Chapter 5:
Cables and Arches
CIVL3310 STRUCTURAL ANALYSIS
Professor CC Chang
Cables: Assumptions



Cable is perfectly flexible & inextensible
No resistance to shear/bending: same as truss bar
The force acting the cable is always tangent to the
cable at points along its length
Only axial force!
Example 5.1 Under Concentrated Forces
Determine the tension in each segment of the cable. Also, what is the
dimension h?
4 unknown external reactions (Ax, Ay, Dx and Dy)
3 unknown cable tensions
1 geometrical unknown h
8 unknowns
8 equilibrium conditions
Solution
M A  0
TCD (3 / 5)( 2m)  TCD ( 4 / 5)(5.5m)  3kN( 2m)  8kN ( 4m)  0
TCD  6.79kN
Joint equilibriu m at C
6.79kN (3 / 5)  TBC cos  BC  0
6.79kN ( 4 / 5)  8kN  TBC sin BC  0
 BC  32.3o and TBC  4.82kN
Similarly, Joint equilibriu m at B
 BA  53.8o and TBA  6.90kN
Finally , h  ( 2m) tan 53.8o  2.74m
BA
BC
Cable subjected to a uniform distributed load


Consider this cable under distributed vertical load wo
The cable force is not a constant.
wo
FH
Fx  0
 T cos  (T  T ) cos(   )  0
T cos   T cos  T cos(   )  cos 
  Fy  0
 T sin   wo (x)  (T  T ) sin(   )  0
Wit hant i- clockwise as  ve
M 0  0
wo (x)(x / 2)  T cosy  T sin x  0
d(T cos)
 0 eqn 1
dx
d(T sin )
 w o eqn 2
dx
dy
 tan eqn 3
dx
Cable subjected to a uniform distributed load
d(T cos)
 0 eqn 1
dx
d(T sin )
 w o eqn 2
dx
dy
 tan eqn 3
dx

Tsin
From Eqn 1 and let T = FH at x = 0:
T cos  constant  FH eqn 4

FH
Integrating Eqn 2 realizing that Tsin = 0 at x = 0:
T sin   wo x eqn 5

Eqn 5/Eqn 4:
dy wo x
tan 

dx FH
eqn 6
T
FH
Cable subjected to a uniform distributed load
tan 

dy wo x

dx FH
eqn 6
FH
Performing an integration with y = 0 at x = 0 yields
wo 2
y
x
2 FH
eqn 7
Cable profile:
parabola
y = h at x = L
h 2
y 2 x
L
eqn 9
wo L2
FH 
2h
eqn 8
Cable subjected to a uniform distributed load
Tmax

Where and what is the max tension?
max
T cos  FH eqn 4
T sin   wo x eqn 5
T  FH 2  ( wo x)2

T is max when x=L
FH
Tmax  FH 2  ( wo L )2
wo L2
FH 
2h
eqn 10
eqn 8
Tmax  wo L 1  ( L / 2h )2
eqn 11
Cable subjected to a uniform distributed load
T cos  FH
T sin   wo x
dy wo x
tan 

dx FH
T

2
wo L
FH 
2h
y
h 2
x
2
L
Tmax  wo L 1  ( L / 2h ) 2
FH
Cable subjected to a uniform distributed load



Neglect the cable weight which is uniform along the length
A cable subjected to its own weight will take the form of a catenary
curve y  a cosh x 
a
This curve ~ parabolic for small sag-to-span ratio
Wiki catenary
Hangers are close and
uniformly spaced
If forces in the hangers are known
then the structure can be analyzed
1 degree of indeterminacy
Determinate structure
hinge
Example 5.2
The cable supports a girder which weighs 12kN/m. Determine the tension
in the cable at points A, B & C.
12kN/m
Solution
The origin of the coordinate axes is established at point B, the lowest
point on the cable where slope is zero,
y
wo 2 12kN/m 2
6 2
x 
x 
x (1)  0.0389x 2
2 FH
2 FH
FH
Assuming point C is located x’ from B:
6
6 2
x'  FH  1.0 x' 2 (2)  154 .4kN
FH
FH
From B to A:
6
[(30  x' )]2
FH
6
2
12 
[

(
30

x
'
)]
1 .0 x ' 2
12 
x' 2 60x'900  0  x'  12.43m
FH
Solution
y  0.0389x2
T cos  FH
TA
TC
A
C
dy
tan  
 0.0777 x
dx
FH=154.4kN
17.57m
tan C 
dy
 0.966
dx x 12.43
C  44.0o
TC 
FH
154.4

 214.6kN
o
cos C cos 44.0
tan  A 
12.43m
dy
 1.366
dx x 17.57
 A  53.79o
TA 
FH
 261.4kN
cos  A
Example 5.3

Determine the max tension in the cable IH
Assume the cable is parabolic
(under uniformly distributed load)
 MC  0  I y  Ay  0.667FH
FH  28.13 kN
 M B  0  I y  Ay  18.75
Example 5.3
wo L2
FH 
2h
FH  28.13 kN
wo  3.13 kN / m
Tmax  wo L 1  ( L / 2h )2  46.9 kN
Cable and Arch
FH
flip
FH
What if the load direction reverses?
Arches


An arch acts as inverted cable so it receives compression
An arch must also resist bending and shear depending
upon how it is loaded & shaped
Arches

Types of arches
indeterminate
indeterminate
indeterminate
determinate
Three-Hinged Arch
Bx
Ax
Ay
By
Problem 5-30
Determine reactions at A and C and the cable force
3 global Eqs
1 hinge condition
Ax
Bx
Ay
By
Cy
Ax
Ay
T
Example 5.4
The three-hinged arch bridge has a parabolic shape and supports the
uniform load. Assume the load is uniformly transmitted to the arch ribs.
Show that the parabolic arch is subjected only to axial compression at an
intermediate point such as point D.
Solution
y
8 kN/m
y
=160 kN
 10
202
x
10 m
x2
=160 kN
=160 kN
=160 kN
tan  D 
160 kN =
dy  20

x
dx ( 20 )2
 0.5
x 10m
 D  26.6o
0=
N D  178.9kN
=160 kN
=160 kN
VD  0
MD  0
Reflection: What Have You Learnt?