static force on surfaces...

Chapter (3)

Static Force on Surfaces (Buoyancy)

Page (43) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Static Forces on Surfaces-Buoyancy Fluid Mechanics

Introduction When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. We also know that (from previous chapter) the pressure will vary linearly with depth as shown in figures below.

The following figures exhibits the pressure distribution and the resultant force of this pressure for different planes:

Note that the resultant force(F ) always acting on the center of pressure (c.p) of the surface.



Resultant Force and Center of Pressure on a Plane Surface Immersed in a Liquid (General Case)

1. Determining the resultant force (퐅퐑):

F = dF but, dF = dP × dA and dP = ρg × h

→ F = ρg × h × dA but, ρg is constant →→

F = ρg × h × dA

The term h × dA is the first moment of 퐚퐫퐞퐚 about the 퐟퐫퐞퐞 퐬퐮퐫퐟퐚퐜퐞

h × dA = Object Area × Distance from centroide to the free surface

h × dA = A × y

y

c.g

dA

c.p

D



→ F = ρgAy

2. Determining the location of resultant force:

The resultant force act perpendicular to the immersed surface at the center of pressure (c.p) at some depth (D) as shown in figure above, the following is the derivation of the formula that used to calculate (D):

Moment = 0.0

What is moment!!! (Environmental engineering students ask ) Moment = Force × Arm Arm = is the distance 퐩퐞퐫퐩퐞퐧퐝퐢퐜퐮퐥퐚퐫 to the force

Moment = 0.0 →→→

F × y = dF × y but, dF = dP × dA and dP = ρg × h

→ F × y = ρg × h × dA × y but h = y sin(θ)

F × y = ρg × y sin(θ) × dA × y but, ρg sin(θ) is constant →

F × y = ρg sin(θ) × y × dA

The term y × dA is the second moment of area about tha free surface

and it is also called (moment of inertia "I") →→ F × y = ρg sin(θ) × I but → F = ρgyA →→ (ρgyA) × y = ρg sin(θ) × I

A × y × y = I sin(θ) → y =I sin(θ)

A × y

D = y sin(θ) → D = sin (θ) ×I

A × y

I = I + Ad (Parallel axis theorem) I = Moment of inertia about the center of gravity of the object A = Area of the object



d = distance from center of gravity to point O

d = y (as shown in figure above) → d = y =y

sin(θ) → Substitute in D

D = sin (θ) ×

⎣⎢⎢⎡I + A y

sin(θ)A × y

⎦⎥⎥⎤

→ D = sin (θ) ×I

A × y+ sin (θ) × A ×

yA × y × sin (θ)

→ D = sin (θ) ×I

A × y+ y

I advise you to use this equation when calculating (D).

The following figure shows the moment of inertia (I ) for different shapes:



Pressure Diagram For vertical or inclined walls of constant width it is possible to find the resultant force and the location of center of pressure using a Pressure Diagram. Pressure Diagram: Is the graphical representation of the gauge pressure change with vertical depth along the wall and this change is increasing linearly from zero (at free surface) to the maximum at the base of the wall.

As we know, the relation of calculating gauge pressure at any depth is: P = ρgh

How to calculate resultant force and center of pressure using pressure diagram:

Case I:

Always, resultant force equals the volume of pressure diagram: F = volume of pressure diagram

F =12

× γh × h × b =12

× γ × h × b (N) (γ = ρg)

If the width (b) is not known, we can represent the resultant force as force per unit area:

F =12

× γ × h (N/m)



Location of F : In the pressure diagram above is a triangle and the resultant force will be in the center of this triangle:

D = h −h3

=2h3

Case II:

Note that the object is fully immersed in fluid at distance h from the free surface. Assume the width (b) is 1m inside the page: F = F + F Let the vertical length of the object is L = (h − h ) F = Volume of the rectangle = γ × h × L × 1

F = Volume of the triangle =12

× γ × L × L × 1

Location of (F ): D = h + y h is given, but how we can find y

Take Moment about point A:

F at the center of rectangle L2

from point A



F at the center of Triangle 2L3

from point A

F × y = F ×L2

+ F ×2L3

→ y = ✓ → D = h + y = ✓

Case III: If the object is the same at the second case above, but is inclined in angle (θ) with the horizontal, do the same in case two, but in finding the location of center of pressure, find the inclined distance firstly (by taking moment about a specified point ) and then calculate the vertical distance.

Case IV: If there are two fluids or more, the pressure diagram will be as following:

F = F + F + F Assume the width (b) is 1m inside the page

F = Volume of upper triangle =12

× ρ × g × h × h × 1

F = Volume of rectangle = ρ × g × h × h × 1

F = Volume of lower triangle =12

× ρ × g × h × h × 1



To find the location D, take moment about the free surface:

F × D = F ×2h

3+ F × h +

h2

+ F × h +2h

3→ D = ✓

Important Note: Before solving any problem, if the width of the object is uniform (i.e. the object is rectangle), it is preferable to use pressure diagram method. However, if the object is not rectangle (width is not uniform) like triangle, trapezoidal and circle you are strongly advise to use equations derived in the beginning of this chapter.

Problems 1. An open tank has a vertical partition and on one side contains gasoline with a density 700 kg/m3 at a depth of 4 m, as shown in the figure below. A rectangular gate that is 4m high and 2 m wide is hinged at one end and is located in the partition. Water is slowly added to the empty side of the tank. At what depth (h) will the gate start to open?



2. The 4-m-diameter circular gate in the shown figure below is located in the inclined wall of a large reservoir containing water. The gate is mounted on a shaft along its horizontal diameter, and the water depth is 10 m above the shaft. Determine: a) the magnitude and location of the resultant force exertedon the gate by the water. b) the moment that would have to be applied to the shaft toopen the gate.



3. Two sides of the container are filled with two different liquids. Find the resultant force exerted on the gate AB and the position of center of pressure.



4. Calculate the minimum force F to keep the cover AB closed (the cover is rectangular with 1.5m wide).



5. For the system shown, water tank is pressurized to 90 cm Hg and 35 kPa.air pressure. Determine the resultant hydrostatic force per meter width on panel AB.



6. A 1.2 m wide gate of negligible weight shown in the figure pivots about the frictionless hinge O. the gate is held in position shown by the 1000 kg counter mass M. determine the water depth h.

Width = 1.2 m 1m 0.6m

M = 1000 kg



7. The gate in figure below is 1.5 m wide, is hinged at point B, and rests against a smooth wall at point A. Compute: (a) The force on the gate due to water pressure. (b) The horizontal force P exerted by the wall at point A. (c) The reactions at the hinge B.

1.8 m

2.4m

4.5m

Water

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