Advanced Methods of Structural Analysis

114 5Cables

RA

H

NA- 1

A

k

k

RA

H

NA- 1

b

c P=20kN

A B

120kNm

C

a

P=20kN

RA

x

H

RB

H

y

a=10m

l=25m

a 1

a 0

a 0

a 0

A B

1

f

k

k

b=15m

Fig. 5.3 (a) Design diagram of the cable; (b) free-body diagram; (c) reference beam and corre-
sponding bending moment diagram

5.2.1.1 Direct Problem

Determine a shape of the cable, if the thrust of the system is givenH D 24 kN.
Vertical reactions of the cable

RA!

X

MBD 0 W RAD

P.la/

l

D

20.2510/

25

D 12 kN;

RB!

X

MAD 0 W RBD

Pa

l

D

20  10

25

D 8 kN:

We can see that the vertical reactions do not depend on the value of the trustH.It
happens because supportsAandBare located on the same elevation. The forces
acting on the segment at supportAand corresponding force polygon is shown in
Fig.5.3b. For assumedx–ycoordinate system the angles ̨ 0 and ̨ 1 belong to the
fourth and first quadrant, respectively, so a shape of the cable is defined as follows

tan ̨ 0 D

RA

H

D

12

24

D

1

2

!cos ̨ 0 D

2

p

5

;

tan ̨ 1 D

RB

H

D

8

24

D

1

3

!cos ̨ 1 D

3

p

10

:

They-ordinate of the cable at the location of the loadPis

yDatan ̨ 0 D 10 0:5D 5 m:

The negative sign corresponds to the adoptedx–ycoordinate system. The sag at the
pointCisfD 5 m.
Tensions in the left and right portions of the cable may be presented in terms of
thrust as follows:

NA 1 D

H

cos ̨ 0

D

24 

p

5

2

D26:83kN;NB 1 D

H

cos ̨ 1

D

24 

p

10

3

D25:30kN: