CHAPTER 3:

CABLES
AND
ARCHES

Dr. ZaynabTarhini +
Dr.Hanadi Elkhansa
Introduction • Cables and arches often form the main load-
carrying element in many types of structures,
and in this chapter we will discuss some of the
important aspects related to their structural
analysis. The chapter begins with a general
discussion of cables, followed by an analysis
of cables subjected to a concentrated load
and to a uniform distributed load. Since most
arches are statically indeterminate, only the
special case of a three-hinged arch will be
considered. The analysis of this structure will
provide some insight regarding the
fundamental behavior of all arched 2

structures.
Outline 1. Cables
o Definition
o Cables subjected to concentrated loads
o Cables subjected to a uniform distributed load

2. Arches
o Definition
o Three hinged arches
3

Reference: structural_analysis_hibbeler_8th - Chapter 5, P. 181

CABLES
o Definition

Cables are used in engineering

structures for support and to
transmit loads from one member
to another.
examples: used to support
suspension roofs, bridges,….

In our case, the weight of the cable

itself maybe neglected.
5

Cables
o Definition
We will make the assumption that the cable is perfectly flexible and
inextensible:
• Due to its flexibility, the cable offers no resistance to shear or bending
(The force acting in the cable is always tangent to the cable at points
along its length).
• Inextensible, the cable has a constant length both before and after the
load is applied.

Cables
o Cables subjected to concentrated loads
• When a cable of negligible weight
supports several concentrated loads,
the cable takes the form of several
straight-line segments.

• Each of these segments is subjected

to a constant tensile force.

 specifies the angle of the cable’s

cord AB
L is the cable’s span 7

Cables
o Cables subjected to concentrated loads
• If L1 , L2 , L3 and P1 , P2 are known, the problem is
→to determine the 9 unknowns:
- the tension in each of the 3 segments,
- the 4 components of reaction at A & B
- the sags yC & yD
• For the solutions,
we can write 2 equations of equilibrium at each of 4 points A, B, C & D
→ Total 8 equations
+ 1 equation comes from the geometry of the cable 8

Cables
Exercise 1
Determine the tension in each
segment of the cable.
Also, what is the dimension h?

Cables
By inspection, there are
→ 4 unknown external reactions (𝑅𝐴𝑥 , 𝑅𝐴𝑦 , 𝑅𝐷𝑥 and 𝑅𝐷𝑦 )
→ 3 unknown cable tensions (𝑇𝐴𝐵 , 𝑇𝐵𝐶 , 𝑇𝐶𝐷 )
→ sag “h”

These unknowns can be determined by equilibrium

equations at A, B, C & D ( σ 𝑭𝒙 = 𝟎 ; σ 𝑭𝒚 = 𝟎 )
𝜃𝐶𝐷
*Direct approach to the solution:
slope of cable CD is specified
→ 𝑇𝐶𝐷 can be determined
Tan 𝜃𝐶𝐷 = 2/1.5 = 4/3 → 𝜃𝐶𝐷 = 53.13° 10

Cables
▪ Section in cable CD at D → M A = 0 → 𝑻𝑪𝑫

M A = 0 +

+ 3(2) + 8(4) − TCD (cos 53.13)(2) − TCD (sin 53.13)(5.5) = 0

R
TCD = 6.79kN R TCD sin 𝜽𝑪𝑫

𝜃𝐶𝐷

TCD cos 𝜽𝑪𝑫

11

Cables
▪ Equilibrium at point C: 𝑻𝑪𝑫

+Fx = 0
6.79(cos 53.13) − TBC cos  BC = 0 𝜃𝐶𝐷

+  Fy = 0
6.79(sin 53.13) − 8 + TBC sin  BC = 0

2 equations , 2 unknowns (1)TBC cos  BC = 4.074

(2)TBC sin  BC = 2.568

(2)/(1) →  BC = 32.3o and TBC = 4.82kN 12

Cables
▪ Equilibrium at point B:

+Fx = 0
4.82(cos 32.3) − TBA cos  BA = 0 𝜃𝐵𝐶

+  Fy = 0
𝑻𝑩𝑪
− 4.82(sin 32.3) − 3 + TBA sin  BA = 0

2 equations , 2 unknowns (1)TBA cos  BA = 4.078

(2)TBA sin  BA = 5.575

(2)/(1) →  BA = 53.8o and TBA = 6.9kN 13

Cables
 tan 𝜃𝐵𝐴 = h/2

h = (2) tan 53.8 = 2.74m

o 𝜃𝐵𝐴

14

Cables
Exercise 2
1. Determine the equilibrium
equations.
2. Determine the force P that
can support cables ABCD.

15

Cables
 There are 3 unknown tensile forces (𝑇𝐴𝐵 , 𝑇𝐵𝐶 & 𝑇𝐶𝐷 )
𝜃𝐴𝐵 , 𝜃𝐵𝐶 , 𝜃𝐶𝐷 are known

tan 𝜃𝐶𝐷 = 6/5 → 𝜃𝐶𝐷 = 50.2°

tan 𝜃𝐶𝐵 = 3/8 → 𝜃𝐶𝐵 = 20.55° 𝜃𝐵𝐴

tan 𝜃𝐵𝐴 = 3/2 → 𝜃𝐵𝐴 = 56.31° 𝜃𝐶𝐵 𝜃𝐶𝐷

16

Cables
 ▪ The equilibrium equations:
M A = 0 +
+ 6( 2m) + p (10m) − TCD (sin 50.2)(15m) = 0
TCD = (12 + 10 p ) / 11.52 TCD
TCD sin 𝜽𝑪𝑫

TCD cos 𝜽𝑪𝑫

17

Cables
▪ Equilibrium at point C:
+Fx = 0
𝑇𝐶𝐷
TCD (cos 50.2) − TBC (cos 20.55) = 0

TCD = 1.46TBC 𝜃𝐶𝐷

+  Fy = 0 P
TCD (sin 50.2) − p + TBC (sin 20.55) = 0
1.46TBC (sin 50.2) − p + TBC (sin 20.55) = 0

TBC = 0.68 p Then, TCD = 0.99 p 18

Cables
▪ Equilibrium at point B:
+Fx = 0
− TBA (cos 56.31) + TBC (cos 20.55) = 0

TBA = 1.688TBC TBA = 1.147 p

𝜃𝐵𝐶
6
+  Fy = 0 𝑇𝐵𝐶

TBA (sin 56.31) − 6 − TBC (sin 20.55) = 0

1.147 p (sin 56.31) − 6 − 0.68 p (sin 20.55) = 0 TBA = 9.6kN
p = 8.38kN TBC = 5.7 kN 19

TCD = 8.3kN
Cables
o Cables subjected to a uniform distributed load

Cables provide a very

effective means of
supporting the dead weight
of girders or bridge decks
having very long spans.

20
“Verrazano-Narrows” bridge (NewYork city)
Cables
o Cables subjected to a uniform distributed load

What is the shape of a cable

subjected to uniform horizontally
distributed vertical load (w0) ?

The x,y axes have their origin located

at the lowest point on the cable such
that the slope is zero at this point.

21

Cables
o Cables subjected to a uniform distributed load

The tensile force in the cable changes

continuously in both magnitude &
direction along the cable’s length
→this change is denoted by T.

The distributed load is represented by

its resultant force ox which acts at
x/2 from point O.
22

Cables
o Cables subjected to a uniform distributed load
Equilibrium equations:
+Fx = 0
− T cos  + (T + T ) cos( +  ) = 0 (1)

+  Fy = 0
− T sin  − wo (x) + (T + T ) sin( +  ) = 0 (2)

M O = 0 +
− wo (x)(x / 2) + T cos y − T sin x = 0 (3)
23

Cables
o Cables subjected to a uniform distributed load
Divide (1),(2),(3)/ x
and taking the limit as x →0, hence, y →0 ,  →0 and T →0
d (T cos  )
(1)→ =0 → T cos  = const = FH (at x = 0, T = FH ) (4)
dx
The horizontal component of force at any point
along the cable remains constant
d (T sin  )
(2)→ = wo → T sin  = wo x (at x = 0, T sin  = 0) (5)
dx

dy dy wo x
(3)→ = tan  → Divide (5)/(4) tan  = = (6)
dx dx FH 24

Cables
o Cables subjected to a uniform distributed load

Integration of (6) wo 2
y= x Equation of parabola
with y=0 at x=0 2 FH

FH ?? using the boundary condition y = h at x = L

wo L2
FH =
2h
h 2
Then, y= 2 x
L 25

Cables
o Cables subjected to a uniform distributed load

Maximum tension in the cable: Tmax ??

From (4), the max tension in the cable occurs when  is max

For x= L 2
(4) +(5) 2 Tmax = F 2
H + ( wo L) 2

wo L2 L 2
or FH = Tmax = wo L 1 + ( )
2h 2h

26

Cables
Exercise 3
The cable supports a girder
which weighs 12kN/m.
1. Determine the conditions of
equilibrium of this cable.
2. Determine the forces in the
cable at points A, B & C.
(𝑇𝐴 , 𝛳𝐴 , 𝑇𝐵 , 𝛳𝐵 , 𝑇𝐶 , 𝛳𝐶 )

27

Cables
1st step: The origin of the coordinate axes is established at point B
(the lowest point on the cable where slope ϴ =0)

Using the equation of parabola, where w0=12kN/m

wo 2 12kN/m 2 6 2
y= x = x = x (1)
2 FH 2 FH FH

→Use equation (1) at points A & C:

*At point C : 𝑥𝐶 = 𝑥 ′ ; 𝑦𝐶 = 6𝑚
6 2
6= x'  FH = x'2 (2)
FH 28

Cables
 *At point A : 𝑥𝐴 = −(30 − 𝑥 ′ ); 𝑦𝐴 = 12𝑚
6 6
12 = [ −(30 − x ' )] = 2 [ −(30 − x ' )]
2 2

FH x'

12 x '2 = 6 x '2 −60 x '+900 
x ' +60 x '−900 = 0
2
 x'1 = −72.42m  x'2 = 12.43m
Rejected Accepted

FH = (12.43) = 154.4kN
2
For x’ = 12.43m
dy w0 x 12
= = x = 0.7772 x (3)
dx FH 154.4 29

Cables
 • At point A: x A = −(30 − 12.43) = −17.57m FH = 154.4kN
dy
tan  = = 0.7772 x
dy
tan  A = = −1.366   A = −53.79o dx
F
dx x = −17.57 T = H
cos 
FH 154.4
TA = = = 261.4kN
cos A cos(−53.79 )
o

• At point B: xB = 0 • At point C: xC = 12.43m

dy
tan  B = = 0   B = 0o tan  C =
dy
= 0.9657   C = 44 o
dx x =0 dx x =12.43

FH 154.4
TB =
FH
=
154.4
= 154.4kN TC = = = 214.6kN
cos B cos 0 o cos  C cos 44 o
30

Cables
Exercise 4
The suspension bridge is
constructed using the 2 stiffening
trusses that are pin connected at
C, and supported by a pin at A and
a rocker (roller) at B.
Determine the maximum tension
in the cable IH.

The cable has a parabolic shape and the

bridge is subjected to the single load of
31
50kN.

Cables
 2
L w
Maximum tension: Tmax = wo L 1 + ( ) 2  wo ?? → from FH ( FH = o ) L
2h 2h
h = 8m; L = 12m (Distance from the lowest point (origin)
*Draw the free body diagram of the cable-truss system
TI TH
THy
TIy

M B = 0 +

FH (14) − 50(9) + RAy (24) + TIy (24) − FH (14) = 0

14
R RAy + TIy = 18.75 (1)
R
RBy
32

Cables
*Take only half of the suspended structure:
TI TIy M C = 0 +
FH (6) + R Ay (12) + TIy (12) − FH (14) = 0
Only FH, because
the lowest point
R Ay + TIy = 0.667 FH (2)

(1) = (2) 0.667 FH = 18.75

FH = 28.125kN
RCx
RAx
RAy
Then,
RCy
wo L2 2 FH h
FH =  wo = 2
= 3.125kN / m
2h L
33
12 2
Tmax = 3.125  12 1 + ( ) = 46.9kN
Cables 28