9.6 fluid pressure according to pascal’s law, a fluid at rest creates a pressure ρ at a point...

9.6 Fluid Pressure9.6 Fluid Pressure

According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions

Magnitude of ρ measured as a force per unit area, depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface

ρ = γz = ρgz Valid for incompressible fluids Gas are compressible fluids and thus the

above equation cannot be used


Consider the submerged plate3 points have been specified


Since point B is at depth z1 from the liquid surface, the pressure at this point has a magnitude of ρ1= γz1

Likewise, points C and D are both at depth z2 and hence ρ2 = γz2

In all cases, pressure acts normal to the surface area dA located at specified point

Possible to determine the resultant force caused by a fluid distribution and specify its location on the surface of a submerged plate


Flat Plate of Constant Width Consider flat rectangular plate of

constant width submerged in a liquid having a specific weight γ

Plane of the plate makes an angle with the horizontal as shown


Flat Plate of Constant WidthSince pressure varies linearly with depth,

the distribution of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ1= γz1

at depth z1 and

ρ2 = γz2 at depth z2


Magnitude of the resultant force FR = volume of this loading diagram and FR has a line of action that passes through the volume’s centroid, C

FR does not act at the centroid of the plate but at point P called the center of pressure

Since plate has a constant width, the loading diagram can be viewed in 2D


Flat Plate of Constant Width Loading intensity is measured as

force/length and varies linearly from w1 = bρ1= bγz1 to w 2 = bρ2

= bγz2

Magnitude of FR = trapezoidal area

FR has a line of action that passes through the area’s centroid C

Curved Plate of Constant Width When the submerged plate is curved, the pressure

acting normal to the plate continuously changes direction

For 2D and 3D view of the loading distribution,

Integration can be used to determine FR and location of center of centroid C or pressure P



Curved Plate of Constant WidthExample Consider distributed loading acting on

the curved plate DB


Curved Plate of Constant WidthExampleFor equivalent loading


Curved Plate of Constant Width The plate supports the weight of the liquid

Wf contained within the block BDA This force has a magnitude of

Wf = (γb)(areaBDA) and acts through the centroid of BDA

Pressure distributions caused by the liquid acting along the vertical and horizontal sides of the block

Along vertical side AD, force FAD’s magnitude = area under trapezoid and acts through centroid CAD of this area


Curved Plate of Constant Width The distributed loading along horizontal side

AB is constant since all points lying on this plane are at the same depth from the surface of the liquid

Magnitude of FAB is simply the area of the rectangle

This force acts through the area centroid CAB or the midpoint of AB

Summing three forces, FR = ∑F = FAB + FAD + Wf


Curved Plate of Constant Width Location of the center of pressure on the

plate is determined by applying MRo = ∑MO

which states that the moment of the resultant force about a convenient reference point O, such as D or B = sum of the moments of the 3 forces about the same point


Flat Plate of Variable Width Consider the pressure distribution

acting on the surface of a submerged plate having a variable width


Flat Plate of Variable Width Resultant force of this loading = volume

described by the plate area as its base and linearly varying pressure distribution as its altitude

The shaded element may be used if integration is chosen to determine the volume

Element consists of a rectangular strip of area dA = x dy’ located at depth z below the liquid surface

Since uniform pressure ρ = γz (force/area) acts on dA, the magnitude of the differential force dF

dF = dV = ρ dA = γz(xdy’)


Flat Plate of Variable Width

Centroid V defines the point which FR acts The center of pressure which lies on the

surface of the plate just below C has the coordinates P defined by the equations

This point should not be mistaken for centroid of the plate’s area

V

V

V

V

A VR

dV

dVyy

dV

dVxx

VdVdAF

'~'

~


Example 9.13Determine the magnitude and location of the resultant hydrostatic force acting on the

submerged rectangular plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3.


Solution The water pressures at depth A and B are

Since the plate has constant width, distributed loading can be viewed in 2D

For intensities of the load at A and B,

mkNkPambw

mkNkPambw

kPamsmmkggz

kPamsmmkggz

BB

AA

BwB

AwA

/58.73)05.49)(5.1(

/43.29)62.19)(5.1(

05.49)5)(/81.9)(/1000(

62.19)2)(/81.9)(/1000(23

23


Solution For magnitude of the resultant force FR created by

the distributed load

This force acts through the centroid of the area

measured upwards from B

mh

N

trapezoidofareaFR

29.1)3(58.7343.29

58.73)43.29(2

3

1

5.154)6.734.29)(3(2

1


Solution Same results can be obtained by

considering two components of FR defined by the triangle and rectangle

Each force acts through its associated centroid and has a magnitude of

HencekNkNkNFFF

kNmmkNF

kNmmkNF

RR

t

5.1542.663.88

2.66)3)(/15.44(

3.88)3)(/43.29(

Re

Re


Solution Location of FR is determined by

summing moments about B

mh

h

MM BBR

29.1

)1(2.66)5.1(3.88)5.154(

;


Example 9.14Determine the magnitude of the resultant hydrostatic force acting on the surface of a

seawall shaped in the form of a parabola. The wall is 5m long and ρw = 1020kg/m2.


Solution The horizontal and vertical components of

the resultant force will be calculated since

Then

ThuskNmkNmF

mkNkPambw

kPamsmmkggz

x

BB

BwB

1.225)/1.150)(3(3

1

/1.150)02.30(5

02.30)3)(/81.9)(/1020( 22


Solution Area of the parabolic sector ABC can be

determined For weight of the wafer within this

region

For resultant force

kN

FFF

kNmmmsmmkg

areagbF

yxR

ABCwy

231

)0.50()1.225(

0.50)]3)(1(3

1)[5)(/81.9)(/1020(

))((

2222

22


Example 9.15Determine the magnitude and location of the resultant force acting on the triangular end plates of the wafer of the water trough. ρw = 1000 kg/m3


Solution Magnitude of the resultant force F =

volume of the loading distribution Choosing the differential volume element,

For equation of line AB

Integrating

kNNdzzz

dzzzdVVF

zx

zxdzxdzgzdAdVdF

V

w

64.11635)(9810

)]1(5.0[)19620(

)1(5.0

19620)2(

1

02

1

0

View Free Body Diagram


Solution Resultant passes through the centroid

of the volume Because of symmetry

For volume element

mdzzz

dzzzz

dV

dVzz

x

V

V

5.01635

)(9810

1635

)]1(5.0[)19620(~

0

1

032

1

0

Chapter Summary Chapter Summary

Center of Gravity and Centroid Center of gravity represents a point where

the weight of the body can be considered concentrated

The distance to this point can be determined by a balance of moments

Moment of weight of all the particles of the body about some point = moment of the entire body about the point

Centroid is the location of the geometric center of the body


Center of Gravity and Centroid Centroid is determined by the moment

balance of geometric elements such as line, area and volume segments

For body having a continuous shape, moments are summed using differential elements

For composite of several shapes, each having a known location for centroid, the location is determined from discrete summation using its composite parts


Theorems of Pappus and Guldinus Used to determine surface area and

volume of a body of revolution Surface area = product of length of the

generating curve and distance traveled by the centroid of the curve to generate the area

Volume = product of the generating area and the distance traveled by the centroid to generate the volume


Fluid Pressure Pressure developed by a fluid at a point on

a submerged surface depends on the depth of the point and the density of the liquid according to Pascal’s law

Pressure will create a linear distribution of loading on a flat vertical or inclined surface

For horizontal surface, loading is uniform Resultants determined by volume or area

under the loading curve


Fluid PressureLine of action of the resultant force

passes through the centroid of the loading diagram

Chapter Review Chapter Review

9.6 fluid pressure according to pascal’s law, a fluid at rest creates a pressure ρ at a point...

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