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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
9.6 Fluid Pressure9.6 Fluid Pressure
� Consider the submerged plate
� 3 points have been specified
9.6 Fluid Pressure9.6 Fluid Pressure
� 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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
Flat Plate of Constant Width
� 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
9.6 Fluid Pressure9.6 Fluid Pressure
� 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
9.6 Fluid Pressure9.6 Fluid Pressure
Flat Plate of Constant Width
� 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,
9.6 Fluid Pressure9.6 Fluid Pressure
� Integration can be used to determine FR and location of center of centroid C or pressure P
9.6 Fluid Pressure9.6 Fluid Pressure
Curved Plate of Constant Width
Example
� Consider distributed loading acting on the curved plate DBcurved plate DB
9.6 Fluid Pressure9.6 Fluid Pressure
Curved Plate of Constant Width
Example
� For equivalent loading
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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 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
9.6 Fluid Pressure9.6 Fluid Pressure
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 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
9.6 Fluid Pressure9.6 Fluid Pressure
Flat Plate of Variable Width
� Consider the pressure distribution acting on the surface of a submerged plate having a variable widthvariable width
9.6 Fluid Pressure9.6 Fluid Pressure
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’)
9.6 Fluid Pressure9.6 Fluid Pressure
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
'~'
~
9.6 Fluid Pressure9.6 Fluid Pressure
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.
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
=
++
=
=+=
9.6 Fluid Pressure9.6 Fluid Pressure
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
=+=+=
==
==
9.6 Fluid Pressure9.6 Fluid Pressure
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(
;
=
+=
∑=∑
9.6 Fluid Pressure9.6 Fluid Pressure
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.
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
=−
=
−==
=
∫
∫∫∫