User:Eml4500.f08.qwiki.bishop/Lecture 13

=Meeting 13, Friday September 26, 2008= Similar to our objective in Meeting 12, today we would like to transform from a 4x2 matrix to a 2x2. The transformation relationship

\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} =T_{(2x4)}^{(e)} \begin{bmatrix} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\\ \end{bmatrix} $$ derived in Meeting 12 can also be expressed in its most general form:
 * $$\mathbf{P}^{(e)}_i=\mathbf{T}^{(e)}\mathbf{f}^{(e)}_i$$, where


 * $$\mathbf{P}^{(e)}_{i}$$ is the axial force of element e at local node $$i$$
 * $$\mathbf{T}^{(e)}_{(2x4)}$$ is the matrix used to transform force components to an axial force
 * $$\mathbf{f}^{(e)}_{i}$$ is the $$i$$th force component of element e

The overall goal is to find the relationship between: Let us consider the force transformation relationship:
 * $$\mathbf{q}^{(e)}_{(2x2)}$$ and $$\mathbf{d}^{(e)}_{(4x2)}$$, (axial displacements of each element node and their respective displacement component)
 * $$\mathbf{P}^{(e)}_{(2x2)}$$ and $$\mathbf{f}^{(e)}_{(4x2)}$$, (axial forces at each element node and their respective force components)
 * $$\mathbf{P}^{(e)}_{(2x2)}=\mathbf{T}^{(e)}_{(2x4)}\mathbf{f}^{(e)}_{(4x2)}$$

Recall that the element axial FD relationship is given by:
 * $$\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{q}^{(e)}_{(2x2)}=\mathbf{P}_{(2x2)}^{(e)}$$

By introducing a transformation matrix into the FD relationship, we obtain the following relationship:
 * $$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\mathbf{d}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

The ultimate goal is to express the above equation in the form $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$. If $$\mathbf{T}^{(e)}$$ were an nxn matrix, then one could simply "move" $$\mathbf{T}^{(e)}$$ from the right hand side to the left hand side by premultiplying the entire equation by the inverse $$\mathbf{T^{(e)}}^{-1}$$. Unfortunately, $$\mathbf{T}^{(e)}$$ is a rectangular 2x4 matrix, which means it is non-invertible.

Since we cannot solve for the inverse of the transformation ($$\mathbf{T}^{(e)}$$ is not an nxn matrix), we must instead make use of the transpose:
 * $$\mathbf{T}^{(e)T}_{(4x2)}\hat{k}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$

Note that $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}_{(4x2)}\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}$$ Therefore, $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

The relationship $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$ can be justified by the principle of virtual work, which was first introduced in Meeting 10. Additionally, applying the principle of virtual work to the relationship $$K_{(6x6)}d_{(6x1)}=F_{(6x1)}$$ reduces it to $$K_{(2x2)}d_{(2x1)}=F_{(2x1)}$$.

One may wonder why we cannot simply solve $$d=K^{(-1)}F$$ Answer : The equation cannot be solved in the current form due to the singularity of $$K$$, i.e, $$detK=0$$ and thus $$K$$ is not invertible. (See above for definition of interbile matrix) Recall that in order to find $$K^{(-1)}$$ we must first compute $$1/{(detK)}$$. If a matrix is singular then $$1/{(detK)}$$ is undefined and cannot be solved. For our case of an unconstrained structural system, there are three possible rigid body motions in 2-dimensions (2 translations and 1 rotation) for each node.

References