Advanced Methods of Structural Analysis

5.2 Cable with Neglected Self-Weight 121

Reaction of supportAequals

RA!

X

MBD 0 WRAlHcC

ql^2

2

D 0 !RAD

ql

2



Hc

l

Free-body diagram of the left part of the cable and corresponding force triangle are
presented in Fig.5.5b. Since shearQ.x/DRAqx, then tension at any section
according to (5.6) equals

N.x/D

p

H^2 C.RAqx/^2 D

s

H^2 C



ql

2



Hc

l

qx

 2

:

The tension in the lowest point

N.x 0 /D

s

H^2 C



ql

2



Hc

l

qx 0

 2

D

s

H^2 C



1:8 30

2



40  3

30

1:812:78

 2

D

p

H^2 C.2727/^2 DH:

Maximum tension occurs at supports

N.0/D

s

402 C



1:8 30

2



40  3

30

1:8 0

 2

D46:14kN;

N.l/D

s

402 C



1:8 30

2



40  3

30

1:8 30

 2

D50:60kN

(b)The reaction of the reference beam isRA^0 Dql=2. The bending moment of the
reference beam and parameterf.x/measured from the inclined chordABare

M^0 .x/D

ql

2

x

qx^2

2

;

f.x/D

M^0 .x/

H

D

1

H



ql

2

x

qx^2

2



:

Distancey^0 between the horizontal linexand cable becomes

y^0 .x/Df.x/xtanD

1

H



ql

2

x

qx^2

2





c

l

x:

Condition dy^0 =dxD 0 leads to the parameterx 0 D12:78m obtained above. The
distance between the cable in the lowest point and horizontal line becomes

y^0 .x 0 /D

1

40



1:8 30

2

12:78

1:812:78^2

2





3

30

12:78D3:674m: