Equilibrium of Floating Bodies -...

Page (77) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Static Forces on Surfaces-Buoyancy Fluid Mechanics

Equilibrium of Floating Bodies: To be the floating body in equilibrium, two conditions must be satisfied:

The buoyant Force (Fb) must equal the weight of the floating body (W). Fb and W must act in the same straight line.

So, for equilibrium: F = W

The equilibrium of a body may be: Stable. Unstable. Neutral (could be considered stable) Stability of a Submerge Bodies Stable equilibrium: if when displaced, it returns to its original equilibrium position.

Unstable equilibrium: if when displaced, it returns to a new equilibrium position

Notes: In this case (body is fully immersed in water) when the body is tilted, the shape of the displaced fluid doesn’t change, so the center of buoyancy remains unchanged relative to the body. The weight of the body is located at the center of gravity of the body (G) and the buoyant force located at the center of buoyancy (B).

Stable Equilibrium: A small angular displacement υ from the equilibrium position will generate a moment equals: (W x BG x υ). The immersed body is considered Stable if G is below B, this will generate righting moment and the body will tend to return to its original equilibrium position.



Unstable Equilibrium: The immersed body is considered Unstable if G is above B, this will generate an overturning moment and the body will tend to be in new equilibrium position.

Stability of Floating Bodies Here, the volume of the liquid remains unchanged since Fb=W, but the shape of this volume changes and thereby its center of buoyancy will differ.

When the body is displaced through an angle υ the center of buoyancy move from B to B1 and a turning moment is produced.

Metacenter (M): The point at which the line of action of the buoyant force (Fb) cuts the original vertical line through G. So, Moment Generated is (W x GM x 훖). GM is known as a metacentric height.

Stability: Stable If M lies above G, a righting moment is produced, equilibrium is stable and GM is regarded as positive. (GM=+VE)



Unstable If M lies below G, an overturning moment is produced, equilibrium is unstable and GM is regarded as negative. (GM= ـــVE).

Neutral: If M coincides with G, the body is in neutral equilibrium. Determination of the Position of Metacenter Relative to Centre of Buoyancy:

BM =I

V

I = the smallest moment of inertia of the waterline plane

Procedures for Evaluating the Stability of Floating Bodies



1. Determine the position of the floating body (Draft) using the principles of buoyancy (Total Weights = Buoyant Force). 2. Locate the center of buoyance B and compute the distance from some datum to point B (yB). The bottom of the object is usually taken as a datum. 3. Locate the center of gravity G and compute (yG) measured from the same datum. 4. Determine the shape of the area at the fluid surface (plane view) and compute I for that shape. 5. Compute the displace volume (Vd) 6. Compute BM distance (BM = I / Vd). 7. Compute (yM = yB+BM) 8. If (yM > yG) >> the body is stable.(GM = +VE) 9. If (yM < yG) >> the body is unstable.(GM = ـــVE)

Important Note: If yM = yG (GM = 0), this case is called neutral and the object could be considered stable.



Problems 1. For the shown figure below, a cube of wood of side length (L) is float in water. If the specific gravity of the wood is 0.88. Determine if this cube is stable or not.

Solution Assume the datum is at the bottom of the cube. Firstly we calculate the draft (h=D) in terms of L from principles of buoyancy: F = Total weight F = γ × V = 9810DL

Total weight = ρ ( ) × g × V Total weight = 0.88 × 1000 × 9.81 × L = 8632.8 L

→ 9810DL = 8632.8 L → D = h =8632.8 L

9810L= 0.88L

→ V = DL = 0.88L

→ y =D2

=0.88L

2= 0.44L (from datum)

y =L2

= 0.5L

Calculation of BM:

BM =I

V

I =L × L

12=

L12

→ BM =L

12 × 0.88L= 0.0947L

→ y = y + BM → y = 0.44L + 0.0947L = 0.534L

GM = y − y = 0.534L − 0.5L = 0.034 L (+VE) → 퐒퐭퐚퐛퐥퐞✓.



2. A barge is 4.5m wide and 12m long and floats with a draft of 1.2m. It is piled so high with gravel so that its center of gravity became 1m above the waterline. Is it stable?

Solution Firstly, we draw a neat sketch for this problem: Assume the datum is at the bottom of the barge.

y =1.22

= 0.6m

y = 2.2m Calculation of BM:

BM =I

V

I =12 × 4.5

12= 91.125m , V = 4.5 × 12 × 1.2 = 64.8m

→ BM =91.1564.8

= 1.4m → y = 0.6 + 1.4 = 2m

GM = y − y = 2 − 2.2 = −0.2 (−VE) → 퐔퐧퐬퐭퐚퐛퐥퐞✓.

3. The figure below shows a cross section of boat. Show if the boat is stable or not. If the boat is stable compute the righting moment if the angle of heel is 10o. The boat is float in water and its 6m long.



Solution

Assume the datum is at the bottom of the boat.

y =1.52

= 0.75m

y =1.5 + 0.3

2= 0.9m

Calculation of BM:

BM =I

V

I =6 × 3

12= 13.5 , V = 3 × 6 × 1.5 = 27m

BM =13.527

= 0.5m → y = 0.75 + 0.5 = 1.25m

GM = y − y = 1.25 − 0.9 = 0.35 (+VE) → 퐒퐭퐚퐛퐥퐞✓. Important Note: If the boat is orientated with any angle, the value of GM remains unchanged.

Now when the boat is orientated with an angle of 10o the cross section will be as following:



The righting Moment (M)= W × X

W = F = γ × V = 9810 × 27 = 264870N

X = GM × sin(10) = 0.35 × sin(10) = 0.0607 m

→ M = 264870 × 0.0607 = 16077.6 N. m = 16.07KN. m✓.

4. A wooden cone floats in water in the position shown in figure below. The specific gravity of the wood is 0.6. Would the cone be stable?

Solution

Assume the datum is at the tip of the cone. The center of gravity of the cone is located at from the tip of the cone

Firstly we calculate the draft from principles of buoyancy: F = Total weight F = γ × V = 9810V Total weight = ρ × g × V

V =πr h

3=

π × 0.182 × 0.25

3

= 0.00212m

→ Total weight = 0.6 × 1000 × 9.81 × 0.00212 = 12.48 N.

→ 9810V = 12.48 → V = 0.001272m →→ (1)

Let the diameter at the level of water is X→ By interpolation→ .

.= → D = 1.388X

V =π × X

2 × 1.388X3

= 0.363X → substitute from (1)



0. 363X = 0.001272 → X = 0.152m → D = 1.388 × 0.152 = 0.21m

y =3 × 0.21

4= 0.1575m (from tip of cone)

y =3 × 0.25

4= 0.1875m (from tip of cone)

Calculation of BM:

BM =I

V

I = Moment of inertia at the plane at the water line The plane (top view) of the cone at the water line is a circle of diameter X

I =π

64D =

π64

× 0.152 = 2.62 × 10 m

V = 0.001272m

→ BM =2.62 × 10

0.001272= 0.02m → y = 0.1575 + 0.02 = 0.1775m

GM = y − y = 0.1775 − 0.1875 = −0.01 (−VE) → 퐔퐧퐬퐭퐚퐛퐥퐞✓.

Equilibrium of Floating Bodies -...

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