SlideShare a Scribd company logo
2011 ME Graduate Student Conference
                                                                                           April 30, 2011



                             MODELING OF DYNAMICS OF GYROSCOPIC SYSTEMS

                                                           Siddharth Sharma
                                                            M.S. Candidate

                                                  Faculty Advisor: Yitshak M. Ram

ABSTRACT

Critical Speed of Flexible Rotating Shaft

The stability of a flexible shaft rotating in bearings is             kυ and k ζ which is further divided equally between the two
investigated to avoid large vibrations at the stability
                                                                      springs of the same set. The stiffness of the shaft is
boundaries of the system. A stability boundary is
                                                                      considered to be acting in two mutually perpendicular
encountered when the poles of the system becomes purely
imaginary. Subsequently, the effects of external damping              directions. The origin A represents the centre of the cross
provided by the bearings and internal damping offered by              section of the shaft in clamped roller bearings. The point B
the shaft are also analyzed for their impact on the stability         locates the centre of mass of the rotor which is deflected
of shaft. The well known counter-intuitive phenomenon,                from its equilibrium position when the shaft is rotating at
e.g. [1], [2] and [3], that internal damping destabilizes the         constant angular velocity ω . The equilibrium position of the
system is demonstrated.                                               rotor coincides with point A .

Modeling, Transformation of Coordinates and                                                                 z
                                                                                                    ζ
Equations of Motion
                                                                                                                     kζ
The simplest system consists of a rotor mounted on a                                                                 2
                                                                                                                          kυ
flexible shaft which rotates at an angular velocity ω in                                       kυ                         2
clamped roller bearing.                                                                        2                B              υ
                                                   Rotor                                                             θ
                         z                                                                              A                          y
            ζ                     υ                                                                             kζ
                                                                                                                2
                                                                                       ω
                             θ = ωt   y

                                                                 χ
                    A
                                                                 x
                                                                          Figure 2 Model of shaft rotating in clamped roller
                                                                                                    bearing
Clamped Roller Bearing                    Shaft
                                                                      The motion of shaft in the χυζ rotating coordinates is
                                                                      governed by the following set of equations, see e.g., [3]
   Figure 1 A system of shaft rotating in clamped roller
                         bearing                                           &              (
                                                                          mυ& − 2mωζ& + kυ − mω 2 υ = 0    )                          (2)
                                                                           &&
The rotating coordinate system χυζ sweeps an angle
                                                                                           (     2
                                                                          mζ + 2mωυ + k ζ − mω ζ = 0
                                                                                    &                      )
θ = ωt from the stationary coordinate system xyz . The
                                                                      Symmetrical Shaft, Symmetrical Bearings (SSSB) with
transformation of coordinates is governed by
                                                                      displacement of shaft at bearing
      υ   cos ωt sin ωt  y                           (1)        A symmetrical shaft has equal stiffness in two mutually
       ς  − sin ωt cos ωt  z 
       =                   
                                                                perpendicular directions and the same is true for
                                                                      symmetrical bearings. The shaft has finite displacement at
The model represents the stiffness of shaft by two sets of            bearing.
two springs each. The stiffness constant of the two sets is
Im(s )
              z                                                                                        z
                                υ                                                           ς
ς                                                                                                                            B
                                                                                  B                                                                                  s-plane
                                                                                                                                                                                             ω =0        ω = 10
                                                                                                                                        υ
     A                                                                                A(υ A ς A )          A                            ωt                          * Our poles
     O                                         x χ                                     O(0 0 )         O                                     y                      . Smith’s poles
                                                                                                                                                                      m = 10
                                             σ ′′(ς − ς A )
                                                  & &                                                                                                                 σ ′ = 0. 2
                                                                                                                     σ ′′(ς& − ς& A )
                                                                                                    κ ′′(ς − ς A )                                                    σ ′′ = 0.3
                                                 κ ′′(υ − υ A )                                                            κ ′′(υ − υ A )
                    z       κ ′′(ς − ς A )                                                 ς           z
     ς                                                                                                                                                                κ ′ = 50
                                                B                                                                        B       σ ′′(υ − υ A )
                                                                                                                                       & &
                                                      σ ′′(υ − υ A )
                                                            & &                                                                                                       κ ′′ = 100
                                                 υ                                                                                υ                                   0 ≤ ω ≤ 10
     κ ′y A
                            A                   ωt                                                                               ωt
     σ ′y A
        &                                             y                                                                                 y
                                O                                                                    O
                  κ ′z A        σ ′z A
                                   &                                                                                                                                                                               Re(s )

      Free Body diagram for massless                                                                     Free Body diagram for rotor
        shaft (moments not shown)
                                                                                                                                                                                         Root locus
                           (a)                                                                                       (b)

      Figure 3 (a) Free body diagram for massless shaft                                                                                               Im(s )
    (moments not shown), and (b) Free body diagram for                                                                                                             s-plane

                            rotor
                                                                                                                                                                  * Our poles
                                                                                                                                                                                                         ω = 400
                                                                                                                                                                  . Smith’s poles            ω=0
The free body diagrams results equations of the form
                                                                                                                                                                    m = 10

                           M&& + Cx + (K + ωG )x = 0
                            x     &                                                                                                     (3)                        σ ′ = 0. 2                                      Re(s )
                                                                                                                                                                   σ ′′ = 0.3
                                                                                                                                                                   κ ′ = 50
where M , C , K , G are the mass, damping, stiffness and                                                                                                           κ ′′ = 100
                                                                                                                                                                   0 ≤ ω ≤ 400
gyroscopic matrices respectively, each of order 4 and

                                                                             T
                           x = ( yA                    zA              y   z) .                                                         (4)

The First-Order realization of the above equation gives the                                                                                                                           Root Locus ∆ω=10

roots s of the characteristic equation φ (ω , s ) of the system.
                                                                                                                                                  Acknowledgements
We denote
                                                          6
                           φ (ω , s ) = ∑ α k s k .                                                                                     (5)       I would like to mention regards to my advisor Dr. Yitshak
                                                      k =0                                                                                        M Ram for his guidance. I convey my sincere thanks to
                                                                                                                                                  Siemens Energy and Automation, Inc., for sponsoring the
The characteristic polynomial corresponding to Smith’s                                                                                            project.
formulation, see e.g., [2] is defined as
                                                                                                                                                  References
                                                          4
                           φ (ω , s ) = ∑ α k s k .
                            ˆ             ˆ                                                                                             (6)
                                                      k =0
                                                                                                                                                      1.       S. H. Crandall. Physical explanations of the
                                                                                                                                                               destabilizing effect of damping in rotating parts,
Results                                                                                                                                                        Texas A & M University Instability Workshop,
                                                                                                                                                               369-383 (1980).
For our system, eight real solutions appear                                                                                                           2.       D. M. Smith. The motion of a rotor carried by a
                                                                                                                                                               flexible shaft in flexible bearings, Proceedings of
                           {ω = 6.69551004 s = ±1.82583925}                                                                             (7)                    the Royal Society (London) A 142, 92-118 (1933).
                                                                                                                                                      3.        H. Zeigler. Principles of structural stability, 1968
                           {ω = −6.69551004 s = ±1.82583925}                                                                            (8)
                                                                                                                                                               (Blaisdell Publishing Company, Waltham).
                           {ω = 18632.40800 s = ±2.23597851i}                                                                           (9)
                           {ω = −18632.40800 s = ±2.23597851i}                                                                          (10)

The system is unstable for 6.695510048 < ω < 18632.40800
and stable for other positive values of ω .
                                                                                                                                                  `

More Related Content

What's hot

Recent advances of MEIS for near surface analysis
Recent advances of MEIS for near surface analysis Recent advances of MEIS for near surface analysis
Recent advances of MEIS for near surface analysis
Instituto Nacional de Engenharia de Superfícies
 
Lec10
Lec10Lec10
1404.7369
1404.73691404.7369
1404.7369
ion tudor
 
Momentum &amp; Energy
Momentum &amp; EnergyMomentum &amp; Energy
Momentum &amp; Energy
matcol
 
Lesson1
Lesson1Lesson1
Lesson1
Teja Ande
 
call for papers, research paper publishing, where to publish research paper, ...
call for papers, research paper publishing, where to publish research paper, ...call for papers, research paper publishing, where to publish research paper, ...
call for papers, research paper publishing, where to publish research paper, ...
International Journal of Engineering Inventions www.ijeijournal.com
 
1990-2010 - Magnetismo Cuántico
1990-2010 - Magnetismo Cuántico1990-2010 - Magnetismo Cuántico
1990-2010 - Magnetismo Cuántico
oriolespinal
 
Sam Session
Sam SessionSam Session
Sam Session
Teja Ande
 
7 - Momentum & energy
7 - Momentum & energy7 - Momentum & energy
7 - Momentum & energy
guest924dbe
 
Jet in crossflow mixing
Jet in crossflow mixingJet in crossflow mixing
Jet in crossflow mixing
Samsung Techwin
 

What's hot (10)

Recent advances of MEIS for near surface analysis
Recent advances of MEIS for near surface analysis Recent advances of MEIS for near surface analysis
Recent advances of MEIS for near surface analysis
 
Lec10
Lec10Lec10
Lec10
 
1404.7369
1404.73691404.7369
1404.7369
 
Momentum &amp; Energy
Momentum &amp; EnergyMomentum &amp; Energy
Momentum &amp; Energy
 
Lesson1
Lesson1Lesson1
Lesson1
 
call for papers, research paper publishing, where to publish research paper, ...
call for papers, research paper publishing, where to publish research paper, ...call for papers, research paper publishing, where to publish research paper, ...
call for papers, research paper publishing, where to publish research paper, ...
 
1990-2010 - Magnetismo Cuántico
1990-2010 - Magnetismo Cuántico1990-2010 - Magnetismo Cuántico
1990-2010 - Magnetismo Cuántico
 
Sam Session
Sam SessionSam Session
Sam Session
 
7 - Momentum & energy
7 - Momentum & energy7 - Momentum & energy
7 - Momentum & energy
 
Jet in crossflow mixing
Jet in crossflow mixingJet in crossflow mixing
Jet in crossflow mixing
 

Similar to Modeling of Dynamics of Gyroscopic System

Ab32189194
Ab32189194Ab32189194
Ab32189194
IJERA Editor
 
Structural design and non linear modeling of a highly stable
Structural design and non linear modeling of a highly stableStructural design and non linear modeling of a highly stable
Structural design and non linear modeling of a highly stable
Alexander Decker
 
Engineering science lesson 1
Engineering science lesson 1Engineering science lesson 1
Engineering science lesson 1
Shahid Aaqil
 
Engineering science lesson 1
Engineering science lesson 1Engineering science lesson 1
Engineering science lesson 1
Shahid Aaqil
 
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
IAEME Publication
 
v2chap2.pdf
v2chap2.pdfv2chap2.pdf
v2chap2.pdf
RicardoGulapa1
 
Chapter 4: Torsion
Chapter 4: TorsionChapter 4: Torsion
Chapter 4: Torsion
Monark Sutariya
 

Similar to Modeling of Dynamics of Gyroscopic System (7)

Ab32189194
Ab32189194Ab32189194
Ab32189194
 
Structural design and non linear modeling of a highly stable
Structural design and non linear modeling of a highly stableStructural design and non linear modeling of a highly stable
Structural design and non linear modeling of a highly stable
 
Engineering science lesson 1
Engineering science lesson 1Engineering science lesson 1
Engineering science lesson 1
 
Engineering science lesson 1
Engineering science lesson 1Engineering science lesson 1
Engineering science lesson 1
 
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
COMPARISON OF RESPONSE TO UNBALANCE OF OVERHUNG ROTOR SYSTEM FOR DIFFERENT SU...
 
v2chap2.pdf
v2chap2.pdfv2chap2.pdf
v2chap2.pdf
 
Chapter 4: Torsion
Chapter 4: TorsionChapter 4: Torsion
Chapter 4: Torsion
 

Modeling of Dynamics of Gyroscopic System

  • 1. 2011 ME Graduate Student Conference April 30, 2011 MODELING OF DYNAMICS OF GYROSCOPIC SYSTEMS Siddharth Sharma M.S. Candidate Faculty Advisor: Yitshak M. Ram ABSTRACT Critical Speed of Flexible Rotating Shaft The stability of a flexible shaft rotating in bearings is kυ and k ζ which is further divided equally between the two investigated to avoid large vibrations at the stability springs of the same set. The stiffness of the shaft is boundaries of the system. A stability boundary is considered to be acting in two mutually perpendicular encountered when the poles of the system becomes purely imaginary. Subsequently, the effects of external damping directions. The origin A represents the centre of the cross provided by the bearings and internal damping offered by section of the shaft in clamped roller bearings. The point B the shaft are also analyzed for their impact on the stability locates the centre of mass of the rotor which is deflected of shaft. The well known counter-intuitive phenomenon, from its equilibrium position when the shaft is rotating at e.g. [1], [2] and [3], that internal damping destabilizes the constant angular velocity ω . The equilibrium position of the system is demonstrated. rotor coincides with point A . Modeling, Transformation of Coordinates and z ζ Equations of Motion kζ The simplest system consists of a rotor mounted on a 2 kυ flexible shaft which rotates at an angular velocity ω in kυ 2 clamped roller bearing. 2 B υ Rotor θ z A y ζ υ kζ 2 ω θ = ωt y χ A x Figure 2 Model of shaft rotating in clamped roller bearing Clamped Roller Bearing Shaft The motion of shaft in the χυζ rotating coordinates is governed by the following set of equations, see e.g., [3] Figure 1 A system of shaft rotating in clamped roller bearing  & ( mυ& − 2mωζ& + kυ − mω 2 υ = 0 ) (2)  && The rotating coordinate system χυζ sweeps an angle ( 2 mζ + 2mωυ + k ζ − mω ζ = 0  & ) θ = ωt from the stationary coordinate system xyz . The Symmetrical Shaft, Symmetrical Bearings (SSSB) with transformation of coordinates is governed by displacement of shaft at bearing υ   cos ωt sin ωt  y  (1) A symmetrical shaft has equal stiffness in two mutually  ς  − sin ωt cos ωt  z   =        perpendicular directions and the same is true for symmetrical bearings. The shaft has finite displacement at The model represents the stiffness of shaft by two sets of bearing. two springs each. The stiffness constant of the two sets is
  • 2. Im(s ) z z υ ς ς B B s-plane ω =0 ω = 10 υ A A(υ A ς A ) A ωt * Our poles O x χ O(0 0 ) O y . Smith’s poles m = 10 σ ′′(ς − ς A ) & & σ ′ = 0. 2 σ ′′(ς& − ς& A ) κ ′′(ς − ς A ) σ ′′ = 0.3 κ ′′(υ − υ A ) κ ′′(υ − υ A ) z κ ′′(ς − ς A ) ς z ς κ ′ = 50 B B σ ′′(υ − υ A ) & & σ ′′(υ − υ A ) & & κ ′′ = 100 υ υ 0 ≤ ω ≤ 10 κ ′y A A ωt ωt σ ′y A & y y O O κ ′z A σ ′z A & Re(s ) Free Body diagram for massless Free Body diagram for rotor shaft (moments not shown) Root locus (a) (b) Figure 3 (a) Free body diagram for massless shaft Im(s ) (moments not shown), and (b) Free body diagram for s-plane rotor * Our poles ω = 400 . Smith’s poles ω=0 The free body diagrams results equations of the form m = 10 M&& + Cx + (K + ωG )x = 0 x & (3) σ ′ = 0. 2 Re(s ) σ ′′ = 0.3 κ ′ = 50 where M , C , K , G are the mass, damping, stiffness and κ ′′ = 100 0 ≤ ω ≤ 400 gyroscopic matrices respectively, each of order 4 and T x = ( yA zA y z) . (4) The First-Order realization of the above equation gives the Root Locus ∆ω=10 roots s of the characteristic equation φ (ω , s ) of the system. Acknowledgements We denote 6 φ (ω , s ) = ∑ α k s k . (5) I would like to mention regards to my advisor Dr. Yitshak k =0 M Ram for his guidance. I convey my sincere thanks to Siemens Energy and Automation, Inc., for sponsoring the The characteristic polynomial corresponding to Smith’s project. formulation, see e.g., [2] is defined as References 4 φ (ω , s ) = ∑ α k s k . ˆ ˆ (6) k =0 1. S. H. Crandall. Physical explanations of the destabilizing effect of damping in rotating parts, Results Texas A & M University Instability Workshop, 369-383 (1980). For our system, eight real solutions appear 2. D. M. Smith. The motion of a rotor carried by a flexible shaft in flexible bearings, Proceedings of {ω = 6.69551004 s = ±1.82583925} (7) the Royal Society (London) A 142, 92-118 (1933). 3. H. Zeigler. Principles of structural stability, 1968 {ω = −6.69551004 s = ±1.82583925} (8) (Blaisdell Publishing Company, Waltham). {ω = 18632.40800 s = ±2.23597851i} (9) {ω = −18632.40800 s = ±2.23597851i} (10) The system is unstable for 6.695510048 < ω < 18632.40800 and stable for other positive values of ω . `