User:Eml4500.f08.bottle.loschak/HW5

Homework 5
The next section of HW 5 will be discussing truss systems in three dimensional space, which had previously not been considered. The axial force displacement relation, commonly referred to as the FD relation, is very similar when compared to the 2-D truss element. Note that the equation of the 2-D truss element is: $$\mathbf{kd}=\mathbf{F}$$ We also recall that the FD relation for axial displacement of each element in the notation used in class is as follows: $$\frac{EA}{L}\begin{pmatrix} 1 & -1\\ -1 & 1 \end{pmatrix} \begin{pmatrix} d_{1}\\d_{2} \end{pmatrix} = \begin{pmatrix} P_{1}\\P_{2} \end{pmatrix} \Rightarrow \mathbf{k^{(e)}q^{(e)}}=\mathbf{p^{(e)}}$$ The main difference for the 3-D truss is that each matrix is a bit larger to account for the movement in the z-direction. The local degrees of freedom become the 6x1 matrix $$\mathbf{d^{(e)}}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \\ d_{5}^{(e)} \\ d_{6}^{(e)} \end{Bmatrix}$$ and the local forces become the 6x1 matrix below. $$\mathbf{f^{(e)}}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \\ f_{5}^{(e)} \\ f_{6}^{(e)} \end{Bmatrix} $$ Now the transform matrix T(e) must contain two extra columns, but the axial displacement q(e) and axial load p(e) will remain as 2x1 matrices. The transform matrix will become: $$\mathbf{T^{(e)}}=\begin{pmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{pmatrix}$$ The transform matrix for the 3-D truss system was derived similarly to the 2-D transform matrix in lecture 12. The figure below displays the unit vectors in the normal x-y coordinate axis along with the rotated coordinate axis. The displacement vector for a local node is defined as follows. $$ \bar{d}_{1}^{(e)}=d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k} $$ It is possible to take the dot product of the displacement vector and the rotated unit vector and carry out the steps listed below. $$ q_{1}^{(e)}=(d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k})\hat{\tilde{i}} $$ $$ q_{1}^{(e)}=d_{1}^{(e)}(\hat{i}\cdot\hat{\tilde{i}})+d_{2}^{(e)}(\hat{j}\cdot\hat{\tilde{i}})+d_{3}^{(e)}(\hat{k}\cdot\hat{\tilde{i}}) $$ $$ q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}+n^{(e)}d_{3}^{(e)} $$ $$ q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)} \end{Bmatrix} $$ And Similarly, for local node 2, the following can also be said. $$ q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ Combining the axial displacements into one expression results in the following expression: $$ \begin{Bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ This expression can be written in its most simple form as the following: $$\mathbf{q}^{(e)}=\mathbf{T}^{(e)}\mathbf{d}^{(e)}$$ Similarly, the local forces can be multiplied by the same transform matrix to equal the axial forces. $$\mathbf{p}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$