Static Surface Forces hinge water ? 8 m 4 m . Static Surface Forces ä Forces on plane areas ä...

Static Surface ForcesStatic Surface Forces

hingehinge

waterwater

??

8 m8 m

4 m4 m

Static Surface ForcesStatic Surface Forces

Forces on plane areas

Forces on curved surfaces

Buoyant force

Stability submerged bodies

Forces on plane areas

Forces on curved surfaces

Buoyant force

Stability submerged bodies

Forces on Plane AreasForces on Plane Areas

Two types of problems Horizontal surfaces (pressure is _______) Inclined surfaces

Two unknowns ____________ ____________

Two techniques to find the line of action of the resultant force Moments Pressure prism

Two types of problems Horizontal surfaces (pressure is _______) Inclined surfaces

Two unknowns ____________ ____________

Two techniques to find the line of action of the resultant force Moments Pressure prism

constantconstant

Total forceTotal force

Line of actionLine of action

dzdp dzdp

Side view

Forces on Plane Areas: Horizontal surfaces

Forces on Plane Areas: Horizontal surfaces

pAdAppdAFR pAdAppdAFR

Top view

A

p = h

F 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?

RF hAg=RF hAg=

weight of overlying fluid!weight of overlying fluid!FR =

centroid

h = _____________ _____________

Vertical distance to free surface

= volume

P = 500 kPa

What is p?

FR

Forces on Plane Areas: Inclined Surfaces


Direction of force Magnitude of force

integrate the pressure over the area pressure is no longer constant!

Line of action Moment of the resultant force must equal the

moment of the distributed pressure force

Direction of force Magnitude of force

integrate the pressure over the area pressure is no longer constant!

Line of action Moment of the resultant force must equal the

moment of the distributed pressure force

Normal to the plane



OO

xx yy

RxRyRy

R cF p A=

cp

centroid

center of pressurecenter of pressure

The coordinate system origin is at the centroid (yc=0)

Where could I counteract pressure by supporting potato at a single point?

g

Magnitude of Force on Inclined Plane Area

Magnitude of Force on Inclined Plane Area

pdAFR pdAFR

ApF cR ApF cR pc is the pressure at the __________________centroid of the area

g y

coscp p gyr q= - coscp p gyr q= -

cosR c

A A

F p dA gy dAr q= -ò ò cosR c

A A

F p dA gy dAr q= -ò ò

cosR c

A

F p A g ydAr q= - òcosR c

A

F p A g ydAr q= - ò 0A

ydA =ò 0A

ydA =ò

First MomentsFirst Moments

Ac xdA

Ax

1

Ac xdAA

x1

AxdAAxdA

1c A

y ydAA

= ò1

c Ay ydA

A= ò

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 axis

h3

1h

3

1

Second MomentsSecond Moments

Also called _______________ of the area

Ax dAyI 2Ax dAyI 2

2yAII xcx 2yAII xcx Ixc is the 2nd moment with respect to an axis passing through its centroid and parallel to the x axis.

The 2The 2ndnd moment originates whenever one computes the moment originates whenever one computes the moment of a distributed load that varies linearly from the moment of a distributed load that varies linearly from the moment axis.moment axis.

moment of inertia

Product of InertiaProduct of Inertia

A measure of the asymmetry of the area A measure of the asymmetry of the area

xycccxy IAyxI xycccxy IAyxI

If x = xc or y = yc is an axis of symmetry then the product of inertia Ixyc is zero.______________________________________

Axy xydAI Axy xydAI

y

x

y

x

Product of inertia

Ixyc = 0Ixyc = 0

(the resulting force will pass through xc)

Properties of AreasProperties of Areas

yc

baIxc

yc

b

aIxc

A ab=2c

ay =

3

12xc

baI =

2ab

A =

3c

b dx

+=

3

36xc

baI =

2A Rp=4

4xc

RI

p=R

ycIxc

0xycI =

( )2

272xyc

baI b d= -

0xycI =

3c

ay =

d

cy R=

2

12xcI aA=

2

18xcI aA=

2

4xcI RA=

Properties of AreasProperties of Areas

3

4xc

baI

p=A abp=

43c

Ry

p=

ayc

b

Ixc

2

2R

Ap

=43c

Ry

p=

4

8xc

RI

p=

ycR

Ixc

0xycI =

0xycI =

4

16xc

RI

p=

2

4R

Ap

=Ryc

cy a=

2

4xcI RA=

2

4xcI aA=

2

4xcI RA=

Forces on Plane Areas: Center of Pressure: xR

Forces on Plane Areas: Center of Pressure: xR

The center of pressure is not at the centroid (because pressure is increasing with depth) x coordinate of center of pressure: xR

The center of pressure is not at the centroid (because pressure is increasing with depth) x coordinate of center of pressure: xR

ARR xpdAFx ARR xpdAFx

A

RR xpdA

Fx

1

AR

R xpdAF

x1

( )1cosR cA

c

x x p gy dAp A

r q= -ò ( )1cosR cA

c

x x p gy dAp A

r q= -ò

R cF p A=R cF p A=

Moment of resultant = sum of moment of Moment of resultant = sum of moment of distributed forcesdistributed forces

coscp p gyr q= - coscp p gyr q= -

1 1cosR c

c cA A

x xp dA x gy dAp A p A

r q= -ò ò1 1

cosR cc cA A

x xp dA x gy dAp A p A

r q= -ò ò

Center of Pressure: xRCenter of Pressure: xR

1 cosR

cA A

gx xdA xydA

A p Ar q

= -ò ò1 cos

RcA A

gx xdA xydA

A p Ar q

= -ò ò

10

AxdA

A=ò

10

AxdA

A=ò

For x,y origin at centroidxyc A

I xydA=òxyc AI xydA=ò

cos xycR

c

Igx

p Ar q

=-cos xyc

Rc

Igx

p Ar q

=-

Center of Pressure: yRCenter of Pressure: yR

ARR ypdAFy ARR ypdAFy

A

RR ypdA

Fy

1

AR

R ypdAF

y1

R cF p A=R cF p A= coscp p gyr q= - coscp p gyr q= -

( )1cosR cA

c

y y p gy dAp A

r q= -ò ( )1cosR cA

c

y y p gy dAp A

r q= -ò

Sum of the moments

You choose the pressure datum to make the problem easy21 1

cosR cA Ac c

y yp dA gy dAp A p A

r q= -ò ò 21 1cosR cA A

c c

y yp dA gy dAp A p A

r q= -ò ò

21 cosR A A

c

gy ydA y dA

A p Ar q

= -ò ò 21 cosR A A

c

gy ydA y dA

A p Ar q

= -ò ò

Center of Pressure: yRCenter of Pressure: yR

2xc A

I y dA=ò 2xc A

I y dA=ò1

0A

ydAA

=ò1

0A

ydAA

=ò

cos xcR

c

Igy

p Ar q

=-cos xc

Rc

Igy

p Ar q

=-

21 cos 1R A A

c

gy ydA y dA

A p Ar q

= -ò ò 21 cos 1R A A

c

gy ydA y dA

A p Ar q

= -ò ò

For y origin at centroid

Location of line of action is below centroid along slanted surface.yR is distance between centroid and line of action

RyRy

g

FR

cosR R xcy F gIr q=-

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

The center of pressure is at the centroid for horizontal surfaces

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

The center of pressure is at the centroid for horizontal surfaces

coincide

below

decreases

0

>0

cosxcR

c

Igy

p Ar

q=- cosxcR

c

Igy

p Ar

q=-

cos xycR

c

Igx

p Ar q

=-cos xyc

Rc

Igx

p Ar q

=-

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 SchemeMagnitude of the force applied by the waterMagnitude of the force applied by the water

Example using MomentsExample using Moments

Location of the resultant forceLocation of the resultant force

Find F using moments about hingeFind F using moments about hinge

teams

Depth to the centroid

Magnitude of the ForceMagnitude of the Force

ApF cR ApF cR

abA abA

R cF gh abr p=R cF gh abr p=

( ) ( ) ( )3 2

kg m1000 9.8 10 m π 2.5 m 2 m

m sRF æ öæ ö=è øè ø

( ) ( ) ( )3 2

kg m1000 9.8 10 m π 2.5 m 2 m

m sRF æ öæ ö=è øè ø

b = 2 m

a = 2.5 mpc = ___

FR= ________

hc = _____

hingehingewaterwater

FF

8 m

4 m

FRFR

cg hr cg hr

10 m

1.54 MN

yyPressure datum? Y axis?

Location of Resultant ForceLocation of Resultant Force


FF

8 m

4 m

FrFr

__Rx __Rxb = 2 m

a = 2.5 m

cp

cosxcR

c

Igy

p Ar

q=- cosxcR

c

Igy

p Ar

q=-

cosq =cosq =45

pc = ___ cg hr cg hr

2 44 5R

c

g ay

ghrr

æö=-è ø

2 44 5R

c

g ay

ghrr

æö=-è ø

02

5Rc

ay

h=- =-

2

5Rc

ay

h=- =- 0.125

2

4xcI aA=

Force Required to Open GateForce Required to Open Gate

How do we find the required force?How do we find the required force?

0hingeM 0hingeM

F = ______ b = 2 m

2.5 mlcp=2.625 m

m 5

m 2.625N 10 x 1.54 6

F

m 5m 2.625N 10 x 1.54 6

F

tot

cpR

l

lFF

tot

cpR

l

lFF

ltot


FF

8 m

4 m

FrFr

Moments about the hinge=Fltot - FRlcp=Fltot - FRlcp

809 kN809 kN

cpcp

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 xc (WHY?) ____________________ ___________________________________

The vertical location of the pressure force is below the centroid. (WHY?) ___________ ___________________

The average magnitude of the pressure force is the pressure at the centroid

The horizontal location of the pressure force was at xc (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.

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

Horizontal component Vertical component Tensile Stress in pipes and spheres

Forces on Curved Surfaces: Horizontal Component

Forces on Curved Surfaces: Horizontal Component

What is the horizontal component of pressure force on a curved surface equal to? (Prove it!)

The center of pressure is located using the moment of inertia technique.

The horizontal component of pressure force on a closed body is _____.

What is the horizontal component of pressure force on a curved surface equal to? (Prove it!)

The center of pressure is located using the moment of inertia technique.

The horizontal component of pressure force on a closed body is _____.zerozero

teams

Forces on Curved Surfaces: Vertical Component


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

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

r

h

p = hp = h

F = hr2 =W!

F = pA

What if the cup had sloping sides?



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

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

surface where the pressure is equal to the reference pressure

I need to change this…

water= (3 m)(2 m)(1 m) + (2 m)2(1 m)= (3 m)(2 m)(1 m) + (2 m)2(1 m)

Example: Forces on Curved Surfaces


Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.

FV =

FH = cp Acp A

2 m

2 m

3 m W1

W2

W1 + W2W1 + W2

= 58.9 kN + 30.8 kN= 58.9 kN + 30.8 kN= 89.7 kN= 89.7 kN

= (4 m)(2 m)(1 m)= (4 m)(2 m)(1 m)= 78.5 kN= 78.5 kN y

x

= 0.948 m (measured from A) with magnitude of 89.7 kN= 0.948 m (measured from A) with magnitude of 89.7 kN

Take moments about a vertical axis through A.Take moments about a vertical axis through A.



The vertical component line of action goes through the centroid of the volume of water above the surface.

c V 1 2

4(2 m)x F (1 m)W W

3p= +c V 1 2

4(2 m)x F (1 m)W W

3p= + water 2 m

2 m

3 m

A

W1

W2( ) ( )

( )c

4(2 m)(1 m) 58.9 kN 30.8 kN

3x89.7 kN

p+

=( ) ( )

( )c

4(2 m)(1 m) 58.9 kN 30.8 kN

3x89.7 kN

p+

=

43

Rp

43

Rp

Expectation???



water 2 m

2 m

3 m

A

W1

W2

The location of the line of action of the horizontal component is given by

b

a

ch =ch = y

x

4 m4 m

0.083Ry m=0.083Ry m=

cosxcR

c

Igy

p Ar

q=- cosxcR

c

Igy

p Ar

q=- cosq =cosq = 1

cp =cp = cghr cghr2

12Rc

ay

h=-

2

12Rc

ay

h=-

2

12xcI aA=



78.5 kN78.5 kN

89.7 kN89.7 kN

4.083 m

0.94

8 m

119.2 kN119.2 kN

horizontalhorizontal

verticalvertical

resultantresultant

C

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 00

0.948 m

1.083 m

89.7kN

78.5kN

Cylindrical Surface Force CheckCylindrical Surface Force Check

All pressure forces pass through point C.

The pressure force applies no moment about point C.

The resultant must pass through point C.

All pressure forces pass through point C.

The pressure force applies no moment about point C.

The resultant must pass through point C.

Curved Surface TrickCurved Surface Trick

water 2 m

3 m

A

W1

W2FFOO

W1 + W2W1 + W2

Find force F required to open the gate.

The pressure forces and force F pass through O. Thus the hinge force must pass through O!

Hinge carries only horizontal forces! (F = ________)

Find force F required to open the gate.

The pressure forces and force F pass through O. Thus the hinge force must pass through O!

Hinge carries only horizontal forces! (F = ________)

Tensile Stress in Pipes: High Pressure

Tensile Stress in Pipes: High Pressure

pressure center is approximately at the center of the pipe

pressure center is approximately at the center of the pipe

T1

T2

FH

b

r

FH = ___

T = ___

= ____

(pc is pressure at center of pipe)

2rpc

(e is wall thickness)

rpc

pcr/e

is tensile stress in pipe wall

per unit length

Tensile Stress in Pipes: Low pressure

Tensile Stress in Pipes: Low pressure

pressure center can be calculated using moments

T2 __ T1

pressure center can be calculated using moments

T2 __ T1

T1

T2

FH

b

rd

>

d

b

Projected area

FH = ___ 2pcr

cosxcR

c

Igy

p Ar

q=- cosxcR

c

Igy

p Ar

q=-

2

12Rc

g dy

pr

=-2

12Rc

g dy

pr

=-

2

12xcI dA=

Solution SchemeSolution Scheme

Determine pressure datum Set pressure datum equal to pressure on the other side

of the surface of interest Usually the pressure datum is atmospheric pressure

Determine total acceleration vector (a) including acceleration of gravity

Determine if surface is normal to a, inclined, or curved

Determine pressure datum Set pressure datum equal to pressure on the other side

of the surface of interest Usually the pressure datum is atmospheric pressure

Determine total acceleration vector (a) including acceleration of gravity

Determine if surface is normal to a, inclined, or curved

Static Surface Forces SummaryStatic Surface Forces Summary

Forces caused by gravity (or _______________) on submerged surfaces horizontal surfaces (normal to total

acceleration) inclined surfaces (y coordinate has origin at

centroid) curved surfaces

Horizontal component Vertical component (________________________)

Forces caused by gravity (or _______________) on submerged surfaces horizontal surfaces (normal to total

acceleration) inclined surfaces (y coordinate has origin at

centroid) curved surfaces

Horizontal component Vertical component (________________________)

total acceleration

R cF p A=R cF p A=

weight of fluid above surface

A is projected area

cosxcR

c

Igy

p Ar

q=- cosxcR

c

Igy

p Ar

q=-R cF p A=R cF p A=

R cF p A=R cF p A=

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 is

always ____.

The vertical components of pressure on the top and bottom surfaces are _________

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 is

always ____.

The vertical components of pressure on the top and bottom surfaces are _________

zero

different

(Two surfaces cancel, net horizontal force is zero.)

Buoyant Force: Thought Experiment

Buoyant Force: Thought Experiment

FB

zero

no

Weight of water displaced

FB=V

Place a thin wall balloon filled with water in a 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? _____________ _________

Place a thin wall balloon filled with water in a 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? _____________ _________

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)

The buoyant force acts through the centroid of the displaced volume of fluid (center of buoyancy)

c

V

Vx xdVg g= òc

V

Vx xdVg g= ò1

c

V

x xdVV

= ò1

c

V

x xdVV

= ò

= volume = volume

d= distributed forced= distributed force

xc = centroid of volumexc = centroid of volume

Definition of centroid of volume

Moment of resultant = sum of moments of Moment of resultant = sum of moments of distributed forcesdistributed forces

If is constant!

Buoyant Force: ApplicationsBuoyant Force: Applications

1 1F V Wg+ =1 1F V Wg+ = 2 2F V Wg+ =2 2F V Wg+ =

F1F1

WW

11

F2F2

WW

22

WeightWeight

VolumeVolume

Specific gravitySpecific gravity

1 21 2

Force balance

Using buoyancy it is possible to determine: _______ of an object _______ of an object _______________ of

an object

Using buoyancy it is possible to determine: _______ of an object _______ of an object _______________ of

an object

>

Buoyant Force: ApplicationsBuoyant Force: Applications

1 1F V Wg+ =1 1F V Wg+ = 2 2F V Wg+ =2 2F V Wg+ =

( )

( )

1 1 2 2

1 2 2 1

2 1

1 2

F V F V

V F F

F FV

g g

g g

g g

+ = +

- = -

-=

-

( )

( )

1 1 2 2

1 2 2 1

2 1

1 2

F V F V

V F F

F FV

g g

g g

g g

+ = +

- = -

-=

-

1 2

1 2

2 1 2 1 2 1

1 2 2 1

2 1

W F W FV

W F W F

F FW

g g

g g g g

g gg g

- -= =

- = -

-=

-

1 2

1 2

2 1 2 1 2 1

1 2 2 1

2 1

W F W FV

W F W F

F FW

g g

g g g g

g gg g

- -= =

- = -

-=

-

1 2

2

F FV

g-

= 1 2

2

F FV

g-

=1FW 1FW

Suppose the specific weight of the first fluid is zero

(force balance)

Equate weightsEquate weights Equate volumes

----------- ________----------- ________

Buoyant Force (Just for fun)Buoyant Force (Just for fun)

The anchor displaces less water when it is lying on the bottom of the lake than it did when in the boat.

The anchor displaces less water when it is lying on the bottom of the lake than it did when in the boat.

A sailboat is sailing on Cayuga Lake. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Cayuga Lake increase or decrease?

Why?_______________________________ ____________________________________ ____________________

A sailboat is sailing on Cayuga Lake. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Cayuga Lake increase or decrease?

Why?_______________________________ ____________________________________ ____________________

Rotational Stability of Submerged Bodies

Rotational Stability of Submerged Bodies

B

G

BG

A completely submerged body is stable when its center of gravity is _____ the center of buoyancy

A completely submerged body is stable when its center of gravity is _____ the center of buoyancy

belowbelow

ReviewReview

How do the equations change if the surface is part of an aquarium on a jet aircraft during takeoff? (accelerating at 4 m/s2)

How do the equations change if the surface is part of an aquarium on a jet aircraft during takeoff? (accelerating at 4 m/s2)

sinAyF cR sinAyF cR

xcR c

c

Iy y

y A= +

g

ajet

atotal

sinc totaly A ar qº sinc totaly A ar qº

p a Use total acceleration

The jet is pressurized…

atotal surface = angle between and

No change!

End of Lecture QuestionEnd of Lecture Question

Write an equation for the pressure acting on the bottom of a conical tank of water.

Write an equation for the total force acting on the bottom of the tank.

Write an equation for the pressure acting on the bottom of a conical tank of water.

Write an equation for the total force acting on the bottom of the tank.

L

d1

d2

Side viewSide view

End of LectureEnd of Lecture

What didn’t you understand so far about statics?

Ask the person next to you Circle any questions that still need answers

What didn’t you understand so far about statics?

Ask the person next to you Circle any questions that still need answers

Team WorkTeam Work

How will you define a coordinate system? What are the 3 major steps required to solve

this problem? What equations will you use for each step?

How will you define a coordinate system? What are the 3 major steps required to solve

this problem? What equations will you use for each step?

hingewater

F

8 m

4 m

GatesGates

Radial GatesRadial Gates

QuestionsQuestions

Why does FR = Weight?

Why can we use projection to calculate the horizontal component?

How can we calculate FR based on pressure at the centroid, but then say the line of action is below the centroid?

Why does FR = Weight?

Why can we use projection to calculate the horizontal component?

How can we calculate FR based on pressure at the centroid, but then say the line of action is below the centroid?

Side viewh

What is p?

FR

Location of average pressure vs. line of action

Location of average pressure vs. line of action

What is the average depth of blocks?

Where does that average occur?Where is the resultant?

0

1 4 3 8 5 12 7 16 9 20R Ry F m blocks m blocks m blocks m blocks m blocks= × + × + × + × + ×1 4 3 8 5 12 7 16 9 20R Ry F m blocks m blocks m blocks m blocks m blocks= × + × + × + × + ×

380R Ry F m blocks= ×380R Ry F m blocks= ×380

6.33360R

m blocksy m

blocks×

= =380

6.33360R

m blocksy m

blocks×

= =

1 2 3 4 5 6 7 8 9 10

3 blocks

5Use moments

Static Surface Forces hinge water ? 8 m 4 m . Static Surface Forces ä Forces on plane areas ä...

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