Understanding 3D Stress Analysis Techniques

So today's lecture is going to be on 3D stress analysis and what we are going to do today is learn two things. One is how to transform a stress matrix in three dimensions from one coordinate system to another and the second thing that we'll be doing is learning how to compute principal stresses at a point for a given state of stress. So for 3D stress transformations let's start with two different coordinate systems x, y, z and x prime, y prime, z prime. The only requirement here is that both these coordinate systems need to be orthogonal coordinate systems. So now let's look at a stress at a point based on the x, y, z coordinate system. And this stress matrix is given by sigma with respect to the x, y, z coordinate system is sigma x, x, tau x, y, tau x, z and so on. Okay. And what we are interested. is finding the stress matrix at the exact same point, okay, but in the x prime, y prime, z prime coordinate system. This is something that I want to emphasize. We are not changing the point, we are not changing the body, we are not changing the loading conditions or the boundary conditions. We are looking at the exact same point, but we are just changing our reference axis from x, y, z to x prime, y prime, z prime. So, How do we do that? So this process is actually very similar to how you do stress transformations in two dimensions. So if Sigma in the x prime y prime z prime axis is given as Sigma x prime y prime z prime, that is going to be a transformation matrix T times Sigma in the original x y z coordinate systems times another transformation matrix but which is the transpose of the initial transformation matrix. So if you think about this particular equation, essentially sigma prime, which is the stress in the new coordinate system, is T times the sigma in the old coordinate system times 3 transpose. Okay, so if you are not familiar with all the transposes, a transpose is essentially you are taking each element and taking it to the, essentially reflecting it about the diagonal. So if it is aij. it becomes aji in the new matrix. Okay, so how do we get this transformation matrix T? Okay, this is an important one and I want you to pay attention. So if you think about the initial coordinate system x, y, z, let's define three unit vectors a1, a2, and a3. Those are along the x, y, and z axis. So I want to emphasize that this a1, a2, and a3 are unit vectors. Okay, so if you are given a vector that is not normalized, you first have to normalize these vectors to get the unit vectors. And now, let's think about the x prime, y prime, z prime coordinate system. And let's define b1, b2, and b3 to be the unit vectors along x prime, y prime, and z prime direction. Again, in this case, if the initial vectors that are given are not normalized, you must first normalize them to get the unit vectors. And once you have a1, a2, a3 and b1, b2, b3, then we can form the transformation matrix T. Okay, so the transformation matrix T is usually denoted in the form L1, m1, n1, L2, m2, n2 and L3, m3, n3. Okay, this L1, m1 and n1 are essentially formed or derived by taking the dot products of the unit vectors in the new coordinate system. with the unit vectors in the old coordinate system. So if you think about L1, it is b1 dot a1. And m1 is b1 dot a2, and n1 is b1 dot a3. So if you look at L1, m1, and n1, what you see is that b1 is common to all three of them, okay? But the unit vectors that you take the dot product with are changing. So if you are... If you can recall what you have learned in math previously, what this L1, M1, and N1 are essentially are the direction cosines of B1 with respect to the old coordinate system A1, A2, and A3. So similarly, you can find L2, M2, and N2, and L3, M3, and N3. So once you get these direction cosines, then you can form the transformation matrix T. Okay, so once you know the Tonson-Mittal metric, You can also find the transpose of the transformation matrix T. So how do you get principal stresses? So essentially we looked at how do you transform stresses from one coordinate system to another. What I'm going to do next is trying to find out how to find principal stresses in 3D. Okay so let's start. with the stress matrix at a point sigma in the x, y, z coordinate system. And the method that we are going to use for finding principal stresses in 3D is actually going to be very similar to the approach that we took. for the 2D stress transformations and 2D principle stresses. So we start by setting sigma minus sigma p times i, where i is the identity matrix, times lp, mp, np to be equal to zero. So this essentially is the equation that we use for finding sigma p. Now, if we expand this equation number one, essentially it becomes sigma xx minus sigma p, tau xy tau xz, tau xy sigma yy minus sigma p, and tau yz and so on, times Lp, Mp, and Np is equal to 0. Here, this Lp, Mp, and Np are essentially the components of the unit vectors corresponding to the three principal stress directions. Or you can say that Lp, Mp, and Np are essentially the components of the principal directions of the principal stresses in 3D. Okay? So... how do you go about solving for sigma p from these equations? Now, if you look back at equation number 2, it is a homogeneous set of equations, and essentially, these set of equations will have a non-trivial solution only if the determinant of the matrix is equal to 0. So essentially, that is what this equation here represents. Now, if you take the determinant of this particular matrix, what you will end up is a... cubic equation in sigma p. So it will be of the form sigma p cube minus i1 times sigma p square plus i2 times sigma p minus i3 equal to zero. So this i1, i2, and i3 have a special significance in the stress analysis, and those are called stress invariants. And we will talk about this later. But the main thing that I want you to know is that once you have this cubic equation, you can essentially solve for the roots of the cubic. equation and because it's a cubic equation you will get three roots and the three roots are going to be sigma p1, sigma p2 and sigma p3 which are the three principal stresses. So essentially the process is very similar to finding principal stresses in 2D except that now instead of looking at a quadratic equation we are going to be looking at a cubic equation and solving it. And this cubic equation in sigma p is sometimes called as the characteristic equation or something like that. sometimes just called the cubic equation. And by solving it, you get the three principal stresses, sigma p1, sigma p2, and sigma p3. Now once you get the principal stresses, you can also find the principal directions corresponding to those principal stresses. And again, the process that we are going to use is going to be very similar to what we did in 2D stress analysis. So what we do is we take any one of these principal stresses that we have found, say let's say sigma p1, and then you substitute. that Sigma P1 in the place of Sigma P in equation number two. Okay? And then essentially solve this equation that is on a screen. Okay? But if you look at this particular set of equations, this is a set of homogeneous equations. Okay? And the equations are not completely independent of each other. So of the three equations, only two of them are independent. So you can't explicitly solve for LP1, MP1, and using the first set of equations. However, we know something more about LP1, MP1, and NP1. We know that these are the components of a unit vector, which means that the sum of the squares of LP1 squared plus MP1 squared plus NP1 squared should be equal to 1. So we use that as the third equation, along with the two independent equations that we get from the matrix to solve for LP1, MP1, and NP1. Okay, so now you substituted sigma p1 instead of sigma p and you get the direction cosines corresponding to sigma p1. You can do the same thing for sigma p2 and get lp2, mp2 and np2, which will be the direction cosines corresponding to sigma p2. And similarly, you can do the same thing for sigma p3. So this way, you can get the direction cosines or essentially the unit vector. that correspond to the three principal directions corresponding to sigma p1, sigma p2, and sigma p3. Okay? Thanks for watching, and I'll see you next time.

The only requirement here is that both these coordinate systems need to be orthogonal coordinate systems. So now let's look at a stress at a point based on the x, y, z coordinate system. And this stress matrix is given by sigma with respect to the x, y, z coordinate system is sigma x, x, tau x, y, tau x, z and so on. Okay.

And what we are interested. is finding the stress matrix at the exact same point, okay, but in the x prime, y prime, z prime coordinate system. This is something that I want to emphasize.

We are not changing the point, we are not changing the body, we are not changing the loading conditions or the boundary conditions. We are looking at the exact same point, but we are just changing our reference axis from x, y, z to x prime, y prime, z prime. So, How do we do that?

So this process is actually very similar to how you do stress transformations in two dimensions. So if Sigma in the x prime y prime z prime axis is given as Sigma x prime y prime z prime, that is going to be a transformation matrix T times Sigma in the original x y z coordinate systems times another transformation matrix but which is the transpose of the initial transformation matrix. So if you think about this particular equation, essentially sigma prime, which is the stress in the new coordinate system, is T times the sigma in the old coordinate system times 3 transpose. Okay, so if you are not familiar with all the transposes, a transpose is essentially you are taking each element and taking it to the, essentially reflecting it about the diagonal. So if it is aij.

it becomes aji in the new matrix. Okay, so how do we get this transformation matrix T? Okay, this is an important one and I want you to pay attention.

So if you think about the initial coordinate system x, y, z, let's define three unit vectors a1, a2, and a3. Those are along the x, y, and z axis. So I want to emphasize that this a1, a2, and a3 are unit vectors. Okay, so if you are given a vector that is not normalized, you first have to normalize these vectors to get the unit vectors. And now, let's think about the x prime, y prime, z prime coordinate system.

And let's define b1, b2, and b3 to be the unit vectors along x prime, y prime, and z prime direction. Again, in this case, if the initial vectors that are given are not normalized, you must first normalize them to get the unit vectors. And once you have a1, a2, a3 and b1, b2, b3, then we can form the transformation matrix T. Okay, so the transformation matrix T is usually denoted in the form L1, m1, n1, L2, m2, n2 and L3, m3, n3. Okay, this L1, m1 and n1 are essentially formed or derived by taking the dot products of the unit vectors in the new coordinate system.

with the unit vectors in the old coordinate system. So if you think about L1, it is b1 dot a1. And m1 is b1 dot a2, and n1 is b1 dot a3.

So if you look at L1, m1, and n1, what you see is that b1 is common to all three of them, okay? But the unit vectors that you take the dot product with are changing. So if you are...

If you can recall what you have learned in math previously, what this L1, M1, and N1 are essentially are the direction cosines of B1 with respect to the old coordinate system A1, A2, and A3. So similarly, you can find L2, M2, and N2, and L3, M3, and N3. So once you get these direction cosines, then you can form the transformation matrix T.

Okay, so once you know the Tonson-Mittal metric, You can also find the transpose of the transformation matrix T. So how do you get principal stresses? So essentially we looked at how do you transform stresses from one coordinate system to another.

What I'm going to do next is trying to find out how to find principal stresses in 3D. Okay so let's start. with the stress matrix at a point sigma in the x, y, z coordinate system.

And the method that we are going to use for finding principal stresses in 3D is actually going to be very similar to the approach that we took. for the 2D stress transformations and 2D principle stresses. So we start by setting sigma minus sigma p times i, where i is the identity matrix, times lp, mp, np to be equal to zero. So this essentially is the equation that we use for finding sigma p.

Now, if we expand this equation number one, essentially it becomes sigma xx minus sigma p, tau xy tau xz, tau xy sigma yy minus sigma p, and tau yz and so on, times Lp, Mp, and Np is equal to 0. Here, this Lp, Mp, and Np are essentially the components of the unit vectors corresponding to the three principal stress directions. Or you can say that Lp, Mp, and Np are essentially the components of the principal directions of the principal stresses in 3D. Okay?

So... how do you go about solving for sigma p from these equations? Now, if you look back at equation number 2, it is a homogeneous set of equations, and essentially, these set of equations will have a non-trivial solution only if the determinant of the matrix is equal to 0. So essentially, that is what this equation here represents.

Now, if you take the determinant of this particular matrix, what you will end up is a... cubic equation in sigma p. So it will be of the form sigma p cube minus i1 times sigma p square plus i2 times sigma p minus i3 equal to zero. So this i1, i2, and i3 have a special significance in the stress analysis, and those are called stress invariants.

And we will talk about this later. But the main thing that I want you to know is that once you have this cubic equation, you can essentially solve for the roots of the cubic. equation and because it's a cubic equation you will get three roots and the three roots are going to be sigma p1, sigma p2 and sigma p3 which are the three principal stresses. So essentially the process is very similar to finding principal stresses in 2D except that now instead of looking at a quadratic equation we are going to be looking at a cubic equation and solving it.

And this cubic equation in sigma p is sometimes called as the characteristic equation or something like that. sometimes just called the cubic equation. And by solving it, you get the three principal stresses, sigma p1, sigma p2, and sigma p3.

Now once you get the principal stresses, you can also find the principal directions corresponding to those principal stresses. And again, the process that we are going to use is going to be very similar to what we did in 2D stress analysis. So what we do is we take any one of these principal stresses that we have found, say let's say sigma p1, and then you substitute.

that Sigma P1 in the place of Sigma P in equation number two. Okay? And then essentially solve this equation that is on a screen. Okay? But if you look at this particular set of equations, this is a set of homogeneous equations.

Okay? And the equations are not completely independent of each other. So of the three equations, only two of them are independent. So you can't explicitly solve for LP1, MP1, and using the first set of equations.

However, we know something more about LP1, MP1, and NP1. We know that these are the components of a unit vector, which means that the sum of the squares of LP1 squared plus MP1 squared plus NP1 squared should be equal to 1. So we use that as the third equation, along with the two independent equations that we get from the matrix to solve for LP1, MP1, and NP1. Okay, so now you substituted sigma p1 instead of sigma p and you get the direction cosines corresponding to sigma p1. You can do the same thing for sigma p2 and get lp2, mp2 and np2, which will be the direction cosines corresponding to sigma p2. And similarly, you can do the same thing for sigma p3.

So this way, you can get the direction cosines or essentially the unit vector. that correspond to the three principal directions corresponding to sigma p1, sigma p2, and sigma p3. Okay?

Thanks for watching, and I'll see you next time.

Transcript for:Understanding 3D Stress Analysis Techniques

Transcript for:
Understanding 3D Stress Analysis Techniques