Advanced Methods of Structural Analysis

298 8 The Displacement Method

Solution.The primary system of the displacement method is presented in
Fig.8.11b. The primary unknowns are the angle of rotation Z 1 and linear
displacementZ 2.
The canonical equations of the displacement method are

r 11 Z 1 Cr 12 Z 2 CR 1 sD0;

r 21 Z 1 Cr 22 Z 2 CR 2 sD0:

The unit reactionsrikwere obtained in Example 8.2; they are

r 11 D2:4EI.kN m=rad/;r 12 D0:093EI.kNm=m/;

r 21 D0:093EI.kN=rad/;r 22 D0:207 .kN=m/:

Free termsR 1 sandR 2 srepresent reactions in both introduced constraints in the
primary system due to the settlement of supportA. They are calculated using the
bending moment diagrams caused by the displacement of supportA.DiagramsMa,
Mb,andM'(Fig.8.11c) present the bending moments in the primary system due to
the separate displacements of supportAbasedongivenvaluesaD 2 cm,bD 1 cm,
and'D0:01radD 3403000 , respectively. Figure8.5a contains the numeration of the
specified points of the frame.
In the case of the verticala-displacement of supportA, only member 6-8 of the
frame in the primary system deforms. The elastic curve is shown by a dashed line;
vertical members 1-3 and 4-5 do not deform. To construct the bending moment
diagramMa,weuseTableA.3.
In the case of the horizontalb-displacement, only member 1-3 deforms; in order
to construct bending moment diagramMb,weuseTableA.4.
In the case of the angular displacement of supportA, only member 1-3 of the
frame in the primary system deforms. This happens because introduced constraint
1 prevents distribution of the deformation. Bending moment diagramM'is shown
according to TableA.4.
The values presented in TablesA.3andA.4correspond to unit displacements,
therefore in order to obtain the reactionsaccording to the given displacements, it
is necessary to multiply the tabulated values by corresponding valuesa,b,and'.
Bending moment diagramsMa,Mb,andM'as well as their sum are presented in
Fig.8.11c.
To calculate the free terms, it is necessary to consider the free-body diagrams for
the rigid joint and for the horizontal element as shown in Fig.8.11d, e.
The equilibrium condition of the rigid joint yields

R 1 sD0:0064EIC0:0012EID0:0076EI:

The shear force within portion 1-3 is equals to

M 3  1 CM 1  3

l 1  3

D

0:0064C0:0104

5

EID0:00336EI:

These forces rotate the member counterclockwise. Force 0.00336EIis transmit-
ted to horizontal member 6-8 in the opposite direction, i.e., from left to right.