User:Eml4500.f08.wiki1.schaet/hw3

Meeting 14 - 26 September 2008
Similarly, (same argument): $$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = I^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$ b Where P is a 2x1 matrix, I is a 2x4 matrix, and f is a 4x1 matrix. This relationship is the same as saying $$ \underline{P}^{(e)}= I^{(e)} \underline{f}^{(e)} $$ Recall the axial Fd relationship: $$ \hat k^{(e)} q^{(e)} = P^{(e)} $$ where k is a 2x2 matrix, q is a 2x1 matrix, and P is a 2x1 matrix. $$ \hat k^{(e)} \underbrace{(\underline{T}^{(e)} \underline{d}^{(e)} )} = \underbrace{(\underline{T}^{(e)} \underline{f}^{(e)})} $$ $$ - \qquad \underline{q}^{(e)} \qquad \qquad \underline{P}^{(e)} $$ Goal: We want to have k (e) d (e)= f (e) so "move" T e from right side to the left side by pre multiplying equation by T (e)-1, the inverse of T (e). Unfortunately, T(e) is a rectangular matrix of the size 2x4 and cannot be inverted. Only square matricies can be inverted. To solve this issue, transpose T. Ans: $$ (\underline {T}^{(e)T} \underline {\hat k}^{(e)} \underline {T}^{(e)})\underline {d}^{(e)} = \underline {f}^{(e)} $$ $$ \underline{k}^{(e)} \underline{d}^{(e)} = \underline{f}^{(e)} $$ $$ \underline{k}^{(e)} = \underline{T}^{(e)T} \underline{\hat k}^{(e)} \underline{T}^{(e)} $$ where k is on p6.1, k hat is on 12-2, and T is on 12-5. Justification for (1): PVW see 10-1 for 1st applications of PVW, reduction of global FD relationship. $$ \underline {K}_{6x6} d_{6x1} = \underline{F}_{6x1} \rightarrow \underline{\bar K}_{2x2} \underline{\bar d}_{2x1} = \underline {\bar F}_{2x1} $$ Remember: Why not solve as follows? $$\underline {d} = \underline{K}^{-1} F$$ Used mathcad to try to solve K-1 and could not, due to singularity of K ., i.e. the determinant of K =0 and thus K is not invertible. Recall that you need to computer $$\begin{matrix} \frac{1}{detK} \end{matrix}$$ to find K -1. Why? For an unconstrained structure system, there are 3 possible rigid body mostion in 2-D (2 translational and 1 rotational). HW: Find the eigenvalues of K and make observations about the number of eigenvalues. Dyanmic eigenvalue problem K v = λ M  v  Where K is the stiffness matrix, M is the mass. Zero evaluation corresponds to zero stored elastic energy which means rigid body motion.