User:Eml4500.f08.bottle.loschak/HW7

HW problems
from 11/21 notes

Revisitng the matrix below we must verify the dimensions for all terms multiplied by $$\tilde{d}$$. $$\tilde{k} =\! $$ $$\begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} &0 &0 \\ \\ & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{-12EI}{L^{3}} &\frac{6EI}{L^{2}} \\ \\ & &  \frac{4EI}{L}&0  & \frac{-6EI}{L^{2}} & \frac{2EI}{L}\\ \\ & &  & \frac{EA}{L} &0  & 0\\ \\ & &  &  &\frac{12EI}{L^{3}}  &\frac{-6EI}{L^{2}} \\ \\ symmetric & &  &  &  & \frac{4EI}{L} \end{bmatrix} \! $$ Verify all $$\tilde{k}_{ij}\tilde{d}_{j}$$ for i = 1, ..., 6 and j = 1, ..., 6 We know from the lecture notes that the dimensions of the first term are

$$[\tilde{k}_{11}\tilde{d}_1] = \frac{[E][A]}{[L]}[\tilde{d}_{1}] = \frac{(\frac{F}{L^2})L^2}{L}(L) = F$$

Calculating the dimensions of the four other terms results in the dimensions

$$[\tilde{k}_{22}\tilde{d}_2] = \frac{[12][E][I]}{[L]^3}[\tilde{d}_{2}] = \frac{(\frac{F}{L^2})L^4}{L^3}(L) = F$$

$$[\tilde{k}_{23}\tilde{d}_2] = \frac{[6][E][I]}{[L]^2}[\tilde{d}_{2}] = \frac{(\frac{F}{L^2})L^4}{L^2}(L) = FL$$

$$[\tilde{k}_{33}\tilde{d}_3] = \frac{[4][E][I]}{[L]}[\tilde{d}_{3}] = \frac{(\frac{F}{L^2})L^4}{L}(1) = FL$$

$$[\tilde{k}_{36}\tilde{d}_6] = \frac{[2][E][I]}{[L]}[\tilde{d}_{6}] = \frac{(\frac{F}{L^2})L^4}{L}(1) = FL$$

After the dimensions of all terms have been determined, the results can be displayed in the dimension matrix below.

$$ \begin{bmatrix} F & 0 & 0 & -F & 0 & 0\\ 0 & F & FL & 0 & -F & FL\\ 0 & FL & FL & 0 & -FL & FL\\ -F & 0 & 0 & F & 0 & 0\\ 0 & -F & -FL & 0 & F & -FL\\ 0 & FL & FL & 0 & -FL & FL \end{bmatrix} $$ To solve for the global stiffness matrix k(e) in $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$ from $$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}$$ we use the equation $$\mathbf{k}^{(e)}=\tilde{\mathbf{T}}^{(e)^{T}}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}$$ The stiffness matrix is first given as $$\tilde{k} = \begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} &0 &0 \\ \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{-12EI}{L^{3}} &\frac{6EI}{L^{2}} \\ \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L}     & 0 & \frac{-6EI}{L^{2}}  & \frac{2EI}{L}\\ \\ \frac{-EA}{L} & 0     & 0                 & \frac{EA}{L}  &0        & 0\\ \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} \! $$ The Transfer matrix and its transpose are below. The R matrix used to assemble the Transfer matrix was discussed in Lecture 19. $$\tilde{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0\\ 0 & 0 & 0 & -m^{(e)} & l^{e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ $$\tilde{T}^{(e)^{T}} = \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 & 0 & 0\\ m^{(e)} & l^{e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & -m^{(e)} & 0\\ 0 & 0 & 0 & m^{(e)} & l^{e)} & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ Multiply the three matrices according to the equation listed above for the global stiffness matrix

We know from previously in the lecture notes that the units of displacement are length (L) and the units of rotation are simply 1. For the transverse displacement term N5 and the rotation of N6, dimensional analysis gives us the following results. $$ \left[N_{5} \right]\left[\tilde{d}_{5} \right] = (1)(L) = L $$ $$ \left[N_{6} \right]\left[\tilde{d}_{6} \right] = (L)(1) = L $$