Advanced Methods of Structural Analysis

5.6 Effect of Axial Stiffness 139

This equation may be rewritten in equivalent form

PD

2

p

2

1 

H

EA

r

H^3

EA

r

1 

H

2EA

(5.42a)

SinceH=EA 1 ,then(5.42a) may be presented as

PD

r

8H^3

EA

or

HD

1

2

p 3

P^2 EA (5.43)

Corresponding sag is

fD

Pl

4H

D

l

2

3

r

P

EA

(5.44)

Equations (5.43)and(5.44) show that the relationshipsP–H andP–f are
nonlinear.

5.6.2 Elastic Cable with Uniformly Distributed Load

Now let us consider a gentile cable of the spanl. In this case a distributed load may
be considered as sum of external load and the weight of a cable itself. Since the
cable is gentile, we assume that the tension in the cable is constant and equal to the
thrust

NDH

s

1 C



dy

dx

 2

H: (5.45)

According to (5.16b)wehave

L

l

D 1 C

1

24

q^2 l^2

H^2

:

Solution of this equation leads to the following expression for a thrust (in terms of
spanl, total lengthLof the cable, and loadq)

HD

ql

2

p

6

1

r

L

l

 1

: (5.45a)

The lengthLof the cable under load isLDL 0 .1C"/,whereL 0 is initial length
of the cable and"DH=EA,so

LDL 0 .1C"/DL 0



1 C

H

EA



: