Advanced Methods of Structural Analysis

342 10 Influence Lines Method

Since the structure is symmetric, the elastic loadsW 4 DW 2 andW 5 DW 1.
Fictitious beam presents the simply supported beam, which is loaded by elastic
loads (Fig.10.10g). The elastic loadW 1 is positive, so this load must be directed as
forces at support 1, i.e., upward (Fig.10.10e). Similarly, all the elastic loads should
be directed upward (Fig.10.10g).

+

0.803 1.0 0.803

0.443 0.443

Inf. line X 1

Fictitious

beam

1.776 3.555 8.525 3.555 1.776

−

38.374 38.374

69.648 86.7 69.648

Mf=dP 1

(Factor 1/EA)

d 11

Rf

A

Rf

B

A B

P= 1

d=4m

3m

Design

diagram

(^12345)
g
h
P 1 =60kN
d=4m
3m
Design
diagram
1
2
3
4
C 5 6
A B
RA RC=X 1 P 2 =20kN
a
i a
Fig. 10.10 (g) Fictitious beam subjected to elastic loads and corresponding bending moment di-
agram; (h) Influence line for primary unknownX 1 .(i) Truss subjected to fixed loadP 1 D 60 kN
andP 2 D 20 kN
Reactions at the left and right supports of fictitious beam are9:5935=EA.The
corresponding bending moment diagram is presented in Fig.10.10g; factor1= EA
for all ordinates is not shown. The ordinates of this diagram present the displace-
ments of corresponding joints of the truss due to unit primary unknownX 1 D 1 .It
is obvious that the unit displacement becomesı 11 D86:7=EA.
10.1.3.1 Influence Line for Primary UnknownX 1
This key influence line is obtained by dividing all ordinates of diagramıP1
(Fig.10.10g) for fictitious beam by.ı 11 /. Corresponding influence lineX 1 is
presented in Fig.10.10h.