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The principal stress in the textbook that I'm following is defined as

The maximum or minimum normal stress that acts on an element.

This makes sense when I take a plane state of stress and I get one maximum value and one minimum value of normal stress - $\sigma_1$, $\sigma_2$


However when we consider a 3D state of stress, there are three principal stresses - $\sigma_1$, $\sigma_2$ and $\sigma_3$.

Doesn't that contradict the definition of principal stress? I mean, if principal stress is the max or min normal stress then even in 3D case we should get only two values , one min and one max, how 3 values can be maximum and minimum?

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4 Answers 4

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Principal stresses are observed in an orientation of an 3d orthonormal coordinate system that there are no shear stresses.

The fact that usually 2 values are considered is because the 2d plane stress is easier to visualize.

In each plane that is considered you could rotate the stresses about its normal but you'd still obtain intermediate values than the principal stresses on that plane.

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  • $\begingroup$ Thank you, So for example I take an element at a point where a 3D stress state existed, I could obtain an orientation of an element at the same point, on which the shear stresses are zero and the normal stresses are principal stresses, somewhat like this - drive.google.com/file/d/1DgVCLwuLVqYPabF5T9YRaCzj9LJy8VxY/… , Contd... $\endgroup$ Apr 6, 2022 at 15:38
  • $\begingroup$ Say that s1>s2>s3, are the three principal stresses, if we're talking about the entire element in 3d, then the max principal stress is s1 and the min principal stress is s3. However, if we talk about the in plane principal stress, for say 1-2 plane, then the max principal stress is s1 and min principal stress in s2, if we talk about 1-3 plane the max principal stress is s1 and min principal stress is s3. Am I right on this? $\endgroup$ Apr 6, 2022 at 15:40
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    $\begingroup$ Yes that s right $\endgroup$
    – NMech
    Apr 7, 2022 at 1:36
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"Principal stresses may be defined as -

  • The extreme values of the normal stresses possible in the material."

For every point inside a body under static equilibrium, there are three planes, called the principal planes, where the stress vector is normal to the plane and there is no shear component (see the right-side figure below). These normal stress vectors are called principal stresses, and two of them are the absolute maximum and minimum stresses (extremes), which usually bear significance for homogeneous and isotropic materials.

enter image description here

ADD: The 2D interpretation of 3D principal stresses is graphically presented below. Note that on each of the principal planes, there are two major/principal stresses distinguished by their numerical values rather than the subscriptions, which depend on the coordinate system in use, and can be assigned arbitrarily.

enter image description here

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  • $\begingroup$ Thanks r13, in the figure to the right, $\sigma_1$ $\sigma_2$, $\sigma_2$ $\sigma_3$, $\sigma_3$ $\sigma_1$ are the in plane principal stresses, and $\sigma_1$ $\sigma_3$ are the principal stresses in 3D. (Assuming $\sigma_1$ is largest and $\sigma_3$ the smallest), right? $\endgroup$ Apr 7, 2022 at 8:55
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    $\begingroup$ Let's try this: On the right figure, we squeeze the cube to a plane bounded by x2 & x3 axes, since there is no thickness, ignore the out-plane normal stress (S1). Now the things left on the plane are S2 & S3, which are the maximum and minimum (principal) stresses of the plane. This is to show that the plane principal stress is essentially the same as 3D principal stress with one dimension ignored. $\endgroup$
    – r13
    Apr 7, 2022 at 13:35
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If we want to transfer the axis and rotate the the 3d element to such a 3d angle to get the maximum normal stresses with no shear, Mohr 3d circle is not the correct tool because it allows rotation about only one axis at a time.

We need to compile the tensor of the state of the 3d element's stress and get its derivative and equate it to zero.

It is kind of impractical thing to do by hand, a good weekend's worth of work.

If you have time this is a 4 hour lecture that walks you through each step.

3D stress transformation

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  • $\begingroup$ Thank You kamran for the video. $\endgroup$ Apr 7, 2022 at 5:58
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The definition on your textbook is almost correct. It's totally correct in $2D$, but in $3D$ it should read something like "the principal stresses include the minimum and maximum normal stresses.

By this I mean, if you get the principal stresses in $3D$, $\sigma_1,\, \sigma_2$ and $\sigma_3$ and rename them such that $\sigma_1 \geq \sigma_2 \geq \sigma_3$ then $\sigma_1$ is the maximum normal stress and $\sigma_3$ is the minimum normal stress, with $\sigma_2$ just existing there in-between.

Let's work out the proof of this statement. First, recall that by definition, and regardless of the coordinates used to define it, the stress tensor $\boldsymbol{\sigma}$ has the property of mapping a unit normal to the traction acting on the plane defined by that normal, i.e. $$ \boldsymbol{\sigma}\mathbf{n}=\mathbf{t} \tag{1} $$ This is illustrated in the following picture.

Traction associated with surface normal

The normal stress, which I'll call $\sigma_n$ is the projection of $\mathbf{t}$ on $\mathbf{n}$. Let the coordinate system be defined as the principal directions, call them $\mathbf{e}_1,\,\mathbf{e}_2$ and $\mathbf{e}_3$, under these coordinates the stress tensor is given by $$ \tag{2} \boldsymbol{\sigma}=\left[\begin{matrix}\sigma_1 & 0 & 0 \\0 & \sigma_2 & 0\\0 & 0 &\sigma_3 \end{matrix}\right] $$ (I'll show how to calculate the principal stresses and principal directions afterward). With this, the traction $\mathbf{t}$ is $$ \mathbf{t}=\boldsymbol{\sigma}\mathbf{n}=\sigma_1n_1\mathbf{e}_1+\sigma_2n_2\mathbf{e}_2+\sigma_3n_3\mathbf{e}_3 $$ and, since the normal stress is the projection (i.e. the dot product) of the traction on the normal, it's $$ \sigma_n=\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2. $$ Since $\mathbf{n}$ is a unit normal, by definition $n_1^2+n_2^2+n_3^2=1$, and then $$ \sigma_1=\sigma_1\left(n_1^2+n_2^2+n_3^2\right)\geq \sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2 = \sigma_n $$ We see that the first principal stress (which, if you recall from before, we chose to be the largest) is greater or equal to any other possible normal stress.

By the same argument $$ \sigma_3=\sigma_3\left(n_1^2+n_2^2+n_3^2\right)\leq \sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2 = \sigma_n $$ then the third principal stress is smaller or equal than any other possible normal stress.

The only step left to clarify is how to determine the principal stresses and the principal directions, so the meaning of writing the stress tensor as in $(2)$ is clear.

First, keeping in mind that the stress tensor is defined by the property written in $(1)$, the principal directions are those vectors $v$ such that $$ \boldsymbol{\sigma}\mathbf{v}=\lambda\mathbf{v}. \tag{3} $$ Physically, this means that if you choose $v$ as the normal to the plane you are looking at, the traction on that surface is parallel to that normal. You'll immediately recognize $(3)$ as an eigenvalue problem. From this you can conclude that the principal directions are the eigenvectors of the stress tensor and, by extension, the principal stresses are its eigenvalues. That's all there is to it.

Some examples to illustrate this better. First, the most typical example from textbooks and courses, a body subject to simple shear, as shown in the figure below.

Simple shear and principal stresses

As you'll usually see in textbooks, specially when studying Mohr's circle, a simple shear configuration, as shown, has principal stresses oriented at a 45$^\circ$ rotation (adapted from here). In $2D$ the principal stresses would be $$ \sigma_{max},\sigma_{min}=\frac{\sigma_{xx}+\sigma_{yy}}{2}\pm\sqrt{\frac{\sigma_{xx}-\sigma_{yy}}{2}+\sigma_{xy}^2} $$ For simple shear this reduces to $\sigma_{max}=\sigma_{xy}$ and $\sigma_{min}=-\sigma_{xy}$. The stress tensor for simple shear is $$ \boldsymbol{\sigma}=\left[\matrix{0 & \tau & 0\\ \tau & 0 & 0\\ 0&0&0}\right] $$ The eigenvalues of this stress tensor are $\lambda_1=\tau$, $\lambda_2=0$ and $\lambda_3=-\tau$ and the eigenvectors are $$ \mathbf{v}_1=\left[\matrix{\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0}\right]\qquad \mathbf{v}_2=\left[\matrix{-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0}\right]\qquad \mathbf{v}_3=\left[\matrix{0\\0\\1}\right] $$ As you can see, the eigenvalues and eigenvectors correspond precisely to the results from Mohr's circle, with the unit vector in the third dimension (as all vector are eigenvectors to a $0$ eigenvalue.

Also, notice that to get this coordinate system from the canonical one you need to rotate the axes by 45$^\circ$ around the $z-$axis, to do this you construct the rotation matrix around said axis, given by $$ \mathbf{Q}=\left[\matrix{\text{cos}(\theta)& -\text{sin}(\theta)& 0\\\text{sin}(\theta)&\text{cos}(\theta)&0\\0&0&1} \right] $$ Rotating the stress tensor by the typical relation $\boldsymbol{\sigma}^\prime=\mathbf{Q}\boldsymbol{\sigma}\mathbf{Q}^T$ and you get $$ \boldsymbol{\sigma}^{\prime}=\left[\matrix{\tau&0&0\\0&0&0\\0&0&-\tau}\right] $$

You can apply this same procedure to other stress configuration, probably the only difference will be that $\mathbf{Q}$ will be a rotation in multiple axes, the rotation matrices compose as shown in the wiki.

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