2 Fluid Statics - Submerged Surfaces

Fluid Statics:Submerged Surfaces

Moments of AreaArea:

A0

A

1 ⅆa

First Moment of Area about the y axis

Sy 0

A

xa ⅆa

Position of centroid

Cx ∫0Axa ⅆa

∫0A1 ⅆa

Cx ∫0Axa ⅆa

A

Cx Sy

A

Second Moment of Area about the y axis

Jyy x2 ⅆy ⅆx

Jyy 0

A

xa2 ⅆa

Parallel Axis Theorem

Jyy ⩵ A hc2 + Jcc

Parallel

Axis

Theorem

Product Moment of Area

Jxy x y ⅆx ⅆy

Jxy 0

A

xa ya ⅆa

Various lists of moments of area have been compiled, e.g. http://en.wikipedia.org/wiki/List_of_area

_moments _of _inertia

2 2 Fluid Statics - Submerged Surfaces.nb

Pressure on Submerged SurfacesFig 4.1 shows examples of forces on sumberged areas in engineering systems

Example of pressure exceeding design limits (video).

2 Fluid Statics - Submerged Surfaces.nb 3

Resultant ForcePressure varies over the surface depending on depth (elevation)

hahc

pc

pa

FR

hR

We want to calculate:

◼ Resultant Force, FR

◼ location of Centre of Pressure

The net force is the integral of the pressure over the area

FR ⩵ ∫0Apa ⅆa 5.31

We can substitute for pressure using the hydrostatic relationships

pa → γ ha

FR ⩵ ∫0Aγ ha ⅆa 5.32

We note that γ is constant and the remaining integral is the first moment of area

FR ⩵ γ Sy 5.33

Substituting for the first moment of area in terms of the centroid


Sy → A hc

FR ⩵ A γ hc Thrust

We note that γ Cy is merely the hydrostatic pressure at the centroid

γ hc → pc

FR ⩵ A pc 5.35


Centre of PressureThe Resultant Force does not act through the centroid!

This is because the pressure distribution is not constant with elevation.

It acts through the “Centre of Pressure”, which is at an elevation hR

Taking moments for elements of A, and summing over the entire area

FR hR ⩵ ∫0Aha pa ⅆa 7.1

We can replace pressure by hydrostatic head:

pa → γ ha

FR hR ⩵ ∫0Aγ ha2 ⅆa 7.2

The specific weight is constant so can come out of the integral

FR hR ⩵ γ ∫0Aha2 ⅆa 7.3

We note the integral is just the second moment of area

0

Aha2 ⅆa → Jyy

FR hR ⩵ γ Jyy 7.4

We have an expression for resultant (thrust) force:

FR → A γ hc

A γ hc hR ⩵ γ Jyy 7.5

Solving this for the height to the resultant, hR

hR ⩵JyyA hc

Centre of

Pressure

We can substitute for JYY using the parallel axis theorem to write this in terms of the second moment

of area about the centroid


Jyy → A hc2 + Jcc

hR ⩵A hc2 + Jcc

A hc

hR ⩵ hc +JccA hc

Centre of

Pressure

2

which is the result in eq 4-5


Moments of Area: Summarynth Moment of Area Common Name Common Symbol Integral Definition

0th Area A ∫x0 ⅆA

1st First Moment of Area (about y) Sy ∫x1 ⅆA

2nd Second Moment of Area (about y) Jy, Iy ∫x2 ⅆA

Key Relationships

◼ Distance of Centroid from y axis

Cx ⩵Sy

A

◼ First Moment of Area about an axis, y, offset from a parallel axis through the centroid

Sy ⩵ SC +ACx

◼ Second Moments of Area about an axis, y, offset from a parallel axis through the centroid (Parallel Axis Theorem)

Jy JC +ACx2

Application to Fluid Statics

◼ Resultant force on submerged, plane surface

FR⩵ pC A

◼ Centre of Pressure on submerged, plane surface, measured from the surface

LR ⩵J

A hC

LR ⩵ LC +JC

A LC


Exercise: Vertical Dam

First calculate the resultant force

FR ⩵ A γ hc

The centroid is halfway down the wall

hc →d

2

FR ⩵A d γ

28.1

Substituting known values:

{A → b d, b → 6.1, d → 3.7, γ → 9810.} 8.2

FR ⩵ 409612. 8.3

Now calculate the centre of pressure


hR ⩵Jyy

A hc

Jyy →b d3

3, A → b d, hc →

d

2

hR ⩵2 d3

8.4

Substituting known vales from (8.2)

hR ⩵ 2.46667 8.5

Alternatively, using the formula involving the second moment about the centroid

hR ⩵ hc +Jcc

A hc

JCC →b d3

12, hc →

d

2, A → b d

hR ⩵d2+2 Jccb d2

8.6

Which is the same result as (8.4)


Inclined PlanFor an inclined plane, the calculation of magnitude of the resultant force is the same.

However, we introduce an additional coordinate system for calculation of centre of pressure.

Fig 4-4

Taking moments about S, which is the intersection of the plane through the surface and the fluid

surface

FR Lp ⩵ ∫0ALa pa ⅆa 9.1

The pressure can be written as a function of head

pa → γ ha

FR Lp ⩵ ∫0Aγ ha La ⅆa 9.2

We can write the distance ha in terms of distance along the plane containing the surface

ha → Sin[θ] La

FR Lp ⩵ ∫0Aγ Sin[θ] La2 ⅆa 9.3

Both γ and Sin[θ] are constant with respect to the integral:

FR Lp ⩵ γ ∫0ALa2 ⅆa Sin[θ] 9.4

We note that the integral is merely the second moment of area about the intersection of the surface

plane and the fluid surface


0

ALa2 ⅆa → Jyy

FR Lp ⩵ γ Sin[θ] Jyy 9.5

We can expand the resultant thrust force FR as per the following:

FR → A γ hc

A γ hc Lp ⩵ γ Sin[θ] Jyy

hc → Sin[θ] Lc

A γ Sin[θ] Lc Lp ⩵ γ Sin[θ] Jyy 9.6

Solving this for the distance to the centre of pressure, Lp

Lp ⩵JyyA Lc

Centre of

Pressure

3

We can also expand (Centre of Pressure 3) using parallel axis theorem

Jyy → Jcc + A Lc2

Lp ⩵Jcc + A Lc2

A Lc

Lp ⩵JccA Lc

+ Lc

Centre of

Pressure

4

We can write this in terms of vertical heights:

Lc →hc

Sin[θ], Lp →

hp

Sin[θ]

Csc[θ] hp ⩵ Csc[θ] hc +Sin[θ] Jcc

A hc

hp ⩵A hc2 + Sin[θ]2 Jcc

A hc

hp ⩵ hc +Sin[θ]2 Jcc

A hc

Centre of

Pressure

5


Exercise: Inclined DamExample 4.5: Figure 4.7 shows a dam 30.5 m long that retains 8 m of fresh water and is inclined at

an angle of 60°. Calculate the magnitude of the resultant force on the dam and the location of the

center of pressure.

Answer

First calculate magnitude

FR ⩵ A γ hc

hc →h

2, A → L w

FR ⩵12h L w γ 10.1

We can write this in terms of vertical dimensions:

L → h Csc[θ]

FR ⩵12h2 w γ Csc[θ] 10.2

Substituting known values

{h → 8, w → 30.5, γ → 9810., θ → 60 °} 10.3

FR ⩵ 1.10557 × 107

The calculate location of the effective thrust force, using the equation where Jyy is the second

moment of area about the surface line


Lp ⩵Jyy

A Lc

Jyy →L3 w

3, A → L w, Lc →

L

2

Lp ⩵2 L3

10.4

Writing this in terms of vertical heights, and substituting values from (10.3

L → h Csc[θ]

Lp ⩵2

3h Csc[θ]

Lp ⩵ 6.1584 10.4


Exercise: Submerged Plane AreaExample 4.20: Calculate the resultant force and location of centre of pressure

Answer

First calculate the thrust force

FR ⩵ A γ hc

A →d2 π

4, γ → 1000 g sg, sg → 1.1, g → 9.81, hc → 3, d → 2.4 12.1

FR ⩵ 146452. 12.2

We now calculate the centre of pressure


Lp ⩵Jcc

A Lc+ Lc

A →π d2

4, JCC →

1

4π

d

2

4, Lc →

3

Cos[30 °]

Lp ⩵ 2 3 +2 Jcc

3 d2 π

Lp ⩵ 2 3 + 0.0638112 Jcc 12.3

We can also write this in terms of a vertical distance

hp ⩵3 Lp

2

hp ⩵1

23 2 3 + 0.0638112 Jcc

which, as expected, is just below the centroid of the circle.


Pressure DiagramsPressure diagrams show the pressure distribution over a submerged surface. They are most appro-

priate/helpful for rectangular submerged areas.


Pressure on Submerged Rectangular PlaneFor a rectangular surface, write the free body diagram equations

FP

FH

Fw

FR

Lp

hp

θ

hc

h

d

The forces are given by products of pressures and areas. Assume the rectangular surface has a

width (normal to the drawing) of y, and spans a distance x in the horizontal plane of the drawing.

Fp ⩵ h x y γ

FH ⩵ d d2+ h y γ

Fw ⩵12d x y γ

FR ⩵ FH2 + (Fp + Fw)2

d2+ h ⩵ hc

A ⩵ d2 + x2 y

11.1

We can eliminate FH, Fp, Fw, d, x, y from the equations and solve for the resultant force:

A2 γ2 hc2 ⩵ FR2

FR ⩵ A γ hc 11.2

Note the relationship between the second moment for the plane area, and the second moment for

the vertical projection:


JC,A ⩵112

d3 y Csc[θ]3

JC,V ⩵d3 y12

11.3

Csc[θ]3 JC,V ⩵ JC,A

We can then solve for the centre of pressure by summing moments about the point on the surface

FR LP ⩵ FH hP + FP xP + FW xW 11.4

We can make the following substitutions for the forces:

FR → Csc[θ] FHFP → A h γ Cos[θ]FH → A γ Sin[θ] hC

FW →12A d γ Cos[θ]

11.5

which gives:

A γ hC LP ⩵ A γ Sin[θ] hC hP + A h γ Cos[θ] xP +12A d γ Cos[θ] xW 11.6

We can also substitute for the various displacements

xP → Cot[θ] hC, xW →2 d

3+ h Cot[θ],

2 d

3+ h →

d

6+ hC, hP → hC +

Csc[θ] JC,V

A hC

A γ hC LP ⩵

A h γ Cos[θ] Cot[θ] hC +1

2A d γ Cos[θ] Cot[θ]

d

6+ hC + A γ Sin[θ] hC hC +

Csc[θ] JC,V

A hC

hC → Sin[θ] LC, JC,V → Sin[θ]3 JC,A

A γ Sin[θ] LC LP ⩵

A h γ Cos[θ]2 LC +12A d γ Cos[θ] Cot[θ]

d6+ Sin[θ] LC +

A γ Sin[θ]2 LC Sin[θ] LC +Sin[θ] JC,A

A LC

11.7

Solving (11.7) for LP gives:

LP ⩵ A d2 Cot[θ]2 + 6 A d Cos[θ] Cot[θ] LC + 12 A h Cos[θ] Cot[θ] LC +

12 A Sin[θ]2 LC2 + 12 Sin[θ]2 JC,A (12 A LC)

11.8

Simplifying, knowing that d>0, A>0, h>0, γ>0


LP ⩵1

2(d + 2 h) Cos[θ] Cot[θ] + Sin[θ]2 LC +

112

d2 Cot[θ]2 + Sin[θ]2 JC,AA

LC

LP ⩵12d Cos[θ] Cot[θ] + h Cos[θ] Cot[θ] +

d2 Cot[θ]2

12 LC+ Sin[θ]2 LC +

Sin[θ]2 JC,AA LC

11.9

For a rectangle we know that

d2 Cot[θ]2

12 LC→

Cos[θ]2 JC,A

A LC

LP ⩵1

2d Cos[θ] Cot[θ] + h Cos[θ] Cot[θ] + Sin[θ]2 LC +

Cos[θ]2 JC,A

A LC+Sin[θ]2 JC,A

A LC

LP ⩵12(d + 2 h) Cos[θ] Cot[θ] + Sin[θ]2 LC +

JC,AA LC

11.11

Finally, we make the substitution

1

2(d + 2 h) → Sin[θ] LC

LP ⩵ Cos[θ]2 LC + Sin[θ]2 LC +JC,A

A LC

LP ⩵ LC +JC,A

A LC

which verifies the centre of pressure result from integration.


Exercise (2013 Test)An engineering student loses control of a car and runs off the road into an alpine river.

(a) If the car has a mass of 2200 kg and encloses a volume of 1.8 m3, will it remain submerged?

(b) If no water has yet entered the passenger compartment, and given the dimensions shown

above, what is the force of water keeping the door closed?

(c) At what depth below the water surface does that force act?

Answer

If the car has a mass of 2200 kg and encloses a volume of 1.8 m3, will it remain

submerged?

List our known parameters

Arect ⩵ 0.9 × 1.1, Atri ⩵1

2(1.1 - 0.6) (0.9 - 0.45), yrect ⩵ 0.1 +

0.9

2,

ytri ⩵ 0.1 +1

3(0.9 - 0.45), g ⩵ 9.81, ρ ⩵ 1000, Vd ⩵ 1.8, m ⩵ 2200

{Arect ⩵ 0.99, Atri ⩵ 0.1125, yrect ⩵ 0.55,

ytri ⩵ 0.25, g ⩵ 9.81, ρ ⩵ 1000, Vd ⩵ 1.8, m ⩵ 2200}

13.1

Buoyant force is equal to the weight of fluid displaced. Weight force is due to mass and gravity

eq[Buoyancy]

eq[Buoyancy]

eq[Buoyancy] 13.2

Weight force is given by:


Fw ⩵ g m

Fw ⩵ 21582. 13.3

The weight force is greater than the buoyant force, so it will remain submerged.

If no water has yet entered the passenger compartment, and given the

dimensions shown above, what is the force of water keeping the door closed?

Calculate the area of the door, substituting values using (13.1)

A ⩵ Arect - Atri

A ⩵ 0.8775 13.4

Calculate the depth of the centroid, substituting values using (13.1)

hc ⩵Arect yrect - Atri ytri

Arect - Atri

hc ⩵ 0.588462 13.5

Calculate the force at that depth, substituting values using (13.1), (13,4) and (13.5)

FR ⩵ A g ρ hc

FR ⩵ 5065.64 13.6

At what depth below the water surface does that force act?

Calculate the second moment of area about the top of the door

(Note, hc is the depth beneath the water, which includes 0.1 m to the top of the door)

Jdoor ⩵ Jrect - Jtri

Jrect →1.1 × 0.93

3, Jtri →

1

12(1.1 - 0.6) (0.9 - 0.45)3

Jdoor ⩵ 0.263503 13.7

Offset the second moment of area to the centroid, by rearranging the parallel axis theorem

Jdoor ⩵ A (-0.1 + hc)2 + Jcc

Jcc ⩵ -1. 1. A (-0.1 + hc)2 - Jdoor 13.8

Substituting values from (13.1), (13.4), (13.5) and (13.7)


Jcc ⩵ 0.0541363 13.9

Now use the formula to calculate the centre of pressure (substituting from (13.9), (13.5) and (13.4)

hR ⩵ hc +Jcc

A hc

hR ⩵ 0.693301 13.11


2 Fluid Statics - Submerged Surfaces

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