Advanced Methods of Structural Analysis

132 5Cables

Length of the portionCBisLCB D asinh.xB=a/D 4:48sinh.8:103=4:48/D
13:303m
Curve AC. Equation of the curve and slope are

y.x/D4:48cosh.0:2232x/;

tanD

dy

dx

Dsinh.0:2232x/:

Ordinateyand slope at the pointAare

y.11:897/D4:48cosh.0:2232.11:897//D32:034m;

tan.11:897/Dsinh.0:2232.11:897//D7:0796;

AD81:96ıI sinAD0:990:

Calculated coordinatesyfor support pointsAandBsatisfy to given design diagram

yAyBD32:03414:034D 18 m

Tension at the pointAis

NADH

r

1 Csinh^2

q 0

H

xAD5:6

q

1 Csinh^2 .0:223211:897/D40:04kN

Length of the portionCAand total length of the cable are

LCADasinh

xA

a

D4:48sinh

11:897

4:48

D31:725m

LDLACCLCBD31:725C13:303D45:028m

Control. The tension at the support points and total weight of the cable must satisfy
to equation

P

YD 0. In our case we have

NAsinACNBsinBLq 0

D40:040:990C17:5440:947645:0281:25D56:26456:285:

The relative error equals 0.035%.

5.4.3.2 Saddle Point Outside of the Span

Now let us consider the cable with total lengthL 0 , which is suspended between two
pointsAandB, as shown in Fig.5.11. Peculiarity of this design diagram, unlike the
previous case, is that the curveABof the cable has no point, for which tanD 0 ,