PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE 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 load ing 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 Number of bearings included However, the results of analysis suggests that the number of M/E bearings in the calculation model has appreciable e f fects 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 have to be calculated precisely, then seven M/E bearings in total need to be included in the calculation model. 6 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE 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 numbe r Reaction force (kgf) Three Fou r 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 cal culations. 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. 7 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE - Equivalent Journal - Reaction force (kgf) - - - 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 t he 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. 0 1 2 3 4 5 6 0 1 2 3 4 5 6 (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 crankshaft, as shown in Fi g . 3.5. 8 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE Fig. 3.5 Comparison of bearing reaction between crank and circul ar bars of dif ferent diameters. Bearing number Reaction force (kgf) Crank Circular bar of diameter equal to journal Round bar of diameter equal to 60% of j ournal - - - - (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 dj r B Fig. 3.6 Necessary dimensions of a crank throw for determining equivalent diameter. 9 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE j pww eq d BBA d 4 1 1 1 1 1 2 1 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ++ + + = (3.1) 2 2 1 6501 1 2 3 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = L d . L l L rq I I A j w j w () 2 2 2 6501 1 13 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = L d . L l L rq I I B j wp j w ν ( ) 23 2 6501 1 13 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = 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 I π = 32 4 j jp d I π = 12 3 Bt I w = 3 BtI wp β = 14496230060580900082857000040570 23 . t B . t B . t B . + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = β ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= 3 4 3 4 1 101 10 1 Bt d . d r Bt d . q j j j ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= 3 4 3 4 2 4501 450 1 Bt d . d r Bt d . q j j j 10 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE (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 buo yancy 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. 11 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE 4. Determination of Initial Bearing Offsets When aligning the shafting, the initial bearing offsets should be determ ined 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 sc ientific approach to deter mining 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 incorp orating 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 12 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 stif fness matrix. PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE 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 No. 2 4.6300E+03 0.000 -214398.670 No. 3 8.4650E+03 0.000 81956.652 No. 4 1.5295E+04 0.000 -9075.998 No. 5 2.2375E+04 0.000 8643.455 No. 6 2.3375E+04 0.000 -6589.810 No. 7 2.4875E+04 0.000 213.664 No. 8 2.6375E+04 0.000 -53.416 No. 9 2.7875E+04 0.000 8.903 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 stif fness 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 (1 mm) δ2 (1mm) δ3 (1mm) δ4 (1mm) δ5 (1mm) δ6 (1mm) δ7 (1mm) δ8 (1mm) δ9 (1mm) R 13 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 of fsets 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) PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE For simplicity, Eq.(4.1) may be rewritten in matrix and vector notation as shown in Eq.(4.2): AδR = (4.2) where R={R 1 R 2 …R 9 } 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=｛δ 1 δ 2 …δ 9 ｝ 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. 14 PART A GUIDELINES ON SHAFTING ALIGNMENT TAKING INTO ACCOUNT VARIATION IN BEARING OFFSETS WHILE IN SERVICE The calculated bearing reactions are shown in Fig. 4.3. -150000 -100000 -50000 0 50000 100000 No. 1 No. 2 No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 Bearing Number Reaction force (kgf) 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 aftm ost 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) ΔR = 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 fro m the required reactions, ΔR by Eq. (4.3) which is a rewritten form of Eq. (4.2), solving for δ. 15 . -459651. 520 45 127 1.610 -26 4078.490 6 922 3.181 -53.416 1 52. 215 -21 1.686 431.110 -51 127 . 028 1 527 28.030 -26 4078.490 25 5095.360 - 929 36.091 8.903 -25 .369 35 .28 1 -71.8 52 8 521 .171 -25 454.671 6 922 3.181 - 929 36.091. 895 929 . 320 -459651. 520 1 527 28.030 -25 454.671 R6 21 3.664 -608.860 846.745 -1 724 .440 20 4508.110 -459651. 520 45 127 1.610 -26 4078.490 6 922 3.181 R7 -53.416 1 52. 215 -21 1.686 431.110 -51 127 . 028 1 527 28.030. -71.8 52 8643.455 -24 630.4 82 3 425 3.719 -64 621 .26 5 48 726 1.830 -6 028 09.510 20 4508.110 -51 127 . 028 8 521 .171 -6589.810 18778.394 -26 115.195 53184.964 -6 028 09.510 895 929 . 320 -459651. 520 1 527 28.030 -25 454.671 21 3.664