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Home , Hydrostatics, Mechanical equilibrium

Module 2:

. Hydrostatic and devices: 2 lectures

. on surfaces: 2.5 lectures

. , , stability: 1.5 lectures

Mech 280: Frigaard Lectures 1-2: Hydrostatic pressure

. Should be able to:

. Use common pressure terminology

. Derive the general form for the pressure distribution in static

. Calculate the pressure within a constant

. Calculate forces in a hydraulic press

. Analyze manometers and barometers

. Calculate pressure distribution in varying density fluid

. Calculate pressure in fluids in rigid body motion in non-inertial frames of reference

Mech 280: Frigaard Pressure

. Pressure is defined as a normal exerted by a fluid per unit area . SI Unit of pressure is N/m2, called a pascal (Pa). . Since the unit Pa is too small for many encountered in practice, kilopascal (1 kPa = 103 Pa) and mega-pascal (1 MPa = 106 Pa) are commonly used . Other units include bar, atm, kgf/cm2, lbf/in2=psi . 1 psi = 6.695 x 103 Pa . 1 atm = 101.325 kPa = 14.696 psi . 1 bar = 100 kPa (close to )

Mech 280: Frigaard Absolute, gage, and pressures

. Actual pressure at a give point is called the absolute pressure . Most pressure-measuring devices are calibrated to read zero in the . . Pressure above atmospheric is called gage pressure:

Pgage=Pabs - Patm . Pressure below atmospheric pressure is called vacuum pressure:

Pvac=Patm - Pabs.

Mech 280: Frigaard Pressure at a Point

. Pressure at any point in a fluid is the same in all directions . Pressure has a magnitude, but not a specific direction, and thus it is a scalar quantity . Later we will talk about the tensor: . Used to define the stress vector (traction) across any surface . Pressure is (minus) the mean normal stress

σ xx σ xy σ xz  − p 0 0  τ xx τ xy τ xz      σ = σ σ σ =  −  + τ τ τ = − δ +τ ij  yx yy yz   0 p 0   yx yy yz  p ij ij σ σ σ   −  τ τ τ   zx zy zz   0 0 p  zx zy zz 

. 1st index = direction of unit normal, 2nd index = direction of force . In a fluid at rest the shear stresses are zero . Pressure is the part of the stress that remains

Mech 280: Frigaard Trapezoidal paradox: fluids & ?

. Is the pressure at the bottom larger h in A or B?

. If material is ?

. If material is ?

W W

. Paradox was resolved by . Shown is a Pascal vase from Yale University

Mech 280: Frigaard Pascal - 1623–1662

. Blaise Pascal was a French , physicist, and religious philosopher. He was a child prodigy who was educated by his father. . Pascal's earliest work was in the natural and applied sciences where he made important contributions to the construction of mechanical calculators, the study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli. Pascal also wrote powerfully in defense of the scientific method. . Inventions include the hydraulic press (using hydraulic pressure to multiply force) and syringe. . Following a mystical experience in late 1654, he abandoned his scientific work and devoted himself to philosophy and theology. His two most famous works date from this period: the Lettres provinciales and the Pensées. However, he had suffered from ill-health throughout his life and his new interests were ended by his early death two months after his 39th birthday.

Mech 280: Frigaard Variation of Pressure with Depth

. In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers.

. To obtain a relation for the variation of pressure with depth, consider rectangular element

. Force balance in z-direction gives

∑ Fzz= ma = 0

P21∆− x Px ∆−ρ gxz ∆∆=0 . Dividing by ∆x and rearranging gives

∆P = P21 − P =ργ gz ∆=s ∆ z . We can also write this in differential form, dividing by ∆z: ∂P = ρg = γ or more generally, ∇P = ρge = γ e ∂z s g s g

Mech 280: Frigaard and Hydrostatic Pressure

Mech 280: Frigaard Example 1:

1 a) Compute the pressure on a scuba diver at 100 ft.

b) What is the danger of 100 ft ascending too quickly?

2

PV= PV If you hold your breath on ascent, your lung 11 2 2 V P3.95 atm volume would increase by a factor of 4, which 12= = ≈ 4 would result in embolism and/or death. V21 P1 atm Mech 280: Frigaard Example 2:

The glass flask is filled with water as shown. The column has inner diameter 3mm and the bulb of the flask has burst pressure of 3kPa.

Is it safe to add 1ml to the flask? Column

8cm

10cm Bulb

Mech 280: Frigaard Variation of Pressure with Depth

. Pressure in a fluid at rest is independent of the shape of the container . See the Trapezoid paradox . Pressure is the same at all points on a horizontal plane in a given fluid. . Consider force balance in x-direction

Mech 280: Frigaard Pascal’s Law

. Pressure applied to a confined fluid increases the pressure throughout by the same amount.

. If the pistons are at same height:

FF12 FA 22 PP12=→=→= AA12 FA 11 . Pascal observed: a small force could generate a large force

. Ratio A2/A1 is called ideal mechanical advantage Example 3: Hydraulic lift

What is the height h of oil required in the narrow tube to lift the load in the

3 large tube? (ρoil = 900 kg/m )

Mech 280: Frigaard Fluids in the news:

Mech 280: Frigaard Review: hydrostatic pressure basics

. From last lecture:

. General equation of hydrostatics: ∇P = ρgeg = γseg • Usually we align z with eg: ∂P = ρg = γ ∂z s . Constant density (and ): ∆P = P2 − P1 = ρg∆z = γs∆z

z2 . Variable density: ∆P = ∫ ρg dz z1

Mech 280: Frigaard The Manometer

. An elevation change of ∆z in constant density fluid at rest corresponds to ∆P/ρg . A device to measure pressure based on this is called a manometer . A manometer consists of a U-tube containing one or more fluids, e.g. mercury, water, alcohol, oil . Heavy fluids such as mercury are used if large PP12= pressure differences are = + ρ anticipated P2 Patm gh . Pressure range limited by height and fluid density

Mech 280: Frigaard Measuring Pressure Drops

. Manometers are well suited to measure pressure drops across valves, pipes, heat exchangers, etc. . Relation for pressure P1-P2 is obtained by starting at point 1 and adding or subtracting ρgh terms until we reach point 2. . If fluid in pipe is a , then ρ2>>ρ1 and P1-P2= ρgh

Mech 280: Frigaard Multi-fluid Manometer

. For multi-fluid systems . Pressure change across a fluid column of height h is ∆P = ρgh . Pressure increases downward, and decreases upward.

. Multi-fluid manometer . Two points at the same elevation in a continuous fluid are at same pressure. . Pressure can be determined by adding and subtracting ρgh terms.

P2++ρρ 11 gh 2 gh 2 + ρ 33 gh = P 1

Mech 280: Frigaard Example 1:

Calculate the air pressure in the pressurized tank, if h1 = 0.18 m, h2 = 0.2m and h3 = 0.25m. The density of the mercury, water and oil are 13,600 kg/m3,1000 kg/m3 and 850 kg/m3, respectively.

Mech 280: Frigaard The Barometer

. Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure.

. PC can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to Patm. . Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance.

PC +=ρ gh Patm

Patm = ρ gh

Mech 280: Frigaard in the atmosphere?

. Throughout atmosphere density varies significantly.

. How does the pressure vary?

. Assume : d P P 2 d P P g 2 d z = −ρ g = − g or ∫ = ln 2 = − ∫ d z RT 1 P P1 R 1 T

. For an isothermal atmosphere with T = To?  g(z2 − z1) P2 = P1 exp −   RTo  . How realistic is constant ?

. How realistic is constant gravitational acceleration?

Mech 280: Frigaard

Example 2:

Up to an altitude of approximately 36,000ft (l l,000 m), the mean atmospheric

temperature decreases nearly linearly and can be represented by: T = To - B z, where B is the lapse rate. The following values are assumed to apply for air, from

2 2 sea level to 36,000 ft: To = 288.16˚K (15 ˚C); B = 6.5˚K/km; R = 287 m /(s ˚K) Calculate atmospheric pressure as a function of z.

g /(RB)  Bz g P = Pa  1 −  where = 5.26(forair)  To  RB

Mech 280: Frigaard

Example 3:

Determine the pressure and density as a function of depth for a fluid in which P=Cρn

Mech 280: Frigaard Accelerating fluids in rigid-body motion

. In rigid body motion there is no shearing in the fluid . Pressure is again only component of the stress . There are 2 special cases where a body of fluid can undergo rigid-body motion: . linear acceleration, and . rotation of a cylindrical container.

. Newton's 2nd law of motion can be used to derive an equation of motion for a fluid that acts as a rigid body

∇P = ρgeg − ρa . In Cartesian coordinates: ∂∂∂PPP =−=−=−+ρρρa,, a( ga) ∂∂∂xyzxy x

Mech 280: Frigaard Example 4: Linear Acceleration

The truck, length 12m is 1/4 full with milk. It accelerates at constant 1m/s2.

How much higher is the level of milk at the back of the tanker than at the front?

ax≠==0, aa yz 0 ∂P ∂∂ PP =ρρag, = 0, = − ∂xx ∂∂ yz Total differential of P dP=−−ρρ ax dx gdz

P21−=− Pρρ ax ( x 21 − x) − gz( 21 − z) Pressure difference between 2 points a ∆=zz − z =−x ( xx −) ss2 s 1g 21 Find the rise by selecting 2 points on free

surface P2 = P1 Rotation in a Cylindrical Container

The container is rotating about the z-axis with constant angular velocity ω. Derive an equation for the z=h(r)

2 arz=−== raaω ,0θ ∂P ∂∂ PP Total differential of P =ρωrg2 , = 0, = −ρ ∂rz ∂∂θ dP=ρω r2 dr − ρ gdz

dz rωω22 isobar =→=z rC2 + dr g isobar 2g 1 On an isobar, dP = 0

2 ω 22 zhs =−−0 ( R2 r) 4g Equation of the free surface What we covered

. Pressure as part of the stress tensor – the mean normal stress, isotropic . General equations for hydrostatics:

∆P = P21 − P =ργ gz ∆=s ∆ z ∂P = ρg = γ or more generally, ∇P = ρge = γ e ∂z s g s g . Mechanical advantage using Pascals Law

FF12 FA 22 PP12=→=→= AA12 FA 11

. E.g. hydraulic lifts for auto repair, large elevators . Used manometers and barometers to measure pressure . Calculated pressure in a variable density fluids and applied it the atmosphere . Calculated pressure in a non-inertial frame of reference, for fluids in rigid motion

Mech 280: Frigaard Fluids in the news:

Mech 280: Frigaard Lectures 3-4: Forces on surfaces

. Should be able to:

. Calculate force & on a submerged planar surface

. Understand centre of pressure

. Understand how to use table of planar sections

. Apply planar surface techniques to computing forces and moments on curved surfaces

. Understand how the pressure force and the act on submerged objects

. Derive the buoyancy relationship and ‘Archimedes Principle’

Mech 280: Frigaard Motivation: structures like Hoover

Mech 280: Frigaard Oct. 4, 2010 - Kolontár, Hungary

. Bauxite tailings dam failed

. 700,000 m3 caustic red mud

. Several towns flooded, 10 people killed, approx. 120 people injured, 8 square kilometres flooded Basic dam structure . 1-2 similar tailings dam disasters each year

Mech 280: Frigaard The force the pressure exerts on a surface?

. In static fluid, stress tensor is isotropic: − p 0 0 contains only p   σij = 0 − p 0 . No shear forces     . From definition of stress tensor, the force per  0 0 − p unit area exerted on a fluid across a surface

with outward normal n is n.σ = Σi=1,3σij ni . Fluid exerts opposite force on walls: . Determines magnitude and direction of F = − ⋅ = . To determine point of action, we calculate F ∫ n σ dA ∫ pndA moment exerted by pressure A A . Similar to calculating static force balances in solid : the pressure is simply a distributed force Hydrostatic Forces on Plane Surfaces

. On a plane surface, the hydrostatic forces form a system of parallel forces . For many problems the magnitude of surface forces and location of the point of application of the surface forces must be determined. . This location is called the center of pressure.

. Atmospheric pressure Patm can be neglected when it acts on both sides of the surface.

Mech 280: Frigaard Force on Submerged Plane, A

. Consider a plane panel, A, completely submerged in fluid

. Force of magnitude F acts normal to A.

. Global coordinates (η,ξ)

. Find centroid of A: (ηCG,ξCG)

. Compute magnitude F F = ∫ p dA A Computation of force magnitude

= = + ρ ξ θ F ∫ p dA ∫ pa g sin dA A A = + ρ θ ξ pa ∫ dA g sin ∫ dA A A  ξ dA  ∫  = + ρ θ A  pa g sin ∫ dA  ∫ dA  A  A  = + ρ ξ θ [pa g CG sin ]∫ dA Magnitude of force is A given by Area x Pressure = p dA evaluated at the centroid CG ∫ A

Mech 280: Frigaard Where does force act?

. Compute moments of force about the centroid

. Define local coordinates:

(x,y) = (ηCG -η,ξCG -ξ) . Point of action CP defined by: = FxCP ∫ p xdA A = FyCP ∫ p ydA A . CP = centre of pressure

Mech 280: Frigaard Location of centre of pressure

= = + ρ ξ θ yCP F ∫ pydA ∫ (pa g sin )ydA A A = + ρ ξ − θ = − ρ θ 2 ∫ (pa g[ CG y]sin )ydA pCG ∫ ydA g sin ∫ y dA A A A y 2dA ρg sinθ ∫ I y = − A = − xx,CG CP p ξ + ρ θ CG ∫ dA [ CG (pa / g sin )]∫ dA A A

= = + ρ ξ θ xCP F ∫ pxdA ∫ (pa g sin )xdA A A = + ρ ξ − θ = − ρ θ ∫ (pa g[ CG y]sin )xdA pCG ∫ xdA g sin ∫ xydA A A A ∫ xydA ρg sinθ I xy,CG x = − A = − CP p ξ + ρ θ CG ∫ dA [ CG (pa / g sin )]∫ dA A A Mech 280: Frigaard Summary: forces on planar surfaces

. Force acts through centre of pressure (CP)

. In general CP lies under CG

. If we neglect: pa << ρgξcgsinθ

I I x = − xy,CG y = − xx,CG CP ξ CP ξ CG ∫ dA CG ∫ dA A A

. Recall Ixx,CG & Ixy,CG are the moments of inertia about CG

. These are tabulated for common shapes Mech 280: Frigaard Centroid and moments of inertia for some common shapes

Mech 280: Frigaard Example 1:

An above ground swimming pool is filled with water. Calculate the hydrostatic force on each wall and the line of action of the force. What is the effect of doubling the wall height on the static forces?

Mech 280: Frigaard Considerations in building

. Hydrostatic forces tend to cause a dam to: Toe . Slide horizontally . Overturn about the toe Sliding resistance . Safety factor against sliding: = Sliding force

Total resisting moment . Safety factor against overturning: = Total overturning moment

. Pressure intensity at toe

Mech 280: Frigaard Example 2: Dam

2m A concrete dam retaining 6m of water is as shown. The unit weight of the concrete is 23.5kN/m3 and the ground is impermeable. 7m Calculate: a) the safety factor against sliding; 6m b) the safety factor against overturning.

4m

Question 3.20 in Schaum

Mech 280: Frigaard Example 3:

Pa Water is held back in the tank as illustrated, by a 1.5m gate AB, that is 1.2m wide and has weight 20 kN. P A a Calculate the vertical force that must be applied to 1.5m the centre of gravity of the gate to keep it in 45° B equilibrium if: a) the gate is hinged at A; b) the gate is hinged at B.

Mech 280: Frigaard Curved surfaces and Archimedes principle

. Recall from planar surface theory: . Force exerted by fluid F = −∫ n ⋅σ dA = ∫ pndA A A . Acts through centre of pressure (CP)

= FxCP ∫ p xdA A = FyCP ∫ p ydA A . Can we use these to solve problems with curved surfaces?

Mech 280: Frigaard Hydrostatic Forces on Curved Surfaces

Fz

. FR on a curved surface requires integration of the pressure forces that change direction along the surface. = − ⋅ = = F ∫ n σ dA ∫ pndA (FH ,FV ) A A

. Determine horizontal and vertical components FH and FV separately. . Exploit mechanical equilibrium to evaluate these more simply Horizontal Component FH

. Project curved surface onto vertical plane . Balance horizontal forces = − = FH ∫ pnx dA Fx AAC Surface ABC A = = = AC Fx ∫ p(x xC , y, z) dA pCG ABC ABC x=xC pCG = pressure at centroid of ABC; |ABC| = area of ABC

. Repeat to find depth zCP of location where FH acts − = − − = − (zCP zCG )FH ∫ (z zCG )pnx dA ∫ (z zCG )p dA AAC ABC = − − ρ − = −ρ ∫ (z zCG )(pCG g(z zCG ))dA gI xx,CG ABC

ρgI xx,CG (zCP − zCG ) = − i.e. depth of center of pressure of ABC pCG ABC Vertical Component FV

. Project AAC onto horizontal plane AAB

Surface AAB . Balance vertical forces z=zA Volume V F = − pn dA = F + ρgV V ∫ z z h(x,y) AAC Surface ABC A = = = = AC Fz ∫ p(x, y, z z A ) dA p(z z A ) AAB AAB x=xC Horizontal location where FV acts: = − = = + ρ xCP FV ∫ xpnz dA ∫ x(p(x, y, z z A ) gh(x, y))dA AAC AAB = p(z = z )x A + ρgVx A CG,AAB AB CG,V p(z = z )x A + ρgVx A CG,AAB AB CG,V xCP = p(z = z A ) AAB + ρgV

i.e. weighted average of x-coordinates of centroids of AAB and V Adding centroids

= . Remember that in general: Vxcg ∫ xdv V

. But this is just an integral

. So we can add centroids for different volumes...

Vx = V x + V x +... cg 1 cg 2 cg 1 2

Mech 280: Frigaard Example 1:

Compute the vertical component F p=P A C V atm • of the force exerted on the curved • surface and find the horizontal c.g. location where FV acts

F v b B •

Mech 280: Frigaard Example 2:

A C - Hinge a) Determine and locate the components of 6m the force (per meter length) due to the water acting on the curved surface AB. B b) Calculate the net moment about the hinge at C. What do you observe?

Mech 280: Frigaard Summary: Forces on Curved Surfaces

. Horizontal force component on Fz curved surface: FH=Fx. . Line of action on vertical plane gives z coordinate of center of pressure on curved surface. . Vertical force component on curved surface: FV=Fz+W . W is the weight of the liquid in the enclosed block W=ρgV. . The x coordinate of the center of pressure is a combination of line of action on the horizontal plane (centroid of area) and line of action through volume (centroid of volume). . Magnitude of force: 2 2 1/2 FR=(FH +FV ) -1 . Angle of force: α = tan (FV /FH )

Mech 280: Frigaard What we learned

. Planar surfaces: . Learnt about centre of pressure and how to calculate forces and moments exerted by the fluid on plane surfaces . Learnt about safety factors for dams . Curved surfaces/shapes: easiest is to break net force into horizontal and vertical forces: . Horizontal forces • Project onto vertical plane and calculate as planar surface . Vertical forces • Project onto horizontal plane • Equal to weight of fluid enclosed plus vertical force through plane • Calculate x-location of action from x-location of centroids . Recall that we can add centroids

Mech 280: Frigaard Lectures 5-6: Buoyancy

. Buoyancy is due to the fluid displaced by a body.

FB=ρf gV . Archimedes principal: The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body and it acts upward through the centroid of the displaced volume.

Mech 280: Frigaard Buoyancy – Archimedes principle

. Buoyancy force FB is equal only to the displaced volume

ρf gVdisplaced. . Three scenarios possible

1. ρbody<ρfluid: Floating body

2. ρbody=ρfluid: Neutrally buoyant

3. ρbody>ρfluid: Sinking body Vertical forces on a cube?

. Submerged in water . Floating in water

h h H

Mech 280: Frigaard Forces on curved surfaces revisited

Mech 280: Frigaard Archimedes 287 BC – 212 BC

. Archimedes was a mathematician, physicist, engineer, astronomer, and philosopher, born in what is now Sicily. . Among his advances in are the foundations of hydrostatics and the explanation of the principle of the lever. His early use of calculus included the first known summation of an infinite series with a method that is still used today. He is also credited with designing innovative machines, including weapons and the screw pump that bears his name. He is best known for allegedly exclaiming "Eureka!" after discovering what is known today as Archimedes' principle. . Archimedes also shaped the fields of physics and practical engineering, and has been called "The greatest scientist ever”

Mech 280: Frigaard Example: Floating Drydock

Auxiliary Floating Dry Dock Resolute Submarine undergoing repair work on (AFDM-10) partially submerged board the AFDM-10

Using buoyancy, a submarine with a displacement of 6,000 tons can be lifted!

Mech 280: Frigaard Submarine Buoyancy and Ballast

. Submarines use both static and dynamic depth control. Static control uses ballast tanks between the pressure hull and the outer hull. Dynamic control uses the bow and stern planes to generate trim forces. Example

An elastic air balloon of diameter D is submerged in water and attached to the bottom of a tank by a cable. Determine the change in force in the cable when the tank pressure is increased from 100kPa to 200kPa, if the balloon diameter decreases according to: P = CD-2

Mech 280: Frigaard Fluids in the news:

Mech 280: Frigaard Golden Crown of Hiero II, King of Syracuse

. Hiero, 306-215 B.C.

. Hiero learned of a rumor where the goldsmith replaced some of the gold in his crown with silver. Hiero asked Archimedes to determine whether the crown was pure gold.

. Archimedes had to develop a method Mech 280: Frigaard Golden Crown of Hiero II, King of Syracuse

. The weight of the crown and nugget are the same in air:

Wc = ρcVc = Wn = ρnVn. . If the crown is pure gold, ρc = ρn which means that the volumes must be the same:

Vc=Vn.

. In water, the buoyancy force is

Β=ρH2OV.

. If the scale becomes unbalanced, this implies that the Vc ≠ Vn, which in turn means that the ρc ≠ ρn . Goldsmith was shown to be a fraud! Stability of floating objects

. Last lecture we considered Archimedes principle

. This implies a form of stability in the vertical plane

. Apply this to a floating body

• Move body up, reduce Vdisplaced, reduce FB, accelerate downwards

• Move body down, increase Vdisplaced, increase FB, accelerate upwards . After this lecture you should be able to:

. Analyze rotational stability of floating objects

. But, only objects that are axisymmetric are considered, e.g. the hull of a long boat

Mech 280: Frigaard Rotational stability of Immersed bodies

. Intuitively, rotational stability of immersed bodies depends upon relative location of center of gravity G and center of buoyancy B (centroid of displaced volume)

. W acts through G

. FB acts through B

Mech 280: Frigaard Stability of Symmetric Floating Bodies

MG >0: stable . If body is bottom heavy (G lower MG <0: unstable MG =0: neutrally stable than B), it is always stable . Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment.

. Metacentre M is where the line of Notes: action through B, before and after . M changes with rotation, so a body can rotation intersect be conditionally stable depending on degree of rotation . A measure of stability is the . For many practical hulls used, M is not very sensitive for rotations < 20° metacentric height MG The Vasa

Vasa from the port side

Career

Laid down: 1626 Launched: 1628 Sunk on her Fate: maiden voyage. General Characteristics Displacement: 1210 metric ton Toal Length: 69 m Beam: 11.7 m Draft: 4.8 m Height, keel to 52.5 m mast: Propulsion: 10 Sails, 3 Masts Sail area: 1,275 m2 Armament: 64 guns Sailors: 145 Soldiers: 300

Mech 280: Frigaard Vasa - The Story

. Regalskeppet Vasa (also Wasa) is a Swedish 64-gun of the line built for King Gustavus Adolphus of Sweden of the House of Vasa, between 1626 and 1628. . The ship's construction was plagued by interference from the King. . Shortly after the keel had been laid the king insisted that it be lengthened significantly. . The master shipwright who had been overseeing the construction became ill and died leaving his inexperienced apprentice in charge of construction. . The king then insisted on the addition of a second gun deck. . The resulting vessel was the best equipped and most heavily armed warship of its day, but one that was too long and tall for its width. . The standard stability test of the day . was thirty sailors running from side to side trying to rock the boat. When this was attempted on Vasa, the ship tilted significantly and the test was canceled. As none dared inform the king, the ship readied for sea. . On August 10 1628 Vasa set sail on her maiden voyage to the harbor of Stockholm. In the harbor a gust of wind forced the ship on her portside, after which water started flowing in through her open gun ports, and she soon sank, killing about 50 sailors. . The ship can be seen in the Vasa Museum in Stockholm, Sweden.

Mech 280: Frigaard FBd= ρ gV I MG=0 − GB Vsub

Where,

Vsub = the volume submerged, I0 = is the moment of inertia of the area of intersection of the water line and floating body, i.e., = 2 I0 ∫ x dA Awaterline

Note: Stability of non symmetric bodies is much tougher… Mech 280: Frigaard Derivation ?

Mech 280: Frigaard Mech 280: Frigaard Example

A barge has uniform rectangular cross section of width 2L and vertical draft of height H, as illustrated. Determine: a) the metacentric height for a small tilt angle; b) the range of ratios L/H for which the barge is statically stable if G is exactly at the waterline

Mech 280: Frigaard What we learned: buoyancy & stability

. Buoyancy: . Studied how the pressure force and weight act on submerged objects . Derived the buoyancy relationship and ‘Archimedes Principle’ . A body is . Stable if MG > 0 I0 . Unstable if MG < 0 MG= − GB Vsub . Meta Stable if MG = 0; . MG is the meta-centric height

. I0 is the area moment of inertia of surface defined by the intersection of the plane of the fluid with the object. . G is the centre of gravity of the floating object . B is the centroid of the displaced volume Mech 280: Frigaard