Maths_Catenaries, Parabolas and Suspension Bridges

SUSPENSION BRIDGES / D A GRIFFIN

CATENARIES, PARABOLAS AND SUSPENSION BRIDGESDAVID GRIFFIN


A Brief, Pictorial History of Suspension Bridges. Famous Examples Problems and Solutions.

Galileo The Equations for the Chain Curve of a Suspension Bridge. The Equation for a Hanging Chain. Parabola or Catenary?

The Catenary

and Jakob

Bernoulli’s Challenge. Leibniz Hugens Bernoulli

The Methods for Deriving the Equation for the Catenary.Calculus.Differential Equation.The Calculus of Variations.

The Relationship Between the Parabola and the Catenary. The Inverted Parabola.

Arch BridgesThe Inverted Catenary.

ArchesCan Catenaries Help You to Cycle with Square Wheels?

TALK CONTENTS


A “Suspended-deck Suspension Bridge”


The remains of the Maya Bridge at Yaxchilan, Mexican/Guatemalan border.The earliest known suspension-deck suspension bridge. 100m in three spans. 7th

Century.


Faust Vrančić

In 1595 the Croation

bishop Faust Vrančić

designed asuspension bridge, but it was never constructed.


James Creek Suspension Bridge, Pennsylvania.

James Finlay, 1801.

Bridge demolished 1833.

The first modern suspension bridge.

It used wrought-iron cables.


Dryburgh

Bridge, River Tweed.Opened1817. Collapsed 1818.

Union Bridge, River Tweed, 1820.The oldest suspension bridge stillcarrying traffic.


The Menai

Suspension BridgeThomas Telford, Completed 1826


The Clifton Suspension BridgeI K Brunel, Completed1864.



Two cablessupport onedeck.


BROOKLYN BRIDGEJohn Augustus Roebling,1883.


Four cables, two decks.


A light walkway is suspended between the two decks of the Brooklyn Bridge.


The Golden Gate Suspension BridgeIrving Morrow, Charle

Alton Ellis, Leon Moissieff, 1937.


Tacoma Narrows BridgeLeon Moissieff. Opened1940.

PC LT


New Tacoma Narrows Bridge (Background: Mt Rainier)Charles E Andrew and Dexter R Smith,1950

Map


A second bridge at Tacoma Narrows was built in 2007.


The Severn BridgeWilliam Brown, 1966


Brown designed the Severn Bridge to avoidthe problems of the Tacoma Bridge. It hasa slender, aerodynamic deck.


The Humber BridgeJohn Hyatt, Douglas Strachan

and others, 1981.


THE TOP TEN SUSPENSION BRIDGESSuspension bridge are typically ranked by the length of their main span.

Akashi-Kaikyo Bridge

(Japan) 1991

m 1998

Xihoumen Bridge

(China) 1650

m 2007

Great Belt Bridge

(Denmark) 1624

m 1998

Runyang Bridge

(China)

1490

m 2005

Humber Bridge

(England) 1410

m

1981

(The longest span from 1981 until 1998.)

Jiangyin Suspension Bridge

(China)

1385

m

1997

Tsing Ma Bridge

(Hong Kong),

1377

m

1997

(Longest span with both road and metro.)

Verrazano-Narrows Bridge

(USA) 1

298 m 1964(The longest span from 1964 until 1981.)

Golden Gate Bridge

(USA)

1280

m 1937

(The longest span from 1937 until 1964.)

Yangluo Bridge

(China)

1280

m

2007

http://en.wikipedia.org/wiki/Akashi-Kaikyo_Bridge

http://en.wikipedia.org/wiki/Japan

http://en.wikipedia.org/wiki/Xihoumen_Bridge

http://en.wikipedia.org/wiki/China


Millenium

Bridge 2000Arup, Foster and Partners

LTPC

Synchronous Lateral Excitationhttp://www2.eng.cam.ac.uk/~den/ICSV9_06.htm


Dampers on Millenium

Bridge to prevent synchronous lateral excitation.


The first person to study the physics and mathematics of thesuspension bridge was Galileo.

Galileo Galilei1564 -

1642



SUSPENSION BRIDGE: FORCES

Tension at O.

y

T

To

W

Deck

0

P

Section of deck supported bycables segment OP. Length = x.

θ

Cable

Weight of redsection of deck.

Tension at P.

Section of cable between O and P.P has horizontal coordinate x.


W

To

T

The three forces T0

, T

and W are in equilibrium.

They form a triangle of forces with tanθ

= W / To

.

T is tangential to the chain at P.

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

P

y

T

To

0

P θ

W

x

x

To

TW

θ


y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

T

P θ

y

P

y

T

To

0

P θ

W

x

x

To

T

θ

W

δx

θ

δy

The triangle of forces is similar to the differential

triangle at P.

They both have gradient tanθ

= W / To

.


μ

has the dimensions of mass/length: m.l-1

g

has the dimensions of acceleration: l.t

-2

To

has the dimensions of force: m.l.t

-2

Thus k

has the dimensions of length-1: l-1

lengthmass /xgW

x = horizontal distance from the point O.

kxdxdy

To

T

θ

W

δx

θ

δy

xyGradient

OT

WGradient

OTW

xy

OTxg

xy

kxxy

In the limit:

OTgk


Since y = 0 when x = 0 C must = 0

k has the dimensions of length-1.x

has the dimensions of length.Thus:y has the dimensions of length.

THE EQUATION FOR THE CHAIN-CURVE OF A SUSPENSION BRIDGE

kxdxdy

dxkxy . Ckxy 2

2

A Parabola2

2kxy


To

T

θ

μxg

222OTxgT

Tension in the Cable

The tension in the cable is greatest at the towers. The tension in the cable is a minimum at the lowest point and =

To

.


BRASSIERES: An Engineering Miracle

From Science and Mechanics, February, 1964

By Edward Nanas“There is more to brassiere design than meets the eye. In many respects, the challenge of

enclosing and supporting a semi-solid mass of variable volume and shape, plus its adjacent mirror image -

together they equal the female bosom -

involves a design effort comparable to that of building a bridge or a cantilevered skyscraper. “

http://www.firstpr.com.au/show-and-tell/corsetry-1/nanas/engineer.html

Other Applications of Suspension-bridge Technology


THE CATENARY

Catenary, Alysoid, Chainette.



A spider’s web: multiple catenaries.


“Simple suspension bridges”or rope bridges are catenaries,not

parabolas.

Söderskär

Bridge, Finland.


Galileo believed that a catenary

had the equation of a parabola.

He had studied the parabola in various contexts and was the first

to state that a projectile would follow the path of a parabola.

In 1669 a posthumous publication by Joachim Jungius proved that

the function describing a catenary

could not be algebraic and could not therefore be a parabola.

Joachim Jungius1587 –

1657.

GALILEO AND THE CATENARY


In 1690 Jakob

Bernoulli issued a challenge to Leibniz, Christiaan

Huygensand Johann Bernoulli to derived the equation for the catenary.The solutions were presented in 1691.

Newton also solved the problem: anonymously. The Age of Big Hair.

THE JAKOB BERNOULLI CHALLENGE.

Gottfried Leibniz Christiaan

HuygensJohann Bernoulli

Jakob

Bernoulli


Several generations of mathematical geniuses.

Jakob

Johann (I)

Daniel

Nicolaus

(II)

Johann (II)

Johann (III)

Nicolaus

(III)

The Bernoulli Family Tree


Jakob

Bernoulli (1654-1705)

First studied to be a minister.

Studied at Basel University.

Received degree in theology.

Fascinated by mathematics.

Furthered the calculus he had learned from Leibniz.

Studied catenaries.

Worked on the design of bridges.

Studied the brachistochrone problem with Johann.

Was a professor at Basel until

his death.


THE SOLUTIONS TO THE BERNOULLI CHALLENGE.

Leibniz used calculus, but did not show his method.

Johann Bernoulli used the calculus of variations.This involves finding the shape which minimizes

the potential energy of the system.

Huygens used a complicated geometric proof.

A solution using differential equations can also used.

In 1691, when the derivation of the equation for the catenary

was published, the Jesuit priest Ignace Gaston Paradies published a

text-book on forces and geometry which included the derivation of the equations for the suspension-bridge cable and the catenary.




y

T

To0

Pθ

Cable

Tension at O.

Tension at P.

Tsinθ

Tcosθ

W = μsg

s


µsg

To

T

The three forces T0

, T and µsg

are in equilibrium.

They form a triangle of forces with tanθ

= µsg

/ To

.

Since T is tangential to the curve formed by the chain

tanθ

is equal to the gradient at the point P.

Gradient = µsg

/ To

.

To

T

θ

µsg

y

T

To0

P θ

Cable

Tension at O.

Tension at P.

Tsinθ

Tcosθ

W = μsg

s


OTsg tan

tang

Ts O

gT

k O

tanks

Derivation of the Catenary

Equation Method 1

We have shown:

Define:

To

has the dimension of force: m.l.t-2μ

has the dimension of mass/length: m.l-1g

has the dimension of acceleration: l.t-2Thus:k has the units of length: l


Consider the differential triangle.In the limit δs

approaches the value

of the hypotenuse.

δS

δx

δy

θ

The above triangle is similar to the

triangle of forces.

To

T

θ

µsg

cosdsdx


Equation Method 1



Equation Method 1

2seckdds

dds

dsdx

ddx

cosdsdx

δS

δx

δy

θ

seccos.sec. 2 kkddx

seckddx

tanks

(1)

(2)From (1) and (2):



Equation Method 1

dds

dsdy

ddy

sindsdy

sin.sec2kddy

δS

δx

δy

θ

tan.seckddy

2seckdds

tanks

(1)

(2)From (1) and (2):



Equation Method 1

0,0 x

Ck 01ln.0 0C

tansecln. kx

Deriv

Parametric equation of the catenary

(1)

seckddx

y

T

0

P

θ

x

sec .x dx k d

.ln sec tanx k C

Separate variables and integrate


tan.seckddy


Equation Method 1

dkdyy .tan.sec

Cky sec

We have not defined where the axis y = 0 is. Define C = 0. Thus when θ

= 0 y = k.

secky

Parametric equation of the catenary

(2)

x} k

Separate variables and integrate


tansecln. kx secky


Equation Method 1

We now have the two parametric equations for the catenary.We need to eliminate θ

to obtain the cartesian

x-y

equation.

ky

sec

22 sectan1

1ln.2

ky

kykx

(1) (2)

From (2): (3)

1sectan 2

From (1), (3) and (4):

(4)

Recall the identity: 1lncosh 21 uuu



Equation Method 1

1ln.2

ky

kykx 1lncosh 21 uuu

kykx 1cosh

ky

kx

cosh

kxky cosh

Jungius

was correct. The catenary

is not

described

by an algebraic function; and is thus not a parabola.

Rearranging and taking the

cosh function of both sides

of the equation gives:

If x has the dimensions of length: x/k

is dimensionless. y

has the dimensions of length.

(1)(2)

From (1) and (2):


tanks

δS

δx

δy

θ

We have previously shown:

The Derivation of the Catenary

by Differential Equations.

tandxdy

dxdyks .

2

2

.dx

ydkdxds

(1)

(2)

From (1) and (2):


δS

δx

δy

θ

222 dydxds



2

1

dxdy

dxds

2

2

.dx

ydkdxds

We have just shown:

2

2

2

1.

dxdy

dxydk

(1)

(2)

From (1) and (2):



2

2

2

1.

dxdy

dxydk



dxdyy '

2'1'. ydxdyk

(1)

Let:

From (1) and (2):

Separate variables

kdxyyd .'1' 2

2'1

'

y

ydkdx

2'1

'

y

ydkdx

(2)


2'1

'

y

ydkdx

Recall the standard integral:

uu

du 1

2sinh

1

'sinh ykx

Take sinh function of both sides:

kx

dxdy sinh

dx

kxdy .sinhC

kxky

cosh

kxky coshOnce again we can define the

coordinate axes so that C = 0.

(1)

(2)

From (1) and (2): Cykx

'sinh 1 0x 0'y0C

When

Thus:

'sinh 1 ykx

Separate the variables


The Relationship Between the Parabola and the Catenary


2.cosh 22xy

2

2xy

A Comparison of a Parabola and a CatenaryO

nline function plotter: http://ww

w.m

athe-fa.de/en#anchor


The Relationship Between the Parabola and the Catenary.The MacLaurin

Series for a Catenary

kxk cosh

0

2

2

)!2.(cosh

nkx

kx

n

n

........)!2.(

........!6.!4.!2.

1cosh 2

2

6

6

4

4

2

2

nkx

kx

kx

kx

kx

n

n

........))!2.(

........!6.!4.!2.

1(cosh 2

2

6

6

4

4

2

2

nkx

kx

kx

kxk

kxk n

n

........))!2.(

........!6.!4.!2.

cosh 12

2

5

6

3

42

nkx

kx

kx

kxk

kxk n

n

kk

xkxk

2cosh

2If k is >1 the catenary

can be approximated

by a parabolic function for small values of x.

SUSPENSION BRIDGES / D A GRIFFIN2.cosh

2xy

2

24xy

If k is >1 the catenary

can be approximated by a parabolic function for small values of

x.

(k = 2)


The Relationship Between the Parabola and the Catenary

If the parabola y = x2

is rolled along the x-axis the locusof its focus is the catenary:

xy 4cosh41


The Construction of the Brooklyn BridgeParabola and Catenary

During the construction of the Brooklyn Bridge it was possible at one stage to contrast a laden and an un-laden cable: a parabola and a catenary.








Inverted Parabolas and Catenaries

Arch Bridges

Free-standing Arches


The New River Gorge Bridge, Virginia.A supported-deck bridge.


The Tyne BridgeA compression-arch suspended-deck bridge


CO

W

C

θ

x

y

CO

W = μxgC

SUPPORTED ARCH BRIDGE

The forces acting on a section of the arch are compressive. An analysis of the triangle of forces leads once again to a parabola. The arch is rigid. I does not assume the shape of an inverted parabola.

It should be constructed as an inverted parabola if it is to have a uniform

deck supported at regular intervals.

The triangle of forces acting on a segment is analogous to that

for a suspension bridge.


THE INVERTED CATENARY

The inverted catenary

is the ideal curve for

an arch which supports only its own weight.

It is the minimum energy structure.

The forces are primarily of compression.


St Louis Gateway Arch,Eero

Saarinen, Completed 1965.

.


St Louis Gateway Arch


This formula is inscribed on the arch.

Thus when x = 0 and y is at a maximumy = 630 ft.This also gives a separation of 630 ft

for the bases.



Taq–i-Kisra, Ctesiphon, Mesopotamia / Irak


Casa Milà, Barcelona.Antoni

Gaudí



Gaudí



Gaudí


Can you ridea cycle withsquare wheels?

http://www.maa.org/mathland/mathtrek_04_05_04.html


For the rolling square the shape of the road is a series

of inverted, truncated catenary

curves.

PC LT


For regular n-sided polygonal wheels the curve of the road is made from inverted catenaries with the equation:

y = -

Rcot(/n).cosh(x/A)


In practice a triangular wheel would get stuck.The vertices puncture the road.

Road Bottom of triangular wheel“Spoke”


www.exploratorium.edu/texnet/exhibits/motion/square.../square_cbk.pdf


www.maa.org/pubs/mathmag.html (General study on “roads”

and non-circular “wheels”.)

www.macalester.edu/mathcs/documents/catenaries.pdf

(Uses two coordinate systems: polar and cartesian.)

http://www.maplesoft.com/applications/view.aspx?SID=6322

(Method using several differential equations.)

http://www.snc.edu/math/squarewheelbike.html

Follow hypertext link on the website mathematics.

(Uses standard geometry and calculus: but long.)

See: Wikipedia

page on Roulette (curve). (Generates the catenary

road as a roulette in the complex plane.)

Proving that the locus of the centre of a square as it rolls over an inverted catenaries is a straight line.

There are several, diverse proofs. Some are long and complicated.

http://www.macalester.edu/mathcs/documents/catenaries.pdf

http://www.maplesoft.com/applications/view.aspx?SID=6322

http://www.snc.edu/math/squarewheelbike.html


Consider the square resting

on a vertex and symmetrically

poised between two of the

humps.We want the centre of the

circle to remain on the line

y = a√2 as the square rolls

along the road.

2a

Road, y = f(x)

a√2Y = a√2

x

y


2a

Road, y = f(x)

y = a√2

x

y

θ

θ

A

B

C


2a

Road, y = f(x)

y = a√2

x

y

θ

θ

A

B

C

a

y

a.secθ

.sec 2a y a

tandydx

Appendix


.sec 2a y a

tandydx

sec 2 ya

2 21 tan sec

221 secdy

dx

secu

(1)

(2)

Let:

221 dy u

dx

(3)

(4)From (2) and (3):

2 1dy udx

2 yua

From (1) and (3):


2 1dy udx

2 yua

1dudy a

du du dydx dy dx

21 1du udx a

2

11

dudxa u

Separate variables.


2

11

dudxa u

2

11

dudxa u

1coshx C ua

cosh xu Ca

2 coshy xCa a

2 .cosh xy a a Ca

2 yua

00 2 .cosha a Ca

2 cosh C 1cosh 2C

12 cosh cosh 2 xy a aa

Take cosh function of both sides.

When x = 0 y = f(x) = 0. Therefore:

The equation of the road.


Plots of the Inverted Catenary

“Road”

and Related Functions.

cosh( )y x

1cosh cosh 2y x

12 cosh cosh 2y x


(Road equationtion

with a = 1)

Road equation



Solve for y = 0 and dy/dx

= 0

(0,0)

1.cosh 2, 2a a a

y

x

(0.8814, 0.4142)

(1.7627, 0)

(Values when a = 1)

12 .cosh 2,0a


PC PC

LT LT


Proof that the Arc Length of the Inverted Catenaries is Equal to the Length of a Side of the Square: i.e. 2a.

2

1 .dxs dydy

tandy

dx cotdx

dy

21 cot .s dy

(1) (2) (3)

From (1) and (3): cos .s ec dy

.sec 2a y a 2 secy a a

sec . tandy ad

.cos .sec . tan .s a ec d

(4)

(5)

From (4) and (5): 2.sec .s a d

APPENDIX 1


2a

a√2

x

y

45o 45o 135o


2.sec .s a d

2

1

2.sec .s a d

2

1tans a C

1 45o

2 135o

tan 45 1

tan135 1

1 1 2s a a


APPENDIX 2

TNB


APPENDIX 3

Conversion of FLV (e.g. Youtube) files to downloadable files (e.g. MP4).


APPENDIX 4

http://www.mathe-fa.de/en


1coshy x coshx y

cosh sinh yy y e 2 2cosh sinh 1y y

2cosh cosh 1ye y y

2 1ye x x 2ln 1y x x

1 2cosh ln 1x x x

1 2cosh ln 1x x x

APPENDIX 5

cosh2

x xe ex

sinh2

x xe ex

2 2cosh sinh 1x x cosh sinh xx x e

Documents

Maths_Catenaries, Parabolas and Suspension Bridges