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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 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 plate3 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 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 γ
Plane of the plate makes an angle with the horizontal as shown
9.6 Fluid Pressure9.6 Fluid Pressure
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
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 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 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
9.6 Fluid Pressure9.6 Fluid Pressure
9.6 Fluid Pressure9.6 Fluid Pressure
Curved Plate of Constant WidthExample Consider distributed loading acting on
the curved plate DB
9.6 Fluid Pressure9.6 Fluid Pressure
Curved Plate of Constant WidthExampleFor 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) 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 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 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 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 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’)
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 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
'~'
~
9.6 Fluid Pressure9.6 Fluid Pressure
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.
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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
9.6 Fluid Pressure9.6 Fluid Pressure
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(
;
9.6 Fluid Pressure9.6 Fluid Pressure
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.
9.6 Fluid Pressure9.6 Fluid Pressure
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
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 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
9.6 Fluid Pressure9.6 Fluid Pressure
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
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
View Free Body Diagram
9.6 Fluid Pressure9.6 Fluid Pressure
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
Chapter Summary Chapter Summary
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
Chapter Summary Chapter Summary
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
Chapter Summary Chapter Summary
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
Chapter Summary Chapter Summary
Fluid PressureLine of action of the resultant force
passes through the centroid of the loading diagram
Chapter Review Chapter Review
Chapter Review Chapter Review
Chapter Review Chapter Review
Chapter Review Chapter Review
Chapter Review Chapter Review
Chapter Review Chapter Review