User:Eml4500.f08.RAMROD.F/HW6

= Mtg. 33 Wed November 12 =

Element stiffness matrix for element i
The Element stiffness matrix originates from the following equation:

$$\int_{x_{x}}^{x_{i+1}}{\left[N^1_iW_i+N^1_{i+1}W_{i+1} \right](EA)}\left[N^1_id_i+N^1_{i+1}d_{i+1} \right]dx$$

$$N^1_i=\frac{dN_i(x)}{dx}$$ Likewise for N_{i+1}

Note: $$u(x)=\begin{vmatrix} N_i(x) & N_{i+1}(x) \end{vmatrix}\begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix}$$ and $$\frac{du(x)}{dx}=\begin{vmatrix} \dot{N}_i(x) & \dot{N}_{i+1}(x) \end{vmatrix}\begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix}$$ where $$\mathbf{N}(x)_{1x2}=\begin{vmatrix} N_i(x) & N_{i+1}(x) \end{vmatrix}$$ and $$\mathbf{B}(x)_{1x2}=\begin{vmatrix} \dot{N}_i(x) & \dot{N}_{i+1}(x) \end{vmatrix}$$

Similiarly, the matrix $$\mathbf{W}(x)$$ can be written as:

$$\mathbf{W}(x)= \mathbf{N}(x)\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix}$$

$$\frac{d\mathbf{W}(x)}{dx}= \mathbf{B}(x)\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix}$$

Recalling the Degree of Freedom to establish these two in a matrix conventions:



$$ \mathbf{d}^{(i)}=\begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix}\begin{Bmatrix} d^{(1)}_1\\ d^{(1)}_{2} \end{Bmatrix}$$ and $$ \mathbf{W}^{(i)}=\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix}\begin{Bmatrix} W^{(1)}_1\\ W^{(1)}_{2} \end{Bmatrix}$$

$$B=\int_{x_i}^{x_{i+1}}{\left[ (BW^{(i)})(EA)(Bd^{(i)})\right]dx}$$ $$B=W^{(i)}\cdot (K^{(i)}d^{(i)})$$

Using the transpose of matrix B to prove that the global stiffness matrix K will be listed as follows:

$$\mathbf{K}^{(i)}_{2x2}=\int_{x_i}^{x_{i+1}}{\left[ (B^{\tau } (x)(EA)(B(x)\right]dx}$$

which leads to the formation of the matrix K when EA is a constant being:

$$\mathbf{K}^{(i)}_{2x2}=\frac{EA}{L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}$$

Transformation of Variable (coordinate) from $$x$$ to $$\tilde{x}$$
$$\tilde{x}=x-x_i$$ and $$d\tilde{x}=dx$$

so the B matrix in the new coordinate system can be expressed this way:

$$ \mathbf{K}^{(i)}_{2x2}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}(B^{\tau } (\tilde{x})(EA){B(\tilde{x})d\tilde{x}}$$

which leads to the formation of the matrix K when EA and L is the length of element i is a constant being:

$$\mathbf{K}^{(i)}_{2x2}=\frac{EA}{L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}$$



Here the area matrix and the elasticity matrix are:

$$A(\tilde{x})=N^{(i)}_1(\tilde{x})A_1+N^{(i)}_2(\tilde{x})A_2 $$

$$E(\tilde{x})=N^{(i)}_1(\tilde{x})E_1+N^{(i)}_2(\tilde{x})E_2$$ Giving us the Stiffness matrix K equaling: $$\mathbf{K}^{(i)}(\tilde{x})=\frac{\left[N^{(i)}_1(\tilde{x})E_1+N^{(i)}_2(\tilde{x})E_2 \right]\left[N^{(i)}_1(\tilde{x})A_1+N^{(i)}_2(\tilde{x})A_2 \right]}{L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1& 1 \end{bmatrix}$$