GUIDELINES ON SHAFTING ALIGNMENT phần 2 docx

Modelling of Shafting for Alignment Calculation 3.1 Number of Engine Bearings in Model Since the emphasis in traditional shafting alignment calculations has been placed on avoiding edge

3 Modelling of Shafting for Alignment Calculation

3.1 Number of Engine Bearings in Model

Since the emphasis in traditional shafting alignment calculations has been placed on avoiding edge loading in sterntube bearings, little attention has been paid to main engine bearings Figure 3.1 shows the number of engine bearings taken into account in the alignment calculations with respect to the number of engine cylinders

in the actual practice of various shipyards As can be seen from Fig 3.1, in some cases only the aftmost three M/E bearings were incorporated into the calculation model, although this varies from yard to yard

Fig 3.1 The numbers of main bearings taken into account in current alignment calculations by builders

Number of engine cylinders

However, the results of analysis suggests that the number of M/E bearings in the calculation model has appreciable effects on the calculated reactions of the aftmost three M/E bearings, while no noticeable effects can

be seen on the sterntube and intermediate bearing reactions, as shown in Fig 3.2, when the number of bearings is changed from three to nine for a seven cylinder engine In particular, the reactions of the second and third aftmost M/E bearings will change remarkably when the number of M/E bearing changes from three to four Supposing the reactions are true when nine (full number) M/E bearings are taken into account, the reaction of the second aftmost M/E bearing is overestimated, while the reaction of the third aftmost M/E bearing is underestimated when only three M/E bearings are incorporated into the calculation model When the number

of M/E bearings is increased to four, the result is reversed, namely, the reaction of the second aftmost M/E bearing is underestimated, while the reaction of the third aftmost M/E bearing is overestimated The result is close to the true value when the number of M/E bearings is increased to five or more

Although the above result was based on a seven-cylinder engine, the result is virtually independent of the number of cylinders, although it was used as an example here That is to say that when the reaction of a M/E bearing needs to be accurately evaluated, two more M/E bearings beyond the bearing have to be incorporated in the calculation model For example, if the reactions of five aftmost M/E bearings of a nine-cylinder engine

For a four-cylinder engine, if the reactions of the five aftmost M/E bearings have to be calculated precisely, then all six M/E bearings need to be taken into consideration Since the aftmost three M/E bearings are most likely

to be affected by the alignment, at least five M/E bearings must be used in the alignment calculation

Fig 3.2 Effect of numbers of main bearings taken into account in alignment calculations on bearing reactions

Bearing number

Three Four Five Six Seven Eight Nine

3.2 Equivalent Circular Bar Representing Crankshaft

3.2.1 Necessity of Determining Equivalent Circular Bar Representing the Crankshaft

It is necessary to replace the crankshaft with a circular bar in the alignment calculations However, a suitable method for determining the equivalent circular bar of a crankshaft has still not yet been clearly established, and

in some cases, the crankshaft is simply replaced by a circular bar with a diameter equal to that of the crank journal However, it is commonly considered that the bending stiffness of a crankshaft, especially in the case of long-stroke crankshafts, is generally lower than that of a circular bar with a diameter equal to the crankshaft due

to the effect of webs, although it will vary slightly with crankshaft position In order to prevent M/E bearings from alignment related damage, it is necessary to calculate the reactions of the M/E bearings accurately Therefore, a circular bar equivalent in bending stiffness to the crankshaft has to be determined In one example considered here, the diameter of the equivalent circular bar of a long-stroke crankshaft is determined as being about 60% of the journal diameter Figure 3.3 shows a comparison of the calculated aftmost three bearing reactions between cases in which a circular bar represents the crankshaft with 100% and 60% of the journal diameter, respectively

As can been seen from Fig 3.3, when a circular bar with a diameter of 60% of the crank journal is adopted, the reaction of the aftmost M/E bearing decreases to almost half of that when the journal diameter is used, while the reaction of the second aftmost M/E bearing is expected to increase That is to say, how the crankshaft is modelled will have a significant effect on the accuracy of determining the reactions of the aftmost M/E bearings

- Equivalent

Journal

-Bearing number Fig 3.3 Effect of crank equivalent diameter on bearing reactions

3.2.2 Method for Determining Equivalent Diameter

(1) Numerical Calculation by FEM

The purpose of determining the equivalent diameter of the crankshaft is to ensure that the reaction of the bearings can be calculated correctly when bearing offsets change Therefore, a detailed FE model of a crankshaft was first developed, as shown in Fig 3.4(a) In the model, the left end was completely restrained, and an enforced vertical displacement was given at each bearing supporting point The reactions of the supports were calculated using this model At the same time, a FE model of a circular bar with identical boundary conditions in the detailed model was also developed, as shown in Fig 3.4(b) The process of determining the equivalent diameter consists of gradually decreasing the diameter of the circular bar model until the calculated reactions become almost equal to those obtained from the detailed model

(a)

All degrees of freedom restrained

(b)

Enforced vertical displacement

Fig 3.4 (a) FE model of crankshaft, (b) FE model of circular bar

The results show that a circular bar with a diameter of 60% of the journal diameter behaves in the same as the

Fig 3.5 Comparison of bearing reaction between crank and circular bars of different diameters.

Bearing number

Crank Circular bar of diameter equal to journal Round bar of diameter equal to 60% of journal

-(2) Approximate Analytical Expression

Assuming that the diameter of crankpin is equal to that of the crank journal, the equivalent diameter of a crankshaft can be calculated from the dimensions shown in Fig 3.6 using Eq (3.1)

t

l

L

B

Fig 3.6 Necessary dimensions of a crank throw for determining equivalent diameter

j p w w

B B A

d

4 1

1

1 1

1 2

1

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ +

+ +

2

2 1

65 0 1

1 2

3

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +

⎟

⎠

⎞

⎜

⎝

⎛ +

=

L

d L l L

r q I

I A

j w

j w

2

2 2

65 0 1

1 1

3

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +

⎟

⎠

⎞

⎜

⎝

⎛ + +

=

L

d L l L

r q I

I B

j wp

j w

ν

2 3

2

65 0 1

1 1

3

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +

+

=

L

d L

lr I

I B

j jp

j p

ν

ν: the Poisson's Ratio of the crankshaft material

64

4

j j

d

=

32

4

j jp

d

=

12

3

Bt

3

Bt

1449623 0

0605809 0

0082857 0

0004057 0

2 3

t

B

t

B

t

B

⎠

⎞

⎜

⎝

⎛ +

⎟

⎠

⎞

⎜

⎝

⎛

−

⎟

⎠

⎞

⎜

⎝

⎛

=

β

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ +

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

=

3 4 3 4

1

1 0 1

1 0 1

Bt

d d

r Bt

d q

j

j j

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ +

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

=

3 4 3 4

2

45 0 1

45 0 1

Bt

d d

r

Bt

d q

j

j

j

(3) Recommended Values by Engine Makers

Where recommended equivalent diameters by the engine manufacturer are available, the recommended values are to be used

3.3 Loads

The selfweight of an equivalent circular bar should be compensated to the same extent as the crankshaft by increasing its specific gravity or by adding the selfweight differential to loads representing the piston weight Loads representing the propeller weight should be determined taking into account the buoyancy corresponding to the draft predicted in the condition in which the shafting alignment calculation is performed Loads representing the selfweight of the propeller shaft should also be modified by reducing the specific gravity to reflect the effect of the buoyancy of lubricating oil or sea water, depending on the type of lubricating system used for the sterntube bearings

4 Determination of Initial Bearing Offsets

When aligning the shafting, the initial bearing offsets should be determined first In other words, a decision has to be made regarding whether a simple straight alignment is acceptable or appropriate offsets are necessary

In practice, the initial bearing offsets are determined based on past experience at most shipyards

Herein，a scientific approach to determining the bearing offsets based on a shafting stiffness matrix is proposed This approach is specifically designed to allow the offsets to be calculated directly from a straight shafting line based on the target bearing reactions

4.1 Construction of Shafting Stiffness Matrix

When a unit offset is given at a given bearing supporting point (positive downward, negative upward), reaction forces will be generated at each bearing supporting point These reaction forces are referred to by a bearing reaction influential number (hereafter referred to as "reaction influential number") The shafting stiffness matrix is constructed by arraying these reaction influential numbers in a certain order It is important to understand that reactions used to construct the shafting stiffness matrix must be the effect only from the given offset When the result contains some components resulting from external loads or selfweight, these components must be eliminated before the result can be used Therefore, it is strongly recommended that the reactions are calculated without any external loads, and that selfweight is neglected

Figure 4.1 shows an example of a shafting alignment calculation model incorporating five M/E bearings and representing the shafting of a VLCC Cross-sectional data of the shafting in different sections are shown in Table 4.1 Table 4.2 shows the bearing reactions at each bearing supporting point when a downward offset of 1.0 mm is applied to the No 1 bearing

Table 4.1 Sectional Parameters of Sample Shafting

Arrangement Did (mm)

Outer Inner Location (mm) I value (mm^4) 675.0 - 5.6000E+02 1.019025E+10 874.6 - 1.4000E+03 2.872155E+10 896.0 - 1.4235E+03 3.166574E+10 918.4 - 2.2870E+03 3.492186E+10 940.0 - 7.7850E+03 3.832492E+10 942.0 - 9.5730E+03 3.865214E+10 903.5 1.0003E+04 3.271015E+10 865.0 - 1.0203E+04 2.748111E+10 1180.0 - 1.0483E+04 9.516953E+10 725.0 - 1.4850E+04 1.356194E+10 730.0 - 1.5740E+04 1.393995E+10 725.0 - 2.1155E+04 1.356194E+10 1270.0 - 2.1294E+04 1.276982E+11 980.0 - 2.1515E+04 1.276962E+11 980.0 80 2.3375E+04 4.527463E+10 536.0 - 2.7875E+04 4.051623E+09 Fig 4.1 A shafting arrangement used to demonstrate

how to create a shaft stiffness matrix

Table 4.2 Reaction on Each Bearing Resulting from Lowering the No 1 bearing by 1.0 mm

Bearing

No. Location (mm) Offset (mm) Reaction (kgf)

No 1 2.8300E+03 1.000 139295.220

By repeating the above calculation for each bearing in turn, a matrix as shown in Table 4.3 can be obtained This result is precisely a shafting stiffness matrix that clearly takes the form of a square and symmetric matrix

Table 4.3 Shaft Stiffness Matrix for Sample Shaft Arrangements

Offset (mm)

eaction (kgf)

δ1 (1mm)

δ2 (1mm)

δ3 (1mm)

δ4 (1mm)

δ5 (1mm)

δ6 (1mm)

δ7 (1mm)

δ8 (1mm)

δ9 (1mm)

R

139295.220 -214398.670 81956.652 -9075.998 8643.455 -6589.810 213.664 -53.416 8.903

R1

-214398.670 342737.790 -147868.080 25863.061 -24630.482 18778.394 -608.860 152.215 -25.369

R2

81956.652 -147868.080 86843.213 -29740.647 34253.719 -26115.195 846.745 -211.686 35.281

R3

-9075.998 25863.061 -29740.647 25755.066 -64621.265 53184.964 -1724.440 431.110 -71.852

R4

8643.455 -24630.482 34253.719 -64621.265 487261.830 -602809.510 204508.110 -51127.028 8521.171

R5

-6589.810 18778.394 -26115.195 53184.964 -602809.510 895929.320 -459651.520 152728.030 -25454.671

R6

213.664 -608.860 846.745 -1724.440 204508.110 -459651.520 451271.610 -264078.490 69223.181

R7

-53.416 152.215 -211.686 431.110 -51127.028 152728.030 -264078.490 255095.360 -92936.091

R8

8.903 -25.369 35.281 -71.852 8521.171 -25454.671 69223.181 -92936.091 40699.447

R9

That is to say that the bearing reactions (without the effect of external loads and selfweight) can be calculated from the known set of offsets through the shafting stiffness matrix expressed in Eq (4.1)

139295.220 -214398.670 81956.652 -9075.998 8643.455 -6589.810 213.664 -53.416 8.903

-214398.670 342737.790 -147868.080 25863.061 -24630.482 18778.394 -608.860 152.215 -25.369

81956.652 -147868.080 86843.213 -29740.647 34253.719 -26115.195 846.745 -211.686 35.281

-9075.998 25863.061 -29740.647 25755.066 -64621.265 53184.964 -1724.440 431.110 -71.852

8643.455 -24630.482 34253.719 -64621.265 487261.830 -602809.510 204508.110 -51127.028 8521.171 -6589.810 18778.394 -26115.195 53184.964 -602809.510 895929.320 -459651.520 152728.030 -25454.671

213.664 -608.860 846.745 -1724.440 204508.110 -459651.520 451271.610 -264078.490 69223.181 -53.416 152.215 -211.686 431.110 -51127.028 152728.030 -264078.490 255095.360 -92936.091 8.903 -25.369 35.281 -71.852 8521.171 -25454.671 69223.181 -92936.091 40699.447

=

R1

R2

R3

R4

R5

R6

R7

R8

R9

δ 1

δ 2

δ 3

δ 4

δ 5

δ 6

δ 7

δ 8

δ 9

(4.1)

For simplicity, Eq.(4.1) may be rewritten in matrix and vector notation as shown in Eq.(4.2):

Aδ

R = (4.2) where

R={R1 R2…R9}T

139295.220 -214398.670 81956.652 -9075.998 8643.455 -6589.810 213.664 -53.416 8.903 -214398.670 342737.790 -147868.080 25863.061 -24630.482 18778.394 -608.860 152.215 -25.369 81956.652 -147868.080 86843.213 -29740.647 34253.719 -26115.195 846.745 -211.686 35.281 -9075.998 25863.061 -29740.647 25755.066 -64621.265 53184.964 -1724.440 431.110 -71.852 8643.455 -24630.482 34253.719 -64621.265 487261.830 -602809.510 204508.110 -51127.028 8521.171 -6589.810 18778.394 -26115.195 53184.964 -602809.510 895929.320 -459651.520 152728.030 -25454.671 213.664 -608.860 846.745 -1724.440 204508.110 -459651.520 451271.610 -264078.490 69223.181 -53.416 152.215 -211.686 431.110 -51127.028 152728.030 -264078.490 255095.360 -92936.091 8.903 -25.369 35.281 -71.852 8521.171 -25454.671 69223.181 -92936.091 40699.447

A =

δ=｛δ1 δ2…δ9｝T

4.2 Target Bearing Reactions

The bearing reactions are to be calculated in the straight shafting line condition first For small ships, even a straight shafting alignment may in some cases meet the relevant requirement, but for large ships, the results from the straight shafting line condition usually do not meet the required acceptance criteria Figure 4.2 shows the calculated deflection line, bending moment and shear force of the shafting model shown in Fig 4.1

Fig 4.2 Deflection, bending moment and shear force for sample shaft arrangement under straight offset of bearings condition

The calculated bearing reactions are shown in Fig 4.3

-150000

-100000

-50000

0 50000

100000

Bearing Number

Bearing reaction Negative values indicate upward reactions while positive values indicate downward reactions

Fig 4.3 Bearing reaction forces for sample shaft arrangement under straight offset

of bearings condition

As can be seen in Fig 4.3, the aft end bearing of the aftmost stern tube is overloaded, while the fore end bearing of the aftmost stern tube is loaded in the wrong direction Therefore, the alignment must be modified Assuming the target bearing reactions are known as shown in Table 4.4, an appropriate set of bearing offsets should be determined so that the difference in the reaction forces between the target reactions and the reactions

under a straight condition, ΔR, can be generated

Table 4.4 Target Bearing Reaction Forces for Sample Shaft Arrangement

Bearing No

Reactions of straight offset (kgf)

Target reactions (kgf)

Δ R = Target - Straight

(kgf)

No 1 -143537 -88333 55204

No 2 62921 -17079 -80000

No 3 -35917 -11917 24000

No 4 -21989 -23989 -2000

No 5 -48172 -28172 20000

No 6 -8661 -26661 -18000

No 7 -42226 -41726 500

No 8 -51097 -51277 -180

No 9 -15295 -15265 30

4.3 Calculation of Initial Bearing Offsets

The necessary initial offsets, δ, can be calculated backward from the required reactions, Δ R by Eq (4.3) which is

a rewritten form of Eq (4.2), solving for δ