Global to local transformation matrix in 2D frame structures

Question

In section 3.2 of this paper [1], where 2D planar frame structures are being analyzed, the authors mentioned a transformation matrix to be used in extracting the element displacement vector from the displacement vector of the ground structure in global coordinates.

$u_i^e = (u_x^{(1)},u_y^{(1)}, \theta^{(1)},u_x^{(2)},u_y^{(2)}, \theta^{(2)})$ is the element $i$'s displacement vector in local coordinates. $$u_i^e=T_i u$$ where $T_i \in \mathbb{R}^ {6\times d}$. What should be the matrix $T_i$?

Note: The same matrices are used to assemble the global stiffness matrix from elements stiffness matrix in local coordinates using the following formula: $$K= \sum_{i\in E}\sum_{j=1}^3 k_{ij}b_{ij}b_{ij}^\top$$ where $b_{i1}, b_{i2},b_{i3}\in \mathbb{R}^d$ are defined by ($\hat{b_{ij}}$ are $6\times 1$ constant vectors):

$$b_{ij}=T_i^\top \hat{b_{ij}}$$

I am not an expert in statics or mechanical concepts but I need these relations for the optimization problem that I am using for my research.

[1] Kureta, Rui, and Yoshihiro Kanno. "A mixed integer programming approach to designing periodic frame structures with negative Poisson’s ratio." Optimization and Engineering 15.3 (2014): 773-800.

Answer 1

Let us consider that you already know how to obtain the stiffness matrix in the local configuration,

$$[K]\{U\} = \{F\}\, .$$

Now, we want to express our systems of equations in a system rotated by an angle $\alpha$.

The (generalized) displacement components would be given by

\begin{align} &u_1' = u_1 \cos\alpha + v_1 \sin\alpha\, ,\\ &v_1' = -u_1 \sin\alpha + v_1 \cos\alpha\, ,\\ &\theta_1' = \theta_1\, ,\\ &u_2' = u_2 \cos\alpha + v_2 \sin\alpha\, ,\\ &v_2' = -u_2 \sin\alpha + v_2 \cos\alpha\, ,\\ &\theta_2' = \theta_2\, ,\\ \end{align}

or, in matrix form,

$$\begin{Bmatrix}u_1'\\ v_1'\\ \theta_1'\\ u_2'\\ v_2'\\ \theta_2'\end{Bmatrix} = \underbrace{ \begin{bmatrix} \cos\alpha &\sin\alpha &0 &0 &0 &0\\ -\sin\alpha &\cos\alpha &0 &0 &0 &0\\ 0 &0 &1 &0 &0 &0\\ 0 &0 &0 &\cos\alpha &\sin\alpha &0\\ 0 &0 &0 &-\sin\alpha &\cos\alpha &0\\ 0 &0 &0 &0 &0 &1\\ \end{bmatrix}}_{T} \begin{Bmatrix}u_1\\ v_1\\ \theta_1\\ u_2\\ v_2\\ \theta_2\end{Bmatrix}\, .$$

That we can write es

$$\{U'\} = [T]\{U\}\, .$$

Thus, we have the following

\begin{align} &[K] \{U'\} = \{F\}\, ,\\ &[K] [T] \{U\} = [T] \{F\}\, . \end{align}

And, multiplying both sides by $[T]^T$, we obtain

\begin{align} &[T]^T [K] [T] \{U\} = \underbrace{[T]^T [T]}_{[I]} \{F\}\, ,\\ &[T]^T [K] [T] \{U\} = \{F\}\, . \end{align}