# transformation matrix from the original reference to the main reference

Consider the state of tension in the neighborhood of the point $$P$$ represented in the Figure.

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?

• surely if you understand the answer to this you can work this new question... engineering.stackexchange.com/q/38237/10902 – Solar Mike Oct 18 '20 at 19:04
• @student is the value of $\theta_P$ which provided in the solution manual equal to $\theta_P \; = \; \frac{1}{2}\arctan({2 \tau_{xy} \over \sigma_{xx} - \sigma_{yy} }) = 28.07\deg$ – NMech Oct 19 '20 at 6:23
• also you state "I already got $\sigma_{yy}=42.67$". Is that another part of the exercise? – NMech Oct 19 '20 at 8:40
• @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 . – user28922 Oct 19 '20 at 9:03

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 $$\theta_P$$

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 $$\epsilon_{x'}= \cos\theta \epsilon_{x} + \sin\theta \epsilon_{y}$$.

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 $$\sigma_I>\sigma_{II}>\sigma_{III}$$

• 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 – NMech Oct 19 '20 at 9:35
• 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 – NMech Oct 19 '20 at 10:15
• As far as I can tell yes (the equation is not rendered properly). – NMech Oct 19 '20 at 11:04
• May I ask which book do you get this exercises from? – NMech Oct 19 '20 at 11:19