1. 7.4 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
2. 7.4 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
3. 7.4 Cables
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
4. 7.4 Cables
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
11. 7.4 Cables
Solution
Point A
+ → ∑ Fx = 0;
TAB cos θ AB − 6.33kN = 0
+ ↑ ∑ Fy = 0;
− TAB sin θ AB + 12kN = 0
θ AB = 62.2o
TAB = 13.6kN
12. 7.4 Cables
Solution
Point C
+ → ∑ Fx = 0;
TCD cosθCD − 10.2 cos 51.6o kN = 0
+ ↑ ∑ Fy = 0;
TCD sin θCD + 10.2 sin 51.6o kN − 15kN = 0
θCD = 47.9o
TCD = 9.44kN
13. 7.4 Cables
Solution
Point E
+ → ∑ Fx = 0;
6.33kN − TED cosθ ED = 0
+ ↑ ∑ Fy = 0;
10kN − TED sin θ ED = 0
θ ED = 57.7o
TED = 11.8kN
14. 7.4 Cables
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
15. 7.4 Cables
Cable Subjected to a Distributed Load
Consider weightless cable subjected to a
loading function w = w(x) measured in the x
direction
17. 7.4 Cables
Cable Subjected to a Distributed Load
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
18. 7.4 Cables
Cable Subjected to a Distributed Load
+ → ∑ Fx = 0;
− T cos θ + (T + ∆T ) cos(θ + ∆θ ) = 0
+ ↑ ∑ Fy = 0;
− T sin θ − w( x)(∆x) + (T + ∆T ) sin(θ + ∆θ ) = 0
∑ M O = 0;
w( x)(∆x)k (∆x) − T cos θ∆y + T sin θ∆x = 0
19. 7.4 Cables
Cable Subjected to a Distributed Load
Divide by ∆x and taking limit,
d (T cos θ )
=0
dx
d (T sin θ )
− w( x) = 0
dx
dy
= tan θ
dx
Integrating,
T cos θ = cos tan t = FH
20. 7.4 Cables
Cable Subjected to a Distributed Load
Integrating,
T sin θ = ∫ w( x)dx
Eliminating T,
dy 1
tan θ = =
dx FH ∫ w( x)dx
Perform second integration,
∫ (∫ w( x)dx )dx
1
y=
FH
21. 7.4 Cables
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
the maximum force developed in the cable and the
cable’s required length. The span length L and, sag
h are known.
22. 7.4 Cables
Solution
Note w(x) = wo
∫ (∫ wodx )dx
1
y=
FH
Perform two integrations
1 wo x 2
y= + C1x + C2
FH 2
Boundary Conditions at x = 0
y = 0, x = 0, dy / dx = 0
23. 7.4 Cables
Solution
Therefore,
C1 = C2 = 0
Curve becomes
wo 2
y= x
2 FH
This is the equation of a parabola
Boundary Condition at x = L/2
y=h
24. 7.4 Cables
Solution
For constant,
wo L2
FH =
8h
8h
4h 2
y= 2 x
L
Tension, T = FH/cosθ
Maximum tension occur at point B for 0 ≤ θ ≤
π/2
25. 7.4 Cables
Solution
Slope at point B
dy w
= tan θ max = o
dx x = L / 2 FH x=L / 2
Or −1wo L
θ max = tan
2 FH
Therefore
FH
Tmax =
cos(θ max )
Using triangular relationship
4 FH + wo L2
2 2
Tmax =
2
26. 7.4 Cables
Solution
2
= o 1+
w L L
Tmax
2 4h
For a differential segment of cable length ds
2
= 1 + dx
dy
ds = (dx ) + (dy )
2 2
dx
Determine total length by integrating
2
L/2 8h
l = ∫ ds = 2 ∫ 1 + 2 x dx
0 L
Integrating yields
L 4h + L sinh −1 4h
2
l = 1+
2
L 4h L
27. 7.4 Cables
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 x
For loading function w = w(s)
acting along the cable,
29. 7.4 Cables
Cable Subjected to its Own Weight
Apply equilibrium equations to the force
system
T cos θ = FH
T sin θ = ∫ w( s )ds
dy 1
=
dx FH ∫ w(s)ds
Replace dy/dx by ds/dx for direct integration
31. 7.4 Cables
Example 7.15
Determine the deflection curve, the length,
and the maximum tension in the uniform
cable. The cable weights wo = 5N/m.
32. 7.4 Cables
Solution
For symmetry, origin located at the center of
the cable
Deflection curve expressed as y = f(x)
ds
x=∫
(
1 + 1 / F 2
H )(∫ wods )
2 1 / 2
Integrating term in the denominator
ds
x=∫
[1 + ( 2
1 / FH )(wo s + C1 ) ]
2 1/ 2
33. 7.4 Cables
Solution
Substitute
u = (1 / FH )(wo s + C1 )
So that
du = ( wo / FH )ds
Perform second integration
x=
FH
wo
{
sinh −1 u + C2 }
or
FH −1 1
x= sinh (wo s + C1 ) + C2
wo FH
34. 7.4 Cables
Solution
Evaluate constants
dy 1
=
dx FH ∫ wods
or
dy 1
= wo s + C1
dx FH
dy/dx = 0 at s = 0, then C1 = 0
To obtain deflection curve, solve for s
FH w
s= sinh o x
wo FH
35. 7.4 Cables
Solution
dy w
= sinh o
F x
dx H
Hence
FH wo
y= cosh
+ C3
x
wo FH
Boundary Condition y = 0 at x = 0
FH
C3 = −
wo
For deflection curve,
F w
y= H cosh o
F x − 1
wo H
This equations defines a catenary curve
36. 7.4 Cables
Solution
Boundary Condition y = h at x = L/2
F w
h= H cosh o
F x − 1
wo H
Since wo = 5N/m, h = 6m and L = 20m,
FH 50 N
6m = cosh
F − 1
5N / m H
By trial and error,
FH = 45.9 N
37. 7.4 Cables
Solution
For deflection curve,
y = 9.19[cosh (0.109 x ) − 1]m
x = 10m, for half length of the cable
l 45.9 5N / m
= sinh (10m ) = 12.1m
2 5N / m 45.9 N
Hence
l = 24.2m
Maximum tension occurs when θ is maximum at
s = 12.1m
38. 7.4 Cables
Solution
dy 5 N / m(12.1m )
= tan θ max = = 1.32
dx s =12.1m 45.9 N
θ max = 52.8o
FH 45.9 N
Tmax = = = 75.9 N
cos θ max cos 52.8 o
