Hydrostatic Forces on Plane & Curved Surfaces
Hydrostatic Forces on Plane & Curved Surfaces
CVEN 212Spring 2015
Riyadh Al-Raoush, PhD, PE
Static Surface ForcesStatic Surface Forces
hinge
water
?
8 m
4 m
Static Surface ForcesStatic Surface Forces
Forces on plane areas Forces on curved surfaces Buoyant force
Forces on Plane AreasForces on Plane Areas
Horizontal surfaces Inclined surfaces Techniques to find the line of action of the
resultant force Center of pressure
Statics Review: First MomentsStatics Review: First Moments
A xdAAx 1A xdA
A ydAAy 1
For a plate of uniform thickness the intersection of the centroidal axes is also the center of gravity
Moment of an area A about the y axis
Location of centroidal axisy
x
Statics Review: Second MomentsStatics Review: Second Moments
Also called ____________________
Ao dAyI 2
AyII o2 is the 2nd moment with respect to the horizontal centroidal axis of the
area
area moment of inertia
I
by the transfer equation
Area and Center of GravityArea and Center of Gravity
Side view
Forces on Plane Areas: Horizontal surfaces
Forces on Plane Areas: Horizontal surfaces
pAdAppdAF
Top view
A
p = hF is normal to the surface and towards the surface if p is positive.
F passes through the ________ of the area.
h
What is the force on the bottom of this tank of water?
centroid
Forces on Plane Areas: Horizontal surfaces
Forces on Plane Areas: Horizontal surfaces
Forces on Plane Areas: Inclined surfaces
Forces on Plane Areas: Inclined surfaces
Direction of force Normal to the plane
Magnitude of force integrate the pressure over the area pressure is no longer constant!
Line of action
Forces on Plane Areas: Inclined Surfaces
Forces on Plane Areas: Inclined Surfaces
A
B
O
O
x
y
xpx
yAh h
Free surface
centroid
center of pressure
py
Magnitude of Force on Inclined Plane Area
Magnitude of Force on Inclined Plane Area
pdAF sinyhp ydAF sin A ydAAy 1 sinAyF
AhF ApF p is the pressure at the __________________centroid of the area
Forces on Plane Areas: Center of pressure: xcpForces on Plane Areas: Center of pressure: xcp
The center of pressure is not at the centroid (because pressure is increasing with depth) x coordinate of center of pressure: xcp Acp xpdAFx
Acp xpdAFx 1 Acp dAxyAyx sinsin1
sinAyF sinyp
Acp xydAAyx 1
Moment of resultant = sum of moment of distributed forces
____ ____
Center of pressure: xcpCenter of pressure: xcp
AyI
x xycp xyxy IAyxI
AyIAyx
x xycp
Acp xydAAyx 1
AyI
xx xycp
If x = x or y = y is an axis of symmetry then the product of inertia Ixy is zero.
Axy xydAI
y
x
y
x
Product of inertia
Ixy = 0_
Ixy = 0_
Center of pressure: ycpCenter of pressure: ycp
Acp ypdAFy Acp ypdAFy 1 sinAyF sinyp
Acp dAyAyy sinsin1 2 Acp dAyAyy 21
AyIy ocp
Ao dAyI 2AyIIo
2y
AyI
AyAyIycp
2
Solve for ycp
________
Transfer equation
yAyIycp
Inclined Surface FindingsInclined Surface Findings
The horizontal center of pressure and the horizontal centroid ________ when the surface has either a horizontal or vertical axis of symmetry
The center of pressure is always _______ the centroid
The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid
What do you do if there isnt a free surface?
yAy
Iy xcp
AyI
xx xycp coincide
below
decreases
Example using MomentsExample using Moments
An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force F applied at the bottom of the gate is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of the gate.
hingewater
F
8 m
4 m
Solution Scheme Magnitude of the force
applied by the water Location of the resultant force Find F using moments about
hinge
Magnitude of the ForceMagnitude of the Force
ApFr abA
abhFr m 2m 2.5m 10
mN 9810F 3r
hingewater
F
8 m
4 m
b = 2 m
a = 2.5 m
p = ___
Fr= ________
h = _____
Fr
10 m
hDepth to the centroid
1.54 MN
Location of Resultant ForceLocation of Resultant Force
4
3baI x
hingewater
F
8 m
4 mhy
yAy
Iy xcp
abybayycp
4
3
)5.12(4)5.2(
4
22
mm
yayycp
abA
_______ yycp
________y
Fr
0.125 m
Slant distance to surface12.5 m
b = 2 m
a = 2.5 mcp
Force Required to Open GateForce Required to Open Gate
How do we find the required force?
hingewater
F
8 m
4 m
0hingeM
F = ______ b = 2 m
2.5 mlcp=2.625 m
Fr
m 5
m 2.625N 10 x 1.54 6Ftot
cpr
llF
F ltot
=Frlcp-FltotMoments about the hinge
809 kN
cp
Forces on Plane Surfaces ReviewForces on Plane Surfaces Review
The average magnitude of the pressure force is the pressure at the centroid
The horizontal location of the pressure force was at x (WHY?) ____________________ ___________________________________
The vertical location of the pressure force is below the centroid. (WHY?) ___________ ___________________
The gate was symmetrical about at least one of the centroidal axes.
Pressure increases with depth.
_
Forces on Curved SurfacesForces on Curved Surfaces
Horizontal component Vertical component Tensile Stress in pipes and spheres
Forces on Curved Surfaces: Horizontal Component
Forces on Curved Surfaces: Horizontal Component
The horizontal component of pressure force on a curved surface is equal to the pressure force exerted on a vertical ______ of the curved surface
The horizontal component of pressure force on a closed body is always ___.
The center of pressure is located on the projected area using the moment of inertia
projection
zero
Forces on Curved Surfaces: Vertical Component
Forces on Curved Surfaces: Vertical Component
What is the magnitude of the vertical component of force on the cup?
r
hp = h F = r2h =W!
Forces on Curved Surfaces: Vertical Component
Forces on Curved Surfaces: Vertical Component
The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the (virtual or real) free surface
Streeter, et. al
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.
FV =
FH = Ap
water 2 m
2 m
3 m W1
W2= 58.9 kN + 30.8 kN
= (4 m)(2 m)(1 m)= 78.5 kN
W1 + W2= (3 m)(2 m)(1 m) + (2 m)2(1 m)
= 89.7 kN
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
The vertical component line of action goes through the centroid of the volume of water above the surface.
21V W3)m2(4W)m 1(Fx water 2 m
2 m
3 m
A
W1
W2 kN 89.7
kN 30.83
)m2(4kN 58.9)m 1(x
Take moments about a vertical axis through A.
= 0.948 m (measured from A) with magnitude of 89.7 kN
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
water 2 m
2 m
3 m
A
W1
W2
The location of the line of action of the horizontal component is given by
yAy
Iy xp
12
3bhIx b
h
xIy
m 4.083m 4m 1m 2m 4m 0.667 4 py
(1 m)(2 m)3/12 = 0.667 m4
4 m
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
78.5 kN
89.7 kN4.083 m
0
.
9
4
8
m
119.2 kN
horizontal
vertical
resultant
Buoyant ForceBuoyant Force
The resultant force exerted on a body by a static fluid in which it is fully or partially submerged The projection of the body on a vertical plane
has two sides and the pressure on the two sides is equal, thus ________________________________
The projection of the body on a horizontal plane also has two sides, but the pressures on the two surfaces are _________
the horizontal forces always balance
different
Buoyant Force: Thought Experiment
Buoyant Force: Thought Experiment
Place a thin wall balloon filled with water in the tank of water.What is the net force on the balloon? _______Does the shape of the balloon matter? ________What is the buoyant force on the balloon? _____________ _________
FB
zero
no
Weight of water displaced
FB=V
Buoyant Force: Line of ActionBuoyant Force: Line of Action
The buoyant force acts through the centroid of the displaced volume of fluid (center of buoyancy)
xxd
xdx
1