7.4 cables flexible cables and chains are used to support and transmit loads from one member to...

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 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

7.4 Cables7.4 Cables 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 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 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 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 7.13Determine the tension in each segment

of the cable.


SolutionFBD for the entire cable


Solution

kNE

EkNkNkNkN

F

kNA

mknmkNmkNmAM

EA

F

y

y

y

y

yE

xx

x

10

0315412

;0

12

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

0

;0


SolutionConsider leftmost

section which cuts cable BC since sag yC = 12m


Solution

kNT

TkNkN

F

kNT

F

kNEA

mkNmkNmAM

BCBC

BCBC

y

BCBC

x

xx

xC

2.10,6.51

0sin412

;0

033.6cos

;0

33.6

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


SolutionConsider point A, C and E,


SolutionPoint A

kNT

kNT

F

kNT

F

AB

AB

ABAB

y

ABAB

x

6.13

2.62

012sin

;0

033.6cos

;0


SolutionPoint C

kNT

kNkNT

F

kNT

F

CD

CD

CDCD

y

CDCD

x

44.9

9.47

0156.51sin2.10sin

;0

06.51cos2.10cos

;0


SolutionPoint E

kNT

TkN

F

TkN

F

ED

ED

EDED

y

EDED

x

8.11

7.57

0sin10

;0

0cos33.6

;0


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 LoadConsider weightless cable subjected to a

loading function w = w(x) measured in the x direction


Cable Subjected to a Distributed Load

For FBD of the cable having length ∆



Since the tensile force in the cable changes continuously in both 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



0sincos)())((

;0

0)sin()())((sin

;0

0)cos()(cos

;0

xTyTxkxxw

M

TTxxwT

F

TTT

F

O

y

x



Divide by ∆x and taking limit,

Integrating, HFtT

dx

dy

xwdx

Tddx

Td

tancoscos

tan

0)()sin(

0)cos(



Integrating,

Eliminating T,

Perform second integration,

dxdxxwF

y

dxxwFdx

dy

dxxwT

H

H

)(1

)(1

tan

)(sin


Example 7.14The 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

the maximum force developed in the cable and the cable’s required length. The span length L and, sag h are known.


SolutionNote w(x) = wo

Perform two integrations

Boundary Conditions at x = 0

0/,0,0

21

1

21

2

dxdyxy

CxCxw

Fy

dxdxwF

y

o

H

oH


SolutionTherefore,

Curve becomes

This is the equation of a parabola

Boundary Condition at x = L/2hy

xF

wy

CC

H

o

2

21

2

0


SolutionFor constant,

Tension, T = FH/cosθ

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

22

2

48

xL

hy

h

LwF oH

7.4 Cables7.4 CablesSolutionSlope at point B

Or

Therefore

Using triangular relationship

2

4

)cos(

2tan

tan

222

max

maxmax

1max

2/max

2/

LwFT

FT

FLw

Fw

dxdy

oH

H

H

o

LxH

o

Lx

7.4 Cables7.4 CablesSolution

For a differential segment of cable length ds

Determine total length by integrating

Integrating yields

Lh

hL

LhL

dxxL

hds

dxdxdy

dydxds

hLLw

T

L

o

4sinh

44

12

812

1

41

2

12

2/

0

2

2

222

2

max


Cable Subjected to its Own WeightWhen weight of the cable is considered,

the loading function becomes a function of the arc length 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


Cable Subjected to its Own Weight Apply equilibrium equations to the force

system

Replace dy/dx by ds/dx for direct integration

dsswFdx

dy

dsswT

FT

H

H

)(1

)(sin

cos


Cable Subjected to its Own Weight

Therefore

Separating variables and integrating

2/12

2

2/12

2

2

22

)(1

1

)(1

1

1

dsswF

dsx

dsswFdx

ds

dxds

dxdy

dydxdsT

H

H


Example 7.15Determine the deflection curve, the length, and the maximum tension in the uniform 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)

Integrating term in the denominator

2/121

2

2/122

/11

/11

CswF

dsx

dswF

dsx

oH

oH


SolutionSubstitute

So that

Perform second integration

or

211

21

1

1sinh

sinh

)/(

/1

CCswFw

Fx

CuwF

x

dsFwdu

CswFu

oHo

H

o

H

Ho

oH

7.4 Cables7.4 CablesSolutionEvaluate constants

or

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

To obtain deflection curve, solve for s

xFw

wF

s

CswFdx

dy

dswFdx

dy

H

o

o

H

oH

oH

sinh

1

1

1


Hence

Boundary Condition y = 0 at x = 0

For deflection curve,

This equations defines a catenary curve

1cosh

cosh

sinh

3

3

xF

w

w

Fy

w

FC

CxF

w

w

Fy

xF

w

dx

dy

H

o

o

H

o

H

H

o

o

H

H

o

7.4 Cables7.4 CablesSolutionBoundary Condition y = h at x = L/2

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

By trial and error,NF

F

N

mN

Fm

xF

w

w

Fh

H

H

H

H

o

o

H

9.45

150

cosh/5

6

1cosh

7.4 Cables7.4 CablesSolutionFor deflection curve,

x = 10m, for half length of the cable

Hence

Maximum tension occurs when θ is maximum at

s = 12.1m

m

mmN

mN

mN

mxy

2.24

1.12109.45

/5sinh

/5

9.45

2

1109.0cosh19.9


NNF

T

N

mmN

dx

dy

H

ms

9.758.52cos

9.45

cos

8.52

32.19.45

1.12/5tan

maxmax

max

max1.12

Chapter Summery Chapter Summery

Internal Loadings If a coplanar force system acts on a

member, then in general a resultant normal force N, shear force V and bending moment M will act at any cross-section along the member

These resultants are determined using the method of sections

The member is sectioned at the point where the internal loadings are to be determined


Internal Loadings A FBD of one of the sectioned parts is

drawn The normal force is determined from the

summation of the forces normal to the cross-section

The shear force is determined from the summation of the forces tangent to the cross section

The bending moment is found from the summation of the moments about the centroid of the cross-sectional area


Internal Loadings If the member is subjected to 3D

loading, in general, a torisonal loading will also act on the cross-section

It can be determined by the summation of the moments about an axis that is perpendicular to the cross section and passes through its centroid


Shear and Moment Diagrams as Functions of x Section the member at an arbitrary point

located distance x from one end to construct the shear and moment diagrams for a member

Unknown shear and moment are indicated on the cross-section in the positive direction according to the established sign convention

Applications of the equilibrium equations will make these loadings functions of x which then, can be plotted


Shear and Moment Diagrams as Functions of x If the external loadings or concentrated

forces and couple moments act on the member, then different expressions of V and M must be determined within regions between these regions

Graphical Methods for Establishing Shear and Moment Diagrams Plot the shear and moment diagrams using

differential relationships that exist between the distributed loading w and V and M


Graphical Methods for Establishing Shear and Moment Diagrams Slope of the shear diagram = distributed

loading at any pointdV/dx = -w

Slope of the moment diagram = shear at any point

V = dM/dx Change in shear between an y two points

= area under the shear diagram between points, ∆M = ∫Vdx


Cables When a flexible and inextensible cable is

subjected to a series of concentrated forces, the analysis of the cable can be performed by using the equations of equilibrium applied to the FBD of wither segments or points of application of the loading

If the external distributed loads or weight of the cables are considered, forces and shape of the cable must be determined by analysis of the forces on a differential segment of the cable and integrating the result

Chapter Review Chapter Review

7.4 cables flexible cables and chains are used to support and transmit loads from one member to...

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