# Strength of Materials II: Review of Strength of Materials I (Torsion, Bending, etc.) (1 of 19)

Okay. So, let’s start. So, to start, we get
chapter one. No, I have this instrument. Everybody knows that I’m using this to show you the
state of a stress. Now, many of you, as I said, you have learned this idea, and everything
that we have given in ME 218. But, let’s put it in perspective here because this is what
we need for ME 218. Let’s say we have a rod here. Look at this two point. One of them
has a blue dot. One of them doesn’t. I can go on top. I can go on the side. Either way,
or can show both of them. Doesn’t matter. So, let’s say we have a rod here like that.
I go from chapter one to chapter six. All of them. Cover it very briefly. Let’s say
we have here a p which is applied to the centroid of this rod. Centroid because this supposed
to be axially loaded. When we said axially loaded means the load is at the centroid of
this rod. Otherwise, it would be eccentric. Notice if the load is here, it’s not set axially
loaded. Therefore, the [inaudible] is going to [inaudible]. So, everything is start from
the centroid in chapter one, two, three, four, etc., etc. So, therefore, this is the [inaudible].
Look at this point here, and this point. Let’s call it this point, for example, b. Okay.
And, I’m looking at that point, and I want to find the state of stress at that point
throughout the book. We are going chapter by chapter to see what happen. First, this
is chapter one is axially loaded member, and I want to calculate the stress at that point.
So, what you do, this is the point, no dimension here. This is just showing that [inaudible]
point there. This is the dot here. This is the head of the needle there. But, in order
show the direction, we show it a little bit larger so at least this plane is perpendicular
to the x. That plain is perpendicular to the y. And, here is your element, and we work
to find out the state of stress at that point b. This is point b, and depending on the what
type of load we have. First, we have this guy. Okay, everybody knows that when I cut
this stream into half. So, this is chapter one. So, we have p here. We have p here. Of
course, you have that point here either on this side or that side. Notice the point could
be on the left or could be on the right. Doesn’t make any difference. This is the static. Now,
what do we have there, inside the body? Everybody says there is a p. Wrong. There is no p there.
There is stress inside the body. Notice it’s very important to recognize this. Many people
go through ME 218 without this recognition. This is an actual load. There is a hook here.
Somebody put a cable and put it. [inaudible] Is there inside, also, a p? Well, sure the
p there, but that’s not what’s happening here. What’s happening here is stress. That’s right.
So, if I show that in a three-dimensional way, like that, and the point is sitting here.
That’s the same point here. What you see inside this body is this. Actually, it is in reverse.
I want you to know, actually. Because of this p outside, you have here a sigma uniformly
being distributed over the entire area. Is that correct or not? Yes? Correct? Now, since
this is the uniform, and you call the area equal to a. If this area is called a, and,
as everybody knows from physics classes, if you want to calculate. If you put the pressure,
uniform pressure on some area, you could multiply the pressure by the area to get the force.
Yes or no? Correct? So, this is the same thing, but it’s the pressure this time is a vacuum.
If it was compression, actually, it was there. Pressure, yes or no? So, it is the [inaudible]
vacuuming, uniformly distributed. Actually, sigma multiplied by area, but that, when I
do it together should be equal to what? It should be equal to p. So, in reverse, actually.
This is the sigma. The inside is sigma, so this is what you see. And, here, you see equal
at. Oh, but I’m sure you have got through that, and you know what I’m talking about.
This p, well generally we don’t write that because we don’t want to show it every time
like that. Generally, we’re going in reverse. We say, as soon as I see a p there, I know
this is happening. So, what we write, we write sigma equal to p over a. that’s what we do
that. But, that’s why internally, when I cut something, I always call it something else.
So, if this is p, I call this one n. why I call it n? Because that is normal to the surface.
That’s right. Perpendicular to the surface, but the value of m is equal to p. It is imaginary
force. I usually put it dash down through it. That means it’s not in this format. It
is in this format. Everybody understand what I’m saying there? Yes or no? So, it is a stress
is not there, but since this is very difficult to show. This is what we do. However, what
happened to this point now? When we go here, sigma is in which direction? Sigma is in there.
That direction. What’s that direction? You have to call that one. Let’s say this is x
and this is y. Is that correct or not? That’s how I call it that. Is that correct? Therefore,
in this side of element on the right hand side of element, sigma is going toward the
right. Is that correct or not? So, when I look at that element on the right hand side
of element, this is the sigma. Yes or no? Correct? On the left hand side, look. This
is the left hand side, but I don’t have to do that. I know the two are equal and opposite.
But, what I’m saying that, be careful. Some of you may choose this left equilibrium. Some
of you choose this one, but the end result is the same. Is that correct or not? Look
what happen here. In this one, you are on the right side of element. On this one, you
are on the left side of element. Everybody with me? So, therefore, doesn’t matter. We
know from the static. I can use the left hand side or I can use the right hand side. The
stress should be the same. Stress is the same, and the stress at, in what we put here. Sigma
x equal to what? Plus p over a, plus because it is in tension point. Yes or no? Correct?
So, that was chapter one, and then we went through some other detail which is not required
at this time because I just want to cover with you the highlight of what we did here.
Then, what did we do in chapter three? What was chapter three? Those of you who’ve just
taken ME 218. [inaudible] chapter three. After we went through temperature change of determinant,
indeterminant, Hooke’s Law, etc., etc. The major part of stress analysis was the sheer
stresses due to the torsion or torque. Is that correct or not? Now, let’s look, this
is the same. I’m changing now the format of the load. Notice what I’m doing there. I’m
changing, now. This one is the same [inaudible] or the same shaft. Brown again. The same point.
This time, I’m going to put here a cutter clockwise torque on that outside here. Obviously,
on the other side, I should have [inaudible]. Equal and operative. Again, this is what I
expect from you to understand because these are type of work we are going to do in this
class is this. If I cut it here, you have done it many time yourself. If I cut it here,
everybody does that automatically. You set a t here, so look at this t. This torque is
going counterclockwise. The internal torque here, we should be going clockwise. On the
other side, should be going? Counterclockwise. So, as you see everything is in equilibrium.
Is that correct or not? However, what happens because of this? What is happening inside?
Again, this is external torque. I can take this item. We’ll do this to it. Everybody
understand. But, if somebody else here doing inside? No. Inside there’s nothing there.
What’s, again, inside? Stresses. Again. But, the stresses are in the form. If I sum it
up, it should be equal to this torque. So, that gives you the idea where that the stresses
should come from and how should they behave. So, is this the left hand side equilibrium?
Yes or no? So, here you have at the end of that, you have here a torque going like that,
external with the reaction there. So, what you have inside here? That’s right. First
of all, the torque is going counterclockwise. Yes or no? Because I’m looking at this side,
not the other side. If it’s going like that, this is what you learn. This is what the review
is all about. First of all, there are stresses there, and the stresses are in the form of
the sheerest stress. It is in the circular format, yes or no? Remember what you learn
there? At the center of the shaft, the stress is? Zero as we’re going outside the stress
cape, increasing when we go on the outside. The stress is the highest magnitude. Is that
correct or not? So, that is what you want to understand, that because of this torque
that we have there, this torque, that this is what we get. We got tau equal to tc over
j on the outside. This is tau max. We call it tau max. That means that stress will be
here, and it is in circular format. Everybody see that. Correct? And, if this is my point
here, which way does it go now? Look. Which way does it go? Is it going down or is it
going up? Going down. On the right side of element, which is this
side. Everybody [inaudible], the sheerest stress is going, yeah. That’s right. So, this
is what I want you to do. Okay, this is really what is expected from you to do both in ME
218 and here. Of course, as you see on the other side will be equal opposite. So, therefore,
on the other side will being going up. Is that correct or not? And then, as everybody
knows, tau xy equal to tau yx. So, I have to go toe to toe and head, I mean toe to toe
Now, then, I have here a tau. This is a state of a stress combining the previous one to
this one. Now, you can have all, the two of them together. Is that correct or not? Yes.
So, now, what is the tau? First, look at it. What is this subscript for tau? It is xy or
yx, okay, equal to what? Tc over j. Is that plus or minus? Everybody’s familiar with that.
That’s [inaudible] ask. If not, I will, I’m going to explain that. You know, many people,
they take a look at that, they think this is the, this is the sheerest stress. The sheerest
stress in this format, you want to put somewhere in your note. If you have a stress element
like that, if sheer stress goes like this, that is plus. This is tau we are talking about.
And, if it goes like that, that is minus. And, in this class is very important to assign
sign to your [inaudible]. In ME 218, sometime we let it go because it was not important
until we get to chapter seven which was the sheer, more [inaudible]. But, generally, we
should assign a sign to that. However, this one is negative. So, I’m going to put a minus
here. Is that understood? Yes? Yes, question? All right. This is chapter three. Now, if
the point was here at the distance of rho, of course, that equation changes to what?
Tau equal to that’s a t rho over j. Just for review. But, usually, because the maximum
is on the outside, we use this equation more than the other equation. This was the highlight
of chapter, of course, three. Of course, there were a deformation. The deformation was phi.
Phi was the angular deformation, etc., etc. We get to that later on. Is that understood?
Now, the last part of discussion is this. What has happened if I can put this rod under
the bending? Now, what kind of bending do you have, or what? In this class, now, [inaudible]
219 especially. Notice we have this as the bending. Everybody understand that. At this
as a bending is that way. See, these are the three component of the moment. This moment
is about which axis? What’s that? What is this moment is? This moment is? Usually this
is xyz, by the way. Do you have [inaudible]. I always call this the direction xyz is standard
one. So, this moment is about the? Z axis. So, if I put here a moment like that. So,
let’s erase that. Now, if this is the same, right? The same point. Nothing has changed
except the load has changed. xy is the same thing. If I put a bending movement like this,
I’m going to create sigma equal to what? Sigma plus minus mzc over iz. Is that correct or
not? Yes? This is what you learn in chapter four. This is mz. Yes or no? Now, if I do
that, the top point, every top point on the top would be in, right, on top. Compression
on the bottom will be tension. Look at this point, guys. If I am putting that in tension,
which direction is the direction of tension. This is very important. It is going this way.
It still is in the direction of x. It still is perpendicular to the cross section. Everybody
see what I’m talking about. There is no mystery there. As soon as you say sigma, sigma must
be normal to the cross section. If your cross section here normal to that is this direction.
So, it must be x. Is that correct or not? However, if the point is here, look what happen.
If I bend it like that, the point on top is compression. The point on bottom is tension.
How about this point which is on the neutral axis? Does it get anything there? No, it [inaudible],
so it doesn’t get. However, if I bend it about the y axis, look what happen. Then, this side
would be in tension. This side would be in compress. [inaudible] that point on here,
so as far as mz, there is nothing there. If there was an my, I put it that, and that’s
sigma. I would add or subtract from that. Is that understood? Yes? Did you see how this
behaves. So, now, from now on, any problem that come in combination, which we see a lot
of that in this class, you can add chapter one and three and our together which is all
you have to do. And, you will see that result in the second page of my handout. Then, you
go to the second page. The first, and this handout is just for review as well. You go
to the second page. You see, for the beam analysis. Second page there, you see the beam
analysis there. You see the sigma and tau is given there. Right? In 3D, that’s what
I was talking about. Just look at there. Sigma and tau. Notice if, let’s do one more thing
actually. Let’s let me do one more thing, then I will go back again to that one. Next
time, if this is the cross section, let’s do chapter six as well. This is the same cross
section. This is the same point b. This time, let’s put there a sheer force of magnitude
[inaudible]. What happen there? Which chapter is this, by the way? Sheer stress in its cross
section. No. We studied that twice. One at the very early. No, I’m changing now. I’m
going from now to sheer forces. So, this was for the bending. I talk about sigma u today,
We had mzc over iz and plus myc over iz. That depends where the point is. Some of them could
be zero. Some of them not zero. Everybody understand that I’m saying that? Depend on
the location of the point. I cannot do it for this problem. It’s difficult unless we
go to the actual problem. However, there is another sheer stress, and that’s the sheerest
stress to you in that beam cross section which was chapter six. Is that correct or not? Yes.
Now, what was the highlight of chapter six? I’m going to put it here. Was this one? Usually
we start it with a rectangular cross section. Yes or no? Is that correct or not? That there
was a sheer force here equal to something, 80 [inaudible] or whatever, or 10 kips, 10,000
pound. Then, we decided that there is a tau average. I put the highlight here. Tau average
was equal to force over the area. We saw that in chapter one. Yes or no? That means from
top to bottom, the sheerest stress is uniformly there. This is the sheer force. I have to
divide that by the area to get a uniform sheer stress from top to bottom. Is that correct
or not? However, everybody knows from discussion of ME 218, that didn’t work. Why didn’t work?
Because on the top, on this direction, notice when I have a stress going vertically, I must
have a stress going horizontally as well. Because tau xy equal to tau yx. Is that correct?
But, on the top, there is no stress going in this direction, so there is no reason a
stress going to be there. So, it is decided that this stress along that line must be equal
to zero. And the same thing here because the lower part of beam doesn’t have any stress.
It’s not dropping against anything, so the stress. So, that formula didn’t work. So,
[inaudible] that formula, we end up with what? Tau equal to plus minus vq over it. So, you
have to remember that. This was the result of chapter four. That’s why I gave you two
example there, too. Look at the third line. That’s there, this reviewing this material.
So, I’m giving you all the highlight of what you have to do in all this homework. Everybody
understand what I’m saying that. Now, what happened there? So, everybody knows what i
is. i is there for this, i is the moment of inertia of this cross section with respect
to neutral axis that passes through the centroid. Yes or no? Of course, sheer force is sheer
force. They will give it to you. So many number of pound or newton or kilonewton. Is that
correct or not? What’s the q? What’s the t? How do we determine the q? How do you determine
the t? That’s the question that many people generally forget about that, to which we need
it for this class. So, what are those? If I give you the, no, that’s it. So, if you
are interested to find out the sheerest stress at this level. Let’s see. We are not on top.
We are coming down little bit at a distance of y here, let’s call it there. q, I don’t,
I’m not going through the prove of. The q, you make a cut here. This area above your
cut is the q. You need the q of that area with respect to neutral axis. Yes or no? And
then, what’s the value of t? Okay. So, are you okay with me? All right. What’s the value
of t? [laughter] Notice guys. I’m watching you all all the time. Remember, I would, I
would explain that to you later on. I hope you don’t mind. I never let you go after this
class. So, if you are in this class, you need to listen to me. And, if you listen, you probably
will learn a little bit here and there. If it is too boring for you, let me know because
I can go much faster if you want, or I can go much slower. But, the point is, you need
to know this material. I have done this enough to know the significance. Although, you may
important. Therefore, the q is q of that area with respect to neutral axis. And the t is?
The length of the cut or the thickness of the beam in this. That’s the t. So, that changes
from problem to problem. So, you put it there. Therefore, you may have here vq over it which
add it to that one. Notice these are normal stresses. These are the sheer stresses. Now,
in this scenario, as you see, if you are this scenario. Since it is like that, that’s the
same problem. It’s still the sheerest stress here is zero. The sheer stress here is zero,
but the sheerest stress at the middle which is the neutral axis at [inaudible] correctly.
But this is supposed to be half half circle, half a circle. The stress going downward,
therefore, I have to put add it to this. Is that correct or not? But, they are going in
this direction. Therefore, it is? Is negative. so, that’s it. So, we are covered the whole
thing. Is that correct or no? Yes? Now, let’s do some example. So, if this is the idea of
stress element. Now, what we do with it and why we do the things we do in chapter seven
which is the more [inaudible], that all [inaudible] going to be explained on Tuesday class. Everybody
understand that? For time being, you should do this sort of thing until we get there.
And now, these are the two equation that you showed there. So, in general this is plus
minus plus minus. Depends on how it work there. Now, let’s go to problem. I usually if I have
time, I do problem number one and two on the handout. They are class exercises. Go on the
next page, or I don’t know which page it is. I can tell you. Go to page four. Okay. There
are four example there, but I’m going to bypass number one and two and go to number three
now for time being. Which is similar to what you have there. I’m going to do one example
of each one of this. So, in order for you to get into the picture. So, let’s do that
together. Everybody brings that page up, and I’m going to put the problem on the board
for you guys to see. This is the problem that it is a plate. There is a hole here. This
is symmetrical, of course. A hole here, and there is a here p and p. Then, somewhere here
we have put a gate. This is a gate there, and we are measuring this train at that point.
And, at that point, the strain is measured. You, everybody knows what the strain is. It
measure to be 100 micro. What’s micro? Micro is 10 to the power or minus 6. So, please,
when you come with the strain in your problem, do not come with .000000. That’s very. I mean,
that’s not readable correctly. You always express your strain in macro. If you go to
the lab to do. Some of you may already taking the lab this quarter. Anybody taking the lab
this quarter? Yeah, so you go to see the machine? The machine always give the strain in micro.
That’s the first test you are going to do, measuring all the strain in micro. This is
the unit of the strain, so please write it like that, which is 10 the power or minus
6. So, anyhow. That we measure. Then, this distance from here to here is given 30 millimeters,
and the cross section here, somewhere. The cross section of this plate here even at,
I believe this is 10 millimeter, and this is 50 millimeter. And, also e, [inaudible]
of elasticity is given equal to 100 g Paschal. G is 10 to the power of 9. That means this
material is still [inaudible]. Yes? What’s the value of p? That’s the question? Couple
of your problem, I believe 121 and under 204 is in the same form I got this one. Is that
correct or not? So, I, what I’m doing, I’m not doing it standard eccentric loading. I’m
just doing one that is a little bit mixing the ideas together. But, this is a review
after all. How do you calculate the p? What should I do first? Let’s put it this way. What? Cut this. Every problem start with that. So,
don’t forget that, guys. You cannot go through this first without first doing the static
or cutting. Is that correct or not? I have to cut this one somewhere, but however for
this problem does not make any difference whether I cut it here and cut it here or cut
it here. That’s why I did not give you any particular section. You will see in a minute,
does not make any effect. Is that correct or not? So, let’s go to this point, actually,
what you want to do, you want to go to the section where there your gauge is because
you are measuring the strain gauge at that point. I did not give you any particular location.
So, that means it doesn’t make any difference whether you are a little bit to the right
or to the left, if that makes sense. That is purposely done this way. Then, I cannot
do any of this. See, I already explain what you have to do. You have to come up with the
sigma at tau, but that is irrelevant unless first you do the static of the problem which
means a cut coming with internal action. The one that supposed to be a stress. Is that
correct or no? Which you show it in the form of a force and a [inaudible]. Is that correct
or no? What do we have there? Now, I can cut it there. I can show the left hand side or
can show the right hand side. Is that correct? The rest show there, left hand side. So, if
I go like this, this is the p. Of course, the p is not given. Remember that. We are,
the question is what’s the value of the p. Okay? What [inaudible] is given to be. This
is the data. If I give you p, this is the whole thing. If I give you p, you go ahead
and calculate the stresses. Yes or no? Correct? This the reverse. Strain is given because
most of problem in ME 219, by the way, is like that. That’s why I want to change your
mindset a little bit. Everybody understand that? We measure the strain. We can never
measure the stress. Everybody understand measurement. You measurement always true to strain. Anyhow,
so problem is [inaudible]. We’ll see that in later discussion. Anyhow, now, do I have
a normal force here? The normal force equal to p. Where should I put it. Centroid, right.
That’s right. You don’t go to the centroid. You put here, this is internal, so I should
use a different color. So, I use blue this time to show you that’s the n. But, of course,
n equal [inaudible]. I put that dashed line to tell you that’s stress. Is that correct
or not? Yes? And, what’s the value of, is there, look at this. This is a couple. Yes
or no? So, it is not in equilibrium because it’s a couple. Therefore, there is action
here in the form of a couple. The action of this one is going that way, so [inaudible]
reaction going like that. That’s the value of n. Is that correct or no? I gave this problem
in the final, actually, to both my classes. I mean, similar to that. Not like that. Many
people, they make a mistake. When they were calculating the m, they calculate the m p
times what? They calculate p times .30. Is that correct? That is a big, big mistake.
Everybody understand that. Since the point is not here. Because you are ruining all the
idea here. I just keep stressing to you, everything is start from neutral axis. Didn’t I? Is that
correct or not? So, you don’t find it here. If I can’t calculate what’s the value of n.
Everybody says p. There’s no problem. Well, I can’t ask you what’s the value of m. Everybody
says, or some people said .03 or 30 millimeter times p. That is incorrect. Everybody understand.
Some of you already answered me correctly. Everything should go to the centroid. Remember
that. So, everything should go to the centroid, but where is the centroid? The centroid is
here. [inaudible] this centroid of that is very simple. It is here, so it is 30 millimeter
plus 20. So, m value actually is equal to .030 plus .025 times p, and so therefore equal
to .055p. Remember, every chapter that we were talking about, that was the key. But,
this was sort of loss among all the other things. Everything is start from neutral axis.
Actually, what happen here. Look. If I am above the neutral axis, the action of the
moment is what? It is tension. If I’m below the neutral axis, the action is. How can you
understand what I’m saying that. Correct? The reason we are deciding on tension or compression
is because we are above the neutral axis or below the neutral. Now, where is our point?
Now, our point is here. Yes or no? And, this is where the gauge sitting here. Now, I already
discussed that. What you have there? You can now, everybody should answer that question.
What is sigma, correct? The sigma equal to what? Sigma equal to no. [inaudible] start
with a state of stress. You see you need there? State up a stress for that point. That’s what
I was talking about. We get to that later on for that. We know there is a, in this section,
there in an internal axial force, and there is an internal moment. Yes or no? On, due
to the axial force, what do we have? Sigma equal to what? [inaudible] is that correct
or not? Yes, because it is tension. Now, up and down determine whether the rest of this
formula is plus or minus. If you are on the bottom of the beam because of m, you have?
Compression. On the top of the beam, you have? Tension. Where are we? We are on top. Therefore,
I need to put here plus what? mc over correct? That determines my state of stress, and that
determine the [inaudible]. Is that correct? Now, we go to there. Is that correct or no?
However, if I was on the bottom, then as I know you appreciate that. That would have
been plus p over 8 minus mc over i. Is that correct or no? Yes. Okay. So, why don’t you push back your chair or
come here or something like that. I’m sorry. I don’t want to spend time doing anything.
Next time, I’ll draw it somewhere else. Thank you for [inaudible]. Can you see it now? Okay.
I do it here. So, it’s easy to. Here is, this is what we are talking about. This is the
p. This is the n. Is that correct? So, since we are here, therefore, [inaudible] to this,
we have p over nu. To that one, we have mc over i. Yes or no? Now, okay. What is given
now, or what’s not given? First off, look at this cross section. This is our cross section.
So, let’s put our now data. I need a. I need i. I need every one to finish this one. Now,
everybody knows from previous chapter [inaudible] 218 previous work that we have done that there
is a relationship between a stress and strain. We get that to that in a minute, but let’s,
first, calculate the area. Of course, the area is, this is the cross section. I’ll put
it here one more time. [inaudible] So, it was 10 and this was 50. Yes or no? So, area
equal to 10 times 50 equal to 500 millimeters squared equal to 500 times 10 to the power
of minus 6 meters. [cough] So, notice time, 10 to the power of minus 6 because meter square.
Is that correct or not? Yes. Many people, again, by mistake, they put here 10 to the
power of minus 3, for whatever reason. What this is meters. Anyhow. Then, you calculate
i. I is equal to 112. Okay, this is the neutral axis. [inaudible] i equal to 112 of base time
high cubic. Everybody can do that, so I’m not going to do that part of it because it
is simple enough. Therefore, let’s see. Where is my solution? Where did I put it here? Nope.
No, here it is. Okay. There, i become equal to. .104 times 10 to the power of minus 6
meter to the power of 4. So, I’m expressing everything in the meter. And, of course, c
in your equation is half at 50 millimeter or 25 millimeter or .25 meter. So, as you
see, ca given. This is p, and this is .055p. So, everything now, I have [inaudible], p
and sigma. So, I’m looking for p. So, somehow, I have to get through my information that
I have from the strength of material. One, I should be able to calculate sigma. What’s
the sigma equal? What? So, I want you to remember this. This is, again, a very, very important
that we are going to make a reference to this curve many, many time in this class. Remember
that sigma versus epsilon, which we call it Hooke’s Law. Remember? All that spent about
a week discussing that in every ME 2 [inaudible]. That goes for straight and goes like that
and goes like that. Everybody understand that? Slope of this line is modulus of elasticity.
This is the equation we are going to use extensively in this course, in this, through this Hooke’s
law we end up to be epsilon to be equal to sigma over modulus of elasticity. So, we use
that a lot in this class. So, that’s the one that’s our purpose [inaudible] of this problem
because we have to use that in this equation. [inaudible] Is that correct or no? Now, to
make this a little bit more for you to review that, then we have another curve similar to
that for tau versus what? Tau versus gamma. What’s gamma? The sheerest strain which is
the angular deformation. Everybody understand? Epsilon is changed in the length. Gamma is
angular deformation because sheerest stress calls angular deformation. But, this time,
when we get the slope, this time the slope is not modulus of elasticity. This time is
modulus of rigidity. That’s right. So, the same ratio works. So, here you’ve tried gamma
equal to tau divided by. This is another equation for a reminder. Of course, we are not there
in this problem. We don’t need it, but I just refresh your memory of some of the things
that you must learn in ME 218. Is that correct or not? Yes. However, for this problem, you
need only the first one. So, here, what I’m saying that. Since sigma, since epsilon equal
to sigma over e, therefore sigma must be equal to, sorry. Sigma must be equal to e times
epsilon. What’s the value of e? It’s 200 times 10 to the power of 9 Paschal which is newton
per meter square. Is that correct or not? Yes. Multiplied by epsilon. [cough] Epsilon
was how much? 1000 times 10 to the power of negative 6. You simplify that. That become
equal to 200 mega Paschal. Because epsilon was given to me. Epsilon is given the type
of material. That’s exactly what you do in the lab. Some of you think you are great in
lab. They are going to give you a piece of metal. They color it. You don’t know whether
it is aluminum or it’s steel or [inaudible]. They ask you to measure the epsilon through
the test, and to tell their guy whether this is steel or this [inaudible] go to the table.
You find out modulus of elasticity, or you do the testing on the sheer stress, or finding
the g, etc., etc. Those are some of the testing will be done in the lab. Is that correct or
no? Anyhow, so, what I’m saying that now, you put in this one, that’s all we have to
do. So, the final phase of the problem is putting the two and two together. Notice what
was important here is these two and the state of stress at that point. Everybody understand
that because that’s the state of stress. That is the eternal forces which was two action
now here. So, finally, all you have to do now, all you have to do here. Sigma, which
is 200 times 10 to the power of 6 Paschal or newton per meter square, equal to p divided
by area. The area is 500 times 10 to the power of minus plus m. m is .055p times c. c was
.025. Divide that by i. i is .104 times 10 to the power of minus 6. Keep these two for
your mega because [inaudible] go to the numerator, it become plus. So, it become your mega through
the rest of the analysis. Calculate for p anyhow. p ends up to be equal, for this problem
become 30, or 13,158 newton. So, you can write it [inaudible] 13.2 kilonewton. Roughly they
are using three digit [inaudible]. For that amount of the p applied to here. You get that
reading there. You increase the p, of course that changes. Is that correct or no? Did you
see what happened? So, we cover lots of things. Now, let’s go quickly to problem number five.
I don’t have time to do that in totality, but I give you the highlight of problem. Sorry.
Problem number four. Problem number three was this one. Problem number four on the same
page. Notice I give you another state of stress. There is a reason I’m doing that, because
everything that you learn through this example in ME 218, a highlight of stresses. Now, look
what we have there. This is the scenario there. So, please write it down. There is a beam
there, or cantilever beam like that. There is a force of 80 kilonewton at this end. Somewhere
at a distance of 102 millimeter at the point here which is 16 millimeter. This is 1 6.
16 millimeter below the neutral axis. You want to find this data with stress for this
problem, knowing the following. The cross section of the beam is, which I’m showing
here a little bit larger. The cross section of the beam is 18, 18 millimeters y 80 millimeter.
This is where the point is. The centroid, of course, is at the middle, and there’s where
your points. So, if I call this one .8, your .8 is here. Again, exactly what you asked
me last time. If I’m about to find the state of a stress at that point, the first thing
is to determining their internal forces, okay? So, I have to make a cut. Notice I do not
need even to do the reaction here. All I have to do it, do it this way. You see? It’s, you
go like this. Sorry for picture. I’m brushing with so the picture doesn’t come out good,
but you understand what I’m talking about. Again, where is the centroid? The centroid
is at the middle of 40 and 40. Yes or no, because it’s rectangular shape? Yes or no?
Therefore, what do we have here? Do I have axial force? No. There is no p over a. Actually,
look what happen here. In ME, one person, right, in particle. I remember it. I put here
a p in one of the problem, and if you end up here saying sigma equal to p over a because
you see that p, that means you haven’t learned anything from ME 218. Is that correct or not?
p over a is only when force is along the this way. Is that correct or not? This force p
causes this to bend downward. Is that correct or no? See, obvious. But, as I said, occasionally,
very occasionally, people make mistake there. So, do I have a sheer force there? There is
a force here. There is a sheer force there. Do I have a bending moment there? Of course,
I’m going faster. Again, the action is here. The reaction is there. Is that correct or
no? Where is my point? This is the centroid. My point is here. This is my point a. Is that
correct or not? A little bit below it. So, this is where you are. You can show it as
a little square again. Now, all that little square looks like that, this is what it’s
going to looks like. I gave it to you in the handout. In the handout, it is there, but
let’s look at it. Let’s go through the sheer. If I want to go through the sheer, this sheer
force is going upward. Therefore, the sheer stress must be going upward. Where? On the
left of element. Is that correct or not? So, where are we? We are on there, this side.
Yes or no? Everybody with me? Okay, so if this happen to be, again, x and y. So, you
define your axis first. Sometime they are not in x and y. If you are here, x and y,
that one will be y and z, etc., etc. Everybody understand that? I just used that for just
explanation. Then, I have a sheer stress. I show the stresses in color. So, I have a
sheer stress going upward like that. Yes. So, on the other side, it will be coming down,
then toe to toe, head to head. So, again, here, I have a sigma x, and I have a tau xy
exactly like previous problem. The various standard there. Now, let’s see what we get
from tau xy. Tau xy is minus vq over it. Yes or no? Yes. For which point? For point a.
So, we have to determine the value of that. Now, as far as the is there a normal stress
there? Yes or no? Look at this moment. This moment, [inaudible] on top is? Tension. On
the bottom is? Compression, and it is linear. Remember that? So, this was the idea was presented
to you. Let’s put it this way. This is the tension. This is the compression. Yes or no?
On the neutral axis is? Zero, and between those two are? Linear. Where are you? You
are here, somewhere below the neutral axis. Is that [inaudible]. You are in compression.
So, you have stress going like that in the direction. And, that is [cough] negative m,
not c over i. y over i, because you are not on top and you are not on the bottom. Yes
or no? Good. So, let’s again. Like last time, I don’t need the area. All I need the i. So,
you calculate the i of this cross section. So, this is the rest of the data. I’m going
very quickly over that. The i become equal to what is problem which is problem number.
No, I’m just looking for that one. Yeah, here it is. So, i become, did you already calculate
that? Yeah. .768 times 10 to the power of minus 6 meter to the power for. That’s the
value of i. The value of m. Let’s calculate the m. The value of m is 80 times 102 is that
guy. So, 80 times 10 to the power of 3 meter times .102 meter. So, it become 8,160 meter.
The value of i here. The value of y, of course, is equal to 16 millimeter [inaudible]. You
don’t have to worry about the sign. So, therefore, it become [inaudible] to .016 millimeter.
So, all of that’s given. You calculate all of that. End up to be 170 mega Paschal. You
see it in the picture there when you put all those numbers together become 170 mega Paschal.
Is that correct or no? Calculating the tau was my major concern here because now we go
to the tau. Notice the sheer force is 80 kilonewton. Yes or no? Correct because the sheer force
is 80 kilonewton, and then, I have the i. I need the q and I need the t. Is that correct
or not? What q and what t. You already answered me previously. So, I have to cut it here.
Yes or no? But, don’t come from top because the q of above and the q of below are equal
and opposite. So, you go, you need only this q. Is that correct or no? Yes. q of that area.
So, I’m going to calculate q here. q for that .8 is equal to this area which represent the
mass which this distance was 80. This distance is 16, and this distance is 40. So, what is
left here is 24. Yes or no? So, it is 80 times 24. So, from here to here is 24. So, the area
is 80 24 multiplied by y-bar. Is that correct? Everybody knows that q is equal to ay-bar.
Is that correct or not? What y-bar are we talking about? Where is the mass center? You
put this mass here, yes, at the center of this rectangular shape? And this is your y.
Is that correct? Which is 16 plus 12, so that’s 28. So, this is 28. So, you put all, where
is the t? The t is equal to 18 millimeter. Yes or no? So, you put all those numbers together,
and the answer for that one is also minus 70 mega Paschal. Notice where I use this one
or two because I wanted to get this round number. And, you see those answer in your,
in your handout. Everybody see that? Okay. Now, one more problem I want to quickly go
over. So, and that is problem number one on the homework assignment. So, please go to
problem number one because your problem is two and three, to do two and three. I want
to give you a little highlight of the sheer and moment diagram. Of course, you are going
to spend a lot of time here with sheer and moment diagram in this class. We determine
a problem indeterminant problem. Lots of other ideas is going to be there. So, let me quickly,
everybody should bring that problem up. Problem number one in the homework assignment. One.
Homework assignment two is for next week. So, please, I gave you the handout for week
and half. So, it’s because [inaudible]. No, it go to the homework assignment one. The
last page, the last page before the last. There is three beam there. Homework assignment
one. Don’t make a mistake there. Yeah. Homework assignment which is the pages before the last
page. Yes. And, there are three problem there, number two and three are your homework. Remember
that. Number one, I’m going to do that. So, this is the problem you will see there. So,
please write it down. There is a roller here. There is a pin there. You have divide that
into here like that. This is point a, point b, point c, point d, and the distances are
5 feet, 5 feet, and 10 feet. And, at point b, you have, I believe, 1500. Is it 1500 load?
Yes. Yeah, 1500 pound load. Then, you have a moment of. 1000, 1000 pound foot. Be careful about the
unit. And then, here, you have a uniform distributed load of 250 pound per foot. And, we want to
draw the sheer and moment diagram quickly. Notice I am about few minute behind, but I
also hopefully I will manage the whole thing for you guys. Is that correct or not? Yes? Yes. All right. Of course, the first thing to find,
the reaction here and there. That should be no problem for any of you. Is that correct
or not? You take the moment about a or b, so that, I’m going to put it here. 1800 pound
here, and 2200 pound there. Is that would be the reaction. Step one. Step two, sheer
diagram. Sheer diagram usually is very simple for you guys, but let’s go over that just
quick. So, here we go. So, what you do, you go with the x axis here, sheer in pound. So,
you put your unit there. So, you put all your point there. a, b, c, and d. Okay? Now, what’s
the value of the sheer at point a? 1800. Plus or minus? Plus. Everybody knows that, know
that? Okay. Good. Why? Because you are talking about this. I’m going to say it anyway. So,
if the format is like that, if the deformation is like that, the right hand side is going
down. The left hand side going up. That is plus. Everybody understand. Is the deformation
is like this, means the right hand side is going up, and the left hand side going down.
Now, get me this. It’s very important. Statically, this is negative. This is? Positive. Statically,
this is? Positive. This is that. This format gives me the sheer and moment diagram the
way [inaudible] because the sheer and moment diagram is all about the deformation of the
beam. The formation of the beam comes in a week or two, and we are going to spend about
two, three weeks on that those deformation in this class. Now, here it is. Of course,
since this is going up, the sheer is going down. So, it will be this format. Obviously,
it is 1800 plus. Absolutely. Yes or no? Then, we go, nothing happened until an epsilon [inaudible].
Is that correct or no? So, we go straight. Then, you don’t go to b, you pass the b. Is
that correct? After you go past the b, suddenly 1500 pound appear there. Yes? 1500 is going
down. The sheer is going up. Is it plus or minus? Look at my hand. It is minus. Then,
I should be going down. Therefore, suddenly, I drop this to 1500, so I end up to how much?
I end up to? 300. Nothing happening until point c. After point c, I have again to subtract,
but what do I subtract? Is the linear? Is that correct? The slope of that would be 250
pound per foot. Is that correct or not? So, it means that from here, which is 300, the
total of this area is 2500 [inaudible] plus 300. I should end up to minus 2200. Then,
I add that to it, so therefore, I go back to zero. That is your sheer diagram. Many
people do that correctly. You don’t have any problem, but when we come to the moment, more
or less, some people have a little problem there. It shouldn’t be because it’s the same
idea. If you so nicely did your sheer diagram, why not do the? Moment diagram, but consider
this system. Consider the sign convention right here. I’m going to put the sign convention
down. If you have a piece of [inaudible] which as you know, I call it a smiley face. Is that
correct or not? Yes. This is positive. Again, remember that. Statically, this is negative.
Statically, this is positive. But, this format is plus. The frowny face which means it gets
this kind of action. This is plus. This is minus. Everybody understand that? That is
a negative deformation. So, both of base of that. We are going to go forward now. Here,
again, you go to x, and here you go to m. m is in pound foot. Don’t forget that. Some
people make a mistake that pound foot, pound inch, makes a different when you design the
beam. So, you are starting here at zero. Yes or no? Yes. Now, you have two option there. Two, three
option there. First, I have to [inaudible] then on, the sheer is constant. The moment
must be linear. Yes or no? The only question is up or down. Obviously, it is? Up because
this is plus. So, therefore, I have to go up to point b. So, this is point b. This is
point a. How far up should I go? You can go the area. You can cut it here. You can do
anything [inaudible]. One thing I didn’t ask you to do, I don’t want you to do, is write
the equation and use the equation. Everybody understand that. You can cut it here, use
equilibrium or use the area method which is much simpler. This is area equal to how much?
This area equal to, as you see, 1800 times 5. So, that is plus 9000. So, I have to go
from 0 to plus 9000. That’s no problem, but what do I do after that? Do I go down or do
I go up? Subtract. Everybody up or down? You see, I
hear some people not saying that. Up or down? It’s very important. Down. Correct. Now, look
what happened here. This moment statically is? Positive. No, it’s statically this is
positive. It, no doubt about it. It’s going to go counterclockwise is positive. That we
didn’t change. What you should look at is not that. What do I look at? Look at this
is? Positive. After that point, the beam behave like that. Is that correct? To balance that,
this is the system. [inaudible] The right hand side look. After that, you are here.
This is going this way. So, in other words, this is that way the beam should be behaving
like that to make it in equilibrium. Therefore, it’s not a positive mode. It is a? Negative
mode. So, I should be going down. So, how much down? 1000. So, this is 8000. Is that
correct or not? Don’t make that mistake. You look at this format or this format. If it
was like that, actually it’s in reverse, as you say. Everybody got that point? All right.
Good. Because that’s the only point. Then, from then on, I should add again how much?
This area. That area is 300 times 5 which is 1500. I should be going up. This is point
c. So, I should be going up to 9500. Yes or no? Look how simple this has been done. Easily,
it’s being done with the same idea. Up, of course, everybody. I mean, up the [inaudible].
Is that correct or not? After that, what happen? Notice this is plus. This is a little plus,
then suddenly, it become minus. So, I must be from this point c, I must be going a little
bit up, right? And then, I must go to point b which is 0 because that’s [inaudible] as
well? Is that correct? Where is that point, and how much that is? Give me one more unit.
So, I can use this idea of similar triangle. This is 300, and that is 2200. This length
is 10. It’s very easy to find that. It’s become 1.2. Is that correct or not? Yes? Then, what’s
this area now? No, this area was 1500 plus that little area is 300 times 1.2 divided
by 2. 360 divided by 2, so it is 180. So, I have to go another 80 down, 9680. And, [inaudible]
that is your moment diagram. This is the very important point. Some of you guys in the final,
you missed that. I don’t know whatever reason. When sheer equal to 0, moment either is maximum
or is minimum. In this case, in this beam, actually it was the maximum amount of moment.
9680 was maximum amount of sheer is? 2200. That’s here, that’s there. The next time,
we are going to design this beam. Everybody understand that? So, we have done some of
the design. So, you are all good for your.

## 14 Replies to “Strength of Materials II: Review of Strength of Materials I (Torsion, Bending, etc.) (1 of 19)”

1. Muhammed HASSAN says:

Strength of Materials I ? WHERE CAN I FIND ???

How long will it take to be recorded? And is strength of materials II the continuation of strength of materials I??

3. subrat kumar choudhury says:

Sir plz add full advance strength of material like thick cylinder,unsymmetrical bending& shear center etc

4. Ujjwal says:

Which book you are referring in this lecture??

thanks for this effort but can you share the handouts here?

6. AB says:

He is so educative, is that correct or not?

7. Husnain Hyder says:

What book is being followed

8. pramod sarode says:

Sir when u upload first part

9. Khoa Huynh Xuan says:

Has the Strength of Materials I been recorded?

Dr Mohammad …am a structural engineer…am really very thankful for the way you are interpreting things..you are super….

11. how to reach from -1to1to -1 says:

budha boar kr rha hai!!

12. Vedant Barot says:

Good explanation… Thank you

13. Nima Atashpareh says:

Who did record this video?!!! He/She was so lazy to adjust the camera to the lecturer !!! missed most of the parts!!!

14. Safeer Haider Zaidi says:

Can you please share the handouts too?