Rotation and transformation of coordinates of stiffness matrix

Question

I face a problem where I know the beam's local stiffness matrix and I want to find the global stiffness matrix. My problem involves the rotation of this matrix and I did not find any solution to this over the internet. Can anyone help me?

Question 1: How to find the expanded rotation matrix $\left[R_{exp}\right]$?

$$ \left[K_{glo}\right]_{12 \times 12} = \left[R_{exp}\right]_{12 \times 12} \cdot \left[K_{loc}\right]_{12 \times 12} \cdot \left[R_{exp}\right]_{12 \times 12}^{T} $$

Question 2: If $\left[R_{exp}\right]$ has the same properties of the rotation matrix $\left[R\right]_{3 \times 3}$, we would have

$$\left[R_{exp}\right]^{T} = \left[R_{exp}\right]^{-1}$$

But if $\left[R_{exp}\right]$'s inverse it's not the same as it's transpose, which must I use?

$$ \left[K_{glo}\right] = \left[R_{exp}\right] \cdot \left[K_{loc}\right] \cdot \left[R_{exp}\right]^{T} $$

or

$$ \left[K_{glo}\right] = \left[R_{exp}\right] \cdot \left[K_{loc}\right] \cdot \left[R_{exp}\right]^{-1} $$

Problem description

Let's say that we have two system of coordinates:

• global: $(\vec{x}, \vec{y}, \ \vec{z})$
• local: $(\vec{n}, \ \vec{r}, \ \vec{t})$

I have a beam with the extremities at $\vec{0} = (0, 0, 0)$ and $\vec{p} = (p_x, p_y, p_z)$.

• Compute $R$

My idea was inspired by the rotations in 3D. My first objective is to bring $\vec{p}$ to the $x$-axis. That means, get $R$ such that

$$ \left(L, \ 0, \ 0\right) = L \cdot \vec{x} = \left[R\right] \cdot \vec{p} = \left[R\right] \cdot L \cdot \vec{n} $$

where $L = \|\vec{p}\| = \sqrt{p_x^2 + p_y^2 + p_z^2}$.

As the general equation for rotation, around the unit vector $\vec{u}$ by an amount of $\theta$ is given by

$$ \left[R\right] = \left[I\right] \cdot \cos \theta + \left[u\right]_{\times} \cdot \sin \theta + \vec{u} \otimes \vec{u} \cdot \left(1-\cos \theta\right) $$

To bring $\vec{p}$ into $L \cdot \vec{x}$, we just need to define $\vec{u}$ the vector perpendicular to $\vec{x}$ and $\vec{p}$

$$ \vec{u} = \dfrac{\vec{p} \times \vec{x}}{L \cdot \sin \theta} = \dfrac{\left(0, \ p_z, -p_y\right)}{L \cdot \sin \theta} $$

And the rotation angle $\theta$ is the angle between $\vec{p}$ and $\vec{x}$

$$ \cos \theta = \dfrac{\vec{p} \cdot \vec{x}}{L} = \dfrac{p_{x}}{L} $$

Calculating $\left[R\right]$:

$$ \left[u\right]_{\times} = \begin{bmatrix} & -u_{z} & u_{y} \\ u_{z} & & -u_{x} \\ -u_{y} & u_{x} & \end{bmatrix} = \dfrac{1}{L \cdot \sin \theta} \begin{bmatrix} & p_{y} & p_{z} \\ -p_{y} & & \\ -p_{z} & & \end{bmatrix} $$

$$ \left[R(\vec{p})\right] = \dfrac{1}{L} \begin{bmatrix} p_{x} & p_{y} & p_{z} \\ -p_{y} & p_{x} & \\ -p_{z} & & p_{x} \end{bmatrix} + \dfrac{1}{p_y^2 + p_z^2} \left(1-\dfrac{p_{x}}{L}\right) \begin{bmatrix} & & \\ & p_{y}^2 & -p_y p_z \\ & -p_y p_z & p_{z}^2 \end{bmatrix} $$ As we can see $R(\vec{p})^{T} \ne R(-\vec{p})$

• Local Stiffness Matrix:

The local stiffness matrix in the space $(\vec{n}, \ \vec{r}, \ \vec{t})$ is well-known and it's like $$ \underbrace{\begin{bmatrix} \square & & & & & & \square & & & & & \\ & \square & & & & \square & & \square & & & & \square \\ & & \square & & \square & & & & \square & & \square & \\ & & & \square & & & & & & \square & & \\ & & \square & & \square & & & & \square & & \square & \\ & \square & & & & \square & & \square & & & & \square \\ \square & & & & & & \square & & & & & \\ & \square & & & & \square & & \square & & & & \square \\ & & \square & & \square & & & & \square & & \square & \\ & & & \square & & & & & & \square & & \\ & & \square & & \square & & & & \square & & \square & \\ & \square & & & & \square & & \square & & & & \square \\ \end{bmatrix}}_{\left[K_{loc}\right]} \underbrace{\begin{bmatrix} u_{n1} \\ u_{r1} \\ u_{t1} \\ \theta_{n1} \\ \theta_{r1} \\ \theta_{t1} \\ u_{n2} \\ u_{r2} \\ u_{t2} \\ \theta_{n2} \\ \theta_{r2} \\ \theta_{t2} \end{bmatrix}}_{\left[U_{loc}\right]} = \underbrace{\begin{bmatrix} F_{n1} \\ F_{r1} \\ F_{t1} \\ M_{n1} \\ M_{r1} \\ M_{t1} \\ F_{n2} \\ F_{r2} \\ F_{t2} \\ M_{n2} \\ M_{r2} \\ M_{t2} \end{bmatrix}}_{\left[F_{loc}\right]} $$ $u$ in this case is the displacement, $\theta_{ij}$ is the tangent's direction, $F$ is the force and $M$ the momentum.

• Attempt to compute $\left[K_{glo}\right]$

My attempt to compute $\left[K_{glo}\right]$ is mounting the matrix $\left[R_{exp}\right]$ with the matrix $\left[R\left(\vec{p}\right)\right]$ in the diagonal, as we can transform

$$ \begin{bmatrix} u_{x1} \\ u_{y1} \\ u_{z1} \end{bmatrix} = \left[R\left(\vec{p}\right)\right] \cdot \begin{bmatrix} u_{n1} \\ u_{r1} \\ u_{t1} \end{bmatrix} $$ $$ \begin{bmatrix} \theta_{x1} \\ \theta_{y1} \\ \theta_{z1} \end{bmatrix} = \left[R\left(\vec{p}\right)\right] \cdot \begin{bmatrix} \theta_{n1} \\ \theta_{r1} \\ \theta_{t1} \end{bmatrix} $$ $$ \begin{bmatrix} u_{x1} \\ u_{y1} \\ u_{z1} \\ \theta_{x1} \\ \theta_{y1} \\ \theta_{z1} \\ u_{x2} \\ u_{y2} \\ u_{z2} \\ \theta_{x2} \\ \theta_{y2} \\ \theta_{z2} \end{bmatrix} = \left[R_{exp}\right] \begin{bmatrix} u_{n1} \\ u_{r1} \\ u_{t1} \\ \theta_{n1} \\ \theta_{r1} \\ \theta_{t1} \\ u_{n2} \\ u_{r2} \\ u_{t2} \\ \theta_{n2} \\ \theta_{r2} \\ \theta_{t2} \end{bmatrix}$$ $$ \left[R_{exp}\right] = \begin{bmatrix} \left[R(\vec{p})\right] & & & \\ & \left[R(\vec{p})\right] & & \\ & & \left[R(\vec{p})\right] & \\ & & & \left[R(\vec{p})\right] \\ \end{bmatrix}$$

Problem 1: When I test it with $\vec{p} = (p_x, \ p_y, \ 0)$ it works fine. $K_{glo}$ is correct. But for any $p_z \ne 0$ the results are wrong.

Problem 2: If $p_z = L \cdot \vec{z}$, the received matrix $\left[K_{glo}\right]$ it's the negative of it should be.

Problem 3: When I rotate $\vec{p}$ to $L \vec{x}$, it's not guaranteed that $\vec{r}$ becomes $\vec{y}$ and $\vec{t}$ becomes $\vec{z}$.

Does anyone know how to solve it?

PS: I'm testing with python with random values. As I don't know a formula to the global matrix, to verify if the global matrix is correct, I solve a system: Given the known values of $U_{glo}$ and $F_{glo}$:

• if $\left[K_{glo}\right] \cdot \left[U_{glo}\right] = \left[F_{glo}\right]$,
  • $\left[K_{glo}\right]$ is correct
• else
  • $\left[K_{glo}\right]$ is not correct

Answer 1

After a while, I found the solution. I will divide it into two parts:

1. For circular symmetry around the vector $\vec{n}$, the vectors $\vec{r}$ and $\vec{t}$ doesn't matter.

For example, for truss or circular beams this is valid.

In that case, we can define the rotation matrix $\left[R\right]_{3\times 3}$ like above

$$ \left[R\right] = \dfrac{1}{L} \begin{bmatrix} p_x & p_y & p_z \\ -p_y & p_x & \\ -p_z & & p_x \end{bmatrix} + \dfrac{1}{p_y^2 + p_z^2} \left(1-\dfrac{p_x}{L}\right)\begin{bmatrix} & & \\ & p_z^2 & -p_y p_z \\ & -p_y p_z & p_y^2\end{bmatrix} $$

The matrix $\left[R\right]$ has the nice properties of a rotation matrix.

With this matrix, we can transform the space $(\vec{n}, \ \vec{r}, \ \vec{t})$ into a space $(\vec{x}, \ \vec{v}, \ \vec{w})$ such that

$$ \begin{cases} \vec{x} = \left[R\right] \cdot \vec{n} \\ \vec{v} = \left[R\right] \cdot \vec{r} \\ \vec{w} = \left[R\right] \cdot \vec{t} \end{cases} $$ The vectors $\vec{v}$ and $\vec{w}$ are linear combinations of $\vec{y}$ and $\vec{z}$, and not necessarily equal.

This rotation matrix brings the point $\vec{p}$ into $L\vec{x}$ $$ \begin{bmatrix} L \\ 0 \\ 0 \end{bmatrix} = \left[R\right] \cdot \begin{bmatrix}p_x \\ p_y \\ p_z \end{bmatrix} $$

Then the global stiffness matrix is given by

$$ \left[K_{glo}\right] = \left[R_{exp}\right]^{T} \cdot \left[K_{loc}\right] \cdot \left[R_{exp}\right]$$

With

$$ \left[R_{exp}\right] = \begin{bmatrix}\left[R\right] & & & \\ & \left[R\right] & & \\ & & \left[R\right] & \\ & & & \left[R\right]\end{bmatrix} $$

2. For any element, the orientations must be preserved: $(\vec{n}, \ \vec{r}, \ \vec{t})$ is transformed into $(\vec{x}, \ \vec{y}, \ \vec{z})$

In this case, we don't need any transformation using the rotations as shown above.

We can describe $[R]$ using the coordinates of $\vec{n}, \ \vec{r}, \ \vec{t}$, which is given by

$$\left[R\right] = \begin{bmatrix} n_{x} & r_{x} & t_{x} \\ n_{y} & r_{y} & t_{y} \\ n_{z} & r_{z} & t_{z} \\ \end{bmatrix} $$

With this matrix we still get

$$ \begin{bmatrix} L \\ 0 \\ 0 \end{bmatrix} = \left[R\right] \begin{bmatrix} p_x \\ p_y \\ p_z \end{bmatrix}$$

Then the global stiffness matrix is given by

$$ \left[K_{glo}\right] = \left[R_{exp}\right]^{T} \cdot \left[K_{loc}\right] \cdot \left[R_{exp}\right]$$

With

$$ \left[R_{exp}\right] = \begin{bmatrix}\left[R\right] & & & \\ & \left[R\right] & & \\ & & \left[R\right] & \\ & & & \left[R\right]\end{bmatrix} $$