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The paper presents a simplified 3D-model for active vibration control of rotating machines with active machine foot mounts on soft foundations, considering static and moment unbalance. After the model is mathematical described in the time domain, it is transferred into the Fourier domain, where the frequencies response functions regarding bearing housing vibrations, foundation vibrations and actuator forces are derived. Afterwards, the mathematical coherences are described in the Laplace domain and a worst case procedure is presented to analyze the vibration stability. For special controller structures in combination with certain feedback strategies, a calculation method is shown, where the controller parameters can be directly implemented into the stiffness matrix, damping matrix and mass matrix. Additionally a numerical example is presented, where the vibration stability and the frequency response functions are analyzed.

In praxis, large rotating machines are often fixed directly on soft foundations. e.g. elastic steel frame foundations, which sometimes lead to problems regarding vibrations, caused by resonances and instability [

The model, which is used here, is a simplified 3D model (

· The mass m sr of the machine consists of the mass m s of the stator and of the mass m r of the rotor and is included in the mass matrix, as a single mass.

m sr = m s + m r (1)

· The moments of inertia θ sry and θ srz , consisting of θ sy and θ sz (the moments of inertia of the stator at the y-axis and z-axis) and of θ ry = θ rz = θ rzy (the moments of inertia of the rotor at the y-axis and z-axis), are also included in the mass matrix.

θ sry = θ sy + θ rzy ; θ srz = θ sz + θ rzy (2)

· However, the moment of inertia of the stator at the x-axis θ sx and the moment of inertia of the rotor at the x-axis θ rx have to be used separately, because θ sx has to be included in the mass matrix, and θ rx in the gyroscopic matrix.

· Additionally m r and θ rzy and θ rx have to be handled separately, because they have to be considered in the excitation vectors, which are described by f u ( t ) and m u ( t ) .

At the centre of gravity S the two excitation vectors f u ( t ) and m u ( t ) are positioned. The vector f u ( t ) represents the rotating force cause by a static unbalance of the rotor, due to a mass eccentricity e ^ with phase shift of φ e . The vector m u ( t ) represents the rotating moment caused by a moment unbalance of the rotor, due to a tilt—angle α ^ with phase shift of φ α —of the rotor mass on the shaft. The center of the bearing housings are described by point B D (on the drive side) and by point B N (on the non-drive side). The boreholes of the machine feet are defined by the points A DL , A DR , A NL and A NR . Index D stands for drive side, index N for non-drive side and index L for left side and index R for right side:

i = D , N ; j = L , R (3)

The mass m aaDL , m aaDR , m aaNL and m aaNR present the mass of each amateur of the actuators and m asDL , m asDR , m asNL and m asNR the mass of each stator of the actuators. In the stiffness and damping matrix of each actuator only translational stiffness and damping is considered, referring to [

C aij = [ c azij 0 0 0 c ayij 0 0 0 c axij ] ; D aij = [ d azij 0 0 0 d ayij 0 0 0 d axij ] (4)

The coefficients c azij , c ayij and c axij are the stiffness coefficients of the actuators and the coefficients d azij , d ayij and d axij are the damping coefficient of the actuators. For each foundation point, also only translational stiffness and damping of the foundation is considered in the stiffness and damping matrix:

C fij = [ c fzij 0 0 0 c fyij 0 0 0 c fxij ] ; D fij = [ d fzij 0 0 0 d fyij 0 0 0 d fxij ] (5)

The coefficients c fzij , c fyij and c fxij are the stiffness coefficients of the foundation and the coefficients d fzij , d fyij and d fxij are the damping coefficient of the foundation. Nine global coordinate systems are used here, one at the centre of gravity S and one for each machine foot point ( A DL , A DR , A NL , A NR ) and one for each foundation point ( F DL , F DR , F NL , F NR ). By using a vibration sensor for each machine foot, the vertical machine foot displacements or velocities or accelerations can be detected and then lead back to separate controllers ( C rDL , C rDR , C rNL and C rNR ). These controllers create then suitable signals for producing vertical actuator forces ( f azDL , f azDR , f azNL and f azNR ). The controller structure of each controller may be different. The structure of each controller is described in the Laplace domain, by a transfer function, with the Laplace variable s:

Controller C rij : G cij,γ ( s ) = ∑ μ = 0 m b μ,ij,γ ⋅ s μ ∑ ν = 0 n a ν,ij , γ ⋅ s ν (6)

The constants b μ,ij,γ and a ν,ij , γ are hereby the constants of the polynomial functions. To describe different feedback strategies the index γ is used, referring to [

γ = { 0 : No feedback (open control loops) z : Feedback of the vertical motor feet displacements z aDL , z aDR , z aNL , z aNR v : Feedback of the vertical motor feet velocities v z,aDL , v z,aDR , v z,aNL , v z,aNR a : Feedback of the vertial motor feet accelerations a z,aDL , a z,aDR , a z,aNL , a z,aNR (7)

The mechanical damping coefficients of the actuators d azij , d ayij , d axij and of the foundation d fzij , d fyij , d fxij can be derived by the corresponding mechanical loss factor tan δ aij for the actuators and tan δ fij for the foundation, by the corresponding stiffness c aqij and c fqij and by the whirling frequency ω F , referring to [

d aqij = c aqij ⋅ tan δ aij ω F ; d fqij = c fqij ⋅ tan δ fij ω F with : q = x , y , z (8)

For analysis of the forced vibration due to unbalance, the whirling angular frequency is equal to the rotor angular frequency ω F = Ω . For analysis of natural vibrations with marginal decay, the whirling angular frequency is defined as the correspondent natural angular frequency, as a simplification, referring to [

A linearization regarding the machine feet displacements is possible, because only small displacements occur due to the excitations. Therefore, kinematic constraints can be used to express the movement of the motor feet, as well as the displacements of the bearing housing points B D and B N , by the movement of the center point S, referring to [

M ⋅ q ¨ + ( D + G ) ⋅ q ˙ + C ⋅ q = f u + m u + f a (9)

The vectors f u and m u are the vectors of excitation and the vector f a is the vector of the actuator forces. The vector q contains the coordinates for displacements and rotations of the machine centre point S and for displacements of the foundation points.

q ( t ) = [ z s ; y s ; x s ; φ sz ; φ sy ; φ sx ; z fDL ; z fDR ; z fNL ; z fNR ; y fDL ; y fDR ; y fNL ; y fNR ; x fDL ; x fDR ; x fNL ; x fNR ] T (10)

The matrix M represents the mass matrix, the matrix D the damping matrix and the matrix C the stiffness matrix:

M = [ m 1 , 1 ⋯ m 1 , 18 ⋮ ⋱ ⋮ m 18 , 1 … m 18 , 18 ] ; D = [ d 1 , 1 ⋯ d 1 , 18 ⋮ ⋱ ⋮ d 18 , 1 … d 18 , 18 ] ; C = [ c 1 , 1 ⋯ c 1 , 18 ⋮ ⋱ ⋮ c 18 , 1 … c 18 , 18 ] (11)

The coefficients of the matrices are presented in the Appendix. The gyroscopic matrix G can be written by:

G = [ 0 0 0 0 0 0 0 1 × 12 0 0 0 0 0 0 0 1 × 12 0 0 0 0 0 0 0 1 × 12 0 0 0 0 Ω ⋅ θ rx 0 0 1 × 12 0 0 0 − Ω ⋅ θ rx 0 0 0 1 × 12 0 0 0 0 0 0 0 1 × 12 0 12 × 1 0 12 × 1 0 12 × 1 0 12 × 1 0 12 × 1 0 12 × 1 0 12 ] (12)

The vector 0 12 × 1 , represents the zero-vector with 0 12 × 1 ∈ ℝ 12 × 1 , the vector 0 1 × 12 the zero-vector with 0 1 × 12 ∈ ℝ 1 × 12 and the matrix 0 12 the zero-matrix with 0 12 ∈ ℝ 12 × 12 . The unbalance force vector f u ( t ) , caused by static unbalance u e , can be described by:

f u ( t ) = [ 1 ; − j ; 0 ; 0 ; 0 ; 0 ; 0 1 × 12 ] T ︸ P e ⋅ Ω 2 ⋅ m r ⋅ e ^ ⋅ e j ⋅ φ e ︸ u e = u ^ e ⋅ e j ⋅ φ e ⋅ e j ⋅ Ω ⋅ t (13)

and the unbalance moment vector m u ( t ) , caused by the moment unbalance u α , by:

m u ( t ) = [ 0 ; 0 ; 0 ; 1 ; − j ; 0 ; 0 1 × 12 ] T ︸ P α ⋅ Ω 2 ⋅ ( θ rzy − θ rx ) ⋅ α ^ ⋅ e j ⋅ φ α ︸ u α = u ^ α ⋅ e j ⋅ φ α ⋅ e j ⋅ Ω ⋅ t (14)

The actuator force vector f a is split into the actuator force vector for each machine foot:

f a ( t ) = f aDL ( t ) + f aDR ( t ) + f aNL ( t ) + f aNR ( t ) (15)

with the actuator force vectors:

f aDL ( t ) = f azDL ( t ) ⋅ [ 1 ; 0 ; 0 ; 0 ; a D ; − b L ; − 1 ; 0 ; 0 ; 0 ; 0 1 × 8 ] T ︸ P aDL (16)

f aDR ( t ) = f azDR ( t ) ⋅ [ 1 ; 0 ; 0 ; 0 ; a D ; b R ; 0 ; − 1 ; 0 ; 0 ; 0 1 × 8 ] T ︸ P aDR (17)

f aNL ( t ) = f azNL ( t ) ⋅ [ 1 ; 0 ; 0 ; 0 ; − a N ; − b L ; 0 ; 0 ; − 1 ; 0 ; 0 1 × 8 ] T ︸ P aNL (18)

f aNR ( t ) = f azNR ( t ) ⋅ [ 1 ; 0 ; 0 ; 0 ; − a N ; b R ; 0 ; 0 ; 0 ; − 1 ; 0 1 × 8 ] T ︸ P aNR (19)

The vector 0 1 × 8 represents the zero-vector with 0 1 × 8 ∈ ℝ 1 × 8 . The vectors P aDL , P aDR , P aNL and P aNR are the actuator force transmission vectors. Now, a state space formulation is used (

For avoiding mix-up with the stiffness matrix C and damping matrix D , index “st” is used for the matrices of the state space. According to

x ( t ) = [ q ( t ) ; q ˙ ( t ) ] T ; y ( t ) = [ q ( t ) ; q ˙ ( t ) ; q ¨ ( t ) ] T (20)

The system matrix A st , the input matrix B st , the output matrix C st , and the straight-way matrix D st can be written as:

A st = [ 0 18 I 18 − M − 1 ⋅ C − M − 1 ⋅ ( D + G ) ] ; B st = [ 0 18 M − 1 ] C st = [ I 18 0 18 0 18 I 18 − M − 1 ⋅ C − M − 1 ⋅ ( D + G ) ] ; D st = [ 0 18 0 18 M − 1 ] (21)

with the zero-matrix 0 18 ∈ ℝ 18 × 18 and the unit-matrix I 18 ∈ ℝ 18 × 18 . Therefore, the state space equations can be described by:

x ˙ ( t ) = A st ⋅ x ( t ) + B st ⋅ [ f u ( t ) + m u ( t ) + f a ( t ) ] (22)

y ( t ) = C st ⋅ x ( t ) + D st ⋅ [ f u ( t ) + m u ( t ) + f a ( t ) ] (23)

By using a controller matrix T st , γ (

f a ( t ) = − T st , γ ⋅ y ( t ) (24)

For deriving the frequency response matrix of the system, Equations (22), (23) and (24) are transferred in the Fourier-domain:

X ( j ω ) ⋅ j ω = A st ⋅ X ( j ω ) + B st ⋅ [ F u ( j ω ) + M u ( j ω ) + F a ( j ω ) ] (25)

Y ( j ω ) = C st ⋅ X ( j ω ) + D st ⋅ [ F u ( j ω ) + M u ( j ω ) + F a ( j ω ) ] (26)

F a ( j ω ) = − T st , γ ⋅ Y ( j ω ) (27)

With these equations follows the output vector:

Y ( j ω ) = [ I 54 + ( C st ⋅ ( I 36 ⋅ j ω − A st ) − 1 ⋅ B st + D st ) ⋅ T st , γ ] − 1 ⋅ [ C st ⋅ ( I 36 ⋅ j ω − A st ) − 1 ⋅ B st + D st ] ⋅ [ F u ( j ω ) + M u ( j ω ) ] (28)

with the unit-matrices I 54 ∈ ℝ 54 × 54 and I 36 ∈ ℝ 36 × 36 . With the definition

Y ( j ω ) = Y γ ( j ω ) (29)

it is highlighted, that the output vector Y ( j ω ) dependents on the different feedback strategies(7), represented by the controller matrix T st , γ . The controller matrix T st , γ can now be derived, based on [

T st , γ = { [ 0 18 0 18 0 18 ] for γ = 0 [ T z 0 18 0 18 ] for γ = z [ 0 18 T v 0 18 ] for γ = v [ 0 18 0 18 T a ] for γ = a (30)

with the matrix T γ , also described in the Fourier-domain with G cij , γ ( j ω ) , which are the frequency response functions of the controllers, based on the controller transfer functions (6) with s → j ω :

With: T z ( j ω ) , T v ( j ω ) , T a ( j ω ) ∈ ℂ 18 × 18 and T 0 ( j ω ) = 0 18 ∈ ℝ 18 × 18 and with the zero-matrix 0 8 × 12 ∈ ℝ 8 × 12 . Now, the output vector Y γ ( j ω ) can be calculated by:

Y γ ( j ω ) = G γ ( j ω ) ⋅ [ F u ( j ω ) + M u ( j ω ) ] (32)

and the frequency response matrix G γ ( j ω ) can be written by:

G γ ( j ω ) = [ I 54 + ( C st ⋅ ( I 36 ⋅ j ω − A st ) − 1 ⋅ B st + D st ) ⋅ T st , γ ] − 1 ⋅ [ C st ⋅ ( I 36 ⋅ j ω − A st ) − 1 ⋅ B st + D st ] (33)

The frequency response vector for each single kind of excitation can now be derived, with following Fouier-transformations, with the Dirac-delta function δ :

F u ( j ω ) = F { f u ( t ) } = F { P e ⋅ Ω 2 ⋅ m r ⋅ e ^ ⋅ e j ⋅ φ e ︸ u e ⋅ e j ⋅ Ω ⋅ t } = P e ⋅ Ω 2 ⋅ m r ⋅ e ^ ⋅ e j ⋅ φ e ︸ u e ⋅ 2 π ⋅ δ ( ω − Ω ) (34)

M u ( j ω ) = F { m u ( t ) } = F { P α ⋅ Ω 2 ⋅ ( θ rzy − θ rx ) ⋅ α ^ ⋅ e j ⋅ φ α ︸ u α ⋅ e j ⋅ Ω ⋅ t } = P α ⋅ Ω 2 ⋅ ( θ rzy − θ rx ) ⋅ α ^ ⋅ e j ⋅ φ α ︸ u α ⋅ 2 π ⋅ δ ( ω − Ω ) (35)

Further on, the index κ is used for both kinds of unbalance, representing the excitations:

κ = e , α (36)

The output vector Y γ , κ ( j ω ) for each kind of unbalance can be calculated by:

Y γ , κ ( j ω ) = G γ ( j ω ) ⋅ P κ ⋅ Ω 2 ︸ G γ , κ ( j ω ) ⋅ u κ ⋅ 2 π ⋅ δ ( ω − Ω ) (37)

Using the sifting property of the Dirac delta function δ ( ω − Ω ) , the Fourier-transformed output vector for each kind of unbalance follows:

Y γ , κ ( j ω ) = G γ ( j Ω ) ⋅ P κ ⋅ Ω 2 ︸ G γ , κ ( j Ω ) ⋅ u κ ⋅ 2 π ⋅ δ ( ω − Ω ) (38)

Now, the inverse transformation of this Fourier-transformed output vector Y γ , κ ( j ω ) back into the time-domain can be done:

y γ , κ ( t ) = F − 1 { Y γ , κ ( j ω ) } = F − 1 { G γ ( j Ω ) ⋅ P κ ⋅ Ω 2 ︸ G γ , κ ( j Ω ) ⋅ u κ ⋅ 2 π ⋅ δ ( ω − Ω ) } = G γ ( j Ω ) ⋅ P κ ⋅ Ω 2 ︸ G γ , κ ( j Ω ) ⋅ u κ ︸ y ^ γ , κ ⋅ e j Ω t (39)

It is useful to relate the amplitude output vector on the respective unbalance:

y ^ γ , κ , ref = y ^ γ , κ u κ = G γ ( j Ω ) ⋅ P κ ⋅ Ω 2 ︸ G γ , κ ( j Ω ) (40)

At this point it is important to mention, that beside the gyroscopic matrix G also the damping matrix D depends on the rotary angular frequency W. The reason is that for forced vibration due to the unbalance, the whirling angular frequency ω F for the mechanical damping coefficients (8), is equal to the rotary angular frequency W ( ω F = Ω ).Therefore, the matrices A st and C st are also functions of the rotor angular frequency W. The frequency response matrix G γ ( j Ω ) can now be formulated by:

G γ ( j Ω ) = [ I 54 + ( C st ( Ω ) ⋅ ( I 36 ⋅ j Ω − A st ( Ω ) ) − 1 ⋅ B st + D st ) ⋅ T st , γ ( j Ω ) ] − 1 ⋅ [ C st ( Ω ) ⋅ ( I 36 ⋅ j Ω − A st ( Ω ) ) − 1 ⋅ B st + D st ] (41)

With this matrix, the frequency response vector G γ , κ ( j Ω ) for both kinds of unbalance can be described:

G γ , κ ( j Ω ) = y ^ γ , κ u κ = G γ ( j Ω ) ⋅ P κ ⋅ Ω 2 (42)

With this frequency response vector G γ , κ ( j Ω ) and the kinematic constraints in [

Due to the use of active vibration control, a stability analysis is very important, therefore the poles of the system have to be derived. Based on [

det [ I 36 ⋅ s − A st + B st ⋅ T st , γ ( s ) ⋅ ( I 54 + D st ⋅ T st , γ ( s ) ) − 1 ⋅ C st ] = 0 (43)

However, this direct procedure is only possible if the matrices A st and C st are independent of the whirling angular frequency ω F , which is here not the case, because the damping coefficients depend here on ω F . The whirling angular frequency ω F corresponds here—when calculating the pols—to the natural angular frequency and therefore to the imaginary part of the complex poles. Therefore, the use of the damping coefficient lead here to a causality problem. Therefore, a worst case procedure is here derived to investigate the stability of the system. The basic requirements are:

· The mechanical damping of the actuators and the foundation is low, which should be usually the case, so that the mechanical loss factors for the actuators and the foundation fulfill following conditions:

tan δ aij < 0.2 and tan δ fij < 0.2 (44)

· The modal mass of the foundation is so low, that only the first six natural vibration modes have to be taken into account.

· The first six natural vibration modes have conjugate complex poles.

Therefore, following worst case procedure is derived to calculate the pols s n , which is shown in

Then, the natural angular frequency of the 6^{th} conjugate complex pole pair ω 6 —which is set to be equal to the whirling angular frequency ω F —is considered, for calculating the damping coefficients d aqij and d fqij . Afterwards, the pols are calculated again, considering damping ( D ≠ 0 ) and the closed control loop operation ( T st , γ ( s ) ≠ 0 ). The corresponding conjugate complex pol pair s 6 * is again taken into account for deriving the new damping coefficients. Finally the pols are calculated with the modified damping matrix D . If the change of the whirling angular frequency is too large (e.g. more than 5%) this approach can be repeated in loops. This procedure presents a worst case scenario regarding instability, because the natural angular frequency of the highest considerable mode (here mode 6) is used for calculating the damping coefficients, which lead to the

lowest damping coefficients.

In this section, special controllers—standard controllers (P-, I-, PI-, PD-(ideal), PID-(ideal) controllers)—are analyzed. Therefore, the control parameter can now be directly implemented into the mass matrix and/or the damping matrix and/or the stiffness matrix, which can be seen in

Therefore, the differential equation can be described now by:

M ˜ ⋅ q ¨ + ( D ˜ + G ) ⋅ q ˙ + C ˜ ⋅ q = f u + m u (45)

By implementing the control parameters into the mass matrix, M ˜ becomes M con . When they are implemented into the damping matrix, D ˜ becomes D con , and if they are implemented into the stiffness matrix, C ˜ becomes C con . The structure of these matrices is shown in (49). Following definitions are used, referring to [

For the mass matrix C con : M con = Ψ con and ψ = m (46)

For the damping matrix D con : D con = Ψ con and ψ = d (47)

For the stiffness matrix C con : C con = Ψ con and ψ = c (48)

The matrices with the integrated controller parameters are formulated by:

The controller parameters K ψij depend on the controller structure and the chosen feedback strategy (

Now, a numerical example is presented, where the vibration stability, as well as the forced vibrations of a rotating machine are analyzed.

The data of the rotating machine are shown in _{L} = b_{R}; a_{D} = a_{N}).

The foundation is a simplified steel frame foundation, which consists of two I-beams (green), which are stiffened by additional welded steel sheets (brown) in the area of the machine feet (cyan), and fixed (red) to ground (

Mass of the stator | m s = 7200 kg |
---|---|

Mass of the rotor | m r = 1900 kg |

Mass inertia of the stator at the x-axis | θ sx = 1500 kg ⋅ m 2 |

Mass inertia of the stator at the y-axis | θ sy = 2800 kg ⋅ m 2 |

Mass inertia of the stator at the z-axis | θ sz = 2800 kg ⋅ m 2 |

Mass inertia of the rotor at the x-axis | θ rx = 50 kg ⋅ m 2 |

Mass inertia of the rotor at the y-axis | θ ry = 100 kg ⋅ m 2 |

Mass inertia of the rotor at the z-axis | θ rz = 100 kg ⋅ m 2 |

Height of the centre of gravity S | h = 560 mm |

Horizontal (y-direction) distance from A DL and A NL to S | b L = 530 mm |

Horizontal (y-direction) distance from A DR and A NR to S | b R = 530 mm |

Axial (x-direction) distance from A DL and A DR to S | a D = 700 mm |

Axial (x-direction) distance from A NL and A NR to S | a N = 700 mm |

Axial (x-direction) distance from B D to S | l D = 1000 mm |

Axial (x-direction) distance from B N to S | l N = 1000 mm |

stiffening effect in the area of the machine feet is considered, so that a low stiffness of the steel frame foundation is considered here, as a worst case.

The substitute foundation stiffness, which is necessary to use the model in

In this example only three actuators are used, two on drive-side, left and right (DL, DR) and only one actuator on non-drive side left (NL) (

The data of the stiff element is shown in

For the control system, I-controllers with feedbacks of the vertical machine feet accelerations are chosen, so that following controller transfer function is used.

G cji , a ( s ) = b 0 , ij , a s (50)

Kind of beam | I-Profil DIN 1025 —S235JR-IPB 180 |
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Number of stiffening sheets | 16 |

Thickness of stiffening sheets | t = 14 mm |

Modal mass of foundation ( m f = m fDL = m fDR = m fNL = m fNR ) | m f = 10 kg |

Mechanical loss factor ( tan δ f = tan δ fDL = tan δ fDR = tan δ fNL = tan δ fNR ) | tan δ f = 0.04 |

Subsitute stiffness of foundation (machine directly mounted on foundation): | |

· c fx = c fxDL = c fxDR = c fxNL = c fxNR | c fx = 1000 kN / mm |

· c fy = c fyDL = c fyDR = c fyNL = c fyNR | c fy = 146 kN / mm |

· c fz = c fzDL = c fzDR = c fzNL = c fzNR | c fz = 1440 kN / mm |

Subsitute stiffness of foundation (machine mounted with actuators on foundation): | |

· c fx = c fxDL = c fxDR = c fxNL = c fxNR | c fx = 42 kN / mm |

· c fy = c fyDL = c fyDR = c fyNL = c fyNR | c fy = 20.5 kN / mm |

· c fz = c fzDL = c fzDR = c fzNL = c fzNR | c fz = 1440 kN / mm |

Mass of the armature | m aaDL = m aaDR = m aaNL = 15 kg |
---|---|

Mass of the stator | m asDL = m asDR = m asNL = 35 kg |

Vertical stiffness | c azDL = c azDR = c azNL = 10.5 kN / mm |

Horizontal stiffness | c ayDL = c ayDR = c ayNL = 26.2 kN / mm |

Axial stiffness | c axDL = c axDR = c axNL = 26.2 kN / mm |

Mechanical loss factor | tan δ aDL = tan δ aDR = tan δ aNL = 0.04 |

Mass (top) | m aaNR = 3 kg |
---|---|

Mass (bottom) | m asNR = 5 kg |

Vertical stiffness | c azNR = 2600 kN / mm |

Horizontal stiffness | c ayNR = 560 kN / mm |

Axial stiffness | c axNR = 560 kN / mm |

Mechanical loss factor | tan δ aNR = 0.04 |

Controller parameter for DL | b 0 , DL , a = 140000 kg / s |
---|---|

Controller parameter for DR | b 0 , DR , a = 140000 kg / s |

Controller parameter for NL | b 0 , NL , a = 140000 kg / s |

Controller parameter for NR | b 0 , NR , a = 0 kg / s |

The data of the control parameters are shown in

Afterwards three different cases are analyzed:

· Case 1: The rotating machine is directly mounted on the steel frame foundation.

· Case 2: Three actuators at DL, DR, NL and a stiff element at NR are placed between machine feet and foundation. The actuators are only operating passively (open control loops).

· Case 3: Same setting as case 2, but the actuators are now operating actively (closed control loops).

To analyze the stability of the system, the poles are calculated for the three different cases, based on the procedure, described in

When comparing the figures, it is obvious that the damping of the poles can be strongly increased by the active vibration control system (case 3), compared to case 1 and case 2.

Now, the amplitudes of the frequency response functions of bearing housing vibration velocities and of foundation vibration velocities are computed, as well as the frequency response functions of the actuator forces, all related to the respective unbalance. The amplitudes of the frequency response functions are calculated in [dB] with the reference gauge in Si-Unit, in a frequency range from 1 Hz to 250 Hz. An additional index d is now introduced for case 1, so that now following definitions are used:

γ = 0 , z , v , a , d ; i = D , N ; j = L , R ; q = x , y , z ; κ = e , α (51)

Related bearing housing vibration velocities:

| G γ , v ^ biq , κ ( j Ω ) | dB = 20 ⋅ log ( | G γ , v ^ biq , κ ( j Ω ) | | G γ , v ^ biq , κ ( j Ω ) | ref ) (52)

with: | G γ , v ^ biq , κ ( j Ω ) | ref = { 1 m / s kg ⋅ m for κ = e 1 m / s kg ⋅ m 2 for κ = α

Related foundation vibration velocities:

| G γ , v ^ fijq , κ ( j Ω ) | dB = 20 ⋅ log ( | G γ , v ^ fijq , κ ( j Ω ) | | G γ , v ^ fijq , κ ( j Ω ) | ref ) (53)

with: | G γ , v ^ fijq , κ ( j Ω ) | ref = { 1 m / s kg ⋅ m for κ = e 1 m / s kg ⋅ m 2 for κ = α

Related actuator forces:

| G γ , f ^ azij , κ ( j Ω ) | dB = 20 ⋅ log ( | G γ , f ^ azij , κ ( j Ω ) | | G γ , f ^ azij , κ ( j Ω ) | ref ) (54)

with: | G γ , f ^ azij , κ ( j Ω ) | ref = { 1 N kg ⋅ m for κ = e 1 N kg ⋅ m 2 for κ = α

The amplitude response functions for the bearing housing vibration velocities at the drive side and at the non-drive side, related to static unbalance are shown in

The amplitude response functions for the foundation vibration velocities at the drive side, related to static unbalance are shown in

The amplitude response functions for the foundation vibration velocities at the non-drive side, related to static unbalance are shown in

The frequency response functions for the actuator forces, related to static

unbalance are shown in

The amplitude response functions for the bearing housing vibration velocities at the drive side and at the non-drive side, related to moment unbalance are shown in

The amplitude response functions for the foundation vibration velocities at the drive side, related to moment unbalance are shown in

The amplitude response functions for the foundation vibration velocities at the non-drive side, related to moment unbalance are shown in

The frequency response functions for the actuator forces, related to moment unbalance are shown in

The amplitude response functions for the bearing housing vibration velocities

for excitation by a static unbalance (

is obvious in

Therefore, it could be shown, that the method “vibration mode coupling by asymmetry”, which was developed and mathematically described in [

The paper presents a simplified 3D-model for active vibration control of rotating machines with active machine foot mounts on soft foundations, considering static and moment unbalance. After the model was mathematical described in the time domain, it was transferred into the Fourier domain, where the frequencies response functions regarding bearing housing vibrations, foundation vibrations and actuator forces have been derived. Afterwards, the mathematical coherences have been described in the Laplace domain and a worst case procedure was derived to analyze the vibration stability. For special controller structures in combination with certain feedback strategies, a calculation method was shown, where the controller parameters can directly be implemented into the stiffness matrix, damping matrix and mass matrix. Additionally a numerical example was presented, where the vibration stability and the frequency response functions have been analyzed. It could be shown, that with the active vibration control system all vibration modes can be damped well, so that an operation in the rotational frequency range from 0 Hz to 80 Hz is possible without off-limits areas.

Finally it could be demonstrated, that the method “vibration mode coupling by asymmetry”, which was developed and mathematically described in [

The author declares no conflicts of interest regarding the publication of this paper.

Werner, U. (2021) Active Vibration Control of Rotating Machines with Active Machine Foot Mounts on Soft Foundations Based on a 3D-Model. Journal of Applied Mathematics and Physics, 9, 57-87. https://doi.org/10.4236/jamp.2021.91006

Coefficients of the mass matrix M:

m 1 , 1 = m 2 , 2 = m 3 , 3 = m sr + m aaDL + m aaDR + m aaNL + m aaNR (55)

m 1 , 5 = m 5 , 1 = a D ⋅ ( m aaDL + m aaDR ) − a N ⋅ ( m aaNL + m aaNR ) (56)

m 1 , 6 = m 6 , 1 = b R ⋅ ( m aaDR + m aaNR ) − b L ⋅ ( m aaDL + m aaNL ) (57)

m 2 , 4 = m 4 , 2 = − a D ⋅ ( m aaDL + m aaDR ) + a N ⋅ ( m aaNL + m aaNR ) (58)

m 2 , 6 = m 6 , 2 = − h ⋅ ( m aaDL + m aaDR + m aaNL + m aaNR ) (59)

m 3 , 4 = m 4 , 3 = − b R ⋅ ( m aaDR + m aaNR ) + b L ⋅ ( m aaDL + m aaNL ) (60)

m 3 , 5 = m 5 , 3 = h ⋅ ( m aaDL + m aaDR + m aaNL + m aaNR ) (61)

m 4 , 4 = θ srz + b L 2 ⋅ ( m aaDL + m aaNL ) + b R 2 ⋅ ( m aaDR + m aaNR ) + a D 2 ⋅ ( m aaDL + m aaDR ) + a N 2 ⋅ ( m aaNL + m aaNR ) (62)

m 4 , 5 = m 5 , 4 = h ⋅ b L ⋅ ( m aaDL + m aaNL ) − h ⋅ b R ⋅ ( m aaDR + m aaNR ) (63)

m 4 , 6 = m 6 , 4 = h ⋅ a D ⋅ ( m aaDL + m aaDR ) − h ⋅ a N ⋅ ( m aaNL + m aaNR ) (64)

m 5 , 5 = θ sry + a D 2 ⋅ ( m aaDL + m aaDR ) + a N 2 ⋅ ( m aaNL + m aaNR ) + h 2 ⋅ ( m aaDL + m aaDR + m aaNL + m aaNR ) (65)

m 5 , 6 = m 6 , 5 = a D ⋅ ( b R ⋅ m aaDR − b L ⋅ m aaDL ) + a N ⋅ ( b L ⋅ m aaNL − b R ⋅ m aaNR ) (66)

m 6 , 6 = θ sx + b L 2 ⋅ ( m aaDL + m aaNL ) + b R 2 ⋅ ( m aaDR + m aaNR ) + h 2 ⋅ ( m aaDL + m aaDR + m aaNL + m aaNR ) (67)

m 7 , 7 = m 11 , 11 = m 15 , 15 = m fDL + m asDL (68)

m 8 , 8 = m 12 , 12 = m 16 , 16 = m fDR + m asDR (69)

m 9 , 9 = m 13 , 13 = m 17 , 17 = m fNL + m asNL (70)

m 10 , 10 = m 14 , 14 = m 18 , 18 = m fNR + m asNR (71)

The other coefficients are zero.

Coefficients of the stiffness matrix C:

c 1 , 1 = c azDL + c azDR + c azNL + c azNR (72)

c 1 , 5 = c 5 , 1 = a D ⋅ ( c azDL + c azDR ) − a N ⋅ ( c azNL + c azNR ) (73)

c 1 , 6 = c 6 , 1 = b R ⋅ ( c azDR + c azNR ) − b L ⋅ ( c azDL + c azNL ) (74)

c 1 , 7 = c 7 , 1 = − c azDL (75)

c 1 , 8 = c 8 , 1 = − c azDR (76)

c 1 , 9 = c 9 , 1 = − c azNL (77)

c 1 , 10 = c 10 , 1 = − c azNR (78)

c 2 , 2 = c ayDL + c ayDR + c ayNL + c ayNR (79)

c 2 , 4 = c 4 , 2 = a N ⋅ ( c ayNL + c ayNR ) − a D ⋅ ( c ayDL + c ayDR ) (80)

c 2 , 6 = c 6 , 2 = − h ⋅ ( c ayDL + c ayDR + c ayNL + c ayNR ) (81)

c 2 , 11 = c 11 , 2 = − c ayDL (82)

c 2 , 12 = c 12 , 2 = − c ayDR (83)

c 2 , 13 = c 13 , 2 = − c ayNL (84)

c 2 , 14 = c 14 , 2 = − c ayNR (85)

c 3 , 3 = c axDL + c axDR + c axNL + c axNR (86)

c 3 , 4 = c 4 , 3 = b L ⋅ ( c axDL + c axNL ) − b R ⋅ ( c axDR + c axNR ) (87)

c 3 , 5 = c 5 , 3 = h ⋅ ( c axDL + c axDR + c axNL + c axNR ) (88)

c 3 , 15 = c 15 , 3 = − c axDL (89)

c 3 , 16 = c 16 , 3 = − c axDR (90)

c 3 , 17 = c 17 , 3 = − c axNL (91)

c 3 , 18 = c 18 , 3 = − c axNR (92)

c 4 , 4 = b L 2 ⋅ ( c axDL + c axNL ) + b R 2 ⋅ ( c axDR + c axNR ) + a D 2 ⋅ ( c ayDL + c ayDR ) + a N 2 ⋅ ( c ayNL + c ayNR ) (93)

c 4 , 5 = c 5 , 4 = h ⋅ b L ⋅ ( c axDL + c axNL ) − h ⋅ b R ⋅ ( c axDR + c axNR ) (94)

c 4 , 6 = c 6 , 4 = h ⋅ a D ⋅ ( c ayDL + c ayDR ) − h ⋅ a N ⋅ ( c ayNL + c ayNR ) (95)

c 4 , 11 = c 11 , 4 = c ayDL ⋅ a D (96)

c 4 , 12 = c 12 , 4 = c ayDR ⋅ a D (97)

c 4 , 13 = c 13 , 4 = − c ayNL ⋅ a N (98)

c 4 , 14 = c 14 , 4 = − c ayNR ⋅ a N (99)

c 4 , 15 = c 15 , 4 = − c axDL ⋅ b L (100)

c 4 , 16 = c 16 , 4 = c axDR ⋅ b R (101)

c 4 , 17 = c 17 , 4 = − c axNL ⋅ b L (102)

c 4 , 18 = c 18 , 4 = c axNR ⋅ b R (103)

c 5 , 5 = a D 2 ⋅ ( c azDL + c azDR ) + a N 2 ⋅ ( c azNL + c azNR ) + h 2 ⋅ ( c axDL + c axDR + c axNL + c axNR ) (104)

c 5 , 6 = c 6 , 5 = a D ⋅ ( b R ⋅ c azDR − b L ⋅ c azDL ) + a N ⋅ ( b L ⋅ c azNL − b R ⋅ c azNR ) (105)

c 5 , 7 = c 7 , 5 = − c azDL ⋅ a D (106)

c 5 , 8 = c 8 , 5 = − c azDR ⋅ a D (107)

c 5 , 9 = c 9 , 5 = c azNL ⋅ a N (108)

c 5 , 10 = c 10 , 5 = c azNR ⋅ a N (109)

c 5 , 15 = c 15 , 5 = − c axDL ⋅ h (110)

c 5 , 16 = c 16 , 5 = − c axDR ⋅ h (111)

c 5 , 17 = c 17 , 5 = − c axNL ⋅ h (112)

c 5 , 18 = c 18 , 5 = − c axNR ⋅ h (113)

c 6 , 6 = b L 2 ⋅ ( c azDL + c azNL ) + b R 2 ⋅ ( c azDR + c azNR ) + h 2 ⋅ ( c ayDL + c ayDR + c ayNL + c ayNR ) (114)

c 6 , 7 = c 7 , 6 = c azDL ⋅ b L (115)

c 6 , 8 = c 8 , 6 = − c azDR ⋅ b R (116)

c 6 , 9 = c 9 , 6 = c azNL ⋅ b L (117)

c 6 , 10 = c 10 , 6 = − c azNR ⋅ b R (118)

c 6 , 11 = c 11 , 6 = c ayDL ⋅ h (119)

c 6 , 12 = c 12 , 6 = c ayDR ⋅ h (120)

c 6 , 13 = c 13 , 6 = c ayNL ⋅ h (121)

c 6 , 14 = c 14 , 6 = c ayNR ⋅ h (122)

c 7 , 7 = c fzDL + c azDL (123)

c 8 , 8 = c fzDR + c azDR (124)

c 9 , 9 = c fzNL + c azNL (125)

c 10 , 10 = c fzNR + c azNR (126)

c 11 , 11 = c fyDL + c ayDL (127)

c 12 , 12 = c fyDR + c ayDR (128)

c 13 , 13 = c fyNL + c ayNL (129)

c 14 , 14 = c fyNR + c ayNR (130)

c 15 , 15 = c fxDL + c axDL (131)

c 16 , 16 = c fxDR + c axDR (132)

c 17 , 17 = c fxNL + c axNL (133)

c 18 , 18 = c fxNR + c axNR (134)

The other coefficients are zero.

Coefficients of the damping matrix D:

The coefficients of the damping matrix can be easily derived by just replacing “c” in the formulas (72)-(134) by “d”.

A2. Response Functions for the Bearing Housing Vibration Velocities· In vertical direction:

Drive side: G γ , v ^ bDz , κ ( j Ω ) = G γ , κ ( j Ω ) 19 + l D ⋅ G γ , κ ( j Ω ) 23 (135)

Non-drive side: G γ , v ^ bNz , κ ( j Ω ) = G γ , κ ( j Ω ) 19 − l N ⋅ G γ , κ ( j Ω ) 23 (136)

· In horizontal direction:

Drive side: G γ , v ^ bDy , κ ( j Ω ) = G γ , κ ( j Ω ) 20 − l D ⋅ G γ , κ ( j Ω ) 22 (137)

Non-drive side: G γ , v ^ bNy , κ ( j Ω ) = G γ , κ ( j Ω ) 20 + l N ⋅ G γ , κ ( j Ω ) 22 (138)

· In axial direction:

Drive side: G γ , v ^ bDx , κ ( j Ω ) = G γ , κ ( j Ω ) 21 (139)

Non-drive side: G γ , v ^ bNx , κ ( j Ω ) = G γ , κ ( j Ω ) 21 (140)

The element G γ , κ ( j Ω ) n is the n^{th} element of the frequency response vector G γ , κ ( j Ω ) .

· In vertical direction:

Drive side, left: G γ , v ^ fDLz , κ ( j Ω ) = G γ , κ ( j Ω ) 25 (141)

Drive side right: G γ , v ^ fDRz , κ ( j Ω ) = G γ , κ ( j Ω ) 26 (142)

Non-drive side, left: G γ , v ^ fNLz , κ ( j Ω ) = G γ , κ ( j Ω ) 27 (143)

Non-drive side, right: G γ , v ^ fNRz , κ ( j Ω ) = G γ , κ ( j Ω ) 28 (144)

· In horizontal direction:

Drive side, left: G γ , v ^ fDLy , κ ( j Ω ) = G γ , κ ( j Ω ) 29 (145)

Drive side right: G γ , v ^ fDRy , κ ( j Ω ) = G γ , κ ( j Ω ) 30 (146)

Non-drive side, left: G γ , v ^ fNLy , κ ( j Ω ) = G γ , κ ( j Ω ) 31 (147)

Non-drive side, right: G γ , v ^ fNRy , κ ( j Ω ) = G γ , κ ( j Ω ) 32 (148)

· In axial direction:

Drive side, left: G γ , v ^ fDLx , κ ( j Ω ) = G γ , κ ( j Ω ) 33 (149)

Drive side right: G γ , v ^ fDRx , κ ( j Ω ) = G γ , κ ( j Ω ) 34 (150)

Non-drive side, left: G γ , v ^ fNLx , κ ( j Ω ) = G γ , κ ( j Ω ) 35 (151)

Non-drive side, right: G γ , v ^ fNRx , κ ( j Ω ) = G γ , κ ( j Ω ) 36 (152)

A4. Response Functions for the Actuator ForcesFeedback of the motor feet displacements γ = z :

· For drive side, left:

G γ , f ^ azDL , κ ( j Ω ) = − [ G γ , κ ( j Ω ) 1 − b L ⋅ G γ , κ ( j Ω ) 6 + a D ⋅ G γ , κ ( j Ω ) 5 ] ⋅ G cDL , γ ( j Ω ) (153)

· For drive side, right:

G γ , f ^ azDR , κ ( j Ω ) = − [ G γ , κ ( j Ω ) 1 + b R ⋅ G γ , κ ( j Ω ) 6 + a D ⋅ G γ , κ ( j Ω ) 5 ] ⋅ G cDR , γ ( j Ω ) (154)

· For non-drive side, left:

G γ , f ^ azNL , κ ( j Ω ) = − [ G γ , κ ( j Ω ) 1 − b L ⋅ G γ , κ ( j Ω ) 6 − a N ⋅ G γ , κ ( j Ω ) 5 ] ⋅ G cNL , γ ( j Ω ) (155)

· For non-drive side, right:

G γ , f ^ azNR , κ ( j Ω ) = − [ G γ , κ ( j Ω ) 1 + b R ⋅ G γ , κ ( j Ω ) 6 − a N ⋅ G γ , κ ( j Ω ) 5 ] ⋅ G cNR , γ ( j Ω ) (156)

For feedback of the motor feet velocities, index γ gets v ( γ = v ) and the index number n of the elements G γ , κ ( j Ω ) n in (153)-(156) has to be changed to n → n + 18 . For feedback of motor feet accelerations index γ gets a ( γ = a ) and the index number n of the elements G γ , κ ( j Ω ) n in (153)-(156) has to be changed to n → n + 36 . For open loop operation index γ gets 0 ( γ = 0 ), and the response functions of the actuator forces are zero.