Page 1 of 17 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 2 of 17 Answer 2
Given:-
1. Dimensions of the vessel = 60m * 10m * 10m 2. RD of DW1 = 1.020 3. Draft = 6m 4. 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
Page 3 of 17 Answer 3
Given:-
1. Shape of vessel is triangular 2. Displacement of vessel = 650t 3. RD of DW = 1.015
4. Dimensions of water plane rectangle = 30m * 8m
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
A
B
C
30m
D
C
Page 4 of 17 640.394 = 1 * 8 * AD * 30 2 640.394 * 2 = h 8 * 30 5.337m = h Draft of vessel = 5.337m
KB = 2 *Draft (Triangular shaped vessel)
3 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
A
D
E
40m
F
B
D
C
12m
Page 5 of 17 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 6 of 17 Answer 5
Given:-
1. Dimensions of log = 3m * 0.75m * 0.75m 2. RD of log = 0.8
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
3m
0.75m
B
0.75m
B
RD of log = 0.8
Page 7 of 17 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. Square section of log = 0.5m2 2. RD of water = 1.005 3. Draft = 0.4 4. RD2 = 1.020 5. RD of log = 0.8
0.5m
2B
RD of log = 0.8
RD of water = 1.005
0.4m
B
0.5m
2B
RD of log = 0.8
RD of water = 1.020
Page 8 of 17 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 = draft 0.5 0.394m = draft COB = draft 2 COB = 0.394 2 COB = 0.197m
RD of log = Mass of log Total volume of log
Page 9 of 17 RD of log = Mass of log
Area of square section * height 0.8 = 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. Diameter of drum = 0.8m 2. Height of drum = 1.5m 3. Weight of drum = 10kgs
4. 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
Page 10 of 17 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 = draft 22 * 0.4 * 0.4 0.994m =draft
1.5m
0.8m
490 KG
10 KG
Page 11 of 17 KB = Draft 2 KB = 0.994 2 KB = 0.497m KB of the vessel is 0.497m Answer 8 Given:-
1. Side of triangular bow = 12m
2. Dimensions of mid part = 80m * 12m 3. Radius of semicircular stern = 6m 4. Light displacement of barge = 500t 5. Cargo loaded = 5000t 6. Π = 3.142 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
12m
12m
12m
80m
6m
A
B
C
D
E
F
Page 12 of 17 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
Page 13 of 17 Answer 9
Given:-
1. Breath of bow triangle = 12m
2. Length in fore & aft direction of Δ = 12m 3. Dimensions of rectangle = 50m * 12m 4. Displacement = 3444t
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
A
B
C
E
D
G
H
12m
12m
50m
Page 14 of 17 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 Weight (t) LCB(m) Moments(tm) Rectangle 3075 25 76875 Triangle 369 54 19926 Total 3444 96801
Page 15 of 17 LCB = Total moments Total weight LCB = 96801 3444 LCB = 28.107m KB is 2.5m LCB is 28.107m Answer 10 Given:- 1. Length of barge = 45m 2. Dimensions of rectangle = 8m * 4m 3. Breath of Δ = 8m 4. Depth of Δ = 3m 5. Displacement of barge = 1620t RD of FW = 1.000
B
A
C
D
E
F
3m
8m
45m
Page 16 of 17 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
Page 17 of 17 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 Weight (t) KB (m) Moments (tm) Δ ABC 540 2 1080 Rectangle BCED 1080 4.5 4860 Total 1620 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-