Chapter 8 (Centre of Buoyancy)

CHAPTER – 8 CENTRE OF BUOYANCY Chapter –8 Centre of Buoyancy Exercise -8

Answer 1 Given:1. Displacement = 1640t 2. Dimensions of vessel = 50m * 10m * 8m RD of SW = 1.025 RD of SW

= Displacement Underwater Volume

1.025

= 1640 Underwater Volume

Underwater Volume

= 1640 1.025

Underwater Volume

= 1600m3

Underwater Volume = L * B * draft 1600

= 50 * 10 * draft

1600 = draft 50*10 3.2m

= draft

KB

= draft 2

KB

= 3.2 2

KB

= 1.6m

KB of vessel in SW in 1.6m

Page 1 of 17

CHAPTER – 8 CENTRE OF BUOYANCY Answer 2 Given:1. 2. 3. 4.

Dimensions of the vessel = 60m * 10m * 10m RD of DW1 = 1.020 Draft = 6m RD of DW2 = 1.004

RD of DW1

= Displacement of vessel Underwater Volume

RD of DW1

= Displacement of vessel L * B * draft

1.020

= Displacement of vessel 60 * 10 * 6

1.020 * 60 * 10 * 6 = Displacement of vessel 3672t = Displacement of vessel We know that displacement of vessel remains constant RD DW2

= Displacement of vessel Underwater Volume

RD DW2

= Displacement of vessel L * B * draft

1.004

= 3672 60 * 10 * draft

Draft

= 3672 60 * 10 * 1.004

Draft

= 6.096m

KB = Draft 2 KB = 6.096 2 KB = 3.048m KB of the vessel is 3.048m

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CHAPTER – 8 CENTRE OF BUOYANCY Answer 3 Given:1. 2. 3. 4.

Shape of vessel is triangular Displacement of vessel = 650t RD of DW = 1.015 Dimensions of water plane rectangle = 30m * 8m

30m

B

D C

C

A

RD of DW

= Displacement of vessel Underwater Volume

1.015

= 650 Underwater Volume

Underwater Volume = 650 1.015 Underwater Volume = 640.3943 Underwater Volume = (Surface Area of triangle ABC) * (length) Underwater Volume = (1 * Base * Height) * (Length) 2 Underwater Volume = (1 * BC * AD) * (Length) 2

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CHAPTER – 8 CENTRE OF BUOYANCY 640.394

= 1 * 8 * AD * 30 2

640.394 * 2 8 * 30

=h

5.337m

=h

Draft of vessel = 5.337m KB = 2 *Draft 3

(Triangular shaped vessel)

KB = 2 *5.337 3 KB = 3.558m KB of the vessel is 3.558m

Answer 4 Given:1. Shape of vessel is triangular 2. Water plane area = 40m * 12m 3. KB = 3.6m RD of SW = 1.025

40m B D

C D

12m F

E

A

Page 4 of 17

CHAPTER – 8 CENTRE OF BUOYANCY KB

= 2 * draft 3

3 * KB = draft 2 3 * 3.6 = draft 2 5.4m

= draft

Area of triangle Δ ADE = 1 * Base * Height 2 Area of triangle Δ ADE = 1 * DE * AF 2 Area of triangle Δ ADE = 1 * 12 * 5.4 2 Area of triangle Δ ADE = 32.4m2 Underwater Volume = Area of Δ ABE * length Underwater Volume = 32.4m2 * 40m Underwater Volume = 1296m3 RD of SW

= Displacement Underwater Volume

1.025

= Displacement 1296

1.025 * 1296

= Displacement

1328.4t

= Displacement

Displacement of vessel is 1328.4t

Page 5 of 17

CHAPTER – 8 CENTRE OF BUOYANCY Answer 5 Given:1. Dimensions of log = 3m * 0.75m * 0.75m 2. RD of log = 0.8

RD of log = 0.8

3m

0.75m B 0.75m B

COG of log = height 2 COG of log = 0.75 2 COG of log = 0.375m RD of log = Mass of log Volume of log 0.8

= Mass of log 3m * 0.75m * 0.75m

0.8 * 3 * 0.75 * 0.75 = Mass of log 1.35t

= Mass of log

We know that Mass of log = Mass of water displaced by log RD of SW

= Mass of log Underwater Volume of log

RD of SW

= Mass of log L * B * draft

1.025

= 1.35 1.025 * 3 * 0.75 Page 6 of 17

CHAPTER – 8 CENTRE OF BUOYANCY Draft

= 0.585m

CB

= draft 2

CB

= 0.585 2

CB

= 0.293m

Difference between COG & COB

= COG – COB

Difference between COG & COB

= 0.375m – 0.293m

Difference between COG & COB

= 0.082m

Difference between its COG and COB is 0.082m

Answer 6 Given:1. 2. 3. 4. 5.

Square section of log = 0.5m2 RD of water = 1.005 Draft = 0.4 RD2 = 1.020 RD of log = 0.8

0.5m2 B

0.5m2 B

RD of log = 0.8

RD of log = 0.8

RD of water = 1.005

RD of water = 1.020

0.4m B

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CHAPTER – 8 CENTRE OF BUOYANCY

Underwater Volume at RD 1.005

= square section * draft

Underwater Volume at RD 1.005

= 0.5 * 0.4

Underwater Volume at RD 1.005

= 0.20m3

RD of DW

= Mass of log Underwater Volume

1.005

= Mass of log 0.20

1.005 * 0.20

= Mass of log

0.201t

= Mass of log

We know that mass of log will remain constatnt RD2

= Mass of log Underwater Volume

Underwater Volume = Mass of log RD2 Underwater Volume = 0.201 1.020 Underwater Volume = 0.197m3 Underwater Volume = square section * draft 0.197m3

= 0.5m2 * draft

0.197 0.5

= draft

0.394m

= draft

COB

= draft 2

COB

= 0.394 2

COB

= 0.197m

RD of log

= Mass of log Total volume of log

Page 8 of 17

CHAPTER – 8 CENTRE OF BUOYANCY RD of log

0.8

= Mass of log Area of square section * height

= 0.201 0.5 * height

Height = 0.201 0.5 * 0.8 Height = 0.503m COG = Height 2 COG = 0.503 2 COG = 0.252m Difference between COG & COB

= COG – COB

Difference between COG & COB

= 0.252m – 0.197m

Difference between COG & COB

= 0.055m

Difference between its COG and COB is 0.055m

Answer 7 Given:1. 2. 3. 4.

Diameter of drum = 0.8m Height of drum = 1.5m Weight of drum = 10kgs Weight of steel loaded = 490kgs

Radius of drum = Diameter 2 Radius of drum = 0.8 2 Radius of drum = 0.4m RD of FW = 1.000

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CHAPTER – 8 CENTRE OF BUOYANCY

0.8m 10 KG

490 KG

1.5m

Total weight = Weight of drum + weight loaded Total weight = 10kg + 490 kg Total weight = 500kg RD of FW

= Total Weight Underwater Volume

1.000 = 0.5 Underwater Volume Underwater Volume

= 0.5 1000

Underwater Volume = 0.5m3 Underwater volume of drum = πr2 * draft 0.5 = 22 * (0.4)2 * draft 7 0.5 = 22 * 0.4 * 0.4 * draft 7 0.5 * 7 22 * 0.4 * 0.4

= draft

0.994m

=draft Page 10 of 17

CHAPTER – 8 CENTRE OF BUOYANCY KB = Draft 2 KB = 0.994 2 KB = 0.497m KB of the vessel is 0.497m

Answer 8 Given:1. 2. 3. 4. 5. 6.

Side of triangular bow = 12m Dimensions of mid part = 80m * 12m Radius of semicircular stern = 6m Light displacement of barge = 500t Cargo loaded = 5000t Π = 3.142

D

B 12m

6m

12m

A 12m C

80m

E

Area of Δ ABC = √3 * a2 4 Area of Δ ABC = √3 * (12)2 4 Area of Δ ABC = √3 * 144 4 Area of Δ ABC = 62.354m2 Area of rectangle BCED = L*B Area of rectangle BCED = 80 * 12 Page 11 of 17

F

CHAPTER – 8 CENTRE OF BUOYANCY Area of rectangle BCED = 960m2 Area of semicircle EFD = 1 * π * r2 2 Area of semicircle EFD = 1 * 3.142 * (6)2 2 Area of semicircle EFD = 1 * 3.142 * 36 2 Area of semicircle EFD = 56.556m2 Total surface area of barge = Area of Δ ABC + Area of rectangle BCED + Area of semicircle EFD Total surface area of barge = 62.354 + 960 + 56.556 Total surface area of barge = 1078.91m2 Total weight of barge = Light displacement + cargo Total weight of barge = 500t + 5000t Total weight of barge = 5500t RD of SW = Total weight of barge Underwater volume RD of SW = Total weight of barge Total Surface area * draft 1.025 = 5500 1078.91 * draft Draft

= 5500 1078.91 * 1.025

Draft

= 4.973m

KB = Draft 2 KB = 4.973 2 KB = 2.487m KB of the vessel is 2.487m

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CHAPTER – 8 CENTRE OF BUOYANCY Answer 9 Given:1. 2. 3. 4.

Breath of bow triangle = 12m Length in fore & aft direction of Δ = 12m Dimensions of rectangle = 50m * 12m Displacement = 3444t

B

A

50m

E

12m H

12m

G

C

D

Area of Δ ABC = 1 * Base * height 2 Area of Δ ABC = 1 * 12 *12 2 Area of Δ ABC = 72m2 Area of rectangle BEDC = L * B Area of rectangle BEDC = 50 * 12 Area of rectangle BEDC = 600m2 Total surface area of barge = Area of Δ ABC + Area of rectangle BEDC Total surface area of barge = 72m2 + 600m2 Total surface area of barge = 672m2 RD of SW

= Displacement of barge Underwater Volume

RD of SW

= Displacement of barge Total surface area of barge * draft Page 13 of 17

CHAPTER – 8 CENTRE OF BUOYANCY 1.025 = 3444 672 * draft Draft

= 3444 1.025 * 672

Draft

= 5m

KB = Draft 2 KB = 5 2 KB = 2.5m Total weight of Δ ABC = Area of Δ ABC * draft * RD of SW Total weight of Δ ABC = 72m2 + 5m * 1.025 Total weight of Δ ABC = 369t Total weight of rectangle BEDC = Area of rectangle BEDC * draft * RD SW Total weight of rectangle BEDC = 600m2 * 5 * 1.025 Total weight of rectangle BEDC = 3075t LCB of rectangle BEDC = Length of rectangle 2 LCB of rectangle BEDC = 50 2 LCB of rectangle BEDC = 25m LCB of Δ ABC = Length of rectangle + ( 1 * length in fore & aft direction) 3 LCB of Δ ABC = 50m + (1 * 12) 3 LCB of Δ ABC = 50m + 4m LCB of Δ ABC = 54m

Rectangle Triangle Total

Weight (t) 3075 369 3444

LCB(m) 25 54

Moments(tm) 76875 19926 96801

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CHAPTER – 8 CENTRE OF BUOYANCY LCB

= Total moments Total weight

LCB

= 96801 3444

LCB

= 28.107m

KB is 2.5m LCB is 28.107m

Answer 10 Given:1. 2. 3. 4. 5.

Length of barge = 45m Dimensions of rectangle = 8m * 4m Breath of Δ = 8m Depth of Δ = 3m Displacement of barge = 1620t

45m

D

B

E

8m

F

C 3m

A RD of FW = 1.000

Page 15 of 17

CHAPTER – 8 CENTRE OF BUOYANCY RD of FW = Displacement of barge Underwater Volume 1.000 =

1620 Underwater Volume

Underwater Volume = 1620m3 Surface area of Δ ABC = 1 * Base * Height 2 Surface area of Δ ABC = 1 * BC * AF 2 Surface area of Δ ABC = 1 * 8 * 3 2 Surface area of Δ ABC = 12m2 Surface area of rectangle BCED = L * B Surface area of rectangle BCED = 8m * 4m Surface area of rectangle BCED = 32m2 Underwater volume of Δ ABC = surface area * length Underwater volume of Δ ABC = 12 * 45 Underwater volume of Δ ABC = 540m3 Volume remaining to be occupied = Volume to be displaced – volume occupied by Δ ABC Volume remaining to be occupied = 1620 – 540 Volume remaining to be occupied = 1080m3 Volume to be displaced by cuboid = L * B * h 1080 = 45 * 8 * h 1080 = h 45*8 3m = h Draft of vessel = height of Δ ABC + height of rectangle BCED Draft of vessel = 3m + 3m Draft of vessel = 6m Page 16 of 17

CHAPTER – 8 CENTRE OF BUOYANCY Mass of water displaced by Δ ABC = Volume of Δ ABC * RD of FW Mass of water displaced by Δ ABC = 540 * 1 Mass of water displaced by Δ ABC = 540t Mass of water displaced by rectangle BCED = Volume * RD of FW Mass of water displaced by rectangle BCED = 1080 * 1 Mass of water displaced by rectangle BCED = 1080t LCB of rectangle BCED = Length or rectangle KB of Δ ABC = 2 * draft 3 KB of Δ ABC = 2 * 3 3 KB of Δ ABC = 2m

Δ ABC Rectangle BCED Total

Weight (t) 540 1080 1620

KB (m) 2 4.5

Moments (tm) 1080 4860 5940

KB = Total moments Total weight KB = 5940 1620 KB = 3.667m LCB = Length of barge 2 LCB = 45 2 LCB = 22.5m 1. KB of the vessel is 3.667m 2. LCB of the vessel is 22.5m

-o0o-

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