User:Eml4500.f08.team.guan/hw4

Derivation of Element Stiffness Matrix and Use of the Invertible Transformation Matrix
Seeing that the goal of finding a transformation matrix that is invertible and can transform a local set of degrees of freedom into another local set of degrees of freedom was achieved from the previous lecture, let us consider the following case to further examine the the new local force displacement relationship marked by the tilda. $$ \tilde{f}^{(e)}_{4x1}$$=$$ \tilde{k}^{(e)}_{4x4}\tilde{d}^{(e)}_{4x1}=k^{(e)} \begin{bmatrix} 1 & 0 & -1 &  0 \\ 0 &  0 & 0 &  0 \\  -1 &  0 & 1 &  0 \\  0 &  0 & 0 &  0 \\ \end{bmatrix} \begin{bmatrix} \tilde{d}^{(e)}_1 \\ \tilde{d}^{(e)}_2 \\ \tilde{d}^{(e)}_3 \\ \tilde{d}^{(e)}_4 \\ \end{bmatrix} $$ This is the force displacement relationship in which the local displacement degrees of freedom in global coordinates have be transformed into local displacement degrees of freedom with transverse and axial displacements. Let's consider in case in which

$$\tilde{d}^{(e)}_4 \ne 0, \tilde{d}^{(e)}_1=\tilde{d}^{(e)}_2=\tilde{d}^{(e)}_3=0$$. Thus, the force displacement relationship is now

$$ \tilde{f}^{(e)}_{4x1}$$=$$ \tilde{k}^{(e)}_{4x4}*\tilde{d}^{(e)}_{4x1}=0$$ where zero here is a 4x1 vector. The transverse displacements can be interpreted as follows. Even though there is a non-zero transverse displacement, the resulting force vector is still zero. This is because transverse displacements induce no force on the member due to the fact that the transverse displacements will not "stretch the spring". It is seen in the matrix equation above where each zero in the stiffness matrix denotes a transverse displacement with no effect on the resulting force. Now, from lecture 19.3 we have derived

$$\tilde{d}^{(e)}=\tilde{T}^{(e)}{d}^{(e)}$$ $$\tilde{f}^{(e)}$$=$$ \tilde{k}^{(e)}\tilde{d}^{(e)}$$ which then leads to $$\tilde{k}^{(e)}\tilde{T}^{(e)}{d}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$ If the transformation matrix here is invertible, then $$[\tilde{T}^{(e)-1}\tilde{k}^{(e)}\tilde{T}^{(e)}]{d}^{(e)}=f^{(e)}$$ The term in the square brackets is the term k that we are looking for.

Consider a general block diagonal matrix A where A can be denoted as $$\displaystyle A= diag[D_1,...,D_s]$$ and $$A^{-1}=diag[D_1^{-1},...D_s^{-1}]$$ where the D's are block matrices of arbitrary nxn dimensions.  $$A=\begin{bmatrix} D_1 & ... &...&...&...&0 \\ 0 & D_2 &...&...&...&0 \\ 0 & ... &...&...&...&0 \\ 0 & ... &...&...&...&0 \\ 0 & ... &...&...&D_{s-1}&0 \\ 0 & ... &...&...&...& D_s\\ \end{bmatrix}$$ What, then, is the inverse of A? Let's consider a simpler example.

 $$B=\begin{bmatrix} d_{11} & ... &...&...&...&0 \\ 0 & d_{22} &...&...&...&0 \\ 0 & ... &...&...&...&0 \\ 0 & ... &...&...&...&0 \\ 0 & ... &...&...&d_{n-1,n-1}&0 \\ 0 & ... &...&...&...& d_{nn}\\ \end{bmatrix}$$ Here, the diagonal is not made up of block matrices but of scalar values that are not zero. What might the inverse of matrix B be? $$B^{-1}=diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$ For a block diagonal matrix A: $$\displaystyle A= diag[D_1,...,D_s]$$ and $$A^{-1}=diag[D_1^{-1},...D_s^{-1}]$$. Now, the transformation matrix is a block diagonal matrix. so it can be denoted as $$\tilde{T}^{(e)-1}=diag[R^{(e)-1},R^{(e)-1}]$$ So the next step is to find $$R^{(e)-1}$$. Recall from 19.2 that $$R^{(e)T}= \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \\ \end{bmatrix} $$ $$R^{(e)T}R^{(e)}= \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \\ \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \\ \end{bmatrix}= \begin{bmatrix} l^{(e)2}+ m^{(e)2}& 0 \\ 0 & l^{(e)2}+ m^{(e)2} \\ \end{bmatrix}$$ Now recall that l is cosine of theta and m is sine of theta. Cosine theta squared please sine theta squared is, by trigonometric identity, one. Thus, the resulting matrix from multiplying R and its transpose is the identity matrix.  $$R^{(e)T}R^{(e)}= \begin{bmatrix} 1& 0 \\ 0 &  1 \\ \end{bmatrix}=I $$ This means that the transpose of the R matrix is actually its inverse. $$\displaystyle R^{(e)T}=R^{(e)-1}$$ With that known, the inverse of the transformation matrix can be found but replace the diagonal with the transposes of the R block matrix. $$\tilde{T}^{(e)-1}=diag[R^{(e)T},R^{(e)T}]=(diag[R^{(e)},R^{(e)})^T=(\tilde{T}^{(e)-1})^T$$ This means that the inverse of the transformation matrix is equal to its transpose. $$\tilde{T}^{(e)-1}=\tilde{T}^{(e)T}$$ This verifies that $$[\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)}]{d}^{(e)}=f^{(e)}$$ where $$[\tilde{T}^{(e)-1}\tilde{k}^{(e)}\tilde{T}^{(e)}]$$ is $$\displaystyle k^{(e)}$$.