User:Eml4500.f08.wiki1.aguilar/hw5

=== Notes 10/27: also show eqn(1) equals eqn(2). Do not do the 6 bar truss coding ===

We want to derive $$\mathbf{k}^{(e)}_{\,4x4}= \mathbf{T}^{(e)T}_{4x2} \,\hat{\mathbf{k}}^{(e)}_{2x2}\, \mathbf{T}^{(e)}_{2x4}$$, an equation that can be found on p14.3 and p23.3 of our class notes.

Recall the force-displacement relationship with axial degrees of freedom, $$q^{(e)}$$.

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} = \mathbf{p}^{(e)}$$

Moving some things around we can get

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)} = \mathbf{0}_{2x1} $$     (1)

Plugging this in for the Principle of Virtual Work:

$$ PVW\Rightarrow \mathbf{\hat{w}}\cdot \underbrace{(\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)})}_{2x1\mathrm{\;for\;all\;} \mathbf{\hat{w}}_{2x1}} = {0}_{1x1} $$         (2)

We showed that equations (1) and (2) are equivalent on p24.1 of our class notes.

Recall,

Here we can see the similarities between the equation for the real displacement and the virtual displacement.

