User:EML4500.f08.delta 6.ramirez/112108

Continued examination of dimensions for various variables.

$$ [\varepsilon ] = \frac{[d_4]}{[dx]} = \frac{L}{L} = 1 $$

Further,

$$ [\sigma] = [E] = \frac{F}{L^2} $$

$$ [A] = L^2 $$

$$ [I] = \int x^2dx = L^4 $$

$$ [\frac{EA}{L}] = [\tilde{k}_{11}]= \frac{L^2\frac{F}{L^2}}{L} = \frac{F}{L} $$

$$ [\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = \frac{F}{L}L = F $$

$$ [\tilde{k}_{23}\tilde{d}_3] = [\tilde{k}_{23}][\tilde{d}_3] = \frac{[6][E][I]}{[L^2]} = \frac{1 \frac{F}{L^2} L^4}{L^2} 1 = F $$

Note: $$ [\tilde{d}_3] $$ = 1

Goal: Derive element force displacement relationship in global coordinate system from element force displacement relationship in local coordinates.

In elaboration, obtaining

$$ k^{(e)}_{6x6} d^{(e)}_{6x1} = f^{(e)}_{6x1} $$

and



from

$$ \tilde{k}^{(e)}_{6x6} \tilde{d}^{(e)}_{6x1} = \tilde{f}^{(e)}_{6x1} $$

In order to move on $$ \tilde{T}^{(e)}_{6x6} $$ must be analyzed, which has a block diagonal structure.

Note: It is easy to expand the $$ \tilde{T}^{(e)}_{6x6} $$ matrix to account for the frame problem since the rotational axis $$ z = \tilde{z} $$

Examining the relationship between $$ \tilde{d}_i $$ and $$ d_i $$ gives,



Note: $$ \tilde{d}_3, \tilde{d}_6, d_3, d_6 $$ represent rotational degrees of freedom.

Furthermore, we will now derive $$ \hat{k}^{(e)} $$ from PVW, focusing only on bending effect.

$$ \frac{\partial^2}{\partial x^2}((EI) \frac{\partial^2v}{\partial x^2}) - f_t(x) = m(x) \ddot{v} $$

Note: EI is a function of x, $$ f_t(x) $$ represents the distributive transverse load, and v represents the transverse displacement.