User:Eml4500.f08.delta 6.guzman/Lecture Notes 09/26

Similarly: $$\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}_{2x1}= T^{(e)}\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}_{4x4} => P^{(e)}=T^{(e)}*F^{(e)}$$ $$\hat{K}^{(e)}_{2x2}*q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ where $$ q^{(e)}_{2x1}=T^{(e)}*d^{(e)}$$ and $$ P^{(e)}_{2x1}= T^{(e)}*F^{(e)} $$ therefore $$\Rightarrow \hat{K}^{(e)}_{2x2}*T^{(e)}*d^{(e)}= T^{(e)}*F^{(e)}$$

The goal with these matrices is to obtain  $$K^{(e)}*d^{(e)}=F^{(e)}$$  in order to do that we need to move   $$T^{(e)}$$  from the right hand side to the left hand side and this usually can be done by multiplying by the inverse   $$T^{(e)-1}$$  but since   $$T^{(e)}$$  is a rectangular matrix with (2x4) dimensions it can not be inverted therefore what needs to be done is the transpose $$T^{(e)^(T)}$$

$$[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]$$$$ * d^{(e)}_{4x1}$$ = $$F^{(e)}_{4x1}$$ where $$K^{(e)}_{4x4}=[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]\Rightarrow $$ $$K^{(e)}*d^{(e)}=F^{(e)}\Rightarrow K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

By the principle of virtual work (PVW) the reduction of the global force displacement relation is $$\Rightarrow$$

$$K_{6x6}*d_{6x1}=F_{6x1}\Rightarrow K_{2x2}*d_{2x1}=F_{2x1}$$

Then why not solve these matrices at this stage as follows   $$d=K^{-1}*F$$   ? The reason why this can not be done is due to the matrix singularity of  $$ K $$   Since the determinant of matrix   $$ K $$   is equal to zero its inverse will be $$ \frac{1}{det [K]} $$ which will be undefined since the denominator is equal to zero.