Advanced Methods of Structural Analysis

11.4 Resolving Equations 385

acb

1 2

Z-P

diagram

ABC^3

h

M 2

M 4

M 3 M^5 M 6

M 1

S-e

(^1) diagram
2
34
5
6
d
∑ hh−
M 1 + M 2 M 5
X = 0: P 3 = –
Joint A Joint B
M 2
M 3 M 4
M 5
P M^6 P^2 =^ M^4 +^ M^5 +^ M^6
1 =M 2 +M 3
P 1
P 3
P 2
M 1 /h+M 2 /h M 5 /h
Cross bar A-C
Fig. 11.19 (a–c) Design diagram of the frame and ancillary diagrams; (d) Construction of static
matrix for frame
Example 11.3.TheframeisshowninFig.11.19. Construct the static matrix.
Solution.This structure has two angular displacements of the rigid joints and one
linear displacement. The arrows 1 and 2 show possible angular displacements and
corresponding possible external loads (moments, kNm); the arrow 3 shows possible
linear displacements and corresponding possible external load (force, kN). The vec-
tor of possible joint loads isPED
P 1 P 2 P 3
̆T

. The vector of unknown internal

moments isSED
M 1 M 2 M 6

̆T

.

Equilibrium equations for rigid joints are shown in Fig. 11.19d.

In matrix form the static equations can be presented as follows

2 6 6 6 6 6

P 1

P 2

P 3

3 7 7 7 7 7

D

2

6

6

4

0 11000

0 00111

1=h 1=h 0 0 1=h 0

3

7

7

5

„ ƒ‚ ...

A.36/



2 6 6 6 6 6 6 6 6 6 6 6 6 6 6

M 1

M 2

M 3

M 4

M 5

M 6

3 7 7 7 7 7 7 7 7 7 7 7 7 7 7

Note again, that the static matrix defines the structure itself (supports, connection of
the members, etc.) and does not depend on the type of external exposures.