# 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

k11(1) = k(1) · ( l(1) )2 = (3/4) · (square root 3 divided by 2)2 = 9/16

k12(1) = k(1) · l(1) · m(1) = (3/4) · (square root 3 divided by 2)· (1/2) = 3 square root 3 over 16

k42(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., kij(1) = kji(1); just interchange the row and column indices

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

In general,

```    kij(e) = kji(e) or kij(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) = [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)kij(2)Т = k(2)

Review

Element force displacement relationship: k4x4(e)·d4x1(e)= f4x1(e)

e = 1,2

d(e)= MATRIX

f(e)= MATRIX

Global force displacement relationship:

Knxn·d nx1= Fnx1

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