112 A.N. Kounadis
Case 5 (f = 0, |c|≠0)
Moreover, it was found
symbolically that the 5th degree polynomial f(|c|) pos-
sesses
three real roots (one double and one single) and two pure imaginary ones.
Discussing their nature, one can find that the double root of f(|c|), being equal to
|c| = −(c
12
−c
22
)
2
−
2
22
c
m<0, yields
()
[]
221112
c1mc
2
1
c ++=
(77)
Then, the double root of eq.(72) becomes
()
2211
2211
H
1
cc
2mkc2c
λ
+
+++
=
(78)
which is always
greater than
c
1
λ
and hence of minor importance for the present
analysis.
The
third real root of f(|c|), if substituted in eq.(72), yields again a double root
in λ, always less than zero, which is rejected. Thus, only the case of a positive
semi-definite damping matrix may lead to an
acceptable value of the correspond-
ing load (i.e., 0<λ
H
<
c
1
λ
) associated with a degenerate Hopf bifurcation as theo-
retically was shown by Kounadis [4, 13].
Case 6 (λ = 0)
If λ=0, eq.(72) implies Γ=0, which after symbolic manipulation of eq.(73c) can
be written in the following form
()
0AcAcAc
c
1
Γ
01
2
2
3
2
22
=+++=
,
(79)
where A
i
(i = 0, 1, 2) are given in the appendix of ref.[18] . It is evident that A
2
>0
and A
0
≥0, a fact implying that eq.(79) can be satisfied only for |c| ≤ 0 if also A
1
≥0;
otherwise (i.e., if A
1
<0) the system may be dynamically locally stable or unstable.
For |c|=0, one can find the corresponding values of c
ii
(i = 1, 2), given in the appendix
of ref. [18] which are always positive. This
special case, for which the trivial
(unloaded) state is associated with a pair of
pure imaginary eigenvalues (necessary
condition for a Hopf bifurcation) implies local dynamic asymptotic instability.
Conditions for a double imaginary root
For a double imaginary root the first derivative of the secular eq.(63) must be also
zero which yields
4ρ
3
+3α
1
ρ
2
+2α
2
ρ
2
+α
3
= 0. (80)
Inserting into eq.(80) ρ = μi, we obtain μ
2
= 0.5α
2
= α
3
/3α
1
and thus α
3
=3α
1
α
2
/2.
Introducing this expression of α
3
into eq.(71), it follows that α
1
=0, which also im-
plies that α
3
=0 and hence eq.(80) becomes ρ
2
= −0.5α
2
. If ρ
2
= −0.5α
2
is inserted
into the secular equation ρ
4
+α
2
ρ
2
+α
4
= 0, for a double imaginary root, it follows
that α
22
−4α
4
= 0 which due to relations (64) yields