Advanced Methods of Structural Analysis

5.4 Cable with Self-Weight 131

Solution.Let a verticaly-axis pass through the lowest pointC. Location of this
point is defined by parametersl 1 and l 2 which are unknown before head, while
l 1 Cl 2 D 20 m. For this direct problem, the parameter of catenary (Fig.5.10)may
be calculated immediately, i.e.,aDH=q 0 D5 : 6=1 : 2 5D4:48m.

Shape of the cable is defined by equation

y.x/D

H

q 0

cosh

q 0

H

xD4:48cosh.0:2232x/: (a)

Conditions of the catenary passing through two pointsAandBare as follows:

AtxDl 2 (pointB) the ordinateyDaChD4:48Ch.
AtxDl 1 D.20l 2 /(pointA) the ordinateyDaChC 18.
Therefore, (a) for pointsBandAbecomes
PointB:4:48cosh.0:2232l 2 /D4:48Ch(b)
PointA:4:48coshŒ0:2232.20l 2 /D4:48ChC 18.
These equations contain two unknownshandl 2. Expressing from both equa-
tions’ unknown parameterhand after that equating the two expressions, we obtain

4:48cosh.0:2232l 2 /D4:48cosh.0:2232l 2 4:464/18:

The root of this equation isl 2 D8:103m, which leads tol 1 Dll 2 D11:897m.
Therefore, vertical distancehbetween pointsCandBaccording to formula (b)
equals

hD4:48cosh.0:22328:103/4:48D9:555m:

Now we can consider the right and left parts of the cable separately.
Curve CB.Equation of the curve and slope are

y.x/D4:48cosh.0:2232x/;

tanD

dy

dx

Dsinh.0:2232x/:

Ordinateyand slope at the pointBare

y.xD8:103/D4:48cosh.0:22328:103/D14:034m;

tan.xD8:103/Dsinh.0:22328:103/D2:969;

BD71:38ıI sinBD0:9476:

The tension at the pointBis

NBDH

r

1 Csinh^2

q 0

H

xBD5:6

q

1 Csinh^2 .0:22328:103/D17:544kN: