6161103 7.4 cables

7.4 Cables7.4 Cables

� Flexible cables and chains are used to support and transmit loads from one member to another

� In suspension bridges and trolley wheels, they carry majority of the loadcarry majority of the load

� In force analysis, weight of cables is neglected as it is small compared to the overall weight

� Consider three cases: cable subjected to concentrated loads, subjected to distributed load and subjected to its own weight


� Assume that the cable is perfectly flexible and inextensible

� Due to its flexibility, the cables offers no resistance to bending and therefore, the tensile force acting in the cable is always tangent to the force acting in the cable is always tangent to the points along its length

� Being inextensible, the length remains constant before and after loading, thus can be considered as a rigid body


Cable Subjected to Concentrated Loads

� For a cable of negligible weight supporting several concentrated loads, the cable takes several concentrated loads, the cable takes the form of several straight line segments, each subjected to constant tensile force


Cable Subjected to Concentrated Loads

� Known: h, L1, L2, L3 and loads P1 and P2

� Form 2 equations of equilibrium at each point A, B, C and DB, C and D

� If the total length L is given, use Pythagorean Theorem to relate the three segmental lengths

� If not, specify one of the sags, yC and yD and from the answer, determine the other sag and hence, the total length L


Example 7.13

Determine the tension in each segment of

the cable.the cable.


Solution

FBD for the entire cable


Solution

EA

F

xx

x

0

;0

=+−

=∑→+

kNE

EkNkNkNkN

F

kNA

mknmkNmkNmAM

EA

y

y

y

y

yE

xx

10

0315412

;0

12

0)2(3)10(15)15(4)18(;0

0

=

=+−−−

=∑↑+

=

=+++−=∑

=+−


Solution

Consider leftmost section

which cuts cable BC which cuts cable BC

since sag yC = 12m


Solution

kNEA

mkNmkNmAM

xx

xC

33.6

0)5(4)8(12)12(;0

=∑→+

==

=+−=∑

kNT

TkNkN

F

kNT

F

BCBC

BCBC

y

BCBC

x

2.10,6.51

0sin412

;0

033.6cos

;0

==

=−−

=∑↑+

=−

=∑→+

oθ

θ

θ


Solution

Consider point A, C and E,


Solution

Point A

kNT

Fx

033.6cos

;0

=−

=∑→+

θ

kNT

kNT

F

kNT

AB

AB

ABAB

y

ABAB

6.13

2.62

012sin

;0

033.6cos

=

=

=+−

=∑↑+

=−

oθ

θ

θ


Solution

Point C

kNT

Fx

06.51cos2.10cos

;0

=−

=∑→+oθ

kNT

kNkNT

F

kNT

CD

CD

CDCD

y

CDCD

44.9

9.47

0156.51sin2.10sin

;0

06.51cos2.10cos

=

=

=−+

=∑↑+

=−

o

o

o

θ

θ

θ


Solution

Point E

TkN

Fx

0cos33.6

;0

=−

=∑→+

θ

kNT

TkN

F

TkN

ED

ED

EDED

y

EDED

8.11

7.57

0sin10

;0

0cos33.6

=

=

=−

=∑↑+

=−

oθ

θ

θ


Solution

� By comparison, maximum cable tension is in segment AB since this segment has the greatest slope

� For any left hand side segment, the horizontal component Tcosθ = Ax

� Since the slope angles that the cable segment make with the horizontal have been determined, the sags yB and yD can be determined using trigonometry


Cable Subjected to a Distributed Load

� Consider weightless cable subjected to a loading function w = w(x) measured in the x loading function w = w(x) measured in the x direction



� For FBD of the cable having length ∆


Cable Subjected to a Distributed Load� Since the tensile force in the cable

changes continuously in both the magnitude and the direction along the magnitude and the direction along the cable’s length, this change is denoted on the FBD by ∆T

� Distributed load is represented by its resultant force w(x)(∆x) which acts a fractional distance k(∆x) from point O where o < k < 1



0)cos()(cos

;0

=∆+∆++−

=∑→+

TTT

Fx

θθθ

0sincos)())((

;0

0)sin()())((sin

;0

0)cos()(cos

=∆+∆−∆∆

=∑

=∆+∆++∆−−

=∑↑+

=∆+∆++−

xTyTxkxxw

M

TTxxwT

F

TTT

O

y

θθ

θθθ

θθθ



Divide by ∆x and taking limit,

dx

Td= 0

)cos( θ

Integrating,HFtT

dx

dy

xwdx

Tddx

==

=

=−

=

tancoscos

tan

0)()sin(

0

θ

θ

θ



Integrating,

∫= dxxwT )(sinθ

Eliminating T,

Perform second integration,

( )∫ ∫

∫

∫

=

==

dxdxxwF

y

dxxwFdx

dy

H

H

)(1

)(1

tanθ


Example 7.14

The cable of a suspension bridge supports half of

the uniform road surface between the two columns

at A and B. If this distributed loading wo, determine at A and B. If this distributed loading wo, determine

the maximum force developed in the cable and the

cable’s required length. The span length L and, sag

h are known.


Solution

Note w(x) = wo

( )1= ∫ ∫ dxdxw

Fy o

H

Perform two integrations

Boundary Conditions at x = 0

( )

0/,0,0

21

21

2

===

++=

∫ ∫

dxdyxy

CxCxw

Fy

F

o

H

oH


Solution

Therefore,

Curve becomesw

CC == 21 0Curve becomes

This is the equation of a parabola

Boundary Condition at x = L/2hy

xF

wy

H

o

=

= 2

2


Solution

For constant, 2

8h

LwF o

H =

Tension, T = FH/cosθ

Maximum tension occur at point B for 0 ≤ θ ≤π/2

22

48

xL

hy

hH

=


Solution

Slope at point B

Or

tan2/

max2/

Lw

F

w

dx

dy

LxH

o

Lx

====

θ

Or

Therefore

Using triangular relationship

2

4

)cos(

2tan

222

max

maxmax

1max

LwFT

FT

F

Lw

oH

H

H

o

+=

=

= −

θ

θ


Solution

For a differential segment of cable length ds

( ) ( )

+=+=

+=

dxdy

dydxds

h

LLwT o

1

41

2

222

2

max

Determine total length by integrating

Integrating yields

( ) ( )

+

+=

+==

+=+=

−

∫∫

L

h

h

L

L

hL

dxxL

hds

dxdx

dydydxds

L

4sinh

44

12

812

1

12

2/

0

2

2

22

l

l


Cable Subjected to its Own Weight� When weight of the cable is considered, the

loading function becomes a function of the arc length s rather than length xlength s rather than length x

� For loading function w = w(s)

acting along the cable,


Cable Subjected to its Own Weight

� FBD of a segment of the cable



� Apply equilibrium equations to the force system

= FT cosθ

� Replace dy/dx by ds/dx for direct integration

∫

∫=

=

=

dsswFdx

dy

dsswT

FT

H

H

)(1

)(sin

cos

θ

θ



−

=

+=

2

22

1dxds

dxdy

dydxdsT

Therefore

Separating variables and integrating

( )

( )∫

∫

∫

+

=

+=

2/12

2

2/12

2

)(1

1

)(1

1

dsswF

dsx

dsswFdx

ds

dxdx

H

H


Example 7.15

Determine the deflection curve, the length,

and the maximum tension in the uniform

cable. The cable weights w = 5N/m.cable. The cable weights wo = 5N/m.


Solution

� For symmetry, origin located at the center of the cable

� Deflection curve expressed as y = f(x)� Deflection curve expressed as y = f(x)

Integrating term in the denominator

( )( )

( )( )[ ]∫

∫∫

++=

+

=

2/121

2

2/122

/11

/11

CswF

dsx

dswF

dsx

oH

oH


Solution

Substitute

So that

( )( )+= 1/1 CswFu oH

So that

Perform second integration

or

{ }

( )

+

+=

+=

=

−

−

211

21

1sinh

sinh

)/(

CCswFw

Fx

Cuw

Fx

dsFwdu

oHo

H

o

H

Ho


Solution

Evaluate constants

or

= ∫dy

dswFdx

dyo

H

1

1

or

dy/dx = 0 at s = 0, then C1 = 0

To obtain deflection curve, solve for s

=

+=

xF

w

w

Fs

CswFdx

dy

H

o

o

H

oH

sinh

11


Solution

Hence+

=

=

cosh

sinh

3CxF

w

w

Fy

xF

w

dx

dy

H

o

o

H

H

o

Boundary Condition y = 0 at x = 0

For deflection curve,

This equations defines a catenary curve

−

=

−=

1cosh

3

xF

w

w

Fy

w

FC

Fw

H

o

o

H

o

H

Ho


Solution

Boundary Condition y = h at x = L/2

xF

w

w

Fh oH 1cosh

−

=

Since wo = 5N/m, h = 6m and L = 20m,

By trial and error,NF

F

N

mN

Fm

Fw

H

H

H

Ho

9.45

150

cosh/5

6

=

−

=


Solution

For deflection curve,

x = 10m, for half length of the cable( )[ ]mxy 1109.0cosh19.9 −=

Hence

Maximum tension occurs when θ is maximum at

s = 12.1m

( )

m

mmN

mN

mN

2.24

1.12109.45

/5sinh

/5

9.45

2

=

=

=

l

l


Solution

( )N

mmN

dx

dy

ms

8.52

32.19.45

1.12/5tan max

1.12

=

====

oθ

θ

NNF

T H 9.758.52cos

9.45

cos

8.52

maxmax

max

===

=

o

o

θ

θ

6161103 7.4 cables

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