User:Eml4500.f08.delta 6.krueger/HW K^i

Finding the element stiffness matrices for the 2 bar system The following is an equation that was derived earlier. It will ultimately be used to find the two element stiffness matrices for the system. Eqn (0) $$\mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\mathbf{B^T(\tilde{x})}}EA(\tilde{x})\mathbf{B(\tilde{x}})d(\tilde{x}) $$ Eqn (1)

$$ N_{i}(x)=N_1^{(i)}(\tilde{x}) $$

Eqn (2)

$$ N_{i+1}(x)=N_2^{(i)}(\tilde{x}) $$ Eqn (3)

$$ A(\tilde{x})=N_1^{(i)}(\tilde{x})A_1+N_2^{(i)}(\tilde{x})A_2 $$

Eqn (4)

$$ E(\tilde{x})=N_1^{(i)}(\tilde{x})E_1+N_2^{(i)}(\tilde{x})E_2 $$

Eqn (5) $$B(x)=\begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix}$$ Eqn (6) $$ B(x)^{T}=\begin{Bmatrix} N_{i}(x)\\ N_{i+1}(x) \end{Bmatrix}$$ To transform the matrices of B(x) and $$B(x)^{T}$$ into a function of $$\tilde{x}$$ Eqn (1) and (2) must be substituted into the B(x) and the $$B(x)^T$$ equation. Then the $$B(\tilde{x})$$, $$B(\tilde{x})^T$$, Eqn (3) and Eqn (4) have to be substituted into Eqn (0).This gives the following equations: Eqn (7) $$k^{(i)}=(N_{1}^{(i)}(\tilde{x})E_{1}+N_{2}^{(i)}(\tilde{x})E_{2})\begin{bmatrix}N_{1}^{(i)}(\tilde{x}) & N_{2}^{(i)}(\tilde{x})\end{bmatrix}\begin{Bmatrix}N_{1}^{(i)}(\tilde{x}) \\ N_{2}^{(i)}(\tilde{x}) \end{Bmatrix}(N_{1}^{(i)}(\tilde{x})A_{1}+N_{2}^{(i)}(\tilde{x})A_{2})$$ $$-(N_{1}^{(i)}(\tilde{x})E_{1}+N_{2}^{(i)}(\tilde{x})E_{2})\begin{bmatrix}N_{1}^{(i)}(\tilde{x}) & N_{2}^{(i)}(\tilde{x})\end{bmatrix}\begin{Bmatrix}N_{1}^{(i)}(\tilde{x}) \\ N_{2}^{(i)}(\tilde{x}) \end{Bmatrix}(N_{1}^{(i)}(\tilde{x})A_{1}+N_{2}^{(i)}(\tilde{x})A_{2})$$ The following values are known for the given variables. $$ (1)N_1^{(i)}(\tilde{x})= 1 $$ at $$\tilde{x}=0$$ $$ (1)N_2^{(i)}(\tilde{x})= 0 $$ at $$\tilde{x}=0$$ $$ (2)N_2^{(i)}(\tilde{x})= 1 $$ at $$\tilde{x}=L^{(i)}$$ $$ (2)N_1^{(i)}(\tilde{x})= 0 $$ at $$\tilde{x}=L^{(i)}$$ The first term in the element stiffness formula, Eqn (7), is evaluated at$$\tilde{x}=L^{(i)}$$ and the second term is evaluated at $$\tilde{x}=0$$. The above values are substituted into Eqn (7) and the $$B(\tilde{x})$$ and $$B(\tilde{x})^{T}$$ matrices are multiplied out to simplify the equation. This gives: Eqn (8) $$k^{(i)}=(E_{2}^{(i)}A_{2}^{(i)})\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}-(E_{1}^{(i)}A_{1}^{(i)})\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ For element 1 the following values are substituted into the element stiffness matrix, Eqn (8). $$k^{(i)}$$. $$E_{1}^{(1)}=2$$ $$E_{2}^{(1)}=4$$ $$A_{1}^{(1)}=.5$$ $$A_{2}^{(1)}=1.5$$ This gives the element stiffness matrix for element 1. $$k^{(1)}=\begin{bmatrix}-1 & 0 \\ 0 & 6\end{bmatrix}$$ For element 2 the same process is used, but the following is substituted into Eqn (8). $$E_{1}^{(2)}=3$$ $$E_{2}^{(2)}=7$$ $$A_{1}^{(2)}=1$$ $$A_{2}^{(2)}=3$$

This gives the element stiffness matrix for element 2. $$k^{(1)}=\begin{bmatrix}-3 & 0 \\ 0 & 21\end{bmatrix}$$