User:Eml4500.f08.delta 6.krueger/HW

Verify the element stiffness matrix. Start with equation 1: $$k^{(e)}={\tilde{T}^{(e)T}}{\tilde{k}^{(e)}}{\tilde{T}^{(e)}}$$ The substitute in with: equation 2 '''$$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix}$$''' and equation 3 '''$$R^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}$$''' and equation 4 '''$${\tilde{T}^{(e)T}}=Diag\begin{bmatrix} R^{(e)T} & R^{(e)T}\end{bmatrix}$$''' and equation 5 '''$${\tilde{T}^{(e)}}=Diag\begin{bmatrix} R^{(e)} & R^{(e)}\end{bmatrix}$$''' also '''$${\tilde{k}^{(e)}}=k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$''' From previous work $$l^{(1)}=cos(30)=.866025$$ and $$m^{(1)}=sin(30)=.5$$. so '''$$R^{(e)T}=\begin{bmatrix} .866025 & -.5\\ .5 & .866025 \end{bmatrix}$$''' and '''$$R^{(e)}=\begin{bmatrix} .866025 & .5\\ -.5 & .866025 \end{bmatrix}$$''' and '''$$k^{(e)}=\begin{bmatrix} .5625 & 3248 & -.5625 & -.3248\\ .3248 & .1875 & -.3248 & -.1875\\ -.5625 & -.3248 & .5625 & .3248\\ -.3248 & -.1875 & .3248 &.1875 \end{bmatrix}$$''' Now the known matrices are substituted into equations 2, 3, 4 and 5. Then equations 2, 3, 4 and 5 are substituted into equation 1 to verify that: '''$$k^{(e)}=\begin{bmatrix} .5625 & 3248 & -.5625 & -.3248\\ .3248 & .1875 & -.3248 & -.1875\\ -.5625 & -.3248 & .5625 & .3248\\ -.3248 & -.1875 & .3248 &.1875 \end{bmatrix}$$''' Prove equation 1 $${\tilde{f}^{(e)}}={\tilde{T}^{(e)}}f^{(e)}$$ known formulas: equation 2 $${\tilde{f}^{(e)}}={\tilde{k}^{(e)}}{\tilde{d}^{(e)}}$$ equation 3 $${\tilde{d}^{(e)}}={\tilde{T}^{(e)}}d^{(e)}$$ equation 4, where I is the identity matrix $${\tilde{k}^{(e)}}=k^{(e)}I$$ equation 5 $$f^{(e)}=k^{(e)}d^{(e)}$$ Substituting equation 2 into equation 1: $${\tilde{f}^{(e)}}={\tilde{k}^{(e)}}{\tilde{T}^{(e)}}d^{(e)}$$ The substituting equation 3 and equation 4 into equation 1: $${\tilde{f}^{(e)}}=k^{(e)}I{\tilde{T}^{(e)}}d^{(e)}$$ Finally, equation 5 is substituted into equation 1 and the Identity matrix is dropped. This gives the wanted results. $${\tilde{f}^{(e)}}={\tilde{T}^{(e)}}f^{(e)}$$