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CHAPTER 3: ANALYSIS OF STATICALLY DETERMINATE

CABLES

What is cable?

- Cable is a tensile structure carrying only tension and no compression or bending.- Cable is used in engineering structures for support and to transmit loads from one

member to another.

- Always used to support suspension roofs, bridges and trolley wheels.- In the force analysis, the weight of the cable itself may be neglected; however, when

cables are used as guys for radio antennas, electrical transmission lines and derricks, the

cable weight may become important and must be included in the structural analysis.- the cable is perfectly flexible and inextensible.

Flexible – cable offers no resistance to shear or bending.

Inextensible – cable has constant length both before and after the load is applied

and geometry of the cable remains fixed.

-They are 3 cases will be considered in the cable system:

1. A cable subjected to concentrated loads.

-same level of support-different level of support

2. A cable subjected to a uniformly distributed load.-same level of support

-different level of support

3. A cable subjected to the combination of both concentrated and uniformlydistributed load

Example:

Figure 1: A cable subjected to a concentrated load supported at same level.

1

 P 

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Figure 2: A cable subjected to 3 concentrated loads supported at different level.

Figure 3: A cable subjected to a uniformly distributed load supported at same level.

Length of cable

1. Cable loaded with concentrated load (supported at same and different level)

S = length of segment AB + length of segment BC + length of segment CD

+ length of segment EF

2

C

E

D

B

A

 P 

 P 

 P 

w

 P 

 P 

 P 

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2. Cable loaded with a uniformly distributed load supported at same level.

 

S = L + 8h2 3L

where, L = Horizontal span between 2 supports

h = Vertical distance from the supports A & B to the lowest point, C

3. Cable loaded with a uniformly distributed load supported at different level.

S = l 1 + l 2 + 2d12 + 2d2

2

3 l 1 3 l 2

Cable loaded with a uniformly distributed load supported at same level.

3

l 1

l 2

d2

d1

w

C

B

A

C

h

A B

w

L

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Reaction for cable subjected to UDL and supported at same level are:

i) Vertical reactive forces at the support:VA = W L  VB = W L 

2 2ii) Horizontal force in the cable at the midspan section:

HC = W L 2 

8h

iii) Vertical force in the cable at the midspan section:VC = W L - Wx

2

iv) Tensile force, T in the cable at C:

T2 = HC2 + VC

2

 T = HC

2 + VC2

  =   W L 2  2 + W L  - Wx 2

  8h 2

 Note: Maximum tensile force in the cable occurs at x=0 (at support)  Tmax  =   W L 2  2 + W L  2

  8h 2Minimum tensile force in the cable occurs at x=L/2 (at the lowest point)

  Tmin  =   W L 2  2 + 0 = H  8h

Cable loaded with a uniformly distributed load supported at different level.

4

WL2

WL2

HBH

A

C

h

A B

W 

L

TV

C

WL2

HC

HA

C

h

A

W 

 X = L

2

HA

HB

VA

VB

l 1

l 2

d2

d1

w

C

B

A

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+↑Σ Fy = 0, VA + VB – wL = 0VA + VB = wL ------------------------ (1)

+ → Σ Fx = 0, -HA + HB = 0

HA = HB ------------------------ (2)

LEFT

+↑Σ Fy = 0, VA – wl 1 = 0

VA = wl 1 --------------- (3)

+ → Σ Fx = 0, -HA + HC = 0

HA = HC --------------- (4)

+ Σ MA = 0, - HC(d1) + wl 1 l 1 = 0

2

HC = wl 12

2d1 

∴ HA = wl 12 --------------- (5)

2d1 

RIGHT

+↑Σ Fy = 0, VB – wl 2 = 0

VA = wl 2 --------------- (6)

+ → Σ Fx = 0, -HC + HB = 0

HB = HC --------------- (7) 

5

TV

C=0

VA

HC

HA

C

d1

A

W 

l 1

C

d2

T

VC=0

VB

HC

B

W 

l 2

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+ Σ MB = 0, HC(d2) – wl 2 l 2 = 0

2HC = wl 2

2

2d2 

∴ HB = wl 22 --------------- (8)

2d2 

HC ( LEFT ) = HC ( RIGHT )

  wl 12 =  wl 2

2

2d1 2d2

  wl 12 = 2d1

  wl 22

2d2

  l 12 = d1

  l 22 d2

  l 1 = d1

  l 2  d2

Clifton Bridge, Bristol, UK Munich Olympic Stadium, Germany

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