# Global to local transformation matrix in 2D frame structures

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.

• Have you looked at this reference? – nicoguaro Jan 27 '20 at 2:32
• @nicoguaro I looked at it unfortunately it’s all about truss structures in which stiffness matrix and degrees of freedom are different from frame structures. – user32251 Jan 27 '20 at 2:59
• It's not that different. You have to transform displacements, and they transform in the same fashion and rotations that doesn't transform in this case. – nicoguaro Jan 27 '20 at 3:12
• @nicoguaro Can you please write as an answer the transformation matrix for an Equilateral triangle frame structure with 9 degrees of freedom(not all of them are free)? As explained in the question, T should be a 6x9 matrix... – user32251 Jan 27 '20 at 3:24

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}

• Thanks for your answer. In the mentioned paper, the authors mentioned the T matrix will be a 6xd in which d is total degrees of freedom in the structure. I still don't understand how to generate such a T matrix (with the mentioned dimensions) from T(6x6) matrix of cosines that is mentioned in your answer. – user32251 Jan 28 '20 at 16:12
• Sorry, it is not clear to me either. I would suggest that you understand the assembly process for Finite Elements, even though your research topic is centered on optimization. – nicoguaro Jan 28 '20 at 16:17
• Thank you for your suggestion and answer. – user32251 Jan 28 '20 at 16:18