Fluid Static Forces 2.2

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

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:

Closed Duct
Pipe
Dam
Direction of Fluid Pressure on Boundaries


 In a fluid at rest, pressure acts equally in all directions.
 Where a fluid is in contact with a surface, the pressure
gives rise to a force acting perpendicular to the surface.
 In a fluid at rest, pressure is constant along a horizontal
plane.
 In a fluid at rest, pressure increases with depth
according to the relation ∆𝑃 = 𝜌 𝑔 ∙ ∆ℎ.
 The pressure at the base of a column of fluid of depth
“h” is equal to the pressure at the top + 𝜌 𝑔 ∙ ∆ℎ

 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.

Hydrostatic Force on an Inclined Plane Surfaces
 - Magnitude,
 - Location

X
dA
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

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

A
P
A
h
g
A
)
sin
y
(
g
F 







Substituting into Eq. (1) gives:
y
h 

 sin
y
h
 - Magnitude

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:
 - Note
Hydrostatic Force on Plane Surfaces
𝐻 = 𝑦 sin 𝜃

Center of Pressure on an Inclined Plane Surface
 - Line of Action of F
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 
 





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

 - Line of Action of F (center of pressure)
 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 


19
 - Hoover Dam

20
 - Hoover Dam

21
 - Moment of Inertia for Common Shapes

23

C.G
C.P
Free Surface
A Completely SubmergedTilted 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




24

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 SubmergedTilted Rectangular Plate is at the
Free Surface and thus (S = 0)



h

25
 =90o
C.G
C.P
Free Surface
A Completely SubmergedVertical 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











26
 =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 SubmergedVertical 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 








 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???

Hydrostatic Force on a Curved Surface
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)
A

D
B
C
yCp
FV(1)
FH
FV(2)
E
 
w
EA
2
EA
DE
g
Fx 











A
  w
ABE
area
BCDE
area
g
FV 




S
B

 On an inclined flat or curved surface, the horizontal force
and its line of action is equal to the horizontal force on the
vertical projection of the inclined or curved surface.
 On an inclined flat or curved surface, the vertical force is
equal to the weight of the volume of water vertically
above the surface and its line of action passes through
the center of gravity of that volume.
 A body immersed in a fluid experiences a vertical
upwards force equal to the weight of the volume of fluid
displaced.
 A floating body displaces its own weight in liquid.

31
 Determine the volume of fluid above the curved surface,
 Compute the weight of the volume above it,
 The magnitude of the vertical component of the resultant
force is equal to the weight of the determined volume. It
acts in line with the centroid of the volume,
 Draw a projection of the curved surface onto a vertical
plane and determine its height called “S”,

 Determine the depth to the centroid of the projected area
=DE+(S/2), where DE is the depth to the top of the projected area
 Compute the magnitude of the horizontal component of
the resultant force, FH =  g h- A =  g (DE+S/2)x(Sx w)
 Compute the depth to the line of action of the horizontal
force,
h
A
I
h
h
cg
cp 


 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 


34
 Compute the angle of inclination of the resultant force
relative to the horizontal.
Thus,
11- Show the resultant force acting on the curved surface
in such a direction that its line of action passes through
the center of curvature of the surface.
 
H
V
1
F
F
tan



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)

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.

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

42
Grand Coulee Dam - Columbia River

43
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?

44
Hydrostatic forces on a vertical dam face

45
Because the area is (H x w), 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
/
wH
(
)
12
/
H
w
(
h
y 2
3
p
.
c 


The hydrostatic pressure is , where is the depth at
which the centroid of the dam is located.
h
g
P 

 2
H
h 

46
The direction of the force is normal and compressive to the dam face as
shown .
The center of pressure is therefore located at a distance H/6 directly below
the centroid of the dam, or a distance 2H / 3 of below the water surface.

47
 What is the net horizontal force acting on a constant
radius arch-dam face due to hydrostatic forces?

49
The projected area is , while
H
sin
R
2
A projected 
 
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 zero

55

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

56
Worked Examples
Static Forces
H.I.T May 2010

57
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

58
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

59
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

60
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


61
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
 


62

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.

63

 



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.

64
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
Pivot

1.0 m
4.0 m

65
Cont.
Water
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.

66
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 




67
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”)

68
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

69

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.

70

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”

72
Water
Pivot

1.0 m
4.0 m

Fluid Static Forces 2.2

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Fluid Static Forces 2.2

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