[email protected] ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...

[email protected] • ENGR-36_Lec-20_Cables.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 7:Flex

Cables



Recall Chp10 Introduction

Examine in Detail Two Important Types Of Engineering Structures:

1. BEAMS - usually long, straight, prismatic members designed to support loads applied at various points along the member

2. CABLES - flexible members capable of withstanding only tension, designed to support concentrated or distributed loads



Load-Bearing Cables

Cables are applied as structural elements in suspension bridges, transmission lines, aerial tramways, guy wires for high towers, etc.

Göteborg, Sweden

http://www.nmt.edu/~armiller/bridgefu.htm

CurvedStraight

WHY the Difference?



Concentrated Loads on Cables To Determine the Cable

SHAPE, Assume:a) Concentrated vertical loads

on given vertical lines

b) Weight of cable is negligible

c) Cable is flexible, i.e., resistance to bending is small

d) Portions of cable between successive loads may be treated as TWO FORCE MEMBERS• Internal Forces at any point reduce to

TENSION Directed ALONG the Cable Axis



Concentrated Loads (2) Consider entire cable

as a free-body Slopes of cable at A

and B are NOT known FOUR unknowns

(i.e., Ax, Ay, Bx, By) are involved and the equations of equilibrium are NOT sufficient to determine the reactions.



Concentrated Loads (3) To Obtain an additional equation • Consider equilibrium of

cable-section AD• Assume that CoOrdinates of SOME

point D, (x,y), on the cable have been Determined (e.g., by measurement)

Then the added Eqn: 0DM

33221103 xPxPxPLBdBM yxA

With pt-D info, the FOUR Equilibrium Eqns xxx BAF 01̀ 12102 PPPBAF yyy

xAyAxxPM yxD 1104



Concentrated Loads (4) The 4 Eqns Yield Ax & Ay

Can Now Work our Way Around the Cable to Find VERTICAL DISTANCE (y-CoOrd) For ANY OTHER point

2 yields02

yMC yxyx TTFF , yields 0,0

Example Consider Pt C2

constantcos xx ATT

known

known

known

UNknown

known

knownknown



Example Concentrated Loads

The cable AE supports three vertical loads from the points indicated. Point C is 5 ft below the left support

For the Given Loading & Geometry, Determine:a) The elevation of

points B and D b) The maximum

slope and maximum tension in the cable.




Solution Plan• Determine

reaction force components at pt-A from solution of two equations formed from taking entire cable as a

free-body and summing moments about E, and from taking cable portion ABC as a free-body and summing moments about C.

• Calculate elevation of B by considering AB as a free-body and summing moments B. Similarly, calculate elevation of D using ABCD as a free-body.

• Evaluate maximum slope and maximum tension which occur in DE.

KnownCoOrds




Determine two reaction force components at A from solution of two equations formed from taking the

entire cable as a free-body and summing moments about E:

333

OR

06606020

041512306406020

:0

yx

yx

yx

E

AA

AA

AA

M




Next take Cable Section ABC as a Free-Body, and Sum the Moments about Point C

Recall from ΣME

126OR

0610305

:0

yx

yx

C

AA

AA

M

333 yx AA Solving 2-Eqns in

2-Unknowns for Ax & Ay

kips 5

kips 18

y

x

A

A



Example Concentrated Loads Determine elevation of B

by considering AB as a free-body and summing moments about B.

Similarly, Calc elevation at D using ABCD as a free-body

020518:0 BB yM

ft 56.5By

0121562554518

:0

D

D

y

M

ft83.5Dy



Maximum Tension Analysis

By the ΣFx = 0

T

cos

but

OR0

TT

AT

TAF

x

xx

xxx

Solving

for T cosxAT

Thus T is Maximized by Maximum θ



Find Maximum Segment Angle

Use the y-data just calculated to find the cable segment of steepest slope

Pt x (ft) y (ft) Seg ∆x (ft) ∆y (ft) θ by aTan(∆y/∆x) (°)A 0 0

AB 20 -5.56 -15.54B 20 -5.56

BC 10 0.56 3.21C 30 -5

CD 15 10.83 35.83D 45 5.83

DE 15 14.17 43.37E 60 20




Use yD to Determine Geometry of tanθ

Evaluate maximum slope and maximum tension which occur in the segment with the STEEPEST Slope (large θ); DE in this case

15

17.14tan 4.43

Employ the Just-Determined θ to Find Tmax

cos

kips 18max T

kips 8.24max T



Distributed Loads on Cables For a negligible-

Weight Cable carrying a Distributed Load ofArbitrary Profile

a) The cable hangs in shape of a CURVE

b) INTERNAL force is a tension force directed along the TANGENT to the curve

x

y



Distributed Loads (2) Consider the Free-Body for a

portion of cable extending from LOWEST point C to given point D. Forces are T0 (FH in the Text Book) at Lo-Pt C, and the tangential force T at D

From the Force Triangle

0

220

0

tan

sincos

T

WWTT

WTTT



Distributed Loads (3) Some Observations based on

0

220

0

tan

sincos

T

WWTT

WTTTTT yx

• Horizontal component of T is uniform over the Cable Length

• Vertical component of T is equal to the magnitude of W

• Tension is minimum at lowest point (min θ), and maximum at A and B (max θ)



Parabolic Cable Consider a cable supporting a

uniform, horizontally distributed load, e.g., support cables for a suspension bridge.

With loading on cable from lowestpoint C to a point D given by internal tension force, T, and Vertical Load W = wx, find:

02

:0 0 yTx

wxMD

0

2

2T

wxy

0

2220 tan

T

wxxwTT

Summing moments about D

The shape, y, is PARABOLIC:



T0 for Uniform Vertical Load

Consider the uniformly Loaded Cable• In this case: w(x) = w

– w is a constant

L is the Suspension Span:

From Last Slide

Or Then

xB & xA

Thus L

L

AB xxL

0

2

2T

wxy

wyTx 02

wyTx

wyTx

AA

BB

0

0

2

2

wyTwyTL AB 00 22



T0 for Uniform Vertical Load

Factoring Out 2T0/w

Isolating T0

If WE design the Suspension System, then we KNOW• L (Span)• w (Load)• yA & yB (Dims)

Example• L= 95 m (312 ft)• w = 640 N/m

(44 lb/ft)• yA = 19m

• yB = 37m

Then T0 = 26 489 N (5955 lb)

Tmax by

AB yywTL 20

22

02 AB yy

wLT

2max

220max xwTT



Tmax for Uniform Vertical Load

Find xmax from

In this case xmax = 55.3 m (181 ft)

And finally Tmax = 44 228 N(9 943 lbs)• Buy Cable rated to

20 kip for Safety factor of 2.0

wyTx BB 02

>> L = 95L = 95 >> w = 640w = 640 >> yA = 19yA = 19 >> yB = 37yB = 37 >> TO =L^2*w/(2*(sqrt(yB)+sqrt(yA))^2)TO = 2.6489e+04 >> xB = sqrt(2*TO*yB/w)xB = 55.3420 >> Tmax = sqrt(TO^2 + (w*xB)^2)Tmax = 4.4228e+04

MATLABCalcs



Load-Bearing Cables

Tension in the Straight-Section is roughly Equal to the Parabolic-Tension at the Tower-Top.• So the Support Tower does Not Bend

Göteborg, Sweden

http://www.nmt.edu/~armiller/bridgefu.htm

VerticallyLoaded

End Loads

NO Deck-Support Cables

Deck-Support COLUMNS



UNloaded Cable → Catenary Consider a cable uniformly

loaded by the cable itself, e.g., a cable hanging under its own weight.

222220

22220 scwswTwswTT

• With loading on the cable from lowest point C to a point D given by W = ws, the Force Triangle reveals the internal tension force magnitude at Pt-D:

mor ft 0 wTc – Where



UNloaded Cable → Catenary (2) Next, relate horizontal

distance, x, to cable-length s

cosdsdx But by Force

Balance Triangle

Thus

T

T0cos

Also From last slide recall

wcTscwT 022 and

dssc

c

scw

wcds

T

Tdsdsdx

2222

0cos



UNloaded Cable → Catenary (3) Factoring Out c in DeNom

Integrate Both Sides using Dummy Variables of Integration: • σ: 0→x η: 0→s

dscsccc

cds

sc

cdx

222222

Finally the Differential Eqn

dscs

dx221

1



UNloaded Cable → Catenary (4) Using σ: 0→x η: 0→s

Now the R.H.S. AntiDerivative is the argSINH

Noting that

sxd

cd

0 220 1

1

s

sxx

ccd

cd

0

0 2200sinharg

1

1

00sinh0sinharg 1



UNloaded Cable → Catenary (5) Thus the Solution to the Integral Eqn

Then

Solving for s in terms of x by taking the sinh of both sides

0sinhsinharg0 1

0

0

c

sc

ccx

sx

c

s

c

x

c

scx 11 sinhsinh

c

x

c

ssinh



UNloaded Cable → Catenary (6) Finally, Eliminate s in favor

of x & y. From the Diagram

So the Differential Eqn

From the Force Triangle

tandxdy

0

tanT

W

And From Before

wcTwsW 0and

dxc

sdx

wc

wsdx

T

Wdxdy

0

tan



UNloaded Cable → Catenary (7) Recall the Previous Integration

That Relates x and s

Integrating the ODE with Dummy Variables: • Ω: c→y σ: 0→x

x

xyc

y

c ccd

cd

00

coshsinh

c

x

c

ssinh

Using s(x) above in the last ODE

dxc

xdxc

sdx

wc

wsdx

T

Wdxdy

o

sinhtan



UNloaded Cable → Catenary (8) Noting that cosh(0) = 1

• Where– c = T0/w

– T0 = the 100% laterally directed force at the ymin point

– w = the lineal unit weight of the cable (lb/ft or N/m)

Solving for y yields theCatenary Equation:

cxcy cosh

cc

xc

cccy

xyc

coshcosh

0



Catenary Comments

With Hyperbolic-Trig ID: cosh2 – sinh2 = 1

Recall From the Differential Geometry

Or:

cxcy cosh

222222

222222

sinhcosh

sinhcosh

ccxcxcsy

cxccxcsy

222222 sycandscy

yTwyywscwscT 222,

wTc 0

or

00

0

0 coshcoshT

wxT

T

wx

w

TwwyxT



Loaded & Unloaded Cable Profiles

0

50

100

150

200

250

300

350

400

450

500

550

-100 -75 -50 -25 0 25 50 75 100

Relative Horizontal Displacement, x

Rel

ativ

e V

erti

cal

Dis

pla

cem

ent,

y

Parabola (Loaded)

Catenary (Unloaded)

file = Catenary-Parabola_0407.xls

PARAMETERS• w/T0 = 0.1 (Parabola)• c = 50 (Catenary)

Loaded and Unloaded Cables Compared

cxcy cosh

0

2

2T

wxy



y = 0 at Cable Minimum

Translate the CoOrd System Vertically from Previous:

Oy

cyycyy

xx

OO Change No

Recall Eqn for y−c

cc

xccy

cosh

Sub y = yO+c

Thus with Origin at cable Minimum

cc

xcccyO

cosh

cc

xcyO

cosh



y = 0 at Cable Minimum (2)

ThenOy

Recall c = T0/w

Thus Next, Change the Name of the Cable’s Lineal Specific Weight (N/m or lb/ft)

1c

xcyO cosh

1

0

0

wT

x

w

TyO cosh

w

L



y = 0 at Cable Minimum (3)

With µ replacing w

Oy

1

0

0

T

xTyO

cosh

In Summary can Use either Formulation based on Axes Origin:

0

0 coshT

wx

w

Ty

1

0

0

T

xTyO

cosh

wT0

L



Cable Length, S, for Catenary Using this Axes Set

With Cable macro-segment and differential-segment at upper-right

The Force Triangle for the Macro-Segment

OyL

W = µs

W = µs



Cable Length, S , for Catenary By Force

Triangle But by Differential

Segment notice

By Transitive Property

Now ReCall

Then dy/dx

Subbing into the tanθ expression

0

tanT

s

dx

dytan

0

tanT

µs

dx

dy

1cosh

0

0

T

xTy

00

0 sinh1coshT

x

T

xT

dx

d

dx

dy

00

sinhT

µs

T

µx

dx

dy



Cable Length, S , for Catenary Solving the last Eqn

for s

From the CoOrds

So Finally

Now find T(y) Recall T = f(x)

0

0 sinhT

µx

µ

Ts

OyL

AB

OAOB

xsxs

ssS

00

0 sinhsinhT

µx

T

µx

µ

TS AB

00 cosh

T

xTT



T(y) for Catenary Also ReCall

Solve above for cosh

Sub cosh into T(x) Expression

1cosh

0

0

T

xTy

11cosh000

T

µy

T

µy

T

x

00

0

00

1

cosh

TµyT

µyTT

T

µxTT



Catenary Summary y(x)

T(x)T(y)

S(x)

Slope at any pt

Angle θ at any pt

1cosh

0

0

T

xTy

0

00 cosh

TµyT

T

µxTT

00

0 sinhsinhT

µx

T

µx

µ

TS AB

00

sinhT

µs

T

µx

dx

dy

00

sinhtanT

µs

T

µx



WhiteBoard Work

Let’s WorkThese NiceProblems



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix00

sinhT

µs

T

µx

dx

dy



WhiteBoard Work

Let’s WorkThese NiceProblems



w

T0

[email protected] ENGR-36_Lec-20_Cables.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...

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