Structural Analysis Methods II: Structural Systems - Mechanics of Structures: Variational and Computational Methods - page 285

The full displacement vector and corresponding load vector can be split into parts with
given displacement boundary conditions at nodes a and d to which the reactions P a and
P d are conjugates, a nd into another part with the unknown nodal displacements to which
the applied loads P are conjugates.

0
0
0
...
U Xb
U Zb
b
...
U Xc
U Zc

P Xa
P Za
M a
...
0
P Z
0
...
P X
0
0
...
P Xd
P Zd
M d

V a P a

P

V

⇐⇒ (10)

c
...
0
0
0

P d

V d

If the columns corresponding to the zero displacements at the supports are deleted and the
rows corresponding to the reactions are temporarily ignored, the equilibrium equations
become

260 . 54 − 79 . 412 2 . 0404 − 213 . 33 0 0 U Xb

U Zb

0
0 − 79 . 412 140 . 99 − 1 . 9632 0 − 2 . 0942 − 3 . 1413 . 02

2 . 0404 − 1 . 9632 11 . 724 0 3 . 1413 3 . 1413 0

0

b
U Xc
U Zc

=

− 213 . 33 0 0 217 . 99 0 6 . 9933 . 01

0 − 2 . 0942 3 . 1413 0 442 . 09 3 . 1413 0

0 0 − 3 . 1413 3 . 1413 6 . 9933 3 . 1413 20 . 270

c

(11)

The displacements found by solving these relations are

U Xb
U Zb
b
U Xc
U Zc

3 . 660 mm

2 . 1828 mm

− 29 . 127 µ rad

= (12)

3 . 6575 mm

0 . 0171 mm

− 921 . 73 µ rad

c

Now that the global displacements are known, we can calculate a considerable amount of
useful information. The reactions are found by returning to the rows of the full system matrix
(9) which corresponds to P a and P d of (10). This gives the reactions (in global coordinates)

P Xa = 0 . 597 kN P Xd =− 10 . 606 kN

P Za =− 12 . 485 kN P Zd =− 7 . 524 kN

(13)

M a = 9 . 961 kN · m M d = 19 . 132 kN · m