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
Page 7 of 17
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
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
Page 12 of 17
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
Page 14 of 17
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
