User:Eml4500.f08.lulz.abcd/HW6

HW: What if EA is constant?
If EA is constant, prove the below stiffness matrix to be true:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

We are given this general form of the stiffness matrix:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

We also can solve $$\mathbf{B}$$ and $$\mathbf{B}^T$$ into terms of $$L^{(i)}=x_{i+1}-x_i$$:

$$\mathbf{B}(x) = \begin{bmatrix} N'_i(x) & N'_{i+}(x)\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x_i-x_{i+1}} & \frac{1}{x_{i+1}-x_i}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

$$\mathbf{B}^T(x) = \begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

From here it is simple substitution and algebra to solve the proof since (EA)(x) can be removed from the integral as a constant.

$$\mathbf{k}^{(i)}=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}\begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}dx=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{{L^{(i)}}^2} & -\frac{1}{{L^{(i)}}^2}\\ -\frac{1}{{L^{(i)}}^2} & \frac{1}{{L^{(i)}}^2}\\ \end{bmatrix}dx=(EA)\begin{bmatrix} \frac{x_{i+1}-x_i}{{L^{(i)}}^2} & -\frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ -\frac{x_{i+1}-x_i}{{L^{(i)}}^2} & \frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ \end{bmatrix}=(EA)\begin{bmatrix} \frac{L^{(i)}}{{L^{(i)}}^2} & -\frac{L^{(i)}}{{L^{(i)}}^2}\\ -\frac{L^{(i)}}{{L^{(i)}}^2} & \frac{L^{(i)}}{{L^{(i)}}^2}\\ \end{bmatrix}$$

Dividing out $$L^{(i)}$$ from this and reducing the fractions yields the target equation:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

HW: Find an expression for $$\mathbf{k}^{(i)}(\tilde{x})$$
We are given the general formula for the stifness matrix and want to find a more useful way to express it in the $$\tilde{x}$$ coordinate system:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

Remembering how to convert the coordinate systems we can convert the general formula to the $$\tilde{x}$$ coordinate system:

$$\tilde{x}:=x-x_i$$

$$d\tilde{x}=dx$$

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$

Next we consider the known solutions for A and E from the notes:

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

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

Substituting these yields:

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})[(N_1^{(i)}(\tilde{x})E_1+N_2^{(i)}(\tilde{x})E_2)(N_1^{(i)}(\tilde{x})A_1+N_2^{(i)}(\tilde{x})A_2)]\mathbf{B}(\tilde{x})d\tilde{x}$$