# Strength of Materials II: Strain Energy (15 of 19)

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[Inaudible] energy is an energy being when

you — an item or any structure go — you load this structure. The load — the structure

goes under the formation, like to be. So, you put the structure load on it, whatever

it is, uniform load or single load. The structure goes down. As soon as you remove the load,

the structure goes back to its original position. That means while it was loaded or during the

loading there are going to be stored some energy into the system, because when you take

all the weight of the beam back to the original location, that requires some work. And, work

is not free, of course. Everybody understand that it doesn’t come free. So, therefore you

load it, the energy’s stored. That energy takes you back to the original location. So,

we are going to talk about that. So, the first three problem is about energy. We call it

strain energy actually, so you’ll want to add that to that one. I forgot to put it there

in the other class, but this is due to the strain or the due to the stress. Doesn’t matter,

or due to the load, but we call it strain energy. So, that is the subject we’re going

to talk about. Then of course after we do that, you — you will ask yourself, what’s

the purpose of this? Why do we calculate strain energy in a system? Obviously, the next part

of the homework will answer you that question, because, true, this strain energy, which is

stored in the system — please — we are going to calculate the deformation, or displacement.

Notice we’ve used many method to calculate the deformation. We use Hooke’s law to calculate

the formation for an axial member, which was delta equal to PL over E. Yeah, everybody

remembered that. We use many methods, including, singularity method, superposition method,

this method, that method to calculate the displacement on there. Load in a beam, which

was [inaudible] and why is that calculated? We also in ME 18 — 218, we calculated the

formation at relative angular deformation of the shaft one side with respect to other

side, which we call it phi, is that — you recall all of that. We are going to repeat

all of that with the energy this time. So, this is a new method. Now, of course, we are

not going to use it for the — for any problem. We use — usually use this method for type

of problem. We — we cannot handle it with the old method, and some of them you will

see in your handout. For example, if you want to have a displacement of a truss member,

which has not submember in it, will be very difficult because you cannot use geometry

to calculate the displacement of node A or node B. Everybody with this method, you can

easily calculate that displacement of a truss, which we call it work and energy method. But,

unfortunately that method has lots of limitation. So, I’m giving you the idea of what we’re

going to do next in the next two hours or so. So, then it has a limitation, and as we

go through the limitation, we find out that that limitation does not let us to do many

problem. It’s only limited to certain problem. There’s some certain restriction. Then we

have to go through a theory which we call it Castellano Theory, which shows you how

to — how to use for every problem. Everybody understand that? So, that theory will be discussed

actually next Tuesday. So, today we are going to concentrate on the two force line. That’s

why I draw a line there to understand what’s the strain energy and how do we calculate

that? And, then to — to use it to — in order to find the displacement, which we call it

work energy method. Is that understood? Yes or no. Okay, so let’s go back again, categorize

everything that we learn in ME 218 and 219 — 219 so far. We should — I would like you

to make a table in your note because you are going to complete that table later. So, here

put cases or loading. And, then I can call it case one, two, three, and four. So, I changed

it. For here, I put that loading system, loading. You can understand what I’m saying what I’ve

got through there. Then perhaps I want to put there maybe next line I want to put the

stress. Then I want to put in the table the strain probably. Yes, the strain or the deformation.

The deformation — let’s put the deformation. And, then I want to put here strain. And,

then here I want to do add — this is new now — energy or strain energy. I put here

S, strain energy, and that has two columns. We’re going to fill out that column today.

There is constant, of course, you don’t know what that means yet until I go through the

whole system. And, variable, and I define as we go there. So, this is from this column

only is today’s lecture. This one, we already know that. Case one, loading, we call it axial.

So, everybody knows what axial loading is, which was lecture one ME 218, yes or no? Any

rod on there [inaudible] — tension of [inaudible], compression, or P — sigma equal two. In other

words, sigma equal two plus minus N over eight. I put N that you do not mistake it with the

P. N was the internal verses the external. Is that correct or not? Then we had of course

deformation for the rod, the formation of — putting there because we needed to discuss

it today. Therefore mention delta was equal to what? Delta was equal to PL over AE or

EA. Yes, correct. And, the strain, of course, is normal strain, which is epsilon that you

are working in different format. Epsilon is the amount of deformation per unit of length.

Is that correct or not? Yes — that you are using there. And, we get to the energy [inaudible].

Then of course, as we go follow the ME 218 system from the axial, we went to the torsion,

yes or not? But, the torsion gave us the tau. So, I’m going to put sigmas together and taus

together. Everybody understand that? So, the next item that gave us our sigma was bending,

yes or no? Okay, the bending give us the sigma equal to what? Sigma equal to plus minus what? M — what — I don’t want to hear MC over

I, because MC over I is a constant. It is on top and on the bottom. MY means the stress

in the cross-section changes, depends on the location of the point. Is that correct? That’s

what was some of your homework were about. Yes, you used it in just the homework that

I gave you. Of course, MY over I. Is that correct or not? Which reminds you the stresses

in the cross-section is not constant. It’s variable. Is that correct or not? Then of

course we had the deformation. What was the deformation of the beam? Of course, you should

remember that one because that was last week or a week ago discussion. What’s the deformation

of a beam? You spent two and a half week on that one,

guys. [Inaudible] and Y, yes or no? Yes? Yes. Yeah. Why did nobody answer me? All those calculation–. –Deformation of this [inaudible] we spent

two and a half week besides there — using singularity method, double integration method.

This is — this is very interesting for me. Quite frankly, that’s the way all the student

think. They just remember the procedure without knowing what’s this all about. We spend three

weeks calculating theta and Y. I asked for this class, the other class, what’s the deformation

in the beam? Everybody’s quiet. What’s the reason? I was going to say because we were thinking

about how to put it into a formula. No, no. If that was your thinking, no. I — I

said what is the displacement. I didn’t say give me, because [inaudible] — this theta

is different. It depend what load you have it at. All I ask you, what is the — yeah,

I’m sorry. If that was the case, now everybody knew. Is that correct or not? That’s what

they say. Okay. So, that’s theta and Y, and what as the strain? Of course, the strain

is epsilon because we basically — in many [inaudible] we neglect that the effect of

transfer shear. Remember that? So, we said that’s minimum compared to the other one.

Anyhow, so that — that was epsilon. It could have gamma but I didn’t put it in. Then these

two are sigma and these two are tau. Remember how you were calculating tau? Tau equal to

TC over J plus minus EQ over — I’m just referring to that. So, we had the torsion, yes or no?

yes? So, the torsion, what was the stress? Stress tau equal to plus minus — you know

that. T what? Y. Y. B. Not Y. Yeah, just this one had Y. This doesn’t

have Y. That’s exactly again here. T what? TC over J? Wrong. Okay, it’s a variable. It

is not a constant. TC over J is an all [inaudible]. T–. –One. It’s — what was C? Actually, C was with outside

loads, zero is inside loads. C equal outside. It was a circular [inaudible]. We have to

go to — okay. So, we’ve got — so, you remember it now. Okay. T load — exactly if you are

calculating a stress here inside the shaft, the stress is like that. Is that correct?

Remember we went from rho and rho was between zero to C. One — those was the derivation

of equation. Nevertheless, you — these are — I’m talking about not outside only. You

are using outside. You know why, because as a designer, you’re interested to find the

highest value. But, in general, that’s the format of a stress. Is that correct or not?

And, the beam [inaudible] but of intention, it has a parabola format, or linear format.

That we all have one through ME two eighteen. Anyway, I don’t have to mention that. And,

what was the deformation? Phi, that’s correct. Phi equal to TL over GJ. Of course, you had

plus minus because it goes clockwise, counterclockwise, etcetera, etcetera. Obviously, because this

is the shear stress, you will have the shear strain, which is gamma. And, as I said, there

is case four, which is transfer shear–. –Which, that was chapter six. Tau equal two plus minus

VQ over IT. But, for this system, although there was a gamma there, etcetera, etcetera.

Remember to calculate that one, but you are going to bypass that. We are — we are going

to assume they are negligible. So, please write it down. Actually, they are not, but

it has a little bit more difficult, so we are not going through that because with the

type of problem you have, which is really very long and narrow shaft or a beam. The

effect of transfer shear generally is negligible. I don’t know whether you notice it in some

of your problems. TC over J was always 20 times larger than VQ over IT unless you have

a short column, which does not quality — if short load. Is that correct or not? So, that’s

— let’s leave it at that. Anyhow, this one we are not going to cover. Let’s put it this

way, but it is there. So, we are going to go through this one and then we are going

to fill out the — this. Now, to do that we start with the axial members. So, I’m going

to put an X axial member here as I did for other class. Then I’m going to put here a

load here. Let’s do it a little bit larger. These are an exaggeration. So, I put here

a load P one and I will get delta one. Everybody knows that. And, then I’m going to plot here,

P verses delta in general. This is not Hooke’s law because in Hooke’s law, remember, we were

plotting epsilon verses sigma. Here is load verses deformation. So, that’s a different

thing. However, the same principle applies. Everybody understand that when P one equal

to zero, delta one equal to zero. So, we start it here, and then as we increase the P one,

delta is going to increase. But, if we are below the sigma yield, everything is linear,

yes or no? Correct. So, with linear, we’re going up like that. Is that correct or not?

Let’s say at this scenario, this is our P one and this is our delta one. Is that correct?

Whatever it is, well, we can plot it, so we can — then at this point suddenly we are

going to add to our P one. Let’s say that P one is 10,000 pound and we have a rod here

with the length and cross-section, we put the load here. Notice, as soon as the load

hits, theirs is an energy being stored in the system, because it is a stress and it’s

the strain. We don’t know yet how much it is. We are going to find out in a minute.

However, if I add to this load another delta P. So, let’s say with 10,000 pound suddenly

become 10,010 pound. So, delta P is ten pound. Obviously, the deformation would increase

a little bit as well, yes? Sort of. So, we are going down a little bit. This is much

smaller than the other one, of course. We are not at this level. That means I have added

to my load here, and I have of course, my increase. So, I’m going to call that delta

delta. So, this one is delta delta. A little bit more, whatever. Although these are all

small number, but that’s the scenario. Is that correct or not? Yes? And, we are there.

This is added a little P here, added a little delta there. Is that correct or not? Now,

at his — in order for you to understand what is going on, at this time suddenly — we should

do the same thing. Remove the delta P. Yes or not? Yes? Correct? Notice, P one is hanging

here. it’s going to be moved up. We are going from the purple to the black. Is it work being

done here? Of course, that is the energy that was [inaudible]. If I didn’t have energy it

would not go up again. Everybody understand that? If this was a plastic material, probably

that would be the case. If it stretches, it never goes back again. Is that correct? This

is elastic material. So, I need some energy which was stored here, to take how much is

that energy, write it down. How much that — you call it delta U. So, we shoot — we

show the energy by capital U. So, all of this will be U. So, I’m putting here the strain

energy, U, or capital U, which shows the total energy stored in the system. In this case,

I have a little bit energy stored here which is what? P one times — look, look. P one

is here, from here goes to here. That’s the amount of work. Work you call to force times

distance. So, it is P one times what? What this? Delta delta. So, delta U was equal because

of that adding delta P and removing it I have restored this much energy. Delta P times delta,

of which delta delta — is that correct or not? Yes? Correct? P one times delta delta.

P one times delta delta. Can you show me area wise what is that number? Is this P one? Is

this delta theta, yes or no? So, that there U is this area, yes or no? Yes? So, that area

that you see there, that band represent the total energy that I stored in this system

when I went from P one to P one plus delta P, yes or no? Therefore, if suddenly I change

the whole thing, and suddenly I put here the general load, because this goes up linearly

until we get to the elastic range, because you always load the structure before we get

to the sigma yield. Is that correct or not? So, if I go to any P now — so let’s put that

any P, which I put it blue. So, if I load this structure to point P delta. This is point

P and that is delta, what’s the total energy? Isn’t that the total energy I have stored

in that because there’s — I have to add all those little strip together, yes or no? Because,

this is one or at least they are not constant. Actually they are variables. So, you go this

one, that one, that one, that one. All this little strip added together. So, the total,

which is the total U for entire system if this structure is loaded to the load P, which

is there and delta there, will become this area, the blue area, yes or no? Which is equal

to what? Equal to–. –P times–. –Delta–. –Delta divided by two because it is a triangle.

So, it would be one half of — that’s what you want to know. P times delta. So, that

is the total energy stored in a axial member. Harbor [assumed spelling], what’s the value

of delta? You already know what’s the value of delta is. Is that correct or no? What’s

the value of delta for axial member? Delta equal to PL over EA or AE as you say. I always

say EA. You say AE. That’s alright. We’re talking the same. So, total energy, U. Of

course, remember this equation, because you are using — you use that for the work of

energy. But, total, now, if I’ve substitute delta in that one, it become P square L about

the Y there by two EA, because this becomes PL over EA. Two is there out of the P. So,

it’s become P square divided by two E. So, I’m going to put that one here. Energy equal

to total energy. I don’t need to write it here because I already wrote it there. But,

there it is. Equal to P square L over that — L — when everything is constant, that

means the area should be constant, the load be constant, because — because this is the

length, it’s area, etcetera, etcetera. Is that correct or not? Then we get into a little

bit of problem if we progress to the beam. Look at happened in this system, right? In

this system, sigma here and here and here and here, here, here is all the same, yes

or no? Agreed? Epsilon here, here, here, is all the same. Is that correct? So, everything

is constant. So, every part of this structure or this rod contribute the same amount of

energy, yes or no? Now, if I change this problem to the second system, this is what I cannot

use this equation for second system unless I modify that a little bit. If I use the beam

now, which you used it in your homework. Couple of people, I noticed some of you were here

and asking questions before the class start. Yes or not? Remember the beam I gave you is

— is [inaudible] etcetera, etcetera. If I give you a beam like that, and I put here

at the end a load which was your — one of your handout problem, yes or no? Do you think

every part of this structure contribute evenly like here? No. Actually, it’s totally different.

Here if I color code it, let’s say all of this become dark red, it’s because all of

them are participating the same amount except of course stress concentration. We are regulating

that. Everybody understand that? Yes? Here, as you — everybody appreciate that you have

done all your homework. You know that not even stresses changes this space [inaudible]

changes that way as well. Is that correct or not? So, where do you think this one will

get, if I color it and red is the highest energy, where I get the maximum redness for

example? At the end and on the top and on the bottom, because on the neutral axis the

stress is zero. The strain is zero. Everybody understand what I’m saying that? There are

for the distribution of a strain or a strain energy is not uniform because it depends on

the sigma or it depends on the strain. Is that correct or not? It depend on the stress

or the strain. So, based on that we are going to — everybody understand what I’m doing

here? The difference here is everything is constant. In other words, if I’m saying that,

if I use one cubic inch of material here and one cubic inch of material here, both of them

give me the same amount of energy because they have the same sigma. They have the same

epsilon. Is that understood, because if you don’t understand that I cannot go to the next

case of the operation. So, based on that, I’m going to define the strain energy density.

Please write it down. So, a strain energy density is this, which we call it a small

U and I put difficulty at U — E in front of that — that you see. And, that is — write

it down — that’s the meaning of that, the strain energy density is the amount of — first

— first of all, it’s energy density. What? It’s for everything, but that’s the definition.

This is the — has nothing to do with that or the other. I use it for that one then I

expand it, yes. But, this is for every problem. Doesn’t make any difference. Energy density

— see all I’m saying that energy density in this rod is constant. Energy density in

the beam is not constant. Somewhere is zeros and somewhere is the highest value, but I

have to calculate total energy. So, as you see I have to use integration. Everybody understand

what I’m saying there, because it’s not constant. Therefore, I’m defining energy density, which

is what? See, I want you to understand that. This very important. For the other class I

explain that. What’s the difference between delta and epsilon? Everybody should know that

by this time because you’ve used [inaudible] if you pay attention. So, what’s the delta? Delta. Any delta. Delta is total axial movement

of deformation, yes or no? What’s the epsilon? The amount of the deformation pair unit of

length. So, that is the unit. Is that — here is — this is total [inaudible] energy, though

I’m called energy density, which is similar. This is — write it down. Energy density is

— this is the definition of that — amount of energy stored in one unit of volume, because

it’s not epsilon. This is energy. Energy depends on the volume, yes or no? I cannot have a

flat plate and have energy. I must have a volume. Yes? Correct? So, that is the unit

again. Remember, this will be going all the time like that P is total load. Sigma is the

amount of force pair unit of area. In a way it’s unit. Delta is total deformation. Epsilon

is the amount of deformation pair, because this is the system you are using. That’s why

sometimes you have integration, sometimes you don’t have. Sometime there are constants,

sometime there are not. But, this is typically when you get in the higher class courses are

more mathematical. That’s how you have to think about everything. Is that correct? Actually,

it’s become much more complex than that, later on mathematically. Now, we are defining total

U and unit density — energy — which is the amount of energy stored only in one cube — one

unit. It could be millimeter by millimeter by millimeter, or inch by inch by inch. Everybody

understand that, yes? We should — some of us are going to sleep because we don’t want

even to listen to that. So, you have to write that right there. Amount of energy is stored

in one unit of volume. This is the cube for — key for entire understanding of the rest

of the problem. So, if you write that, what will be our energy density for this rod. Notice,

it’s true that all of them are participating. What’s the total energy? This is the total

energy. If I — because everything is constant, can I divide that by the volume and find out

the energy density, yes or no? So, let’s do that. For the rod, that’s what you asked me,

but it’s applied to anything with the sigma. So, for energy density, we come for — since

it is the amount of energy — paired unit of volume, I’m put the total energy and divide

that by my volume. Is that correct or not? Yes? So, therefore I have P square L divided

by two EA divided by volume. The volume of this rod, of course, I don’t have to go take

this etcetera. We go L and the cross-section A, because we never defined a cross-section

A. Is that correct? A is area. Could be square. Could be [inaudible] etcetera. So, will be

divided by what? What’s the volume? Volume is–. –LA. Not LA. Right, not LA. RLA — it is

length times area. Is that correct or not? Yes? Good. So, therefore as you see A and

L, it drops out, you get P squared over A squared. So, you pick up equal to P squared

over A — A squared. So, it becomes sigma squared divided by two E. So, as you see,

it’s related to sigma. You said sigma is — the sooner you understand this — sigma is constant.

Therefore energy is constant everywhere. Is that — for the beam, sigma is not constant

and we cannot why zero energy zero. Is that correct or not? Yes? Okay, beautiful. So,

that’s it. So, this, you want to put that equation also in the system. I’m going to

keep that here. Energy density equal to also — doesn’t matter. No, the first problem or

second problem. This is based on [inaudible] not PL etcetera. So, it’s a sigma squared

divided by two E. However, since we know from the Hooke’s law that epsilon equal to sigma

over E or sigma equal to E over epsilon, I can put E squared epsilon squared there. Eliminate

part of that one, and that one become equal to — like that. Or, become equal to E times

epsilon squared divided by two. Either of them is acceptable by the use of [inaudible]

because that’s the one you have. As — as you see, I can use the stress or I can use

the strain. That’s why we call it the strain energy. Any time you load the structure, the

structure grows through sigma, yes or no? Structure goes through epsilon there, and

it’s going to store energy. Then it go, that energy do the work, and we want to calculate

that energy. [Inaudible] the primary formula you want to have that, of course these two,

[inaudible] four drops with these two, it’s in general. Is that correct or not? That’s

what we’ve been talking [inaudible] use that for tests, general [inaudible]. Now, look

what would happen if I do that and those are [inaudible], I’m going to have to do a series

of problem that we could not handle previously. Obviously, this [inaudible] for use this method

for a rob because we already — we want of ME 218, we explored the [inaudible], yes or

no? Who wants to use energy [inaudible]. Nobody who wants to use that method? That’s the reason.

You are using this method if I give you this problem. If I give you this problem like that–.

–And, I put here a P, then just solve. See that? What is its single constant yet? No,

because area changes, yes or no? That’s [inaudible] write it down. Take this constant. The area

keeps changing, yes or no? Therefore, sigma here, sigma there, sigma there, sigma. So,

I cannot use that constant here. It has become variable, especially area, yes or no? However,

I can use this technique now to — this is the technique you are going to use. Total

energy, so you want to write always equal to energy density. But, energy density is

how much? Is the amount of energy stored within [inaudible] and BV actually integrated over

the entire one. So, that will be our general equation, in general no matter what I have,

I have to go to the integration which is X, Y, and Z, yes or no? In many cases we avoid

that. Of course, like as usual, we don’t go X, Y, Z because some — some of the problem

are constant. Look at this problem. It’s signal in Y direction is variable if I got to this

line. Its sigma here in this direction variable, or the same? Same. There’s — this way’s the

same, this way is the same. In other words, if I look at this — this other thing that’s

here, is that — if I look at this, this way and this way is regular constant, yes or no?

So, only I have to integrate it all over X. All of the general equation is one [inaudible]

but I don’t need to integrate that over the [inaudible]. You will see what I do in a minute.

So, this is my problem, yes or no? So, A is a variable. A is a function of X, so A can

change. That depends what parabola you have here or what line you have here. I will give

it you. I’m saying that we’re — our problem, we already discuss that, yes or no, because

[inaudible]. But, it’s — we put the variable, in this case A variable. Also, for the other

class I mentioned a problem that P is variable, that we have never mentioned, but we should

have mentioned it. But, we are going all the mathematical problems. But, if I had it out

here with this, and here is let’s say 10 foot low, and inside the cross-section unless they

each put this weight on material which has about three pound pair for the [inaudible].

How big do I have there? Is it going to stretch because of its weight? Yes. So, how do I reverse that? There is the weight

constant. It’s ten times [inaudible]–. –We don’t put the weight at the middle here.

It’s not [inaudible]. But, here how much weight do I have? This one here, how much weight

do I have? Everybody see what I’m talking about? The height, the weight, or the force

acting in this member as P is a variable depending [inaudible] parabola. That problem by itself

requires integration. Everybody understand that? Even for delta, calculating the delta,

because P is not constant. So, we are really when we say P is at the end, we are simplifying

lots of equation. So, get ready for graduate work, if you are not going to do this type

of work, because everything as it appears [inaudible], the one that you assume. There

is [inaudible]. We only change everything to a concentrated load at the [inaudible],

yes or no? Correct? So, that’s not the truth, so anyhow. So, what that’s saying that even

this simple problem requires integration. What I’m saying that many time that this [inaudible]

put it in integration. You — you [inaudible] total energy. That’s different. First notice, if A’s constant, right now you

have to put A as a function of X. Everybody understand? Very — two and half X to the

power of one half or two or what do I do along those line? You have to do a difficult integration

as well. Everybody understand how we are not going to — but if A is constant, E is constant,

B is constant, everything comes out, yes or no? Yes, and then we come [inaudible]. It’s

X to the evaluated between zero to L. So, it become zero to L, so it become P squared.

L divided by two here, everybody [inaudible] this one. Is that correct? So, this is constant

of that one. Everybody see that, because integral of DS is L. But, if any other item, like D

or like A was a variable, I have to put in that equation or I have to use a — this is

one example. It’s that original way to 24 torsion as well, because you are going as

that mechanical engineer, you are going to design item. One of the item that I like,

[inaudible] you can put it, you know, this type of item, this is just like that. And,

put that axial load or torsion and that sort, but this is the method to use. Is that correct

or not? Also, remember that when we go to the higher classes like the design classes,

the machine design, especially I’ve noticed that many time, the lab instructor gives you

lots of problem that you have to reach — to solve the deformation because part of you

[inaudible]. We should not distend our beam and column. Everybody understand. So, this

method cost heavy. [Inaudible], is that correct or not? We’ll see when we go to the example.

We can do lots of thing with this method for particular — we use it for this to explain

to you what happening. Everybody understand? Then, we’ll be expanding into more difficult

problems. Now, let’s go to problem number two, which is [inaudible]. Is that correct

or not? Let’s see whether we have [inaudible] equation for value, and the torsion and then

let’s see what happen afterward. We don’t know — if I want to calculate strain energy

in the rod, either on this one or that one. Which one is more common? The one — more

common one I put it in the box. The more common one according to this one, because you have

it all crunched. So, let’s say you have a problem here like this. You are explain that

at the end [inaudible]. Let’s go down the page. Now, let’s go to bending this one just

a little bit. [Inaudible] to stay right here. So, this one, I just put one over here, P.

Doesn’t matter. So, that’s [inaudible]. Doesn’t matter how many loads you have here. But,

in this section, when, again, you are [inaudible] X, and delta X, the sigma changes, yes or

no? Well, this time I cannot take a vertical volume over here, because on the Y direction

its stress changes as well, yes or no? That’s why I wanted you to say if Y — everybody

understand that? If Y equal to zero, you are on the — if Y equal to zero you are the neutral

axis. You go up for example, you relieve the tension or compression, depend what kind of

moment you are. Is that correct or not? So, one side is tension, one side is compression.

And, to change is linear. You know that from ME 218, correct? Here’s what I cannot take

the [inaudible] if I know in the depth of the beam is the stress is constant. So, that

is the only two integration, X and Y. Not Z. Everybody — because Z [inaudible] is the

same. And, the [inaudible] this direction always were constant. We recall that if you

are doing this, so here was call compression. Here was all tension or Y stress. So, Z does

not come into the picture. Is that correct? So, for that reason, I can choose my element

the same way as before. So, let’s say a piece at the cross-section like that. So, then this

is the neutral axis, and we are here and we call this one PA. Is that correct? And, this

tension’s Y. Is that correct or not? So, I can choose my DA, a strip like that, which

gives me, again, it’s a neutral surface. And, this is my element now. My element now is

something like that. Yes or no? What is this dimension? DS. What is this dimension? DA. DA, but this could be any shape. Everybody

— I’m just [inaudible]. So, my body over there like that, but my unit volume will be

what? [Inaudible] will be what? Will be DA times DX. DX. DA — yes. So, that means I

have to integrate over X and integrate over Y. Depends on [inaudible], yes? Because, Y

[inaudible] maybe I should not draw that because I can’t show — draw it like that. Everybody

see what I’m talking about? This DA keep changing. [Inaudible] could be round, could be circle.

It’s all written out like that. This one doesn’t make sense, but that’s the general rule. Is

that parabolic? Yes? So, there — or that’s [inaudible]. Then you get the formula, I think.

Total energy — write it down — equal to what? Equal to over the volume, which is X,

now Y, that XD, so [inaudible] energy density. What’s the energy density? Energy density

— I’m erasing. I’m sorry. It was sigma squared divided by two E. Is that correct or not?

So, let’s write it down again one more time. Sigma squared divided by two E times BE, which

in this case becomes equal [inaudible] sigma. Sigma is M squared, yes or no? Yes? Y squared

over I squared, correct? MY over I. But, doesn’t matter whether plus or minus, because we have

plus minus the [inaudible] square, Y squared, I squared. Is that correct or not? Yes? Divided

by two E. That is from this form, yes or no? Both are divided by a little volume. That’s

why volume DA at DX. DA at DX. Now, everybody knows from the previous discussion that we

had that M is not a function of Y. M is a function of X. Is that correct or not? Y is

a constant, yes or no? E is a constant. But, Y is a variable and DA is a variable. Is that

correct or not? So, this formula can be written like this. [Inaudible] M squared divided by

I squared times two E. There’s a pair. I squared times two E [inaudible] squared. And, then

Y squared DA integrated over the area. This is [inaudible] — Y squared is DA, because

I have to go over the volume. Volume is X and A, and this one will be zero to L. So,

for the X part this would be this and then you have BX here and left it there. So, this

BX, Y squared DA, that is over the volume. I separate that area and the length. So, what

is Y squared [inaudible] was it integral of Y squared DA? Moment of — first, second moment, and moment

of inertia or second moment, yes or — that’s I. Is that correct or not? So, Y squared DA

becomes, again, equal to — all right, what’s the definition of the I? Is that correct or

not? [Inaudible] you? I hope not, because I have mentioned it many times. Y squared

DA is moment of inertia. One more time [inaudible]. It doesn’t have that to bring everything together,

but if you are missing, some of you, not all of you. So, if this is DA, like, I mentioned

that in this class two, three times. DA times Y is what? DA times Y is Q, yes or not? PA

times Y squared is — move it off — inertia with the [inaudible] axis, horizontal axis.

And, if this belongs to the larger area, all you have to do, write EAY squared, and integrate

that over the entire area. That becomes the definition of movement up, energy — all you

need for chapter — I don’t know, two chapters you study. Separate and move it up inertia.

EAY is QX. EAX is Q what? EAY prime is IXEA. X [inaudible] is by Y and DA the [inaudible]

of the square is — pull that moment up. That’s it. That’s the entire two chapter that you’re

looking for. But, it is that simple to go based from the idea. The rest is formulation.

Is that correct or not? You have to use that integration to come to moment of inertia for

rectangular shape, which end up to be 112 of base time. IQ, so that is the principle.

Nevertheless, look what happened. One of these I and one of these Y drops out and finally

this become integral of zero to L, N squared divided by two EIDX which is very similar

to the other one. So, therefore U here become equal to [inaudible]. Total energy move to

the [inaudible] of zero L. M is 12, this [inaudible] of two EA, is two EI. [Inaudible]. And, some

of you may think that this is very difficult to remember. But, actually it is very easy

to remember. I give you the key, I gave it to the other class as well. Look at the key.

Everybody that [inaudible] over EA, yes or no? So, all you have to do is make the note,

remember it’s make the doughnut squared and then put it through there. In other words,

add it to there and put it through here. Everybody see what I’m talking about? Look [inaudible].

P is squared, L divided by two E. Of course, EA belongs to the axial node. EI belongs to

the beam. Anytime you are calculating theta and Y you are using EI, yes or no? Correct?

Okay, now if F — now this is M. Is this — in this problem, M is constant or M is variable? Variable. Right, so you have to write M as a function

of X all the time. So, this is more common. See, this is more common for us and this is

also more common. So, F no matter how many times you read this chapter, write F as a

function of X. Here you have to do the same thing. You have to be able to write M as a

function of X. Of course, we do simplify that a little bit. We’ll see that later. One [inaudible]

put the M constant. Yes. Yes, of course everybody know that. [Inaudible].

Give me that example of M constant. If we were just [inaudible]. Absolutely. Is this what you are saying? Is

that correct? Is that M constant? See here? M, M, M, M, you are seeing it in the pattern.

So, M is become constant [inaudible]. So, U become what? It’s exactly the same format.

U become integral to what? M square or M over square — M over square can now become–. –Constant. But, it is M, which is usually

a function of X. This is what we do all the time. W is function of X. When WO, it is a

constant. I don’t know if you know this in all your problem or not. We never got [inaudible].

That’s the case. All right. Welcome to the class. So — so therefore, MO squared L divided

by what? Two DI. Is that correct or not? Yes. Okay, now there are similar formulations for

the torque. I have to do the same thing. I do that, I do that. This time it’s a sigma

verses force. There’s a delta I have to write. T verses T, but the same thing apply. You

don’t have to do that thing again with as for this. By looking at what happened here,

that will give me the answer for–. Of course, yes. U equals — as I said, it’d

typical — T squared L divided by what? Two what? Two — what’s that? JG. G. [Inaudible] J. Is that correct, because that’s

the formula check. Is that correct or not? Yes, so I don’t have to draw that for you

so you can see it by similarity. But, if this become for integration, if the variable — what

should I do? TA squared divided by–. Two G. Two G. Two G. JDX. Now, remember T could be a variable,

G could — J would be a variable. As I said, G could be one, but we’ve never use it in

this class, of course, or undergraduate. Is that part correct? So, we are assuming G always

is constant, which is [inaudible] in every direction. Everybody understand that? And,

then there are — J could be because I can give you a cone-shaped item and I put it on

there then. You can design something like that. Yes or no? Then you can calculate the

P equal distance. This is for special problem. Everybody understand that? Harvard, they use

it for regular problem as well for you to get used to it. It’s that kind. So, if I give

you a regular beam, or regular shaft to calculate all of this, it’s just for the purpose of

learning. But, in real-life scenario, you use it and things become a little bit more

object, is that correct or not? So, I want you to know why we put this for last. And,

anything that’s straight now, you can use one of the other methods. [Inaudible] solution

for the back of the book, yes? Okay, so that’s it. Now — now, and that’s it. So, we are

not using there, so that’s the system. Now, you have to say what’s in this group. So,

I give you a stereotype example because I don’t have time to go through the whole thing

but I want you to over the weekend to really look at all the problem and try to do it,

because I’m going to do coming up here. But, you should be able to, but I’m more interested

in this one, because you know what this one is? It is like knowing the moment of inertia

of the rectangular is one 12th of base height by cubic and you don’t know parallaxes to

it. Is this useful to you? No, it’s very limited, use its own parallaxes story, this is very

limited to you. When I explain, you want to work of energy is, you will find out there

is loss of lineation here and this Castellano Theory makes it more general. Everybody out

there? Now, here is the problem. If I give you this problem — this is a simpler problem

I can think of, can you calculate work of energy there? Yes, but you have to put in

two forms, yes or no? Like, before, because everything is constant [inaudible] V two,

and then from V two C. So, you use this formula depend what L I give you, what E I give you,

what P I give you. You can calculate total energy stored in this beam, yes or no? Or,

if I give you this, the same problem, but this is a similar shape, and I put here torque,

then you can calculate total energy using this process without the method. And, if I

give you this beam, I’m putting all of that. And, then over here P, that said P [inaudible].

This is let’s say A and this is BA and E and P will be given. And, then I can find the

total energy stored in this beam. However, in this beam I have to use the integration,

because F is a function of X. Everybody — and for this one? I have used constant, constant,

added together. Constant, constant, added together. This one I have to use from A to

B, separate from B two. See, I show you the technique next time, but I want you to get

this weekend to — hopefully you will do some of these by yourself. If you get into difficulty,

I’ll tell you how this works. Is that correct or not? Yes? And, then you will ask, okay,

what’s all of this good for? So, okay, [inaudible] total energy. I’ve got that much of energy.

So, what? Yes? So, what is work and energy method? All of them, you — I forgot one more

thing. It was important. Let’s say that I have here a truss. Remember, this is a truss.

So–. –I make it the same before the other class.

Let’s say the truss, a five-membered truss. Let’s say here is a pin, and here is a rod,

yes or no? Correct? And, I put here a P. How do I solve this problem? Which one is this

one? Is it case one, case two, or case three? That’s what I’m asking. Case one. Those are all [inaudible], right? All P — I said truss. When I say truss, it

is all P. Yes or no? I said this is not the frame. I said this is a truss, not a structure.

Truss means all the members are two force — remember actually some of you are probably

not like that. I have couple of problem like that. So, I’ll show you that. So, in this

case you have one, two, three, four, and five members. Let’s say I provide E and P for example,

equal to 5000 pound. You have to find force in each member. Doesn’t matter if you are

tension or [inaudible] because it’s a square. Remember that, because both are — the both

tension member and compression member contributed energy, yes or no? Doesn’t matter whether

it is plus or minus. Okay, I calculate force in this member AB, so this is ABCE — DE.

Five member, have to add all of them together. Is that correct or not? So, what? Okay, I’ll

find that in. I haven’t done anything yet. Everybody [inaudible] to come to the board

and energy. You work out energy, the equation I have written down here [inaudible] to look

at that [inaudible] before the axial member, total energy was close to one half of P times

delta. Yes or no? So, if I calculate energy somehow with that method, then I can — that

make it equal to one half as of P times delta. So, I can calculate the delta. So, if I calculate

total energy at this structure, I can calculate something, delta depth. But, then I have to

tell you what the work and energy ideal is. It is some theory that I’m not going to go

through that, but I give you answer to solve. So, please write it down. All of this total

energy, part here or part here, part here, part here, is summed up into this by — by

this idea there. So, I’m going to put it in general format, but I’m going to expect you

to give numerical example when class next start, everybody, because I don’t have time,

and I want you to do it. So, every time you calculate, you — in any system. Doesn’t matter,

this system, this system, this system. But, look at the which system there is. Always

do equal one half. We’ve got one half comes from [inaudible], yes or no? Times the load,

whatever I told you have. The load, you change. If it is the torque, you put T. If it is M,

you put M here. P here, whatever you have. The loading can change. Everybody understand,

because what kind of load you have here? [Inaudible] times what? What — what at end? At the end

was displacement, yes or no? But, that displacement, you have to qualify that. Times displacement

— but this is through the [inaudible] can be proven. I’m bypassing that, but that displacement

is in the direction — I put this in capital, in the direction–. –Of load. Put under the

load, yes, under the load. But, that’s the same meaning. In the direction of the load,

and corresponding–. –To the load. In the direction of the load and corresponding to

the load. Now, let’s see what happen. Okay, that’s why it has a limitation. Now, if I’m

calculating total energy here for this member, this member is [inaudible] this much for one,

yes or no? So, therefore, what I’m calculating is here — it’s delta. Everybody understand?

Delta must be in the direction of the P, is that correct? That’s a direction, and corresponding

to P is delta, because that is appropriate therefore mentioned for axial member. Everybody

understand that? So, then he stops and he comes and asks me, what is the formation here.

Everybody understand that, because B goes over two, yes or no? Can I answer that question?

No. Not at all. I’ll tell you to go through there with Castellano method, because this

is — write it down — this method, because this is one load not two load. I cannot have

two load. If I have load here, a load here, I cannot write that because I have only one

load and one displacement, yes or no? Please write it down. So, this is limited to one

load. This is very important. And, one displacement only. And, that displacement has to be corresponding

— has to be similarity with the load. So, if you are talking about P or M you are talking

about delta, yes or no? If this point is up and down, I cannot calculate that. Everybody

understand that? We come to here or are calculating P and C. I cannot calculate P at P. When I

comes here, now what am I calculating? If you answer me that question, then we go to

the quiz. If I calculate that one, we will do that. This beam is going to bear like that. In the direction of the load, yes or no, would

be? One. One. Actually, somebody come and tell me theta,

I got to do it. Is that correct or not? However, if I have a beam like this that’s chambered,

put that one in your note, too — if I have a beam like that, and there is an M and MO

is involved here, and this being bent like that for example, which is not true. Is that

correct or not? What am a calculating here? Not what — see that corresponding to the

load. What’s corresponding to the load? Under this, what are we supposed to [inaudible]?

Angular formation [inaudible] theta. Is that correct or not? So, please write it down in

your note, because this is something — you’re still not getting it. Look guys, there is

a R and there is an M there? R is for what? R is for Y. M is for theta. It’s not–. –If you don’t touch that M, you don’t touch

the theta there. Everybody understand that? We have gone through that ten times. So, corresponding

to this load is what? Corresponding to that load is theta. Everybody understand that?

So, if somebody was — take the theta here, I cannot do it. If somebody come Y here, I

cannot do it. Is that correct? So, there’s a way around it. We’ll see that next time.

Is that correct or not? However, that’s what we are going to do. One load, one displacement.

That’s important. The other problem is this. If there’s AB here and there is another load

P two here, which displacement? I cannot tell if that’s correct or not. So, this is [inaudible]

displacement in that scenario, so you go there. You calculate that. So, you should be able

all to calculate. Do all your homework. Do the first two line–. –For example, but–. –Is that correct or not? Yes? So, it should

not be that difficult because these are all we did, but [inaudible] most appears [inaudible]

this one also. This is the most common. So, actually that this one is good too because

of the direction. It’s going to be the integration of [inaudible]. Okay, guys. So, that’s the

due date.

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##### Written by Valentin Lakin

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very nice thanks

plz wrk engineering mechanics mathematics

lectures are great but the editor is stupid . you idiot you skipped important section of lecture at 31:07 while editing. what's the point of a lecture you moron.

31:09 to 31:11. There is a sudden gap in the video

Delta U equals delta P times delta delta, right? 🧐