Advanced Methods of Structural Analysis

13.2 Stability of Structures with Finite Number Degrees of Freedom 455

The total energy of the system is

UDU 0 CWDkrot

^2

2

Pl

^2

2

:

Condition (13.2)
@U
@

DkrotPlD 0

leads to the same critical loadPcrDkrot=l.

Example 13.1.The structure contains two absolutely rigid bars.EID1/,which
are connected by hingeC, and supported by elastic support at this point; the rigidity
of elastic supports isk. The structure is subjected to axial compressed forceP
(Fig.13.4a). Calculate the critical force by the static and energy methods.

R=ka

P P

C a

ka/2 ka/2

P P

k

l l

C

U 0 = kf^2 /2

P P

f

D

q k W = −PΔ

DC

EI=•

a

b c

Fig. 13.4 Absolutely rigid 2-elements structure with elastic support: (a) Design diagram;
(b) Static method; (c) Energy method

Solution.Static method(Fig.13.4b). The structure has one degree of freedom. Let
a vertical displacementaof the hingeCbe a generalized coordinate, so the reaction
of elastic support isRDkaand reactions of pinned and rolled supports areka=2.
Bending moment at hingeCequals zero, therefore stability equation becomes

MCleftDPa

ka

2

lD0:

This equation immediately leads to the critical force becomes

PcrD

kl

2

:

Energy method(Fig.13.4c). Let an angular displacementat the support points be a
generalized coordinate. The vertical displacement of hingeCequalsfDlsinŠ
l, while the horizontal displacement at the same pointCisCDl.1cos/,so
horizontal displacement at the point of applicationPbecomes