User:Eml4500.f08.ateam.nobrega/hw 2



It is important to note that the matrix, k(1), is symmetric along the diagonal. This means that values with interchangeable row and column indices are equal; for example, k(13) is equal to k(31) and k(12) is equivalent to k(21). Thus, only three values need to be computed. The remaining coefficients have the same absolute value and only differ by a positive or negative sign. In general, the transpose of matrix k(ij)(e) is equal to the original matrix.

The yellow highlighted portion of matrix k(1) is referred to as the upper triangular portion; whereas the lower triangular portion is not highlighted. The blue highlighted portion provides the line of symmetry.

Element 2:

$$ k^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}} = 5(2)/2=5 $$

$$ \theta^{(2)} = -\frac{4} $$

$$ \theta = \cos{\frac{-\pi}{4}} = \frac{\sqrt{2}}{4} $$

$$ m^{(e)} = \sin{\frac{-\pi}{4}} = -\frac{\sqrt{2}}{2} $$ $$ k^{(2)} = k_{(ij)}^{(e)} $$

$$ K_{(11)}^{(2)} = K^{(1)} \cdot (l^{(1)})^{2} $$

Observations:

1) The absolute values of all coefficients k(ij)(e), e=2 (I,j) =1,…,4 are the same =>  compute1 coefficient For other coefficients the corresponding sign (positive or negative).

2) (k(2))^T = k(2) . Thus k(2) is symmetric.

Elem. FD rel: [k(e)][d(e)]= [F(e)] $$ d^{(e)} = \begin{bmatrix} d_{(1)}^{(e)}\\ d_{(2)}^{(e)}\\ d_{(3)}^{(e)}\\ d_{(4)}^{(e)}\\ \end{bmatrix} F^{(e)} = \begin{bmatrix} F_{(1)}^{(e)}\\ F_{(2)}^{(e)}\\ F_{(3)}^{(e)}\\ F_{(4)}^{(e)}\\ \end{bmatrix} $$ The column matrices d and F are both 4x1.

The global FD relation is [K][d] = [F], where K is an nxn matrix, d is an nx1 matrix and F is a nx1. In the example provided n is equal to six. Professor Vu-Quoc also mentioned a book titled "The Dragons of Eden" by [| Carl Sagan].