MAAE 2300: Fluid Mechanics I
Hydrostatics – Forces on Surfaces
Carleton University - Winter 2019
MAAE 2300 - Fluid Mechanics I
1
Hydrostatic Forces on Planar Surfaces
Rideau
Canal Locks
boomervoice.ca
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MAAE 2300 - Fluid Mechanics I
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Hydrostatic Forces on Planar Surfaces
Rideau
Canal Locks
en.wikipedia.org
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MAAE 2300 - Fluid Mechanics I
3
Hydrostatic Forces on a Horizontal Surface
In a container with a flat, horizontal bottom, the resultant
force equals the pressure at the bottom times the area.
H
ܨ = ܣ
= γܣܪ
What about the non-horizontal planar surfaces ?
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MAAE 2300 - Fluid Mechanics I
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Hydrostatic Forces on Inclined Surfaces
Hydrostatic pressure is perpendicular to all surfaces, and it
increases with depth.
howitworksdaily.com
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Hydrostatic Forces on Inclined Surfaces
• The objective is to find:
• the total hydrostatic force, called: “Resultant Force”
• the line of action of the resultant force, called: “Center of Pressure”
Actual condition
Equivalent model
F
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MAAE 2300 - Fluid Mechanics I
Center of
pressure
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Resultant Force
( ∫ = ܣ݀ ∫ = ܨ + ߛℎ) ݀ܣ
= ܣ+ ߛ∫ ℎ ݀ܣ
(take: ߦ = ℎ/ sin ߠ)
= ܨ ܣ+ ߛ sin ߠ ∫ ߦ ݀ܣ
ଵ
the centroid: ߦେୋ = ∫ ߦ ݀ܣ
= ܨ ܣ+ ߛ sin ߠ ߦେୋ ܣ
( = ܨ + ߛℎେୋ ) = ܣେୋ ܣ
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Center of Pressure
To find the line of action (Center
of Pressure) of the Resultant
Force with respect to the centroid,
we balance its moment with the
moment of the elemental force
“ ”ܣ݀ about the centroid.
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Center of Pressure
ݕܨେ = ∫ ܣ݀ ݕ
= ∫ (ݕ + ߛߦ sin ߠ)݀ܣ
= ∫ ݕ ݀ ܣ+ ߛ sin ߠ ∫ ܣ݀ߦݕ
(note: ߦ = ݕେୋ − ߦ →
ߦ = ߦେୋ − )ݕ
ݕܨେ = ߛ sin ߠ (ߦେୋ ∫ ܣ݀ ݕ− ∫ ݕଶ ݀)ܣ
= −ߛ sin ߠ ܫ௫௫
ߛ sin ߠ ܫ௫௫
ݕେ = −
େୋ ܣ
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Distance of Center of Pressure from centroid
MAAE 2300 - Fluid Mechanics I
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Center of Pressure
Likewise in horizontal direction:
ߛ sin ߠ ܫ௫௬
ݔେ = −
େୋ ܣ
ߛ sin ߠ ܫ௫௫
ݕେ = −
େୋ ܣ
The Center of Pressure is always below the centroid and towards the
high-pressure side of the surface. The negative sign in the equation
corresponds to this fact.
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Resultant Force and Center of Pressure
When the fluid surface is open to the atmosphere, = 0 (gauge), then
for the Resultant Force and the Center of Pressure we have:
ߛ = ܨℎେୋ ܣ
ܫ௫௫ sinߠ
ݕେ = −
ℎେୋ ܣ
ܫ௫௬ sinߠ
ݔେ = −
ℎେୋ ܣ
Example:
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Resultant Force and Center of Pressure
Example:
The gate in the figure is 5 ft wide, it is hinged at point
B, and it rests against a smooth wall at point A.
Compute:
a) the force on the gate due to the seawater pressure,
b) the horizontal force P exerted by the wall at point
A, and
c) the reactions at the hinge B.
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MAAE 2300 - Fluid Mechanics I
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Resultant Force and Center of Pressure
Solution:
a) By geometry, the length of the gate from A to B is
= ܮ8ଶ + 6ଶ = 10ft, and the depth ℎେ is 15 − 3 = 12 ft.
12 ft
The resultant force:
CG
= ܨେୋ ߛ = ܣℎେୋ ܣ
= (64 lbf/ ft ଷ )(12 ft)(50 ft ଶ ) = 38,400 lbf
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Resultant Force and Center of Pressure
b) We need to find the center of pressure.
For a rectangular gate of 5 ft⨯10 ft:
ܫ௫௬ = 0,
ܾ ܮଷ (5 ft)(10 ft)ଷ
ܫ௫௫ =
=
= 417 ft ସ
12
12
b
ݔ
ݔ
ܮ
The distance between CP (Center of Pressure) and
CG (centroid):
6
ସ
417
ft
ܫ௫௫ sinߠ
10
݈ = −ݕେ =
=
= 0.417 ft
ℎେୋ ܣ
(12 ft)(50 ft ଶ )
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Resultant Force and Center of Pressure
b) The distance from point B to force F is:
10 ft − 5 ft − ݈ = 4.583 ft.
Balancing the moment about point B : (∑ ܯ = 0 )
ܲ ܮsinߠ − ܨ5 ft − ݈ =
ܲ 6 ft − 38,400 lbf 4.583 ft = 0.
Therefore:
ܲ = 29,300 lbf
Carleton University - Winter 2019
MAAE 2300 - Fluid Mechanics I
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Resultant Force and Center of Pressure
c) Balancing the forces in two directions:
∑ ܨ௫ = ܤ௫ + ܨsinߠ − ܲ.
= ܤ௫ + 38,400 lbf 0.6 − 29,300 lbf = 0.
Therefore:
ܤ௫ = 6,300 lbf
∑ ܨ௭ = ܤ௭ − ܨcosߠ.
= ܤ௭ − 38,400 lbf (0.8) = 0.
Therefore:
ܤ௭ = 30,700 lbf
Carleton University - Winter 2019
MAAE 2300 - Fluid Mechanics I
16
Resultant Force and Center of Pressure
Example:
Gate AB is 5 ft wide into the screen, it is hinged at A,
and it is restrained by a block stop at B. The depth is
ℎ = 9.5 ft, and the water is at 20°C. Compute:
a) the resultant hydrostatic force on the gate,
b) the force on stop B, and
c) the reactions at A.
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Resultant Force and Center of Pressure
Solution:
a) The centroid of AB is 2 ft below A:
ℎେୋ = ℎ − 4 + 2 = 7.5 ft
ℎେୋ
The resultant force:
= ܨେୋ ߛ = ܣℎେୋ ܣ
= (64.4 lbf/ ft ଷ )(7.5 ft)(20 ft ଶ ) = 9,360 lbf
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CG
MAAE 2300 - Fluid Mechanics I
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Resultant Force and Center of Pressure
b) To find the reactions, we need to find the center of
pressure.
b
For a rectangular gate of 5 ft⨯10 ft:
ܫ௫௬ = 0,
ܾ ܮଷ (5 ft)(4 ft)ଷ
ܫ௫௫ =
=
= 26.67 ft ସ
12
12
ݔ
ݔ
The distance between the CP (Center of Pressure)
and the CG (centroid):
ߠ = 90°
ܮ
CG
l
CP
ܫ௫௫ sinߠ
26.67ft ସ 1
݈ = −ݕେ =
=
= 0.178 ft
ℎେୋ ܣ
(7.5 ft) 20 ft ଶ
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Resultant Force and Center of Pressure
b) The distance from point A to the force F is:
4ft /2 + 0.178 ft = 2.178 ft.
Balancing the moment about point A : (∑ ܯ = 0 )
ܣ௫
ܤ௫ 4 ft − 9,360 lbf 2.178 ft = 0.
Therefore:
ܤ௫ = 5,100 lbf
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F
MAAE 2300 - Fluid Mechanics I
CG
CP
ܤ௫
20
Resultant Force and Center of Pressure
c) Balancing the forces in two directions:
∑ ܨ௫ = −ܤ௫ + ܨ− ܣ௫ = 0.
Therefore:
ܣ௫ = 9,360 lbf − 5100 lbf = 4,260 lbf
ܣ௫
Note: the vertical reaction force of the hinge (ܣ௬ )
depends on the weight of the gate and this is not
studied here as not enough information is available.
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F
MAAE 2300 - Fluid Mechanics I
CG
CP
ܤ௫
21
End of Lecture
Thank You
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