Advanced Methods of Structural Analysis

460 13 Stability of Elastic Systems

R 1 Dkf 1 Dklˇ 1 andR 2 Dkf 2 Dklˇ 2. The energy accumulated in elastic

supports is

XRifi

2

D

k

2

.lˇ 1 /^2 C

k

2

.lˇ 2 /^2 :

The total energy

UDNC

XRifi

2

DNl



ˇ 12 Cˇ 22 ˇ 1 ˇ 2



C

k

2

l^2 ˇ^21 C

k

2

l^2 ˇ^22 :

Derivative of the total energy with respect to generalize coordinate leads to the fol-

lowing equations

@U

@ˇ 1

D 2 Nlˇ 1 CNlˇ 2 Ckl^2 ˇ 1 D 0

@U

@ˇ 2

D 2 Nlˇ 2 CNlˇ 1 Ckl^2 ˇ 2 D 0

or

.kl2N / ˇ 1 CNˇ 2 D^0

Nˇ 1 C.kl2N / ˇ 2 D 0

(c)

Nontrivial solution of homogeneous system (c) occurs if

ˇ

ˇ

ˇ

ˇ

kl2N N

Nkl2N

ˇ

ˇ

ˇ

ˇD^0

Stability equation becomes.kl2N /^2 N^2 D 0. This equation leads to the same

critical loads (b).

Each critical load corresponds to a specified shape of equilibrium. Both critical

loads should be considered.

1.LetNDN1crD

kl

3

. Substituting it in thefirstequation of system (a), we obtain

a 1



3

kl

3

2kl



a 2 klD0;

and relationship between generalized coordinates isa 1 a 2 D 1 , which de-
termines the first form of a loss of stability; considering thesecondequation of
system (a) we will get the same result. Corresponding equilibrium form is pre-
sented in Fig.13.6d.
2.LetNDN2crDkl. Substituting it in the first equation of system (a), we obtain

a 1 .3kl2kl/a 2 klD0;

and the second form of the loss of stability is defined by relationship between

generalized coordinates asa 1    a 2 DC 1. Corresponding equilibrium form is

presented in Fig.13.6e. Note that for the each critical load, we cannot define the

displacementsa 1 anda 2 separately.However,theshapeof the loss of stability

is defined by their relationships.