User:EML4500.f08.Ateam.Miller/homework6

11/14/08
Our goal is to use linear interpolation to determine $$N_i(\tilde{x})$$. If $$\tilde{x}= x-x_i$$, then

$$N_i(\tilde{x})=\frac{\tilde{x}-\tilde{x}_{i+1}}{\tilde{x}_{i}-\tilde{x}_{i+1}}$$

Now, the shape function can be written as follows:

$$ N_2^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}= \begin{cases} & \text{0 at } \tilde{x}= 0\\ & \text{1 at } \tilde{x}= L \end{cases} $$

Our next goal is to compare the general $$\mathbf{k}^{(i)}$$ to the stiffness matrix obtained by using

$$\frac{1}{2}(A_1+A_2)$$ and $$\frac{1}{2}(E_1+E_2)$$, where $$E_1 \neq E_2$$. The resulting stiffness matrix is

$$ \frac{(E_1+E_2)(A_1+A_2)}{4L^{(i)}} \begin{bmatrix} 1 & -1\\ 1 & -1\\ \end{bmatrix} =\mathbf{k}_{avg}^{(i)} $$

This matrix is also known as the average stiffness matrix.

Now, we want to determine $$\mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}$$

$$ \mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}= \frac{(A_1+A_2)}{4L^{(i)}}(2E-E_1-E_2) $$

These results show that the average value of the stiffness matrix is not equal to the value of the stiffness matrix evaluated at the average value of x. This can be verified with the Mean Value Theorem.

Mean Value Theorem
Consider the 2D mass located in the xy plane. Its centroid is located at $$(\bar{x},\bar{y})$$.



The Mean Value Theorem (MVT) states

$$\int_{x=a}^{x=b}{f(x)dx}=f(\bar{x})[b-a]$$

for $$\bar{x}\epsilon[a,b]$$ or $$a \leq \bar{x} \leq b$$.

Similarly,

$$\int_{x=a}^{x=b}{f(x)g(x)dx}=f(\bar{x})g(\bar{x})[b-a]$$

for $$a \leq \bar{x} \leq b$$.

However,

$$f(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of f.

Similarly,

$$g(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of g.