Structural Analysis Methods II: Structural Systems - Mechanics of Structures: Variational and Computational Methods - page 299

In order to compute p 1 for the case with th e distrib ut ed load located on bar 1, i.e., for load
case 1, it is necessary to use p 1

p 10 with p 10

k 1 v 1 given by (2). For load case 2, p 1

= − is

simply equal to k 1 v 1 Thus, p 1 . becomes

Load Case 1 Load Case 2

.

− 1 . 2277 − 2 . 9998 0 . 6043

. −

− 0 . 8192 − 1 . 7320 12 . 4831

.

2 . 3183 − 1 . 9999 9 . 9429

p 1

=

.

1 . 2277 − 2 . 9998 − 0 . 6043

.

0 . 8192 − 1 . 7320 12 . 4831

.

2 . 7853 − 1 . 9999 9 . 8748

Load Case 1 Load Case 2

.

− 4 . 2275 kN 0 . 6043 kN

. −

− 2 . 5512 kN 12 . 4831 kN

.

4 . 3182 kN · m 9 . 9429 kN · m

= (10)

.

− 1 . 7721 kN − 0 . 6043 kN

.

− 0 . 9128 kN 12 . 4831 kN

.

0 . 7854 kN · m 9 . 8748 kN · m

p 1

=

T 1 p 1

In terms of the local coordinates, the forces are found from to be

Load Case 1 Load Case 2

.

N a
V a
M a
N b
V b
M b

0 . 09559 kN 11 . 1125 kN

.

− 4 . 9366 kN − 5 . 7182 kN

.

4 . 3182 kN · m 9 . 9429 kN · m

= (11)

. −

− 0 . 09557 kN 11 . 1125 kN

.

− 1 . 9910 kN 5 . 7182 kN

.

. · . ·

0 7854 kN m 9 8748 kN m

The local forces on bar 1 for load case 1 are pictured in Fig. 5.29, along with other responses
on all of the bars.

5.3.15 Symmetry

Many structures such as bridges, buildings, and ships exhibit some form of symmetry.
For symmetrical structures, an analysis of the whole structure can often be avoided by
considering only a portion of the structure. This will simplify the problem and decrease the
cost of the computation.
There are three steps involved in taking advantage of symmetry in the analysis of a
structure: (1) Recognition of the type of symmetry, (2) Use of superposition to reorganize