1
$\begingroup$

Consider the state of tension in the neighborhood of the point $ P $ represented in the Figure. enter image description here

Determine the principal directions of the stress state in $P$ and write the transformation matrix from the original $S(O,\vec{\imath},\,\vec{\jmath},\,\vec{k})$ reference to the main reference $S(O,\vec{e}_\mathrm{I},\vec{e}_\mathrm{II},\vec{e}_\mathrm{III})$.

I already got that $\sigma_{yy}=42.67$.

The solution is \begin{aligned} \text{}\\[2.75ex] \begin{cases} \vec{e}_\mathrm{I} &= \cos\theta_\mathrm{P} \vec{\imath} + \sin\theta_\mathrm{P} \vec{\jmath} + 0 \vec{k} \\ \vec{e}_\mathrm{II} &= -\sin\theta_\mathrm{P} \vec{\imath} + \cos\theta_\mathrm{P} \vec{\jmath} + 0 \vec{k} \\ \vec{e}_\mathrm{III} &= 0 \vec{\imath} + 0 \vec{\jmath} + 1 \vec{k} \\ \end{cases} \end{aligned}

I know that 1$\ \vec{k}$ is one of the main directions of the state of tension because we got that $\sigma_{zx}=\sigma_{zy}=0$ and therefore \begin{cases} \vec{e}_\mathrm{III} &= 0 \vec{\imath} + 0 \vec{\jmath} + 1 \vec{k} \\ \end{cases}

However i can't understand why $\vec{e}_\mathrm{I} $ and $\vec{e}_\mathrm{II} $ are that expression. Could someone explain it to me?

So the rotation matrix about x and y axis (not needed in this exercice) will be: \begin{alignedat}{1}R_{x}(\theta )&={\begin{bmatrix}1&0&0\\0&\cos \theta &\sin \theta \\[3pt]0&-\sin \theta &\cos \theta \\[3pt]\end{bmatrix}}\\[6pt]R_{y}(\theta )&={\begin{bmatrix}\cos \theta &0&-\sin \theta \\[3pt]0&1&0\\[3pt]\sin \theta &0&\cos \theta \\\end{bmatrix}}\end{alignedat}

right?

$\endgroup$
4
  • $\begingroup$ surely if you understand the answer to this you can work this new question... engineering.stackexchange.com/q/38237/10902 $\endgroup$ – Solar Mike Oct 18 '20 at 19:04
  • $\begingroup$ @student is the value of $\theta_P$ which provided in the solution manual equal to $\theta_P \; = \; \frac{1}{2}\arctan({2 \tau_{xy} \over \sigma_{xx} - \sigma_{yy} }) = 28.07\deg$ $\endgroup$ – NMech Oct 19 '20 at 6:23
  • $\begingroup$ also you state "I already got $\sigma_{yy}=42.67$". Is that another part of the exercise? $\endgroup$ – NMech Oct 19 '20 at 8:40
  • $\begingroup$ @NMech but my question was how do i know that e1=cos(θP)i +sin(θP)j +0k and why e2=−sin(θP)i +cos(θP)j +0k . $\endgroup$ – user28922 Oct 19 '20 at 9:03
1
$\begingroup$

It's really very simple. As you state, $\sigma_{zz}$ is one of the principal stresses, that means that the plane x-y contains the other two stresses. So essentially this is a problem of finding the principal stresses in the plane xy.

As you may already know, if you got $\sigma_{xx}, \sigma_{yy}, \tau_{xy}$, then the orientation of the principal plane is at an angle $\theta_p$ which satisfies the following equation.

$$\tan(2\theta_P )\; = \; {\tau_{xy} \over \sigma_{xx} - \sigma_{yy} }$$

So you need to rotate the x,y axis by $\theta_P $

enter image description here

A way to do that is by using a rotation matrix.

When the rotation is about axis z, then the transformation/rotation matrix is given by:

$$Q_z(\theta) = \begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0& 1\\ \end{bmatrix}$$

To break it down even further, in following 2D image you see that by rotating $\theta$, the frame of reference, then new $\epsilon_{x'}= \cos\theta \epsilon_{x} + \sin\theta \epsilon_{y}$.

enter image description here

So the way you use it:

$$ \begin{bmatrix} \epsilon_{I}\\ \epsilon_{II}\\ \epsilon_{III} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0& 1\\ \end{bmatrix} \begin{bmatrix} \epsilon_{x}\\ \epsilon_{y}\\ \epsilon_{z} \\ \end{bmatrix} $$

This is equivalent to the solution $$\begin{aligned} \text{}\\[2.75ex] \begin{cases} \vec{e}_\mathrm{I} &= \cos\theta_\mathrm{P} \vec{\imath} + \sin\theta_\mathrm{P} \vec{\jmath} + 0 \vec{k} \\ \vec{e}_\mathrm{II} &= -\sin\theta_\mathrm{P} \vec{\imath} + \cos\theta_\mathrm{P} \vec{\jmath} + 0 \vec{k} \\ \vec{e}_\mathrm{III} &= 0 \vec{\imath} + 0 \vec{\jmath} + 1 \vec{k} \\ \end{cases} \end{aligned}$$

However, please note that in this case the index $I, II, III$, does not indicate that $\sigma_I>\sigma_{II}>\sigma_{III}$

$\endgroup$
4
  • $\begingroup$ You are right, that its the symmetric matrix. This is because of the direction of the rotation angle. The wikipedia text is generic, I've added a text which has the proper transformation matrix however this is only 2d. see Coordinate transform $\endgroup$ – NMech Oct 19 '20 at 9:35
  • $\begingroup$ To answer to your comment , it might seem counterintuitive but when expressing the new axis in terms of the old ones, you need to take the symmetric matrix (i.e. rotate by $-\theta$.I've added another graph to illustrate that. It might make it easier to understand $\endgroup$ – NMech Oct 19 '20 at 10:15
  • $\begingroup$ As far as I can tell yes (the equation is not rendered properly). $\endgroup$ – NMech Oct 19 '20 at 11:04
  • $\begingroup$ May I ask which book do you get this exercises from? $\endgroup$ – NMech Oct 19 '20 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy