Advanced Methods of Structural Analysis

134 5Cables

or in equivalent form

sinh

D 0

2a

D

1

2a

q

L^20 h^20 (5.33a)

CoordinatesxAandxAof the pointsAandBare

xADatanh^1



h 0

L 0





D 0

2

;xBDxACD 0 : (5.34)

The thrust of the cable equalsH D q 0 a. Also, equation of the curve and slope
according to (5.27)are

y.x/Dacosh

x

a

I tan™.x/Dsinh

x

a

Slopes at support pointsAandBare

tanADsinh

xA

a

I tanBDsinh

xB

a

(5.35)

Tension at pointsAandBaccording (5.1), (5.2), and (5.34)are

NADH

r

1 Csinh^2

xA

a

I NBDH

r

1 Csinh^2

xB

a

: (5.36)

Example 5.5.Let us consider the cable, which is presented in Fig.5.11. The cable
has following parameters:L 0 D117:7mI D 0 D 100 mI h 0 D57:7mI q 0 D
0:014kN=m. Determine the shape of the cable, thrust, and tension.

Solution.Parameteraof the catenary can be calculated from (5.33a)

sinh

100

2a

D

1

2a

p

117:7^2 57:7^2 !aD127:41m:

Coordinatexof the supportAisxAD127:41tanh^1 .57:7=117:7/. 1 0 0=2 /D
18:34m.
Slopes at supportsAandB

tanADsinh

18:34

127:34

D0:1446I sinAD0:1431I cosAD0:9897

tanBDsinh

18:34C 100

127:34

D1:0694I sinBD0:7304I cosBD0:6830:

Thrust of the cable isHDq 0 aD0:014127:41D1:7837kN.