User:Eml4500.f08.ateam.rivero/w5

Principle of Virtual Work Continued
The following is analogy between the weighing factor used in the principle of virtual work and how the course grade for this course, eml 4500 is calculated.

$$Course Grade = \alpha_0 + HW Grade + \sum_{i=1}^3 \alpha_i * Exam 1 $$

Deriving :$$ \mathbf k^{(e)} = \mathbf T^{(e)T} * \mathbf \hat k^{(e)} * \mathbf T^{(e)} $$ Recall Force Displacement relation with axial displacement degrees of freedom $$ \mathbf q^{(e)}$$ $$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$ $$ \rightarrow \mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)} = \mathbf 0 $$ : $$ (Eq.1) $$ $$ \mathbf \hat w_{(2x1)} * (\mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)})_{(2x1)} = 0 $$ for all $$ \mathbf \hat w_{(2x1)} $$ : $$ (Eq.2) $$ Recall: $$ \mathbf q_{(2x1)}^{(e)} = \mathbf T_{(2x4)}^{(e)} * \mathbf d_{(4x1)}^{(e)} $$ : $$ (Eq.3) $$ Similarly: $$ \mathbf \hat w_{(2x1)} = \mathbf T_{(2x4)}^{(e)} * \mathbf w_{(4x1)} $$ : $$ (Eq.4) $$ Note: $$ \mathbf \hat w_{(2x1)} $$ is the equivalent of the axial virtual displacement