GUIDELINES ON SHAFTING ALIGNMENT phần 10 ppsx

3 Hull Deflection that results in Engine Bearings Becoming Unloaded Using the reaction influence number Si , connecting with hull deformation, we can determine the amount of relative dis

3 Hull Deflection that results in Engine Bearings Becoming Unloaded

Using the reaction influence number Si , connecting with hull deformation, we can determine the amount of relative displacement of the hull that would cause the ith engine bearing to become unloaded There are two cases in which either elastic support (spring support) or rigid support (simple support) are used as the support condition of the bearings to calculate the reaction influence numbers, Cm, n Each case is examined below

3.1 In the case of elastic support

In this case, δB , which is the relative displacement at which the engine bearings become unloaded, can be obtained using a simplified calculation method Let Ri be the reaction of engine bearing No

i (from aft) before hull deformation and let δ B ibe the relative displacement of the hull (at the position of the aftmost bulkhead of the engine room) at which engine bearing No i becomes unloaded after hull deflection, then

Hence,

Since unloading of engine bearings due to hull deflection is often limited to the second or third aftmost bearing in most cases, both δB2 and δB3 are to be calculated and whichever is less is to be compared with the lower limit

3.2 In the case of rigid support

Figure 3.1 shows how to deal with the stiffness of an engine bearing support in the state before hull deflection In cases where rigid support is considered in the calculation, bearing reaction Ri corresponding to the given offset is easily obtained for each respective bearing, and the displacement

of the support point due to the load is zero, as shown in Figure 3.1(a) However, if the effect of the stiffness (spring) is considered here, the support points are in fact displaced downward slightly by the load, as shown in Figure 3.1(b), which actually involves a change in the bearing reaction Ri for each engine bearing

Figure 3.1 Initial condition of engine bearings

Assuming that reaction Ri does not change even if the effect of the stiffness (spring) is considered, the displacement of each support point after loading is expressed with the stiffness K (= constant) of each support point using the following equation:

In the discussion below, we examine the change in bearing reaction when the hull deflects from the initial condition shown in Figure 3.1(b) due to an increase of draught Figure 3.2 shows the change

in the support point (contact point with the shaft) that occurs before and after hull deflection As can be seen from this figure, it is assumed that the change in the vertical position of the support point shown here is not caused by the deflection of hull structure below the engine, but by the elastic deformation of the engine bed due to the increase in the bearing load

∆Ri = 0 – R i = δBi S i

R1

R 2 R 3 R 4 R 5

(a) Rigid support (b) Elastic support

K

h1 h2 h3 h4 h 5

h i =R i / K

Figure 3.2 Displacement of support points due to elastic deformation

of the engine bed

In paragraph 3.1 above, a simple calculation could be used to give the bearing reactions of each bearing after hull deflection However, in order to obtain the bearing reactions after hull deflection using reaction influence numbers based on rigid support, the equations that include the reactions and displacements of the support points as variables need to be solved for specific conditions Setting the number of engine bearings at five and denoting the relative displacement of the hull at the position of the engine room aftmost bulkhead by δB, changes in the reactions of each engine bearing can be expressed as follows:

where,

i : number of support point (engine bearings) counted from the aft of the engine,

Ri : reaction of support point i (counted from the aft of the engine) before hull deflection,

R′i : reaction of support point i (counted from the aft of the engine) after hull deflection,

δ i : elastic displacement of support point i (counted from the aft of the engine),

Ci, j : amount of increase of reaction (reaction influence number) at support point i when support point j is displaced downward by 1 mm (i,j : number of support point counted from the aft of the engine)

(Note that the meaning of the subscripts in Equation (8) differ from those in Equation (2).)

On the other hand, from the relationship:

Equation (8) can be rewritten as:

R′1 – R1 = S1 δB + C1, 1δ1 + C1, 2δ2 + C1, 3δ3 + C1, 4 δ 4 + C1, 5δ5 , R′2 – R2 = S2 δB + C2, 1δ1 + C2, 2δ2 + C2, 3δ3 + C2, 4 δ 4 + C2, 5δ5 , R′3 – R3 = S3 δB + C3, 1δ1 + C3, 2δ2 + C3, 3δ3 + C3, 4 δ 4 + C3, 5δ5, R′4 – R4 = S4 δB + C4, 1δ1 + C4, 2δ2 + C4, 3δ3 + C4, 4 δ 4 + C4, 5δ5 , R′5 – R5 = S5 δB + C5, 1δ1 + C5, 2δ2 + C5, 3δ3 + C5, 4 δ 4 + C5, 5δ5 ,

(8)

(9) R′i – Ri = K δ i ,

S1 δB + (C1, 1 – K )δ1 + C1, 2δ2 + C1, 3δ3 + C1, 4 δ 4 + C1, 5δ5 = 0,

S2 δB + C2, 1δ1 + (C2, 2 – K )δ2 + C2, 3δ3 + C2, 4 δ 4 + C2, 5δ5 = 0,

S3 δB + C3, 1δ1 + C3, 2δ2 + (C3, 3 – K )δ3 + C3, 4 δ 4 + C3, 5δ5 = 0,

S4 δB + C4, 1δ1 + C4, 2δ2 + C4, 3δ3 + (C4, 4 – K )δ 4 + C4, 5δ5 = 0,

S5 δB + C5, 1δ1 + C5, 2δ2 + C5, 3δ3 + C5, 4 δ 4 + (C5, 5 – K )δ5 = 0

(10)

hi Bearing offset input

δi

Ri

R′i Reference line

(After hull deflection) (Before hull deflection)

Tank top of engine room double bottom

In Equation (10), we consider the situation that the second aftmost engine bearing becomes unloaded after hull deflection, which is expressed by the following equation:

This indicates that the support point of the second aftmost engine bearing moves upward so that the displacement, h2 , which is due to the bearing load R 2 before hull deflection return to the original state, in other words, the restoring force of the spring may be zero

Let δB2 be the relative displacement, δB, which causes the second aftmost engine bearing to become unloaded Then, the following simultaneous equations with five variables, i.e., δB2, δ1, δ3 , δ4 , and

δ5, need to be solved in order to find δB2, since δ2 is not a variable:

where, K = 5,000 kN/mm (constant value)

Figure 3.3 illustrates the state of the shafting and relative displacement of the engine bearings before and after hull deflection

Figure 3.3 Reactions and displacements of support points when stiffness

of engine bearings is considered

Simultaneous Equations (12) can be expressed in matrix form as follows:

where,

S1 δB2 + (C1, 1 – K )δ1 + C1, 3δ3 + C1, 4 δ 4 + C1, 5δ5 = C1, 2 R2/K ,

S2 δB2 + C2, 1δ1 + C2, 3δ3 + C2, 4 δ 4 + C2, 5δ5 = (C2, 2 – K )R 2/K ,

S3 δB2 + C3, 1δ1 + (C3, 3 – K )δ3 + C3, 4 δ 4 + C3, 5δ5 = C3, 2 R 2/K ,

S4 δB2 + C4, 1δ1 + C4, 3δ3 + (C4, 4 – K )δ 4 + C4, 5δ5 = C4, 2 R 2/K ,

S5 δB2 + C5, 1δ1 + C5, 3δ3 + C5, 4 δ 4 + (C5, 5 – K )δ5 = C5, 2 R 2/K ,

(12)

(a) Before hull deflection

R1 R2 R3 R4 R 5

(b) After hull deflection

δB2 R’2 = 0 :δ2 = –h2

R’1

R’2

R’3 R’4 R’5

[D 2 ] [δ B2] =h2[C 2], (13)

Hence,

where, [D 2]INV denotes the inverse matrix of [D 2]

Thus, we can obtain δB2 as follows:

where, ([D 2]INV)L1 denotes the elements (vector) of the first row of [D 2]INV

Similarly, the relative displacement δB3 which results in the third aftmost engine bearing to become unloaded can be obtained by solving the following simultaneous equations with five variables (the matrix representation is omitted):

In a shafting alignment with constant offsets of the engine bearings, the calculation of δB2 is sufficient to check the strength of all engine bearings, because the load on the second aftmost engine bearing decreases as draught increases However, in some cases where the engine bearing offsets are unequal, the load on the third aftmost engine bearing could decrease Consequently, both δB2 and δB3, in general, should be calculated and whichever is less is to be compared with the lower limit

[D 2] = ,

S1 C1, 1K C1, 3 C1, 4 C1, 5

S2 C2, 1 C2, 3 C2, 4 C2, 5

S3 C3, 1 C3, 3K C3, 4 C3, 5

S4 C4, 1 C4, 3 C4, 4K C4, 5

S5 C5, 1 C5,3 C5,4 C5,5K

δB2

δ1

δ3

δ4

δ5

[δ B2] = ,

C1, 2

C2, 2K

C3, 2

C4, 2

C5, 2

[C 2] = ,

h2 = R2/ K ,

Ci, i K = Ci i– K ( i =1, 2, 3, 4, 5 )

[δB2] = h2 [D 2]INV [C 2] , (14)

δB2 = h2 ([D 2]INV)L1 [C 2] , (15)

S1 δB3 + (C1, 1 – K )δ1 + C1, 2δ2 + C1, 4 δ 4 + C1, 5δ5 = C1, 3 R3/K ,

S2 δB3 + C2, 1δ1 + (C2, 2 – K )δ2 + C2, 4 δ 4 + C2, 5δ5 = C2, 3 R 3/K ,

S3 δB3 + C3, 1δ1 + C3, 2δ2 + C3, 4 δ 4 + C3, 5δ5 = (C3, 3 – K )R 3/K ,

S4 δB3 + C4, 1δ1 + C4, 2δ2 + (C4, 4 – K )δ 4 + C4, 5δ5 = C4, 3 R 3/K ,

S5 δB3 + C5, 1δ1 + C5, 2δ2 + C5, 4 δ 4 + (C5, 5 – K )δ5 = C5, 3 R 3/K

(16)

4 Calculation Example

Figure 4.1 and Table 4.1 show the dimensions of the shafting prepared to test and verify the calculations and the reaction influence numbers, respectively The values of the reaction influence numbers were obtained based on the assumption of rigid support for the bearings, therefore paragraph 1.3.3-1.(2) of the Annex to the Guidance was applied in the calculation of δB2 and δB3 The reactions and offsets of the bearings under the hot condition are shown in Table 4.2

Figure 4.1 Dimensions of shafting

Table 4.1 Reaction Influence Numbers (kN /mm)

Table 4.2 Bearing Reactions and Offsets (Hot Condition)

Results of Calculations

Table 4.3 shows the results of the calculations of

table that the engine bearings most notably

affected by hull deflection are the aftmost and

second aftmost engine bearings, and further, that

the load on the second aftmost engine bearing

decreases whereas the load on the aftmost engine

bearing increases

The values of δB2 and δB3are shown in Table 4.4

This table shows that the second aftmost engine

bearing will first become unloaded when the

ship’s draught increases and δB reaches 7.26 mm

Figure 4.2 shows a comparison between δB2 and

the allowable limit in the case of L = 10,400 mm

Figure 4.2 Comparison between δB2and

the allowable limit

1370 2555 5665 5540 870 1290

1290 1290

E/R aft bulkhead

10400

SB1 -1094.4 1760.31 -710.86 60.88 -58.32 44.00 -2.00 0.50 -0.08 SB2 1760.31 -2956.05 1338.87 -193.96 185.79 -140.17 6.38 -1.59 0.27 SB3 -710.86 1338.87 -778.71 217.35 -244.91 184.77 -8.41 2.10 -0.35 IB1 60.88 -193.96 217.35 -175.85 436.06 -357.35 16.26 -4.06 0.68

1 -58.32 185.79 -244.91 436.06 -3498.3 4574.60 -1761.8 440.43 -73.40

2 44.00 -140.17 184.77 -357.35 4574.60 -7194.4 4005.52 -1340.2 223.35

3 -2.00 6.38 -8.41 16.26 -1761.8 4005.52 -3991.1 2353.52 -618.22

4 0.50 -1.59 2.10 -4.06 440.43 -1340.2 2353.2 -2283.1 832.48

5 -0.08 0.27 -0.35 0.68 -73.40 223.35 -618.22 832.48 -364.72

Reaction (kN) 446.69 15.16 69.89 95.33 226.39 131.83 256.28 314.09 90.80 Offset (mm) 0.35 0.00 -0.15 0.20 0.40 0.40 0.40 0.40 0.40

73.91 -66.79 3.04 -0.75 0.14

Table 4.3 S i (kN/mm)

7.26 19.14

Table 4.4 δB2 and δB3 (mm)

5 7 9 11 13 15x103

10 8 6 4 2 0

L (mm)

δB2

Allowable limit