Bouyancy and Stability

Bouyancy and Stability - Introduction

• Bouyant force is the force in fluid that tends to lift upward

• Whenever an object is floating or when it is completely submerge in the fluid, it is subjected to a buoyant force.

• Bouyancy is the tendency of fluid to exert a supporting force on a body placed in the fluid

• Net forces can be calculated on floating object on fluid or on object that is submerged in the fluid

• Stability of a floating or submerged body can be determine

• Stability refers to the ability of a body in a fluid to return to its original position after being tilted about horizontal axis

Bo

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t • Scuba diver is an example of submerged body

• The scuba diver will typically tend to float

• But, by adding calculated weight, the diver can swim at whatever depth is desired

• Sailboat uses application of stability• It is also an example of floating body• Sailboat has the wind force acting on it

sail and wave acting on the hull• The boat must remain stable under all

the forces acting on it

Objective

• Write the equation for the buoyant force

• Analyze the case of bodies floating on a fluid

• Use the principle of static equilibrium to solve the forces involved in buoyancy problems

• Define the conditions that must be met for a body to be stable when completely submerged in a fluid

Buoyancy

• A body in a fluid, whether floating or submerged, is buoyed up by a force equal to the weight of the fluid displaced

• Buoyant force acts vertically upward through the centroid of the displaced volume

• Discovered by Archimedes

• Buoyant force: 𝐹𝑏 = 𝛾𝑓𝑉𝑑 where 𝐹𝑏= buoyant force𝛾𝑓=specific weight of the fluid

𝑉𝑑=Displaced volume of the fluid

Buoyancy

•When a body is floating freely, it displaces a sufficient volume of fluid to just balance its own weight

• The analysis of problems dealing with buoyancy requires the application of static equilibrium in the vertical direction 𝐹𝑉 = 0, assuming the object is at rest in the fluid.

Procedure for Solving Buoyancy Problems

1. Determine the objective of the problem solution. What need to be calculated – force, weight, volume, specific weight

2. Draw a free-body diagram of the object in the fluid. Show all forces that act on the free body in the vertical direction, including the weight of the body, the buoyant force and all external forces. If the direction of some force is unknown, assume the most probable direction and show it on the free body

3. Write the equation of static equilibrium in the vertical direction, assuming the positive direction to be upward.

4. Solve for the desired force, weight, volume, or sp. weight. Remember the following concept:• The buoyant force is calculated from 𝐹𝑏 = 𝛾𝑓𝑉𝑑• The weight of a solid object is the product of its total volume and

its sp weight𝑤 = 𝛾𝑉.

• An object with an average sp weight less than that of the fluid will tend to float because 𝑤 < 𝐹𝑏 with the object submerged

• An object with an average sp weight greater than that of the fluid will tend to sink because 𝑤 > 𝐹𝑏 with the object submerged

• Neutral buoyancy occurs when a body stays in a given position wherever it is submerged in a fluid. An object whose average specific weight is equal to that of the fluid is neutrally buoyant

Example

• A cube 0.5m on a side is made of bronze having a specific weight of 86.9kN/m3. Determine the magnitude and direction of the force required to hold the cube in equilibrium when completely submerged (a) in water (b) in mercury. Specific gravity of mercury is 13.54.

• Solution1. Determine magnitude and direction of force to hold the cube (external

force)

2. Draw free-body diagram

Weight = w

𝐵𝑢𝑜𝑦𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑏 = 𝛾𝑓𝑉𝑑

External force that need to be determined, Fe

(assume direction)

3. Write equation for static equilibrium (upward force is positive, downward force is negative)

𝐹𝑉 = 0

𝐹𝑏 + 𝐹𝑒 −𝑤 = 0

4. Solve (for (a) first)

𝛾𝑤𝑉𝑐𝑢𝑏𝑒 + 𝐹𝑒 − 𝛾𝑐𝑢𝑏𝑒𝑉𝑐𝑢𝑏𝑒 = 0

Rearrange: 𝐹𝑒 = 𝛾𝑐𝑢𝑏𝑒𝑉𝑐𝑢𝑏𝑒 −𝛾𝑤 𝑉𝑐𝑢𝑏𝑒𝐹𝑒 = 86.9 0.5

3 − (9.81)(0.53)=9.63kN

The result is positive indicating the assumption was right.

Solve for (b) on your own. Ans. = downward (5.74kN)

Example

Based on apparent weight while submerged in water to be 7kN.

A certain solid metal object weighs10kN in the normal manner in air, butit has such an irregular shape that it isdifficult to calculate its volume bygeometry. Use principle of buoyancy.Determine its volume and specificweight.

Total weight=0.7kN

• FBD

𝐹𝑏 + 𝐹𝑒 −𝑤 = 0𝛾𝑤𝑉𝑠 + 𝐹𝑒 − 𝑤 = 0

𝑉𝑠 =𝑤 − 𝐹𝑒𝛾𝑤

=10 − 7

9.81

= 𝟎. 𝟑𝟎𝟔𝒎𝟑

Example

• A cube 80mm on a side is made of a rigid foam material and floats in glycerine with 60mm of the cube below the surface. Calculate the magnitude and direction of the force required to hold it completely submerged in glycerine, which has a specific gravity of 1.26

60mm below surface

• FBD when float freely

weight

Fb

𝐹𝑏 − 𝑤 = 0𝑤 = 𝐹𝑏

= 𝛾𝑔𝑙𝑦𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑= 9.81 × 1.26 0.60 × 0.80 × 0.80

= 4.746𝑘𝑁

• FBD when fully submerged

weight

Fb

𝐹𝑏 − 𝐹𝑒 −𝑤 = 0𝐹𝑒 = 𝐹𝑏 − 𝑤

𝐹𝑒 = 𝛾𝑔𝑙𝑦𝑉𝑐𝑢𝑏𝑒 − 4.746

𝐹𝑒 = 9.81 × 1.26 0.83 − 4.746

= 𝟏. 𝟓𝟖𝟑𝒌𝑵

Fe = ?

Example

• A brass cube 0.15m on a side weighs 0.3kN. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam has specific weight of 0.7kN/m3, what is the minimum required volume of the buoy?

brass

Light foam (0.7kN/m3)

• FBD

Light foam (0.7kN/m3)

brass

𝐹𝑏

𝐹𝑒 = 𝐹𝑏 𝑓𝑜𝑎𝑚

𝐹𝑏 𝑓𝑜𝑎𝑚

𝑤𝑓𝑜𝑎𝑚

𝑤𝑏𝑟𝑎𝑠𝑠

𝐹𝑒 + 𝐹𝑏 −𝑤 = 0𝐹𝑏 𝑓𝑜𝑎𝑚 + 𝛾𝑤𝑎𝑡𝑒𝑟𝑉𝑏𝑟𝑎𝑠𝑠 − 0.3𝑘𝑁 = 0

𝛾𝑓𝑜𝑎𝑚𝑉𝑓𝑜𝑎𝑚 + (9.81)(0.153) − 0.3 = 0

𝛾𝑓𝑜𝑎𝑚 + (9.81)(0.153) − 0.3 = 0

𝑉𝑓𝑜𝑎𝑚 =0.3𝑘𝑁 − (9.81)(0.153)

𝛾𝑓𝑜𝑎𝑚

𝑉𝑓𝑜𝑎𝑚 =0.2669

0.7= 0.381𝑚3

Buoyancy Materials• Design of floating bodies often requires the use of lightweight

materials – offer high degree of buoyancy

• When a relatively heavy object must be moved while submerged in a fluid, it is often desirable to add buoyancy to facilitate mobility

• The buoyancy material should typically have:• Low specific weight and density• Little or no tendency to absorb fluid• Compatibility with the fluid in which it will operate• Ability to be formed to appropriate shapes• Ability to withstand fluid pressures to which it will be subjected• Abrasion resistance and damage tolerance• Attractive appearance

Stability of Completely Submerged Bodies

• A body in a fluid is considered stable if it willreturn to its original position after being rotateda small amount about a horizontal axis

• Example of completely submerged bodies are submarines and weather balloons

Stability of Completely Submerged Bodies

• It is important for these kinds of object to remain in a specific orientation despite the action of currents, winds or maneuveringforces

• Condition of stability for submerged bodies: the center of gravity of the body must be below the center of buoyancy

• The center of buoyancy of a body is at the centroid of the displaced volume of fluid

• at center of buoyancy, buoyant force acts in a vertical direction

• The weight of the body acts vertically downward through the centerof gravity

Stability of Completely Submerged Bodies• The sketch of this undersea

vehicle has a stable configuration due to its shape and the location of the equipment within the structure

• The design places heavier equipment in lower part of the structure.

• Much of the upper part is filled with light syntactic foam to provide buoyancy

Stability of Completely Submerged Bodies• In the sketch, cb is the center of

buoyancy while cg is the centerof gravity. cg is located lowerthan cb

• In figure (b), the vehicle is having some angular displacement with total weight w acting vertically downward through the cg

• Fb is acting vertically upward through cb

Stability of Completely Submerged Bodies• Because their lines of action are

now offset, these forces create a righting couple that bring the vehicle back to its original orientation, demonstrating stability

• If the cg is above the cb, the couple created when the body is tilted would produce an overturning couple that would cause it to capsize

Stability of Completely Submerged Bodies

• Solid, homogenous objects have the cg and cb coincident and they exhibit neutral stability when completely submerged, meaning that they tend to stay in whatever position they are placed

Stability of Floating Bodies

• Different from submerged bodies

• Metacenter – is the intersection of the vertical axis of a body when in its equilibrium position and a vertical line through the new position of the center of buoyancy when the body is rotated slightly

• Condition of stability for floating bodies : its center of gravity is below the metacenter

Procedure for Evaluating the Stability of Floating Bodies1. Determine the position of the floating body, using the principles of buoyancy

2. Locate the center of buoyancy, cb; compute the distance from some reference axis to cb, called ycb. Usually the bottom of the object is taken as the reference axis

3. Locate the center of gravity, cg; compute ycg measured from the same reference axis

4. Determine the shape of the area at the fluid surface and compute the smallest moment of inertia I for that shape.

5. Compute the displaced volume Vd

6. Compute MB = I/Vd

7. Compute ymc = ycb+MB

8. If ymc>ycg, the body is stable

9. If ymc<ycg, the body is unstable

Example• Figure shows a flatboat

hull that, when fullyloaded, weighs 150kN.

• Note the location of thecenter of gravity, cg.

• Determine whether theboat is stable in freshwater.

𝐹𝑣 = 0

𝐹𝑏 − 𝑤 = 0𝐹𝑏 = 𝑤

Submerged volume:𝑉𝑑 = 𝐵 × 𝐿 × 𝑋𝐹𝑏 = 𝛾𝑓𝑉𝑑 = 𝛾𝑓𝐵𝐿𝑋

Then𝛾𝑓𝐵𝐿𝑋 = 𝑤

𝑋 =𝑤

𝛾𝑓𝐵𝐿=

150𝑘𝑁

(2.4𝑚)(6.0𝑚)(9.81𝑘𝑁𝑚3)

𝑋 = 1.06𝑚

Buoyant force,Fb

Weight, w

X

1. Determine the position of the floating body, using the principles of buoyancy

• ycg=0.8m as given in the question

• Because ycg>ycb, we must locate the metacentre to determine whether the boat is stable

2. Locate the center of buoyancy, cb; compute the distance from some reference axis to cb, called ycb. Usually the bottom of the object is taken as the reference axis

3. Locate the center of gravity, cg; compute ycg measured from the same reference axis

• Center of buoyancy is located at the center of the displaced volume of water

(1.06𝑚

2= 0.53𝑚) from bottom (reference axis)

• ycb=0.53m

Procedure 4-9

ymc (metacentre) =ycb + MB

𝑀𝐵 =𝐼

𝑉𝑑=𝐿𝐵3/12

𝐵𝐿𝑋=(6)(2.4)3/12

6(2.4)(1.06)= 0.45𝑚

ymc =0.53 + 0.45=0.98m

The moment inertia is determined about x-x axis because this would yield the smallest value for I

Hence, the boat is stable because metacentre is above the center of gravity.

What if the center of gravity is changed?

𝐹𝑏 = 𝑤

𝑋 =𝑤

𝐵𝐿𝛾𝑓=

150𝑘𝑁

(2.4𝑚)(6.0𝑚)(9.81𝑘𝑁𝑚3)= 1.06𝑚

𝑦𝑐𝑏 = 𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 =1.06

2= 0.53𝑚

MB = I/Vd=𝐿𝐵3/12

𝐵𝐿𝑋=(6)(2.4)3/12

6(2.4)(1.06)= 0.45𝑚

ymc (metacentre) =ycb + MB = 0.53+0.45 = 0.98m

Hence ymc< ycg (0.98m < 5m)

Therefore the object is unstable

ycg=5m

Weight =150kN

10m

2.4m

ymc

By increasing the height

Example

• A solid cylinder is 3ft in diameter, 6ft in height, 1550lb of weights. If cylinder is placed in oil (sg=0.9) with its vertical, would it be stable?

𝐹𝑏 =𝜋𝐷2

4(𝛾𝑜𝑖𝑙) 𝑋 = 𝑤

Hence: 𝑋 =𝑤 4

(𝛾𝑜𝑖𝑙)𝜋𝐷2=1550 4

(0.9(62.4))𝜋(3)2= 3.9𝑓𝑡

𝑦𝑐𝑏 = 𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 =3.9

2= 1.95𝑚

MB = I/Vd=

𝜋𝐷4

64

𝜋𝐷2

4𝑋

=𝐷2

16𝑋= 0.144𝑓𝑡

ymc (metacentre) =ycb + MB = 1.95+0.144= 2.09ftHence ymc< ycg (2.09ft < 3ft)

Therefore the object is unstable. It would tend to fall to one side until it reached a stable orientation. (probably with the horizontal axis)

Degree of Stability

• Some objects are more stable than others – relative stability

• To measure relative stability we used term – metacentric height (MG)

• Metacentric height (MG) is defined as the distance to the metacentre above the center of gravity

𝑀𝐺 = 𝑦𝑚𝑐 − 𝑦𝑐𝑔

Degree of Stability

• An object with larger metacentric height is more stable than one with smaller value

• Small sea going vessels should have MG = 1.5ft (0.46m), large ships should have MG > 3.5ft (1.07m) – (reference 1)

• Too large MG will cause uncomfortable rocking motions in the ship (seasick)

Static Stability Curve

• Another measure of stability of a floating object is - the amount of offset between the line of action of the weight of the object acting through the center of gravity and that of the buoyant force acting through the center of buoyancy

• The product of one of theforces and the amount ofthe offset produces therighting couple thatcauses the object toreturn to its originalposition

- and thus to be stable

Static Stability Curve• In this figure, the boat hull in

a rotated position

• Horizontal line is drawn through the center of gravity intersect buoyant force at point H

• The horizontal distance, GH is called the righting arm

• GH is a measure of the magnitude of the righting couple

Static Stability Curve

• Distance GH varies as the angle rotation varies

• This curve shows a characteristic plot of the righting arm versus the angle of rotation for a ship.

• This curve is called static stability curve• As long as the value of GH is positive, the

ship is stable (vice versa)• In this curve, the ship will be unstable at

around 68o

• Because steep slope of the curve after about 50o which can be the recommended limit for rotation