Advanced Methods of Structural Analysis

8.2 Canonical Equations of Displacement Method 285

To calculate coefficients ofr 12 andr 22 it is necessary to construct the bending
moment diagramMN 2 in the primary system due to the linear displacement of in-
duced constraint 2 (Fig.8.7d). The bending moment at the bottom of the vertical
members is3i= hD.30:2EI/=5D0:12EI(TableA.3, row 2). The equilibrium
equation of joint 1 leads tor 12 D 0.
The free-body diagram for the crossbar is presented in Fig.8.7e, where shear in
each vertical member due to unit displacementZ 2 D 1 is

QD

0:12EI

h

D

0:12EI

5

D0:024EI:

Therefore, the equilibrium equation

P
XD 0 for the crossbar leads to the fol-
lowing result:r 22 D 3 0:024EID0:072EI.kN=m/.

Calculation of free terms. The forcePis applied at the level of the crossbar, there-
fore there is no bending of the horizontalelements in the primary system. Since
constraint 2 prevents displacement in a horizontal direction then bending of the ver-
tical members of the frame does not occur either. Thus, in the primary system there
are no elements subject to bending. But this does not imply that all reactions in
induced restrictions 1 and 2 are zero. Indeed,R 1 PD 0 ,butR 2 PDP.Thelast
expression is obtained from the equilibrium of the crossbar subjected to the given
loadP(Fig.8.7f).
The canonical equations become

2 EIZ 1 C 0 Z 2 D0;

0 Z 1 C0:072EIZ 2 PD0: (b)

The roots of these equations areZ 1 D 0 ,Z 2 DP =0:072EI.TheresultZ 1 D 0
means that the crossbar is not deformed, butdisplaced in the horizontal direction as
an absolutely rigid element. This is because the crossbar is connected to the vertical
elements by means of hinges.
The bending moment diagram can be constructed using the principle of super-
position:
MPDMN 1 Z 1 CMN 2 Z 2 CMP^0 : (c)

SinceZ 1 D 0 and acting loadPdoes not cause bending of the members, then
formula (c) becomesMPDMN 2 Z 2. The resulting bending moment diagram is
presented in Fig.8.7g.
The bending moments at the clamped supports are

0:12EI

P

0:072EI

D1:667P .kN m/:

The shear force at vertical members

QD

1:667P

h

D

1:667P

5

D

P

3

.kN/: