Advanced Methods of Structural Analysis

13.4 Stability of Continuous Beams and Frames 479

Solution.The frame has two unknowns of the displacement method. They are the
angle of rotationZ 1 of rigid joint and horizontal displacementZ 2 of the cross bar.
The primary system is presented in Fig.13.16b.

Bending moment diagrams caused by unit displacement of the introduced con-
straints are presented in Fig.13.16c, d; elastic curves are shown by dotted line. The
diagrams within members 1–3 and 2–4 are curvilinear. Direction ofRforM 1 di-
agram is explained for the left column (Fig.13.16c). Similarly may be explained
directions forR 1 andR 2 .M 2 diagram).

Parameters of a critical load are 13 D  D h

r

P

EI

I 24 D h

r

1:4P

EI

D

1:1832.
Canonical equations and unit reactions are

r 11 Z 1 Cr 12 Z 2 D0;

r 21 Z 1 Cr 22 Z 2 D0;

(a)

where

r 11 DŒ0:4' 2 . /C1:2EIIr 21 Dr 12 D0:6EI' 4 . /I

r 22 DŒ0:012

 2 . /C0:003

 1 .1:1832/EI

Stability equation becomes:

ˇ

ˇ

ˇ

ˇ

0:4' 2 . /C1:2 0:6' 4 . /

0:6' 4 ./ 0:012

 2 . /C0:003

 1 .1:1832/

ˇ

ˇ

ˇ

ˇD^0

Solution of this equation leads to parameter of critical loadD2:12. The critical
load is

PcrD

^2 EI

h^2

D

4:49EI

h^2

:

Thus the frame becomes unstable if it will be loaded by two forcesPand 1.4P
simultaneously.

Example 13.6.Design diagram of the multispan frame is presented in Fig.13.17a.
Derive the stability equation and calculate the critical force.

Solution.The primary system is shown in Fig.13.17b. The introduced constraint 1
prevents linear displacement. Canonical equation isr 11 Z 1 D 0 , so stability equation
isr 11 D 0. Bending moment diagram caused by unit displacement of the introduced
support is presented in Fig.13.17b. Within the second and third columns, the bend-
ing moment diagrams are curvilinear.

Free body diagram of a cross-bar is shown in Fig.13.17c. Shear forces for
compressed-bent members are presented in Table A.22. Parameter of critical load

Dh

r

P

EI

. Unit reaction equals