Fluid Statics
? Study of fluid at rest,i.e,not in motion or
not flowing
? On any plane,shear force is zero (no
velocity gradient or shear deformation)
hence only pressure forces exist
Pascal’s Law
? For fluid at rest,pressure at a point is the same in
all direction
P = Px = Py = Pz
Proving Pascal’s Law
As element is in equilibrium,
?Fy = 0
Py ?x ?z = Ps ?x ?s Sin ?
?s Sin ? = ?z
Hence,
Py = Ps
Similarly,for ?Fz = 0
Pz = Ps
Basic Hydrostatic Equation
for Pressure Field
Body of fluid in equilibrium,
? Fy = 0
P.A + W = (P + dP), A
P.A + ? g A dh = (P + dP) A
dP/dh = ? g = ?
dP/dy = - ? g = - ?
?
h
dh
P
P + dP
W
Rectangular free body
of top area A
y
Incompressible Fluid
Pressure Difference between two points in a body of
fluid
dP/dh = ? g
dP = ? g dh
Integrating from P1 to P2,
If fluid is incompressible and homogeneous
? = constant
P2 – P1 = ? g (h2 – h1) = ? g (? h)
?
P1
P2
h1h
2
?h
?? ?
2
1
2
1
h
h
P
P
g d hdP ?
Incompressible Fluid
If P1 is on free water surface,i.e,P1 = Patm,
P2 - P1 = ? g h
P2 - Patm = ? g h
P2,abs = ? g h + Patm
For gauge pressure measurement,Patm = 0
P2,gauge = ? g h
P2,abs = P2,gauge + Patm
Gauge pressure at a point h below free surface = ? g h
?
h
P2
P1=Patm
Definition of Pressure Head
Pressure head at point = P/? g (m)
Piezometric head = P/? g + y (m)
Piezometric pressure = P + ? g y
In fluid statics,gauge pressure head at depth h below free
surface
= P/? g = ? g h / ? g = h
Datum
Pressure = P
y
?
h
Example
P1 = F/A = 1000 / (? 0.0252) = 509.3 kPa
P2 = P1 - ? g (2.5) = 484.77 kPa
F = P2 A2
= 484.77 x ? 0.12 = 15.23 kN
2.5m
1 kN
F?
0.2 m
diam0.05 m
diam P
1
P2
Pressure at liquid interface
P1 = ?1gh1
P2 = ?1gh1 + ?2gh2
P3 = ?1gh1 + ?2gh2 + ?3gh3
?
Liquid 1 ?1
Liquid 2 ?2
Liquid 3 ?3
h1
h2
h3
P1
P2
P3
Pressure Measurement
? Absolute Pressure, Measured relative to perfect
vacuum
? Perfect vacuum, 0 absolute pressure
? Gauge Pressure, Measured relative to local
atmospheric pressure
? If no specified,pressure reading is usually assumed
to be gauge
? Gauge pressure is positive if higher than
atmospheric pressure,and negative if lower than
atmospheric pressure
? Negative gauge pressure is also known as suction or
vacuum pressure
Atmospheric Pressure
Atmospheric pressure is usually measured using a
mercury barometer
Patm = ρHg g h ? 101,300 N/m2 abs
= 0 N/m2 gauge
? 1.013 bar abs
? 760 mm Hg abs
? 10.3 m water abs
Manometers
Devices for pressure measurement,
? Piezometer tube (stand pipe)
? Simple U-tube
? Differential U-tube manometer
? Inclined tube manometer
Other types,
? Micro-manometer
? Barometer
Manometer Calculation
Within a continuous fluid of constant ?
P1 = P3
P2 = P1 - ? g ?h1 = P1- ? ?h1
P4 = P1 + ? g ?h2 = P1 + ??h2
? Start computation at one end,and compute pressure from
meniscus to meniscus till the other end
? Alternatively,select two points along same liquid column at
same level,equate the pressures at the two points
?h1
?h2P1
P2
P3
P4
Piezometer Tube
PA.gauge = ? g h1
Simple U-Manometer
P2 = P3
PA + ?1gh1 = ?2 g h2
PA,gauge = ?2 g h2 - ?1gh1
Differential U-tube
PA + ?1 g h1 - ?2gh2 - ?3gh3 = PB
PA – PB = ?2gh2 + ?3gh3 - ? g h1
Inclined Manometer
Pipes A & B contain gas
PA - ?2 g l2 Sin ? = PB
PA – PB = ?2 g l2 Sin ?
Pressure and force on a small plane area
Pressure at small plane,
P = ? g h
Force on small plane
dF = P x dA = ? g h dA
P depends on vertical depth h below surface
Force dF = P x plane area
P and dF perpendicular to plane
?
Pressure on plane P
h
Small plane area dA
Hydrostatic thrust on a plane (flat) surface
Force on dA,
dF = ? g h dA =? g y Sin ? dA
Resultant thrust F on plane of area A given by
integration of dF over A
Hydrostatic Thrust on a Plane (flat) Surface
dF = ? g h dA =? g y Sin ? dA
Resultant thrust F on plane of area A given by integration of
dF over A
? Thrust F = Resultant of pressure acting on every point of the
plane surface
? F acts perpendicular to plane
? Magnitude of F = ? g A hc
? F does not act through centroid of plane,C,in general
cc
A
A
g A hAyg S in
yd Ag S in
dAg yS indFF
???
??
??
??
?
??
?
? ?
Line of Action of Thrust F
Moment of dF about O,
dM = dF, y = ? g y2 Sin ? dA
c
o
c
A
R
R
A
A
Ay
I
Ay
dAy
y
FydAyg S in
dAS ingydMM
??
??
??
?
?
? ?
2
2
2
??
??
Line of Action of Thrust F
Parallel axes theorem,
Io = Ic + Ayc2
yR = (Ic + Ayc2) / (Ayc)
= yc + Ic / (Ayc)
?y = yR - yc = Ic / (Ayc)
? ?
C
hc
yc
yR
F
P
O
?y = yR-yc
Summary,Thrust on Plane Surface
Magnitude of thrust F,
F = ? g A hc
Acting at a point P ?y below the centroid of plane
C,
?y = Ic / (A yc)
?
?
C
hc
yc
yR
F
P
O
?y = yR-yc
Properties of common shapes
Concept of Pressure Prism
Average pressure Pav = ? g h/2
Resulting force FR = Pav x A
= ? g h/2 x b x h
= ? g A h/2
= Volume of pressure prism and passes
through the centroid of the pressure prism
Pressure Prism
Resulting force FR
= Volume of pressure prism
= ABDE + BCD
= F1 + F2
Location of FR can be determined by taking
about some axis,like A,
FR yA = F1y1 + F2y2
Effect of Atmospheric Pressure
Atmospheric pressure acts on both sides of wall,the
net effects on the wall cancel out
Resultant fluid force on a surface in contact with
fluid is caused by gauge pressure
Pressure on Surfaces
P1 = ? g h1
P2 = ? g h2
P3 = ? g h3
P4 = ? g h4
h1 h
2h3
h4
P1
P2P3
P4
?
Thin weightless plate submerged in water
Hydrostatic thrust F acts on both sides of plate
and cancel each other
?
F
F
Thrust on surface in contact with liquid
?
F
F1
?
?
F2
?
F
P
P
Rectangular Vertical Surface
?
F
h
h/3
P
Inclined Rectangular Surface
?
F
hh/3
?
F
h? h/3
P
P
Horizontal Plane Surface
?
F
C=P
Summary,Thrust on Plane Surface
Magnitude of thrust F,
F = ? g A hc
Acting at a point ?y below the centroid of plane
C,
?y = Ic / (A yc)
?
?
C
hc
yc
yR
F
P
O
?y = yR-yc
Example, Circular Surface
Magnitude,
F = ? g A hc
?y = Ic / Ayc
For circle,
Ic = ? D4 / 64
?y = (? D4 / 64) / (? D2 /4) (D/2) = D/8
or (3/8) D from base
?
F
D
3/8 D
hc
C
P
?y
Example, Circular Gate
Show that moment about hinge is independent of h
F = ? g A hc = ? g (? D2 /4) (h + D/2)
?y = Ic / Ayc = (? D4 /64) / (? D2 /4) (h + D/2)
Moment about hinge = F ?y
= ? g (? D4 /64) = independent of h
1.2 m diameter gate
?
F Hinge
h
?y
Hydrostatic Thrust on Curved Surface
Compute horizontal component Fh and vertical
component Fv of thrust acting on surface,
? Fh = thrust on projected area of curved surface on
a vertical plane
? Fv = weight of liquid vertically above curved
surface up to free liquid surface,acting through
centroid of that volume of liquid
? Resultant = vector sum of Fv and Fh
?
Curved surface
Projection of curved
surface on vertical
plane
Fh
Fv = W
Fh
Projected volume above
curved surface
Inclined Plane
Compute as single resultant FR or
Two components FV and Fh,and sum:
Resultant FR = vector sum of Fv and Fh
?
FR
?
FV=W
Fh
L Gate
Resultant FR = vector sum of F1 and F2
?
F1
?
F2
F1
F2
Direction of Thrust
Direction of Thrust always from liquid onto
surface
?
Fh
Fv=W
?
Fh Fv=W
Example
At what h will gate open up?
F1 = ?gAhc = 1000 x 9.81 x 3 x 3/2
F2 = ?gAhc = 1025 x 9.81 x h x h/2
Moment about hinge,
F1 x (2 + 0.6) = F2 x (3.6 - h/3)
Solve for h
?
F1
?
F2
Hinge
3m
0.6m
h=?
S=1.0 S=1.025
1 h/3
Example
Fv2 = ?g V = 1000 x 9.81 x (2x1x5)
Fv1 = ?g V = 1000 x 9.81 x (1/4 ? x 12 x 5)
Fh = ?g A hc = 1000 x 9.81 x (1x5) (2.5)
?y = Ic / Ayc = (1/12 x 5 x 13)/(1x5x2.5)
Moment about hinge,
M = Fv1 (4R/3?) + Fv2 (0.5) + Fh (0.5 - ?y )
FR = (Fh2 + (Fv1 + Fv2)2)1/2
FR = Resultant of Fh,Fv1,Fv2,passes through centre of circle
?
2m
1m 5 m long
circular gateFh
Fv1
Fv2
4R/3?
?y
FR
Example
What L for 0 moment about hinge?
F = ?g A hc = 1000x9.81x0.5x 2 x 0.25Sin45o
W = (L + 0.5) x 2 x 1000
Take moment about hinge,
F x 0.5 x 2/3 = W x ((L+0.5)/2 -0.5)Sin45o
Solve for L
?
0.5 m
L=?
2 m wide plate,weight
1000 N/m2,inclined at 45o
W
F
0.5/3
Example
What W for 0 moment about hinge?
Fv1 = ?g V =1000x9.81x(1.52-1/4 ?R2)
Fh = ?g Ahc = 1000x9.81x1.5x(1.5/2)
Fv2 = ?g V = 1000x9,81x (1.5x3 + 1/2 ?R2)
Take moment about hinge,
W x 1.5 + Fv1 x (3-0.223R) =
Fh x (1.5/3) + Fv2 x 1.5
?
Weight W
R=1.5mFh1.5/3
Fv1
Fv2
0.223R
Buoyancy Force
Pressure at points
Sum up
Thrusts on surfaces
Sum up
Buoyancy force on a body
Consider a floating cube,
Thrusts acting on 4 vertical surfaces of cube cancel each other
Resulting thrust on body =
Thrust acting on bottom surface of cube,
F3 = ? g h A = ? g Vi = Buoyancy force FB
h W F1F2
F3Area A
Volume immersed below
surface = Vi
?
Buoyancy Force
For a Submerged Cube,
Thrusts acting on 4 vertical surfaces of cube cancel out
Thrust acting on bottom surface of cube,
F3 = ? g h1 A
Thrust acting on top surface,
F4 = ? g h2 A
Resulting vertical thrust acting on cube,
Buoyancy force Fb = F4-F3
= ? g A(h2 – h1) = ? g Vi
h2 W F1F2
F4Area A
Volume immersed below
surface = Vi
?
h1 F3
Buoyancy Force, Summary
Resultant fluid force acting on a completely submerged or
floating body is a fluid is called buoyancy force FB
Archimedes’ Principle states magnitude of FB = weight of
fluid displaced by body,is directed vertically upward,and
passes through centroid of displaced fluid.
The point where FB passes through is called Centre of
Buoyancy,B,centroid of displaced fluid
Fb
Fb
Vi B
B
Example
Density of pole in N/m?
Fb = ?wg Vi = 1000 x 9.81 x (AxL/3)
A = cross section area of pole
W = ?pg Vp = ?p x 9.81 x (AxL)
Take moment about hinge,
Fb x (L - L/6) Cos 30o =
W x (L/2) Cos 30o
Solve ?p,density of pole in kg/m3
?
Hinge
water Uniform pole
30o
FbW
2/3 L
1/3 L
Example
Volume of concrete for zero moment at hinge?
(neglect weight of plate)
F1 = ?g Ahc = 1000x 9.81 x (2x2) x 2
F2 = ?g Ahc = 1000x 9.81 x (2x2) x 1
T + Fb = W
T + ?wg Vc = ?cg Vc
T + 1000x9.81xVc = 2400x9.81xVc
Moment about hinge,
F1 x 1 + F2 x (2/3) = T x (3),solve for Vc
? 2 m wide L
gate
Concrete block
?=2400 kg/m3
2 m
2m
1 m
HingeF
1
1F2
2/3
T
T
WFb
Stability, Fully submerged body
? Stable equilibrium, Body returns to its equilibrium
position when displaces
? Unstable equilibrium, Body moves to a new
equilibrium position when displaced
? A fully submerged body is in stable equilibrium if its
centre of gravity CG is below the centre of buoyancy c
Stability, Floating body
? For floating body,the centre of buoyancy shifts as body is
displaced
? In upper example,CG is higher than c (B) but body is
stable
? In lower example,CG is higher than c (B) but body is
unstable
Stability, Floating body
For floating body,
W= Fb
As body is displaced,Centre of Buoyancy shifts from B to B’
M = Meta-centre = intersection of vertical line through B’
and BG
? BM > BG ?Stable equilibrium
? BM = BG ?Neutral equilibrium
? BM < BG ?Unstable equilibrium
W
Fb
B
G
B’B
G
W
Fb
M
Stability, Floating body
For small displacement,
BM ? Iyy / Vi
Iyy = Moment of inertia of plane of rotation about
its longer centroidal axis
For a rectangular plane of rotation,
Iyy = 1/12 b h3
?
y
y
b
h
Example
Is buoy stable?
W = Fb
10,000 = ? g Vi = 1025 x 9.81x (? 1.82/4) h
h = 0.391 m
BG = 0.5 - h/2 = 0.305
BM ? Iyy / Vi = (? 1.84/64) / (? 1.82/4) h
= 0.518
As BM > BG
Body is stable
?
10 kN buoy
1 m
1.8 m diam
Seawater
1025 kg/m3
Fb
h GB
? Study of fluid at rest,i.e,not in motion or
not flowing
? On any plane,shear force is zero (no
velocity gradient or shear deformation)
hence only pressure forces exist
Pascal’s Law
? For fluid at rest,pressure at a point is the same in
all direction
P = Px = Py = Pz
Proving Pascal’s Law
As element is in equilibrium,
?Fy = 0
Py ?x ?z = Ps ?x ?s Sin ?
?s Sin ? = ?z
Hence,
Py = Ps
Similarly,for ?Fz = 0
Pz = Ps
Basic Hydrostatic Equation
for Pressure Field
Body of fluid in equilibrium,
? Fy = 0
P.A + W = (P + dP), A
P.A + ? g A dh = (P + dP) A
dP/dh = ? g = ?
dP/dy = - ? g = - ?
?
h
dh
P
P + dP
W
Rectangular free body
of top area A
y
Incompressible Fluid
Pressure Difference between two points in a body of
fluid
dP/dh = ? g
dP = ? g dh
Integrating from P1 to P2,
If fluid is incompressible and homogeneous
? = constant
P2 – P1 = ? g (h2 – h1) = ? g (? h)
?
P1
P2
h1h
2
?h
?? ?
2
1
2
1
h
h
P
P
g d hdP ?
Incompressible Fluid
If P1 is on free water surface,i.e,P1 = Patm,
P2 - P1 = ? g h
P2 - Patm = ? g h
P2,abs = ? g h + Patm
For gauge pressure measurement,Patm = 0
P2,gauge = ? g h
P2,abs = P2,gauge + Patm
Gauge pressure at a point h below free surface = ? g h
?
h
P2
P1=Patm
Definition of Pressure Head
Pressure head at point = P/? g (m)
Piezometric head = P/? g + y (m)
Piezometric pressure = P + ? g y
In fluid statics,gauge pressure head at depth h below free
surface
= P/? g = ? g h / ? g = h
Datum
Pressure = P
y
?
h
Example
P1 = F/A = 1000 / (? 0.0252) = 509.3 kPa
P2 = P1 - ? g (2.5) = 484.77 kPa
F = P2 A2
= 484.77 x ? 0.12 = 15.23 kN
2.5m
1 kN
F?
0.2 m
diam0.05 m
diam P
1
P2
Pressure at liquid interface
P1 = ?1gh1
P2 = ?1gh1 + ?2gh2
P3 = ?1gh1 + ?2gh2 + ?3gh3
?
Liquid 1 ?1
Liquid 2 ?2
Liquid 3 ?3
h1
h2
h3
P1
P2
P3
Pressure Measurement
? Absolute Pressure, Measured relative to perfect
vacuum
? Perfect vacuum, 0 absolute pressure
? Gauge Pressure, Measured relative to local
atmospheric pressure
? If no specified,pressure reading is usually assumed
to be gauge
? Gauge pressure is positive if higher than
atmospheric pressure,and negative if lower than
atmospheric pressure
? Negative gauge pressure is also known as suction or
vacuum pressure
Atmospheric Pressure
Atmospheric pressure is usually measured using a
mercury barometer
Patm = ρHg g h ? 101,300 N/m2 abs
= 0 N/m2 gauge
? 1.013 bar abs
? 760 mm Hg abs
? 10.3 m water abs
Manometers
Devices for pressure measurement,
? Piezometer tube (stand pipe)
? Simple U-tube
? Differential U-tube manometer
? Inclined tube manometer
Other types,
? Micro-manometer
? Barometer
Manometer Calculation
Within a continuous fluid of constant ?
P1 = P3
P2 = P1 - ? g ?h1 = P1- ? ?h1
P4 = P1 + ? g ?h2 = P1 + ??h2
? Start computation at one end,and compute pressure from
meniscus to meniscus till the other end
? Alternatively,select two points along same liquid column at
same level,equate the pressures at the two points
?h1
?h2P1
P2
P3
P4
Piezometer Tube
PA.gauge = ? g h1
Simple U-Manometer
P2 = P3
PA + ?1gh1 = ?2 g h2
PA,gauge = ?2 g h2 - ?1gh1
Differential U-tube
PA + ?1 g h1 - ?2gh2 - ?3gh3 = PB
PA – PB = ?2gh2 + ?3gh3 - ? g h1
Inclined Manometer
Pipes A & B contain gas
PA - ?2 g l2 Sin ? = PB
PA – PB = ?2 g l2 Sin ?
Pressure and force on a small plane area
Pressure at small plane,
P = ? g h
Force on small plane
dF = P x dA = ? g h dA
P depends on vertical depth h below surface
Force dF = P x plane area
P and dF perpendicular to plane
?
Pressure on plane P
h
Small plane area dA
Hydrostatic thrust on a plane (flat) surface
Force on dA,
dF = ? g h dA =? g y Sin ? dA
Resultant thrust F on plane of area A given by
integration of dF over A
Hydrostatic Thrust on a Plane (flat) Surface
dF = ? g h dA =? g y Sin ? dA
Resultant thrust F on plane of area A given by integration of
dF over A
? Thrust F = Resultant of pressure acting on every point of the
plane surface
? F acts perpendicular to plane
? Magnitude of F = ? g A hc
? F does not act through centroid of plane,C,in general
cc
A
A
g A hAyg S in
yd Ag S in
dAg yS indFF
???
??
??
??
?
??
?
? ?
Line of Action of Thrust F
Moment of dF about O,
dM = dF, y = ? g y2 Sin ? dA
c
o
c
A
R
R
A
A
Ay
I
Ay
dAy
y
FydAyg S in
dAS ingydMM
??
??
??
?
?
? ?
2
2
2
??
??
Line of Action of Thrust F
Parallel axes theorem,
Io = Ic + Ayc2
yR = (Ic + Ayc2) / (Ayc)
= yc + Ic / (Ayc)
?y = yR - yc = Ic / (Ayc)
? ?
C
hc
yc
yR
F
P
O
?y = yR-yc
Summary,Thrust on Plane Surface
Magnitude of thrust F,
F = ? g A hc
Acting at a point P ?y below the centroid of plane
C,
?y = Ic / (A yc)
?
?
C
hc
yc
yR
F
P
O
?y = yR-yc
Properties of common shapes
Concept of Pressure Prism
Average pressure Pav = ? g h/2
Resulting force FR = Pav x A
= ? g h/2 x b x h
= ? g A h/2
= Volume of pressure prism and passes
through the centroid of the pressure prism
Pressure Prism
Resulting force FR
= Volume of pressure prism
= ABDE + BCD
= F1 + F2
Location of FR can be determined by taking
about some axis,like A,
FR yA = F1y1 + F2y2
Effect of Atmospheric Pressure
Atmospheric pressure acts on both sides of wall,the
net effects on the wall cancel out
Resultant fluid force on a surface in contact with
fluid is caused by gauge pressure
Pressure on Surfaces
P1 = ? g h1
P2 = ? g h2
P3 = ? g h3
P4 = ? g h4
h1 h
2h3
h4
P1
P2P3
P4
?
Thin weightless plate submerged in water
Hydrostatic thrust F acts on both sides of plate
and cancel each other
?
F
F
Thrust on surface in contact with liquid
?
F
F1
?
?
F2
?
F
P
P
Rectangular Vertical Surface
?
F
h
h/3
P
Inclined Rectangular Surface
?
F
hh/3
?
F
h? h/3
P
P
Horizontal Plane Surface
?
F
C=P
Summary,Thrust on Plane Surface
Magnitude of thrust F,
F = ? g A hc
Acting at a point ?y below the centroid of plane
C,
?y = Ic / (A yc)
?
?
C
hc
yc
yR
F
P
O
?y = yR-yc
Example, Circular Surface
Magnitude,
F = ? g A hc
?y = Ic / Ayc
For circle,
Ic = ? D4 / 64
?y = (? D4 / 64) / (? D2 /4) (D/2) = D/8
or (3/8) D from base
?
F
D
3/8 D
hc
C
P
?y
Example, Circular Gate
Show that moment about hinge is independent of h
F = ? g A hc = ? g (? D2 /4) (h + D/2)
?y = Ic / Ayc = (? D4 /64) / (? D2 /4) (h + D/2)
Moment about hinge = F ?y
= ? g (? D4 /64) = independent of h
1.2 m diameter gate
?
F Hinge
h
?y
Hydrostatic Thrust on Curved Surface
Compute horizontal component Fh and vertical
component Fv of thrust acting on surface,
? Fh = thrust on projected area of curved surface on
a vertical plane
? Fv = weight of liquid vertically above curved
surface up to free liquid surface,acting through
centroid of that volume of liquid
? Resultant = vector sum of Fv and Fh
?
Curved surface
Projection of curved
surface on vertical
plane
Fh
Fv = W
Fh
Projected volume above
curved surface
Inclined Plane
Compute as single resultant FR or
Two components FV and Fh,and sum:
Resultant FR = vector sum of Fv and Fh
?
FR
?
FV=W
Fh
L Gate
Resultant FR = vector sum of F1 and F2
?
F1
?
F2
F1
F2
Direction of Thrust
Direction of Thrust always from liquid onto
surface
?
Fh
Fv=W
?
Fh Fv=W
Example
At what h will gate open up?
F1 = ?gAhc = 1000 x 9.81 x 3 x 3/2
F2 = ?gAhc = 1025 x 9.81 x h x h/2
Moment about hinge,
F1 x (2 + 0.6) = F2 x (3.6 - h/3)
Solve for h
?
F1
?
F2
Hinge
3m
0.6m
h=?
S=1.0 S=1.025
1 h/3
Example
Fv2 = ?g V = 1000 x 9.81 x (2x1x5)
Fv1 = ?g V = 1000 x 9.81 x (1/4 ? x 12 x 5)
Fh = ?g A hc = 1000 x 9.81 x (1x5) (2.5)
?y = Ic / Ayc = (1/12 x 5 x 13)/(1x5x2.5)
Moment about hinge,
M = Fv1 (4R/3?) + Fv2 (0.5) + Fh (0.5 - ?y )
FR = (Fh2 + (Fv1 + Fv2)2)1/2
FR = Resultant of Fh,Fv1,Fv2,passes through centre of circle
?
2m
1m 5 m long
circular gateFh
Fv1
Fv2
4R/3?
?y
FR
Example
What L for 0 moment about hinge?
F = ?g A hc = 1000x9.81x0.5x 2 x 0.25Sin45o
W = (L + 0.5) x 2 x 1000
Take moment about hinge,
F x 0.5 x 2/3 = W x ((L+0.5)/2 -0.5)Sin45o
Solve for L
?
0.5 m
L=?
2 m wide plate,weight
1000 N/m2,inclined at 45o
W
F
0.5/3
Example
What W for 0 moment about hinge?
Fv1 = ?g V =1000x9.81x(1.52-1/4 ?R2)
Fh = ?g Ahc = 1000x9.81x1.5x(1.5/2)
Fv2 = ?g V = 1000x9,81x (1.5x3 + 1/2 ?R2)
Take moment about hinge,
W x 1.5 + Fv1 x (3-0.223R) =
Fh x (1.5/3) + Fv2 x 1.5
?
Weight W
R=1.5mFh1.5/3
Fv1
Fv2
0.223R
Buoyancy Force
Pressure at points
Sum up
Thrusts on surfaces
Sum up
Buoyancy force on a body
Consider a floating cube,
Thrusts acting on 4 vertical surfaces of cube cancel each other
Resulting thrust on body =
Thrust acting on bottom surface of cube,
F3 = ? g h A = ? g Vi = Buoyancy force FB
h W F1F2
F3Area A
Volume immersed below
surface = Vi
?
Buoyancy Force
For a Submerged Cube,
Thrusts acting on 4 vertical surfaces of cube cancel out
Thrust acting on bottom surface of cube,
F3 = ? g h1 A
Thrust acting on top surface,
F4 = ? g h2 A
Resulting vertical thrust acting on cube,
Buoyancy force Fb = F4-F3
= ? g A(h2 – h1) = ? g Vi
h2 W F1F2
F4Area A
Volume immersed below
surface = Vi
?
h1 F3
Buoyancy Force, Summary
Resultant fluid force acting on a completely submerged or
floating body is a fluid is called buoyancy force FB
Archimedes’ Principle states magnitude of FB = weight of
fluid displaced by body,is directed vertically upward,and
passes through centroid of displaced fluid.
The point where FB passes through is called Centre of
Buoyancy,B,centroid of displaced fluid
Fb
Fb
Vi B
B
Example
Density of pole in N/m?
Fb = ?wg Vi = 1000 x 9.81 x (AxL/3)
A = cross section area of pole
W = ?pg Vp = ?p x 9.81 x (AxL)
Take moment about hinge,
Fb x (L - L/6) Cos 30o =
W x (L/2) Cos 30o
Solve ?p,density of pole in kg/m3
?
Hinge
water Uniform pole
30o
FbW
2/3 L
1/3 L
Example
Volume of concrete for zero moment at hinge?
(neglect weight of plate)
F1 = ?g Ahc = 1000x 9.81 x (2x2) x 2
F2 = ?g Ahc = 1000x 9.81 x (2x2) x 1
T + Fb = W
T + ?wg Vc = ?cg Vc
T + 1000x9.81xVc = 2400x9.81xVc
Moment about hinge,
F1 x 1 + F2 x (2/3) = T x (3),solve for Vc
? 2 m wide L
gate
Concrete block
?=2400 kg/m3
2 m
2m
1 m
HingeF
1
1F2
2/3
T
T
WFb
Stability, Fully submerged body
? Stable equilibrium, Body returns to its equilibrium
position when displaces
? Unstable equilibrium, Body moves to a new
equilibrium position when displaced
? A fully submerged body is in stable equilibrium if its
centre of gravity CG is below the centre of buoyancy c
Stability, Floating body
? For floating body,the centre of buoyancy shifts as body is
displaced
? In upper example,CG is higher than c (B) but body is
stable
? In lower example,CG is higher than c (B) but body is
unstable
Stability, Floating body
For floating body,
W= Fb
As body is displaced,Centre of Buoyancy shifts from B to B’
M = Meta-centre = intersection of vertical line through B’
and BG
? BM > BG ?Stable equilibrium
? BM = BG ?Neutral equilibrium
? BM < BG ?Unstable equilibrium
W
Fb
B
G
B’B
G
W
Fb
M
Stability, Floating body
For small displacement,
BM ? Iyy / Vi
Iyy = Moment of inertia of plane of rotation about
its longer centroidal axis
For a rectangular plane of rotation,
Iyy = 1/12 b h3
?
y
y
b
h
Example
Is buoy stable?
W = Fb
10,000 = ? g Vi = 1025 x 9.81x (? 1.82/4) h
h = 0.391 m
BG = 0.5 - h/2 = 0.305
BM ? Iyy / Vi = (? 1.84/64) / (? 1.82/4) h
= 0.518
As BM > BG
Body is stable
?
10 kN buoy
1 m
1.8 m diam
Seawater
1025 kg/m3
Fb
h GB
