Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

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StaticsStatics

Surface Forces

Static Surface Forces

Forces on plane areas

Forces on curved surfaces

Buoyant force

Stability submerged bodies

Forces on Plane Areas

Two types of problemsHorizontal surfaces (pressure is _______)Inclined surfaces

Two unknowns________________________

Two techniques to find the line of action of the resultant forceMomentsPressure prism

constant

Total forceLine of action

dp g

dzr=-

p a

Side view

Forces on Plane Areas: Horizontal surfaces

pAdAppdAFR

Top view

A

p = gh

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 g hAr=

weight of overlying fluid!FR =

centroid

h = _____________ _____________

Vertical distance to free surface

= volume

P = 500 kPa

What is p?

FR

p a 0x

pa

xr

¶- = =¶

gage

Forces on Plane Areas: Inclined Surfaces

Direction of forceMagnitude of force

integrate the pressure over the areapressure is no longer constant!

Line of actionMoment of the resultant force must equal the

moment of the distributed pressure force

Normal to the plane

Forces on Plane Areas: Inclined Surfaces

x y

RxRy

R cF p A=

cp

centroid

center 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

pdAFR

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

g y

coscp p gyr q= -

cosR c

A A

F p dA gy dAr q= -ò ò

cosR c

A

F p A g ydAr q= - ò

Change in pressure due to change in elevation

0A

ydA =ò for y origin at centroid

First Moments

Ac xdA

Ax

1

AxdA

1c A

y ydAA

= 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

1c A

y gAt y gtdA

Second Moments

Also called _______________ of the area

Ax dAyI 2

2x xc cI I Ay

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

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

moment of inertia

Could define i as I/A…

Product of Inertia

A measure of the asymmetry of the area

xycccxy IAyxI

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

Axy xydAI

y

x

y

x

Product of inertia

Ixyc = 0Ixyc = 0

(the resulting force will pass through xc)

Properties 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 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

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

ARR xpdAFx

A

RR xpdA

Fx

1

( )1cosR cA

c

x x p gy dAp A

r q= -ò

R cF p A=

Moment of resultant = sum of moment of distributed forces

coscp p gyr q= -

1 1cosR c

c cA A

x xp dA x gy dAp A p A

r q= -ò ò

Center of Pressure: xR

1 cosR

cA A

gx xdA xydA

A p Ar q

= -ò ò

10

AxdA

A=ò

For x,y origin at centroidxyc A

I xydA=ò

cos xycR

c

Igx

p Ar q

=-

xr is zero if the x axis or y axis is a line of symmetry

Center of Pressure: yR

ARR ypdAFy

A

RR ypdA

Fy

1R cF p A= coscp p gyr q= -

( )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 cosR A A

c

gy ydA y dA

A p Ar q

= -ò ò

Center of Pressure: yR

2xc A

I y dA=ò1

0A

ydAA

=ò

cos xcR

c

Igy

p Ar q

=-

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

Ry

g

FR

cosR R xcy F I gr q=- The moment about the centroid is independent of pressure!

cos yg gq =

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= × + × + × + × + ×

380R Ry F m blocks= ×380

6.33360R

m blocksy m

blocks×

= =

1 2 3 4 5 6 7 8 9 10

3 blocks

5Use moments

Inclined Surface Findings

The horizontal center of pressure and the horizontal centroid ________ when the x or y axis is a line of symmetry for the surface

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=-

cos xycR

c

Igx

p Ar q

=-

( )cos 90 0=

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 water

Example using Moments

Location of the resultant force

Find F using moments about hinge

teams

Depth to the centroid

Magnitude of the Force

ApF cR

abA

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

b = 2 m

a = 2.5 mpc = ___

FR= ________

hc = _____

cg hr

10 m

1.54 MN

Pressure datum? _____ Y axis?atmhingewater

F

8 m

4 m

FR

g

y

2

4a

Location of Resultant Force

__Rxb = 2 m

a = 2.5 m

cp

cosxcR

c

Igy

p Ar

q=-

cosq =45

pc = ___ cg hr

2 44 5R

c

g ay

ghrr

æö=-è ø

02

5Rc

ay

h=- =- 0.125 m

xcIA=

hingewater

F

8 m

4 m

FR

g

Force Required to Open Gate

How do we find the required force?

0hingeM

F = ______ b = 2 m

2.5 mlcp=2.625 m

m 5

m 2.625N 10 x 1.54 6

F

tot

cpR

l

lFF

ltot

Moments about the hinge=Fltot - FRlcp

809 kN

cp

hingewater

F

8 m

4 m

FR

g

Forces 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 gate was symmetrical about at least one of the centroidal axes.

Pressure increases with depth.

Forces on Curved Surfaces

Horizontal componentVertical componentTensile Stress in pipes and spheres

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 _____.zero

teams

Forces on Curved Surfaces: Vertical Component

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

r

h

p = gh

F = ghr2 =W!

F = pA

What if the cup had sloping sides?

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 thesurface where the pressure is equal to the reference pressure.

water= (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 A

2 m

2 m

3 m W1

W2

W1 + W2

= 58.9 kN + 30.8 kN= 89.7 kN

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

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

Take moments about a vertical axis through A.

Example: Forces on Curved Surfaces

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= + 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+

=

43

Rp

Expectation???

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

b

a

ch = y

x

4 m

0.083Ry m=-

cosxcR

c

Igy

p Ar

q=- cosq = 1

cp = cghr2

12Rc

ay

h=-

2

12xcI aA=

Example: Forces on Curved Surfaces

78.5 kN

89.7 kN

4.083 m

0.94

8 m

119.2 kN

horizontal

vertical

resultant

C

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0

0.948 m

1.083 m

89.7kN

78.5kN

Cylindrical 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.

Curved Surface Trick

water 2 m

3 m

A

W1

W2FO

W1 + 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 = ________)

Tensile Stress in Pipes: High Pressure

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

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=-

2

12Rc

g dy

pr

=-

2

12xcI dA=

Use moments to calculate T1 and T2.

Solution Scheme

Determine total acceleration vector (a) including acceleration of gravity

Locate centroid of the surface Draw y axis with origin at the centroid (projection of

total acceleration vector on the surface) Set pressure datum equal to pressure on the other side

of the surface of interest Determine the pressure at the centroid of the surface Calculate total force (pcA) Calculate yr

Static Surface Forces Summary

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

acceleration)inclined surfaces (y coordinate has origin at

centroid)curved surfaces

Horizontal componentVertical component (________________________)

total acceleration

R cF p A=

weight of fluid above surface

A is projected area

cosxcR

c

Igy

p Ar

q=-R cF p A=

R cF p A=

Review

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

g

ajet

atotal

p a Use total acceleration

The jet is pressurized…

atotal bottom = angle between and

ApF cR

cosxcR

c

Iy

p Ar

q=- atotal

_____cp =hc

total ca hry

Buoyant Force

The resultant force exerted on a body by a static fluid in which it is fully or partially submergedThe 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

FB

zero

no

Weight of water displaced

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? _____________ _________

BF gr= "

Buoyant Force: Line of Action

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

c

V

Vx x dVg g=ò1

c

V

x xdVV

= ò = volume

d= distributed force

xc = centroid of volume

Definition of centroid of volume

Moment of resultant = sum of moments of distributed forces

If is constant!

Buoyant Force: Applications

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

F1

W

1

F2

W

2

Weight

Volume

Specific gravity

1 2

Force balance

Using buoyancy it is possible to determine:_______ of an object_______ of an object_______________ of

an object

>

Buoyant Force: Applications

1 1F 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 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-

=1FW

Suppose the specific weight of the first fluid is zero

(force balance)

Equate weights Equate volumes

Rotational Stability of Submerged Bodies

B

G

BG

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

below

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

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.

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?_______________________________ ____________________________________ ____________________

End 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.

L

d1

d2

Side view

End of Lecture

What didn’t you understand so far about statics?

Ask the person next to youCircle any questions that still need answers

Team Work

How will you define a coordinate system?

What is the pressure datum? What are the major steps

required to solve this problem?

What equations will you use for each step?

hingewater

F

8 m

4 m

Gates

Gates

Radial Gates

Gates at Itaipu:Why this shape?

Questions

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