2.10 Cyclic Symmetry

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Cyclic symmetry is an important issue in rotating structures such as disks (Ramamurti

and Seshu 1990). It basically enables you to calculate eigenmodes for a complete disk

by modeling a segment only. Look at the deformed disk in Figure 2.22. It exhibits an

eigenmode with a nodal diameter of two. This means that there are two diameters for which

the displacements are zero. This corresponds to four zero crossings or to two complete

waves along the circumference of the disk. In general, a nodal diameter N corresponds to

N waves along the circumference and 2N zero crossings. Suppose only a segment extending

over an angle _S is modeled. For the disk in Figure 2.23, _S can take any value smaller

or equal to 2π. For practical models, the value of _S depends on the size of substructures

such as blades. In general, if there are M identical sectors along the circumference, _S

must be a multiple of 2π/M. For instance, for the structure in Figure 2.24, there are four

identical sectors and _S must be a multiple of π/2.

LINEAR MECHANICAL APPLICATIONS 115

Figure 2.22 Eigenmode of a disk with nodal diameter two

Y

0 180◦

360◦

540◦

A

B

UX

UY

UR

U_

_

_S

R

U

Phase

Phase

Phase

Phase Phase

Phase

Nodal diameter

Nodal diameter

X

x

x + 2_s

Figure 2.23 Phase along the boundary of the disk and modeled segment

Figure 2.23 shows that for the mode shown in Figure 2.22, the displacements in cylindrical

coordinates in any node B on the “clockwise” side are phase shifted with respect to

those of the corresponding node A on the “counter clockwise” side by 2_S. For a mode

with nodal diameter N, this shift takes the value NφS. Taking for _S the smallest possible

value _S = 2π/M, one arrives at

_U_cyl,B

= _U_cyl,A ei 2πN

M . (2.298)

116 LINEAR MECHANICAL APPLICATIONS

Figure 2.24 Structure consisting of four identical sectors

This is the central equation of cyclic symmetry. For modal analysis no other equation is

needed. This basically means the following:

1. The governing equations are not modified but only the boundary conditions are

changed.

2. Because of the complex nature of the boundary conditions, the resulting eigenvalue

problem is a generalized complex eigenvalue problem. It will be shown that it can

be reduced to a generalized real eigenvalue problem twice the size.

3. Because of the presence of N in Equation (2.298), the eigenvalue system is different

for a different nodal diameter. For a given nodal diameter N, the solution of the

eigenvalue system yields all modes having 2N zeros along the circumference. Since

cosine and sine are periodic functions with period 2π, one can write

ei 2πN

M = ei 2π

M [N+kM] (2.299)

where k is an integer. This means that the application of Equation (2.298) will yield

all modes with nodal diameter N + kM. For other nodal diameters, Equation (2.298)

is different and another eigenvalue system results. Accordingly, a cyclic symmetry

calculation takes longer, but not as long as when the complete disk is modeled.

4. Equation (2.299) shows that it is sufficient to perform calculations for nodal diameters

0, 1, . . . ,M/2 if M is even and up to (M − 1)/2 for M odd. For instance, if M is odd,

calculations for N = (M − 1)/2 also yield the eigenmodes for N = |(M − 1)/2 −

M| = (M + 1)/2.

LINEAR MECHANICAL APPLICATIONS 117

Using the cylindrical coordinate system in Figure 2.23, Equation (2.298) is equivalent

to





UR,B = UR,AeiN_S

U_,B = U_,AeiN_S

UZ,B = UZ,AeiN_S .

(2.300)

The cylindrical and rectangular coordinates in Figure 2.23 are related by





X = R cos_

Y = R sin_

Z = Z

(2.301)

or





R =

√

X2 + Y2

_ = tan−1 Y

X

Z = Z

(2.302)

leading to (see Equation (2.212)),





UR = UX cos_ + UY sin_

U_ = −UX sin_ + UY cos_

UZ = UZ

(2.303)

since (Equation (1.7))





G1       = cos_I 1 + sin_I 2

G2       = −r sin_I 1 + r cos_I 2

G3       = I 3

(2.304)

and accordingly,

G1       1          = G3   3          = 1,G2            2          = r2. (2.305)

Inverting Equation (2.303) yields





UX = UR cos_ − U_ sin_

UY = UR sin_ + U_ cos_

UZ = UZ.

(2.306)

Now, Equations (2.300) lead in rectangular coordinates to the following linear complex

equations:





UX,B cos_B + UY,B sin_B = _UX,A cos_A + UY,A sin_A_ eiN_S

−UX,B sin_B + UY,B cos_B = _−UX,A sin_A + UY,A cos_A_ eiN_S

UZ,B = UZ,AeiN_S .

(2.307)

118 LINEAR MECHANICAL APPLICATIONS

Solving for UX,B, UY,B and UZ,B, we get three linear multiple point constraints with UX,B,

UY,B and UZ,B as dependent variables. Notice that Equation (2.300) and Equation (2.306)

lead to

UX,B = UR,B cos_B − U_,B sin_B

= (UR,A cos_B − U_,A sin_B)eiN_S

_= UX,AeiN_S , (2.308)

that is, Equations (2.300) do not apply in rectangular coordinates.

The resulting complex eigenvalue system

_KR + iKI_ _UR + iUI_ = ω2 _M_ _UR + iUI_ (2.309)

where the index R denotes the real part and I the imaginary part is equivalent to

_KR −KI

KI KR

_)UR

UI

* = ω2 _M 0

0 M

_)UR

UI

*

. (2.310)

Since the basic equilibrium equations lead to a real symmetric and consequently a Hermitian

stiffness matrix, and the treatment of the boundary conditions discussed in Section 2.6

conserves the Hermitian character, KR + iKI is Hermitian. Accordingly,

KR + iKI = (KR + iKI)T

⇓

KR − iKI = KTR

+ iKT

I

⇓

) KR

−KI

=

=

KTR

KT

I

(2.311)

which shows that Equation (2.310) is a symmetric eigenvalue problem.

Solving Equation (2.310), we get each eigenvalue twice. Indeed, one can check that if

)UR

UI

* (2.312)

is a solution,

)−UI

UR

* (2.313)

is a solution too with the same eigenfrequency. Recomposing the complex form, the

first solution corresponds to _U1_ = _UR + iUI_, the second to _U2_ = _−UI + iUR_ =

_UR + iUI_ eiπ/2 which shows that the difference between both is a phase shift of 90◦.

The resulting solution amounts to (Equation (2.246))

_U_ = _UR + iUI_ eiωt (2.314)

LINEAR MECHANICAL APPLICATIONS 119

or

_U_ = __UR_ cos ωt − _UI_ sin ωt_ + i __UR_ sin ωt + _UI_ cos ωt_ . (2.315)

Since the governing equation is linear, both the real and imaginary part are a solution.

Taking the real part, one arrives at

_U_ = _UR_ cos ωt − _UI_ sin ωt. (2.316)

This is not a standing wave. However, if ω is a solution, so is −ω and accordingly

_U_ = _UR_ cos ωt + _UI_ sin ωt (2.317)

is a solution too and any linear combination of Equation (2.316) and Equation (2.317) as

well. Consequently, half the sum and half the difference, which are both standing waves,

are also solutions:

_U_ = _UR_ cos ωt (2.318)

_U_ = _UI_ sin ωt. (2.319)

How do we arrive at the solution in the other sectors? Let the solution in a point P

in the primary sector be U. The solution in a point Q exactly K sectors ahead satisfies

(Equation (2.300))





UR,Q = UR,P eiKN_S

U_,Q = U_,P eiKN_S

UZ,Q = UZ,P eiKN_S .

(2.320)

_UR,P,U_,P,UZ,P _ are related to _UX,P,UY,P,UZ,P _ through the relations in Equation

(2.303). _UX,P,UY,P,UZ,P _ are generally complex and so are _UR,P,U_,P,UZ,P _.

Because the Equations (2.303) are linear, they can be applied to the real and imaginary

parts of the solution separately. The first equality in Equation (2.320) now reads

(UR,Q)R + i(UR,Q)I = _(UR,P )R + i(UR,P )I_ eiKN_S (2.321)

which leads to

((UR,Q)R = (UR,P )R cos(KN_S) − (UR,P )I sin(KN_S)

(UR,Q)I = (UR,P )R sin(KN_S) + (UR,P )I cos(KN_S).

(2.322)

The rectangular components of the solution in Q are obtained through Equation (2.306).

Accordingly, the solution in point Q is obtained by

1. converting the solution in P to cylindrical coordinates (Equation (2.303)),

2. applying the mapping in Equation (2.321) to each component to obtain the solution

in Q in cylindrical coordinates,

3. converting this solution into rectangular coordinates (Equation (2.306)).

This also applies to higher-order tensors such as stresses or strains.

Cyclic symmetry properties can also be used in static calculations by expanding the

circumferential loading in its Fourier components.

120 LINEAR MECHANICAL APPLICATIONS

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