Advanced Methods of Structural Analysis

352 10 Influence Lines Method

Solution.(a)Since a loadPis located at point 8, the internal forces at sectionkac-
cording to Figs.10.12cand10.13careMkD0:0384P landQkD0:096P.
Now we can make a cut section of the beam at the point 3.k/and show cor-
responding internal forces at this section (Fig.10.14a). Direction of the bending
moment and shear are shown according to their signs.
The free-body diagram for part 1-3 allows us to find the reaction of support
A .0:096Pdownward/and to perform checking of calculation
P
MAD0:096P0:4l0:0384P lD^0.
Free-body diagram for part 3-11 allows calculating all reactions and construct-
ing internal force diagrams, which correspond to given location of the forceP.
For example

RB!

X

MCD 0

RBlCP0:6lC0:096P .0:6lCl/C0:0384P lD 0 !RBD0:792P ."/:

(b)If forcePis located at point 8, then primary unknownZ 1 of the displacement
method (Fig.10.11e) equalsZ 1 D0:032



Pl^2 =EI



. Now we can consider two
beams. The left pinned–clamped beamABis subjected to angular displacement
Z 1 only and the right clamped–rolled beamBCis subjected to angular displace-
mentZ 1 and forceP(Fig.10.14b).
The vertical reaction at supportAfor the left beam (according to TableA.3)
equals

RAD

3 EI

l^2

Z 1 D

3 EI

l^2

0:032

Pl^2

EI

D0:096P .#/:

The vertical reaction at supportCfor the right beam (according to TableA.3)

equals

RCD

3 EI

l^2

Z 1 C

P

2

.3u/D

3 EI

l^2

0:032

Pl^2

EI

C

P0:4^2

2

.30:4/

D0:096PC0:208PD0:304P ."/

Required reaction at the supportB

RB!

X

YD 0 WRACRBPCRCD 0

!0:096PCRBPC0:304PD 0 !RBD0:792P ."/:

(c)Reaction of the supportBusing influence lineX 1 of the force method is calcu-

lated by the next way. Primary unknown (bending moment at supportB) equals

X 1 DMBDP0:096l(Fig.10.2d). Now we can consider two simply sup-

ported beams (left beam is subjected to momentMBDX 1 only and the right

beam is subjected toMBDX 1 and forceP) as shown in Fig.10.14c.