Wire Ropes Under Tensile Load - Wire Ropes: Tension, Endurance, Reliability - page 121

c 1 m g sin b F ð L x Þ x = 2
c 2 d ð S 0 þ m g sin b F

u ¼

L = 2 Þþ c 3 G d 3 :

ð 2 : 93 Þ

The maximum rotary angle (for x = L/2) is

L 2

u max ¼ 1

c 1 m g sin b F

8 L = 2 Þþ c 3 G d 3 :

ð 2 : 94 Þ

c 2 d ð S 0 þ m g sin b F

Example 2.11 Wire rope supported non-rotated at both ends
Data:
Warr. 8 9 19-NFC-sZ,
rope diameter d = 16 mm or d = 0.016 m
rope length related mass m = 0.89 kg/m
shear module G = 76,000 N/mm 2 or G = 76 9 10 9 N/m 2

rope length L = 500 m
lower tensile load S 0 = 10,000 N
angle b F = 90
constants (Table 2.6 ) c 1 = 0.108; c 2 = 0.222; c 3 = 0.000268
Results:
According to ( 2.87 ) the torque is

M ¼ 40 : 58 þ 61 : 56 ¼ 20 : 98 Nm :

According ( 2.89 ) the maximum angle occurs at the rope length

x ð u max Þ ¼4080 : 8 3835 : 8 ¼ 245 m

The maximum rotary angle—( 2.88 )—is

u max ¼ 232 : 8 rad :

The maximum number of rope turns is then

n max ¼ u max

2 p ¼ 37 :

According to ( 2.85 ) and ( 2.87 ), the maximum twist angle is on the lower rope
end (x = 0)

x lower ¼ x ð v ¼ 0 Þ ¼1 : 94 rad = m ¼ 111 = m ¼ 178 = 100d :

The twist angle on the upper rope end is