Advanced Methods of Structural Analysis

456 13 Stability of Elastic Systems

D2CD2l .1cos/Š2l

^2

2

:

Strain energy accumulated in elastic support

U 0 DR

f

2

where reaction ofelasticsupport isRDkf, and vertical displacement of hingeC

in terms of generalized coordinate isfDlsinŠl,so

U 0 Dk

f^2

2

Dk

l^2 ^2

2

:

The potential of external force isW DPDPl^2 , so the total energy is

UDU 0 CWDk

l^2 ^2

2

Pl^2 :

Derivative of this expression with respect to generalized coordinate equals to zero,

i.e.,

@U

@

Dkl^2 2P lD0;

which leads to the above determined critical load.

Discussion

1.The static method requires calculation of reactions ofallsupports, while energy
method requires calculation of reactions only forelasticsupports.
2.Why potential of external force equalsW DP, while the energy accumu-
lated in elastic support contains coefficient 0.5? If a structure switches into the
new position, a compressed forcePremains the same; that is why potential of
this force equalsWDP. The internal forces in the elastic constrain increase
from zero to maximum values; that is why the expression for accumulated energy
contains coefficient 0.5.
So far we have considered only two types of joints, mainly hinged and rigid joints.
Let us introduce a concept ofelasticjoint. If a structure is subjected to any loading
then for such joint the initial angle between members forming the joint changes.
Each elastic joint is characterized by the rigidity of the elastic hingekrot.Themo-
ment which arise at elastic connectionCisMDkrot,whereis a mutual angular
displacement of the members at the elastic joint.
Let us show an application of static and energy methods for determination of crit-
ical load for following structure: two absolutely rigid rods are connected by elastic
hingeC, as presented in Fig.13.5a. The rigidity of the hinge iskrot; the structure is