Miscelleneous

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

Q1. The TPC of a ship are as follows:
Draft (m)
6.05.04.03.02.0
TPC
22.4522.0421.5320.9119.68
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
6.022.45(100)⨯22.45/1.025122.45x4h89.8xh
5.022.0422.04x488.16x3h264.48xh
4.021.5321.53x243.06x2h86.12xh
3.020.9120.91x483.64x1h83.64xh
2.019.6819.68x119.68x0h0

SSR 01
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 
1.211.2
 9.6438.4
13.2226.4
15.0460.0
15.3230.6
15.6462.4
15.6231.2
14.7458.8
12.9225.8
9.0436.0
0.010.0
  370.8

SSR 03

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
0.10.50.05
7.5215
101.515
12448
12.3224.6
11.4445.6
81.512
5.2210.4
1.00.50.5
 SOP = 171.15

SSR 02

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
80018000h0
76043040h3040h
700214002h2800
600424003h7200h
4501.56754h2700h
18023604.5h1620h
100.5055h25h
  SOP = 8680 SOM = 17385h

SSR 04

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
0585015850    0h   0h
158854235401h23540h
259002118002h23600h
359154236603h70980h
459432118864h47544h
559754239005h119500h
65995159956h35970h
   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
19.2119.20h0
18.0472.03h216
17.1234.26h205.2
16.2464.89h583.2
14.4114.412h172.8
  SOP = 204.6 SOM = 1179.2h

SSR 05

Moment = 3/3 ⨯ 1179.2
= 1179.2

Area = 3/3 ⨯ 204.6
= 204.6

Ordinate
SM
Product
Lever
Moment
14.4114.412h172.8h
12.0336.015h540h
9.3327.918h502.2h
6.016.021h126h
  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
0.40.50.20h0h
4.428.80.5h4.4h
8.81.513.2h13.2h
11.0444.03h8.8h
11.6223.23h69.6h
11.82447.24h181.8h
11.81.523.65h118h
11.6446.46h278h
9.61.514.47h100.8h
7.0214.07.5h105h
0.40.50.28h1.6h
 SOP=235.2 SOM=967.8h

SSR 06

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

SSR 07

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  
01  0
5420
9218
10440
9218
5420
010
  SOP = 116

SSR 08

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
0.00.50  00
 2.024624
3.31.54.951259.4
4.541824432
4.829.636345.6
4.541848864
3.627.260432
2.7410.872777.6
1.511.584126
  74.05 3060.6

SSR 09

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 –

SSR 10

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
8.28.2x18.2x
16.516.5x466x
18.718.7x237.4x
19.419.4x477.6x
20.020.0x240.0x
20.520.5x482.0x
21.121.1x121.1x
   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

manish-mayank

About the author

Manish Mayank

Graduated from M.E.R.I. (Mumbai). A cool, calm, composed and the brain behind the development of the database. The strong will to contribute in maritime education and to present it in completely different and innovative way is his source of inspiration.

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