Advanced Methods of Structural Analysis

138 5Cables

The total length of the cable

LD 2

f

sin ̨

Dl

p

1 Ctan^2 ̨Dl

r

1 C

P^2

4H^2

: (5.37)

Now let us introduce the elastic properties of the cable. Let the stiffness of the cable

beEA,whereEis modulus of elasticity;Ais cross-sectional area of the cable. The

strain of the cable

"D

N

EA

D

H

EA

r

1 C

P^2

4H^2

: (5.38)

If the initial length of the cable isL 0 , then the length of the cable upon the load is

LDL 0 .1C"/: (5.39)

Equations (5.37)–(5.39) lead to one resolving equation

PD2H

vu

u

u

t

ˇ^2



1 

H

EA

ˇ

 2 1; ˇD

L 0

l

: (5.40)

This equation allows to calculate unknown thrustHfor givenP,EA, initial length

L 0 ,andspanl.

The sag of elastic cable is

felD

Pl

4H

D

l

2

vu

u

u

t

ˇ^2



1 

H

EA

ˇ

 2 1: (5.41)

Limiting cases

1.The cable is nondeformable.EAD1;L 0 DL/. In this case, the sag

fndD

l

2

r

L^2

l^2

1:

This result also may be obtained from geometrical consideration of the design
diagram. Let us evaluate increasing of the sag for elastic cable. IfˇD1:5and
HD0:01EA,thenfelD1:35fnd, i.e., a sag increased to 35%.
2.The initial length of a cable equals to the span.L 0 Dl/; it means that the cable
may be treated as a string. In this caseˇD 1 and (5.40) becomes

PD2H

vu

u

u

t

1



1 

H

EA

 2 ^1 (5.42)