User:Eml4500.f08.jamama.justin/HW3

Similarly (same argument)

$$\begin{Bmatrix} P{_{1}}^{(e)}\\ P{_{2}}^{(e)}\\ \end{Bmatrix} = T^{(e)} \begin{Bmatrix} f{_{1}}^{(e)}\\ f{_{2}}^{(e)}\\ f{_{3}}^{(e)}\\ f{_{4}}^{(e)} \end{Bmatrix} $$

2X1 2X4 4X1

$$\Rightarrow P^{(e)}=T^{(e)}f^{(e)}$$

Recall element axial force displacement relationship

$$\hat{k}^{(e)}q^{(e)}=P^{(e)}$$

2X2 2X12X1

$$\Rightarrow \hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)=\left(T^{(e)}f^{(e)} \right)$$

\_________/\__________/

$$q^{(e)}$$$$p^{(e)}$$

Goal: Want to have $$k^{(e)}d^{(e)}=f^{(e)}$$ so "move" $$T^{(e)}$$ from right to left hand side by pre-multiplying equation by $$T^{(e)-1}$$ (inverse of $$T^{(e)}$$ ). Unfortunately, $$T^{(e)}$$ is a rectangular matrix (2X4) so it cannot be inverted.

So the actual answer is:  $$\begin{bmatrix} T{^{(e)T}}\bullet \hat{k}^{(e)}T^{(e)} \end{bmatrix} d^{(e)}=f^{(e)}$$

[4X22X22X4] 4X14X1

[ equal to a 4X4 ]

 $$k^{(e)}d^{(e)}=f^{(e)}$$ 

$$k^{(e)}=T{^{(e)T}} \hat{k}^{(e)}T^{(e)}$$

Justification for the above equation: Principle Virtual Work (PVW)

(See notes from the September1 17th for the first application of PVW. This is shown through Reduction of Global FD relationship.)

$$k d = F \Leftrightarrow  \bar{k} \bar{d} = \bar{F}$$

6X6 6X1 6X62X2 2X1 2X1

Question: Why would you not solve as follows? $$d = k^{-1} F$$ ?

It is because the matrix k is a singularity, and the det k=0. Therefore $$k^{-1}$$ is not invertible. This is because

to find $$k^{-1}$$ you need to compute $$\frac{1}{det k}$$

why? For an unconstrained structural system, there are three possible rigid body motions in 2-D. (2-translational and 1-rotational)

Dynamic eigenvalue problems: $$K v=\lambda Mv $$

K=Stiffness matrix

$$\lambda$$=Eigen value (related to the vibrational frequency)

M=Mass matrix

Zero eigenvalue corresponds to zero stored elastic energy, so rigid body modes. Modes, also called mode shapes, are used to describe the vibrations in an object. These correspond to the number of half wave vibrations in a body.