Advanced Methods of Structural Analysis

128 5Cables

The sag of the cablefat the axis of symmetry.xD0/in terms of ordinatey.l=2/

of the cable at support is

fDy



l

2



aD

H

q 0

cosh

q 0 l

2H



H

q 0

D

H

q 0



cosh

q 0 l

2H

 1



!cosh^2

q 0 l

2H

D



q 0 f

H

C 1

 2

: (5.29)

Since sinh^2 zDcosh^2 z 1 , then expressions (5.28)and(5.29) lead to the following

formula

NmaxDq 0 fCH: (5.30)

Example 5.3.A uniform cable weightingq 0 D1:25kN=m is suspended between

two pointsAandBof equal elevation, which arel D 20 mapartasshownin

Fig.5.9.

1.Thrust–shape problem. Determine shape of the curve assumed by the cable, dis-
tribution of tension, and length of the cable, if a thrust isHD5:6kN
2.Length–thrust problem. Determine thrust of the cable and shape of the cable, if a
total lengthLD 42 m.

Nmax

qmax

H

W

y

O

C

A B

x

a

l=20m

Nmax

qmax

f

l/ 2

Fig. 5.9 Cable carrying a uniformly distributed load along the cable itself. Design diagram and

force triangle forCBportion

Solution.

1.Thrust–shape problem. Parameter of catenaryaDH=q 0 D5:6=1:25D4:48m.
Equation of the curve assumed by the cable and slope are

y.x/D

H

q 0

cosh

q 0

H

xD4:48cosh.0:2232x/

tanDsinh

q 0

H

xDsinh.0:2232x/:

(a)

Coordinates of pointBarex D 10 m;y.10/D4:48cosh.0:223210/D

21:11m.

Therefore, sag of the cable at pointCbecomesf Dy.10/aD21:11

4:48D16:63m: