Chapter 1 Hydrostatic Force

CHAPTER 1 : HYDROSTATIC FORCE

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CHAPTER 1 HYDROSTATIC FORCE 1.1

Introduction This chapter focuses on how to determine hydrostatic forces exerted on a plane or curved surface submerged in a static fluid. It covers the force produced by the pressure in a fluid that acts on the wall and the location of the resultant force, called the centre of pressure. At the end of this chapter, student should be able to: a) Define the terms of ‘Pressure’ and ‘Hydrostatic Force’. b) Understand the terms centroid and center of pressure. c) Understand the basic knowledge of hydrostatic forces acting on plane surface. d) Understand the basic knowledge of hydrostatic forces action on curve surface. e) Understand the basic knowledge of hydrostatic forces action on inclined surface.

1.2

Hydrostatic Force Hydrostatic is the branch of fluid mechanics that related to the fluids at rest. In other word, it's deal with pressures and forces resulting from the weight of fluids at rest. By referring Figure 1.1, the fluid exerts force and pressure against the walls of its container, whether it is stored in a tank or flowing in a pipe. But there is a difference between force and pressure, although they are closely related.

Hydrostatic Forces Figure 1.1: Hydrostatic Forces in Tank In summarize, the formula of hydrostatic forces (F) is: F=P*A [1]


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Where, P = pressure & A = area over which the force is distributed. Unit is in Newton (N). 1.2.1 Hydrostatic Pressure The pressure water exerts is called hydrostatic pressure. These principles always apply to hydrostatic pressure: a) Pressure depends only on the depth of water above the point in question (not on the water surface area). b) Pressure increases in direct proportion to the depth of water. c) Pressure in a continuous volume of water is the same at all points that are at the same depth or elevation. d) Pressure at any point in the water acts in all directions at the same magnitude. Specifically, pressure is defined as force per unit area. In S.I units, pressure is usually expressed in Newton per square meter (N/m2). For convenience, the unit N/m2 is called a Pascal (Pa). In this equation, pressure can be expressed as: P=F/A Where, P = pressure, F = hydrostatic force A = area over which the force is distributed Depending on the benchmark used (with/without atmospheric pressure), pressure can be described as absolute pressure or relative pressure. a)

Atmospheric pressure (ρa) is defined as the pressure at any given point in the earth atmosphere caused by the weight of air above the measurement point. Atmosphere pressure at sea level (standard) is approximately 101.325 kPa or 760 mmHg.

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b)

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Absolute pressure (ρabs) is the pressure with its zero point set at the vacuum pressure.

c)

Relative pressure (ρr) is the terms of pressure with its zero set at the atmospheric pressure. This pressure is more widely use in engineering than absolute pressure. There is the relationship between them is: ρr

= ρabs - ρa

1.2.2 Head Pressure It is often convenient to express pressure in terms of the height of a column of water in meters instead of terms of kPa. This is called pressure head, h. The water that filled into a tank will produce pressure at the sides and bottom of the tank. If the h is high, the hydrostatic pressure is calculated from the bottom of the tank: P

= ρgh

Where is ρ fluid density, g is gravity acceleration and h is head pressure. Problem 1.1 What is the pressure and force at the bottom of the cylindrical containers?

OIL (s.g = 0.9)

2.4 m

WATER (s.g = 1.0)

1.5 m

3.0 m

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Solution: The pressure for cylindrical tank: P

= ρgh(oil) + ρgh(water) = (0.9 x 1000 x 9.81 x 2.4) + (1.0 x 1000 x 9.81 x 1.5) = 21189.6 + 14715 = 35904.6 Pa = 359.046 kPa

The force for cylindrical tank: F

=PxA = 359.046 x (π x 32 / 4) = 253.795 kN

1.2.3 Definition of Centroid, C The centroid gives a definition of the mean position of an area (volume). It is closely related to the center of mass a body. 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: Figure 1.2: Centroid Location

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1.2.4 The First Moment of Area By referring Figure 1.2 above, the 1st moments of areas are the average displacement of an area about an axis of rotation. They are closely related to the centroid. The first moment of area about the y -axis is:

So;

The first moment of area about x –axis is:

The first moment of area have units of m3. 1.2.5 The Second Moment of Area, Ix The 2nd moments of areas are the average (displacement) of an area about an axis of rotation. Have units of m4. The second moment of area about the x -axis is:

It is sometimes called the moment of inertia of the area. The second moment of inertia is always positive since y2 > 0. The second moment of area about the y -axis is:

The products of inertia about a xy coordinate axes:

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Working out the second moments would be troublesome as the axes of rotations moved but for the parallel axes theorem. The moments of many objects through their centroid are known. The 2nd Moment of Area is:

One writes down second moment through centroid, then determines distance of centroid to axis of rotation and finally applies the parallel axis theorem. 1.3

Hydrostatic Force on Plane Area There are several steps to be followed for determine the hydrostatic force on plane area: a)

Specify the magnitude of the force.

b)

Specifying the direction of the force.

c)

Specifying the line of action of the force.

d)

To determine completely the resultant force acting on submerged force.

1.3.1 Hydrostatic Force on Horizontal Plane Area This is the simplest cases to find forces on horizontal plane area. By referring figure above, the tank bottom showed a uniform pressure distribution on the entire plane. So, the pressure at the bottom is: [6]


P

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= γh

Now the Resultant force, FR: FR = PA = γhA Where, A is bottom area of the tank. 1.3.2 Hydrostatic Force on Iclined Plane Surface Consider a plane shown in Figure below: Where:  The origin O is at the free surface.  θ is the angle that plane makes with the free surface.  y is directed along the plane surface.  A is the area of the surface.  dA is a differential element of the surface.  dF is the force acting on the differential element.  C is the centroid.  CP is the center of pressure.

 FR is the resultant force acting through CP.

Then, Differential force acting on differential area dA of plane: dF

= (Pressure) . (Area) = (γh). (dA)

(Perpendicular to plane)

Then, the magnitude of total resulting force, FR acting on the entire surface:

Where, h = y sin θ [7]


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With γ and θ take as constant: Note that the integral part is the first moment area about the x-axis: Where, yc = y coordinate of the centroid for the object.

Then,

Equation 1.1

Where, hc is the vertical distance from the fluid surface to the centroid of area. Now, we must find the location of the center of pressure where the resultant force acts as “The moment

of resultant force must equal to the moment of the distribution force”.

Moment about the x-axis is: *Note that dF = γh dA and h = y sinθ

We note that, F = y Ayc sin θ Then, [8]


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Where, Ix = ∫ y2 dA is 2nd Moment of Inertia

By using Parallel Axis Theorem: Where, Ixc = 2nd Moment of Inertia through centroid

Substitute parallel axis theorem and rearranging: hp = Ixc sin2 + hs hc A

Equation 1.2

where, hp = center of pressure height from fluid surface Ixc= second moment of area

 = angle of the inclined object A= area of the object hc= centroid height from fluid surface Note that for a submerged plane, the resultant force always acts below the centroid of the plane. Summary from above formula to determine Hydrostatic Force on Plane Surface: a)

Find area in contact with fluid.

b)

Locate centroid of that area.

c)

Find hydrostatic pressure Pc at centroid, typically = (generally neglect Patm).

d)

Find force F = Pc A.

e)

The location will not be at the centroid, but at a distance below the centroid.

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1.3.3 Geometric Properties for locating Centroid Coordinates, Areas & Moment of Inertia Shape

Area, A

Centroid Location, C x y

Moment of 2nd Inertia, Ixc

ba

b 2

a 2

ba3 12

ba 2

2b 3

a 3

ba3 36

πR2 4

2

R

R

πR2 2

R

4R 3π

0.1098R4

πR2 4

4R 3π

4R 3π

0.05488R4

πR

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1.3.3.1 Calculation of centroid, C or hc To determine the hydrostatic force, first we must locate centroid of the area. Centroid means the geometric center of the object’s shape, or in other word center of gravity for the object/mass. Below are the some examples to calculate the centroid: Problem 1.2 An object is immersed in the water that has specific weight, γ = 9.81kN/m3. Determine the centroid of object, hc for the following cases below: i.

If object is immersed horizontally.

ii.

If object is immersed vertically.

iii.

If object is immersed at an inclination of θ˚.

Case 1:

1.1m 1.1m

hc hc

60cm 90cm

Solution: Height of rectangular is 0.6m. Centroid of the rectangular, C = 0.6/2 = 0.3m Locate centroid of the object from the surface, hc is: hc = 1.1 + 0.3 = 1.4m Case 2: 1.1m

hc

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d = 1.0m


Solution: Diameter of circle is 1.0m. Centroid of the circle, C = 1.0/2 = 0.5m Locate centroid of the object from the surface, hc is: hc = 1.1 + 0.5 = 1.6m Case 3 0.5m hc

1.5m

1.3m

1.5m

Solution: First, define θ: 1.3m

sin θ = 1.0/1.3

1.0m

= 0.769

θ

θ = sin-1 0.769 = 50.3˚ Then; define centroid of triangle, C: h

2/3h

C = 2/3 (1.3)

1/3h

= 0.8666m

Locate centroid of the object from the surface, hc is: 0.8666 m

Sin 50.3˚ = y/0.8666 50.3˚

y

y = sin 50.3˚ (0.8666) = 0.668m

0.5m

1.168m

So; [12]

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hc = 0.5 + 0.668 = 1.168m Problem 1.3 By referring the Figure below, determine the hydrostatic force on the plate and location of the center pressure.

Solution: The Hydrostatic Force, F = ρgAhc = 1000 (9.81) [(1.2 x 1.0)/2] [3 + (1.0/3)] = 9810 (0.6) (3.333)

Where;

= 19618.038 N

Ixc = 1.2 x (1.0)3 36 = 0.0333 m4

= 19.62 kN The Center of Pressure, hp = [Ixc*sin2θ / Ahc] + hc = [0.0333(1) / (0.6 x 3.333)] + 3.333

Sin 900= 1 Sin2900= 1

= 3.35 m Problem 1.4 A pool has a water gate AB with 60˚diagonal to the water surface and has been fixed to pool wall. The water gate has a rectangular shape and 2m width shown in Figure below. Calculate the hydrostatic force and determine the center of the pressure gate.

A 60˚ 3m B Pool Base

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Solution: Find Hydrostatic Force, F = ρgAhc Where, hc = distance from water surface to the gravity center hc = 3/2 = 1.5m

60˚ Where;

3m

Where, h = 3 / (sin 60˚)

h

= 3.464m

A = 2 x 3.464 = 6.928m2 Ixc = bh3 /12 = 2 x 3.4643 12 = 6.928mm4

F = 1000 (9.81) (3.464 x 2) (1.5) = 101945.52N = 101.95kN

Sin 60= 2

The Center of Pressure, hp = [Ixc sin / Ahc] + hc

Sin260= 0.75m

= [6.928 (0.75)/ (6.928 x 1.5)] + 1.5 = 2.0m 1.4

Hydrostatic Force on Curved Surface Curved surface occurs in many hydraulic structures for example dams, tanks and cross sections of circular pipes. Since this class of surface is curved, the direction of the force is different at each location on the surface. The pressure forces are divided into horizontal and vertical component. Look at forces acting on wedge of water ABC by referring the right figure. Weight force W due to weight of volume of water. F1 and F2 is the hydrostatic force on each planar face. Reaction Forces of FH and FV due to wall of tank. The weight force W passes through the center of gravity of the wedge. To determine the horizontal force on static equilibrium: FH = F2 = ρgAhc

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The vertical component of the force on a curved surface may be determined by considering the fluid enclosed by the BC curved surface and AC vertical projection lines extending to the free surface. Thus: FV = F1 + W = ρgV1 + ρgVABC = ρgV Where, V is the volume on the curve BC. The Resultant or Magnitude Forces, FR is a triangular combination of the horizontal and vertical parts. So: FR = (FV2 + FH2) The direction of Resultant Force is determined by using following formula:  = tan-1 FV FH Problem 1.5 Determine the resultant force on the curved part of the base by referring Figure below.

5m B

12m

Solution:

A

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We know that resultant force, FR = (FV2 + FH2) So, let’s find horizontal force first: FH = ρgAhc = 1000 x 9.81 x (12 x 1) x (5 + 12/2) = 1294920 N = 1294 kN Then, calculate vertical force: FV = ρgV But, the problem is how to find volumes, V? Separate the Figure above to two blocks known as V1 volume for full blocks & V2 volume for quadrant (quarter blocks). V = V1 - V2 = (12 x 17 x 1) - [(π x 122 / 4) x 1] = 204 – 113.112 = 90.888 m3

FV

FV = 1000 x 9.81 x 90.888 FH

= 891611N = 891.6 kN

17 m R

So, the Resultant Force, FR

= (FV2 + FH2) = [(1294)2 + (891.6)2] = 2469386.56 = 1571.4 kN

Therefore, Direction of Resultant Force,  = tan-1 (FV / FH) = tan-1 (891.6/ 1294) = 34.56˚

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1 m long 12 m


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Problem 1.6 Determine the resultant force on the curved part of the base by referring Figure below.

4m

V1

3m

j = 2m

V2

FV

FH



R

Solution: Area, A = 2  4 = 8 m 2 Centroid height, hc= 3 +

2 = 4m 2

Total volume, V = v 1 + v 2 v 1 = 4  3  2 = 24m 3

 3.142(2 2 )    4 = 12.57 m 3 v 2 =  4   Total volume,V = 12.57 + 24 = 36.57 m 3 Horizontal Force, F H = gAhc = 10 3 (9.81)(8)(4) = 313920 N = 313.92 kN Vertical Force, F v = gV = 1000 (9.81)(36.57) = 358751 N = 358.75 kN

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FV

R


Resultant Force, R R

= (FV2 + FH2)

 = tan 1

1.5



= [(313.92)2 + (358.75)2] = 476.7 kN

Direction of Resultant Force, R = tan 1

FV

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FH

FH

358.75 = 48.8 0 313.92

Hydrostatic Force on A Vertical And Inclined Rectangular Wall The dam is a structure that is built to hold and retain water. The dam surface contact

with the water experiences a hydrostatic force. This hydrostatic forces acting perpendicular to the dam wall surface area that contact with water, either square-shaped dam or trapeziumshaped. Dam to withstand the force of gravity, weight or mass of the dam structure and act on the center of gravity (centroid).

1.5.1 Hydrostatic Force Exerted On A Vertical Rectangular Wall Problem 1.7 A concrete wall with 7m high and 5m long has been used to hold water up to 4.5m as shown Figure below. Determine: i. Hydrostatic Force (F) over the dam wall. ii. Center of pressure height from surface water (hp). 3m

7m 4.5m

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Solution: i.

g = 9.81 m/s2

= 1000(9.81)(4.5/2)(4.5*5)

A= 4.5 * 5 = 22.5 m2

= 496631.25 N = 496.63 kN ii.

 = 1000 kg/m3

where;

Hydrostatic Force, F =  ghcA

hs= 4.5/2 = 2.25 m Ixc =

2

Center of pressure, hp = Ixc sin  + hs

bd3

= 5(4.53)

12

hcA

12

= 37.96875 m

= 37.96875(1) + 2.25

4

 = 900

2.25(22.5)

sin2 = 1

= 3 meter

1.5.2 Hydrostatic Force Exerted On An Inclined Trapezoidal Wall

Problem 1.8 A concrete dam with trapezium-shaped are 10 m high and 350 m long has been hold water depth of 8m. Determine hydrostatic force over an inclined dam surfaced and center of pressure. 5m

10m 8m

2m

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7m


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Solution :

W

Determine the values of , A, hc and Ixc.

F

FH



Center of Pressure

FV 

How to define  ………  10/2 = tan  @ tan  = 5

10m

  = tan-1 5



= 78.690 = 78.690

2m

How to define area of section, A ………

x



8m

Sin  = 8/x

x = 8/sin 78.690 x = 8.16 m

 Area, A

= Ldam. * x = 350 * 8.16 = 2856 m2

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How to define hc ………





= 8/2

hc

= 4m

centroid

8m

How to define Ixc ……… 

hc = d/2

L @ b = 350m

Ixc = bd3 x @ d = 8.16 m

12

Front elevation of dam

= 350(8.163) 12 = 196244.7448 12 = 15847.37 m4 i.Hydrostatic Force, F =  ghcA

= 1000(9.81)(8/2)(8.16*350) 

= 112069 kN

ii.Center of pressure, hp = Ixc sin2 + hc

where;

 = 1000 kg/m3 g

hcA = 15847.37(0.96) + 4 4(2856) = 15213.48 + 4 [21]

= 9.81 m/s2

A = 2886.1 m2 hc = 4 m Ixc = 16353.73 m4  = 78.690 sin2 = 0.96


11424 = 5.33 m

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Chapter 1 Hydrostatic Force

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