Matrix Structural Stiffness - Numerical Structural Analysis - page 194

P
P
P
P
P
P
P
P
P
P

0
0
0
0
0
0
12
8
9
8

1 x

1 y

2 x

2 y

− [

] [] =

T 3 x

=

PPM & FEPM R =

g g m

3 y

4 x

−

4 y

5 x

−

6 y

Using any of the methods for solving non-homogenous linear algebraic
equations, the global deformations can be found from the global stiffness
Equation 4.36 (step 4). The rows and columns corresponding to the sup-
port constraint degrees of freedom must be deleted prior to the solution.
This would be both x and y at joints 1 and 3. The solution for the deforma-
tions will be in inches.

−1

=

∆ KP

g g g

−1

∆
∆
∆
∆
∆
∆

1707 0 − 427 320 − 427 − 320 0

0
12
8
9
8

0 0246
0

.

2 x

0 960 320 − 240 − 320 − 240 − 0057

0 0595
0 0022
0 0602
0 0261

2 y

− 427 320 843 − 320 − 417 0 .

.
.

4 x

= =

320 − 240 − 320 796 0 0 −

4 y

− 427 − 320 − 417 0 843 320

5
6

x

− 320 − 240 0 0 320 795 − − .

y

The reactions at the support can be found using the solution of the global
deformation with Equation 4.37 (step 5). Only the terms in the rows cor-
responding to the restrained degrees of freedom and in the columns of
the unrestrained degrees of freedom need to be included. Since there are
no applied loads on the members, the fixed-end forces and moments are
omitted. In addition, there are no applied loads at the support locations so
the global applied forces and moments are omitted. The reaction forces
are in kips (k).

− [

] [] [] =

T

PK PM FEPM R ∆ − &

g g g m