This document discusses cables and arches used in structural engineering. It provides information on how cables carry loads primarily in tension, while arches carry loads in compression. The document outlines various assumptions made in analyzing cables, including that cables are flexible, inextensible, and resist only axial forces. It also discusses analyzing cables under concentrated and uniformly distributed loads. Additional considerations are provided for cable-supported structures, including wind forces that must be considered in design.

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BY:
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2. 2
2. CABLES AND ARCHES
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3. 2.1 INTRODUCTION
2.1 Introduction 3
 Cables carry applied loads & develop mostly tensile stresses - Loads
applied through hangers - Cables near the end supporting structures
experience bending moments and shear forces
 Arches carry applied loads and develop mainly in-plane compressive
stresses; three-hinged, two-hinged and fixed arches - Loads applied through
ribs - Arch sections near the rib supports and and arches, other than three-hinged
arches, experience bending moments and shear forces
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4. 4
2.1 INTRODUCTION
(Cont’d)
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5. 2.1 INTRODUCTION (Cont’d)
5
 In cables, the loads applied through hangers is considered to be
a uniformly distributed load - in the same manner, the loads
distributed to the arches through the ribs are considered to
be uniformly distributed
Cable type structures - Suspension roof, suspension bridges, cable cars,
guy-lines, transmission lines, etc.
Arch type structures - Arches, domes, shells, vaults
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6. 2.2 ANALYSIS OF CABLE
2.2.1 Assumptions 6
 Cable is flexible and in-extensible; hence does not resist any bending
moment or shear force (this is not always true - e.g., fatigue of cables); self
weight of cable neglected when external loads act on the cable
 Since only axial tensile forces are carried by the cable, the force in the
cable is tangential to the cable profile
 Since it is in-extensible, the length is always constant; as a
consequence of the cable profile not changing its length and form, it is
assumed to be a rigid body during analysis
 Even when a moving load is acting on the cable, the load is
assumed to be uniformly distributed over the cable (since the cable
profile is not assumed to change)
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7. 2.2 ANALYSIS O7F CABLE (Cont’d)
2.2.2 Cables subjected to concentrated loads
 When the weight of the cable is neglected in analysis and is subjected to only
concentrated loads, the cable takes the form of several straight
line segments; the shape is called as funicular polygon.
Consider for instance the cable shown in Figure 5.1
q
L
L2
L1
L3 C
B
D
A
yc
yD
P1
P2
Figure 5.1
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8. 2.2 ANALYSIS O8F CABLE (Cont’d)
2.2.2 Cable under concentrated loads (Cont’d)
 In figure 5.1, the known parameters are L1, L2, L3, P1 & P2 - the unknowns are
the four support reactions at A and B, the three cable tensions (in
the three segments) and the two sags (yC and yD) - 9 unknowns
 Eight force equilibrium equations can be written at the four nodes
and we need to have one more condition to solve the problem - This is
met by assuming something about the cable, either its total length, or one of
its sags (say yC or yD)
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9. 2.2 ANALYSIS OF CABLE (Cont’d)
9
2.2.2 Cable under concentrated loads (Cont’d)
Problem 5.1: Determine the tension in each segment of
the cable, shown below, and the total length of the
cable
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10. 2.2 ANALYSIS OF CABLE - FOR CONCENTRATED
LOADS (Cont’d)
10
· AB = (72 +42 ) =8.062 ft; BC = ( y2 +52 )
· cos( q ) = 4 /(8.062) = 0.4962; sin( q ) = 7 /(8.062) =
0.8683 1 1 · cos( q ) = 5/ ( y 2 + 5 2
); sin( q ) =
y
2 2 ( y
2 +
5 2
)
· [(3 ) 3 ]; cos( ) 3 ; sin( )
3 2 2 3
2 2 q q
+ +
[(3 ) 3 ]
= + + =
y
CD y
3
= +y +y 2 +2 q = +y
(3 ) / [(3 ) 3 ]; tan( ) (3 ) 3
· Considering horizontal and vertical equilibrium at
B,
å =0; å=0 H V F F
· cos( ) cos( ) 0.0 (0.4962) / cos( ); 1 2 2 BA q -BC q = BC =BA´ q
and sin( ) 5 sin( ) 0 1 2 BA q - -BC q = ;
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5 /[0.8683 0.4962 tan( )].........................( ) 2 BA= - q I

11. 2.2 ANALYSIS OF CABLE - FOR CONCENTRATED LOADS
(Cont’d)
11
· Considering equilibrium at C, åH =0, & åV =0 F F ;
· cos( ) cos( ) 0; (0.4962) /(cos( )); 2 3 3 BC q -CD q = CD =BA´ q
· sin( ) sin( ) 10 0 2 3 BC q +CD q - = ;
10 /(0.4962 tan( ) 0.4962 tan( )).................( ) 2 3 BA = q + q II
3 2 q
q
· Dividing equation (I) by (II),
· [0.8683 - 0.4962 tan( q )]/[0.4962 ´ (tan( q ) + tan( q )] =
1/ 2 2 2 3 · Substituting for tan( ) and tan( ) in terms of y and solving,
· y = 2.6784 ft
· BA = 8.2988 kips; BC = 4.6714 kips and CD = 8.815 kips;
· Total length of cable = 8.062 + 5.672 + 6.422 = 20.516 ft
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12. 2.3 CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED
LOA12DS
· åF =0; -T cos( ) +(T +DT) cos( +D ) =0.......(A) x q q q
· å =0; - sin( ) - D +( +D )sin( +D ) =0.......( ) 0 F T w X T T B y q q q
å =0; ( D )(D / 2) - cos( )D + sin( )D =0......( ) 0 M w x x T y T x C o School.edhole.com q q

13. 2.3 CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED
LOADS13 (Cont’d)
· Equation (A) reduces to : DT cosq( )-T sinq( )Dq=0;
· ®[T cosq( )]=0;
d integrating T cos( ) Cons tant F ........(D) H =
dx
q=
· Equation (B) reduces to: sin( ) cos( ) 0; 0 DT q+T qDq-wDx=
d q=
this equation can be rewritten as [ sin( )] ....( ) 0 T w E
dx
· Equation (C) reduces to -T cosq( )Dy+T sinq( )Dx=0; this equation
dy=q
reduces to tan( )..................(F)
dx
· From equation (E), one gets sin( ) .......( ) 0 T q=w x G , using the condition that
at x = 0,q =0
· From equation (D) and (G), dividing one by the other (G/D),
one obtains
w x F dy Htan( )=/ = 0 q
dx
from Eqn. (F); and integrating further,
2 /(2 ) .
0 y w x F B H = + At x = 0, y = 0. This leads to the final form given by
Schyo=owl.exdhoFle.2 /(2 )
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0 H

14. 2.3 CABLES SUBJECTED TO UNIFORMLY
14
DISTRIBUTED LOADS (Cont’d)
0 H y = w x F …..This is the equation for a parabola.
· 2 /(2 )
· Using the condition, at x = L, y = h, one obtains that 2 /(2 )
0 F w L h H = ;
hence y = h(x / L)2
· Considering the point B, 2
T = [F 2 +
(w L) mas H 0
[( /(2 )) ( ) ] ( ) [( /(2 )2 1]
0
2
0
2
0 = w L h + w L = w L L h +
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15. 2.4 ADDITIONAL CONSIDERATIONS FOR CABLE
SUPPORTED S1T5 RUCTURES
 Forces on cable bridges: Wind drag and lift forces - Aero-elastic
effects should be considered (vortex-induced oscillations, flutter,
torsional divergence or lateral buckling, galloping and buffeting).
Wind tunnel tests: To examine the aerodynamic behavior
 Precaution to be taken against: Torsional divergence or lateral
buckling due to twist in bridge; Aero-elastic stability caused by geometry
of deck, frequencies of vibration and mechanical damping present;
Galloping due to self-excited oscillations; Buffeting due to unsteady
loading caused by velocity fluctuations in the wind flow
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