C L DUBEY – EXERCISE – 09 (GROUNDING)

Q1. A vessel of length 142m, displacement 16130t, KM 8.24m, KG 7.26m MCTC 192 mt, TPC 23.4t, LCF 71m, FSM 1372 mt, at a draft of F 6.70m, A 8.86m, ground on a rock. The draft then F 5.90m, A 9.30m.
Calculate :

1. The upthrust provided by the rock.

2. The position with respect to AP where the grounding occurred.

3. Her virtual GM (FI.) then

4. The rise of tide required for her to refloat.

Solution –

• Draft before Grounding = 6.70m, Aft = 8.86m

Trim = 2.16m
Correction to aft draft = 2.16 ⨯ 71/142
= 1.08m
∴ Hydrostatic draft = 7.78m

Draft after grounding = 5.90 m fwd

Trim = 3.4m
Correction to aft draft =   3.4 ⨯ 71/142
∴ Hydrostatic draft = 7.60m
Rise = P/ TPC ⨯ 100
P = (0.18 ⨯ 23.4 ⨯ 100)

   = 421.2 tonnes

• Trim before grounding = 2.16m,

Trim after grounding = 3.40m
Change in trim = 1.24m
1.24 = 421.2 ⨯ d/192 ⨯ 100
d = 56.52 m from COF

From AP distance = 71.00 + 56.52 = 127.52m

• GG₁ = (P ⨯ KM)/W

= 421.2 ⨯ 8.24/16130
= 0.215m

New GM = 0.765m

FSC = 1372/(16130 – 421.2)
= 0.0873m

GM (fluid) = 0.678m

• Rise of tide required = P/ TPC + P ⨯ d/ MCTC ⨯ d/ LBP

= 421.2/23.4 + 421.2 ⨯ 56.52 ⨯ 56.52/192 ⨯ 142
= 18 + 49.35

= 67.35cm

Q2. A vessel of length 140m is floating at draft forward 5.23m Aft 5.74m runs aground lightly on a rock 12m abaft the stern. The tide falls by 80cms. Given the following data estimate the virtual GM of the vessel and her draft forward and aft after the fall of tide. KG 5.80m, KM 6.40m, MCTC 130tm, TPC 18t, AF 73m, Displacement 18500t.

Solution –

Distance of grounding point from COF = 55m
We have P = trim ⨯ MCTC/d
Change in draft = P/TPC + P ⨯ d/MCTC ⨯ d/LBP
So, 80 =   P/18 + P ⨯ 55/130 ⨯ 55/140

Or, P = 360.74

GG₁ = (P ⨯ KM)/ W
       = 0.125m

Virtual GM   = GM – Virtual loss of GM( GG1)
= 0.475m

Also,
Trim caused = Tc = (P ⨯ d)/ MCTC ⨯ 100
= 1.526m

Ta = 0.796m,
Tf = 0.730m

Rise = P/ TPC ⨯ 100
= 0.200 tonnes