hydrostatic forces

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


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: 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


What is the magnitude of the vertical component of force on the cup?

r

hp = h F = r2h =W!



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


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



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



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



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

hydrostatic forces

Documents

horizontal center of

plane areas forces

plane areasforces

area pressure

plane magnitude of force

xcpcenter of pressure

ycpcenter of pressure

plane curved surfacescven