Advanced Methods of Structural Analysis

120 5Cables

x

RA

y

H

N(x)

q

x

q

H

N(x)

Q(x)=RA − qx

b

a

l

y

x

q

x

RB

H

RA

H

A j c

x 0

ymax

B

f y^0

Fig. 5.5 (a) Cable under action of uniformly distributed load; (b) free-body diagram of the part of
the cable

Constants of integration should be calculated from the following boundary condi-
tions: atxD 0 (supportA)yD 0 and atxDl(supportB)yDc. Constants of
integration areC 2 D^0 and

C 1 D



c

ql^2

2H



1

l

:

Now the shape of the cable and its slope for anyxmay be presented in the form

y.x/D
ql^2
2H



x^2

l^2



x

l



Cc

x

l

D

1:8 302

2  40



x^2

302



x

30



C 3

x

30

D0:675



x^2

30

x



C0:1x;

tanD
dy
dx
D
ql
2H



2

x

l

 1



C

c

l

D

1:8 30

2  40



2

x

30

 1



C

3

30

D0:675

x

15

 1



C0:1:

Equationy.x/presents nonsymmetrical parabola. In the lowest point the slope of
the cable equals zero

tanD

ql

2H



2

x

l

 1



C

c

l

D0;

This equation leads to correspondingx 0 -coordinate

x 0 D

H

q



ql

2H



c

l



D

40

1:8



1:8 30

2  40



3

30



D12:78m:

The maximumyoccurs atx 0

ymaxDy.12:78/D0:675



12:78^2

30

12:78



C0:112:78D3:674m: