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7 – Static Stability of Floating Bodies, Slides of Fluid Mechanics

7 – Static Stability of Floating Bodies Description: Discusses metacentric height and equilibrium types Applies to ship design and floating structures Includes diagrams and sample computations

Typology: Slides

2021/2022

Available from 06/06/2025

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ENGR. BON RYAN ANIBAN
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ENGR. BON RYAN ANIBAN

ENGR. BON RYAN ANIBAN

3 Types of Stability

Neutral/ Upright BO Bo = center of buoyancy in upright position (center of displaced liquid)

G

G = center of gravity of the body BF W Stable BO

G

BF W ϴ

BO’

Bo’ = center of buoyancy in tilted position (center of displaced liquid)

M

M = metacenter, the point of intersection between the line of action of the buoyant force and the axis of the body Unstable BO

G

BF W ϴ

BO’

M

Righting moment Overturning moment

𝟐

B B (𝑓𝑜𝑟 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑜𝑛𝑙𝑦)

3 Types of Stability

Neutral/ Upright BO

G

BF W Stable BO

G

BF W ϴ

BO’

M

Unstable BO

G

BF W ϴ

BO’

M

Righting moment Overturning moment

𝟐

B

D

GM = Metacentric height, Distance from M to G

A rectangular scow 9 m. wide 15 m. long and 3.6 m high has a draft in sea water of 2.4 m.. Its center of gravity is 2.7 m above the bottom of the scow.

  1. Determine the initial metacentric height.
  2. Determine the final metacentric height when the scow tilts until one side is just at the point of submerged
  3. Determine the RM or Om when the scow tilts until one side is just at the point of submerged Solution for a. 9 m 3.6 m 2.7 m

G

2.4 m (^) Bo 1.2 m 1.5 m

𝐵^2

2 𝜃) =

2 0 ) 𝑀𝐵 0 = 2. 813 m.

= 2. 813 m. − 1. 5 m. 𝑀𝐺 = 1. 313 m.

A rectangular scow 9 m. wide 15 m. long and 3.6 m high has a draft in sea water of 2.4 m.. Its center of gravity is 2.7 m above the bottom of the scow. a. Determine the initial metacentric height. b. Determine the final metacentric height when the scow tilts until one side is just at the point of submerged c. Determine the RM or Om when the scow tilts until one side is just at the point of submerged Solution for b Bo ϴ

𝐵^2

2 𝜃) =

2 ( 14. 931 )) 𝑀𝐵 0 = 2. 913 m. 𝑀𝐺 = 𝑀𝐵 0 − 𝐺𝐵 0 = 2. 913 m.. − 1. 5 m. 𝑀𝐺 = 1. 413 m.

M

A rectangular scow 9 m. wide 15 m. long and 3.6 m high has a draft in sea water of 2.4 m.. Its center of gravity is 2.7 m above the bottom of the scow. a. Determine the initial metacentric height. b. Determine the final metacentric height when the scow tilts until one side is just at the point of submerged c. Determine the RM or Om when the scow tilts until one side is just at the point of submerged Solution for c Bo ϴ

𝑀𝐺 = 1. 413 m.

M

W BF

BO’

Righting moment 𝑅𝑀 = 𝑊 𝑜𝑟 𝐵𝐹 x (distance between the force) x

𝑥 = 𝑀𝐺 sin 𝜃 = 1. 413 sin 𝜃

𝑥 = 0. 364 m. 𝐵𝐹 = 𝑊 𝑉𝑑 x 𝛾𝑙 = 𝑊 9 2. 4 15 x 9. 81 ( 1. 03 ) = 𝑊 𝑊 = 3273. 793 kN 𝑅𝑀 = 𝑊𝑥 = 3273. 793 ( 0. 364 ) 𝑅𝑀 = 1191. 492 kN − m

A solid wood cylinder of sg = 0.6 is 600 mm diameter and 1200 mm high. If placed vertically in oil (sg=0.85). Would it be stable? Solution for c

𝑉𝐷 x 𝛾𝑙 = 𝑉𝑂 x 𝛾𝑂 𝑉𝐷 9. 81 0. 85 =

2 ) 4

𝑉𝐷 = 0. 240 m^3

𝐼𝑐𝑖𝑟𝑐𝑙𝑒 = 6. 362 x 10

− 3

𝑀𝐵 0 = 0. 027 m.

0.847 m

𝐺𝐵 0 = 0. 177 m. 𝑀𝐵 0 <^ 𝐺𝐵 0