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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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