User:Eml4500.f08.RAMROD.F/HW5

Mtg 26 Monday October 27
We are going to try to derive element stiffness matrix by Principle of Virtual Work.

$$K^{(e)}=\underline{T}^{(e)^\tau} \hat{\underline{K}}^{(e)}\underline{T}^{(e)}$$

Recalling the force displacement with axial degree of freedom $$\underline{q}^{(e)}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} = \underline{P}^{(e)}_{2x1}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}= \mathbf{0}_{2x1}$$ Equation $$(1)$$

Principle of Virtual Work applied here: $$\hat{\mathbf{W}}\bullet \left(\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}\right)= \mathbf{0}_{2x1}$$ Equation $$(2)$$

We showed equation $$(1) \Leftrightarrow (2)$$

Recall $$\mathbf{\underline{q}}^{(e)}= T^{(e)}\mathbf{\underline{d}}^{(e)}$$

Similarly $$\hat{\underline{W}}^{(e)}_{2x1}= \bar{T}^{(e)}_{2x4}\mathbf{\underline{W}}^{(e)}_{4x1}$$

FOR KEVIN

$$K^{(e)}=\underline{T}^{(e)^\tau} \hat{\underline{K}}^{(e)}\underline{T}^{(e)}= k^{(e)}\begin{bmatrix} l^{2^{(e)}} & lm^{(e)} & lmn^{(e)} & -l^{2^{(e)}} & -lm^{(e)} & -ln^{(e)}\\ ln^{(e)} & n^{2^{(e)}} & mn^{(e)} & -lm^{(e)} & -m^{2^{(e)}} & -mn^{(e)}\\ lmn^{(e)} & mn^{(e)} & n^{2^{(e)}} & -lmn^{(e)} & -mn^{(e)} & -n^{2^{(e)}}\\ -l^2^ & -lm^{(e)} & -lmn^{(e)} & l^{2^{(e)}} & lm^{(e)} & lm^{2^{(e)}}\\ -lm^{(e)} & -m^{2^{(e)}} & -mn^{(e)} & lm^{(e)} & m^{2^{(e)}} &mn^{(e)} \\ -ln^{(e)} & -mn^{(e)} & -n^{2^{(e)}} & lm^{2^{(e)}} & mn^{(e)} & n^{2^{(e)}} \end{bmatrix}$$