Q1. A vessel displacing 16500 tonnes has KG, 7.50m. Calculations of grain shift give the following data.
Volumetric heeling moment = 3960m^{4}
Stowage factor = 1.2m^{3}/tone
For the given displacement and KG the vessel has the following values of GZ at the angle of heel given.
Heel |
0 |
15 |
30 |
45 |
60 |
75 |
90 |
GZ |
0.00 |
0.267 |
0.645 |
0.571 |
0.163 |
-0.454 |
-1.104m |
Verify whether the vessel satisfies intact stability requirements for grains cargo. Angle of flooding is 40^{0}.
Solution –
Given VHM = 3960m^{4}
Weight heeling moment = 3960/1.2
= 3300m^{4}
λ_{0 }= 0.2m, λ_{40} = 0.16m
- GM = 0.776m
Therefore, condition 1 satisfied
- Angle of heel = 12^{0} = Marginal.
Heel | Ordinate | SM | Product |
0 | 0.0 | 1 | 0 |
10 | 0.15 | 4 | 0.60 |
20 | 0.33 | 2 | 0.66 |
30 | 0.645 | 4 | 2.28 |
40 | 0.610 | 1 | 0.610 |
4.45 |
Area = 10/3 ⨯ 4.45/57.3
= 0.259m radian
Area of triangle = 0.5 ⨯ 12 ⨯ 0.19/57.3
= 0.0199m radian
Area of trapezium = 0.5 ⨯ (0.19 + 0.16) ⨯ 28/57.3
= 0.085 m radian
Residual area = 0.16m radian ˃ 0.075
Therefore, condition 3 satisfied.
Hence, vessel satisfy intact stability requirement for grain cargo or grain loading criteria.
Q2. The values of GZ of a vessel at different angles of heel are as follows:
Heel (^{0}) |
0 |
5 |
12 |
20 |
30 |
40 |
60 |
75 |
GZ (m) |
0.0 |
0.188 |
0.449 |
0.764 |
1.227 |
1.551 |
1.713 |
1.298 |
The heeling arm at 0^{0} = 0.275 metres. The angle of flooding is 400 and maximum difference between the righting arm and heeling arm occurs at an angle greater than 400. Verify whether the vessel satisfies the intact stability requirements for a cargo ship carrying grain in bulk.
Solution –
Given λ_{0} = 0.275, λ_{40} = 0.22
Heel | Ordinate | SM | Product |
0 | 0 | 1 | o |
10 | 0.35 | 4 | 1.4 |
20 | 0.764 | 2 | 1.53 |
30 | 1.227 | 4 | 4.91 |
40 | 1.551 | 1 | 1.55 |
SOP = 9.30 |
Area of triangle = 0.5 ⨯ 7 ⨯ 0.265/57.3
= 0.016m radian
Area of trapezium = 0.5 ⨯ (0.265 + 0.22) ⨯ 33/57.3
= 0.139m radian
Residual area = 0.397m radian > 0.075m radian
Therefore, condition 3 satisfied
Hence, vessel satisfy intact stability requirement for grain cargo or grain loading criteria.
Q3. The values of GZ vessel for various angles of heel are as follows:
Heel (^{0}) |
00 |
05 |
12 |
20 |
30 |
40 |
60 |
75 |
GZ (m) |
0.000 |
0.136 |
0.315 |
0.472 |
0.604 |
0.631 |
0.562 |
0.248 |
The heeling arm at 0^{0} = 0.18m. The angle of flooding is 40^{0} and maximum difference between the righting arm and heeling arm occurs at 43^{0}. Verify whether the vessel satisfied the criteria of stability for a cargo ship carrying grain in bulk (Assume the deck edge immerses at 10^{0 } heel).
Solution –
Given, λ_{0 }= 0.18m, λ_{40} = 0.14m
List | Ordinate | SM | Product |
0 | 0.0 | 1 | 0 |
10 | 0.263 | 4 | 1.052 |
20 | 0.472 | 2 | 0.944 |
30 | 0.604 | 4 | 2.416 |
40 | 0.631 | 1 | 0.631 |
SOP = 5.043 |
Area = 10/3 ⨯ 5.043/57.3
= 0.293 m radian
Area of triangle = 0.5 ⨯ 6.5 ⨯ 0.1735/57.3
= 0.010m radian
Area of trapezium = 0.5 ⨯ (0.1735 + 0.14) ⨯ (4.0 – 6.5)/57.3
= 0.191m radian
Residual area = 0.192m > 0.075m radian
Therefore, All condition satisfied.
Hence, vessel satisfy intact stability requirement for grain cargo or grain loading criteria.
Q4. A vessel loaded with grain in bulk is at a displacement of 35186t, KG 7.809m, KM 11.30m. The stowage factor of the grain is 1.3m3/t and the total volumetric heeling Moments are 21321 m^{4}. AT the displacement her KN values are as follows:
Heel (^{0}) |
5 |
12 |
15 |
30 |
45 |
60 |
GZ (m) |
1.00 |
2.42 |
2.90 |
5.60 |
7.30 |
8.05 |
If her angle of flooding exceeds 400, ascertain whether the vessel complies with the intact stability requirements for such vessels.
Solution –
Given, KM = 11.30
KG = 7.809m
GM = (KM – KG)
= 3.491m > 0.3m
Therefore, Condition 1 satisfied
Now total volumetric heeling moment = 21321
Weight heeling moment = VHM/SF
= 16400.7692
Heeling arm = λ_{0} = 16400.76923/35186
= 0.466m
λ_{40 }= 0.373m
heel | 5 | 12 | 15 | 30 | 45 | 60 |
KN | 1.00 | 2042 | 2.90 | 5.60 | 7.30 | 8.05 |
KG SinӨ | 0.680 | 1.623 | 2.021 | 3.904 | 5.522 | 6.763 |
GZ | 0.320 | 0.797 | 0.879 | 1.696 | 1.778 | 1.287 |
Heel | Ordinate | SM | Product |
0 | 0 | 1 | 0 |
10 | 0.65 | 4 | 2.6 |
20 | 1.05 | 2 | 2.10 |
30 | 1.696 | 4 | 6.784 |
40 | 1.760 | 1 | 1.760 |
SOP = 13.244 |
Area = 10/3 ⨯ 13.244/57.3
= 0.770 m radian
Area of triangle = 0.5 ⨯ 7 ⨯ 0.420/57.3
= 0.026 m radian
Area of trapezium = 0.228 m radian
Therefore, Residual area = 0.516 m radian > 0.075
Condition 3 satisfied.
Hence, vessel satisfy intact stability requirement for grain cargo or grain loading criteria.
Q5. The values of GZ vessel for various angles of heel are as follows:
Heel (^{0}) |
00 |
05 |
12 |
20 |
30 |
40 |
60 |
75 |
GZ (m) |
0.000 |
0.136 |
0.315 |
0.472 |
0.604 |
0.631 |
0.562 |
0.248 |
The heeling arm at 0^{0} = 0.18m. The angle of flooding is 40^{0} and maximum difference between the righting arm and heeling arm occurs at 43^{0}. Verify whether the vessel satisfied the criteria of stability for a cargo ship carrying grain in bulk (Assume the deck edge immerses at 10^{0 } heel).
Solution –
λ_{0 }= 0.18_{, }λ_{40} = 0.144
List | ordinate | SM | Product |
0 | 0 | 1 | 0 |
10 | 0.135 | 4 | 0.54 |
20 | 0.472 | 2 | 0.944 |
30 | 0.604 | 4 | 2.416 |
40 | 0.631 | 1 | 0.631 |
4.531 |
Area = 10/3 ⨯ 4.531/57.3
= 0.2637 m radian
Area of triangle = 0.5 ⨯ 17 ⨯ 0.162/57.3
= 0.024 m radian
Area of trapezium = 0.5 ⨯ (0.163 + 0.144) ⨯ 23/57.3
= 0.062 m radian
Residual area = 0.1811m radian
Therefore, Ship doesn’t comply.
Hence, vessel does not satisfy intact stability requirement for grain cargo or grain loading criteria.