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Chapter 2
FLUID STATICS
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The science of fluid statics :
 the study of pressure and its variation throughout a fluid
 the study of pressure forces on finite surfaces
Special cases of fluids moving as solids are included in the treatment of
statics because of the similarity of forces involved.
Since there is no motion of a fluid layer relative to an adjacent layer,
there are no shear stresses in the fluid
 all free bodies in fluid statics have only normal pressure forces
acting on their surfaces
2.1 PRESSURE AT A POINT
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Average pressure: dividing the normal force pushing against a plane
area by the area.
Pressure at a point: the limit of the ratio of normal force to area as the
area approaches zero size at the point.
At a point: a fluid at rest has the same pressure in all directions  an
element δA of very small area, free to rotate about its center when
submerged in a fluid at rest, will have a force of constant magnitude
acting on either side of it, regardless of its orientation.
To demonstrate this, a small wedge-shaped free body of unit width is
taken at the point (x, y) in a fluid at rest (Fig.2.1)
Figure 2.1 Free-body diagram of wedge-shaped particle
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There can be no shear forces  the only forces are the normal surface
forces and gravity  the equations of motion in the x and y directions
px, py, ps are the average pressures on the three faces, γ is the unit
gravity force of the fluid, ρ is its density, and ax, ay are the accelerations
When the limit is taken as the free body is reduced to zero size by
allowing the inclined face to approach (x, y) while maintaining the same
angle θ, and using
the equations simplify to
Last term of the second equation – infinitestimal of higher of smallness,
may be neglected
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When divided by δy and δx, respectively, the equations can be combined
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θ is any arbitrary angle  this equation proves that the pressure is the
same in all directions at a point in a static fluid
Although the proof was carried out for a two-dimensional case, it may
be demonstrated for the three-dimensional case with the equilibrium
equations for a small tetrahedron of fluid with three faces in the
coordinate planes and the fourth face inclined arbitrarily.
If the fluid is in motion (one layer moves relative lo an adjacent layer),
shear stresses occur and the normal stresses are no longer the same in all
directions at a point  the pressure is defined as the average of any
three mutually perpendicular normal compressive stresses at a point,
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Fictitious fluid of zero viscosity (frictionless fluid): no shear stresses can
occur  at a point the pressure is the same in all directions
2.2 BASIC EQUATION OF FLUID STATICS
Pressure Variation in a Static Fluid
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Force balance:
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The forces acting on an element of fluid at rest (Fig. 2.2): surface forces
and body forces.
With gravity the only body force acting, and by taking the y axis
vertically upward, it is -γ δx δy δz in the y direction
With pressure p at its center (x, y, z) the approximate force exerted on
the side normal to the y axis closest to the origin and the opposite e side
are approximately
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δy/2 – the distance from center to a face normal to y
Figure 2.2
Rectangular
parallelepiped
element of fluid at
rest
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Summing the forces acting on the element in the y direction
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For the x and z directions, since no body forces act,
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The elemental force vector δF
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If the element is reduced to zero size, alter dividing through by δx δy δz
= δV, the expression becomes exact.
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This is the resultant force per unit volume at a point, which must be
equated to zero for a fluid at rest.
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The gradient ∇ is
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-∇p is the vector field f or the surface pressure force per unit volume
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The fluid static law or variation of pressure is then
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For an inviscid fluid in motion, or a fluid so moving that the shear stress
is everywhere zero, Newton's second law takes the form
a is the acceleration of the fluid element, f - jγ is the resultant fluid force
when gravity is the only body force acting
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In component form, Eq. (2.2.4) becomes
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The partials, for variation in horizontal directions, are one form of
Pascal's law; they state that two points at the same elevation in the same
continuous mass or fluid at rest have the same pressure.
Since p is a function of y only,
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relates the change of pressure to unit gravity force and change of
elevation and holds for both compressible and incompressible fluids
For fluids that may be considered homogeneous and incompressible, γ is
constant, and the above equation, when integrated, becomes
in which c is the constant of integration. The hydrostatic law of variation
of pressure is frequently written in the form
h = -y, p is the increase in pressure from that at the free surface
Example 2.1 An oceanographer is to design a sea lab 5 m high to withstand
submersion to 100 m, measured from sea level to the top of the sea lab.
Find the pressure variation on a side of the container and the pressure
on the top if the relative density of salt water is 1.020.
At the top, h = 100 m, and
If y is measured from the top of the sea lab downward, the pressure
variation is
#
Pressure Variation in a Compressible Fluid
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When the fluid is a perfect gas at rest at constant temperature
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When the value of γ in Eq. (2.2.7) is replaced by ρg and ρ is eliminated
between Eqs. (2.2.7) and (2.2.9),
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If P = P0 when ρ = ρ0, integration between limits
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- the equation for variation of pressure with elevation in an isothermal
gas
- constant temperature gradient of atmosphere 
Example 2.2 Assuming isothermal conditions to prevail in the atmosphere,
compute the pressure and density at 2000 m elevation if P = 105Pa, ρ =
1.24 kg/m3 at sea level.
From Eq. (2.2.12)
Then, from Eq. (2.2.9)
#
2.3 UNITS AND SCALES OF PRESSURE MEASUREMENT
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Pressure may be expressed with reference to any arbitrary datum
 absolute zero
 local atmospheric pressure
Absolute pressure: difference between its value and a complete
vacuum
Gage pressure: difference between its value and the local atmospheric
pressure
Figure 2.3 Bourdon gage.
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The bourdon gage (Fig. 2.3): typical of the devices used for measuring
gage pressures
 pressure element is a hollow, curved, flat metallic tube closed at one
end; the other end is connected to the pressure to be measured
 when the internal pressure is increased, the tube tends to straighten,
pulling on a linkage to which is attached a pointer and causing the
pointer to move
 the dial reads zero when the inside and outside of the tube are at the
same pressure, regardless of its particular value
 the gage measures pressure relative to the pressure of the medium
surrounding the tube, which is the local atmosphere
Figure 2.4 Units and scales for pressure measurement
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Figure 2.4: the data and the relations of the common units of pressure
measurement
Standard atmospheric pressure is the mean pressure at sea level, 760
mm Hg.
A pressure expressed in terms of the length of a column of liquid is
equivalent to the force per unit area at the base of the column.
The relation for variation of pressure with altitude in a liquid p = γh
[Eq. (2.2.8)] (p is in pascals, γ in newtons per cubic metre, and h in
metres)
With the unit gravity force of any liquid expressed as its relative density
S times the unit gravity force of water:
Water: γ may be taken as 9806 N/m3.
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Local atmospheric pressure is measured by
 mercury barometer
 aneroid barometer (measures the difference in
pressure between the atmosphere and an
evacuated box or tube in a manner analogous
to the bourdon gage except that the tube is
evacuated and sealed)
Mercury barometer: glass tube closed at one end,
filled with mercury, and inverted so that the open
end is submerged in mercury.
It has a scale: the height of column R can be
determined
The space above the mercury contains mercury
vapor. If the pressure of the mercury vapor hv is
given in millimetres of mercury and R is
measured in the same units, the pressure at A may
be expressed as (mm Hg)
Figure 2.5 Mercury
barometer
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Figure 2.4: a pressure may be located vertically on the chart, which
indicates its relation to absolute zero and to local atmospheric pressure.
If the point is below the local-atmospheric-pressure line and is referred
to gage datum, it is called negative, suction, or vacuum.
Example: the pressure 460 mm Hg abs, as at 1, with barometer reading
720 mm, may be expressed as -260 mm Hg, 260 mm Hg suction, or 260
mm Hg vacuum.
Note:
Pabs = pbar + pgage
Absolute pressures : P, gage pressures : p.
Example 2.3 The rate of temperature change in the atmosphere with
change in elevation is called its lapse rate. The motion of a parcel of air
depends on the density of the parcel relative to the density of the
surrounding (ambient) air. However, as the parcel ascends through the
atmosphere, the air pressure decreases, the parcel expands, and its
temperature decreases at a rate known as the dry adiabatic lapse rate. A
firm wants lo burn a large quantity of refuse. It is estimated that the
temperature of the smoke plume at 10 m above the ground will be 11oC
greater than that of the ambient air. For the following conditions
determine what will happen to the smoke.
(a) At standard atmospheric lapse rate β = -0.00651oC per meter
and t0 =20oC.
(b) At an inverted lapse rate β = 0.00365oC per meter.
By combining Eqs. (2.2.7) and (2.2.14),
The relation between pressure and temperature for a mass of gas expanding
without heat transfer (isentropic relation, Sec. 6.1) is
in which T1 is the initial smoke absolute temperature and P0 the initial
absolute pressure; k is the specific heat ratio, 1.4 for air and other diatomic
gases.
Eliminating P/P0 in the last two equations
Since the gas will rise until its temperature is equal to the ambient
temperature,
the last two equations may be solved for y. Let
Then
For β = -0.00651oC per metre, R = 287 m·N/(kg·K), a = 2.002, and y = 3201
m. For the atmospheric temperature inversion β = -0.00365oC per metre, a =
-0.2721, and y = 809.2 m. #
2.4 MANOMETERS
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Manometers are devices that employ liquid columns for determining
differences in pressure.
Figure 2.6a: the most elementary manometer – piezometer
 It measures the pressure in a liquid when it is above zero gage
 Glass tube is mounted vertically so that it is connected to the space
within the container
 Liquid rises in the tube until equilibrium is reached
 The pressure is then given by the vertical distance h from the
meniscus (liquid surface) to the point where the pressure is to be
measured, expressed in units of length of the liquid in the container.
 Piezometer would not work for negative gage pressures, because air
would flow into the container through the tube
Figure 2.6 Simple manometers.
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Figure 2.6b: for small negative or positive gage pressures in a liquid
With this arrangement the meniscus may come to rest below A, as
shown. Since the pressure at the meniscus is zero gage and since
pressure decreases with elevation,
units of length H2O
Figure 2.6c: for greater negative or positive gage pressures (a second
liquid of greater relative density employed)
It must be immiscible in the first fluid, which may now be a gas
If the relative density of the fluid at A is S1 (based on water) and the
relative density of the manometer liquid is S2, the equation for pressure
at A
hA - the unknown pressure, expressed in length units of water,
h1, h2 - in length units
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1.
2.
3.
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A general procedure in working all manometer problems :
Start at one end (or any meniscus if the circuit is continuous) and
write the pressure there in an appropriate unit (say pascals) or in an
appropriate symbol if it is unknown.
Add to this the change in pressure, in the same unit, from one
meniscus to the next (plus if the next meniscus is lower, minus if
higher). (For pascals this is the product of the difference in elevation
in metres and the unit gravity force of the fluid in newtons per cubic
metre.)
Continue until the other end of the gage (or the starting meniscus) is
reached and equate the expression to the pressure at that point,
known or unknown.
The expression will contain one unknown for a simple manometer or
will give a difference in pressures for the differential manometer. In
equation form,
Figure 2.7 Differential manometers
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A differential manometer (Fig. 2.7) determines the difference in
pressures at two points A and B when the actual pressure at any point
in the system cannot be determined
Application of the procedure outlined above to Fig. 2.7a produces
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For Fig. 2.7b:
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If the pressures at A and B are expressed in length of the water column,
the above results can be written, for Fig. 2.7a,
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For Fig 2.7b:
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Example 2.4 In Fig. 2.7a the liquids at A and B are water and the
manometer liquid is oil. S = 0.80; h1 = 300 mm; h2 = 200 mm; and h3 =
600 mm.
(a) Determine pA - pB, in pacals.
(b) If pB = 50kPa and the barometer reading is 730 mm Hg, find the
pressure at A, in meters of water absolute.
(a)
(b)
(a)
Micromanometers
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For determining very small differences in pressure or determining
large pressure differences precisely – several types of manometers
One type very accurately measures the differences in elevation of two
menisci of a manometer.
By means of small telescopes with horizontal cross hairs mounted
along the tubes on a rack which is raised and lowered by a pinion and
slow motion screw so that the cross hairs can be set accurately, the
difference in elevation of menisci (the gage difference) can be read with
verniers.
Figure 2.8 Micromanometer using two gage liquids
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Fig. 2.8: two gage liquids, immiscible in each other and in the fluid to
be measured  a large gage difference R can be produced for a small
pressure difference.
The heavier gage liquid fills the lower U tube up to 0-0; then the lighter
gage liquid is added to both sides, filling the larger reservoirs up to 1-1.
The gas or liquid in the system fills the space above 1-1. When the
pressure at C is slightly greater than at D, the menisci move as
indicated in Fig. 2.8.
The volume of liquid displaced in each reservoir equals the
displacement in the U tube 
Manometer equation
γ1, γ2 and γ3 are the unit gravity force
Example 2.5 In the micromanometer of Fig 2.8 the pressure difference is
wanted, in pascals, when air is in the system, S2 = 1.0, S3 = 1.10, a/A =
0.01, R = 5 mm, t = 20oC, and the barometer reads 760 mm Hg.
The term γ1(a/A) may be neglected. Substituting into Eq. (2.4.1) gives
#
Figure 2.9
Inclined
manometer
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The inclined manometer: frequently used for measuring small
differences in gas pressures.
Adjusted to read zero, by moving the inclined scale, when A and B are
open. Since the inclined tube requires a greater displacement of the
meniscus for given pressure difference than a vertical tube, it affords
greater accuracy in reading the scale.
Surface tension causes a capillary rise in small tubes. If a U tube is
used with a meniscus in each leg, the surface-tension effects cancel.
2.5 FORCES ON PLANE AREAS
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In the preceding sections variations oF pressure throughout a fluid
have been considered.
The distributed forces resulting from the action of fluid on a finite area
may be conveniently replaced by a resultant force, insofar as external
reactions to the force system are concerned.
In this section the magnitude of resultant force and its line of action
(pressure center) are determined by integration, by formula, and by
use of the concept of the pressure prism.
Horizontal Surfaces
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A plane surface in a horizontal position in a fluid at rest is subjected to
a constant pressure.
The magnitude of the force acting on one side of the surface is
The elemental forces pdA acting on A are all parallel and in the same
sense  a scalar summation of all such elements yields the magnitude
of the resultant force. Its direction is normal to the surface and toward
the surface if p is positive.
Fig. 2.10: arbitrary xy axes - to find the line of action of the resultant,
i.e., the point in the area where the moment of the distributed force
about any axis through the point is zero,
Then, since the moment of the resultant must equal the moment of the
distributed force system about any axis, say the y axis,
x’ – the distance from the y axis to the resultant
Figure 2.10 Notation for determining the line
of action of a force
Momentum
(1) First moment
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The moment of an area A about the y axis
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The moment about a parallel axis, for example, x = k, the moment
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Centroidal axis
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Volume center
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Mass center: center of gravity of a body
Figure A.1 Notation for first and second moments
(2) Second moment
 The second moment of an area A (the moment of inertia of the area)
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The moment about a parallel axis, for example, x = k, the moment
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Figure A.2 Moments of inertia of simple areas about centroidal axes
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The product of inertia Ixy of an area
- the product of inertia about centroidal axes parallel to the xy axes.
Inclined Surfaces
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Fig. 2.11: a plane surface is indicated by its trace A'B‘;it is inclined θo
from the horizontal. x axis: intersection of the plane of the area and the
free surface. y axis: taken in the plane of the area, with origin O in the
free surface. The xy plane portrays the arbitrary inclined area. The
magnitude, direction, and line of action of the resultant force due to the
liquid, acting on one side of the area, are sought.
For δA:
Since all such elemental forces are parallel, the integral over the area
yields the magnitude of force F, acting on one side of the area,
Magnitude of force exerted on one side of a plane area submerged in a
liquid is the product of the area and the pressure at its centroid
The presence of a free surface is unnecessary
Figure 2.11 Notation for force of liquid on one side of a plane inclined area.
Center of Pressure
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Fig. 2.11: the line of action of the resultant force has its piercing point
in the surface at a point called the pressure center, with coordinates (xp,
yp). Center of pressure of an inclined surface is not at the centroid. To
find the pressure center, the moments of the resultant xpF, ypF are
equated to the moment of the distributed forces about the y axis and x
axis, respectively 
- may be evaluated conveniently through graphical integration, for
simple areas they may be transformed into general formulas:
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When either of the centroidal axes is an axis of symmetry for the
surface, I xy vanishes and the pressure center lies on x = x- . Since
I xy may be either positive or negative, the pressure center may lie on
either side of the line x = x-. To determine yp by formula, with Eqs.
(2.5.2) and (2.5.6)
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In the parallel-axis theorem for moments of inertia
in which IG is the second moment or the area about its horizontal
centroidal axis. If IG is eliminated from Eq. (2.5.9)
Example 2.6 The triangular gate CDE (Fig. 2.12) is hinged along CD and is
opened by a normal force P applied at E. It holds oil, relative density 0.80,
above it and is open to the atmosphere on its lower side Neglecting the
weight of the gate, find (a) the magnitude of force exerted on the gate by
integration and by Eq. (2.5.2); (b) the location of pressure center; (c) the
force P needed to open the gate.
Figure 2.12 Triangular gate
(a) By integration with reference to Fig. 2.12,
When y = 4, x = 0, and when y = 6.5, x = 3, with x varying linearly with
y; thus
in which the coordinates have been substituted to find x in terms of y.
Solving for a and b gives
Similarly, y = 6.5, x = 3; y = 9, x = 0; and x = 6/5(9 - y). Hence,
Integrating and substituting for γsinθ leads to
By Eq. (2.5.2)
(b) With the axes as shown,
In Eq. (2.5.8)
I-xyis zero owing to symmetry about the centroidal axis parallel to the x
axis; hence
In Eq. (2.5.11),
i.e., the pressure center is 0.16 m below the centroid, measured in the
plane of the area.
(c) When moments about CD are taken and the action of the oil is
replaced by the resultant,
The Pressure Prism
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Pressure prism: another approach to determine the resultant force and
line of action of the force on a plane surface - prismatic volume with its
base the given surface area and with altitude at any point of the base
given by p = γh. h is the vertical distance to the free surface, Fig. 2.13.
(An imaginary free surface may be used to define h if no real free
surface exists.) (in the figure, γh may be laid off to any convenient scale
such that its trace is OM)
The force acting on an elemental area δA is
(2.5.12)
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- an element of volume of the pressure prism. After integrating, F = ϑ
From Eqs. (2.5.5) and (2.5.6),
(2.5.13)
 xp, yp are distances to the centroid of the pressure prism  the line
of action of the resultant passes through the centroid of the pressure
prism
Figure 2.13 Pressure prism
Effects of Atmospheric Pressure on Forces on Plane Areas
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In the discussion of pressure forces the pressure datum was not
mentioned : p = γh  the datum taken was gage pressure zero, or the
local atmospheric pressure
When the opposite side of the surface is open to the atmosphere, a
force is exerted on it by the atmosphere equal to the product of the
atmospheric pressure P0 and the area, or P0A , based on absolute zero
as datum. On the liquid side the force is
The effect P0A of the atmosphere acts equally on both sides and in no
way contributes to the resultant force or its location
So long as the same pressure datum is selected for all sides of a free
body, the resultant and moment can be determined by constructing a
free surface at pressure zero on this datum and using the above
methods
Example 2.8 An application of pressure forces on plane areas is given in the
design of a gravity dam. The maximum and minimum compressive
stresses in the base of the dam are computed from the forces which act
on the dam. Figure 2.15 shows a cross section through a concrete dam
where the unit gravity force of concrete has been taken as 2.5γ and γ is
the unit gravity force of water. A 1 m section of dam is considered as a
free body; the forces are due to the concrete, the water, the foundation
pressure, and the hydrostatic uplift. Determining amount of hydrostatic
uplift is beyond the scope of this treatment. but it will be assumed to be
one-half the hydrostatic head at the upstream edge, decreasing linearly
to zero at the downstream edge of the dam. Enough friction or shear
stress must be developed at the base of the dam to balance the thrust
due to the water that is Rx = 5000γ. The resultant upward force on the
base equals the gravity force of the dam less the hydrostatic uplift Ry =
6750γ + 2625γ - 1750γ = 7625γ N. The position of Ry is such that the free
body is in equilibrium. For moments around O,
Figure 2.15
Concrete gravity
dam
It is customary to assume that the foundation pressure varies linearly
over the base of the dam, i.e., that the pressure prism is a trapezoid
with a volume equal to Ry; thus
in which σmax, σmin are the maximum and minimum compressive
stresses in pascals. The centroid of the pressure prism is at the point
where x = 44.8 m. By taking moments about 0 to express the position of
the centroid in terms of σmax and σmin,
Simplifying gives
When the resultant falls within the middle third of the base of the dam,
σmin will always be a compressive stress. Owing to the poor tensile
properties of concrete, good design requires the resultant to fall within
the middle third of the base.
2.6 FORCE COMPONENTS ON CURVED SURFACES
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When the elemental forces p δA vary in direction, as in the case of a
curved surface, they must be added as vector quantities
 their components in three mutually perpendicular directions are
added as scalars, and then the three components are added
vectorially.
With two horizontal components at right angles and with the vertical
component - all easily computed for a curved surface - the resultant
can be determined.
The lines of action of the components also are readily determined.
Horizontal Component of Force on a Curved Surface
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The horizontal component pressure force on a curved surface is equal
to the pressure force exerted on a projection of the curved surface. The
vertical plane of projection is normal to the direction of the component.
Fig. 2.16: the surface represents any three-dimensional surface, and δA
an element of its area, its normal making the angle θ with the negative
x direction. Then
Projecting each element on a plane perpendicular to x is equivalent to
projecting the curved surface as a whole onto the vertical plane
force acting on this projection of the curved surface is the horizontal
component of force exerted on the curved surface in the direction
normal to the plane of projection.
To find the horizontal component at right angles to the x direction, the
curved surface is projected onto a vertical plane parallel to x and the
force on the projection is determined.
Figure 2.16 Horizontal
component of force on a
curved surface
Figure 2.17 Projections of
area elements on opposite
sides of a body
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When looking for the horizontal component of pressure force on a
closed body, the projection of the curved surface on a vertical plane is
always zero, since on opposite sides of the body the area-element
projections have opposite signs (Fig. 2.17).
Let a small cylinder of cross section δA with axis parallel to x intersect
the closed body at B and C. If the element of area of the body cut by
the prism at B is δAB and at C is δAC, then
and similarly for all other area elements
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To find the line of action of a horizontal component of force on a
curved surface, the resultant of the parallel force system composed of
the force components from each area element is required. This is
exactly the resultant of the force on the projected area, since the two
force systems have an identical distribution of elemental horizontal
force components. Hence, the pressure center is located on the
projected area by the methods of Sec. 2.5.
Example 2.9 The equation of an ellipsoid of revolution submerged in
water is x2/4 + y2/4 + z2/9 = 1. The center of the body is located 2 m
below the free surface. Find the horizontal force components acting on
the curved surface that is located in the first octant. Consider the xz
plane to be horizontal and y to be positive upward.
Projection of the surface on the yz plane has an area of
Its centroid is located
m below the free surface 
Vertical Component of Force on a Curved Surface
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The vertical component of pressure force on a curved surface is equal
to the weight surface and extending up to the free surface
Can be determined by summing up the vertical components of
pressure force on elemental areas δA of the surface
In Fig.2.18 an area element is shown with the force p δA acting normal
to it. Let θ be the angle the normal to the area element makes with the
vertical. Then the vertical component of force acting on the area
element is p cos θ δA, and the vertical component of force on the
curved surface is given by
(2.6.2)
p replaced by its equivalent γh; cos θ δA is the projection of δA on a
horizontal plane  Eq. (2.6.2):
(2.5.3-4)
in which δϑ is the volume of the prism of height h and base cos θ δA,
or the volume of liquid vertically above the area element
Figure 2.18 Vertical
component of force on a
curved surface
Figure 2.19 Liquid with
equivalent free surface
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Fig. 2.19: the liquid is below the curved surface and the pressure
magnitude is known at some point (e.g., O), an imaginary or equivalent
free surface s-s can be constructed p/γ above O, so that the product of
unit gravity force and vertical distance to any point in the tank is the
pressure at the point.
The weight of the imaginary volume of liquid vertically above the
curved surface is then the vertical component of pressure force on the
curved surface.
In constructing an imaginary free surface, the imaginary liquid must
be of the same unit gravity force as the liquid in contact with the
curved surface; otherwise, the pressure distribution over the surface
will not be correctly represented.
With an imaginary liquid above a surface, the pressure at a point on
the curved surface is equal on both sides, but the elemental force
components in the vertical direction are opposite in sign  the
direction of the vertical force component is reversed when an
imaginary fluid is above the surface.
In some cases a confined liquid may be above the curved surface, and
an imaginary liquid must be added (or subtracted) to determine the
free surface.



The line of action of the vertical component is determined by equating
moments of the elemental vertical components about a convenient axis
with the moment of the resultant force. With the axis at O (Fig.2.18),
in which
is the distance from 0 to the line of action
Since Fv = γϑ
the distance to the centroid of the volume
 the line of action of the vertical force passes through the centroid of
the volume, real or imaginary, that extends above the curved surface
up to the real or imaginary free surface
Example 2.10 A cylindrical barrier (Fig. 2.20) holds water as shown. The
contact between cylinder and wall is smooth. Considering a 1-m length of
cylinder, determine (a) its gravity force and (b) the force exerted against
the wall.
(a) For equilibrium the weight of the cylinder must equal the vertical
component of force exerted on it by the water. (The imaginary free
surface for CD is at elevation A.) The vertical force on BCD is
The vertical force on AB is
Hence, the gravity force per metre of length is
(b) The force exerted against the wall is the horizontal force on ABC minus
the horizontal force on CD. The horizontal components of force on BC
and CD cancel; the projection of BCD on a vertical plane is zero ,
since the projected area is 2 m2 and the pressure at the centroid of the
projected area is 9806 Pa.
Figure 2.20 Semifloating body
Tensile Stress in a Pipe and Spherical Shell




Fig. 2.21: a circular pipe under the action of an internal pressure is in
tension around its periphery; assuming that no longitudinal stress
occurs, the walls are in tension
Consider a section of pipe of unit length (the ring between two planes
normal to the axis and unit length apart). If one-half of this ring is
taken as a free body, the tensions per unit length at top and bottom are
respectively T1 and T2
The horizontal component of force acts through the pressure center of
the projected area and is 2pr, in which p is the pressure at the
centerline and r is the internal pipe radius.
For high pressures the pressure center may be taken at the pipe center;
then T1 = T2, and
T is the tensile force per unit length. For wall thickness e, the tensile
stress in the pipe wall is
Figure 2.21 Tensile stress in pipe
Example 2.11 A 100 mm-ID steel pipe has a 6 mm wall thickness. For an
allowable tensile stress of 70 MPa, what is the maximum pressure?
From Eq. (2.6.6)
2.7 BUOYANT FORCE



Buoyant force: the resultant force exerted on a body by a static fluid in
which it is submerged or floating
 Always acts vertically upward (there can be no horizontal
component of the resultant because the projection of the submerged
body or submerged portion of the floating body on a vertical plane
is always zero)
The buoyant force on a submerged body is the difference between the
vertical component of pressure force on its underside and the vertical
component of pressure force on its upper side
Figure 2.22
Figure 2.22 Buoyant force on floating and submerged bodies

Fig. 2.22: the upward force on the bottom is equal to the gravity force
of liquid, real or imaginary, which is vertically above the surface ABC,
indicated by the gravity force of liquid within ABCEFA. The
downward force on the upper surface equals the gravity force of liquid
ADCEFA. The difference between the two forces is a force, vertically
upward, due to the gravity force of fluid ABCD that is displaced by the
solid. In equation form

FB is buoyant force, V is the volume of fluid displaced, and γ is the unit
gravity force of fluid
The same formula holds for floating bodies when V is taken as the
volume of liquid displaced


Fig.2.23: the vertical force exerted on an element of the body in the
form of a vertical prism of cross section δA is

δV is the volume of the prism. Integrating over the complete body gives



γ is considered constant throughout the volume
To find the line of action of the buoyant force, moments are taken
about a convenient axis O and are equated to the moment of the
resultant, thus,
i is the distance from the axis to the line of action.
This equation yields the distance to the centroid of the volume;  the
buoyant force acts through the centroid of the displaced volume of
fluid; this holds for both submerged and floating bodies.
The centroid of the displaced volume of fluid is called the center of
buoyancy.
Figure 2.23 Vertical force components on element of body



Determining gravity force on an odd-shaped object suspended in two
different fluids yields sufficient data to determine its gravity force,
volume, unit gravity force, and relative density.
Figure 2.24: two free-body diagrams for the same object suspended
and gravity force determined in two fluids, F1 and F2; γ1 and γ2 are the
unit gravity forces of the fluids. W and V, the gravity force and volume
of the object, are to be found.
The equations of equilibrium are written and solved:
Figure 2.24 Free body diagrams for body suspended in a fluid




A hydrometer uses the principle of buoyant force to determine relative
densities of liquids
Figure 2.25: a hydrometer in two liquids with a stem of prismatic
cross section a
Considering the liquid on the left to be distilled water (unit relative
density S = 1.00), the hydrometer floats in equilibrium when
V0 is the volume submerged, γ is the unit gravity force of water, and W
is the gravity force of hydrometer
The position of the liquid surface is marked 1.00 on the stem to
indicate unit relative density S. When the hydrometer is floated in
another 1iquid, the equation of equilibrium becomes
where ΔV = aΔh. Solving for Δh with Eqs. (2.7.2) and (2.7.3)
Figure 2.25 Hydrometer in water and in liquid of relative density
Example 2.12 A piece of ore having a gravity force of 1.5 N in air is found to
have a gravity force 1.1 N when submerged in water. What is its volume,
in cubic centimetres, and what is its relative density?
The buoyant force due to air may be neglected. From Fig. 2.24
2.8 STABILITY OF FLOATING AND SUBMERGED BODIES





A body floating in a static liquid has vertical stability.
A small upward displacement decreases the volume of liquid displaced
 an unbalanced downward force which tends to return the body to
its original position.
Similarly, a small downward displacement results in a greater buoyant
force, which causes an unbalanced upward force.
A body has linear stability when a small linear displacement in any
direction sets up restoring forces tending to return it to its original
position.
A body has rotational stability when a restoring couple is set up by any
small angular displacement.



Methods for determining rotational stability are developed in the
following discussion
A body may float in

stable equilibrium

unstable equilibrium (any small angular displacement sets up a
couple that tends to increase the angular displacement)

neutral equilibrium (any small angular displacement sets up no
couple whatever)
Figure 2.26: three cases of equilibrium
a.
a light piece of wood with a metal mass at its bottom is stable
b.
when the metal mass is at the top, the body is in equilibrium but
any slight angular displacement causes it to assume the position in
a
c.
a homogeneous sphere or right-circular cylinder is in equilibrium
for any angular rotation; i.e., no couple results from an angular
displacement
Figure 2.26 Examples of stable, unstable, and neutral equilibrium


A completely submerged object is rotationally stable only when its
center of gravity is below the center of buoyancy (Fig. 2.27a)
When the object is rotated counterclockwise, the buoyant force and
gravity force produce a couple in the clockwise direction (Fig. 2.27b)
Figure 2.27 Rotationally stable submerged body



Normally, when a body is too heavy to float, it submerges and goes
down until it rests on the bottom.
Although the unit gravity force of a liquid increases slightly with depth,
the higher pressure tends to cause the liquid to compress the body or
to penetrate into pores of solid substances and thus decrease the
buoyancy of the body
Example: a ship is sure to go to the bottom once it is completely
submerged, owing to compression of air trapped in its various parts
Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy
(centroid of displaced volume) floats in stable equilibrium (Fig. 2.26a).
Certain floating objects, however, are in stable equilibrium when their
center of gravity is above the center of buoyancy.

Figure 2.28a: a cross section of a body with all other parallel cross
sections identical. The center of buoyancy is always at the centroid of
the displaced volume, which is at the centroid of the cross-sectional
area below liquid surface in this case.
Figure 2.28 Stability of a prismatic body





 when the body is tipped (Fig. 2.28b), the center of buoyancy is at the
centroid B' of the trapezoid ABCD, the buoyant force acts upward
through B', and the gravity force acts downward through G, the center
of gravity of the body
When the vertical through B' intersects the original centerline above C,
as at M, a restoring couple is produced and the body is in stable
equilibrium
The intersection of the buoyant force and the centerline is called the
metacenter (M)
When M is above G, the body is stable; when below G, it is unstable;
and when at G, it is in neutral equilibrium
The distance MG is called the metacentric height and is a direct
measure of the stability of the body. The restoring couple is
in which θ is the angular displacement and W the gravity force of the
body
Example 2.13 In Fig. 2.28 a scow 6 m wide and 20 m long has a gross
mass of 200 Mg. Its center of gravity is 30 cm above the water surface.
Find the metacentric height and restoring couple when Δy = 30 cm.
The depth of submergence h in the water is
The centroid in the tipped position is located with moments about AB
and BC,
By similar triangles AEO and B'PM,
G is 1.97 m from the bottom; hence
The scow is stable, since
#
is positive; the righting moment is
Nonprismatic Cross Sections



For a floating object of variable cross section (e.g., a ship) (Fig. 2.39a),
a convenient formula can be developed for determination of
metacentric height for very small angles of rotation θ
The horizontal shift in center of buoyancy r (Fig. 2.29b) is determined
by the change in buoyant forces due to the wedge being submerged,
which causes an upward force on the left, and by the other wedge
decreasing the buoyant force by an equal amount ΔFB on the right.
The force system, consisting of the original buoyant force at B and the
couple ΔFB x s due to the wedges, must have as resultant the equal
buoyant force at B'. With moments about B to determine the shirt r,
Figure 2.29
Stability
relations in a
body of variable
cross section



The amount of the couple can be determined with moments about O, the
centerline of the body at the liquid surface
For an element of area δA on the horizontal section through the body at
the liquid surface, an element of volume of the wedge is xθ δA. The
buoyant force due to this element is γxθ δA, and its moment about O is
γx2θ δA, in which θ is the small angle of tip in radians.
By integrating over the complete original horizontal area at the liquid
surface, the couple is determined to be
I is the moment of inertia of the area about the axis y-y (Fig.2.29a)
Substitution into the above equation produces

V is the total volume of liquid displaced
Since θ is very small
and
Example 2.14 A barge displacing 1 Gg has the horizontal cross section at
the waterline shown in Fig. 2.30. Its center of buoyancy is 2.0 m below
the water surface, and its center of gravity is 0.5 m below the water
surface. Determine its metacentric height for rolling (about y-y axis)
and for pitching (about x-x axis).
GB = 2 – 0.5 = 1.5 m
For rolling
For pitching
#
Figure 2.30 Horizontal
cross section of a ship at
the waterline
2.9 RELATIVE EQUILIBRIUM





Fluid statics: no shear stresses  the variation of pressure is simple to
compute
For fluid motion such that no layer moves relative to an adjacent layer,
the shear stress is also zero throughout the fluid
A fluid with a translation at uniform velocity still follows the laws of
static variation of pressure.
When a fluid is being accelerated so that no layer moves relative to an
adjacent one (when the fluid moves as if it were a solid), no shear
stresses occur and variation in pressure can be determined by writing
the equation of motion for an appropriate free body
Two cases are of interest, a uniform linear acceleration and a uniform
rotation about a vertical axis
 When moving thus, the fluid is said to be in relative equilibrium
Uniform Linear Acceleration





Fig. 2.31: a liquid in an open vessel is given a uniform linear
acceleration a
After some time the liquid adjusts to the acceleration so that it moves
as a solid, i.e., the distance between any two fluid particles remains
fixed  no shear stresses occur
By selecting a cartesian coordinate system with y vertical and x such
that the acceleration vector a is in the xy plane (Fig. 2.31a), the z axis is
normal to a and there is no acceleration component in that direction
Fig. 2.31b: the pressure gradient ∇p is then the vector sum of -ρa and jγ
Since ∇p is in the direction of maximum change in p (the gradient), at
right angles to ∇p there is no change in p. Surfaces of constant pressure,
including the free surface, must therefore be normal to ∇p
Figure 2.31 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x, y,
and z, that is, p = p(x, y, z), Eq. (2.2.5) is written in component form :

Since p is a function of position (x, y, z), its total differential is

Substituting for the partial differentials gives
which can be integrated for an incompressible fluid,

To evaluate the constant of integration c: let x = 0, y = 0, p = p0; then c
= p0 and

When the accelerated incompressible fluid has a free surface, its equation
is given by setting p = 0 in the above eq. Solving it for y gives

The lines of constant pressure, p = const, have the slope
and are parallel to the free surface. The y intercept of the free surface is
Example 2.15 The tank in Fig. 2.32 is filled with oil, relative density 0.8,
and accelerated as shown. There is a small opening in the rank at A.
Determine the pressure at B and C; and the acceleration ax required to
make the pressure at B zero.
By selecting point A as origin and by applying Eq. (2.9.2) for ay = 0
At B, x = 1.8 m, y = - 1.2 m, and p = 2.35 kPa. Ft C, x = -0.15 m, y = 1.35 m, and p = 11.18 kPa. For zero pressure at B, from Eq. (2.9.2) with
origin at A,
Figure 2.32 Tank completely filled with liquid
Example 2.16 A closed box with horizontal base 6 by 6 units and a height
of 2 units is half-filled with liquid (Fig. 2.33). It is given a constant
linear acceleration ax = g/2, ay = -g/4. Develop an equation for variation
of pressure along its base.
The free surface has the slope:
hence, the free surface is located s shown in the figure. When the origin
is taken at 0, Eq. (2.9.2) becomes
Then, for y = 0, along the bottom,
Figure 2.33 Uniform linear acceleration of container
Uniform Rotation about a Vertical Axis




Forced-vortex motion: rotation of a fluid, moving as a solid, about an
axis
 Every particle of fluid has the same angular velocity
 This motion is to be distinguished from free-vortex motion, in which
each particle moves in a circular path with a speed varying inversely
as the distance from the center
A liquid in a container, when rotated about a vertical axis at constant
angular velocity, moves like a solid alter some time interval.
No shear stresses exist in the liquid, and the only acceleration that occurs
is directed radially inward toward the axis of rotation.
By selecting a coordinate system (Fig. 2.34a) with the unit vector i in the
r direction and j in the vertical upward direction with y the axis of
rotation, the following equation may be applied to determine pressure
variation throughout the fluid:
(2.2.5)
Figure 2.34 Rotation of a fluid about a vertical axis



For constant angular velocity w, any particle of fluid P has an
acceleration w2r directed radially inward (a = -iw2r)
Vector addition of -jγ and -ρa (Fig. 2.34b) yields ∇p, the pressure
gradient. The pressure does not vary normal to this line at a point  if P
is taken at the surface, the free surface is normal to ∇p
Expanding Eq. (2.2.5)
k is the unit vector along the z axis (or tangential direction). Then

p is a function of y and r only:

For a liquid (γ ≈ const) integration yields
c is the constant of integration

If the value of pressure at the origin (r = 0, y = 0) is p0, then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected
and the above eq. is divided by γ,
: the head, or vertical depth, varies as the square of the radius. The
surfaces of equal pressure are paraboloids of revolution.



When a free surface occurs in a container that is being rotated, the fluid
volume underneath the paraboloid of revolution is the original fluid
volume
The shape of the paraboloid depends only upon the angular velocity with
respect to the axis (Fig. 2.35). The rise of liquid from its vertex to the wall
of the cylinder is w2r02/rg (Eq. (2.9.6)), for a circular cylinder rotating
about its axis.
Since a paraboloid of revolution has a volume equal to one-half its
circumscribing cylinder, the volume of the liquid above the horizontal
plane through the vertex is

When the liquid is at rest, this liquid is also above the plane through the
vertex to a uniform depth of

Hence, the liquid rises along the walls the same amount as the center
drops, thereby permitting the vertex to be located when w, r0, and depth
before rotation are given
Figure 2.35
Rotation of circular cylinder
about its axis
Example 2.17 A liquid, relative density 1.2, is rotated at 200 rpm about a
vertical axis. At one point A in the fluid 1 m from the axis, the pressure
is 70 kPa. What is the pressure at a point B which is 2 m higher than A
and 1.5 m from the axis?
When Eq. (2.9.5) is written for the two points,
Then w = 200 x 2π/60 = 20.95 rad/s, γ = 1.2 x 9806 = 11.767 N/m3, rA =
1 m, and rB = 1.5 m.
When the second equation is subtracted from the first and the values
are substituted,
Hence
Example 2.18 A straight tube 2 m long, closed at the bottom and filled
with water, is inclined 30o with the vertical and rotated about a vortical
axis through its midpoint 6.73 rad/s. Draw the paraboloid of zero
pressure, and determine the pressure at the bottom and midpoint of
the tube. In Fig. 2.36, the zero-pressure paraboloid passes through
point A. If the origin is taken at the vertex, that is, p0 = 0, Eq. (2.9.6)
becomes
which locates the vertex at O, 0.577 m below A. The pressure at the
bottom of the tube is
or
At the midpoint,
=.289 m and
Figure 2.36
Rotation of inclined
tube of liquid about a
vertical axis
Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid
accelerating as a rigid body can be obtained by integrating over the
surface

The nature of the acceleration and orientation of the surface governs the
particular variation of p over the surface
 When the pressure varies linearly over the plane surface (linear
acceleration), the magnitude of force is given by the product of
pressure at the centroid and area, since the volume of the pressure
prism is given by pGA
 For nonlinear distributions the magnitude and line of action can be
found by integration.
과제물










스캐닝된 파일을 텍스트 파일로 만들 것. (2,3,4,5 장, 자습서)
교과서의 모든 예제는 반드시 제출하여야 함.
정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라.
힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라.
기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을
유도하라.
액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라. (단순 액주계와
차분 액주계의 예를 포함하여).
면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을 유도
하라.
1차 2차 모멘트에 대하여 설명하라. 부록에 나와있는 도형에 대하여 도심축과 x
축에 대한 2차모멘트에 대한 식을 유도하라.
부록 B의 수치해석 기법을 설명하라. 프로그램밍 포함. (교과서와 다른 책의 내
용 참조).
예제 2.7의 문제를 Bisection과 Newton-Raphson Method를 사용하여 컴퓨터 프
로그램을 작성하여 풀어라.
과제물

Runge-Kutta Method를 설명하고 수치해석 책에 나와있는 문제를 풀어라(프로
그램밍 포함 : C 및 Fortran).