User:Eml4500.f08.gravy.jad/Notes4

$$\ \,$$ $$\  \,$$ $$\  \,$$

'''Connectivity Array '''

Here we are using a connectivity array "conn" (from p.5-6 2-bar truss system)

$$\ conn = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \,$$

Where the columns represent the local node number, but the numbers inside the matrix state the global node number. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

conn(e,j)= global node # of local node j of element e.

 Location Matrix Master Array

Here we are using a location matrix master array "lmm" (from p.5-6 2-bar truss system)

$$\ lmm = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 5 & 6 \end{bmatrix} \,$$

Where the columns represent the local dof's, but the numbers inside the matrix state the global dof's or equation number in matrix k. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

lmm(i,j)= eq # (global dof's #) for the elem. stiffness coeff. corresponding to the jth local dof #.

Insert HW 18-2

Back to p.12-1: Method 2 to desire k(e)4x4 : transform a system with 4 dofs into a system with also 4 dofs (instead of only 2 dofs) so that the transformation matrix is now 4x4 and inverted. We want a system of equations that follows a coordinate system which correlates to the axial direction of the bar.

Goal : Want to find $$\ T^{(e)}_{4x4} \,$$ that transform the set of local (elem) dofs $$\ d^{(e)}_{4x4} \,$$ to another set of local (elem) dofs $$\ \underline{\tilde{d}}^{(e)}_{4x4} \,$$ such that $$\ \underline{\tilde{T}}^{(e)} \,$$  is invertible.



$$\ \underline{\tilde{d}}^{(e)}_{4x1} = \underline{\tilde{T}}^{(e)}_{4x4} \tilde{d}^{(e)}_{4x1} \,$$

Continuing the example form p.12-4

$$\ \tilde{d}^{(e)}_{1} = \left \lfloor \ l^{e} m^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$  where $$\ \underline{\tilde{ d }}^{(e)}_{1} \,$$ also equals $$\ q^{(e)}_{1} \,$$ (Equation 1)

$$\ \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{1} = \vec{d}^{(e)}_{1} \cdot \vec{\tilde{j}} \,$$ (component of $$\ \tilde{d}^{(e)}_{1} \,$$ along $$\ \vec{\tilde{j}} \,$$ j, ie. $$\ \tilde{y}-axis \,$$) (of the local nodes)

$$\ = - \sin\theta^{(e)} d^{(e)}_{1} + \cos \theta^{(e)} d^{(e)}_{2} \,$$

$$\ = \left \lfloor \ -m^{e} l^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$ (Equation 2)

Insert HW 19-2

Put (EQ 1) and (EQ 2) in matrix form:

$$\ \begin{Bmatrix} \tilde{d}^{(e)}_{1} \\ \tilde{d}^{(e)}_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$

where $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

A complete matrix form of all dofs

$$\ \underbrace{ \begin{Bmatrix} \tilde{d}^{(e)}_{1} \\ \tilde{d}^{(e)}_{2} \\ \tilde{d}^{(e)}_{3} \\ \tilde{d}^{(e)}_{4} \end{Bmatrix} }_{\underline{\tilde{d}}^{(e)}_{4x1}} = \underbrace{ \begin{bmatrix} \underline{R}^{(e)}_{2x2} & \underline{0}_{2x2} \\ \underline{0}_{2x2} & \underline{R}^{(e)}_{2x2} \end{bmatrix} }_{\underline{\tilde{T}}^{(e)}_{4x4}} \underbrace{ \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \\d^{(e)}_{3} \\d^{(e)}_{4} \end{Bmatrix} }_{\underline{d}^{(e)}_{4x1}} \,$$



$$\ \underline {\tilde{f}}^{(e)}_{4x1}= k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \underline{\tilde{d}}^{(e)} \,$$

In conclusion:

$$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} \,$$

Special Case

$$\ \tilde{d}^{(e)}_{4} \ne 0 \,$$, $$\ \tilde{d}^{(e)}_{1} =\tilde{d}^{(e)}_{2} =\tilde{d}^{(e)}_{3} = 0 \,$$

Therefore: $$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} = \underline{0}_{4x1} \,$$ where the zero matrix is the 4th column of $$\ \underline{\tilde{k}}^{(e)} \,$$

Now interpret the transverse of the dofs: $$\ \underline{\tilde{d}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{d}^{(e)} \,$$

Similarly: $$\ \underline{\tilde{f}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{f}^{(e)} \,$$

Insert HW 20-1

Also: $$\ \underline{\tilde{k}}^{(e)} \underline{\tilde{d}}^{(e)} = \tilde{f}^{(e)} \Rightarrow \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} {d}^{(e)} \,$$

Hence, if $$\ \underline{\tilde{T}}^{(e)} \,$$ is invertible then: $$\ \left [ {\underline{\tilde{T}}^{(e)}}^{-1} \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} \right ] \underline{\tilde{d}}^{(e)} = \underline{f}^{(e)} \,$$

$$\ \underline{\tilde{T}}^{(e)} \,$$ is a block diagonal matrix (from p.19-3).

Now consider the general block-diagonal matrix: $$\ A = \begin{bmatrix} \underline{D}_{1} & & \underline{0} \\ & \ddots &  \\ \underline{0} &  & \underline{D}_{s} \end{bmatrix} \,$$

Question: Then what is $$\ A^{-1} \,$$?

Answer: Lets make a simpler case- $$\ B = \begin{bmatrix} d_{11} & & & \underline{0} \\ & d_{22} & &  \\ & & \ddots & \\ \underline{0} &  & & D_{s} \end{bmatrix} = Diag \left [ d_{11},d_{22}, \ldots, d_{nn} \right ] \,$$

$$\ B^{-1}= Diag \left [ \frac{1}{d_{11}},\frac{1}{d_{22}}, \ldots, \frac{1}{d_{nn}} \right ] \,$$ assuming that $$\ d_{ij} \ne 0 \,$$ for $$\ i=1 \ldots n \,$$.

$$\ B \cdot B^{-1} = I \,$$

Now going back to the A-matrix:

$$\ A= Diag \left [ \underline{D}_{1},\underline{D}_{2}, \ldots, \underline{D}_{s} \right ] \,$$

$$\ A^{-1}= Diag \left [ \frac{1}{\underline{D}_{1}},\frac{1}{\underline{D}_{2}}, \ldots, \frac{1}{\underline{D}_{s}} \right ] = Diag \left [ \underline{D}^{-1}_{1},\underline{D}^{-1}_{2}, \ldots, \underline{D}^{-1}_{s} \right ] \,$$

Therefore: $$\ {\underline{\tilde{T}}^{(e)}}^{-1} = Diag \left [ {\underline{R}^{(e)}}^{-1}, {\underline{R}^{(e)}}^{-1} \right ] \,$$

Recall from p.19-2: $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Now: $$\ {\underline{R}^{(e)}}^{T} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Therefore: $$\ {\underline{R}^{(e)}}^{T} \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} = \begin{bmatrix} {l^{(e)}}^{2} + {m^{(e)}}^{2} & 0 \\ 0 & {l^{(e)}}^{2} + {m^{(e)}}^{2} \end{bmatrix} \,$$

since $$\ {l^{(e)}}^{2} + {m^{(e)}}^{2} = 1 \,$$, then $$\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_{2x2} \,$$ the Identity matrix.

In conclusion: $$\ \Rightarrow {R^{(e)}}^{-1} = {R^{(e)}}^{T} \,$$ and $$\ \Rightarrow {\underline{\tilde{T}}^{(e)}}^{-1} = {\underline{\tilde{T}}^{(e)}}^{T} \,$$