Advanced Methods of Structural Analysis

8.2 Canonical Equations of Displacement Method 283

Inourcasewehave

M1AD0:375EI

6:163

EI

 16 D18:31kN m;

M1BD0:3EI

6:163

EI

20:16D18:31kN m;

MkD0:12EI

6:163

EI

C20:736D21:475kN m:

Of course,M1AandM1Bare equal. The negative sign indicates that the extended
fibers at support 1 are located above the neutral line. The final bending moment
diagramMPis presented in Fig.8.6f; the location of the extended fibers is shown
by the dotted line.
Now we will consider the analysis of some frames by the displacement method
in canonical form.

Example 8.1.The crossbar of the frame in Fig.8.7a is connected with vertical
members by means of hinges. The bending stiffness isEIfor all vertical members
and 2 EIfor the crossbar; their relative stiffnesses, 1 and 2, are shown in the circles.
Concentrated forcePacts horizontally at the level of the crossbar. Construct the
internal force diagrams.

Solution.The system has two unknowns of the displacement method: the angular
displacement of joint 1 and the linear displacement of the crossbar. The primary
system is shown in Fig.8.7b.

The introduced constraint 1 is related only to the horizontal member, but not
the vertical one. The primary unknowns of the displacement method are angular
displacementZ 1 of constraint 1 and linear displacementZ 2 of constraint 2. The
bending stiffness per unit length for the vertical and horizontal members areivertD
1 EI
5 D0:2EI;ihorD

2 EI
6 D0:333EI. The canonical equations of the displacement
method are:

r 11 Z 1 Cr 12 Z 2 CR1PD0;

r 21 Z 1 Cr 22 Z 2 CR2PD0: (a)

Calculation of unit reactionsTo calculate coefficientsr 11 andr 21 , we need to con-
struct the bending moment diagramMN 1 in the primary system due to the rotation of
induced constraint 1 (Fig.8.7c).
The bending moment at the clamped support is3i D .3 2 EI/=6D 1 EI
(TableA.3, row 1). Because of the hinged connections of the vertical elements with
the horizontal bar, the bending moments in the vertical elements do not arise. The
positive unknown unit reactions are shown by the dotted arrow. The equilibrium of
joint 1 from bending moment diagramMN 1 leads tor 11 D 2 EI.kN m=rad/.
The equilibrium of the crossbar, considering the bending moment diagramMN 1 ,
leads tor 21 D 0. Indeed, within the vertical members the bending moments do not
arise. Therefore, shear forP ces within these members also do not arise and equation
XD 0 for the crossbar leads to the above result.