€ Hydrostatic Force (= Force due to the pressure of a fluid at rest)
e.g Force exerted on the wall of storage tanks, dams, and ships)
Q. How is Hydrostatic Force on the vertical or inclined planes determined?
Basic conditions for a Plane surface submerged in a fluid
- Force on the surface: Perpendicular to the surface (No τ )
- Pressure: Linearly dependent only to the vertical depth
1. On a Horizontal surface (e.g. the bottom of a tank)
Pressure at the bottom, p = γh
: Uniform on the entire plane
∴ Resultant force FR = pA = γhA
(A: the bottom area of container)
2. On an Inclined surface
Consider a plane shown
y
€
- At surface: p = patm
- Angle θ between free surface
& the inclined plane
x
- y axis: Along the surface
x axis: Out of the plane
y Along the vertical depth h
& Pressure linearly changes &
Hydrostatic force changes
Differential Force acting on the differential area dA of plane,
dF = (Pressure) ⋅ (Area) = (γh) ⋅ (dA)
(Perpendicular to plane)
Then, Magnitude of total resultant force FR
FR = ∫A γhdA = ∫A γ ( y sin θ )dA
= γ sin θ
∫A ydA
where h = y sin θ
1st moment of the area
- Related with the center of area
Ú
∫A ydA = yc A where yc: y coordinate of the center of area (Centroid)
c.f. Center or 1st moment
∫ xdm = MX C &
M
∫A xdA =
xc
&
∫ ydm = MYC
(XC & YC: Center of Mass)
∫A ydA = yc
(xc & yc: Center of Area)
M
Moment of inertia or 2nd moment
2
∫ r dm = I
(2nd moment of Mass)
M
2
∫A y dA = I x
&
2
∫A x dA = I y
(2nd moment of Area)
Then,
FR = γAyc sin θ = (γ hc ) A
where γ hc : Pressure at the centroid
= (Pressure at the centroid) × Area
- Magnitude of a force on an INCLINED plane
- Dependent on γ , Area, and Depth of centroid
- Perpendicular to the surface (Direction)
Ú Where is the point (location) of action of FR?
- Not passing though Centroid!! (Why?)
- Related with the balance of torques due to of FR
i) Position of FR on y-axis
yR
: y coordinate of the point of action of FR
Moment about x axis:
FR y R = (γAyc sin θ ) y R = ∫A ydF = ∫A γ sin θ y 2 dA = γ sin θ ∫A y 2 dA
∴ yR
∫ y
= A
2
dA
yc A
=
Ix
where I x = ∫A y 2 dA : 2nd moment of area
yc A
or, by using the parallel-axis theorem, I x = I xc + Ayc 2
∴ yR =
I xc
+ yc
yc A
(Always below the centroid !)
ii) Position of FR on x-axis
xR
: x coordinate of the point of action of FR
By the similar manner,
FR xR = (γAyc sin θ ) xR = ∫A xdF = ∫A γ sin θ xydA = γ sin θ ∫A xydA
∴
xydA
∫
A
xR =
=
yc A
I xy
yc A
where ∫A xydA = Ixy: Area product of inertia
or, by using the perpendicular-axis theorem,
∴ xR =
I xyc
yc A
+ xc
c.f. For a symmetric submerged area, xR = xc (Ixyc = 0)
Ú Pressure Prism (Graphical interpretation of pressure distribution)
- Especially useful for a rectangular plane (or area)
- Simple method for finding the force and the point of action
Consider the situation shown
Ú Information from the diagram
- Vertical wall of width b and height h
- Contained liquid with specific weight γ
- Pressure: ptop = 0 & pbottom = γ h
From the last section,
⎛h⎞
FR = (γhc ) ⋅ ( A) = pav (at the centroid)×area = γ ⎜ ⎟ A
⎝2⎠
Let’s define a pressure-area space. (See the right figure above]
1. Horizontal axis: Magnitude of the pressure
2. Vertical axis: Height of the area
3. Axis toward the plane: Width of the area
: Resultant volume (Pressure prism)
y How to find the resultant force FR from the pressure prism
1
⎛h⎞
FR = Volume of the pressure prism = (γh)(bh) = γ ⎜ ⎟ A
2
⎝2⎠
y How to find the point of action of FR (the point of action)
From the last section,
yR =
I xc
I xc
1
+ h
+ yc =
yc A
2
⎛1 ⎞
⎜ h ⎟(bh)
⎝2 ⎠
(In case of rectangular plate, I xc =
1
1
Ah 2 = bh3 )
12
12
1 3
bh
1
1
1
2
12
+ h = h + h = h (from the top)
yR =
1
6
2
3
h ⋅ (bh) 2
2
From the pressure prism
h
yR = Centroid of the pressure prism
=
1
2
h (above the base) = h (from the top)
3
3
xR = Horizontal center
h/3
Ú Special case of a plane surface not extending up to the fluid surface
- Completely submerged plane
(See Figure)
Consider the situation shown
Pressure prism
- Trapezoidal cross section
(1) Resultant force FR
= Volume of
the shadow region
FR = Volume of hexahedron + Volume of prism
= F1( ABDE ) + F2( BCD )
1
= (γh1 ) A + [γ (h2 − h1 )] A
2
(2) The location of FR ( y A ) :
Consider the moments again
Moment by FR acting at y A
= Moment by F1 at y1 + Moment by F2 at y2
FR y A = F1 y1 + F2 y2
h
for rectangle
2
2h
for triangle (From the top)
y2 =
3
where y1 =
y The effect of the atmospheric pressure patm
: Increasing Volume of hexahedron (F1), NOT the prism (F2)
3. Curved Surface (Nonplanar surface)
- FR: Change in both magnitude and direction
- Can be determined by integration but not simple
y Special method using equilibrium of fluid
Q. Determine the magnitude and location of
FR exerting on the section BC
Consider the equilibrium situation shown,
Step 1. Isolate the section of fluid
(Section ABC with a unit perpendicular length)
Step 2. Draw a free-body diagram
F1 ( F2 ): Pressure force on Horizontal (Vertical) plane
W : Weight of fluid ( γ × Volume)
FH ( FV ): Horizontal (Vertical) comp. of reaction force by wall
Step 3. Equations of motion
FH = F2
&
FV = F1 + W
(Equilibrium condition)
In equilibrium condition,
r
Hydrostatic force by fluid ( FR )
r r
r
& Balance with reaction force by wall ( F = FH + FV )
∴ FR = ( FH ) 2 + ( FV ) 2 : Magnitude &
Ú Where is the location of FR ?
tan θ =
FV
: Direction
FH
θ : Perpendicular to the wall (Why?)