Advanced Methods of Structural Analysis

13.2 Stability of Structures with Finite Number Degrees of Freedom 457

subjected to axial static forceP. The structure has one degree of freedom. Take an-
gle ̨as a generalized coordinate. A new form of equilibrium is shown by solid line.

P P

a

q= 2 a

a

M=krotq

f

krot

P P

l l

C

DC

P P

a

q= 2 a

a

D W = −PΔ

2

U q^2

bc^0 = krot^

a

Fig. 13.5 Absolutely rigid elements with elastic jointC:(a) Design diagram; (b) Static method;
(c) Energy method

Static method (Fig.13.5b). The vertical displacement of the jointC isf D
lsin ̨Šl ̨. The mutual angle of rotation of the bars isD2 ̨, so the moment,
which arises in elastic connection at pointC, equalsM DkrotD2krot ̨.The
vertical reactions of supports are zero. The bending moment at the hingeCequals
zero, so the stability equation becomes

X

MCleftD 0 W PfkrotD 0 !Pl ̨2krot ̨D 0

Nontrivial solution of this equation is

PcrD

2krot

l

:

Energy method(Fig.13.5c). Horizontal displacement of the point of application of
the forcePequals

D2CD2l .1cos ̨/2l

̨^2

2

;

so potential of the external force isWDPDPl ̨^2. Mutual angle of rotation
at jointCisD2 ̨, the strain energy accumulated in elastic connection is

krot.2 ̨/^2

2

D2krot ̨^2 ;

so the total energy isU DPl ̨^2 C2krot ̨^2. Condition

@U

@ ̨

D 0 leads to the

following stability equation

2P l ̨C4krot ̨D0;