CEE 3310 – Hydrostatics, Sept. 10, 2010 41

2.7 Review

• Compressible fluids ⇒ Isothermal, adiabatic

• The U.S. standard atmosphere is based on the concept of adiabatic conditions

(γ = γ(θ)) for the troposphere (0 < z < 11.0 km) and isothermal conditions

(γ = γ0) for the stratosphere (11.0 < z < 20.1).

• If the surface S is horizontal, P = a constant ⇒ FR = PA

2.8 Pressure Prism

Consider the following example:

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We can solve this directly as just shown, however, for many situations (or just for many

people who prefer to think in a different manner!) a decomposition of the pressures into

a series of pressure prisms is often easier. Consider the decomposition such that

FyR= F1y1 + F2y2

= (γbh2h1)

(

h1 +h2

2

)

+

(

γbh2

2

2

) (

h1

2h2

3

)

Therefore yR =

h1

(

h1 +h2

2

)

+h2

2

(

h1 +2h2

3

)

h1 +h2

2

= h1 +h2

2

h1 +2h2

3

h1 +h2

2

= 4 + 3

(

4 + 4

4 + 3

)

= 7.43′

CEE 3310 – Hydrostatics, Sept. 10, 2010 43

2.9 Review

• Hydrostatic force on an inclined plane surface

FR = PCA

xR =Ixycγ sin θ

PCA

yR =Ixcγ sin θ

PCA

• Pressure prism → break the force up into a part arising from the constant

pressure due to everything that happens to the point just above the surface and

the depth varying part over the surface.

⇒ Most easily applied to shapes with no lateral variation and constant density over

the surface.

2.10 Hydrostatic Force on a Curved Surface

There are two approaches:

1. dFR = P dA ⇒ non-planar hence the integration is hard!

2. Apply a static control volume

∑

Fx = 0 = FAC − FH ⇒ FAC = FH

∑

Fz = 0 = −FAB − W + FV ⇒ FV = FAB + Q

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Where is the center of pressure?

FH must be collinear (no shear) with FAC

FV must be collinear with the resultant of FAB + W

Example – Oil Tanker

FH = FAC = PCA = γhCA = γ(d −r

2)rb

Now assume we are interested in solving for a unit width (i.e., b = 1 m). Therefore

FH = 10kN

m3(24m −

1.5m

2)(1.5m)(1m) = 349kN

2.10.1 Line of Action

How do we determine the line of action for the resultant force on a curved surface? We

could use our moment of inertia based method reviewed above but let’s use pressure

prisms here.

F1 = γ(d − r)rb = 10 · 22.5 · 1.5 · 1 = 337.5kN

y1 = d −r

2= 23.25m

CEE 3310 – Hydrostatics, Sept. 10, 2010 45

F2 = γr

2rb = 10 ·

1.5

2· 1.5 · 1 = 11.25kN

y2 = d −r

3= 23.5m

Therefore

yACFAC = F1y1 + F2y2 ⇒ yAC =337.5kN · 23.25m + 11.25kN · 23.5m

348.75kN= 23.26m

Similarly

FV = 355.3kN and xBC = −0.756m (x = 0 at the tanker’s vertical surface)

2.11 Buoyancy

FB =

∫

body

(P1 − P1) dAH = −γ

∫

(z2 − z1) dAH = γ · (body volume)

Therefore

FB = FV2− FV1

= Fluid weight above S2 − fluid weight above S1

= weight of fluid occupying the volume of the body

2.11.1 Archimedes’ Principal

FB = Weight of fluid displaced by a body or a floating body displaces its own weight of

the fluid on which it floats.

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2.11.2 Stability of Floating Bodies

The stability of a floating body depends on the location of the buoyancy force and the

weight of the body. They each exert a moment – one is a righting moment (the tendency

to rotate the object to an upright position) while the other is an overturning moment

(the tendency to flip the body over).