Advanced Methods of Structural Analysis

140 5Cables

Thus (5.45a)forthrustHmay be presented as

HD

ql

2

p

6

1

r

L 0 .1C"/

l

 1

D

ql

2

p

6

1

s

ˇ



1 C

H

EA



 1

;ˇD

L 0

l

: (5.46)

So for computation of the thrustH the equation (5.46) may be rewritten in the

following form

ˇ

EA

H^3 C.ˇ1/H^2 D

q^2 l^2

24

:

This is noncomplete cubic equation with respect toH. Solution of this equation

for string.ˇD1/is

HD

3

r

q^2 l^2 EA

24

: (5.46a)

The reader is invited to derive the following expression for sag of an gentile cable:

fD

p

3

2

p

2

l

s

ˇ



1 C

H

EA



 1 (5.47)

Hint. Use formulafDql^2 =8H(formula5.10) and expression (5.46).

We can see that consideration of the elastic properties of a cable leads to the

nonlinear relationshipsH–qandf–H.

Limiting cases

1.In case of nondeformable cable.EAD1/,formulas(5.46)and(5.47) coincide
with formulas (5.16c)and(5.16d) for the case of gentile cable carrying uniformly
distributed load.
2.IfL 0 Dlthen formulas (5.46a)and(5.47) lead to the following nonlinear rela-
tionshipsf–qW

fD

l

4

3

r

3ql

EA

(5.48)

Problems

5.1.Design diagram of a cable is presented in Fig.P5.1. SupportsAandBare

located on different elevations. Parameters of the system are:a 1 D 10 m,a 2 D

22 m,cD 3 m,lD 30 m,P 1 D 18 kN,P 2 D 15 kN.

Determine the thrustHof the cable, if total length of the cableLD 34 m.

Ans.HD23:934kN