Advanced Methods of Structural Analysis

5.4 Cable with Self-Weight 129

Slope at pointBis

tanmaxDsinh.0:223210/D4:6056!

sinmaxDsin.tan^1 4:6056/D0:9772;cosmaxDcos.tan^1 4:6056/D0:2122:

Tension at any point is

NDH

r

1 Csinh^2

q 0

H

x;

and maximum tension occurs at pointBand equals

NmaxD5:6

q

1 Csinh^2 .0:223210/D26:39kN:

Since shape of the cable is symmetrical, then total length of the cable is

LD2asinh

x

a

ˇ

ˇ

ˇ

xDl 2

D 2 4:48sinh

10

4:48

D41:27m:

ControlIf total lengthLD41:27m, then total weight of the cable equals41:27

1:25D51:58kN, and forceW, considering portionCB, equals51:587=2D

25:79kN. From a force triangle we can calculate

HDNmaxcosmaxD26:390:2122D5:6kN;

WDNmaxsinmaxD26:390:9772D25:79kN;

tanmaxD

W

H

D

25:79

5:6

D4:605:

The maximum tension according to (5.30) equals

NmaxDq 0 fCHD1:25.21:114:48/C5:6D26:39kN:

2.Length–thrust problem.We assume that a total lengthLof a cable is given and
it is required to find the thrust, shape of the cable, and sag. Since the cable is
symmetrical with respect toy-axis, then total length of the cable equalsLD
2asinh.xB=a /,whereLD 42 mandxBDl=2D 10 m. Unknown parametera
may be calculated from transcendental equation (5.25), i.e.,

42 D2asinh

10

a

:

Solution of this equation leads toaD4:42m. Thrust of the cable isHDaq 0 D

4:421:25D5:525kN.