Understanding Stress Transformations in Solids

Hello everyone, this is Manas, your friend and tutor and guys I'm back with another video on mechanics of solids. The topic for today or rather the chapter is going to be stress transformations, principal stress and strain, something which is also referred to as compound stress and strain. Now before we start talking about as to what we are going to learn in this particular chapter, we first need to touch upon a basic concept of a tensor. Now guys, Until now, the one thing that we are absolutely sure about is that a point inside a rigid body is Subjected to or is characterized by six normal and shear stress components Did I just say six? Wasn't it supposed to be nine? If you just try to make a count of the total number of different components of stress It works out as nine, isn't it? So what is the answer? What is the right value? Is it nine component or six? If you ask me, well, both of them are correct. If you just try to make a count, then that will work out as 9. But effectively, there are only 6 components of stress. 3 of them are normal stress components and the remaining 3 are shear stress components. Well, that can essentially be worked out with the help of this 3-dimensional state of stress at a point inside a rigid body which is subjected to external loadings. Now, just take a look at this. This element over here, small element is acted upon by nine different kinds of stress, isn't it? And if you take a careful look, this is tau xy and this over here is tau yx. Well, to tell you, these both shear stresses are equal in magnitude, but I'll prove it to you right now. And if you watch, if you just try to have a look at this entire arrangement from this side. From the side, let's say we are having the side view, then the arrangement would be something of this sort. Here we go. Well, the face is going to look something like this, isn't it? There is going to be sigma x, sigma x. There is going to be sigma y and sigma y. So let me just make them. This over here is sigma x. And that's sigma x. That's sigma y. This also is Sigma Y done. What about the shear stresses if you watch along this surface? Okay, not this direction but in along this direction. This is tau y X. So this over here is tau y X Just like there is tau y X along this direction in the bottom surface also There is going to be tau y X but in the opposite direction. This is tau y X by This is again tau yx. Again, take a careful look. This is tau xy in the upper direction, this right hand side face, that means this one. This over here is tau xy and somewhere along this face, over to the side, we will have this tau xy in the bottom direction, this way. Tau xy, isn't it? Now, you know, this entire system has attained a state of equilibrium. And whenever any body, any particle is in a state of equilibrium, we always have these two equations working with us. Summation of all the forces along x, y or z direction will be equal to 0 and also summation of rotation, let's say torque, let's say moment of all the forces with respect to x-axis, y-axis or with respect to z-axis is also going to be equal to 0. So, we are viewing it from here. In that case, the rotation or the moment about the z-axis will be equal to 0. So, here it is. Moment about the z-axis, z-axis is equal to Zero, isn't it? Okay. So these are the two forces separated by a distance. Let's say this is the distance A and this over here is a square. So that's A and that's A. Two forces which are parallel to each other. Now these can be converted into forces also. Let's say at the top the area is A. In that case, this force over here, this stress can be changed to a force and I can just write A over here. If I just multiply A over here, this will become a force. All of them. If I just multiply them with A, they would become force. Isn't it? So these are all the force acting on this element. Isn't it? So two forces, right, separated by a distance, two forces are parallel, separated by a distance, they form a couple. So this over here, if you watch this, what sort of a couple is this? It's a clockwise couple. And for a clockwise couple, you need to take the positive rather negative direction or negative sign and for An anti-clockwise couple, you need to take the positive direction. So tau yx, okay. So this is clockwise minus of tau yx dot a. And this over here, tau xy, if you watch this, this is upwards and this is downwards. What sort of an effect is it producing? This is an anti-clockwise effect. For that, you need to put a positive sign. So plus, this is going to be tau. What? xy. xy multiplied by a and all of this is going to be equal to 0. Now, since the object is in equilibrium, you can do this stuff with respect to x-axis also, with respect to y-axis also. That can be done. In that case, a and a will cancel out. Tau xy will be equal to tau yx. So, tau yx is equal to tau xy. And guys, in the same manner, if you take the moment about x-axis, And if you take the moment about y axis, you will realize that tau yz will be equal to tau zy and tau zx will be equal to tau xz. So effectively, these are the three components plus sigma x, sigma y and plus sigma z. So in totality, if you just make a count, there are effectively six components of stress. Acting at a point inside a rigid bond. Done. Okay, so let me just rub this and let us start the main business. Okay. Okay, there is something else which I'll have to tell you what we know and what we don't know and what exactly is plane stress. Right. What we know. Okay, so this over here can be assumed as a cuboid. Now let's say this cuboid is being acted upon by a force. like this. Okay, let's say the force acting is P. Now, if I tell you that, can you tell me how much is the normal stress induced? Well, you can do that very easily. We'll say that's a normal stress can be calculated very easily along any cross section like this. So in this cross section, in this cross section, we can actually calculate the value of stress. Let's say this area is A. In that case, sigma n or the normal stress induced will be equal to B by A. Okay, fine. That means we know how to calculate the value of stress induced on a plane which is perpendicular to the axis. If I can just extend this over here, which I will drop this, the angle made over here will be equal to 90 degrees. So whenever you have a plane perpendicular to the axis there you can calculate the normal stress. But what you don't know is this. Okay, this is what you don't know. If I tell you, please go ahead and calculate the value of normal stress and also shear stress or tangential stress along this oblique plane. Let's say this is making an angle theta with the vertical here also it is making an angle theta with the vertical Can you tell me how much is the maximum normal stress or let's say in general how much is the normal stress at an angle theta on this oblique plane? Can we do that? Well, we can definitely do that but at the end of the video. I'll try to explain you all of this stuff and how exactly you can calculate the value of sigma n and sigma t that is normal stress and angelic stress on a plane which sort of is inclined at an angle theta with the vertical. So all of this and much more coming up in today's session, but before that let us have a discussion on plane stress Okay, so let's get back on track and we now know very well that there are as many as six Distinct stress components acting at a point inside a rigid body this state of stress. However, it's not very often Encountered in engineering applications or engineering practice Instead the loadings are coplanar that means they produce stresses Which which can be analyzed on a single plane and when this is the case We'll say that the member or the material is subjected to planes Let me just explain all of that with the help of a beautiful example. It's something like this. Let's say we've got a material like this, okay Yeah, like this. So these are the three directions. This is let's say x and this right here is the y direction and this over here. Well, let's say this is z. Now there are a bunch of forces acting. Okay, this let's say is f1. Okay, there is a force over here. Also, let's say this is f2. There is a force here. This is let's say f3 and then there is a force here. Let's say F4. Now, because of all these forces, a net force is going to act, right? If you consider a very small element, very small element like this, because of all these forces, right? So, there is going to be a tangential stress along this x direction. This is going to be sigma x, isn't it? Let me just put the arrows first of all. There is going to be a stress in the y direction also. this way and there are going to be shear stresses this way this way and this over here is the complementary shear stress if i were to make this in a proper sense it would be something like this that's the element this is going to be sigma y sigma y this is sigma x so sigma x then we have this done done tau x y Tau xy. That's it. So this is plane stress and this is applicable for thin plates. In the upcoming part of the video, we'll actually be discussing four cases. Case number one is going to be about direct stress in a particular direction. We'll take it as x direction. Then there are going to be two mutually perpendicular stresses. Case three is going to be a purely shear stress case. And case four is basically a combination of two directly mutually perpendicular stresses plus shear stress. So these are the four cases that we are going to take up. But in today's video, I'll only talk about case one. So let's start with case one. Here we go. So now let us start with case one, that is when a member is subjected to direct stress only in one direction. Let's say we are talking only about the x direction. So here is the analysis. So first of all, let us try to make a member and if we rather make a bigger figure, there is only one stress acting. Let's say in the x direction, let me call that sigma x. Okay, that is sigma s. By the way, the one assumption which I'm going to take is that thickness, assume thickness as unity. So guys, actually, the arrangement is something like this. If I try to make this in 3d, it would be something like this. Yeah, right. So this is sigma x. So this is the area. So if you multiply sigma x with area, you'll get the value of load applied in this direction, that is px. So let me just write px also over here, try to have a better picture. Okay, just over there, not to write here. Okay, so this is what we've got a member subjected to an axial tensile stress. And what we wish to do is to find the state of stress along this oblique plane. This is what we wish to find. Okay, we wish to find the normal stress along this plane normal stress along this plane, and we wish to find the tangential stress along this plane. Let's say that this, this is represented by PQ. And let's just say, the angle that this makes with the vertical is theta over here also, the angle is going to be theta. How do we get this? Very interesting. Okay, let me have another figure. So let's let's start from here and this way, this way and this way. The force acting let's say is represented by Px, isn't it? So there are going to be forces acting along this plane P2. Let's call that as Pt and there is going to be a force acting in a direction normal to this Pq. First of all, let me just write this Pq. This over here is Pn. Okay, so there is a bit of geometry which we need to encounter right now. This angle over here by the way is 30. So, the angle over here is going to be 90 minus theta. Sorry, not 30 but theta. So, if that is the case, Then the angle over here, here, just try to work it out. Let me make this properly. Okay, so that's 90 minus theta, this is going to be how much if this is 90 minus theta, this is got to be theta. Okay, that's, that's quite obvious. If this is theta, right, then this also is going to be theta, that's for sure. Just try to work it out. This is fairly simple. If this is theta, this again has got to be 90 minus theta. And if this is 90 minus theta, this is theta. If this is theta, between these two lines, if this is theta, then between these two lines, these two are vertically opposite angles. That's theta and that's theta. Done. That means what we can do is, here we go. Let me just start calculating. What we wish to find is normal stress. Normal stress Pn, not Pn but sigma n. Okay, so guys here it is. Normal stress where? Along this plane Pq. So that is going to be equal to the load acting normally that is Pn divided by area of this oblique plane. Area of this plane let's say is represented by area Pq. Now how can this be worked out? So Pn can essentially be written as component of Px along Pn. Along Pn. Whole divided by, that's area Pq. This is fairly easy. This is going to be fairly easy. Trust me. Okay, just keep watching. So the component of Px along Pn. Here it is. What is this? Px. So we want the component of this Px along. So, if this over here is Px, this is theta, then this has got to be, this direction has got to be Px cos theta. Okay, it's very simple. This is going to be Px cos theta divided by, what is this area Pq? This sounds a bit confusing, but it's not that confusing. Let's say we are considering this triangle. Okay. this triangle. So, it's something like this, right? Something like this, this way. So, we know this area. We want to know this area. That's PQ area. Okay? Now, this angle over here is how much? This is 90 minus theta. So, what you can do is, you can write sin of 90 minus theta is equal to sin of 90 minus theta is equal to a divided by this area a pq. a divided by a of pq. In that sense, you will have this. This is cos theta by the way. And a pq will work out as a over cos theta. Now, this is a known fact. Let me just write this somewhere. Okay, let me just write it over here. Area of the oblique plane is going to work out as a over cos theta. And this is something which we will be using in the upcoming cases also. So better to memorize this. Okay, here we go. So this is going to be how much? A over cos theta. And finally you have the result. Implies that sigma n. Sigma n will be equal to what? Sigma n will be equal to Px over A. Px over A. Here it is. If you try to relate these two. Px over this area. This area. Okay, this area. will be equal to sigma x by the way. So, sigma n finally works out as sigma x times of cos theta shifts to the numerator and this is going to be cos squared. So, this is the first result that we have got. Normal stress onto this oblique plane will work out as sigma x cos square theta. And there are, just to note down, Few more conclusions which can be derived. Number one, what conclusion you can derive is that sigma n will be maximum for a particular value of theta. Sigma n will be maximum when this cos square theta is maximum and when is cos theta maximum when theta is equal to 0 degrees. So, maximum when theta is equal to 0 degrees. And when theta is equal to 0 degrees, this theta will be 0 degrees, clockwise we have taken the angle, then this oblique plane will become absolutely vertical. Okay? So, again, this is the same result which we already know. Normal stress will be maximum on a plane which is perpendicular to the axis. That's for sure. You can note this down. If theta is equal to 0, that means the plane is going to be absolutely vertical like this. It will be perpendicular to the axis. Okay? So, normal stress will be maximum when the plane makes an angle of how much? 90 degrees. That's it. That's the first result which we can write. Second thing is sigma t. This is the shear stress by the way. So, right now we are representing this by sigma t. You can also write this as simply tau. Because here, force along this plane on a direction or should I say force along this tangential direction with respect to the plane. this tangential direction will be taken into account. So this is a fairly simple this is pt over again apq. Now pt if you watch here that's if this is px okay here we wrote component of px along pn okay and here we are going to write component of px along pt. And all of this divided by, well, it's quite basic a, p, q. So, component of px. Component of px, this was along pl and along pt. Well, along pt, the component of px is going to be this way. Well, that's px sine theta. Done. That's it. So, this is px sine theta. Area, we know very well. You already know this. a, p, q is nothing but a over. cosine theta, we just need to simplify this p x over a and this is sine theta cos theta, what we need to do, try to use multiple angle formula, multiply by two and divide by two, this becomes two sine theta cos theta, which is sine two theta. So there you go. This over here represents sigma sigma x pi, axial tensile stress in the x direction. So there you have a final the result Sigma t or the shear stress along this oblique plane. Okay, shear stress. Sigma t will be equal to what? Well, sigma by 2. Let me just write this properly. Sigma x by 2 sine 2 theta. Now, again, we need to think when will we have a maximum value of shear stress? Okay, so Tau t, where shall I write? There is absolutely no space. Okay, let me just write it over here. Tau t is maximum when sine 2 theta is maximum. So, just take a look at this sinusoidal curve. I am pretty much sure that you must have seen this. Let's say this is theta and let's say this is sine theta. So sinusoidal curve is something like this, isn't it? Over here we have the maximum value. Okay, so the maximum values are obtained at two different points. One is plus one, one is minus one. This over here is corresponding to 90 degrees. This is 180, this is 0 degrees and this over here is for 270 degrees. That means the value will be maximum when 2 theta is equal to 90, when 2 theta is equal to 270. which is quite clear 2 theta is equal to 90 or even 2 theta is equal to 270 degrees. You just have to do the math. Theta is going to be equal to what 45 degrees and this theta is going to work out as half of 270 that is 135 degrees. That means there are two planes which depict the value of shear stress maximum and let me just show that those two planes are essentially Like this, it's fairly simple. Okay, here it is. That's the member, okay, subjected to an axial loading in the x direction. Okay, forget about this now. Let me just make a coordinate axis. Okay, here the angle measurement was done in the sort of clockwise sense from here, from here with the vertical. So here also we are going to do in the clockwise sense with the vertical from the vertical only. How much? 45 degrees. So 45 degrees, that's it. Okay, so this angle over here is 45 degrees and this ring represents the first plane of shear. Let me just show you the second plane of shear that will be obtained at an angle of 135 degrees. So starting from here, 0, 90, And this is 180. So between 90 and 180, there is going to be this at extent this. This angle over here guys is going to be equal to 135 degrees. And this over here is the second plane of shear. So that's all you have to know about this case 1 when a member is subjected to a direct stress in the x direction. Now in the next case, I will be talking about a member subjected to two mutually perpendicular stresses, then we'll also take case three that is based on pure shear and then case four is a combination of case two and case three, okay, there are going to be two mutually perpendicular stresses accompanied with a shear stress. So guys, that was all for today. I'll see you again in the next video. Until then, take care. Have a nice day. Keep learning. Thanks.

Wasn't it supposed to be nine? If you just try to make a count of the total number of different components of stress It works out as nine, isn't it? So what is the answer? What is the right value? Is it nine component or six?

If you ask me, well, both of them are correct. If you just try to make a count, then that will work out as 9. But effectively, there are only 6 components of stress. 3 of them are normal stress components and the remaining 3 are shear stress components. Well, that can essentially be worked out with the help of this 3-dimensional state of stress at a point inside a rigid body which is subjected to external loadings. Now, just take a look at this.

This element over here, small element is acted upon by nine different kinds of stress, isn't it? And if you take a careful look, this is tau xy and this over here is tau yx. Well, to tell you, these both shear stresses are equal in magnitude, but I'll prove it to you right now. And if you watch, if you just try to have a look at this entire arrangement from this side. From the side, let's say we are having the side view, then the arrangement would be something of this sort.

Here we go. Well, the face is going to look something like this, isn't it? There is going to be sigma x, sigma x. There is going to be sigma y and sigma y. So let me just make them.

This over here is sigma x. And that's sigma x. That's sigma y.

This also is Sigma Y done. What about the shear stresses if you watch along this surface? Okay, not this direction but in along this direction. This is tau y X. So this over here is tau y X Just like there is tau y X along this direction in the bottom surface also There is going to be tau y X but in the opposite direction.

This is tau y X by This is again tau yx. Again, take a careful look. This is tau xy in the upper direction, this right hand side face, that means this one.

This over here is tau xy and somewhere along this face, over to the side, we will have this tau xy in the bottom direction, this way. Tau xy, isn't it? Now, you know, this entire system has attained a state of equilibrium.

And whenever any body, any particle is in a state of equilibrium, we always have these two equations working with us. Summation of all the forces along x, y or z direction will be equal to 0 and also summation of rotation, let's say torque, let's say moment of all the forces with respect to x-axis, y-axis or with respect to z-axis is also going to be equal to 0. So, we are viewing it from here. In that case, the rotation or the moment about the z-axis will be equal to 0. So, here it is. Moment about the z-axis, z-axis is equal to Zero, isn't it?

Okay. So these are the two forces separated by a distance. Let's say this is the distance A and this over here is a square.

So that's A and that's A. Two forces which are parallel to each other. Now these can be converted into forces also. Let's say at the top the area is A. In that case, this force over here, this stress can be changed to a force and I can just write A over here.

If I just multiply A over here, this will become a force. All of them. If I just multiply them with A, they would become force. Isn't it? So these are all the force acting on this element.

Isn't it? So two forces, right, separated by a distance, two forces are parallel, separated by a distance, they form a couple. So this over here, if you watch this, what sort of a couple is this? It's a clockwise couple. And for a clockwise couple, you need to take the positive rather negative direction or negative sign and for An anti-clockwise couple, you need to take the positive direction.

So tau yx, okay. So this is clockwise minus of tau yx dot a. And this over here, tau xy, if you watch this, this is upwards and this is downwards.

What sort of an effect is it producing? This is an anti-clockwise effect. For that, you need to put a positive sign.

So plus, this is going to be tau. What? xy.

xy multiplied by a and all of this is going to be equal to 0. Now, since the object is in equilibrium, you can do this stuff with respect to x-axis also, with respect to y-axis also. That can be done. In that case, a and a will cancel out. Tau xy will be equal to tau yx.

So, tau yx is equal to tau xy. And guys, in the same manner, if you take the moment about x-axis, And if you take the moment about y axis, you will realize that tau yz will be equal to tau zy and tau zx will be equal to tau xz. So effectively, these are the three components plus sigma x, sigma y and plus sigma z.

So in totality, if you just make a count, there are effectively six components of stress. Acting at a point inside a rigid bond. Done. Okay, so let me just rub this and let us start the main business.

Okay. Okay, there is something else which I'll have to tell you what we know and what we don't know and what exactly is plane stress. Right.

What we know. Okay, so this over here can be assumed as a cuboid. Now let's say this cuboid is being acted upon by a force.

like this. Okay, let's say the force acting is P. Now, if I tell you that, can you tell me how much is the normal stress induced? Well, you can do that very easily. We'll say that's a normal stress can be calculated very easily along any cross section like this.

So in this cross section, in this cross section, we can actually calculate the value of stress. Let's say this area is A. In that case, sigma n or the normal stress induced will be equal to B by A.

Okay, fine. That means we know how to calculate the value of stress induced on a plane which is perpendicular to the axis. If I can just extend this over here, which I will drop this, the angle made over here will be equal to 90 degrees. So whenever you have a plane perpendicular to the axis there you can calculate the normal stress. But what you don't know is this.

Okay, this is what you don't know. If I tell you, please go ahead and calculate the value of normal stress and also shear stress or tangential stress along this oblique plane. Let's say this is making an angle theta with the vertical here also it is making an angle theta with the vertical Can you tell me how much is the maximum normal stress or let's say in general how much is the normal stress at an angle theta on this oblique plane? Can we do that? Well, we can definitely do that but at the end of the video.

I'll try to explain you all of this stuff and how exactly you can calculate the value of sigma n and sigma t that is normal stress and angelic stress on a plane which sort of is inclined at an angle theta with the vertical. So all of this and much more coming up in today's session, but before that let us have a discussion on plane stress Okay, so let's get back on track and we now know very well that there are as many as six Distinct stress components acting at a point inside a rigid body this state of stress. However, it's not very often Encountered in engineering applications or engineering practice Instead the loadings are coplanar that means they produce stresses Which which can be analyzed on a single plane and when this is the case We'll say that the member or the material is subjected to planes Let me just explain all of that with the help of a beautiful example. It's something like this.

Let's say we've got a material like this, okay Yeah, like this. So these are the three directions. This is let's say x and this right here is the y direction and this over here. Well, let's say this is z.

Now there are a bunch of forces acting. Okay, this let's say is f1. Okay, there is a force over here. Also, let's say this is f2.

There is a force here. This is let's say f3 and then there is a force here. Let's say F4.

Now, because of all these forces, a net force is going to act, right? If you consider a very small element, very small element like this, because of all these forces, right? So, there is going to be a tangential stress along this x direction.

This is going to be sigma x, isn't it? Let me just put the arrows first of all. There is going to be a stress in the y direction also.

this way and there are going to be shear stresses this way this way and this over here is the complementary shear stress if i were to make this in a proper sense it would be something like this that's the element this is going to be sigma y sigma y this is sigma x so sigma x then we have this done done tau x y Tau xy. That's it. So this is plane stress and this is applicable for thin plates.

In the upcoming part of the video, we'll actually be discussing four cases. Case number one is going to be about direct stress in a particular direction. We'll take it as x direction. Then there are going to be two mutually perpendicular stresses.

Case three is going to be a purely shear stress case. And case four is basically a combination of two directly mutually perpendicular stresses plus shear stress. So these are the four cases that we are going to take up. But in today's video, I'll only talk about case one.

So let's start with case one. Here we go. So now let us start with case one, that is when a member is subjected to direct stress only in one direction. Let's say we are talking only about the x direction. So here is the analysis.

So first of all, let us try to make a member and if we rather make a bigger figure, there is only one stress acting. Let's say in the x direction, let me call that sigma x. Okay, that is sigma s. By the way, the one assumption which I'm going to take is that thickness, assume thickness as unity.

So guys, actually, the arrangement is something like this. If I try to make this in 3d, it would be something like this. Yeah, right.

So this is sigma x. So this is the area. So if you multiply sigma x with area, you'll get the value of load applied in this direction, that is px. So let me just write px also over here, try to have a better picture.

Okay, just over there, not to write here. Okay, so this is what we've got a member subjected to an axial tensile stress. And what we wish to do is to find the state of stress along this oblique plane.

This is what we wish to find. Okay, we wish to find the normal stress along this plane normal stress along this plane, and we wish to find the tangential stress along this plane. Let's say that this, this is represented by PQ. And let's just say, the angle that this makes with the vertical is theta over here also, the angle is going to be theta.

How do we get this? Very interesting. Okay, let me have another figure.

So let's let's start from here and this way, this way and this way. The force acting let's say is represented by Px, isn't it? So there are going to be forces acting along this plane P2. Let's call that as Pt and there is going to be a force acting in a direction normal to this Pq. First of all, let me just write this Pq.

This over here is Pn. Okay, so there is a bit of geometry which we need to encounter right now. This angle over here by the way is 30. So, the angle over here is going to be 90 minus theta.

Sorry, not 30 but theta. So, if that is the case, Then the angle over here, here, just try to work it out. Let me make this properly.

Okay, so that's 90 minus theta, this is going to be how much if this is 90 minus theta, this is got to be theta. Okay, that's, that's quite obvious. If this is theta, right, then this also is going to be theta, that's for sure.

Just try to work it out. This is fairly simple. If this is theta, this again has got to be 90 minus theta.

And if this is 90 minus theta, this is theta. If this is theta, between these two lines, if this is theta, then between these two lines, these two are vertically opposite angles. That's theta and that's theta. Done. That means what we can do is, here we go.

Let me just start calculating. What we wish to find is normal stress. Normal stress Pn, not Pn but sigma n.

Okay, so guys here it is. Normal stress where? Along this plane Pq. So that is going to be equal to the load acting normally that is Pn divided by area of this oblique plane. Area of this plane let's say is represented by area Pq.

Now how can this be worked out? So Pn can essentially be written as component of Px along Pn. Along Pn.

Whole divided by, that's area Pq. This is fairly easy. This is going to be fairly easy.

Trust me. Okay, just keep watching. So the component of Px along Pn. Here it is. What is this?

Px. So we want the component of this Px along. So, if this over here is Px, this is theta, then this has got to be, this direction has got to be Px cos theta. Okay, it's very simple. This is going to be Px cos theta divided by, what is this area Pq?

This sounds a bit confusing, but it's not that confusing. Let's say we are considering this triangle. Okay.

this triangle. So, it's something like this, right? Something like this, this way.

So, we know this area. We want to know this area. That's PQ area.

Okay? Now, this angle over here is how much? This is 90 minus theta.

So, what you can do is, you can write sin of 90 minus theta is equal to sin of 90 minus theta is equal to a divided by this area a pq. a divided by a of pq. In that sense, you will have this.

This is cos theta by the way. And a pq will work out as a over cos theta. Now, this is a known fact.

Let me just write this somewhere. Okay, let me just write it over here. Area of the oblique plane is going to work out as a over cos theta. And this is something which we will be using in the upcoming cases also.

So better to memorize this. Okay, here we go. So this is going to be how much? A over cos theta.

And finally you have the result. Implies that sigma n. Sigma n will be equal to what?

Sigma n will be equal to Px over A. Px over A. Here it is. If you try to relate these two.

Px over this area. This area. Okay, this area.

will be equal to sigma x by the way. So, sigma n finally works out as sigma x times of cos theta shifts to the numerator and this is going to be cos squared. So, this is the first result that we have got.

Normal stress onto this oblique plane will work out as sigma x cos square theta. And there are, just to note down, Few more conclusions which can be derived. Number one, what conclusion you can derive is that sigma n will be maximum for a particular value of theta.

Sigma n will be maximum when this cos square theta is maximum and when is cos theta maximum when theta is equal to 0 degrees. So, maximum when theta is equal to 0 degrees. And when theta is equal to 0 degrees, this theta will be 0 degrees, clockwise we have taken the angle, then this oblique plane will become absolutely vertical.

Okay? So, again, this is the same result which we already know. Normal stress will be maximum on a plane which is perpendicular to the axis. That's for sure.

You can note this down. If theta is equal to 0, that means the plane is going to be absolutely vertical like this. It will be perpendicular to the axis. Okay?

So, normal stress will be maximum when the plane makes an angle of how much? 90 degrees. That's it.

That's the first result which we can write. Second thing is sigma t. This is the shear stress by the way. So, right now we are representing this by sigma t. You can also write this as simply tau.

Because here, force along this plane on a direction or should I say force along this tangential direction with respect to the plane. this tangential direction will be taken into account. So this is a fairly simple this is pt over again apq. Now pt if you watch here that's if this is px okay here we wrote component of px along pn okay and here we are going to write component of px along pt. And all of this divided by, well, it's quite basic a, p, q.

So, component of px. Component of px, this was along pl and along pt. Well, along pt, the component of px is going to be this way.

Well, that's px sine theta. Done. That's it.

So, this is px sine theta. Area, we know very well. You already know this.

a, p, q is nothing but a over. cosine theta, we just need to simplify this p x over a and this is sine theta cos theta, what we need to do, try to use multiple angle formula, multiply by two and divide by two, this becomes two sine theta cos theta, which is sine two theta. So there you go. This over here represents sigma sigma x pi, axial tensile stress in the x direction. So there you have a final the result Sigma t or the shear stress along this oblique plane.

Okay, shear stress. Sigma t will be equal to what? Well, sigma by 2. Let me just write this properly.

Sigma x by 2 sine 2 theta. Now, again, we need to think when will we have a maximum value of shear stress? Okay, so Tau t, where shall I write? There is absolutely no space.

Okay, let me just write it over here. Tau t is maximum when sine 2 theta is maximum. So, just take a look at this sinusoidal curve. I am pretty much sure that you must have seen this.

Let's say this is theta and let's say this is sine theta. So sinusoidal curve is something like this, isn't it? Over here we have the maximum value. Okay, so the maximum values are obtained at two different points. One is plus one, one is minus one.

This over here is corresponding to 90 degrees. This is 180, this is 0 degrees and this over here is for 270 degrees. That means the value will be maximum when 2 theta is equal to 90, when 2 theta is equal to 270. which is quite clear 2 theta is equal to 90 or even 2 theta is equal to 270 degrees. You just have to do the math.

Theta is going to be equal to what 45 degrees and this theta is going to work out as half of 270 that is 135 degrees. That means there are two planes which depict the value of shear stress maximum and let me just show that those two planes are essentially Like this, it's fairly simple. Okay, here it is. That's the member, okay, subjected to an axial loading in the x direction. Okay, forget about this now.

Let me just make a coordinate axis. Okay, here the angle measurement was done in the sort of clockwise sense from here, from here with the vertical. So here also we are going to do in the clockwise sense with the vertical from the vertical only.

How much? 45 degrees. So 45 degrees, that's it. Okay, so this angle over here is 45 degrees and this ring represents the first plane of shear.

Let me just show you the second plane of shear that will be obtained at an angle of 135 degrees. So starting from here, 0, 90, And this is 180. So between 90 and 180, there is going to be this at extent this. This angle over here guys is going to be equal to 135 degrees. And this over here is the second plane of shear.

So that's all you have to know about this case 1 when a member is subjected to a direct stress in the x direction. Now in the next case, I will be talking about a member subjected to two mutually perpendicular stresses, then we'll also take case three that is based on pure shear and then case four is a combination of case two and case three, okay, there are going to be two mutually perpendicular stresses accompanied with a shear stress. So guys, that was all for today.

I'll see you again in the next video. Until then, take care. Have a nice day.

Keep learning. Thanks.

Transcript for:Understanding Stress Transformations in Solids

Transcript for:
Understanding Stress Transformations in Solids