Wire Ropes Under Bending and Tensile Stresses - Wire Ropes: Tension, Endurance, Reliability - page 302

! p

k ¼ X with X

n n

u i k i u i ¼ 1

i¼1 i¼1

k ¼ 0 : 49 :

Example 3.10 Crane according to Fig. 3.81

Calculation with Seilleb2.xls

Number of hoisting cycles, no more than
10 % of the ropes have to be discarded Z A10

Rope class WS

Number of strands 6

Rope core FC

Nominal load Q = 5,000 kg

Hook, bottom sheave and rope mass U = 100 kg

Diameter of drum and sheave D = 400 mm

Nominal rope diameter d = 16 mm

R = 1,960 N/mm 2

Nominal rope strength

Effective rope bending length l = 2,000 mm

Rel. force difference (S - S u )/S = 0,95

Number of bearing rope traces n T = 2

Force factor for efficiency f S2 = 1.02

Force factor for ac- and deceleration (small, because of the different
start-positions)

f S4 = 1.04

Endurance factor for side deflection f N = 0.94

Collective factor k = 0.51

Analyse
For nominal load and for load collective and

Nominal bendings per hoisting eff. bendings per hoisting

w com ; 0 ¼ 1 w com ¼ w com ; 0 ¼ 1

w sim ; 0 ¼ 2

w sim ¼ w 0 : 7

sim ; 0 ¼ 1.62

w rev ; 0 ¼ 2

w rev ¼ w 0 : 7

rev ; 0 ¼ 1 : 62

Nominal rope tensile force rope collective force

S N ¼ ð Q þ U Þ g
1000 n T

S ¼ k S N

f S2 f S4

S N ¼ ð 5000 þ 100 Þ 9 : 81
1000 2

1 : 02 1 : 04

S N ¼ 26 : 54 kN S ¼ 0 : 51 26 : 54 ¼ 13 : 53 kN

The five dimensioning limits
For the nominal tensile force
1. Rope safety factor m ¼ 165 : 6 = 26 : 54 ¼ 6 : 2 [ 2 : 5

2. Donandt force S D1 = 65.4 [ 26.54 kN

For the collective tensile force
3. Number of hoisting cycles Z A10 ¼ 0 : 94 24300 ¼ 22800

4. Optimal rope diameter d opt = 17.5 mm [ 16 mm

5. Discarding number of wire breaks B A30 =16[ 8