6161103 9.6 fluid pressure

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 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 plate

� 3 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 ρ = γzLikewise, 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 γweight γ

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



� Since pressure varies linearly with depth,

the distribution of pressure over the plate’s

surface is represented by a trapezoidal 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 � 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



� Loading intensity is measured as force/length and varies linearly from

w = bρ = bγz to w = bρ = bγzw1 = 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



Example

� Consider distributed loading acting on the curved plate DBcurved plate DB



Example

� For 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) 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



� 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 liquidat 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



� Location of the center of pressure on the plate is determined by applying

MRo = ∑MO

which states that the moment of the resultant Ro O

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 widthvariable 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 altitudealtitude

� 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

∫ ∫ ===A VR VdVdAF ρ

� 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

dV

dVyy

dV

dVxx

'~'

~


Example 9.13

Determine the magnitude and location of the

resultant hydrostatic force acting on the submerged

rectangular plate AB. The 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 kPamsmmkggz

kPamsmmkggz

BwB

AwA

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

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

23

===

===

ρρ

ρρ

� 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

BB

AA

BwB

/58.73)05.49)(5.1(

/43.29)62.19)(5.1(

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

===

===

===

ρ

ρ

ρρ


Solution

� For magnitude of the resultant force FR created by the distributed load

trapezoidofareaFR =

� This force acts through the

centroid of the area

measured upwards from B

mh

N

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 rectangleand 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

( ) MM ;∑=∑( )

mh

h

MM BBR

29.1

)1(2.66)5.1(3.88)5.154(

;

=

+=

∑=∑


Example 9.14

Determine 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 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

� Thus

kNmkNmF

mkNkPambw

kPamsmmkggz

x

BB

BwB

1.225)/1.150)(3(3

1

/1.150)02.30(5

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

==

===

===

ρ

ρρ


Solution

� Area of the parabolic sector ABC can be determined

� For weight of the wafer within this region� 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

=

+=+=

==

= ρ


Example 9.15

Determine the magnitude and location of

the resultant force acting on the triangular

end plates of the wafer of the water trough. 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

==−=

−===

−=

====

∫

∫ ∫

ρρ


Solution

� Resultant passes through the centroid of the volume

� Because of symmetry� 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

=−

=

−==

=

∫

∫∫∫

6161103 9.6 fluid pressure

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