EXAMPLE - LIGHT BALLASTED VESSEL
250 000 dwt tanker, lightest operational draft 6. t m, trim 1070
14 no. 42 mm wires of MBL 1110 kN
Layout as shown in Figure B.2.Note: AB values of e go over 25 Q and are larger and more variable than those of 0, the method
uses refinement 3 described in tbe [ext.Line a^0 loo. H 80 Iou! + I;n = L cos 0 cos a
plan true1 15.0 (^47) 19.2 22.2 51 + 6 = 51 0.926 0.966
2 17 .3 46 19 .2 22 .6 50 + 6 = S6 0.923 0.955
3 11.2 41.5 18.8 24.4 45.5 + 42 .5 '"' 88 0.911 0.955
4 14 .2 37 .5 18.8 26.6 42 + 44 = 86 0.894 0.969
5 12.4 35.5 18.8 27.9 40 + 35 = 75 0.884 0.977
6 9.0 68 18.0 14.8 70.5 + 35 = 106 0.967 0.988
7 8.8 69.5 18.0 14.5 72 + 35 = 107 0.968 0.988
8 10.2 6O.S 14.7 13.7 62 + 35 = 97 0.972 0.984
9 10.4 59 14.7 14.0 61 + 3S = 96 0.970 0.984
10 11.0 40 13.7 18.9 42.5 + 32 = 74.5 0.946 0.982
11 12.8 41.5 13.7 18.3 43.S + 32 = 75.S 0.950 0.975
12 16 .0^44 13.7 17.3^46 + 7.S z^53 .5 0.955 0.961(^13) 10.2 58.5 13 .3 12 .8 60+6=66 0.975 0 .984
(^14) 10.8 6S 13.3 11.6 66.5 + 6 = 72.5 0.980 9.82
·lndicalCS critical line of group.
R y = 0.55 (MBL) ( L COS2 a cos <
l
8)( L )
L cos 0, cos BcR YF = 0.55 (I J 10)(5.52)(_1_) = 2147 kN
1.57RYA = 0.55 (1110)(6.54)(_1_) = 2321 kN
1.72Rx = N (O.55)(MBL) cos (J cos 8RXF = 2 (0.55)( I11 0)(0.988)(0.967) = 1167 kNRXA = 2 (0.55)(1110)(0.984)(0.970) = 1165 kN
100Comparison of predicted forces and hand ca1cnlaled rcst:mlnt capacitycos Q cos 8
L1.57-
1.57-
0.99
1.01
1.151.25
1.23
I.n-
1045
J.33Appendix 0 predicts the following max.imum forces for lhe assumed conditions:Mal< F x = - 790 kN (in ballast conditions)F YF = + 2253 kN (in ballast conditions)
F YA = + 2499 kN (in full load conditions)'"
- Practically the same as ballasl condirions.
cos^1 a cos^2 0
100
L1.40
1.39
0.86
0.87
0.99
r = 5.521.16
1.14
1.57
1.39
1.28
r = 6.54