User:Eml4500.f08.delta 6.ramirez/classnotes

Matrix 2-Bar Truss System
(continued)  Element 1:

θ(1) = 30° l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ k(1) = E(1)·A(1)/ L(1) = $$\frac{3}{4}$$

$$k^{(1)} = k_{(1)} \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{bmatrix}$$

k11(1) = k(1) · ( l(1) )2 = $$\frac{3}{4}$$ · $$(\frac{\sqrt{3}}{2})^{2}$$ = $$\frac{9}{16}$$

k12(1) = k(1) · l(1) · m(1) = $$\frac{3}{4}$$ · $$\frac{\sqrt{3}}{2}$$· $$\frac{1}{2}$$ = $$\frac{\sqrt{3}}{16}$$

k42(1) = - k(1) · ( m(1) )2 = - $$\frac{3}{4}$$ · $$(\frac{2})^{2}$$ = - $$\frac{3}{16}$$

Observation:

1) Only three numbers need to be computed. Other coefficients have the same absolute value just differ by sign (+ or -) 2) Matrix k(1) is symmetrical i.e., kij(1) = kji(1); just interchange the row and column indices

For example: k13(1) = k31(1)

In general,

kij(e) = kji(e) or k ij(1)Т = k (1)

where Т represents the transpose of the matrix.

$$k^{(e)} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ & & k_{33}^{(1)} & k_{34}^{(1)}\\ & &  & k_{44}^{(1)} \end{bmatrix}$$

Since the matrix is symmetrical only the upper right triangle part needs to be calculated.

$$k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{\sqrt{3}}{16} & -\frac{9}{16} & -\frac{\sqrt{3}}{16}\\ \frac{\sqrt{3}}{16} & \frac{3}{16} & -\frac{\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{\sqrt{3}}{16} & \frac{9}{16} & \frac{\sqrt{3}}{16}\\ -\frac{\sqrt{3}}{16} & -\frac{3}{16} & \frac{\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}$$

Element 2:

k(2) = E(2)·A(2)/ L(2) = 5 θ(2) = $$-\frac{\Pi }{4}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$

k(2) = [kji(2)]4x4

Observation:

1) The absolute value of all coefficients kji(e), e=2 (i,j) = 1,...,4 are the same (|l| = |m|) → Only four coefficients need to be computed; all other coefficients add + or -. 2) k ij(2)Т = k (2)

$$k^{(2)} = \begin{bmatrix} \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} \end{bmatrix}$$