Wire Ropes Under Tensile Load - Wire Ropes: Tension, Endurance, Reliability - page 120

x ð u max Þ ¼ c 2 S 0 þ c 3 G d 2
c 2

L

þ : ð 2 : 89 Þ

m g sin b F c 2 m g L sin b F

c 2 S 0 þ c 3 G d 2 þ 1

ln

The maximum rotary angle is given with x = x(u max ) in ( 2.88 ). The maximum
twist angle x max occurs on the lower rope end, v = 0. It will be calculated with
( 2.85 ) for v = 0 and the torque out of ( 2.87 ).
For the practical calculation of the rotary angle u and the twist angle x, the
Excel-program SEILDRE2.XLS can be used.

2.4.4.2 Wire Rope Supported Non-rotated at Both Ends, Simplified
Calculation

The torque M in the wire rope supported non-rotated at both ends can be set
simplified with only a small failure

M ¼ c 1 d ð S 0 þ m g L = 2 sin b F Þ: ð 2 : 90 Þ

Then with ( 2.85 ) the twist angle is

c 1 m g ð L = 2 x Þ sin b F
c 2 d ð S 0 þ m g x sin b F Þþ c 3 G d 3 :

x ¼ ð 2 : 91 Þ

With that on the upper rope end the twist angle is

c 1 m g L = 2 sin b F
c 2 d ð S 0 þ m g L sin b F Þþ c 3 G d 3

x upper ¼ ð 2 : 91a Þ

and on the lower rope end

x lower ¼ c 1 m g L = 2 sin b F

c 2 d S 0 þ c 3 G d 3 :

ð 2 : 91b Þ

For the integration to evaluate the rotary angle u, the denominator of ( 2.91 ) can be
further simplified with x = L/2. The failure for that is very small if the rope weight
force mgL sinb F is smaller than the rope tensile force S 0 . Then the twist angle is

c 1 m g ð L = 2 Þ sin b F
c 2 d ð S 0 þ m g L = 2 sin b F Þþ c 3 G d 3

x ¼ ð 2 : 92 Þ

and after integration the rotary angle u is