Advanced Methods of Structural Analysis

330 10 Influence Lines Method

Discussion. Obtained influence lines allow calculate bending moment and shear
foranysection of the beam. LetPD 10 kN is located at section 9. In this case, the
moment
MkD0:0336l 10 D0:336l .kNm/:
It means that reaction of the left support isRADMk=0:4lD.0:336l=0:4l/D
0:84kN. Negative sign shows that reaction is directed downward. Now construc-
tion of bending moment and shear diagrams has no problem. For example, the shear
isQ 1 D:::DQleft 6 D0:84kN. The bending moments at specified points are

M 2 D0:840:2lD0:168kNm;:::; M 5 D0:840:8lD0:672kNm

It is obvious that same results may be obtained using influence line forX 1 .Ifload
PD 10 kN is located at section 9 then moment at supportBisX 1 D0:84l.Neg-
ative sign indicates that extended fibers in vicinity of supportBare located above
the neutral line, i.e., this moment acts on the supportBof theleftspanclockwise
andonthesamesupportBof therightspan in counterclockwise direction. In this
case the reaction of supportAequalsRADX 1 =lD.0:84l = l /D0:84kN.
Influence line for primary unknownX 1 should be treated as afundamental char-
acteristicfor given structure. This influence line allows us to construct the bending
moment diagram in case ofany fixedload.

Example 10.1.The uniform continuous beam with two equal spanslD 10 mis
loaded by two forcesP 1 D 10 kN andP 2 D 20 kN, which act at points 2 and 4
(Fig.10.5). It is necessary to construct the bending moment diagram. Use the influ-
ence line for bending momentX 1 at the middle supportB(Fig.10.2d).

Inf. line X 1

(factor l)

0.0480.0840.0960.072 0.0720.0960.0840.048

−− −−

P 1 P 2

RA

24

41.6

24

27.2

M kNm

P 1 =10kNP 2 =20kN

2 m 4 m 10 m

ABC

24

4 m

M=24kNm

Fig. 10.5 Construction of bending moment diagram using influence line for primary unknownX 1