nm0855: at the end of the last lecture [1.1] i'd really set out for you the the 
basic [0.2] ideas of multimode vibration and promised to revise that with [0.2] 
an example [0.4] er [0.4] a-, and that's what this is [1.9] this is [0.2] in 
the [0.2] handout that you had at the start of this section of the course it's 
one of the the two [0.2] worked example sheets in there [0.7] er [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 [0.5] so there 
shouldn't be any need to do more than just have a quick look through this [1.8] 
and it it's more or less the same as we set up in the simple example last week 
[0.3] except that there are now three masses [1.9] four springs [0.2] two rigid 
walls [1.5] er [1.0] in the notes that you have there are r-, [0.2] rather more 
words describing what goes on between this so this is just an edited down 
version of the sheet that you have [1.9] but note the basic principle as before 
[0.4] is that for each mass in turn M-one M-two [0.2] 
et cetera [0.6] we set up [0.4] the Newton's equation of motion [0.9] we have 
[0.2] force [1.6] generated by springs so we have terms like K-two A-two-minus-
A-one [0.3] K-one [0.3] A-one-minus-zero [0.2] in this case [0.5] and that's 
going to equal an acceleration [0.5] because it's a sinusoidal motion [0.3] the 
acceleration becomes just omega-squared [0.5] times the [0.4] amplitude so we 
get [1.1] mass times acceleration as a force M-one A-one omega-N-squared equals 
[0.8] spring constant times the deflection [0.3] spring constant times the 
deflection [0.4] as we go along [0.6] the the chain of events [0.5] so that's 
how we set it up basically and we do that with a separate equation [0.5] for 
each of the masses M-one [0.4] M-two [0.3] and [0.3] M-three just reduced to et 
cetera [0.7] [cough] er excuse me [0.2] on this slide [2.5] the only thing to 
point out about those two again just to reinforce what i said last time [1.0] 
whether you [0.4] do it just the same as this example doesn't matter but do be 
systematic about what you do [0.4] and something that i think works well is to 
start always with the highest [0.7] subscripts [0.5] 
to the left-hand side as you go through so i recommend that [0.4] given we're 
looking at M-one and we'd interested in the spring either side [0.3] start with 
K-two [0.3] write down its extension as A-two-minus-A-one [0.6] then go to K-
one [0.2] and have A-one-minus [0.2] and here [0.4] it's a wall i've written in 
the zero [0.3] formally just to stress that that's a wall that doesn't move at 
that end [0.3] er but obviously you could just write K-one A-one [0.2] if you 
wish [0.4] 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 [0.5] if you systematize it like 
that [0.2] then the signs come out in that nice pattern of [0.4] brackets with 
a negative number inside [0.2] minus the other one with brackets with the 
negative number inside [0.3] and you're less likely just to make the little 
slips [0.4] that that can upset these sort of procedures [1.0] so [0.8] that's 
each equation [0.4] and normally [0.4] we would [0.2] for larger problems then 
put this into [1.5] a matrix notation omega-N-squared is a constant of this so 
we can just 
have [0.6] some set of A [0.2] amplitude vector A [0.9] is some sort of 
stiffness matrix here [0.4] multiplied by the same [0.3] vector [0.2] A [1.2] 
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 [0.2] outside and the Ks inside [0.
5] and you can get this form [0.5] but notice that it's convenient [1.1] to 
divide through in each row by M-one by M-two and so on so that appears [0.5] in 
the denominator of each term inside there [0.7] now that's [0.4] very typically 
of how [1.1] books present it [0.5] you might argue that you're used to seeing 
K [0.2] as the stiffness [0.3] so using K for this stiffness divided by math-, 
mass [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 [0.3] it is a sort of stiffness matrix [0.
4] but it's modified by those masses [1.3] if you get it to that form [0.4] 
then by far the best approach for larger problems [0.3] is then to give it to a 
solver [0.3] and ask for the eigenvalues [0.6] which correspond to the [0.3] 
natural frequencies [0.2] 
squared the omega-N-squareds [0.6] and [0.3] the eigenvectors which give you 
the mode shapes [0.9] if you don't fancy doing that or it's a problem you're 
doing by hand [0.4] then [0.4] you're probably better off just solving this as 
a set of simultaneous equations [0.3] at that level rather than worrying about 
the matrix formality [0.4] the matrices do help on big problems they're not 
essential [1.0] so [0.4] if you do it [0.4] in normal ways you solve for omega-
N [0.2] and the ratio [0.4] of the amplitudes [0.3] to [0.8] say the amplitude 
A-one remember that we always have [0.3] one too many variables we have the 
natural frequency [0.4] plus [0.2] N [0.2] amplitudes [1.5] so that's how you 
solve for that and it's not worth working the details out on it [1.1] if you 
use the eigenvalue eigenvector [0.2] formally through something like Matlab or 
a similar program [0.5] you get the answers directly [0.9] you can of course [0.
7] solve the mode shapes directly as well from the simultaneous equations [0.3] 
but once you have the resonant frequencies the omega-Ns [0.3] you can in fact 
guess what's going on [0.5] and 
that's what i've done with th-, with the arrows i've drawn under the diagram 
there [0.6] they correspond to the three possible motions [0.3] all three could 
be going the same way in phase [0.6] er with the same amplitude [0.5] oh er 
sorry with different amplitudes but in the same [0.3] er phase relationship [0.
5] or we could have the two [1.4] outside ones going the same way [0.2] and the 
other one coming the other way [0.5] er [0.2] which is the bottom one or in 
between those where [0.2] two of them are moving in one way and one the other 
[0.6] 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 [0.4] 
together [0.7] er [0.3] now i i've written [0.6] from this the top downwards in 
the way they are because that corresponds [0.3] to the number of changes of 
direction if you look at that the top one has no change of direction [0.4] the 
next one down has one change [0.4] the one below that [0.4] has two changes of 
direction [0.9] because the change of direction corresponds to there being a 
node [0.4] between [0.7] 
those two displacements [0.5] and as again we discussed last week [0.3] we know 
that ev-, [0.3] the more nodes you have [1.3] the nodes the higher mode shape 
that you have [0.3] corresponds to increasingly high resonant frequencies 
that's the energy link between them [0.4] so once you've got the omega-Ns [0.4] 
just by inspection you can work out [0.2] the general shape of the modes [0.4] 
but you need to so-, and which [0.2] goes with which omega [0.3] but you then 
need to solve formally to get the actual ratio [0.6] of these amplitudes the 
sketching can't tell you that [1.8] so that's the basic example that's really 
all there is to [0.2] most of er [0.8] er [0.3] axial vibration and again let 
just put flash that slide up no [0.2] nothing to copy down there it's just a 
quick [0.5] look at that just s-, sketched out just before i came down today [0.
3] and notice that if it were a torsional problem [0.2] it's exactly the same 
we'd have shafts of some stiffness [0.4] there'd be moments of inertia for the 
rotors [0.2] and a theta [0.3] displacement for each one you just plug in the 
form-, the symbols into the w-, [0.2] example over there [0.9] if you have say 
a building frame a steel framed building with sort of uprights [0.3] and fairly 
some concrete floors something mairly fairly massive there [0.6] then you may 
have a sway system by which you know experimentally [0.5] the values of these 
stiffnesses of these different [0.2] storeys [0.3] and you've just got 
effectively the slab of mass [0.2] and again [0.3] that is the same problem as 
that this just corresponds to the spring experimentally [0.2] this is the mass 
[0.4] and so on and so forth [0.4] so all these problems reduce [0.2] whatever 
variables they're in [0.5] to [0.7] that piece of sort of handle turning sort 
of [0.4] er mathematical analysis 
nm0855: th-, what i'd like to do on though i-, is not sort of work more 
complicated examples [0.3] er [0.6] but move on to [0.3] the slightly different 
conditions [0.7] er [0.3] hinted at by that little frame i just showed on the 
other s-, [0.2] 
projector [0.8] er [0.8] if we know experimentally what the effective lateral 
stiffness of the [0.5] building is [0.2] fine we can use that simple method [1.
2] if we don't and we have to work with beams transverse vibration so we have a 
beam that's vibrating in some [0.2] motion of that sort [0.8] then there are 
one or two tricks which are just a little bit different [0.2] and you'll find 
that nearly all the textbooks do treat them [0.2] as a separate case [0.3] as 
i'm doing in th-, er er in these lectures [1.8] so let's just very briefly [0.
3] look at beam vibration and it's not a thing i want to spend a lot of detail 
[0.4] the the theory here [0.4] again [1.0] is mathematically [0.4] tricky [0.
2] just thumping out the solutions from the basic information er er er is [0.8] 
is a slog it's easy to make mistakes [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 [0.5] that will [0.3] get you to the level by which you can then look up 
solutions with a degree of confidence [0.4] er when you need them [0.5] so [0.
3] we'll deal with beams 
and just to remind you what we mean by a beam [0.3] it's something which is 
long and thin [0.4] er so we've we've got it between supports there's a 
structure across here [0.6] er whatever the depth of that beam is it's small 
compared to its [0.5] its length [0.4] and when we do that we can ignore [0.4] 
generally speaking sheer [0.6] effects and concentrate [0.4] only [0.3] on [0.
2] beam bending and then we can also use classical [0.4] beam bending theory 
that you met last year [0.4] and have met [0.4] earlier in this course in the 
bit that namex [0.2] was dealing with [1.0] [1.4] okay well there's different 
things we might want to do let's start with the very simplest case [0.7] oops 
[2.0] i'm just drawing it [0.3] out like that [0.7] we have [0.2] a light beam 
[1.0] er always the easier case on perfect simple supports as shown there but 
it doesn't [0.2] the end conditions we'll come back to [0.5] a light beam in 
other words we can neglect its mass [0.2] with a single point lumped mass [0.4] 
stuck somewhere [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 [0.3] that means it has to be [0.4] in the centre it can be 
anywhere along the beam [1.8] and this might correspond to the case where we 
have a structural [0.6] I-beam or something like that [0.4] er [0.7] which has 
got [0.6] something maybe ten times the weight of that mounted near the centre 
of it perhaps a machine [0.3] that could be vibrating and may cause the beam to 
vibrate [1.8] so that's the scenario we set up [0.3] and [1.4] generally 
speaking when when you can set up an a [0.9] a coordinate frame [0.3] and if i 
just replace this mass [0.3] by a force and just had [0.2] said there is a 
point force pushing downwards [0.3] on there [0.3] what is the shape of the 
beam how does the beam deflect [0.5] then [0.4] as far as i'm concerned [0.6] 
er that is no problem for you to solve you may worry about it but [0.6] at this 
stage the assumption of this course [0.4] is that you're [0.3] comfortable 
solving problems like that [0.5] so [0.6] we put a force on there [0.9] er how 
would you do it well there's two things you could do [0.4] er [0.2] let's deal 
with the formal one first [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 [0.4] and then we know that the 
deflection shape [0.9] or the second derivative of the of the er [0.3] 
deflection shape [0.4] relates functionally [0.2] directly to that bending 
moment [1.5] and we get the shape out [0.7] okay [0.2] what do we do if it's 
vibrating [0.4] well notice what we've done with everything else i've said i've 
[0.6] got a force from the springs [2.3] i'm now going to generate a force [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 [0.2] an inertia 
force it's the force M-A [0.2] associated with the acceleration [0.6] so all i 
need to do is substitute that inertia force in terms of the fixed F [0.5] that 
you would have had on a a an earlier example [0.3] a-, and we're away we do the 
same thing [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 if you like [0.4] that that [0.2] 
tells you [0.4] that thing [0.9] the inertia force then if we're moving 
transverse vibrations i er tha-, tha-, [0.2] again the mass is i say where my 
hands are joined it is going through like that [0.8] what we have [1.1] is [0.
7] an inertia force mass of the lump thing Y [0.7] is its deflection [0.8] er 
so Y-double-dot would be its acceleration so we've sort of got Y- [0.5] double-
dot or Y D-, [0.2] er sorry [0.2] M-double-dot [0.2] Y-double-dot or M [0.4] D-
two-Y D-ec [0.3] T-squared if you prefer to call it that [0.6] er we have [1.2] 
that form [0.3] again we know that Y is going to be a sinusoidal motion that's 
the assumption of all our vibration systems [0.3] so the Y-double-dot term can 
be replaced by omega-squared-Y [0.3] again just plugging it in the derivative 
[0.7] er approach we discussed a couple of lectures back [0.6] so the inertia 
force then [0.4] has [0.6] that as a a peak value [0.3] er it it's oscillating 
sinusoidally but that is [0.2] perfectly adequate [0.3] amount of data [1.0] 
the bending moment as always is just this formula you should be very familiar 
with [0.4] bending moment it's is equivalent to [0.3] E-I [0.9] D-two-Y D-X-
squared [0.2] 
in other words the the second spatial derivative [0.3] the curvature of the 
beam [0.2] locally [0.6] where E is Young's modulus I is the second moment of 
area of the cross section perfectly normal [0.3] standard beam theory [1.5] and 
[0.2] all you do is equate that [0.2] to that [0.6] er you then need to 
integrate twice with respect to Y [0.5] and you can solve for [0.7] er omega-N-
squared and and for Y from from that [0.4] formula [0.9] so that's formally how 
you go about doing the problem and [0.6] it's [0.3] 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 [0.3] and that you can plug them in to the 
beam bending formula [0.5] that you've been using for a long while [0.9] 
however [0.2] it it is often [0.2] quite sort of a tricky sort of thing to do 
[0.5] and one of the big advantages we have [0.5] is that a lot of the beam [0.
9] structural beam bending formulae and so on [0.3] are tabulated for us [0.3] 
so let can we find a short cut trick that will help and i think that's 
something that's [0.4] of more immediate 
interest [3.0] [cough] [0.2] excuse me again [2.7] so [0.3] an easier way for 
this problem [2.0] always we are dealing with conditions which are linearly 
elastic and we're er let me s-, we said this [0.3] at the start of this section 
of the course but let's repeat it again [0.3] once we lose linear elasticity 
none of the stuff that i'm talking about is going to work properly [0.3] so [0.
2] that that [0.4] is just available to us as a [0.3] a a prerequisite [0.8] if 
it's linear elastic [0.2] the beam [0.2] just acts as a simple spring i've got 
this thing here if i push it downwards [0.5] against its own elastic [0.5] 
bending properties it'll try and push me back up again so it just acts as a 
spring [1.8] 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 [0.4] 
how much force do i need to deflect a certain amount [0.2] i could actually [0.
2] calculate that from first principles [0.3] using the sort of formulae [0.9] 
little bit higher up on that 
page [0.3] or maybe i can do that experimentally i can just s-, put [0.3] 
balance a weight on there and see how much it deflects [0.4] to get an 
equivalent stiffness [1.2] now we know that this is only a single mass and and 
a sort of spring system here because the the beam is light [0.3] so we know 
from [0.2] the elementary theory [0.3] that the natural frequency will be just 
square root [0.3] of [0.5] stiffness over mass [0.2] that is here lambda-over-M 
it was square root of K-over-M [0.3] for the or or helical spring [0.3] example 
we looked at earlier [2.0] so all we need to do now is [0.4] see whether we 
know what er lambda [0.3] is [0.5] we know what the mass is i assume er er in 
all of these problems [0.8] er [0.2] well [0.2] er we could be ex-, 
experimental [0.4] but we can actually get at it [1.0] by some other methods so 
if i can just give you that bit as an extension to the bottom of that sheet [2.
8] what happens if we do just [0.2] as i said experimentally [0.2] sort of 
balance a mass on the beam and see how much it deflects [0.3] well that gives 
us lambda [0.2] and it gives it as from 
that thing the static deflection [1.0] is the force M-G now because we're 
looking at how much it would deflect [0.7] er as a horizontal beam [0.6] 
divided by this same stiffness it's the same thing that causes both effects [0.
6] so experimentally that's how i get at lambda [0.4] but i can also get at it 
because i can i can look up [0.6] that static deflection [0.4] for [1.0] a 
weight [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 [0.8] er [0.2] so what we do actually [0.2] if i can now take that one 
off [2.6] is [1.2] okay let er let me just put it back a second in fact [0.3] 
er [0.6] if we just look at this formula here [0.5] and just plug in [0.6] to 
get rid of lambda there which is the sort of unknown and we don't really want 
to know explicitly [0.5] if i just substitute from this formula in there [0.2] 
then i finish up with this rather magic looking formula [0.5] 
that [1.6] the natural frequency is the square root of G [0.5] upon [0.3] the 
static deflection [2.4] okay [0.4] simple enough formula but it worries quite a 
lot of people and with some reason [1.7] all the while i've been saying to you 
with this stuff that the the vibration [0.3] does not depend at all [0.7] on 
the [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 [0.4] other things may 
change but that stays the same [0.6] so [0.5] why does G appear in this formula 
[1.4] er and the answer is very straightforwardly [0.2] that it's [0.2] because 
[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 [0.5] er in going through it because 
if we now look at a simple example [2.8] so just a very simple example of 
simply supported beam which is what [0.2] i'd actually drawn on the previous 
sheet [cough] [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 [0.9] 
force [0.5] type condition [0.5] and you will find that if you have a simply 
supported beam [0.3] with a force [0.8] W [0.3] a weight W in the centre of it 
and the deflection at under that weight is W-L-cubed [0.5] over forty-eight-E-I 
[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 [1.0] now here the weight is 
simply just M-G [0.5] and i think you can see what's happening now is that 
although i've [0.2] i've lost apparently the mass and have gained a G in there 
that i didn't like [0.4] whenever we look up in the tables [0.7] what the 
deflection's going to be [0.3] ha ha the ma-, the M and G come there that gives 
me the mass back [0.5] and notice the G will always cancel out [0.4] so that 
trick of the G in the formula [0.3] above [0.3] is purely because that's the 
way you will read it from the tables so [0.2] so again the textbooks tend to 
quote the formula to you [0.7] in that form [0.8] but so we 
just plug that Y-S into that formula [0.5] er it's easy to call it omega-N-
squared and lose the square root sign [0.4] er and it just comes out [0.3] of 
that form [0.4] so for a simple [0.2] transverse beam [0.2] we can relate it to 
the normal beam bending [0.4] criteria the the [0.3] second moment of area [0.
4] a-, and so on and so forth [0.4] very straightforwardly [2.3] okay that's 
fine if we have just one mass [0.3] let's [0.3] consider [0.4] still [1.2] a 
light beam [0.2] so we've still got no no significant mass in the beam itself 
[0.4] but i'm now going to put more than one mass along it so i've got the beam 
along here [0.3] lump of mass there lump of mass here [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 [0.7] of them oscillating like that [0.5] i've got 
two masses [0.2] two degrees of freedom [0.3] two different modes of vibration 
so that all fits into the same sort of spring mass pattern that we're seeing 
before [0.7] er but the question is how do we go about solving it [1.2] well 
again formally [0.2] and i'm not 
going to work an example [1.4] at this stage [0.3] but formally if we had two 
or more masses [0.6] we'd do something like that [3.6] there are two points 
along the beam [0.4] we [0.6] obviously each has a mass M-one [0.3] we allocate 
to those a deflection [0.3] at the mass of Y-one Y-two [0.2] et cetera [0.8] er 
[1.1] and again i think of it as a force [1.3] a point force at each of those 
mass points [0.2] and i can again do just what i did before [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 [0.7] now because there are point forces in there [0.2] you may want 
to u-, and more than one of them [0.2] you will need to use one of the special 
notations [0.3] what i've called here a singularity function [0.3] you might 
know it as Macauley's notation it depends which [0.5] textbook you've gone to 
but again from other courses [0.3] you should be quite familiar with the idea 
[0.4] of handling these extra forces coming in [0.2] as we take the bending 
moment [0.3] along the beam [0.8] so we can establish a bending 
moment [0.2] through there [0.8] we can [0.8] know that each of the forces F-
one is M-one Y-one omega-N-squared [0.2] F-two of course just [0.2] F [0.2] 
will be w-, M-two [0.2] Y-two [0.2] omega-N-squared and so on [0.2] so we just 
have a series of forces [0.3] it's a more complicated version of the previous 
problem [0.6] but what we actually do is we will set up [0.6] er [2.0] this F-
one [0.3] equals [0.2] the deflection [0.6] Y-one calculated from the bending 
moment [0.3] equation in the same way [0.7] and that will give us a set of 
equations [0.3] which just the same as they would with the spring mass system 
[1.3] er on the axial case [0.2] will give us [2.3] N unknowns where we have N 
masses [0.3] Y-one [0.2] through to Y-N [0.3] plus omega-N-squared so we're in 
just the same situation as before [0.4] we have one too many variables and we 
solve [0.9] for omega-N-squared [0.4] and the ratio of each of those 
deflections to [0.2] say [0.5] Y-one [0.2] so it's exactly the same [0.4] slog 
it out by hand or give it to a computer and ask it for the eigenvalues [0.7] 
and eigenvectors [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 [0.7] 
tedious mathematics which gain nothing [0.2] or give us nothing for the 
physical understanding of the problem [0.5] there is a question very similar to 
this example [0.7] er [0.2] that sketch there [0.3] on the example sheet [0.4] 
we have a class in week nine when we'll discuss [0.3] the results of those [0.
3] and as always of course well the the expectation is that you'll have had a 
go at those problems for yourself [0.4] and found out how much you know about 
it [0.4] before we discuss the solution [0.3] that day [0.4] so i am expecting 
[0.3] that you will have a fair background to this in the [0.2] week nine 
Thursday examples class [0.2] that i will actually go through [0.6] the worked 
example on that [1.2] now [1.1] just to tell you in advance of that [0.4] the s-
, formal solution that we'll look at in the examples class [0.3] uses this 
method primarily just so that we [0.8] revise that method [0.9] but [1.2] just 
going off the top of the screen the bit er er there is about [0.2] that's the 
trick you can use [0.2] assuming we have a 
simple deflection system [0.2] we can look it up in tables [0.8] i hope at 
least some of you have [0.2] 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 [0.6] 
but this is all linear elasticity [1.0] i can look up in a table what would 
happen if there was only [0.5] that one [0.3] th-, that's in the data book [0.
6] er [1.0] that one's in the data book it's really just a d-, [0.2] different 
numbers into the same problem [0.3] i can s-, [0.4] so i can look up the 
solution for each of those single force systems [0.9] but it's all linear [0.2] 
i can superpose those two solutions to get [0.4] just add the deflections if i 
calculate the deflection Y-one and Y-two [0.3] only for F-one [0.5] and Y-one 
and Y-two only for F-two [0.5] then [0.2] for both forces [0.4] the total 
deflection at Y-one is the sum of those [0.8] two sort of [0.4] partial Y-ones 
[0.2] and similarly for Y-two [0.3] so you can [0.4] get at the system by just 
superposing [0.3] the looked-up solutions er in in the same way [0.2] and you 
again [0.2] then just equate [0.4] plug [0.3] these F values in [0.3] and you 
get the same set of equations [0.2] obviously [0.4] the two methods work the 
same one is doing the formal linear theory [0.3] the other is using tables to 
look up bits of the theory [1.4] [sigh] [0.4] it gets to be a bit of a balance 
it's worth the trick [0.4] for a single system [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 [0.7] if you've got three of these [0.9] i reckon 
there's probably because the superposition means slogging out a lot of terms 
and then adding them together [0.3] i would have thought if you've got three [0.
3] it's probably less actual algebraic slog [0.4] to do it by the formal method 
[0.5] two is about a break point [0.8] it it's [1.5] it's probably easier to 
think about what you're doing [0.5] er [0.2] doing it by superposition [0.2] it 
might actually be slightly less work doing it by formal methods but they're 
about equal in terms of the effort [0.3] it takes [2.6] 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 [0.2] confidence in using the basic 
ideas [1.0] so what i'd like to do is is move on to [0.2] the other case er of 
vibration which is [0.4] where we again to transverse [0.4] but we now [0.5] 
cannot any longer ignore [0.7] the mass [0.6] of the beam itself [0.7] and in 
fact we're going to go the whole hog and look at only one special case this 
year [0.5] er [1.0] we're going to look only at the case where it's er a 
uniform beam [0.4] of some [0.2] mass and there is no point masses on it at all 
so we've gone from one extreme only point masses [0.5] and a light beam [0.3] 
to a uniform beam and no other masses at all [0.5] and this is the other one 
which is a worked example [0.3] so you have a sheet of this again with some of 
the more text written in to just give you a better [0.5] explanation than is on 
these slides [1.7] er and again this time i've given it you because the final 
result [1.0] is really rather a tricky er [0.2] awkward [0.4] looking formula 
[0.9] so [0.4] anyway [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 [0.4] er [0.2] 
it's uniform density [0.3] and with any uniform system [0.3] our normal 
recourse is to looking at an elementary section [0.3] and then attacking it 
with calculus [0.4] so that's what we do here let's take a short section [0.2] 
D-X [0.3] along the length of the beam [1.1] er the beam is defined i've [0.5] 
doesn't really matter but its length happens to be L [0.2] but notice we need 
the mass per unit length now we're using the linear density idea that's so 
commonly used [0.3] through structural mechanics [3.9] right the transverse 
acceleration we're interested in the motion that way [0.3] so we're going to 
have the mass of this little element will just be M-D-X [0.3] okay because it's 
a linear density this thing small-M [2.0] the vibration is going to be along 
this way [0.2] and it's just a Y-double-dot if you like [0.2] or D-two-Y D-T-
squared [0.3] but notice that we're going to be interested in the spatial 
distribution [0.4] of the shape of the beam [0.3] and in the time [0.3] motion 
of it [0.4] so to be strictly 
correct [0.4] we have to be working in partial derivatives at this stage hence 
the [0.3] the symbols used here [0.2] so we have partial D-two-Y [0.4] D-T-
squared [2.6] for the other case you're only er usually [0.4] for these beam 
case you're only usually concerned [0.3] with the displacement directly under 
the mass [0.3] and so you tend to [0.3] to go back to a a a full derivative 
there and not worry about the other plane [0.2] here we have to be more precise 
[1.8] okay [0.6] if we've got this acceleration going on [0.7] force going on 
there [0.7] now what the beam is seeing [0.3] er [0.3] er [0.3] the 
acceleration there has of course has a certain little force [0.3] where does 
that force come from well it comes by [0.8] er elasticity from the next bit of 
the beam along and all the way out [0.2] to the supports [0.3] er just as it 
does in all other [0.2] bending or deflection problems [0.5] so what i 
effectively have here is because of the acceleration [0.4] i have [0.5] that [0.
6] load function [0.8] now that's no more than saying that if this were a 
static problem that i've got a a [0.6] a heavy beam [0.2] hanging between 
supports [0.9] how is it going to sag [0.3] well what i say is its density is M 
per unit length [0.3] therefore i have [0.2] a force [0.4] M-G [0.5] per unit 
length in other words a a linear force function of that sort and you're [0.2] 
again you know b-, [0.5] er in the data book rather than having capital W for a 
point weight you have a s-, [0.2] lower case W for a distributed weight [0.2] 
it's no more than that [0.3] but it's our general acceleration of motion [0.2] 
not just big-G it's obviously dimen-, er sorry little-G [0.3] it's not 
dimensionally [0.4] er any different [3.5] so [1.3] we have [0.2] that as a low 
density [1.5] we know from basic beam theory [0.6] that [1.0] the fourth 
partial derivative of displacement [2.5] is the load function [0.8] er [0.6] 
the second er you remember the second [0.6] derivative gave us the bending 
moment [0.2] and if you think about how you calculate a bending moment from a 
force [0.2] there are two steps in in that as well so you get to the fourth 
derivative [0.4] but these again are equations that should be well known to [0.
2] to you from [0.6] other courses [2.1] so we get this basic [0.3] equation of 
motion 
which occurs for all these types of problems [0.6] er [0.7] if we [0.2] have [0.
2] uniform elastic properties [0.8] for E there if we have a uniform cross 
section [0.6] er all the way along so that I is a constant and if we have 
uniform mass distribution so M is a constant [0.4] then the thing is fairly 
straightforward [0.3] in principle you could solve it [0.2] with those things 
variable along the length [0.4] but in general you will find that you by [0.7] 
classical methods anyway you will not get a solution out under those conditions 
you will have non-linear [0.4] differential equations [0.7] you can solve them 
by computers m-, er finite element methods and such things [0.8] so we're going 
to deal only with this simple case where everything's a constant [0.6] it's a 
separable partial differential equation [0.6] er [0.2] i hope you can remember 
coming across those before [0.5] er but you probably hoped you would never have 
to do [0.2] anything very serious with them [0.5] and that's fine by me [0.3] 
er i'm quite happy to reduce all the [0.2] difficulty of solving that down to 
one 
line [0.5] it can be solved readily [0.2] in quotation marks [0.3] and what i 
mean by that is there's a perfectly standard solution [0.2] you can look up in 
a textbook how to do it [0.3] there's nothing difficult it's a just a routine 
turning the handles [1.0] job [0.3] to churn out the solution to that [0.2] 
it's just a lot of hard work to do it [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 [0.6] er [1.0] comes out to be that the displacement Y as a function of 
[0.2] displ-, [0.2] er [0.5] the transverse displacement Y as a function of the 
distance along the beam X [0.3] and time [0.6] comes out to be this amazing 
thing here [1.0] looking at the far end it's got sine-omega-T-plus-phi [0.4] 
type shape there which is what we always expect to see in these [0.6] but 
whereas before we just sort of sad had sort of A-sine-omega-T for the single 
spring mass [0.4] we now have something which has got four constants [0.5] 
that's no surprise we've got a fourth derivative there [1.1] but it's got a 
sine term [0.2] a 
cos term [0.3] the hyperbolic sine a hyperbolic cosine term there [0.3] all 
those four [0.2] terms can crop up all the things we can do [0.3] with [0.5] 
exponential functions in other words [0.2] er come up in this equation [0.2] 
and in general the solution is like that [0.6] we've stuck an alpha [0.2] in 
here [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 [0.5] all go in there [1.4] and 
all [0.2] transverse uniform beams in vibration [0.3] satisfy [0.2] that 
formula [1.3] we simplify it when we know the [0.6] N conditions we know 
whether the Ns are fixed or pinned [0.3] er [0.2] we know whether [0.4] er 
whether whether we hit it with a hammer in the middle to get it going various 
forcing functions [0.3] so by using boundary and initial conditions [0.3] we 
can simplify [0.2] down for [0.5] er [0.6] real problems and find that not 
usually all these terms [0.3] exist some of those Cs will go to zero [0.5] but 
that's really what what we're dealing with on that level [0.3] and i think you 
can see that [0.4] 
although [0.9] er a sort of er [0.7] an understanding of the principles of 
what's going on is important at this stage [0.4] er [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 [0.9] so anyway just working on 
this is still part of the stuff you've got in the notes [0.8] let's just take a 
specific example which picks up on that formula on the previous page [0.6] 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 [0.6] in other words one that can't [0.5] it holds 
position doesn't transmit a moment [0.5] and we then know [0.6] that there will 
be zero N deflection [0.2] and zero bending moments [0.4] er at both ends of 
the beam underneath each support [0.4] that's that's what [0.2] defines what we 
mean by simply support [0.2] s-, supported beam [1.9] so when we plug those 
numbers in so you just slog 
through that [0.2] horrible formula we had last time work out the the 
appropriate [0.5] put in er er [1.2] X-equals-nought [0.3] er set Y-equals-
nought and see what the constants come out and so on and so forth [0.3] then it 
turns out that that one comes out to be [0.5] like that [0.3] only the sine 
term only one of those four C [0.3] C-one to C-four things [0.3] is actually 
non-zero for that [0.4] classic case [0.2] if you have a cantilever [0.3] 
you'll find you will finish up with two of the terms and there'll be a 
different two [0.2] and so on and so forth but [0.9] we just hammer it through 
in that way [0.9] now the phi [0.2] there [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 [0.2] in the very 
simple spring mass [0.4] so that's the one that's worked from the time initial 
condition [1.7] er [0.4] what we get in here notice is we do have [0.4] a 
subtlety that's come up in this formula [0.3] notice that it's not just sine-pi-
X or something like that 
inside there [0.3] it's sine-N- [1.9] pi-X [0.2] where N is an integer [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 [0.3] well it goes to zero at [0.
3] an angle of zero at an angle of pi at an angle of two-pi at an angle of 
three-pi [0.2] and so on and so forth [0.4] so all of those are possible 
solutions [0.3] to how our string's vibrating or our beam is vibrating [0.2] 
there's no difference between a string vibration and a beam vibration in the 
sense we're talking about them [4.4] so [0.8] we get [0.8] a series of 
solutions here [0.3] if we look at [0.4] N-equals-one in here [0.4] we've got 
[0.6] time varying behaviour but we've got spatially varying [0.7] behaviour 
from that term [0.3] and it's saying that i-, if N is one [0.2] then the thing 
is a half sine wave like that in other words it's zero at both ends [0.3] and 
it's moving through in that sort of motion [0.7] at any point along the beam [0.
6] it's going at a speed [0.3] omega-N [1.3] from that [0.5] time sensitive te-,
sinusoidal term [0.3] but the distribution is like that [0.5] 
if i go up [0.2] to N-equals-two i get something that [0.3] has [0.6] the 
classical sine wave shape a a node in the middle [0.3] it's still any point 
this way it's still going at exactly the same speed of oscillation [1.0] for [0.
4] the relevant solution for omega-N [0.5] and so for each of those modes [0.2] 
here as we go up [0.4] we're getting more [1.1] turning points [0.8] so we're 
getting [0.4] a more difficult e-, [0.3] energetic shape [0.3] to drive [0.3] 
we need a higher omega-N to cope with that [0.3] so we know that we're going to 
get [0.4] in theory an infinite number of these now because there's an infinite 
number of solutions to that equation [0.4] and we're having an infinite number 
of modes [0.6] of increasing [0.8] frequency [1.2] now what's the practical 
side of that well what we know in practice [1.7] is that [1.7] it's more 
difficult to drive the high modes this one goes fairly easily [0.5] er it 
doesn't take a lot of energy to do that to do it with a [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 [0.3] the bending energy you have to put in to make the beam 
go into that shape [0.3] and you can see it's going to be just harder to get it 
going like that [0.5] so in practice [0.4] beams rarely show more than with any 
[0.7] non-negligible am-, amplitude [0.3] more than one or two of these low 
order modes it's fairly rare [0.6] to g-, get something with a very complicated 
pattern [0.2] so we normally finish up with a a fairly simple practical problem 
[0.4] er you can more or less guarantee for for all realistic [0.5] conditions 
that this [0.5] 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 [0.5] and 
so on and so forth [0.3] and you can argue that purely on the this [0.4] er 
notion of the energy that's involved [2.0] so [0.9] if you need to do the 
formal theory [0.4] you can slog it through [0.5] if you have ac-, [0.2] access 
to a textbook [0.5] and you can remember that there is that horrible formula 
with the cos cosh sine and sinh terms in it [0.6] you can look that 
up in the textbook and just do the boundary conditions [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 [0.5] and we could then do [0.3] a guess [0.4] 
type solution rather than do it properly knowing which of these mode shapes and 
knowing the mode shapes [0.4] we can sketch out [0.4] the shape not the 
amplitude but we can sketch out the shapes [0.7] purely by a sort of physical 
intuition [1.2] we could then think in terms of using approximate methods [0.2] 
that can sort of home in on the one that we want [0.4] and there are some very 
sophisticated a-, and very useful techniques [0.3] that that are widely used 
for doing that [0.2] but they going into those and doing those properly [0.3] 
is really er er if you like a topic for the third year [2.6] so [0.6] i'm going 
to leave it the beam example by saying look that's the basic theory [0.5] 
that's how you would get at the solution if you needed to and you ought to be 
aware of that [0.4] but the basic behaviour is this series of 
omega-Ns and when you actually plug in real boundary conditions [0.3] to the 
complicated formula [1.0] life's fairly straightforward [0.6] but er [0.2] i 
mean there was no [0.3] sense in which we expect you to memorize that formula 
for exam [0.5] conditions or anything of that sort it [0.2] it's just not the 
sort of thing that's sensible to do [0.3] you look it up in a textbook if you 
want it [2.0] so we do use approximate methods er [0.5] and er [0.3] again [0.
7] when you look at the er [1.5] example sheets if you haven't done so yet 
you'll find one of them [0.4] has got an asteri-, one of the questions has got 
an asterisk against it [0.5] because it's [0.2] a a sort of [0.7] an extra one 
to do if you're feeling er enthusiastic [0.7] er it is a a case of looking at 
one of the earlier problems and looking how we might [0.4] use [0.3] 
approximate methods that is you know to use an approximate method [0.3] you 
usually are iterating [0.3] and trying to guess roughly what the right answer 
is [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 [0.2] a little bit of 
brain power to get a good guess [0.6] er and there's one example that shows you 
that [0.9] but what i'd like to spend the last few minutes today [0.5] i-, is 
[0.6] just hinting at one w-, how one or two of those [0.4] er [0.2] methods 
are developed [0.4] er [0.4] not b-, so much to [1.8] show you the en-, enough 
to get you into next year's course [0.4] but just to [0.6] show how we can use 
[0.2] energy methods [0.2] alongside these other approaches [0.3] you will 
notice that regularly [0.4] i've given you a formal classical solution in 
Newton's laws [0.4] and then said think about the energy [0.3] and it will [0.
3] it will help you to see why this pattern of sha-, of modes and nodes and so 
on goes together [1.0] well if that's the case then presumably there there 
could be good reasons for using energy [0.3] more formally as the solution 
method and i and i think it's just worth seeing this [0.6] now this first 
little bit it it's nothing very complicated [0.3] it's the sort of stuff that 
is in the first year 
texts so [0.2] you know if again if you look at the reading list [0.3] you'll 
find that the reading associated with this particular little section of the 
course [0.4] is in things like Hibbler and Mariam as well as the the main texts 
[0.5] for [0.2] for this [0.5] er particular session [3.3] i think all we'll do 
here is just look at one example of it so let's just consider [0.3] 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 [0.2] the idea [0.2] that you 
just are aware how the methods work [1.8] if we consider it as a free vibration 
then [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 [0.6] without any 
s-, significant influence from the outside world [0.5] and if that is true [0.
4] then the mechanical energy within that system must be conserved [4.3] now 
what mechanical energy do we have in a vibration problem [0.2] again this is 
where we were 
discussing [0.3] this time last week [0.3] we've got [0.4] strain energy or 
spring [0.2] potential energy if you like [0.3] and we've got kinetic energy 
stored in the the [0.5] the inertial part of it [0.6] so we're saying that [0.
5] again let's take the simple case [0.5] because it's [0.2] thermodynamically 
isolated in its most simple case then we've [0.2] ignored friction because that 
would generate heat [0.5] out of the [0.2] kinetic energy [0.3] and we don't 
want that [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 [0.7] 
that some of them must be a constant [2.4] so [1.3] just [0.5] flogging that 
through [0.2] if the motion is a sine wave [0.7] as we're assuming everywhere 
else [0.3] X equals A-sine-omega-T [1.2] its maximum displacement obviously is 
A [3.8] and when is that maximum displacem-, er [0.9] going to occur [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 [0.8] so [0.2] we'll get the maximum displacement A [0.
4] at the points where the 
speed [0.2] is zero [1.0] 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 [0.3] and it's clearly fastest [0.5] when it's in the middle [0.7] when 
there's no displacement [0.5] so we'll get the maximum speed [0.5] is A-omega 
we've been using A-omega-squared for the [0.6] acceleration a lot so that 
should be [0.5] totally familiar to you [0.6] er [0.5] the speed will be A-
omega at displacement zero [0.6] that statement i think you can see purely by 
physical intuition [0.4] but if you want to be formal about it [0.3] then just 
put [0.6] differentiate this with respect to time and have X-dot [0.4] equals A-
omega-si cos-omega-T [0.5] and you know do do the formal maxima and minima [0.
2] and you will you can prove it like that but er but i think the physical 
intuition [0.5] should be all you need to do that [2.1] now this is useful 
because [1.6] strain energy is to do with displacement [0.8] kinetic energy is 
to do only with speed [0.5] now it happens that [0.5] 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 [0.4] so the maximum strain energy [0.4] and the 
maximum kinetic energy [0.2] must [0.2] actually be the same value [0.5] and we 
can either sum them formally [0.8] er for all values or we can just say look 
just look at the maximum strain look at the maximum speed [1.0] from that we 
get [0.2] strain energy [0.2] kinetic energy and those two maximum values must 
be the same [0.3] so we actually have two slightly different things we can do 
we can use it in that [0.7] strict formulation [0.4] or we can use it in in 
this form [0.2] and in given situations [1.1] thinking out about it one way or 
the other [0.5] tends to be a little easier [3.1] so [0.2] very simple example 
let's go right back to the [0.2] very first example we did as a revision [0.4] 
at at the start of this section [1.4] single mass single spring K [0.5] er and 
let's look at what the strain energy is well the maximum strain energy [0.7] is 
when we have the thing extended A beyond its natural length [0.4] and is [0.5] 
a half-K times the displacement squared [0.6] that's the straightforward spring 
[0.2] Hooke's law [0.6] energy formula [4.0] the 
other side we have [0.5] the maximum kinetic energy which is half-M-V-squared 
[0.2] and here [0.3] is is the maximum V we want which is A-omega from the line 
before [0.3] so we get half-M [0.8] A-omega-all-squared [1.2] reorganize that 
remembering that we're looking for the natural frequencies so putting the 
subscript- [0.3] N back in there [0.5] omega-N-squared [0.3] equals K-upon-N [0.
3] er and of [0.4] it course it would be a terrible shock [0.3] if anything 
other than that result [0.2] came out since it's one that we're so familiar 
with [0.3] but just notice that [0.4] that works you know drops it straight out 
in one line [0.3] under those conditions [0.4] and it will do equivalent things 
[0.2] in more complicated [0.7] situations [0.9] so [1.0] here for this year [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 [0.3] for that simple case [1.0] if you 
like that method there's no reason why you shouldn't use it [0.6] for some more 
complicated cases [0.4] er er of [0.3] complex motions [0.3] we can set up [1.
6] special coordinate 
systems [0.3] we can use these sort of tricks [0.3] including the one that says 
let's formally [0.8] differentiate the energy equation and say if it's 
conserved then the the change with time must be zero [0.4] which is the the 
sort of one just going off the screen at the top [1.5] and [0.4] if you do that 
then you get to a whole series of techniques [0.3] going under the name of 
Lagrangian mechanics [0.4] er and [0.2] later on even if you're really 
desperate Hamiltonian mechanics [0.2] but they're a whole series of very 
powerful [0.2] analytical techniques [0.7] whose underpinning [0.3] is this 
very simple example er and [1.1] all i want to do today is say look if you come 
across these formally next year [0.4] er [0.2] you should go away with enough 
confidence [0.3] if you choose a set of options by which you don't get involved 
with those courses [0.8] but then later i-, early in your professional career 
[0.3] do come across people throwing Lagrangians at you and things like like 
that and expecting you to understand it [0.3] don't panic [0.3] with a bit of 
luck just a 
little reminder will say you know that's how you start [0.3] it's not too much 
of a problem [0.4] given that you need to do it to go and look in a textbook [0.
8] and just build it up from the sorts of things that we've been looking at [3.
5] okay so i think that's all we'll do there can i show you one other slide 
it's in the note [0.3] er in the er copies of the er [0.4] slides in the 
library [0.4] and i don't really want to talk about it in any detail at all [0.
3] it won't come up on an exam paper or anything of that sort [0.2] it is 
purely information for you for the future [0.3] depending on which route [0.3] 
you might choose to go through [0.6] and that is just to point out that 
alongside the energy methods [1.1] that there [0.3] are [0.3] this trick of 
estimating where you are and using [0.3] approximate processes [0.3] er goes 
very closely [0.3] that you can use approximations in all sorts of ways but [0.
2] Rayleigh [0.4] er in p-, [0.2] developed a whole series of techniques [0.3] 
which [0.2] use a mixture of energy [0.3] and an approximate guess [0.2] 
at what's going on [0.3] and you get this sort of pattern [0.2] of behaviour [0.
6] er [2.3] we can [0.4] very often in real vibration problems [0.4] if we know 
omega-N [0.3] that's good enough we n-, [0.3] we know that the initial [0.2] 
the actual ax-, [0.2] amplitude of the vibration depends just how we kick it [0.
3] so omega-N's [0.2] usually the thing we're worried about [0.8] if we [1.5] 
can guess [0.2] say for a [0.2] vo-, [0.2] oscillating beam or something like 
that the shape in which it's oscillating [0.6] reasonably precisely [1.1] we 
can estimate at its maximum deflection by [0.2] standard beam theory say [0.4] 
what the strain energy would be in that [0.2] it's a formulae you can look up 
[0.4] so providing we get just a a reasonable guess of that [0.2] shape [0.3] 
that would give us a maximum strain energy [0.4] we could then e-, [0.5] work 
out [0.3] a maximum kinetic energy [0.6] er as [0.4] a sort of M-omega-squared 
sort of term [0.4] and [0.9] plug in those together it gives us [0.3] an 
estimate of omega-squared [0.3] and that's really what it says on here we we 
can guess at the shape [0.5] er what Rayleigh discovered and what makes the 
method 
powerful [0.6] is that really quite crude guesses of that shape [0.3] get you 
surprisingly close [0.4] to the right answer [0.7] and not only that [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 [0.4] er 
the conditions that work so it's a nice safe [0.3] predictable [0.2] 
approximate technique [2.8] normally the way you find the shape is just to say 
okay [0.3] as i did in that simple example we use a more sophisticated version 
of that [0.7] here's a beam vibrating how how would it sag under gravity [0.3] 
that's a good guess of the shape it will vibrate in for its lowest mode of 
behaviour [0.7] so let's use that as the starting point for the system you know 
so so you use sort of common sense [0.3] guesses intuition whatever you like to 
call it [0.4] to deal 
with that [0.3] but some of you undoubtedly will later on in your courses come 
across a formal treatment of Rayleigh's method [0.8] others of you [0.5] 
perhaps more on the civil side [0.3] may not deal with it directly don't get 
involved perhaps with earthquake engineering [0.2] team at some stage [0.3] and 
find other people using it [0.3] at least you've heard the words to get you 
into the field [0.2] and that's all i'm concerned with here just that you've 
heard the words [0.2] it's not something that's going to get examined but i but 
i thought it was worth spending [0.3] just a couple of minutes at the end of 
this session [1.0] talking about it [0.6] so that's it for today [0.2] tomorrow 
we'll meet up again [0.2] and we're back on examined material like so it is 
stuff that's fair game [0.3] we will look at vibration isolation as a subtopic 
[0.4] er wi-, within the overall vibration picture