Submerged Body, Center of Pressure and Buoyancy

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Hydrostatic Forces and stability

The Joola after capsizing (Senegal 2003)

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Hydrostatics Force p=0

p=0

h

h

p = γh p =γ h

p=0

p=0

• The pressure on the bottom is uniform so the resultant force acts through the centroid. • The pressure of the sides increases with decreasing depth. The force will not act through the centroid of the surface. • The centroid is the geometric mean position center of the surface. The line of action (center of pressure) weights the area integral by the force applied through that area.

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Definition of the centroid The centroid gives a definition of the mean position of an area (volume). It is closely related to the center of mass a body.

y

dAi xi

A

dx dy

xc

C yc yi

x

One adds up position of x for all the little pieces dAi of the Area, A to get average x position, xc . The x and y -coordinates of the centroid are evaluated mathematically as RR P i xi dAi A x dx dy P xc = = A i dAi RR P y dx dy yi dAi i A P yc = = A dA i i

4

First moment of Area The 1st moments of areas are the average displacement of an area about an axis of rotation. They are closely related to the centroid.

y

dAi xi

A

dx dy

xc

C yc yi

x

The first moment of area about the y -axis is ZZ X Qy = xi dAi = x dx dy A

i

Qy = Qy

ZZ

(x − xc ) dx dy + xc A ZZ = 0 + xc dx dy = xc A

ZZ

dx dy

A

A

The first moment of area about the x -axis is ZZ Qx = y dx dy = yc A A

The first moments of area have units of m3 .

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Second moment of Area The 2nd moments of areas are the average (displacement)2 of an area about an axis of rotation. Has units of m4

y

dAi xi

A

dx dy

xc

C yc yi

x

The second moment of area about the x -axis is ZZ X Ix = yi2 dAi = y 2 dx dy i

A

It is sometimes called the moment of inertia of the area. The second moment of inertia is always positive since y 2 > 0 . The second moment of area about the y -axis is ZZ X x2i dAi = x2 dx dy Iy = i

A

The product of inertia about an xy coordinate axes ZZ X xi yi dAi = Ixy = xydx dy i

A

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Parallel axis theorem Working out the second moments would be y dAi dx troublesome as the axes xi of rotations moved but xc for the parallel axes theorem. The moyi ments of many objects through their centroids are known. The second moment of areas are

A dy

C yc x

Ix = Ixc + yc2 A Iy = Iyc + x2c A Ixy = Ixyc + xc yc A One writes down second moment through centroid, then determines distance of centroid to axis of rotation, and finally apply the parallel axis theorem.

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Submerged inclined plane Want to work out forces on inclined surface.

• The x -axis points out of page. • The distance down incline is y . Depth is h . • p = γh (gauge pressure).

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Inclined plane To determine net force, need to add up all contributions over each small piece of the area.

dF FR FR FR

= p dA = γh dA one piece of A ZZ = γh dA adding up Z ZA = γy sin θ dA h = y sin θ A ZZ = γ sin θ y dA if γ and θ constant A

Now the integral over y is the first moment of Area, ZZ y dA = yc A A

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Inclined plane

ZZ

y dA = yc A

A

Since F = γ sin θ

RR

Ay

dA

FR = γAyc sin θ = γhc A

hc = yc sin θ

The net force on the plane depends on the depth of the plate centroid below the surface. The net force is the area multiplied by the pressure at the centroid.

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Center of pressure To determine the center of pressure, add up all contributions to the force over each small piece of the area, but multiplied by y . The center of pressure is essentially a weighted average. P yi δFi i hyi = P i δFi RR A y dF RR yR = A dF RR A yp dA RR yR = A p dA RR yγy sin θ dA A RR yR = γ sin θy dA RR 2 RR 2 y dA y dA A A yR = RR = yc A y dA

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Center of pressure yR =

RR

2 dA y A yc A

The y position of the force is the 2nd moment of the area with respect to the x axis. This is essentially the moment of inertia about the x axis. Ix yR = Ayc Parallel axis theorem Ix = Ixc + Ayc2 . Ixc Ixc + Ayc2 = + yc yR = Ayc Ayc The resultant force FR always passes below the centroid since yR > yc .

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Center of Pressure, x xR =

RR

A xydA

yc A

The mean x-position of the force can be determined by similar technique. This is just the product of inertia for the coordinate system. xR = xR = xR =

RR

A xydA

Ixy = yc A yc A Ixyc + xc yc A yc A Ixyc + xc yc A

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Geometric properties for shapes Rectangle A = ba 3

a/2

c

Ixc = ba /12 Iyc = ab3 /12

x a/2

y

Ixyc = 0

b/2

b/2

Circle A = πR2

R

Ixc = Iyc = πR4 /4

x

c y

Ixyc = 0 Half-circle A = πR2 /2 Ixc = 0.1098R

Iyc = 0.3927R4 Ixyc = 0

c

4

y R

x R

4R 3π

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More shapes Triangle d

A = ba/2

a

c

3

Ixc = ba /36

y (b+d)/3 b

Ixyc = ba2 (b − 2d)/72

a/3

4R 3π

Quarter Circle A = πR2 /4

c

Ixc = Iyc = 0.05488R4 Ixyc = −0.01647R4

x

R

x

y

Ixyc is only non-zero if the shape does not have a bilateral symmetry.

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Worked example Determine the resultant force on the plate and the reaction at the step.

3.0 m

1.2 m

Inlet

1.0 m

step

Hy ρ = 1000 kg/m3 Draw FBD for plate. Ignore the weight force. W = 0 , Hy = 0

Hx Fr

Fstep W

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Pipe-inlet Determine resultant force Fr = ρghc A 1 1 A = ab = 1.2 × 1.0 = 0.60 m2 2 2 a 1 hc = 3.0 + = 3.0 + 1.0 = 3.333 m 3 3 Fr = 1000 × 9.810 × 0.600 × (3 + 0.333) = 19.6 kN Ixc

1.2 × 1.03 = = 0.0333 m4 36

Now for center of pressure yr yr yr

Ixc + yc = Ayc 0.0333 = + 3.333 0.60 × 3.333 = 3.349 m

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Pipe-inlet: Step reaction Hy = 0 Net torque about hinge must be zero.

0.349 m Fr =19.6 kN

Hx Fstep W= 0

Fstep × 1.0 = Fr (0.349) Fstep = 19.6 × 103 (0.349) Fstep = 6.85 kN The force on the hinge determined from Fr − Fstep + Hx = 0 . Hx = Fstep − Fr = 6.85 − 19.6 = −12.7kN The force on the hinge acts to the left (opposes Fr )

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Pressure Prism

This is a intuitive recipe for determining the force on submerged surfaces. Useful for surfaces that are rectangular in shape. • Gauge pressure is zero at top and γh at bottom. • Pressure variation with h is linear. • Average pressure hpi = γh/2 • Resultant force Fr = hpiA = γ(hA/2) • Volume of pressure prism (= γhA/2) . • The center of pressure passes through the centroid of the pressure prism.

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Pressure Prism

The pressure prism can be regarded as arising from 2 parts. Let w be width of surface into page. Force due to rectangle ABDE . F1 = (γh1 )(h2 − h1 )w Force due to triangle BCD.   1 F2 = γ(h2 − h1 ) (h2 − h1 )w 2 The resultant force is FR = F1 + F2

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Pressure Prism

Determination of Center of Pressure done from moments of the forces. FR y R = F1 y 1 + F2 y 2 F1 y 1 + F2 y 2 ⇒ yR = FR Moment of rectangular part about AB level is 1/2 distance apex to base, i.e. y1 = 12 (h2 − h1 ) . Moment of triangular part is 2/3 distance apex to base, i.e. y2 = 23 (h2 − h1 ) .

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The tank problem Want to determine the force on the cover plate. Will also determine the center of pressure.

pg = 50 kPa Air Water

2.0 m 0.6 0.6 m

Useful hint: While the air pressure inside the tank needs to be taken into consideration, the impact of atmospheric pressure can be ignored (the tank pressure is a gauge pressure).

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Tank Problem, pressure prism

pair

γ x 2.0 B

2.0 m

A

C γ x0.6

0.6 m

pair pwater

The over-pressure due to air in the tank, 50 kPa is constant with depth. Pressures and forces due to (A+B) and C are FA+B = (50, 000 + 2.0 × 9, 800) × 0.62 = 25, 060 N FC

= ( 12 0.6 × 9, 800) × 0.62 = 1060 N

Total force is FR = 26, 100 N . The center of pressure is yR =

25060 × 0.3 + 1060 × 0.400 = 0.304 m 26100

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Force on curved surface Curved surfaces occur in many structures, e.g. dams and cross sections of circular pipes. The loads on the surface are all due to pressure forces. Look at forces acting on wedge of water ABC. Weight force W due to weight of wedge of water. Pressure forces F1 , F2 due to water above and from left. Reaction Forces FH , FV due to wall of tank.

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Curved surface: Free body diagram

The weight force W passes through the center of gravity of the wedge.

For static equilibrium, F1 + W

= FV

F2 = FH Also F2 is co-linear with FH and FV is co-linear with the resultant of F1 + W .

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Curved Surface, example Determine the resultant force on the curved part of the base and also determine its line of action. The bottom corner of the tank is a circle of radius 2.0 m . The tank length (out of page) is 8.0 m .

2.0 m 2.0 m 2.0 m

The centroid of the quarter circle wedge is 4R 3π = 0.84883 m

F1 2.0 m

xc = xc

2.0 m

It is 0.84883 m from the left boundary of quarter circle.

W 2.0 m

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Curved Surface, Vertical F1 = γ2.0(2.0 × 8.0) F1 = 313.6 kN

F1 2.0 m W

2.0 m

W

1 = γ π(2.0)2 × 8.0 4 = 246.3 kN

W 2.0 m

The total vertical force is 246.3 + 313.6 = 559.9 kN

F1 line of action 1.0 m from wall. W line of action 2.0 − 0.84883 = 1.151 m from wall. Line of action for F1 and W . xR xR

313.6 × 1.0 + 246.3 × 1.151 = 559.9 = 1.066 m

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Curved Surface, Horizontal Rectangle F = γ2.0(2.0 × 8.0) 2.0 m 2.0 m 2.0 m

F = 313.6 kN

F△ F△

1 = γ 2.0(2.0 × 8.0) 2 = 156.8 kN

The net horizontal force is 313.6 + 156.8 = 470.4 kN F line of action 1.0 m below 2.0 m line. F△ line of action 1.33 m below 2.0 m line. Line of action for horizontal force. yR yR

313.6 × 1.0 + 156.8 × 1.333 = 470.4 = 1.11 m

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Curved Surface, Summary

2.0 m 2.0 m 2.0 m

1.11 m 470.4 kN 731.3 kN 1.06 m 559.9 kN

The net force is p FR = 559.92 + 470.42 = 731.3 kN

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Buoyancy: Archimedes principle When a body is wholly or partially immersed in a fluid there is an upward buoyancy force equal to the weight force of the fluid displaced by the body.

p1

A

h px

px p = p1 +γ h 2

Consider pressure forces on a rectangular slab Fnet:pressure = p2 A − p1 A Fnet:pressure = (p1 + γh)A − p1 A FB = γhA = γV The buoyancy force arises as a result of higher pressure on the bottom surface.

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Archimedes principle: Proof Any body can be decomposed into a number of very small slabs. The buoyancy force on each slab is just δFi = γδVi .

δVi Therefore the proof for a rectangular slab can be generalized to a body of arbitrary shape. X δFi = γV. FB = i

Buoyancy force does not depend on the density of the submerged object. The buoyancy force only depends on the density of the fluid and the volume(shape) of the submerged object.

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Archimedes principle The Buoyancy force of a submerged body passes through its centroid. Called the center of buoyancy.

FB = γ V C

CG

W = mg The buoyancy force for a partially submerged object passes through the centroid of the displaced volume, V′ .

FB = γ V’ C

CG

W = mg The weight force passes through the center of gravity and does not always pass through the centroid.

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Stability The stability of a body depend on what happens when it is displaced from the equilibrium position.

Stable

Unstable

Neutral

• The equilibrium is stable if the forces acting on the object act to return it to its equilibrium position. • The equilibrium is unstable if the forces acting on the object act to send it away from its equilibrium position. • The equilibrium is neutral if there are no net forces acting on the object to return it or remove it from equilibrium.

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Buoyant stability: submerged

The buoyant force acts through the centroid of the object. The gravitational weight force acts through the center of gravity. • A completely submerged body is stable if the center of gravity lies below the centroid. • A completely submerged body is unstable if the center of gravity above the centroid.

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Buoyant stability: floating

• Even though the center of gravity lies above the centroid the resultant torques is a restoring torque. • The centroid (of the displaced volume) can shift as the body has an angular displacement. It is the movement of the centroid to the right that gives this body its stability.

35

Buoyant stability: floating

• In this case the body is unstable. • The buoyancy force and gravitational force act to create an overturning torque.

36

Buoyant stability: Technical Consider a floating body given a small angular deflection. The magnitude of the buoyancy force will stay the same, (weight force does not change) but the location of the centre of buoyancy changes.

M O G

dθ

F

B

Β1 W Β 2 dθ

O = Waterline point about which boat rolls. B1 = original buoyancy point. B2 = buoyancy point after displacement. The line of action of the original buoyancy force (through the center of gravity) and new buoyancy force intersect at the metacenter.

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Buoyant stability: Terminology Metacenter The metacenter M is the point of intersection between the original line of action and new line of action of the buoyancy force G The center of gravity moves as mass is added or removed, so the metacentric height d(GM ) changes as mass is added or removed. d(GM ) The metacentric height d(GM ) is the displacement of M from G . Negative if M below G dθ The angular displacement should be small, e.g. less than 20o . Positive Stability The ship is stable if d(GM ) > 0 , i.e. the metacenter lies above the center of gravity. Negative Stability The ship is unstable if d(GM ) < 0 , i.e. the metacenter lies below the center of gravity. Neutral stability Occurs when GM = 0 .

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The stability conditions M G

FB

Β1 W

FB Β1

Β2

G M W Β2

d(GM ) > 0 The torque applied to the vessel is a restoring torque

d(GM ) < 0 The torque applied to the vessel is an overturning torque

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Pitch and Roll

O

ROLL AXIS

M G

θ

Β1

dz

W FB ROLL AXIS

Β2

dy

y

x

dx

Iy d(B1 M ) = V V = displaced volume PITCH AXIS

Pitch and roll refer to the rotation about long and narrow axes of the vessel.

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Buoyant Stability Recipe • One determines position of center of gravity • Determine water level • One determines center of buoyancy of displaced volume • Then d(BG) is known. • Finally, d(BM ) is evaluated with Iy d(B1 M ) = BM = V

• Stable if d(BM ) − d(BG) > 0 • Unstable if d(BM ) − d(BG) < 0

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Stability example 1:Roll A wooden pine block, specific gravity = 0.500 . (out of page dimension = 2.0 m ) Is it stable?

1.0 m 1.2 m y

Determine water-level. FB = FG ρwater Vsubmerged g = mblock g = ρblock Vblock g ρwater ×1.0×y×2.0 = 0.50×ρwater ×1.0×1.2×2.0 y = 0.50×1.2 = 0.60 m Center of gravity is 0.60 m above base. Center of buoyancy is 0.30 m above base.

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Stability example 1: continued 1.0 m

d(KB) = 0.30 m d(KG) = 0.60 m

M 0.60 m

G B

1.2 m

K

The Metacentric height for roll is d(BM ) = =

1 3 2.0 1.0 Iy 12 = V 1.0 × 0.6 × 2.0 1.0 = 0.1389 m 12 × 0.60

The metacentric height is 0.3000 + 0.1389 = 0.4389 m above the keel. Since M is below G the block is not stable.

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Stability example 1: Pitch d(KB) = 0.30 m d(KG) = 0.60 m Out of page length = 1.00 m

2.0 m M 0.60 m

1.2 m G B K

The Metacentric height for pitch is d(BM ) = =

1 3 1.0 2.0 Iy 12 = V 1.0 × 0.6 × 2.0 4.0 = 0.5556 m 12 × 0.60

The metacentric height is 0.3000 + 0.5556 = 0.8556 m above the keel. Since M is above G the block is stable in pitch.

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Comments about stability 1.0 m

What can be done to improve stability?

M 0.60 m

G

1.2 m

B K

Stability is affected by position of M, B and G . d(BM ) =

Iy V

• Increase width. Makes Iy larger. (Catamaran). • Lower position of G . Add ballast (near the bottom). This also tends to raise B from the keel, and raises the position of M .