User:Eml4500.f08.team.dwyer.jd/HW5 Oct 27 (Meeting 26)

The Principle of Virtual Work - Part II
$$d^{i}_{4x1}\Rightarrow\mathbf{d(lmm(i,:))}$$

element DOFS in the global (x,y) coordinate system for element (i)

Recall: $$\mathbf{q}^{(i)}=\mathbf{Td}^{(i)}$$

The composition of our course grade is an example of the weighting factor shown below.

 GTotal = $$\alpha _{o} \cdot (HW Grade)+\sum_{i=1}^{3}{\alpha _{i}}\cdot (Exam_{i})$$

Our overall goal is to derive $$\mathbf{k}_{4x4}^{(e)} = \mathbf{T}_{4x2}^{(e)T} \hat\mathbf{k}_{2x2}^{(e)} \mathbf{T}_{2x4}^{(e)}$$ via the Principle of Virtual Work Recall: $$\hat\mathbf{k}_{2x2}^{(e)} \mathbf{q}_{2x2}^{(e)} = \mathbf{p}_{2x1}^{(e)}$$

$$ \Rightarrow \hat\mathbf{k}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1} - \mathbf{p}^{(e)}_{2x1} = \mathbf{0}_{2x1} \qquad \rightarrow \qquad (1) $$ By the Principle of Virtual Work (PVW): $$\mathbf{w}_{2x1} * ( \hat\mathbf{k}^{(e)} \mathbf{q}^{(e)} - \mathbf{p}^{(e)} )_{2x1} = 0_{1x1} \quad $$ for all $$\; \mathbf{w}_{2x1}\qquad \rightarrow \qquad (2) $$ $$\therefore \;$$ We showed that $$ (1) \longleftrightarrow (2)$$ Recall:

$$\mathbf{q}{(e)} = \mathbf{T}^{(e)} \mathbf{d}^{(e)} \qquad \rightarrow \qquad (3)$$ Similarly,
 * $$\mathbf{q}^{(e)} \;$$ is the real displacement matrix defined in the axial coordinate system
 * $$\mathbf{d}^{(e)} \;$$ is the real displacement matrix defined in the global coordinate system

$$ \; \hat\mathbf{w}_{2x1} = \mathbf{T}_{2x4}^{(e)} \mathbf{w}_{4x1}^{(e)} \qquad \rightarrow \qquad (4)$$
 * $$\hat\mathbf{w}_{2x1} \;$$ is the virtual displacement matrix defined in the axial coordinate system
 * $$\mathbf{w}_{4x1} \;$$ is the virtual displacement matrix defined in the global coordinate system