User:EML4500.f08.FEABBQ.Hamdan/HW-4

Notes October 10
considering the case where $$ \mathbf{\tilde{d}_{4}^{(e)}} \neq 0     $$,     and       $$  \mathbf{\tilde{d}_{1}^{(e)}}=\mathbf{\tilde{d}_{2}^{(e)}}=\mathbf{\tilde{d}_{3}^{(e)}} = 0   $$

$$ \mathbf{\tilde{f}_{4x1}^{(e)}}= \mathbf{\tilde{k}_{4x4}^{(e)}}\mathbf{\tilde{d}_{4x1}^{(e)}}= \mathbf{0_{4x1}}  $$

The zero matrix $$\mathbf{0_{4x1}} $$ is the fourth col. of matrix$$\mathbf{\tilde{k}_{4x4}^{(e)}}$$

Interpretation of transverse dof's:

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

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)} \\\tilde{d}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$

Similarly for: $$  \mathbf{\tilde{f}^{(e)}}= \mathbf{\tilde{T}^{(e)}}\mathbf{f^{(e)}}  $$

$$ \begin{Bmatrix} \tilde{f}_1^{(e)} \\ \tilde{f}_2^{(e)}\\ \tilde{f}_3^{(e)} \\\tilde{f}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$

Also: $$\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{d}^{(e)}}=\mathbf{\tilde{f}^{(e)}}$$

$$\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\mathbf{\tilde{d}^{(e)}}=\mathbf{\tilde{T}^{(e)}}\mathbf{f^{(e)}}$$

if $$\mathbf{\tilde{T}^{(e)}}$$ is invertible then:

$$\left[\mathbf{\tilde{T}^{(e)^{-1}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\right]\mathbf{\tilde{d}^{(e)}}=\mathbf{f^{(e)}}$$

Where $$\mathbf{\tilde{T}^{(e)}}$$ is block diagonal matrix, consider a general block diagonal matrix A

$$\mathbf{A} = \left[ \begin{matrix} D_{1} &...& 0  \\  ...& D_{2}&...\\ 0&...&D_{n} \end{matrix} \right]$$

$$\mathbf{D}_{1}, \mathbf{D}_{2}, .... , \mathbf{D}_{s}   $$     are matrices

$$\mathbf{A}^{-1}= ?$$

$$\mathbf{B}^{-1}=\left[ \begin{matrix} d_{11} &...& 0 \\  ...& d_22&...\\ 0&...&d_(nn) \end{matrix} \right] $$

$$\mathbf{B}=Diag\left[ \mathbf{d}_{11},\mathbf{d}_{22}, ...... , \mathbf{d}_{nn}  \right] $$

$$\mathbf{B}^{-1}=Diag\left[ \dfrac{1}{d}_{11},\dfrac{1}{d}_{22}, ...... , \dfrac{1}{d}_{nn}  \right] $$

Assuming $$\mathbf{d}_{ii}\neq 0 $$ for i = 1,2, ....,n for a block diagonal matrix A

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

$$\mathbf{A}^{-1}=Diag\left[ \mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1}, ...... , \mathbf{D}_{s}^{-1}  \right] $$

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

$$\mathbf{R}^{(e)^{T}}=\left[ \begin{matrix} l^{(e)} & -m^{(e)} \\  m^{(e)} & l^{(e)} \end{matrix} \right]$$

$$\mathbf{R}_{2x2}^{(e)^{T}}\mathbf{R}_{2x2}^{(e)}=\left[ \begin{matrix} l^{(2)}+m^{(2)} & 0 \\  0 & l^{(2)}+m^{(2)} \end{matrix} \right] = \left[ \begin{matrix} l^{(2)}+m^{(2)} =1 & 0  \\  0 & l^{(2)}+m^{(2) =1} \end{matrix} \right] =\left[ \begin{matrix} 1 & 0  \\  0 & 1 \end{matrix} \right] = \mathbf{I}_{2x2}$$

From this we can conclude that $$\mathbf{R}_{2x2}^{(e)^{(-1)}}=\mathbf{R}_{2x2}^{(e)^{T}}$$

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

$$\left[\mathbf{\tilde{T}^{(e)^{T}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\right]\mathbf{d^{(e)}}=\mathbf{f^{(e)}}$$

The above relation agrees with the fundamental relation $$\mathbf{k^{(e)}}\mathbf{d^{(e)}}=\mathbf{f^{(e)}}$$

Where : $$ \mathbf{k^{(e)}}=\mathbf{\tilde{T}^{(e)^{T}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}$$