# Class Notes 09/10/08

== Matrix 2-Bar Truss System == (continued)

Element 1:

θ^{(1)} = 30°

l^{(1)} = cos θ^{(1)} = cos30° = (SQAURE ROOT OF 3 DIVIDED BY 2)

m^{(1)} = sin θ^{(1)} = sin30° = (1/2)

k^{(1)} = E^{(1)}·A^{(1)}/ L^{(1)} = (3)·(1)/ (4) = 3/4

k^{(1)} = matrix

k_{11}^{(1)} = k^{(1)} · ( l^{(1)} )^{2} = (3/4) · (square root 3 divided by 2)^{2} = 9/16

k_{12}^{(1)} = k^{(1)} · l^{(1)} · m^{(1)} = (3/4) · (square root 3 divided by 2)· (1/2) = 3 square root 3 over 16

k_{42}^{(1)} = - k^{(1)} · ( m^{(1)} )^{2} = - (3/4) · (1/2)^{2} = - 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., k_{ij}^{(1)} = k_{ji}^{(1)}; just interchange the row and column indices

For example: k_{13}^{(1)} = k_{31}^{(1)}

In general,

k_{ij}^{(e)}= k_{ji}^{(e)}ork_{ij}^{(1)Т}=k^{(1)}

where Т represents the transpose of the matrix.

k^{(e)} = matrix

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

Element 2:

k^{(2)} = E^{(2)}·A^{(2)}/ L^{(2)} = (5)·(2)/ (2) = 5

θ^{(2)} = -π/4

l^{(2)} = cos θ^{(2)} = cos-π/4 = (SQAURE ROOT OF 2 DIVIDED BY 2)

m^{(2)} = sin θ^{(2)} = sin-π/4 = - (SQAURE ROOT OF 2 DIVIDED BY 2)

k^{(2)} = [k_{ji}^{(2)}]_{4x4}

Observation:

1) The absolute value of all coefficients k_{ji}^{(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)}

^{Review}

^{Element force displacement relationship: k4x4/sub>(e)·d4x1(e)= f4x1(e)}

e = 1,2

__d__^{(e)}= MATRIX

__f__^{(e)}= MATRIX

Global force displacement relationship:

__K___{nxn}·__d__ _{nx1}= __F___{nx1}

Here, n=6 (6 d.o.f.'s)