Short Notes: Arches & Cables | Short Notes for Civil Engineering - Civil Engineering (CE) PDF Download

 Page 1


 
 
 ARCHES & CABLES 
1. TYPES OF ARCHES 
There are three types of arches depending upon the number of hinges provided. 
(i) Three hinged arch (Determinate) 
(ii) Two hinged arch (Indeterminate to 1 degree) 
(iii) Fixed arch (Indeterminate to 3 degree) 
1.1. Three hinged Arch 
The three hinged arches are statically determinate structure as equations of equilibrium 
alone are sufficient to find all the unknown quantities. 
 
Circular Arch: 
From the properly of a circle the radius r of the circular arch of span 
L and rise h may be found as 
LL
h(2R h)
22
? = - 
2
Lh
R
8h 2
? = + 
Taking origin at A, the coordinates of any pont d on the arch may be defined as 
L
x R sin
2
??
= - ?
??
??
 
y = Rcos? – (R – h) 
y h R(1 cos ) ? = - - ? 
Parabolic Arch: 
Taking spring point as the origin, its equation is given by 
x
2
4h
y (L x)
L
=- 
Bending moment at the section X-X 
BMX – X = +VA × x – HA × y 
? BMX – X = Beam moment – H-moment 
When compared with a beam of similar span, bending moment at any section in a three 
hinged arch is less by an amount of ‘H×y’ or moment dur to horizontal force. 
 
 
Page 2


 
 
 ARCHES & CABLES 
1. TYPES OF ARCHES 
There are three types of arches depending upon the number of hinges provided. 
(i) Three hinged arch (Determinate) 
(ii) Two hinged arch (Indeterminate to 1 degree) 
(iii) Fixed arch (Indeterminate to 3 degree) 
1.1. Three hinged Arch 
The three hinged arches are statically determinate structure as equations of equilibrium 
alone are sufficient to find all the unknown quantities. 
 
Circular Arch: 
From the properly of a circle the radius r of the circular arch of span 
L and rise h may be found as 
LL
h(2R h)
22
? = - 
2
Lh
R
8h 2
? = + 
Taking origin at A, the coordinates of any pont d on the arch may be defined as 
L
x R sin
2
??
= - ?
??
??
 
y = Rcos? – (R – h) 
y h R(1 cos ) ? = - - ? 
Parabolic Arch: 
Taking spring point as the origin, its equation is given by 
x
2
4h
y (L x)
L
=- 
Bending moment at the section X-X 
BMX – X = +VA × x – HA × y 
? BMX – X = Beam moment – H-moment 
When compared with a beam of similar span, bending moment at any section in a three 
hinged arch is less by an amount of ‘H×y’ or moment dur to horizontal force. 
 
 
 
 
2. Two hinged arches 
A two hinged arch is an indeterminate arch. The horizontal thrust is determined using 
Castigliano’s theorem of least energy.  
 
Assuming the redundant to be H, As per Castigliano’s theorem  
????
????
= 0 
Which gives the following condition 
l
x
C
0
l
2
C
0
M ydx
EI
H
y dx
EI
=
?
?
 
Where, 
Mx = beam moment at any section x – x 
IC = Moment of inertia of the cross section of the arch at the crown. 
(i) Horizontal Thrust in case of circular arch subjected to point load  
?? =
?? ?? sin
2
?? 
(ii) Horizontal Thrust in case of circular arch subjected to UDL  
?? =
4
3
????
?? 
(iii) Horizontal Thrust in case of parabolic arch subjected to a point load at centre 
?? =
25
128
????
?? 
(iv) Horizontal Thrust in case of parabolic arch subjected to a UDL  
?? =
?? ?? 2
8h
 
If there is rib shortening, temperature rise by t°C and yielding of supports then horizontal 
thrust is given by 
x
C
2
C
M ydx
tl
EI
H
y dx l
k
EI AE
+?
=
++
?
?
 
Where, 
?? tl = due to increase in temperature 
Page 3


 
 
 ARCHES & CABLES 
1. TYPES OF ARCHES 
There are three types of arches depending upon the number of hinges provided. 
(i) Three hinged arch (Determinate) 
(ii) Two hinged arch (Indeterminate to 1 degree) 
(iii) Fixed arch (Indeterminate to 3 degree) 
1.1. Three hinged Arch 
The three hinged arches are statically determinate structure as equations of equilibrium 
alone are sufficient to find all the unknown quantities. 
 
Circular Arch: 
From the properly of a circle the radius r of the circular arch of span 
L and rise h may be found as 
LL
h(2R h)
22
? = - 
2
Lh
R
8h 2
? = + 
Taking origin at A, the coordinates of any pont d on the arch may be defined as 
L
x R sin
2
??
= - ?
??
??
 
y = Rcos? – (R – h) 
y h R(1 cos ) ? = - - ? 
Parabolic Arch: 
Taking spring point as the origin, its equation is given by 
x
2
4h
y (L x)
L
=- 
Bending moment at the section X-X 
BMX – X = +VA × x – HA × y 
? BMX – X = Beam moment – H-moment 
When compared with a beam of similar span, bending moment at any section in a three 
hinged arch is less by an amount of ‘H×y’ or moment dur to horizontal force. 
 
 
 
 
2. Two hinged arches 
A two hinged arch is an indeterminate arch. The horizontal thrust is determined using 
Castigliano’s theorem of least energy.  
 
Assuming the redundant to be H, As per Castigliano’s theorem  
????
????
= 0 
Which gives the following condition 
l
x
C
0
l
2
C
0
M ydx
EI
H
y dx
EI
=
?
?
 
Where, 
Mx = beam moment at any section x – x 
IC = Moment of inertia of the cross section of the arch at the crown. 
(i) Horizontal Thrust in case of circular arch subjected to point load  
?? =
?? ?? sin
2
?? 
(ii) Horizontal Thrust in case of circular arch subjected to UDL  
?? =
4
3
????
?? 
(iii) Horizontal Thrust in case of parabolic arch subjected to a point load at centre 
?? =
25
128
????
?? 
(iv) Horizontal Thrust in case of parabolic arch subjected to a UDL  
?? =
?? ?? 2
8h
 
If there is rib shortening, temperature rise by t°C and yielding of supports then horizontal 
thrust is given by 
x
C
2
C
M ydx
tl
EI
H
y dx l
k
EI AE
+?
=
++
?
?
 
Where, 
?? tl = due to increase in temperature 
 
 
l/AE= due to rib shortening 
K = yielding of support/unit horizontal thrust. 
In a two hinged parabolic arch as the temperature increase, horizontal thrust increases. If the 
effect of rib shortening and yielding of support are considered, then horizontal thrust 
decreases. 
3. CABLES: 
Suspension Cables are generally used to support suspension bridges, roofs and cable car 
system. They are used to transmit load from one structure to another. Cables are deformable 
structure so they can undergo change in shape according to externally applied load. Thus, 
bending moment and shear force at every point in the cable is zero.   
3.1. CABLES SUBJECTED TO CONCENTRATED LOADS: 
When cables are subjected to concentrated loads, it may take shape of several straight-
line members subjected to tension. The shape cable takes is known as funicular polygon. 
The cable will always be subjected to pure tensile forces having the funicular shape of 
load. 
Consider the cable subjected to concentrated loads P1 and P2 as shown in the figure. Due 
to the loading, cable assumes the shape ACDB. 
 
The tension at each member can be easily determined using the equation of equilibrium 
at each joint and the geometry of the structure. 
3.2 CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD: 
The tension in the cable at any distance x will be 
?? = v(?? ?? )
2
+ (?? 0
?? )
2
 
? ?? =
v
(
?? 0
?? 2
2h
)
2
+ (?? 0
?? )
2
 
Tension will be maximum when x is maximum i.e., x = L. 
?? ?????? = ?? 0
?? v
1 + (
?? 2h
)
2