiate the energy equation and say if it's conserved then the 
the change with time must be zero which is the the sort of one just going off 
the screen at the top and if you do that then you get to a whole series of 
techniques going under the name of Lagrangian mechanics er and later on even if 
you're really desperate Hamiltonian mechanics but they're a whole series of 
very powerful analytical techniques whose underpinning is this very simple 
example er and all i want to do today is say look if you come across these 
formally next year er you should go away with enough confidence if you choose a 
set of options by which you don't get involved with those courses but then 
later i-, early in your professional career do come across people throwing 
Lagrangians at you and things like like that and expecting you to understand it 
don't panic with a bit of luck just a 
little reminder will say you know that's how you start it's not too much of a 
problem given that you need to do it to go and look in a textbook and just 
build it up from the sorts of things that we've been looking at okay so i think 
that's all we'll do there can i show you one other slide it's in the note er in 
the er copies of the er slides in the library and i don't really want to talk 
about it in any detail at all it won't come up on an exam paper or anything of 
that sort it is purely information for you for the future depending on which 
route you might choose to go through and that is just to point out that 
alongside the energy methods that there are this trick of estimating where you 
are and using approximate processes er goes very closely that you can use 
approximations in all sorts of ways but Rayleigh er in p-, developed a whole 
series of techniques which use a mixture of energy and an approximate guess 
at what's going on and you get this sort of pattern of behaviour er we can very 
often in real vibration problems if we know omega-N that's good enough we n-, 
we know that the initial the actual ax-, amplitude of the vibration depends 
just how we kick it so omega-N's usually the thing we're worried about if we 
can guess say for a vo-, oscillating beam or something like that the shape in 
which it's oscillating reasonably precisely we can estimate at its maximum 
deflection by standard beam theory say what the strain energy would be in that 
it's a formulae you can look up so providing we get just a a reasonable guess 
of that shape that would give us a maximum strain energy we could then e-, work 
out a maximum kinetic energy er as a sort of M-omega-squared sort of term and 
plug in those together it gives us an estimate of omega-squared and that's 
really what it says on here we we can guess at the shape er what Rayleigh 
discovered and what makes the method 
powerful is that really quite crude guesses of that shape get you surprisingly 
close to the right answer and not only that 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 er the conditions that work so it's a nice safe 
predictable approximate technique normally the way you find the shape is just 
to say okay as i did in that simple example we use a more sophisticated version 
of that here's a beam vibrating how how would it sag under gravity that's a 
good guess of the shape it will vibrate in for its lowest mode of behaviour so 
let's use that as the starting point for the system you know so so you use sort 
of common sense guesses intuition whatever you like to call it to deal 
with that but some of you undoubtedly will later on in your courses come across 
a formal treatment of Rayleigh's method others of you perhaps more on the civil 
side may not deal with it directly don't get involved perhaps with earthquake 
engineering team at some stage and find other people using it at least you've 
heard the words to get you into the field and that's all i'm concerned with 
here just that you've heard the words it's not something that's going to get 
examined but i but i thought it was worth spending just a couple of minutes at 
the end of this session talking about it so that's it for today tomorrow we'll 
meet up again and we're back on examined material like so it is stuff that's 
fair game we will look at vibration isolation as a subtopic er wi-, within the 
overall vibration picture