User:Eml4500.f08.team.figuerrez/hw5

Meeting 27 - The Principle of Virtual work - Part III
$$\tilde{\boldsymbol{W}}_{2x1}=$$ Virtual axial displacement. $$\tilde{\boldsymbol{W}}_{2x1}$$ corresponds to the axial displacement matrix, $${\boldsymbol{q}}^{(e)}_{2x1}$$.

$${\boldsymbol{w}}_{4x1}=$$ Virtual displacement in global coordinate system. $${\boldsymbol{w}}_{4x1}$$ corresponds to the global displacement matrix, $${\boldsymbol{d}}^{(e)}_{4x1}$$. When you combine equations  (3)  &  (4)  into equation  (2)  it yields:

 (5)  $$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})* \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}}_{4x1}$$

Recall:

 (6)  $$({\boldsymbol{A}}_{pxq}{\boldsymbol{B}}_{qxr})^{T}= {\boldsymbol{B}}^{T}{\boldsymbol{A}}^{T} $$ Equation (6) equates that the transpose of the multiplication of two matrices is equal to the transpose of matrix one multiplied by the transpose of matrix two. Equation (6) is verified below with given values for matrix A and B.

Recall:

 (7)  $${\boldsymbol{a}}_{nx1}{\boldsymbol{b}}_{nx1}= {\boldsymbol{a}}^{T}_{1xn}{\boldsymbol{b}}_{nx1} $$

 Note:  The right side of equation (7) is scalar: $${\boldsymbol{(1xn)}}{\boldsymbol{(nx1)}}={\boldsymbol{(1x1)}}=(Scalar)

$$

When you apply equations  (7)  &  (6)  into equation  (5)  it yields:

$$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})^{T}* \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}}_{4x1}$$

 Note:  $$({\boldsymbol{T}}^{(e)}{\boldsymbol{w}})^{T}$$ comes from equation (7). Applying the transpose rule for matrix multiplication from equation (6) yields:

$${\boldsymbol{w}}^{(T)}{\boldsymbol{T}})^{(e)T} \begin{bmatrix} \tilde{\boldsymbol{k}}^{(e)}({\boldsymbol{T}}^{(e)}{\boldsymbol{d}}^{(e)})-{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$

When you replace the transpose with the dot product it yields:

$${\boldsymbol{w}}* \begin{bmatrix} ({\boldsymbol{T}}^{(e)}\tilde{\boldsymbol{k}}^{(e)}{\boldsymbol{T}}^{(e)}){\boldsymbol{d}}^{(e)}-({\boldsymbol{T}}^{(e)T}{\boldsymbol{p}}^{(e)} \end{bmatrix} =o $$

Therefore, $${\boldsymbol{w}}* \begin{bmatrix} ({\boldsymbol{k}}^{(e)}{\boldsymbol{d}}^{(e)}-{\boldsymbol{f}}^{(e)} \end{bmatrix} =o $$  for all $${\boldsymbol{w}} $$

So far, only discrete cases have been given. Now, we have continuous cases, i.e. PDEs.

Motivational model problem:

We have an elastic bar (shown in Figure #1 below) with varying A(x), Young's Modulus E(x), which is subjected to a varying axial load (distributed), concentrated load, and an inertia force (dynamic force). Both the varying axial load and the concentrated load are both time dependent.