There is a simplest explaination of fluid mechanics

1. 1
Fluid Static Forces
Hydraulic I - CVE 215
Civil Eng. Dept.
By: Dr. Ezzat El-sayedG. SALEH

2. 2

3. Hydraulic I (3) Fluid Static Forces
Hydrostatic Forces on Plane Surfaces
 - Magnitude,
- Direction, it is the easiest! why?
 - Line of action
 To define the force, we need:

4. Closed Duct
Pipe
Dam
Hydraulic I (4) Fluid Static Forces
Direction of Fluid Pressure on Boundaries

5. (5)

6. 6

7.  Pressure can be measured by manometers.
 Surface tension can be affect the readings of
manometers.
 In a fluid at rest, pressure is constant along a horizontal
plane.
 On a submerged horizontal surface the pressure is
constant and the center of pressure is also the center of
area (centroid).
 On a submerged Vertical surface the pressure increases
with depth and the center of pressure is below the
centroid. (7)

9. Hydraulic I (9) Fluid Static Forces
Hydrostatic Force on an Inclined Plane Surfaces
 - Magnitude,
 - Location

10. X
dA
Hydraulic I (10) Fluid Static Forces
Center of Pressure
CP
Center of Gravity
CG
y
ỳ
ycp


F
P = g y sin 
Free
Surface

 

A
A
dA
y
g
dA
P
F

 sin
Hydrostatic Force on Plane Inclined Surfaces

11. Hydraulic I (11) Fluid Static Forces
A
y
sin
g
dA
sin
y
g
dA
P
F
A
A
A


 






Surfaces exposed to fluids experience a force due to
the distribution in the fluid
From solid mechanics the location of the center of
gravity (centroid of the area) measured from the
surface is
A
y
A
1
y
A


..(1)
 - Magnitude
Hydrostatic Force on Plane Inclined Surfaces

12. Hydraulic I (12) Fluid Static Forces
A
P
A
h
g
A
)
sin
y
(
g
F 







Substituting into Eq. (1) gives:
y
h 

 sin
y
h
 - Magnitude
Hydrostatic Force on Plane Inclined Surfaces

13. Hydraulic I (13) Fluid Static Forces
A
P
A
h
g
A
P
A
h
g
A
P
F atm
atm
)
R
(








The net pressure force on the plane, submerged
surface is
 Hydrostatic Force on Plane Surfaces

14. Center of Pressure on an Inclined Plane Surface
 - Line of Action of F
Hydraulic I (14) Fluid Static Forces
 Does the resultant force pass thought the center of
gravity ?
 No ! lies below the centroid, since pressure increases
with depth,
 Moment of the resultant force must equal the moment
of the distributed pressure force.
dA
y
sin
g
y
dF
y
F
A
2
A
cp 
 





15. 15
  2
g
.
c
cp
0
cp
y
A
I
A
y
y
or
I
sin
g
)
A
sin
y
g
(
y









X
dA
Hydraulic I (15) Fluid Static Forces
y
ỳ
ycp


F
P = g y sin 
Free
Surface
C.G
C.P
y
A
I
y
y
g
.
c
cp 


16.  - Line of Action of F (center of pressure)
Hydraulic I (16) Fluid Static Forces
 The location of the center of pressure is
independent of the angle ,
 The center of pressure is always below the
centroid,
 As the depth of immersion increase, the depth of
the center of pressure approaches the centroid.
y
A
I
y
y
g
.
c
cp 


17.  Patm = 0 Free surface
F
h
 g h
h/3
Liquid of
density, 
Pavg.
C. P
C. G
B
h
Pressure and Hydrostatic Force on a Vertical Plane Surface

18. Hydrostatic Force on a Dam

20. (c)

21. Pressure, Forces, Moments on a Submerged Symetric Plate
Determine the force and center of pressure on a rectangular wall
on a tank filled with a fluid (gasoline of s.g = 0.86) with total depth
= 3 m and length of wall = 10 m.

3 m
F
Wall
A
3 m
10 m
dA
Typical Problem
Typical Schematics

22. B
Hydrostatic Force: Vertical Wall
Find the pressure on a vertical wall using Hydrostatic Force Method?
O

H

25. Effect of Hydrostatic Force on Buildings

26. Hydrostatic Force on Curved Surfaces

27. Hydrostatic Force on Curved Surfaces
What force F is required to hold
the plate AB steady?
Plate AB has a width of 1.0m on to the paper

28. • Curved Submerged Surface
 Horizontal Force = Equivalent Vertical Plane Force
 Vertical Force = Weight of Fluid Directly Above
(+ Free Surface Pressure Force)
Hydrostatic Force on Submerged Surfaces

29. The dam shown below has a cylindrical surface with a
radius of 8 meters. If water is built up to the top of the dam,
what is the equivalent point load of the water pressing
against the dam.

8 m
Worked Example

30. Hydrostatic Force on Curved Surface

31. Hydrostatic Force on Curved Surface

32. 23

33. 33 Radial Gate Fabrication
Applications
Diversion of water for irrigation,
On top of dams to increase reservoir
capacity,
In spillways or in drainage canals to
maintain water elevation,
In other locations where wide, clear
waterway openings are necessary and
where economical control of water is
important.

34. 34 Overflow Radial Gate Installation

35. A 5-m long seawall separates a freshwater body from a saltwater body
as shown in the figure. The wall is a total of 4 m high, with the top half of
the wall being semicircular. Under design conditions, the surface of the
freshwater body is at the top of the wall and the surface of the saltwater
body is at the mid-height of the wall.
What is the net hydrostatic force on the wall under the design condition?
Assume that the freshwater and the and saltwater are 1.0 and 1.06 m
respectively.
Worked Example

36. 4 m
2 m
2 m



2 m
2 m
A
B C
D

38. 26

39. 39

40. Hydraulic I (31) Fluid Static Forces
 - Hoover Dam

41. Hydraulic I (27) Fluid Static Forces
 - Hoover Dam

42. 42
Hydraulic I (28) Fluid Static Forces
 - Moment of Inertia for Common Shapes

43. 29

C.G
C.P
Free Surface
A Completely Submerged Tilted Rectangular Plate
  )
b
a
(
sin
2
/
a
S
g
A
h
g
F 








 
  )
b
a
(
2
/
a
S
)
12
/
a
b
(
2
/
a
S
y
3
cp







  


 sin
2
/
a
S
h
h
y
cp
y
S
F
a




44. 30

C.G
C.P
Free Surface


 sin
)
2
/
a
(
h y
a
cp
y
F
  )
b
a
(
sin
2
/
a
g
A
h
g
F 




 


h
 
 
a
3
2
)
b
a
(
2
/
a
)
12
/
a
b
(
2
/
a
y
3
cp 





When the Upper Edge of the Submerged Tilted Rectangular Plate is
Free Surface and thus (S = 0)



h

45. 45
 =90o
C.G
C.P
Free Surface
A Completely Submerged Vertical Rectangular and thus ( = 0)
a
S
1
90
sin
:
where
)
2
/
a
S
(
h
o



cp
cp h
y 
h
y 
F
  )
b
a
(
2
/
a
S
g
A
h
g
F 







 
  )
b
a
(
2
/
a
S
)
12
/
a
b
(
2
/
a
S
y
3
cp











46. 31
 =90o
C.G
C.P
Free Surface
h
y 
cp
cp h
y 
0
S
,
1
90
sin
:
where
2
/
a
h
o



a
F
When the Upper Edge of the Submerged Vertical Rectangular
Plate is at the Free Surface and thus (S = 0 &  = 90o)
  )
b
a
(
2
/
a
g
A
h
g
F 






 
 
a
3
2
)
b
a
(
2
/
a
)
12
/
a
b
(
2
/
a
y
3
cp 








47. 32
 F increases as H increases ?
 decreases as H increases?
 is constant as H increases?
 T increases as H increases?
 T is constant as H increases?
y
ycp 
y
ycp 
True or False???

48. Hydraulic I (21) Fluid Static Forces
D C
E B
A
FH
W1
W2
 Assume Patm= 0 (gage) at the free
surface,
 The hydrostatic force on the surface
EA : is Fx
 The net vertical force on the curve
surface AB is: FV = W1+W2
To find the line of action of the
resultant force, balance the
momentum about some convenient
point
Pressure on Curved Surface (free body)

49. A

D
B
C
yCp
FV(1)
FH
FV(2)
E
Pressure on Curved Surface (free body)
 
w
EA
2
EA
DE
g
Fx 











A
  w
ABE
area
BCDE
area
g
FV 




S
B
Hydraulic I (33) Fluid Static Forces

50. Pressure on Curved Surface (free body)
 Determine the volume of fluid above the curved surfa
 Compute the weight of the volume above it,
 The magnitude of the vertical component of the resu
force is equal to the weight of the determined volume.
acts in line with the centroid of the volume,
 Draw a projection of the curved surface onto a vertica
plane and determine its height called “S”,
Hydraulic I (34) Fluid Static Forces

51. 51
Hydraulic I (35) Fluid Static Forces
 Determine the depth to the centroid of the projected a
=DE+(S/2), where DE is the depth to the top of the projected a
 Compute the magnitude of the horizontal component
the resultant force, FH =  g h- A =  g (DE+S/2)x(Sx w)
 Compute the depth to the line of action of the horizon
force,
h
A
I
h
h
cg
cp 

Pressure on Curved Surface (free body)

52. 52
Hydraulic I (36) Fluid Static Forces
 For regular rectangular cross-section,
Thus,
 Compute the resultant force,
h
A
I
h
h
cg
cp 

 
w
S
h
12
S
DA
g
A
h
g
F
2
H 
















2
V
2
H
R F
F
F 

Pressure on Curved Surface (free body)

53. 53
Hydraulic I (37) Fluid Static Forces
 Compute the angle of inclination of the resultant forc
relative to the horizontal.
Thus,
11- Show the resultant force acting on the curved surfac
in such a direction that its line of action passes throu
the center of curvature of the surface.
 
H
V
1
F
F
tan


Pressure on Curved Surface (free body)

54. 54
Hydraulic I (38) Fluid Static Forces
2.50 m
2.80 m
2. 0 m
water Oil s.g = 0.90
Support
Gate, 0.60 m wide
Hinge
Pressure on Plane Surface (free body)

55. 55
Hydraulic I (39) Fluid Static Forces
The figure shows a gate hinged at its bottom
hinged at its bottom and held by a simple support
at its top. The gate separates two fluids. Compute
 the net force on the gate due to the fluid on each
side, and
 Compute the force on the hinge and on the
support.
Pressure on Plane Surface (free body)

56. Hydraulic I (40) Fluid Static Forces
2.50 m
2.80 m
2. 0 m
water Oil Sg = 0.90
Support
Gate, 0.60 m wide
Hinge “o”
F1
F2
Hcp (1)
Hcp (1)
2.50 m
Pressure on Plane Surface (free body)

58. 31

59. 59
Grand Coulee Dam - Columbia River

60. 60
What are the magnitude and direction of the force on
the vertical rectangular dam (the shown figure) of
height H and width, w, due to hydrostatic loads, and
 At what elevation is the center of pressure?

61. 61
Hydrostatic forces on a vertical dam face

62. 62
Because the area is (Hxw), the magnitude of the force,
2
H
w
g
2
1
A
h
g
F 
 

The center of pressure is found using, )
h
A
/(
I
h
y cg
p
.
c 
 12
/
H
w
I 3
cg 
Thus, 6
/
H
)
2
/
w H
(
)
12
/
H
w
(
h
y 2
3
p
.
c 


The hydrostatic pressure is , where is the dept
which the centroid of the dam is located.
h
g
P 
 2
H
h 

63. 63
The direction of the force is normal and compressive to the dam face a
shown .
The center of pressure is therefore located at a distance H/6 directly b
the centroid of the dam, or a distance 2H / 3 of below the water surfac

64. 64
 What is the net horizontal force acting on a constant
radius arch-dam face due to hydrostatic forces?

65. 65

66. 66
The projected area is , while
H
sin
R
2
A proje cte d 
 
The pressure at the centroid of the projected surface is h
g
P 

The magnitude of the horizontal force is thus
and the center of pressure lies below the water surface on the
line of symmetry of the dam face.


 sin
R
H
g
A
h
g
F 2


3
2 /
H
Because the face is assumed vertical, the vertical force on the dam is ze

67. 67
Radial Gate

68. 68

69. 69

70. 70
A
B
C
FH, U
D
FH, U
FH, D
FV, U FV, D
O
E


)
W
"
AD
("
"
AD
"
g
F U
,
H 



 )
W
"
OA
("
"
OA
"
g
F D
,
H 




  W
BCD
area
ABCD
area
g
F U
,
V 




  W
AOE
area
g
F D
,
V 




3
/
r
4
Curved Surfaces

71. 71
A
B
C
o
D
FH
W
)
W
"
AD
("
"
AD
"
g
FH 




  W
BCD
area
ABCD
area
g
FV 




2
V
2
H
t F
F
F 
 V
F
H
F
t
F

)
F
/
F
(
tan H
V
1



Curved Surfaces


72. 72

73. 73

74. 74

Fluid of s. g =
P =
)
g
g
.
s
/(
P
h w
.
equi 



B
B
A
h
B
h
The tank shown in the figure contains a
fluid of s.g = and is pressurized to P =
How to calculate the forces (horizontal
& vertical on the quarter circle gate
AB????
Convert the pressure to an equivalent head “h” equiv.
)
g
g
.
s
/(
P
h w
.
equi 


 and solve it as before

75. 75

76. 76
Worked Examples
Static Forces
H.I.T May 2010

77. 77
9.0 m
45 m
Water
 Water Surface
9.0 m
A semicircular 9.0 m diameter tunnel is to be built under a 45
m deep, 240 m long lake, as shown. Determine the total
hydrostatic force acting on the roof of the tunnel.
Tunnel
Worked Example

78. 78
9.0 m
45 m
Water
 W
ater Surface
9.0 m
Tunnel
A
B C
D
E
Fx
Fy
Fx
Fy
The hydrostaticforce actingon the roof of the tunnel
The hydrostatic force acting on the roof of the tunnel

79. 79
 It acts vertically downward (see the shown figure
below).
0
Fx 

  w
A E D
a re a
A B C D
a re a
g
Fy 



kN
10
64
.
874
240
9
8
9
45
81
.
9
1000 3
2
















 The vertical force can be computed as

80. 80
R
1

If this weightless quarter-cylindrical gate is in static equilibrium, what is
the ratio between 1 and 1 ?
2

Worked Example “Static Forces on Curved Surfaces”
Pivot
 


81. 81

3
/
R
4
R
U
,
H
F D
,
H
F
U
,
V
F
1
 2

U
,
H
D
,
H
H F
F
F 


W
)
ABC
volume
(
g
FV 



3
/
R 3
/
R
)
1
R
(
2
R
F 1
U
,
H 




)
1
R
(
2
R
F 2
D
,
H 




Both of the horizontal
forces, FH,U and FH,D
act at a vertical
distance of R/3 above
the pivot.
A
B C
For a unite wide of the gate
U
,
H
F
 
Cont.

82. 82

 



3
R
4
F
3
R
F U
,
V
H
Taking the moment about the pivot “o” and equating it to zero gives,
or























3
R
4
0
.
1
4
R
3
R
)
0
.
1
R
(
2
R
)
(
2
1
1
2
 H
F  U
,
V
F
1
1
2 2
)
( 





 the ratio between 1 and 1
3
/
1
2
1 


Cont.

83. 83
Worked Example “Static Forces on Curved Surfaces”
The homogeneous gate shown in the figure consists of one quarter of a
circular and is used to maintain a water depth of 4.0 m. That is, when
the water depth exceeds 4.0 m, the gate opens slightly and lets the
water flow under it. Determine the weight of the gate per meter of length.
Water
Worked Example “Static Forces on Curved Surfaces”
Pivot

1.0 m
4.0 m

84. 84
Cont.
Water
Worked Example “Static Forces on Curved Surfaces”
Pivot

1.0 m
4.0 m
0.5 m
1- 4R/3
FV,2
FV,1

FH
C.p
3
2
/
1
h 

C.G
O
A
B
C D
W
In the shown figure, FH and FV are the
horizontal and vertical components of
the total force on the one quarter of a
circular immersed homogenous gate.

85. 85
Cont.
0.5 m
1- 4R/3
FV,2
FV,1

FH
C.p
O
A
B
C D
W
For the Horizontal Force:
g
5
.
3
)
1
1
(
)
2
/
1
3
(
g
A
h
g
FH 









It acts towards left at a distance hcp from the free surface,
which can be calculate as
m
52
.
3
5
.
3
)
1
1
(
)
12
/
1
1
(
5
.
3
h
A
I
h
h
3
cg
p
.
c 








For the Vertical Force:
The horizontal force FH= Resultant force on the projection
of “OA” on a vertical plane:
The vertical force FV = Weight of the
volume of the water which would lie
vertically above “OA”.
  0
.
1
AODCDA
area
g
width
unit
per
,
FV 




86. 86
Cont.
0.5 m
1- 4R/3
FV,2
FV,1

O
A
B
C D
W
  0
.
1
ODCB
area
AOB
area
g
width
unit
per
,
FV 





2
,
V
1
,
V F
and
F
  g
785
.
0
0
.
1
)
4
/
1
(
g
0
.
1
AOB
area
g
F 2
1
,
V 











  g
3
0
.
1
)
3
1
(
g
0
.
1
ADCB
area
g
F 2
,
V 










The force acts upward, through the centroid of the gate at a
distance of (1- 4R/ 3) = 1- (4x1/ 3 ) = 1- 0.424 =0.576 m (left
to “O”)
The force acts upward, through the centroid of the rectangular
“ODCB” at a distance of 0.50 m (left to “O”)

87. 87
Cont.
0.5 m
1- 4R/3
FV,2
FV,1

O
A
B
C D
W
Since the gate is homogenous, it weight acts downward,
through the centroid at a distance of (1- 4R/ 3) = 1- (4x1/ 3 ) =
1- 0.424 =0.576 m (left to “O”).
FH
3.52 - 3.0 = 0.52m
Taking the moments about pivot “O”
576
.
0
W
50
.
0
F
576
.
0
F
52
.
0
F
0
M 2
,
V
1
,
V
H
"
O
"
about 








576
.
0
50
.
0
F
576
.
0
F
52
.
0
F
W
2
,
V
1
,
V
H 






Substituting the compute values gives,
kN
25
.
64
1000
81
.
9
1000
55
.
6
g
55
.
6
576
.
0
50
.
0
3
576
.
0
g
785
.
0
52
.
0
g
5
.
3
W














W = the weight of the
gate per meter of length

88. 88

1.20 m
1.80
m
2.40 m
P
S.G = 0.95
Air
Air
A
Worked Example “Static Forces on
Plane Surfaces”
•What is the pressure at “A”?
• Draw a free body diagram of the
gate (5.0 m) showing all forces and
the locations of their lines of action.
• Calculate the minimum force “P”
necessary to keep the gate close.

89. 89

1.20 m
1.80
m
2.40 m
P
S.G = 0.95
Air
Air
A
8
.
1
g
PA 


H
F
V
F
Worked Example “Static Forces on
Plane Surfaces”

90. 90
Water
Worked Example “Static Forces on Curved Surfaces”
Pivot

1.0 m
4.0 m