Advanced Methods of Structural Analysis

188 6 Deflections of Elastic Structures

Elastic loadW 3 Two unit couples are applied to members 2-3 and 3-4. Performing
similar procedure we get

W 3 D

128

36

P

EA

:

A fictitious beamis a simply supported one, which is loaded at joints 1,...,5 by
elastic loadsWi,i D1;:::;5. The elastic loadsWiare positive, so they must be
directed downward (Fig.6.26).

15

1

Fictitious 0 263 4

beam

W 1 W 2 W 3 W 4 W 5

M f =Dvert

(factor P/EA)

42.33

77.55 84.66

Rf

0 R 6 f

Fig. 6.26 Fictitious beam with elastic loads and bending moment diagram

Since the truss under consideration is symmetrical, thenW 1 DW 5 andW 2 DW 4.
Reactions of supports of the fictitious beam areRf 0 DR 6 fD.381=36/.P =EA/.
Bending moments at the specific points of the fictitious beam caused by elastic
loads are

M 1 f DRf 0 ^4 D42:33

P

EA

M 2 f DRf 0  8 W 1  4 D77:55

P

EA

(d)

M 3 f DRf 0  12 W 1  8 W 3  4 D84:66

P

EA

Ordinates of the bending moment diagram of the fictitious beam present the vertical
displacement of the joints of the lower chord of the truss,MfDvert.

 1 D42:33

P

EA

; 2 D77:55

P

EA

; 3 D84:66

P

EA

: (e)

Discussion.For calculation of displacement at joint 1 by the Maxwell–Mohr inte-
gral, it is necessary to apply unit force at this joint and calculate all internal forces
(for given truss 21 forces) and then calculate 21 standard term.NN Npl/=EA.
For calculation of displacements of joints 1, 2, and 3 the total number of unknown
forces in the unit states is 3  21 D 63 and the number of standard terms according
to Maxwell–Mohr integral is 63.
Application of elastic loads to each joint of a truss leads to appearance of internal
forces only in the members of two adjacent panels. As a result, for given truss the
total number of unknown forces in the all unit states is 16 and the number of terms
for elastic loads, according to (6.26), equals 4.