C L DUBEY – EXERCISE – 02 : Simpson’s Rule

Q1. The TPC of a ship are as follows:

Draft (m)	6.0	5.0	4.0	3.0	2.0
TPC	22.45	22.04	21.53	20.91	19.68

The displacement and KB at 2m draft are 3280t and 1.2m respectively. Find the displacement and KB at 6m draft in SW.

Solution –

Given,

Draft	TPC	Area (TPC= A/100 ⨯ density)	SM	Product	Lever	Moment
6.0	22.45	(100)⨯22.45/1.025	1	22.45x	4h	89.8xh
5.0	22.04	22.04x	4	88.16x	3h	264.48xh
4.0	21.53	21.53x	2	43.06x	2h	86.12xh
3.0	20.91	20.91x	4	83.64x	1h	83.64xh
2.0	19.68	19.68x	1	19.68x	0h	0

Sum of product of volume = 256.99x
Volume = 1/3 ⨯ 256.99 ⨯ 100/1.025
= 8357.398m³

Sum of product of moment = 524.04 ⨯ 1/3 ⨯ 100/1.025 ⨯ 1
= 17041.95m⁴

Total volume including appendage =?
Given displacement of appendage = 3280 tonnes at draft 2m
We know displacement = ( v/w volume X  density of water displaced)
Volume of appendage = 3280/1.025
= 3200m³

Total volume including appendage = (8357.398 + 3200) m³
= 11557.398 m³

COG of the structure without appendage
= volume/Moment
= 2.039m

COG from 2.0 mtr draft = 2.039 m
Now total moment = moment of appendage + moment of structure
11846.33 ⨯  X   = (3280 ⨯ 2) + (8566.33 ⨯ 4.039)

 X = 3.253m

Q2. The breadths of a ship’s water-plane 120m long, measured at equal intervals from aft, are: 1.2, 9.6, 13.2, 15.0, 15.3, 15.6, 15.6, 14.7, 12.9, 9.0 and 0.0m respectively. Find:

• The water plane area;

• TPC in sea water;

• FWA if displacement is 6690t.

Solution –

Ordinate		SM	Product
1.2		1	1.2
9.6		4	38.4
13.2		2	26.4
15.0		4	60.0
15.3		2	30.6
15.6		4	62.4
15.6	2		31.2
14.7	4		58.8
12.9	2		25.8
9.0	4		36.0
0.0	1		0.0
			370.8

Water plane Area = ( H x SOP )/ 3
= 1483.2m²

We know that – TPC = A/100 ⨯ density
= 15.2 tonnes

FWA =    W/40 ⨯ TPC
=     6690/40 ⨯ 15.2

= 11.00cm

Q3. A ship of length 300m floating at a draft of 20m has half ordinates of water plane as follows, commencing from the after perpendicular.

Station	0	1/2	1	2	3	4	5	5 1/2	6
½ ordinates	0.1	7.5	10	12	12.3	11.4	8	5.2	1.0

There is an appendage forward with an area of 2.8 m².

Find : (a) the area of the water Plane.

           (b) The TPC in SW and

            (c) The FWA at 20m draft.

Solution –

Half Ordinate	SM	Product
0.1	0.5	0.05
7.5	2	15
10	1.5	15
12	4	48
12.3	2	24.6
11.4	4	45.6
8	1.5	12
5.2	2	10.4
1.0	0.5	0.5
	SOP = 171.15

Area = 50/3  ⨯ 171.15
= 2852.5m²

TCP = 29.23 t/cm

Q4. The area of a ship’s water-planes, commencing from the load water-plane and spaced 1m apart, are 800, 760, 700, 600, 450 and 10sq.m. Midway between the lowest two water-planes, the area is 180 sq.m. Find the load displacement in salt water and the height of centre of buoyancy above the keel.

Solution –

Area	SM	Product of volume	Lever	Product of moment
800	1	800	0h	0
760	4	3040	h	3040h
700	2	1400	2h	2800
600	4	2400	3h	7200h
450	1.5	675	4h	2700h
180	2	360	4.5h	1620h
10	0.5	05	5h	25h
		SOP = 8680		SOM = 17385h

Volume = 8680/3
= 2893.33

Displacement = 2965.67 tonnes

COG from Top = 2.00288m
COG from bottom = 2.99m

Q5. A wall-sided vessel of constant water-plane area of length 105m has the following equally spaced half-ordinates of water plane:
1, 7, 3.4, 5.6, 6.7, 6.7, 5.6, 3.4 and 1.7 mtrs. When floating at an even keel draft of 4.5m, an empty compartment with transverse end bulkheads at the 6.7m ordinates is bilged. Calculate the end draft at which she would settle.

Solution –  Will be Uploaded Soon.

Q6. A wall sided double bottom tank of depth 1.5m has half ordinates of breadth as follows at 4m intervals commencing at the after bulkhead:

Station	0	1	2	3	4	5	6
½ ord mtr	6.1	6.0	5.9	5.7	5.4	4.9	4.3

The after bulkhead is 65m forward of the centre of floatation. Calculate the change of trim if the is filled with fuel oil density 0.95, MCTC 200 tm.

Solution –  Will be Uploaded Soon.

Q7. The water plane areas of a ship at 1m intervals, commencing from the keel are as follows:

Draft (m)	0	1	2	3	4	5	6
Areas (sq.m)	5850	5885	5900	5915	5943	5975	5995

Calculate her KB and FWA at a draft of 6m.

Solution –

Draft	Area	SM	Product	Lever	Product of moment
0	5850	1	5850	0h	0h
1	5885	4	23540	1h	23540h
2	5900	2	11800	2h	23600h
3	5915	4	23660	3h	70980h
4	5943	2	11886	4h	47544h
5	5975	4	23900	5h	119500h
6	5995	1	5995	6h	35970h
			SOP = 106631		SOM = 321134h

KB = 3.012m

Volume = 3554.667
W = 36432.258

TPC = A/100 ⨯ density of water
= 61.448

We can calculate – FWA = W/40 ⨯ TPC

= 14.82cm

Q8. The breadths of a bulkhead, at 3.0m intervals from the top, are : 19.2, 18.0, 17.1, 16.2, 14.4, 12.0, 9.3 and 6.0 m. Find the distance of its geometric cente from the top. (use Simpson’s I Rule for first five ordinates and II rule for remaining ordinates).

Solution –

Ordinate	SM	Product	Lever	moment
19.2	1	19.2	0h	0
18.0	4	72.0	3h	216
17.1	2	34.2	6h	205.2
16.2	4	64.8	9h	583.2
14.4	1	14.4	12h	172.8
		SOP = 204.6		SOM = 1179.2h

Moment = 3/3 ⨯ 1179.2
= 1179.2

Area = 3/3 ⨯ 204.6
= 204.6

Ordinate	SM	Product	Lever	Moment
14.4	1	14.4	12h	172.8h
12.0	3	36.0	15h	540h
9.3	3	27.9	18h	502.2h
6.0	1	6.0	21h	126h
		SOP=84.3		SOM = 1341h

Moment = (3 ⨯ 3/8) ⨯ 1341
= 1508.625

Area = 9/8 ⨯ 84.3
= 94.83

Total moment = 2685.825
Area = 299.43

Therefore,
Geometric Centre = ( Total moment / Area)

= 8.97m from top.

Q9. A ship’s water plane is 216m in length. The half ordinates of the waterplane commencing from forward are as follows;
0.4, 4.4, 8.8, 11.0, 11.6, 11.8, 11.8, 11.6, 9.6, 7.0, 0.4 respectively.
The spacing between the first three and last three ordinates is half of the spacing between the other half ordinates. Calculate her waterplane area and the position of the CF with respect to the midlength.

Solution –

Ordinate	SM	Product	Hours	Moment
0.4	0.5	0.2	0h	0h
4.4	2	8.8	0.5h	4.4h
8.8	1.5	13.2	h	13.2h
11.0	4	44.0	3h	8.8h
11.6	2	23.2	3h	69.6h
11.8	24	47.2	4h	181.8h
11.8	1.5	23.6	5h	118h
11.6	4	46.4	6h	278h
9.6	1.5	14.4	7h	100.8h
7.0	2	14.0	7.5h	105h
0.4	0.5	0.2	8h	1.6h
		SOP=235.2		SOM=967.8h

Area = 2 ⨯ 27/3 ⨯235.2
= 4233.6m²

COG from one end = 111.099

COG from m/ship = 3.099.

Q10. A double bottom tank is 1.5m deep. The horizontal Areas of the tank at equal intervals, commencing from the tank top are 192, 186 and 158 sq.m. respectively. The tank is ballasted to a sounding of 0.75m with water of R.D. 1.012. Calculate the weight of ballast and its KG.

Solution –

Using Simpson’s 3 rules

Volume = 0.75/1 [5 ⨯ 158 + 8 ⨯ 186 – 196]
= 130.375m³

Weight = 131.94 tonnes

For moment = 0.75/12 [3 ⨯ 158 + 8 ⨯ 186 – 196]
= 50.203m⁴

KG = 0.385m

Q11. A wall sided vessel, 150m in length of constant waterplane, the semiordinates of which at equal intervals are : 0, 5, 9, 10, 9, 5 and 0m, has a double bottom 1m deep extending from side to side between the 9m and 9m half ordinates given. Calculate the thrust on the tanktop if the tank is run up by opening the sea valve, if her original evenkeel draft was 5.2m.

Solution –

Original even Keel draft = 5.2m

Semi-ordinate	SM	Product area
0	1	0
5	4	20
9	2	18
10	4	40
9	2	18
5	4	20
0	1	0
		SOP = 116

Water Plane area = 25/3 ⨯ 116
= 966.67 ⨯ 2
= 1933.33m²

TPC = 1933.33/100 ⨯ 1.025
= 19.816

Volume between ordinate of 9m & 9m =?
Area between 9m & 9m = 25/12 ⨯ [45 + 80 – 9] ⨯2 ⨯ 2
= 966.67m²

Volume between 9m & 9m = 966.67m²
Weight of water = 990.83 tonnes
Sinkage = W/TPC
= 0.5m

Therefore, final draft = 5.7m = (5.2 + 0.5)
Thrust on tank top = (Thrust due to draft – Thrust due to tank water)

Thrust = Depth ⨯ Density ⨯ Area
= (5.7 ⨯ 1.025 ⨯ 977.07) – (1 ⨯ 1.025 ⨯ 966.07)
= 5647.769 – 990.83

= 4656.939 tonnes

Q12. The half breadths of a ship’s water-plane at 12m intervals from aft are: 0.0, 3.3, 4.5, 4.8, 4.5, 3.6, 2.7 and 1.5m. The half-breadth, midway between the first two from aft is 2m. At the fore end is an appendage by way of a bulbose bow 4.5m long. Its area is 24m² and its GC 2m from forward extremity. Find the area of the water-plane and the position of the COF. 

Solution –

Ordinate	SM	Product	Lever	POM
0.0	0.5	0	0	0
2.0	2	4	6	24
3.3	1.5	4.95	12	59.4
4.5	4	18	24	432
4.8	2	9.6	36	345.6
4.5	4	18	48	864
3.6	2	7.2	60	432
2.7	4	10.8	72	777.6
1.5	1	1.5	84	126
		74.05		3060.6

Area = (2 ⨯ 12/3) ⨯ 74.05
= 592.4m²

Total area = 616.4 m²

COG from aft = 41.33m
COG = (592.4 ⨯ 41.33) + (24 ⨯ 86.5)/616.4

         = 43.08m from aft

Q13. Three semi-ordinate spaced at equal distance of 20m are 7.5, 11.8 and 15.8m. Calculate

• Geometric centre of the area between first two and last two ordinates.

• Amount of cargo that can be loaded on the deck area between first two ordinates, if the load density of the deck is 10t/m² .

Solution –

Area between first two ordinate
= 2 ⨯ (20/12) ⨯ [5 ⨯ 7.5 + 8 ⨯ 11.8 – 15.8]
= 387m²

Area between last two ordinate
= 2 ⨯ (20/12) ⨯ [5 ⨯ 15.8 + 8 ⨯ 11.8 – 7.5]
= 553m²

Moment between first two ordinate
= 2 ⨯ (20 ⨯ 20/24) [3 ⨯ 7.5 + 10 ⨯ 11.8 – 15.8]
= 4156.67 tm

Moment between first last two ordinate from aft
= 2 ⨯ (20 ⨯ 20/24)  [3 ⨯ 15.8 + 10 ⨯ 11.8 – 7.5]
= 5263.335 tm

• Geometric centre between first two = 4156.67/387

                                                                       = 10.74m

• Geometric centre between last two = 5263.335/553

                                                                      = 9.518m

Q14. The TPC values for a ship at 1.2 meters intervals of draught commencing at the keel, are 8.2, 16.5, 18.7, 19.4, 20.0, 20.5, 21.1 respectively. Calculate her displacement at 7.2 metres draught.

Solution –

TPC	Ordinate	SM	Product of volume
8.2	8.2x	1	8.2x
16.5	16.5x	4	66x
18.7	18.7x	2	37.4x
19.4	19.4x	4	77.6x
20.0	20.0x	2	40.0x
20.5	20.5x	4	82.0x
21.1	21.1x	1	21.1x
			332.2x

TPC = (A/100) x density
A = TPC ⨯ (100/density)
X = 100/1.025

V/W volume = (1.2/3) ⨯ 332.3 ⨯ 100/1.025
= 12967.804

Displacement = (v/w volume ⨯ density)
= 13291.9 tonnes