Research ArticleShafting Alignment Computing Method of New MultibearingRotor System under Specific Installation Requirement

Qian Chen Qi Yuan Ming Lei and Mengyao Wang

School of Energy and Power Engineering Xirsquoan Jiaotong University Xirsquoan Shaanxi 710049 China

Correspondence should be addressed to Qi Yuan qyuanmailxjtueducn

Received 30 June 2015 Revised 21 December 2015 Accepted 27 December 2015

Academic Editor Yuri Vladimirovich Mikhlin

Copyright copy 2016 Qian Chen et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The shafting of large steam turbine generator set is composed of several rotors which are connected by couplings The computingmethod of shafting with different structure under specific installation requirement is studied in this paper Based on three-momentequation shafting alignment mathematical model is established The computing method of bearing elevations and loads undercorresponding installation requirements where bending moment of each coupling is zero and there exist preset sag and gap insome couplings is proposed respectively Bearing elevations and loads of shaftingwith different structure under specific installationrequirement are calculated calculation results are compared with installation data measured on site which verifies the validity andaccuracy of the proposed shafting alignment computing method The above work provides a reliable approach to analyze shaftingalignment and could guide installation on site

1 Introduction

The shafting of large rotating machinery is a multibearingrotor system Shafts are manufactured separately and thenconnected by rigid or flexible couplings All shafts havesome form of catenary due to their own weight thusshafts are not straight So in the installation each bearingelevation should be adjusted along the vertical directionto keep the catenary curve smooth when the machine isrunning Unreasonable bearing elevation will bring aboutextra bearing load which causes the vibration of shaft Whenthe bearing load is light the formation of bearing oil film isdifficult which induces self-excited oil film oscillation Whenthe bearing load is heavy grinding pad phenomenon is proneto happen Misalignment is a condition where the centerlinesof coupled shafts do not coincide It causes over 70 ofrotatingmachinery vibration problemsMisalignment causesa decrease in motor efficiency and misaligned machine ismore prone to failure due to increased loads on bearing andcouplings [1] Misalignment conditions are generally classi-fied as angular parallel and the combination of these twoShaft misalignment effects negatively influence the rolling

sealing and coupling parts and can also produce eccentricityin the air gap

Misalignment of rotor causes generation of reactionforces in the coupling which are the major cause of machinevibration Gibbons [2] first evaluated themisalignment forcesgenerated in various types of coupling Patel and Darpe[3] studied coupling characteristics under misalignmentcondition experimentally They measured excitations forcesdue to parallel misalignment and angular misalignmentrespectively Force variation for angular misalignment caseshows significant 1x and 3x harmonic vibration Parallelmisalignment on the other hand has significant 1x 2x and 3xcomponents Sekhar and Prabhu [4] numerically presentedthe effect of coupling misalignment on rotor vibration Itwas shown that the location of coupling with respect to thebending mode shape has a strong influence on vibrationResearch on dynamic response of misalignment rotor is lim-itedMore importantly there are discrepancies in the findingsof different investigators For example vibrations at twicethe rotational speed (ie 2x) and its harmonics are reported[5 6] as a characteristic misalignment feature Contrary tothis Al-Hussain and Redmond [7] reported coupled lateral

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 1647575 12 pageshttpdxdoiorg10115520161647575

2 Mathematical Problems in Engineering

and torsional 1x vibration for parallel misalignment It isalso reported that different types of couplings reveal differentfrequencies composition in the vibration response under thesame misalignment condition

The main investigations of shaft alignment date fromthe late 1960s Early studies demonstrated the feasibility ofaligning two shafts using an arrangement of strain gages andload cells [8 9] However despite many advantages of suchmethods including quickness and accuracy this techniquehas a few limitations in its application for long shafts withseveral inaccessible bearings In general rotating mechanicalsystem shaft alignment is traditionally achieved using someform of dial indicator alignment method [10] with themost common techniques being the rim and face methodand the reverse indicator method [11] The dial indicatoralignment method ensures bearing elevations according todial indicator readings measured at predetermined locationsIn 1980s laser shaft alignment technique became a hotresearch topic Perry [12] presented a high-precision laseralignment system based on two position sensing detectors(PSDs) and showed that the system enabled a simpler andmore accurate alignment of two shafts than using traditionaldial indicator alignment method As a result of their inherentaccuracy laser alignment systems have been widely appliedfor a diverse range of shaft alignment applications includingrotor systems [13] and gas turbines [14] Simmons et al [15]developed laser instrumentation system for reliably measur-ing casing thermal distortion and alignment deviation ona large combustion turbine This system could accuratelymeasure deflection and misalignment which are importantfor cold condition shafting alignment Liao [16] constructeda measurement system comprising a laser light source anda detector to facilitate the alignment of two rotating shaftsThemain purpose of these shafting alignment methods men-tioned above is to decrease the gap and sag of every couplingThese methods are efficient to deal with the alignment ofsmall rotating machinery which does not pay attention tothe influence of bearing elevation for bearing load Howeverfor the shafting alignment of large rotating machinery suchas steam turbine generator set or ground-based heavy-dutygas turbine generator set bearing elevations significantlyinfluence bearing loads distribution So these methods aboveare inefficient and may bring about irrational bearing loaddistribution for large rotating machinery

For large rotating machinery alignment bearing eleva-tions should firstly be ensured through calculation Thereare mainly two assumptions in calculation corresponding totwo different installation requirements the first assumptionis that the bending moment of every bearing is zero andthe second assumption is that the bending moment of everycoupling is zero Taking a 600MW turbo-generator setwith double bearing support as the study object He et al[17] calculated bearing elevations and loads based on thesetwo assumptions respectively It was shown that bearingelevations and loads gained from the second assumption aremore reasonable through comparison The latter calculationmethod based on the second assumption does not apply toshafting in addition to double bearing support because thecatenary curve of every double bearing supported rotor is

calculated firstly and each bearing elevation is adjusted tomake each coupling fulfill the zero to zero condition Withthe rapid development of unit capacity shafting structuresbecome more compact and more complex Besides shaftingwith double bearing support shafting with single bearingsupport and shafting with synchro-self-shifting (3S) clutchare widely introduced in the design of turbo-generator unitThere are various factors such as thermal distortion fluctu-ation of bearing oil film thickness and unequal settlementof foundation which disturb the alignment condition androtor position in operation compared with cold conditionIn cold condition there exist a preset gap and sag in rigidcouplings before connection to compensate for the changein rotor position caused by the above factors in thermalcondition Study for new structure shafting alignment com-puting method under specific installation requirements ismeaningful and necessary

Based on three-moment equation shafting alignmentmathematical model was established by using transfer matrixmethod in this paper Corresponding mechanical boundaryconditions of different installation requirements where thebendingmoment in couplings is zero and there exist a gap andsag in couplings were determined Detailed mathematicalderivation of new structure shafting under differentmechani-cal boundary conditions was conductedThrough calculationby means of programming bearing elevations and loads ofshafting with single bearing support with double bearingsupport andwith synchro-self-shifting clutch under differentinstallation requirements were obtained respectively

2 Shafting Alignment Computing Method

21 Shafting Alignment Mathematical Model The shaftingof large rotating machinery is a multibearing rotor systemSeveral rotors are coupled together which are supported onbearings The following assumptions are made to simplifythe model (1) ignoring the fluctuation caused by bearingoil film shafting elevations at the bearings are regarded asbearing elevations (2) compared with the entire shafting thelength and weight of overhang are negligibly small so twoshafting overhangs are not taken into account in mechanicalequation derivation (3) bearing support is viewed as a hingesupport in modeling Figure 1 shows a typical multisupportbearing system which is composed of rotors and bearings Ithas 119873 bearings and 119873 minus 1 beam spans ranked from the 2ndto the 119873th The weights of rotors and blades are viewed asdistributed load 119902

The multibearing rotor system is a redundant structurethat is difficult to be solved directly In order to facilitatesolving it is separated into several simply supported beamsThere exists unknown value bending moment on each beamas Figure 2 shows

Deflection angles at bearings are caused by three factorsrespectively bending moments at bearings shafting weightand elevation difference between adjacent bearings Take the119868th bearing as study objectThe left deflection angle of the 119868thbearing caused by bending moment 120574

1015840

119894is given by

1205741015840

119894= 119872119894minus1

1205931015840

119894minus1119894+ 1198721198941205931015840

119894119894 (1)

Mathematical Problems in Engineering 3

1 2 3

(3)(2)

middot middot middotmiddot middot middot K

q

N Nminus 1

(N)

Figure 1 Multisupport redundant beam system

120601998400i

120601998400998400i

MI

MIminus1

The Ith span

The (I + 1)th span

MI

MI

+1

I I

I

minus 1

I + 1

Figure 2 Deflection angles at bearings of each span

where 119872119894minus1

and 119872119894are the bending moments at the (119868 minus

1)th and 119868th bearing Flexibility coefficient 1205931015840

119894minus1119894is the left

deflection angle of the 119868th bearing caused by unit momentat the (119868 minus 1)th bearing 120593

1015840

119894119894is the left deflection angle of the

119868th bearing caused by unit moment at the 119868th bearingSimilarly the right deflection angle of the 119868th bearing

caused by bending moment 12057410158401015840

119894is given by

12057410158401015840

119894= 11987211989412059310158401015840

119894119894+ 119872119894+1

12059310158401015840

119894+1119894 (2)

Taking the influence of elevation difference between adjacentbearings and shafting weight into consideration the totaldeflection angles at the left and right side of the 119868th bearingare given respectively by

1206011015840

119894= 1205741015840

119894+ 120579119871

119894119894minus 120573119871

119894119894 (3)

12060110158401015840

119894= 12057410158401015840

119894+ 120579119877

119894+1119894+ 120573119877

119894+1119894 (4)

where 1206011015840

119894is the total deflection angle at the left side of the 119868th

bearing and 12060110158401015840

119894is the total deflection angle at the right side

of the 119868th bearing 120579119871

119894119894is the deflection angle at the left side

of the 119868th bearing caused by the weight of the 119868th span and120579119877

119894+1119894is the deflection angle at the right side of the 119868th bearing

caused by the weight of the (119868 + 1)th span 120573119871

119894119894and 120573

119877

119894+1119894are

the deflection angles caused by elevation difference at the leftand right sides of the 119868th bearing as shown in Figure 3

The left deflection angle of the 119868th bearing caused byelevation difference in the 119868th span is given by

120573119871

119894119894=

119910119894minus1

minus 119910119894

119871119894

(5)

I minus 1

I

I

+ 1

yiminus1 yi+1yiLi Li+1

120573Ri+1i

120573Lii

Figure 3 Deflection angle caused by elevation difference

where 119910119894minus1

and 119910119894are the elevations of the (119868 minus 1)th and 119868th

bearing 119871119894is the length of the 119868th span

Similarly the right deflection angle of the 119868th bearingcaused by elevation difference in the (119868 + 1)th span is given by

120573119877

119894+1119894=

119910119894

minus 119910119894+1

119871119894+1

(6)

Since shafting catenary curve is smooth when the machine isrunning total deflection angles at the left and right sides ofevery bearing are equal

1206011015840

119894= minus12060110158401015840

119894(119894 = 2 3 119873 minus 1) (7)

Thus the moment equilibrium equation can be expressed as

119872119894minus1

1205931015840

119894minus1119894+ 119872119894(1205931015840

119894119894+ 12059310158401015840

119894119894) + 119872

119894+112059310158401015840

119894+1119894+ 120579119871

119894119894+ 120579119877

119894+1119894

minus (120573119871

119894119894minus 120573119877

119894+1119894) = 0

(8)

4 Mathematical Problems in Engineering

This equation for bearings is ranked from 2nd to (119873 minus 1)thand can be written in matrix form

[[[[[[[[[[[[[[

[

1205931015840

22+ 12059310158401015840

2212059310158401015840

32

1205931015840

231205931015840

33+ 12059310158401015840

3312059310158401015840

43

d d d

1205931015840

119894minus11198941205931015840

119894119894+ 12059310158401015840

11989411989412059310158401015840

119894+1119894

d d d

1205931015840

(119899minus3)(119899minus2)1205931015840

(119899minus2)(119899minus2)+ 12059310158401015840

(119899minus2)(119899minus2)12059310158401015840

(119899minus1)(119899minus2)

1205931015840

(119899minus2)(119899minus1)1205931015840

(119899minus1)(119899minus1)+ 12059310158401015840

(119899minus1)(119899minus1)

]]]]]]]]]]]]]]

]

1198722

1198723

119872119894

119872119899minus2

119872119899minus1

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(9)

and simplified form

[120593] sdot 119872 = minus 120573 (10)

where [120593] is the flexibility coefficient matrix 119872 is themoment vector and 120573 is the deflection angle vector

Moment equilibrium equations express the relationshipbetween moments deflection angles and elevations at everybearing which lay the foundation of shafting alignmentcomputation However only moment equilibrium equationsare not enough for calculating the bearing elevations andloads since the number of unknown quantities exceeds thenumber of equationsThus every bearing moment and othermechanical boundary conditions should be calculated firstlyby using transfer matrix method

22 Transfer Matrix Method The transfer matrix method isa fast and easy way to get the parameters such as momentshear force and deflection which are important in shaftingalignment calculation This method could guarantee highcomputational accuracy and save time Although finite ele-mentmethod is versatile it takesmore time to build geometrymodel and requires more computational resources This isthe reason for using transfer matrix method in shaftingalignment calculation

Usually shafting is divided into several segments with dif-ferent diameters in calculation The mechanical relationshipbetween two arbitrary nodes could be represented by usingtransfermatrixmethod Take the 119868th span as the study object

It is divided into 119898 segments and 119898 + 1 node as shown inFigure 4

Figure 5 shows the forces and moments of the 119895th shaftsegment on the 119868th span From the equilibrium of forces andmoments an equation is derived by Timoshenko and Gere[18]

119876119895+1

= 119876119895

minus 119902119895119897119895

119872119895+1

= 119872119895

+ 119876119895119897119895

minus1

21199021198951198972

119895

(11)

In matrix form

119885119895+1

= [119870]119895

119885119895

(12)

where

119885119895

= 119872119895 119876119895 1119879

[119870]119895

=

[[[[

[

1 119897119895

1199021198951198972

119895

2

0 1 minus119902119895119897119895

0 0 1

]]]]

]

(13)

Mathematical Problems in Engineering 5

I Iminus 1

1

2

3 4

middot middot middot

middot middot middot

middot middot middot

middot middot middot

(1)(2)

(3)

n minus 1 n n + 1

m

(m)

m + 1

Figure 4 119868th span of the rotor

y

Mj

Qj

qjQj+1

Mj+1

z

Figure 5 Applied force of the 119895th shaft element

If there is no coupling within the 119868th span the relationshipbetween the (119868 minus 1)th bearing and the 119868th bearing can bewritten as

119885119898+1

= [119863]1198941198851

119872

119876

1

119898+1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]119894

119872

119876

1

1

(14)

Transformation matrix [119863]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]1

If there exists coupling within the 119868th span and thecoupling locates at the node 119899 the relation between the (119868 minus

1)th bearing the coupling and the 119868th bearing meets thefollowing equations

119885119899

= [119892]1198941198851

119885119898+1

= [119866]119894119885119899

(15)

where

[119892]119894

= [119870]119899minus1

[119870]119899minus2

sdot sdot sdot [119870]1

[119866]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]119899

(16)

1 2 3 4 5 6 7 8

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 6 Configuration diagram of single bearing support shafting

23 BearingMoment and Load Calculation Inmoment equi-librium equations the expression of deflection angle vector120573 has the elevation of every bearing which is unknown invalueWithmoment vector the number of unknownqualitiesis more than equations Thus boundary conditions such asbendingmoment of every coupling being zero are introducedto make equations solvable During the installation of shaft-ing at site bearing elevations are determined by using theassumption that bending moment of each coupling is zeroThis method reduces the bending moment on each couplingthus enhancing the stability and reliability of turbo-generatorset Under this assumption bending moment of each bearingis derived in a certain order by using transfer matrix methodEvery bearing elevation can be solved by plugging bearingbendingmoment vector intomoment equilibrium equationsDue to structure difference mechanical boundary conditionsof shafting in different configuration are distinct So bendingmoment calculation order is diverse Take the single bearingsupport shafting as example to illustrate bearingmoment andload calculation procedure based on transfer matrix method

The shafting of a 1000MWultra supercritical turbinewithsingle bearing support is shown in Figure 6 This shaftingis composed of a high pressure rotor (HP) intermediatepressure rotor (IP) two low pressure rotors (LP

1 LP2)

generator (GEN) and exciter rotor (EXC) High pressurerotor is supported with two bearings Intermediate pressurerotor and low pressure rotors are supported with singlebearing Generator and exciter rotor are supported with threebearings Single bearing support shafting is compact in struc-ture which decreases the entire length of shafting and reducesthe negative influence of basefoundation deformation onalignment

Without considering the influence of two overhangssimplified configuration diagram of single bearing supportshafting is shown in Figure 7 Because of intermediatepressure rotors low pressure rotor and exciter rotors aresingle bearing supported there exists shear force in couplingsC1 C2 C3 and C5 to support the rotor Since generator rotoris double bearing support shear force in coupling C4 is zero

From the 8th span spans with couplings are analyzedfirstly in reverse order In the 8th span the bending momentand shear force in bearing 8 and coupling C5 meet thefollowing relation

1198728

1198768

1

=[[

[

1 11986612

11986613

0 1 11986623

0 0 1

]]

]8

1198721198625

1198761198625

1

(17)

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

and torsional 1x vibration for parallel misalignment It isalso reported that different types of couplings reveal differentfrequencies composition in the vibration response under thesame misalignment condition

The main investigations of shaft alignment date fromthe late 1960s Early studies demonstrated the feasibility ofaligning two shafts using an arrangement of strain gages andload cells [8 9] However despite many advantages of suchmethods including quickness and accuracy this techniquehas a few limitations in its application for long shafts withseveral inaccessible bearings In general rotating mechanicalsystem shaft alignment is traditionally achieved using someform of dial indicator alignment method [10] with themost common techniques being the rim and face methodand the reverse indicator method [11] The dial indicatoralignment method ensures bearing elevations according todial indicator readings measured at predetermined locationsIn 1980s laser shaft alignment technique became a hotresearch topic Perry [12] presented a high-precision laseralignment system based on two position sensing detectors(PSDs) and showed that the system enabled a simpler andmore accurate alignment of two shafts than using traditionaldial indicator alignment method As a result of their inherentaccuracy laser alignment systems have been widely appliedfor a diverse range of shaft alignment applications includingrotor systems [13] and gas turbines [14] Simmons et al [15]developed laser instrumentation system for reliably measur-ing casing thermal distortion and alignment deviation ona large combustion turbine This system could accuratelymeasure deflection and misalignment which are importantfor cold condition shafting alignment Liao [16] constructeda measurement system comprising a laser light source anda detector to facilitate the alignment of two rotating shaftsThemain purpose of these shafting alignment methods men-tioned above is to decrease the gap and sag of every couplingThese methods are efficient to deal with the alignment ofsmall rotating machinery which does not pay attention tothe influence of bearing elevation for bearing load Howeverfor the shafting alignment of large rotating machinery suchas steam turbine generator set or ground-based heavy-dutygas turbine generator set bearing elevations significantlyinfluence bearing loads distribution So these methods aboveare inefficient and may bring about irrational bearing loaddistribution for large rotating machinery

For large rotating machinery alignment bearing eleva-tions should firstly be ensured through calculation Thereare mainly two assumptions in calculation corresponding totwo different installation requirements the first assumptionis that the bending moment of every bearing is zero andthe second assumption is that the bending moment of everycoupling is zero Taking a 600MW turbo-generator setwith double bearing support as the study object He et al[17] calculated bearing elevations and loads based on thesetwo assumptions respectively It was shown that bearingelevations and loads gained from the second assumption aremore reasonable through comparison The latter calculationmethod based on the second assumption does not apply toshafting in addition to double bearing support because thecatenary curve of every double bearing supported rotor is

calculated firstly and each bearing elevation is adjusted tomake each coupling fulfill the zero to zero condition Withthe rapid development of unit capacity shafting structuresbecome more compact and more complex Besides shaftingwith double bearing support shafting with single bearingsupport and shafting with synchro-self-shifting (3S) clutchare widely introduced in the design of turbo-generator unitThere are various factors such as thermal distortion fluctu-ation of bearing oil film thickness and unequal settlementof foundation which disturb the alignment condition androtor position in operation compared with cold conditionIn cold condition there exist a preset gap and sag in rigidcouplings before connection to compensate for the changein rotor position caused by the above factors in thermalcondition Study for new structure shafting alignment com-puting method under specific installation requirements ismeaningful and necessary

Based on three-moment equation shafting alignmentmathematical model was established by using transfer matrixmethod in this paper Corresponding mechanical boundaryconditions of different installation requirements where thebendingmoment in couplings is zero and there exist a gap andsag in couplings were determined Detailed mathematicalderivation of new structure shafting under differentmechani-cal boundary conditions was conductedThrough calculationby means of programming bearing elevations and loads ofshafting with single bearing support with double bearingsupport andwith synchro-self-shifting clutch under differentinstallation requirements were obtained respectively

2 Shafting Alignment Computing Method

21 Shafting Alignment Mathematical Model The shaftingof large rotating machinery is a multibearing rotor systemSeveral rotors are coupled together which are supported onbearings The following assumptions are made to simplifythe model (1) ignoring the fluctuation caused by bearingoil film shafting elevations at the bearings are regarded asbearing elevations (2) compared with the entire shafting thelength and weight of overhang are negligibly small so twoshafting overhangs are not taken into account in mechanicalequation derivation (3) bearing support is viewed as a hingesupport in modeling Figure 1 shows a typical multisupportbearing system which is composed of rotors and bearings Ithas 119873 bearings and 119873 minus 1 beam spans ranked from the 2ndto the 119873th The weights of rotors and blades are viewed asdistributed load 119902

The multibearing rotor system is a redundant structurethat is difficult to be solved directly In order to facilitatesolving it is separated into several simply supported beamsThere exists unknown value bending moment on each beamas Figure 2 shows

Deflection angles at bearings are caused by three factorsrespectively bending moments at bearings shafting weightand elevation difference between adjacent bearings Take the119868th bearing as study objectThe left deflection angle of the 119868thbearing caused by bending moment 120574

1015840

119894is given by

1205741015840

119894= 119872119894minus1

1205931015840

119894minus1119894+ 1198721198941205931015840

119894119894 (1)

Mathematical Problems in Engineering 3

1 2 3

(3)(2)

middot middot middotmiddot middot middot K

q

N Nminus 1

(N)

Figure 1 Multisupport redundant beam system

120601998400i

120601998400998400i

MI

MIminus1

The Ith span

The (I + 1)th span

MI

MI

+1

I I

I

minus 1

I + 1

Figure 2 Deflection angles at bearings of each span

where 119872119894minus1

and 119872119894are the bending moments at the (119868 minus

1)th and 119868th bearing Flexibility coefficient 1205931015840

119894minus1119894is the left

deflection angle of the 119868th bearing caused by unit momentat the (119868 minus 1)th bearing 120593

1015840

119894119894is the left deflection angle of the

119868th bearing caused by unit moment at the 119868th bearingSimilarly the right deflection angle of the 119868th bearing

caused by bending moment 12057410158401015840

119894is given by

12057410158401015840

119894= 11987211989412059310158401015840

119894119894+ 119872119894+1

12059310158401015840

119894+1119894 (2)

Taking the influence of elevation difference between adjacentbearings and shafting weight into consideration the totaldeflection angles at the left and right side of the 119868th bearingare given respectively by

1206011015840

119894= 1205741015840

119894+ 120579119871

119894119894minus 120573119871

119894119894 (3)

12060110158401015840

119894= 12057410158401015840

119894+ 120579119877

119894+1119894+ 120573119877

119894+1119894 (4)

where 1206011015840

119894is the total deflection angle at the left side of the 119868th

bearing and 12060110158401015840

119894is the total deflection angle at the right side

of the 119868th bearing 120579119871

119894119894is the deflection angle at the left side

of the 119868th bearing caused by the weight of the 119868th span and120579119877

119894+1119894is the deflection angle at the right side of the 119868th bearing

caused by the weight of the (119868 + 1)th span 120573119871

119894119894and 120573

119877

119894+1119894are

the deflection angles caused by elevation difference at the leftand right sides of the 119868th bearing as shown in Figure 3

The left deflection angle of the 119868th bearing caused byelevation difference in the 119868th span is given by

120573119871

119894119894=

119910119894minus1

minus 119910119894

119871119894

(5)

I minus 1

I

I

+ 1

yiminus1 yi+1yiLi Li+1

120573Ri+1i

120573Lii

Figure 3 Deflection angle caused by elevation difference

where 119910119894minus1

and 119910119894are the elevations of the (119868 minus 1)th and 119868th

bearing 119871119894is the length of the 119868th span

Similarly the right deflection angle of the 119868th bearingcaused by elevation difference in the (119868 + 1)th span is given by

120573119877

119894+1119894=

119910119894

minus 119910119894+1

119871119894+1

(6)

Since shafting catenary curve is smooth when the machine isrunning total deflection angles at the left and right sides ofevery bearing are equal

1206011015840

119894= minus12060110158401015840

119894(119894 = 2 3 119873 minus 1) (7)

Thus the moment equilibrium equation can be expressed as

119872119894minus1

1205931015840

119894minus1119894+ 119872119894(1205931015840

119894119894+ 12059310158401015840

119894119894) + 119872

119894+112059310158401015840

119894+1119894+ 120579119871

119894119894+ 120579119877

119894+1119894

minus (120573119871

119894119894minus 120573119877

119894+1119894) = 0

(8)

4 Mathematical Problems in Engineering

This equation for bearings is ranked from 2nd to (119873 minus 1)thand can be written in matrix form

[[[[[[[[[[[[[[

[

1205931015840

22+ 12059310158401015840

2212059310158401015840

32

1205931015840

231205931015840

33+ 12059310158401015840

3312059310158401015840

43

d d d

1205931015840

119894minus11198941205931015840

119894119894+ 12059310158401015840

11989411989412059310158401015840

119894+1119894

d d d

1205931015840

(119899minus3)(119899minus2)1205931015840

(119899minus2)(119899minus2)+ 12059310158401015840

(119899minus2)(119899minus2)12059310158401015840

(119899minus1)(119899minus2)

1205931015840

(119899minus2)(119899minus1)1205931015840

(119899minus1)(119899minus1)+ 12059310158401015840

(119899minus1)(119899minus1)

]]]]]]]]]]]]]]

]

1198722

1198723

119872119894

119872119899minus2

119872119899minus1

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(9)

and simplified form

[120593] sdot 119872 = minus 120573 (10)

where [120593] is the flexibility coefficient matrix 119872 is themoment vector and 120573 is the deflection angle vector

Moment equilibrium equations express the relationshipbetween moments deflection angles and elevations at everybearing which lay the foundation of shafting alignmentcomputation However only moment equilibrium equationsare not enough for calculating the bearing elevations andloads since the number of unknown quantities exceeds thenumber of equationsThus every bearing moment and othermechanical boundary conditions should be calculated firstlyby using transfer matrix method

22 Transfer Matrix Method The transfer matrix method isa fast and easy way to get the parameters such as momentshear force and deflection which are important in shaftingalignment calculation This method could guarantee highcomputational accuracy and save time Although finite ele-mentmethod is versatile it takesmore time to build geometrymodel and requires more computational resources This isthe reason for using transfer matrix method in shaftingalignment calculation

Usually shafting is divided into several segments with dif-ferent diameters in calculation The mechanical relationshipbetween two arbitrary nodes could be represented by usingtransfermatrixmethod Take the 119868th span as the study object

It is divided into 119898 segments and 119898 + 1 node as shown inFigure 4

Figure 5 shows the forces and moments of the 119895th shaftsegment on the 119868th span From the equilibrium of forces andmoments an equation is derived by Timoshenko and Gere[18]

119876119895+1

= 119876119895

minus 119902119895119897119895

119872119895+1

= 119872119895

+ 119876119895119897119895

minus1

21199021198951198972

119895

(11)

In matrix form

119885119895+1

= [119870]119895

119885119895

(12)

where

119885119895

= 119872119895 119876119895 1119879

[119870]119895

=

[[[[

[

1 119897119895

1199021198951198972

119895

2

0 1 minus119902119895119897119895

0 0 1

]]]]

]

(13)

Mathematical Problems in Engineering 5

I Iminus 1

1

2

3 4

middot middot middot

middot middot middot

middot middot middot

middot middot middot

(1)(2)

(3)

n minus 1 n n + 1

m

(m)

m + 1

Figure 4 119868th span of the rotor

y

Mj

Qj

qjQj+1

Mj+1

z

Figure 5 Applied force of the 119895th shaft element

If there is no coupling within the 119868th span the relationshipbetween the (119868 minus 1)th bearing and the 119868th bearing can bewritten as

119885119898+1

= [119863]1198941198851

119872

119876

1

119898+1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]119894

119872

119876

1

1

(14)

Transformation matrix [119863]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]1

If there exists coupling within the 119868th span and thecoupling locates at the node 119899 the relation between the (119868 minus

1)th bearing the coupling and the 119868th bearing meets thefollowing equations

119885119899

= [119892]1198941198851

119885119898+1

= [119866]119894119885119899

(15)

where

[119892]119894

= [119870]119899minus1

[119870]119899minus2

sdot sdot sdot [119870]1

[119866]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]119899

(16)

1 2 3 4 5 6 7 8

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 6 Configuration diagram of single bearing support shafting

23 BearingMoment and Load Calculation Inmoment equi-librium equations the expression of deflection angle vector120573 has the elevation of every bearing which is unknown invalueWithmoment vector the number of unknownqualitiesis more than equations Thus boundary conditions such asbendingmoment of every coupling being zero are introducedto make equations solvable During the installation of shaft-ing at site bearing elevations are determined by using theassumption that bending moment of each coupling is zeroThis method reduces the bending moment on each couplingthus enhancing the stability and reliability of turbo-generatorset Under this assumption bending moment of each bearingis derived in a certain order by using transfer matrix methodEvery bearing elevation can be solved by plugging bearingbendingmoment vector intomoment equilibrium equationsDue to structure difference mechanical boundary conditionsof shafting in different configuration are distinct So bendingmoment calculation order is diverse Take the single bearingsupport shafting as example to illustrate bearingmoment andload calculation procedure based on transfer matrix method

The shafting of a 1000MWultra supercritical turbinewithsingle bearing support is shown in Figure 6 This shaftingis composed of a high pressure rotor (HP) intermediatepressure rotor (IP) two low pressure rotors (LP

1 LP2)

generator (GEN) and exciter rotor (EXC) High pressurerotor is supported with two bearings Intermediate pressurerotor and low pressure rotors are supported with singlebearing Generator and exciter rotor are supported with threebearings Single bearing support shafting is compact in struc-ture which decreases the entire length of shafting and reducesthe negative influence of basefoundation deformation onalignment

Without considering the influence of two overhangssimplified configuration diagram of single bearing supportshafting is shown in Figure 7 Because of intermediatepressure rotors low pressure rotor and exciter rotors aresingle bearing supported there exists shear force in couplingsC1 C2 C3 and C5 to support the rotor Since generator rotoris double bearing support shear force in coupling C4 is zero

From the 8th span spans with couplings are analyzedfirstly in reverse order In the 8th span the bending momentand shear force in bearing 8 and coupling C5 meet thefollowing relation

1198728

1198768

1

=[[

[

1 11986612

11986613

0 1 11986623

0 0 1

]]

]8

1198721198625

1198761198625

1

(17)

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

1 2 3

(3)(2)

middot middot middotmiddot middot middot K

q

N Nminus 1

(N)

Figure 1 Multisupport redundant beam system

120601998400i

120601998400998400i

MI

MIminus1

The Ith span

The (I + 1)th span

MI

MI

+1

I I

I

minus 1

I + 1

Figure 2 Deflection angles at bearings of each span

where 119872119894minus1

and 119872119894are the bending moments at the (119868 minus

1)th and 119868th bearing Flexibility coefficient 1205931015840

119894minus1119894is the left

deflection angle of the 119868th bearing caused by unit momentat the (119868 minus 1)th bearing 120593

1015840

119894119894is the left deflection angle of the

119868th bearing caused by unit moment at the 119868th bearingSimilarly the right deflection angle of the 119868th bearing

caused by bending moment 12057410158401015840

119894is given by

12057410158401015840

119894= 11987211989412059310158401015840

119894119894+ 119872119894+1

12059310158401015840

119894+1119894 (2)

Taking the influence of elevation difference between adjacentbearings and shafting weight into consideration the totaldeflection angles at the left and right side of the 119868th bearingare given respectively by

1206011015840

119894= 1205741015840

119894+ 120579119871

119894119894minus 120573119871

119894119894 (3)

12060110158401015840

119894= 12057410158401015840

119894+ 120579119877

119894+1119894+ 120573119877

119894+1119894 (4)

where 1206011015840

119894is the total deflection angle at the left side of the 119868th

bearing and 12060110158401015840

119894is the total deflection angle at the right side

of the 119868th bearing 120579119871

119894119894is the deflection angle at the left side

of the 119868th bearing caused by the weight of the 119868th span and120579119877

119894+1119894is the deflection angle at the right side of the 119868th bearing

caused by the weight of the (119868 + 1)th span 120573119871

119894119894and 120573

119877

119894+1119894are

the deflection angles caused by elevation difference at the leftand right sides of the 119868th bearing as shown in Figure 3

The left deflection angle of the 119868th bearing caused byelevation difference in the 119868th span is given by

120573119871

119894119894=

119910119894minus1

minus 119910119894

119871119894

(5)

I minus 1

I

I

+ 1

yiminus1 yi+1yiLi Li+1

120573Ri+1i

120573Lii

Figure 3 Deflection angle caused by elevation difference

where 119910119894minus1

and 119910119894are the elevations of the (119868 minus 1)th and 119868th

bearing 119871119894is the length of the 119868th span

Similarly the right deflection angle of the 119868th bearingcaused by elevation difference in the (119868 + 1)th span is given by

120573119877

119894+1119894=

119910119894

minus 119910119894+1

119871119894+1

(6)

Since shafting catenary curve is smooth when the machine isrunning total deflection angles at the left and right sides ofevery bearing are equal

1206011015840

119894= minus12060110158401015840

119894(119894 = 2 3 119873 minus 1) (7)

Thus the moment equilibrium equation can be expressed as

119872119894minus1

1205931015840

119894minus1119894+ 119872119894(1205931015840

119894119894+ 12059310158401015840

119894119894) + 119872

119894+112059310158401015840

119894+1119894+ 120579119871

119894119894+ 120579119877

119894+1119894

minus (120573119871

119894119894minus 120573119877

119894+1119894) = 0

(8)

4 Mathematical Problems in Engineering

This equation for bearings is ranked from 2nd to (119873 minus 1)thand can be written in matrix form

[[[[[[[[[[[[[[

[

1205931015840

22+ 12059310158401015840

2212059310158401015840

32

1205931015840

231205931015840

33+ 12059310158401015840

3312059310158401015840

43

d d d

1205931015840

119894minus11198941205931015840

119894119894+ 12059310158401015840

11989411989412059310158401015840

119894+1119894

d d d

1205931015840

(119899minus3)(119899minus2)1205931015840

(119899minus2)(119899minus2)+ 12059310158401015840

(119899minus2)(119899minus2)12059310158401015840

(119899minus1)(119899minus2)

1205931015840

(119899minus2)(119899minus1)1205931015840

(119899minus1)(119899minus1)+ 12059310158401015840

(119899minus1)(119899minus1)

]]]]]]]]]]]]]]

]

1198722

1198723

119872119894

119872119899minus2

119872119899minus1

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(9)

and simplified form

[120593] sdot 119872 = minus 120573 (10)

where [120593] is the flexibility coefficient matrix 119872 is themoment vector and 120573 is the deflection angle vector

Moment equilibrium equations express the relationshipbetween moments deflection angles and elevations at everybearing which lay the foundation of shafting alignmentcomputation However only moment equilibrium equationsare not enough for calculating the bearing elevations andloads since the number of unknown quantities exceeds thenumber of equationsThus every bearing moment and othermechanical boundary conditions should be calculated firstlyby using transfer matrix method

22 Transfer Matrix Method The transfer matrix method isa fast and easy way to get the parameters such as momentshear force and deflection which are important in shaftingalignment calculation This method could guarantee highcomputational accuracy and save time Although finite ele-mentmethod is versatile it takesmore time to build geometrymodel and requires more computational resources This isthe reason for using transfer matrix method in shaftingalignment calculation

Usually shafting is divided into several segments with dif-ferent diameters in calculation The mechanical relationshipbetween two arbitrary nodes could be represented by usingtransfermatrixmethod Take the 119868th span as the study object

It is divided into 119898 segments and 119898 + 1 node as shown inFigure 4

Figure 5 shows the forces and moments of the 119895th shaftsegment on the 119868th span From the equilibrium of forces andmoments an equation is derived by Timoshenko and Gere[18]

119876119895+1

= 119876119895

minus 119902119895119897119895

119872119895+1

= 119872119895

+ 119876119895119897119895

minus1

21199021198951198972

119895

(11)

In matrix form

119885119895+1

= [119870]119895

119885119895

(12)

where

119885119895

= 119872119895 119876119895 1119879

[119870]119895

=

[[[[

[

1 119897119895

1199021198951198972

119895

2

0 1 minus119902119895119897119895

0 0 1

]]]]

]

(13)

Mathematical Problems in Engineering 5

I Iminus 1

1

2

3 4

middot middot middot

middot middot middot

middot middot middot

middot middot middot

(1)(2)

(3)

n minus 1 n n + 1

m

(m)

m + 1

Figure 4 119868th span of the rotor

y

Mj

Qj

qjQj+1

Mj+1

z

Figure 5 Applied force of the 119895th shaft element

If there is no coupling within the 119868th span the relationshipbetween the (119868 minus 1)th bearing and the 119868th bearing can bewritten as

119885119898+1

= [119863]1198941198851

119872

119876

1

119898+1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]119894

119872

119876

1

1

(14)

Transformation matrix [119863]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]1

If there exists coupling within the 119868th span and thecoupling locates at the node 119899 the relation between the (119868 minus

1)th bearing the coupling and the 119868th bearing meets thefollowing equations

119885119899

= [119892]1198941198851

119885119898+1

= [119866]119894119885119899

(15)

where

[119892]119894

= [119870]119899minus1

[119870]119899minus2

sdot sdot sdot [119870]1

[119866]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]119899

(16)

1 2 3 4 5 6 7 8

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 6 Configuration diagram of single bearing support shafting

23 BearingMoment and Load Calculation Inmoment equi-librium equations the expression of deflection angle vector120573 has the elevation of every bearing which is unknown invalueWithmoment vector the number of unknownqualitiesis more than equations Thus boundary conditions such asbendingmoment of every coupling being zero are introducedto make equations solvable During the installation of shaft-ing at site bearing elevations are determined by using theassumption that bending moment of each coupling is zeroThis method reduces the bending moment on each couplingthus enhancing the stability and reliability of turbo-generatorset Under this assumption bending moment of each bearingis derived in a certain order by using transfer matrix methodEvery bearing elevation can be solved by plugging bearingbendingmoment vector intomoment equilibrium equationsDue to structure difference mechanical boundary conditionsof shafting in different configuration are distinct So bendingmoment calculation order is diverse Take the single bearingsupport shafting as example to illustrate bearingmoment andload calculation procedure based on transfer matrix method

The shafting of a 1000MWultra supercritical turbinewithsingle bearing support is shown in Figure 6 This shaftingis composed of a high pressure rotor (HP) intermediatepressure rotor (IP) two low pressure rotors (LP

1 LP2)

generator (GEN) and exciter rotor (EXC) High pressurerotor is supported with two bearings Intermediate pressurerotor and low pressure rotors are supported with singlebearing Generator and exciter rotor are supported with threebearings Single bearing support shafting is compact in struc-ture which decreases the entire length of shafting and reducesthe negative influence of basefoundation deformation onalignment

Without considering the influence of two overhangssimplified configuration diagram of single bearing supportshafting is shown in Figure 7 Because of intermediatepressure rotors low pressure rotor and exciter rotors aresingle bearing supported there exists shear force in couplingsC1 C2 C3 and C5 to support the rotor Since generator rotoris double bearing support shear force in coupling C4 is zero

From the 8th span spans with couplings are analyzedfirstly in reverse order In the 8th span the bending momentand shear force in bearing 8 and coupling C5 meet thefollowing relation

1198728

1198768

1

=[[

[

1 11986612

11986613

0 1 11986623

0 0 1

]]

]8

1198721198625

1198761198625

1

(17)

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

This equation for bearings is ranked from 2nd to (119873 minus 1)thand can be written in matrix form

[[[[[[[[[[[[[[

[

1205931015840

22+ 12059310158401015840

2212059310158401015840

32

1205931015840

231205931015840

33+ 12059310158401015840

3312059310158401015840

43

d d d

1205931015840

119894minus11198941205931015840

119894119894+ 12059310158401015840

11989411989412059310158401015840

119894+1119894

d d d

1205931015840

(119899minus3)(119899minus2)1205931015840

(119899minus2)(119899minus2)+ 12059310158401015840

(119899minus2)(119899minus2)12059310158401015840

(119899minus1)(119899minus2)

1205931015840

(119899minus2)(119899minus1)1205931015840

(119899minus1)(119899minus1)+ 12059310158401015840

(119899minus1)(119899minus1)

]]]]]]]]]]]]]]

]

1198722

1198723

119872119894

119872119899minus2

119872119899minus1

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(9)

and simplified form

[120593] sdot 119872 = minus 120573 (10)

where [120593] is the flexibility coefficient matrix 119872 is themoment vector and 120573 is the deflection angle vector

Moment equilibrium equations express the relationshipbetween moments deflection angles and elevations at everybearing which lay the foundation of shafting alignmentcomputation However only moment equilibrium equationsare not enough for calculating the bearing elevations andloads since the number of unknown quantities exceeds thenumber of equationsThus every bearing moment and othermechanical boundary conditions should be calculated firstlyby using transfer matrix method

22 Transfer Matrix Method The transfer matrix method isa fast and easy way to get the parameters such as momentshear force and deflection which are important in shaftingalignment calculation This method could guarantee highcomputational accuracy and save time Although finite ele-mentmethod is versatile it takesmore time to build geometrymodel and requires more computational resources This isthe reason for using transfer matrix method in shaftingalignment calculation

Usually shafting is divided into several segments with dif-ferent diameters in calculation The mechanical relationshipbetween two arbitrary nodes could be represented by usingtransfermatrixmethod Take the 119868th span as the study object

It is divided into 119898 segments and 119898 + 1 node as shown inFigure 4

Figure 5 shows the forces and moments of the 119895th shaftsegment on the 119868th span From the equilibrium of forces andmoments an equation is derived by Timoshenko and Gere[18]

119876119895+1

= 119876119895

minus 119902119895119897119895

119872119895+1

= 119872119895

+ 119876119895119897119895

minus1

21199021198951198972

119895

(11)

In matrix form

119885119895+1

= [119870]119895

119885119895

(12)

where

119885119895

= 119872119895 119876119895 1119879

[119870]119895

=

[[[[

[

1 119897119895

1199021198951198972

119895

2

0 1 minus119902119895119897119895

0 0 1

]]]]

]

(13)

Mathematical Problems in Engineering 5

I Iminus 1

1

2

3 4

middot middot middot

middot middot middot

middot middot middot

middot middot middot

(1)(2)

(3)

n minus 1 n n + 1

m

(m)

m + 1

Figure 4 119868th span of the rotor

y

Mj

Qj

qjQj+1

Mj+1

z

Figure 5 Applied force of the 119895th shaft element

If there is no coupling within the 119868th span the relationshipbetween the (119868 minus 1)th bearing and the 119868th bearing can bewritten as

119885119898+1

= [119863]1198941198851

119872

119876

1

119898+1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]119894

119872

119876

1

1

(14)

Transformation matrix [119863]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]1

If there exists coupling within the 119868th span and thecoupling locates at the node 119899 the relation between the (119868 minus

1)th bearing the coupling and the 119868th bearing meets thefollowing equations

119885119899

= [119892]1198941198851

119885119898+1

= [119866]119894119885119899

(15)

where

[119892]119894

= [119870]119899minus1

[119870]119899minus2

sdot sdot sdot [119870]1

[119866]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]119899

(16)

1 2 3 4 5 6 7 8

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 6 Configuration diagram of single bearing support shafting

23 BearingMoment and Load Calculation Inmoment equi-librium equations the expression of deflection angle vector120573 has the elevation of every bearing which is unknown invalueWithmoment vector the number of unknownqualitiesis more than equations Thus boundary conditions such asbendingmoment of every coupling being zero are introducedto make equations solvable During the installation of shaft-ing at site bearing elevations are determined by using theassumption that bending moment of each coupling is zeroThis method reduces the bending moment on each couplingthus enhancing the stability and reliability of turbo-generatorset Under this assumption bending moment of each bearingis derived in a certain order by using transfer matrix methodEvery bearing elevation can be solved by plugging bearingbendingmoment vector intomoment equilibrium equationsDue to structure difference mechanical boundary conditionsof shafting in different configuration are distinct So bendingmoment calculation order is diverse Take the single bearingsupport shafting as example to illustrate bearingmoment andload calculation procedure based on transfer matrix method

The shafting of a 1000MWultra supercritical turbinewithsingle bearing support is shown in Figure 6 This shaftingis composed of a high pressure rotor (HP) intermediatepressure rotor (IP) two low pressure rotors (LP

1 LP2)

generator (GEN) and exciter rotor (EXC) High pressurerotor is supported with two bearings Intermediate pressurerotor and low pressure rotors are supported with singlebearing Generator and exciter rotor are supported with threebearings Single bearing support shafting is compact in struc-ture which decreases the entire length of shafting and reducesthe negative influence of basefoundation deformation onalignment

Without considering the influence of two overhangssimplified configuration diagram of single bearing supportshafting is shown in Figure 7 Because of intermediatepressure rotors low pressure rotor and exciter rotors aresingle bearing supported there exists shear force in couplingsC1 C2 C3 and C5 to support the rotor Since generator rotoris double bearing support shear force in coupling C4 is zero

From the 8th span spans with couplings are analyzedfirstly in reverse order In the 8th span the bending momentand shear force in bearing 8 and coupling C5 meet thefollowing relation

1198728

1198768

1

=[[

[

1 11986612

11986613

0 1 11986623

0 0 1

]]

]8

1198721198625

1198761198625

1

(17)

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

I Iminus 1

1

2

3 4

middot middot middot

middot middot middot

middot middot middot

middot middot middot

(1)(2)

(3)

n minus 1 n n + 1

m

(m)

m + 1

Figure 4 119868th span of the rotor

y

Mj

Qj

qjQj+1

Mj+1

z

Figure 5 Applied force of the 119895th shaft element

If there is no coupling within the 119868th span the relationshipbetween the (119868 minus 1)th bearing and the 119868th bearing can bewritten as

119885119898+1

= [119863]1198941198851

119872

119876

1

119898+1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]119894

119872

119876

1

1

(14)

Transformation matrix [119863]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]1

If there exists coupling within the 119868th span and thecoupling locates at the node 119899 the relation between the (119868 minus

1)th bearing the coupling and the 119868th bearing meets thefollowing equations

119885119899

= [119892]1198941198851

119885119898+1

= [119866]119894119885119899

(15)

where

[119892]119894

= [119870]119899minus1

[119870]119899minus2

sdot sdot sdot [119870]1

[119866]119894

= [119870]119898

[119870]119898minus1

sdot sdot sdot [119870]119899

(16)

1 2 3 4 5 6 7 8

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 6 Configuration diagram of single bearing support shafting

23 BearingMoment and Load Calculation Inmoment equi-librium equations the expression of deflection angle vector120573 has the elevation of every bearing which is unknown invalueWithmoment vector the number of unknownqualitiesis more than equations Thus boundary conditions such asbendingmoment of every coupling being zero are introducedto make equations solvable During the installation of shaft-ing at site bearing elevations are determined by using theassumption that bending moment of each coupling is zeroThis method reduces the bending moment on each couplingthus enhancing the stability and reliability of turbo-generatorset Under this assumption bending moment of each bearingis derived in a certain order by using transfer matrix methodEvery bearing elevation can be solved by plugging bearingbendingmoment vector intomoment equilibrium equationsDue to structure difference mechanical boundary conditionsof shafting in different configuration are distinct So bendingmoment calculation order is diverse Take the single bearingsupport shafting as example to illustrate bearingmoment andload calculation procedure based on transfer matrix method

The shafting of a 1000MWultra supercritical turbinewithsingle bearing support is shown in Figure 6 This shaftingis composed of a high pressure rotor (HP) intermediatepressure rotor (IP) two low pressure rotors (LP

1 LP2)

generator (GEN) and exciter rotor (EXC) High pressurerotor is supported with two bearings Intermediate pressurerotor and low pressure rotors are supported with singlebearing Generator and exciter rotor are supported with threebearings Single bearing support shafting is compact in struc-ture which decreases the entire length of shafting and reducesthe negative influence of basefoundation deformation onalignment

Without considering the influence of two overhangssimplified configuration diagram of single bearing supportshafting is shown in Figure 7 Because of intermediatepressure rotors low pressure rotor and exciter rotors aresingle bearing supported there exists shear force in couplingsC1 C2 C3 and C5 to support the rotor Since generator rotoris double bearing support shear force in coupling C4 is zero

From the 8th span spans with couplings are analyzedfirstly in reverse order In the 8th span the bending momentand shear force in bearing 8 and coupling C5 meet thefollowing relation

1198728

1198768

1

=[[

[

1 11986612

11986613

0 1 11986623

0 0 1

]]

]8

1198721198625

1198761198625

1

(17)

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

M1 = 0

1 2 3 4 5 6 7 8

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span

M8 = 0

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 ne 0

MC3 = 0

QC3 ne 0

Figure 7 Simplified configuration diagram of single bearing support shafting

With the boundary condition that 1198721198625

= 0 and 1198728

= 0 theexpression of shear force in coupling C5 and bearing 8 can beobtained

1198761198625

= minus1198728

+ 1198668

13

1198668

12

= minus1198668

13

1198668

12

11986588

= 1198768

= 1198761198625

+ 1198668

23

(18)

When calculating bearing load shear force in the right andleft sides of bearing should be ensured firstly It should benoted that when analyzing the 119868th span and the (119868 + 1)thspan there exist two different shear forces 119876

119894in bearing 119868

To distinguish these two different shear forces we define 119865119894119895

to express the shear forces in bearing 119868 when analyzing the119895th span

The bending moment and shear force in bearing 7 andcoupling C5 meet the following equation

1198721198625

1198761198625

1

=[[

[

1 11989212

11989213

0 1 11989223

0 0 1

]]

]8

1198727

1198767

1

(19)

Substituting 1198721198625

= 0 and (18) into (19) moment and shearforce in bearing 7 are obtained

11986578

= 1198767

= 1198761198625

minus 1198928

23

1198727

= 1198721198625

minus 1198928

121198767

minus 1198928

13

(20)

Applying the same method to the 6th 5th 4th and 3rd spanbendingmoments and shear forces in bearings 6 5 4 3 and 2are obtained successively So far the unknownmoment vector119872 in equilibrium equations has been determined Value ofevery bearing elevation can be solved by plugging 119872 intoequations

There is no coupling with the 2nd and 7th span We takethe 7th span as example to illustrate the computing methodof shear force on bearing Bending moments and shear forcesin bearing 6 and bearing 7 meet the following equation

1198727

1198767

1

=[[

[

1 11986312

11986313

0 1 11986323

0 0 1

]]

]7

1198726

1198766

1

(21)

I Ri

Fii Fii+1

Figure 8 Applied force in bearing

1198726and 119872

7have been solved in the analysis of spans

with couplings Substituting them into (3) shear forces areobtained

11986567

= 1198766

=1198727

minus 1198726

minus 1198637

13

1198637

12

11986577

= 1198767

= 1198766

+ 1198637

23

(22)

Figure 8 shows the shearing forces and bearing reaction forceof bearing 119868 According to force balance the bearing reactionforce is given by

119877119894

= 119865119894119894+1

minus 119865119894119894

(23)

3 Calculation of Results and Analysis

31 Zero Bending Moment in Couplings Based on the abovemathematical modeling and derivation work shafting align-ment program is developed to calculate bearing elevationsand loads of shafting system with different structure underthe condition that the bending moment value of eachcoupling is zero Bearing elevations and loads of singlesupport shafting and shafting with 3S clutch are calculatedrespectively and comparedwith the installation data to verifythe accuracy and validity of the computingmethod proposedabove

The first step of shafting alignment calculation is to divideeach rotor into discrete elements Bending stiffnessmass and

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

1

1 2 3

63 154 214

4 5 6 7 8

3273293112812743

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

7 8

32327311281274214154633

1

1 2 3 4 5 6C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 9 The schematic diagram of single support shafting with mass and stiffness diameter of the 1000MW turbine generator shafting

Table 1 Bearing elevations and loads of single support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 7378 7345 045 624592 5553 5367 347 1877113 4075 4147 minus174 4852214 1780 1840 minus326 7173655 0 0 3384376 0 0 3370547 7394 7397 minus004 3284768 11155 11130 022 6338

length of shafting should remain unchanged before and aftermodelingThemass diameter represents themass of each ele-ment which could be obtained from mass conservation Thebending stiffness diameter represents the bending stiffness ofeach element The bending stiffness will be impaired wherethe cross section of rotor abruptly changes 45∘ method andstrain energy method have been widely used in engineeringto calculate the stiffness diameter of rotor Figure 9 shows theschematic diagram of the single bearing support 1000MWturbine generator shafting with both mass diameter andbending stiffness diameter The shafting is divided into 328segments and has 329 nodes

Giving the elevations of the 5th and 6th bearing as 0mmbearing elevations and loads distribution of single bearingsupport shafting are calculated and compared with installa-tion data measured on site The calculation and comparisonresults are shown in Table 1 The maximum relative deviationis 347 at the 2nd bearing

Figure 10 shows the catenary curve calculation resultof single bearing support shafting Section bending stresscalculation result under corresponding elevations is shownin Figure 11 Bending stress of each coupling is 0MPa whichmeets the installation requirement that bending momentof every coupling is zero The maximum bending stress is2668MPa at the end of generator rotor

The configuration diagram of shafting with synchro-self-shifting clutch is shown in Figure 12 High pressure rotor andlow pressure rotors are supported by two bearings Interme-diate pressure rotor is supported by one bearing Generatorrotor and exciter rotor are three-bearing supported Bendingmoment and shearing force of each coupling are shown inFigure 13 The bearing moment and load computing methodof single bearing support shafting applies to shafting with 3Sclutch as well

Figure 14 is the schematic diagram of shafting with 3Sclutch The shafting is divided into 331 segments and has

12

10

8

6

4

2

2

4

3

1

5 6

7

8

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

Figure 10 Catenary curve of single support shafting of the 1000MWturbine generator

332 nodes Giving the elevations of the 7th and 8th bearingas 0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data from the company The calculation andcomparison results are shown in Table 2 The maximumrelative deviation is minus614 at the 1st bearing

Figure 15 shows the catenary curve calculation result ofshafting with 3S clutch Section bending stress calculationresult under corresponding elevations is shown in Figure 16Bending stress of each coupling is 0MPa which meetsthe installation requirement that bending moment of everycoupling is zero as well The maximum bending stress is2682MPa at the end of generator rotorThe elevation relativeerror of both shafting structures is within acceptable rangewhich verifies the validity and accuracy of shafting alignmentcomputing method proposed above

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Table 2 Bearing elevations and loads of shafting with 3S clutch

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 15375 16380 minus614 604182 10536 10700 minus153 2216233 5556 5678 minus215 1394884 4966 5078 minus221 3691765 1614 1661 minus283 3202206 1303 1343 minus298 3691767 0 0 3202208 0 0 3387719 7431 7373 079 33015010 11211 11100 100 6371

30

20

10

0

minus10

minus20

0 10 20 30 40

Axial length (m)

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1 C2 C3 C4 C5

Figure 11 Section bending stress of single bearing support shaftingof the 1000MW turbine generator

Synchro-self-shifting clutch

1 2 3 4 5 6 7 8 9 10

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 12 Configuration diagram of shafting with synchro-self-shifting clutch

32 Gap and Sag in the Coupling In consideration of variousfactors which influence the rotor position in thermal con-dition such as thermal distortion fluctuation of bearing oilfilm thickness and unequal settlement of foundation thereexist preset gap and sag in rigid coupling in cold conditionalignment to compensate for the position change in thermalcondition The gap and sag of coupling are parameters todescribe shafting alignment condition in terms of angularityand offset in the horizontal and vertical view Angularity canbe expressed directly as the angle difference between twocoupling flanges The coupling gap itself has no alignmentmeaning it must be divided by the coupling flange diameter

to describe the angle difference between coupling flangesOffset describes the distance between rotation axes Foreach rotor the catenary bow caused by weight is negligibleand therefore the rotation axis is not straight Couplingsag actually represents the elevation difference between twocoupling flanges

Since there exist preset gap and sag shafting catenarycurve is no longer continuous and smooth Thus momentequilibrium equation does not apply to the calculation ofbearing elevation and load under this installation require-ment For a rotor with single support an auxiliary supportis added in the front-end of rotor as temporary support ininstallation Every rotor in shafting becomes double bearingsupport through adding auxiliary bearing in installation

The alignment procedure under this installation require-ment will be described in the following passage and anexample of shafting with six bearings and three rotorsas shown in Figure 17 will then illustrate this computingmethod

Firstly when the elevation of each bearing is zero thecatenary curve of each double bearing support rotor is cal-culated respectively Before connection bearing elevationsof adjacent rotors are adjusted to make couplings fulfillgeometry boundary condition caused by preset sag and gapThe geometry relationship between two coupling flanges canbe written as

1199101

minus 1199102

= Δ119884

1205791

minus 1205792

=Δ119866

119863

(24)

where Δ119866 = 119866119887 minus 119866119905The elevation of every bearing including auxiliary bearing

is ensured in turnorder according to the geometrical rela-tionship namely deflection and angle between two couplingflanges Tighten bolts of each coupling to connect shafts afteradjusting every bearing to assigned altitude along the verticaldirection

Figure 18 shows the schematic diagram of the doublebearing support 600MW turbine generator shafting Sincethe exciter rotor is single support structure auxiliary support

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

M1 = 0

1 2 3 4 5 6 7 8 9 10

MC4 = 0 MC5 = 0

QC4 = 0 QC5 ne 0

2nd span 3rd span 4th span 5th span 6th span 7th span 8th span 9th span 10th span

MC1 = 0

QC1 ne 0

MC2 = 0

QC2 = 0

MC3 = 0

QC3 = 0

M10 = 0

Figure 13 Simplified configuration diagram of shafting with synchro-self-shifting clutch

1

1 2 3 4

63 155159 216

65 7 8 9 10

330332

3142842772203

C1 C2 C3 C4 C5

(a) Rotor system with mass diameter

2 3 4

63 155159

65 7 8 9 10

33033

314284277216 2201

1

3

C1 C2 C3 C4 C5

(b) Rotor system with bending stiffness diameter

Figure 14 The schematic diagram of shafting with 3S clutch with mass and stiffness diameter

10

8

6

4

2

0

1

2

34

5 67 8

9

10

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

12

14

16

Figure 15 Catenary curve of shafting with 3S clutch

Table 3 Preset sag and gap of each coupling

Coupling number Preset sag valuemm Preset gap valuemm1198621 015 011198622 0 01198623 015 01198624 015 01198625 0 01

will be added in the front-endThe preset sag and gap of eachcoupling are shown in Table 3

Giving the elevations of the 8th and 9th bearing as0mm bearing elevations and loads distribution of singlebearing support shafting are calculated and compared withinstallation data The calculation and comparison results areshown in Table 4Themaximum relative deviation is minus648at the 7th bearing

Figure 19 shows the catenary curve calculation resultof double bearing support shafting Section bending stresscalculation result under corresponding elevations is shown

30

25

20

15

10

5

0

minus10

minus5

minus15

Sect

ion

bend

ing

stres

s (M

Pa)

C1 C2 C3 C4 C5

Max

0 10 20 30 40

Axial length (m)

Figure 16 Section bending stress of shafting with 3S clutch

in Figure 20 As shown in Table 3 preset sag of couplingsC1 and C2 is 0mm bending stress value of C1 and C2approaches zero as well The maximum value of bendingstress is 2670MPa at the end of generator rotor Differentfrom previous analysis about single bearing support shaftingor shafting with 3S clutch bending moment stress at the1st bearing varied intensely The maximal bending stress is2415MPa

4 Summary and Conclusions

This paper has established the mathematical model of shaft-ing alignment and computing methods of bearing elevationsand loads under corresponding installation requirementswhere bending moment of each coupling is zero and thereexist preset sag and gap in every coupling are proposedrespectively Based on the above work shafting alignmentprogram aimed at bearing elevation and load calculationis developed Bearing elevations and loads of shafting with

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Table 4 Bearing elevations and loads of double bearing support shafting

Bearing number Computed bearingelevationmm

Installation bearingelevationmm

Elevation relativeerror

Computed bearingloadN

1 22667 22880 minus093 751502 14819 15190 minus244 1476003 13893 14280 minus271 483504 8220 8551 minus387 1533005 7400 7703 minus393 3701006 2500 2579 minus306 3374007 1948 2083 minus648 3375008 0 0 2958009 0 0 36050010 7601 7754 minus197 32820011 11510 11580 minus060 6251

1

2 3 4 5

C1

ΔY

y1

y2

1205791

1205792

6C2

D

Gb

(a) Shafting composed of double bearing support rotors

(b) Preset sag and gap

Gt

Figure 17 Sketch of computing method with sag and gap in coupling

1 2 3 4 5 6 7 8 9 10 11

HP IP LP1 LP2 GEN EXC

C1 C2 C3 C4 C5

Figure 18 Configuration diagram of double bearing support shaft-ing

different structure under specific installation requirementsare calculated by using this alignment calculation programCompared with installation data measured on site the valid-ity and accuracy of proposed model and computing methodhave been verified Computing method presented in thisstudy provides a comprehensive and reliable approach foranalyzing shafting alignment and guiding the installation onsite

Appendix

The deflection angle vector 120573 is composed of two parts

120573

= minus

120579119871

22+ 120579119877

32minus (120573119871

22minus 120573119877

32)

120579119871

33+ 120579119877

43minus (120573119871

33minus 120573119877

43)

120579119871

119894119894+ 120579119877

119894+1119894minus (120573119871

119894119894minus 120573119877

119894+1119894)

120579119871

(119899minus2)(119899minus2)+ 120579119877

(119899minus1)(119899minus2)minus (120573119871

(119899minus2)(119899minus2)minus 120573119877

(119899minus1)(119899minus2))

120579119871

(119899minus1)(119899minus1)+ 120579119877

119899(119899minus1)minus (120573119871

(119899minus1)(119899minus1)minus 120573119877

119899(119899minus1))

(A1)

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

The first part is the deflection angle caused by weight 120579119871

119894119894+

120579119877

119894+1119894 the mathematical derivation is presented in Figure 21The relationship of force moment bending angle and

deflection between adjacent nodes is

119876119895+1

= 119876119895

119873119895+1

= 119873119895

+ 119876119895119897119895

120579119895+1

= 120579119895

minus

119873119895119897119895

119864119868119895

minus

1198761198951198972

119895

2119864119868119895

119884119895+1

= 119884119895

minus 120579119895119897119895

+

1198731198951198972

119895

2119864119868119895

+

1198761198951198973

119895

6119864119868119895

(A2)

In matrix form

119885119895+1

= [1198651015840]119895

119885119895

(A3)

where

119885119898+1

=

119884

120579

119873

119876

1

119898+1

[1198651015840]119895

=

[[[[[[[[[[[[[

[

1 minus119897119895

1198972

119895

2119864119868119895

1198973

119895

6119864119868119895

minus1

24

1199021198951198974

119895

119864119868119895

0 1 minus

119897119895

119864119868119895

minus

1198972

119895

2119864119868119895

1199021198951198973

119895

6119864119868119895

0 0 1 119897119895

minus1

21199021198951198972

119895

0 0 0 1 minus119902119895119897119895

0 0 0 0 1

]]]]]]]]]]]]]

]119895

(A4)

In this span which is divided into 119898 segments and 119898 + 1

nodes the relationship between node 119898 + 1 and node 1 is

119885119898+1

= [119865] 1198851

(A5)

where

[119865] = [1198651015840]119898

[1198651015840]119898minus1

sdot sdot sdot [1198651015840]2

[1198651015840]1

(A6)

In matrix form

119884

120579

119873

119876

1

119898+1

=

[[[[[[[[

[

1 11989112

11989113

11989114

11989115

0 1 11989123

11989124

11989125

0 0 1 11989134

11989135

0 0 0 1 11989145

0 0 0 0 1

]]]]]]]]

]

119884

120579

119873

119876

1

1

(A7)

Substituting the boundary condition in this case

119884119898+1

= 1198841

= 0

119873119898+1

= 1198731

= 0

(A8)

25

20

15

10

5

0

Defl

ectio

n (m

m)

0 10 20 30 40

Axial length (m)

1

2

3

45

67

8 9

10

11

Figure 19 Catenary curve of double bearing support shafting

30

25

20

15

10

5

0

minus10

minus5

Sect

ion

bend

ing

stres

s (M

Pa)

Max

C1

C2

C3C4 C5

0 10 20 30 40

Axial length (m)

Figure 20 Section bending stress of double bearing support shaft-ing

Shear force and deflection angle caused by weight could beobtained

1205791

=11989114

11989135

minus 11989115

11989134

11989112

11989134

120579119898+1

=11989114

11989135

minus 11989115

11989134

minus 11989124

11989135

11989112

+ 11989125

11989112

11989134

11989112

11989134

1198761

= minus11989135

11989134

119876119898+1

=11989134

11989145

minus 11989135

11989134

(A9)

This part caused by weight in deflection angle vector could bepresented as

120579119877

119894119894minus1= 1205791

120579119871

119894119894= 120579119898+1

(A10)

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Mathematical Problems in Engineering

y

x

Qj

Njq

120579j

Nj+1

Qj+1

Figure 21 Applied force of shaft element

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the National Natural ScienceFoundation of China (no 11372234)

References

[1] S R Bognatz ldquoAlignment of critical and noncritical machinesrdquoOrbit vol 4 pp 23ndash25 1995

[2] C B Gibbons ldquoCoupling misalignment forcesrdquo in Proceedingsof the 5th Turbomachinery Symposium pp 12ndash14 Turbomachin-ery Laboratory Texas AampM University College Station TexUSA October 1976

[3] T H Patel and A K Darpe ldquoExperimental investigations onvibration response of misaligned rotorsrdquo Mechanical Systemsand Signal Processing vol 23 no 7 pp 2236ndash2252 2009

[4] A S Sekhar and B S Prabhu ldquoEffects of couplingmisalignmenton vibrations of rotating machineryrdquo Journal of Sound andVibration vol 185 no 4 pp 655ndash671 1995

[5] M Xu and R D Marangoni ldquoVibration analysis of a motor-flexible coupling-rotor system subject to misalignment andunbalance part I theoretical model and analysisrdquo Journal ofSound and Vibration vol 176 no 5 pp 663ndash679 1994

[6] D L Dewell and L D Mitchell ldquoDetection of a misaligneddisk coupling using spectrum analysisrdquo Journal of VibrationAcoustics Stress and Reliability in Design vol 106 no 1 pp 9ndash16 1984

[7] K M Al-Hussain and I Redmond ldquoDynamic response of tworotors connected by rigid mechanical coupling with parallelmisalignmentrdquo Journal of Sound and Vibration vol 249 no 3pp 483ndash498 2002

[8] R B Grant Shaft Alignment Methods with Strain Gages andLoad Cells Diehl and Lundgaard 1978

[9] AW Forrest Jr and R F Labasky ldquoShaft alignment using straingagesrdquoMarine Technology vol 18 no 3 pp 276ndash284 1981

[10] J Piotrowski Shaft Alignment Handbook CRC Press 2006[11] N Michael P Needham and R Horrell Couplings and Shaft

Alignment Wiley 1991[12] S Perry ldquoAlignment Monitoring and Correction of a Turbine

Driven Feed Water Pumprdquo 2001 httpwwwalignmentcomcasestudypermalignmonitorpdf

[13] E Brommundt and E Kramer ldquoInstability and self-excitationcaused by a gear coupling in a simple rotor systemrdquo Forschungim Ingenieurwesen vol 70 no 1 pp 25ndash37 2005

[14] A Luedeking ldquoLaser alignment verificationmdashon site at thelargest gas turbine in the USrdquo 2005 httpwwwludecacomcasestudyultra uptime0905pdf

[15] H R Simmons A J Smalley RW Frischmuth et al ldquoTools fordiagnosing case deflections and alignment on a power utilitycombustion turbinerdquo in Proceedings of the ASME 1992 Inter-national Gas Turbine and Aeroengine Congress and ExpositionV004T10A004 American Society of Mechanical EngineersCologne Germany June 1992

[16] T-T Liao ldquoModeling and analysis of laser shaft alignmentusing 4 times 4 homogeneous coordinate transformation matrixrdquoMeasurement vol 42 no 1 pp 157ndash163 2009

[17] A He Q Ge and D Li ldquoAdvanced design for shaft system oflarge turbinerdquo Journal of Vibration Engineering vol 17 pp 231ndash233 2004

[18] S Timoshenko and J Gere Mechanics of Materials Van Nos-trand Reinhold Company 1972

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of