Advanced Methods of Structural Analysis

12 1 Kinematical Analysis of Structures

b

c

1

2

d

1

2

a 23

5

6710

1

4 8 9

Fig. 1.13 Different ways of truss formation. For cases (a), (b), and (c) degrees of freedomWD 0 ,

for case (d)WD 1. Only case (a) may be adopted as engineering structure

Degrees of freedomWfor these cases according to formula (1.2) equal to zeros.

Indeed:

Case aWJD10; SD17; S 0 D3; so WD 2  10  17  3 D0;

Case bWJD10; SD17; S 0 D3; so WD 2  10  17  3 D0;

Case cWJD10; SD16; S 0 D4; so WD 2  10  16  4 D0:

The degrees of freedom for trusses may be calculated also by the Chebushev for-

mula. The truss in Fig.1.13a contains 17 bars, which are considered as the rigid

discs, and 3 support constraints. The number of theequivalent simple hingesH

can be calculated as follows: only two hinges are simple (joints 1 and 9) and

all other hinges are multiple. The each multiple hinge at joints 2, 3, 5, 7, 10 are

equivalent to two simple hinges; the each multiple hinge 4, 6, 8 are equivalent

to four simple hinges. Thus, the total number of equivalent simple hinges is

HD 2  1 C 5  2 C 3  4 D 24. The Chebushev formula leads to the fol-

lowing resultWD3D2HS 0 D 3  17  2  24  3 D 0. The same results

may be obtained for Cases (b) and (c).

Even ifWD 0 for systems in Fig.1.13a–c, only system (a) may be used as en-

gineering structure. System (b) has a rigid disc (shown as solid) and geometrically

changeable right part. System (c) consists of two rigid discs, which are connected

bytwomembers 1 and 2 (while for generation of geometrically unchangeable struc-

tures two rigid discs must be connected bythreenonconcurrent members), and

therefore it is geometrically changeable system.

3.W<0. The system has redundant constraints. However, existence of redun-
dant constraints still does not mean that the structure can resist load, because the
structure can be generated incorrectly. The system in Fig.1.13d contains one re-
dundant constraint; indeed, the degrees of freedom isW D 2  10  17  4 D 1.