WIRE ROPES
Find the total tensile load when a 1” – 6 x 19 MPS wire rope carrying a load of
3 metric tons is bent around an 1140 mm diameter sheave. The rope weighs
300 lb. Acceleration of the load is about 2 fps^2.
Solution:
a
F T = ( W L + W R )( 1 + )
g
2205
W L= 3 ( ) = 6615 lbs
1
2
=( 6615 + 300) ( 1 + ) = 7349.5 lbs
32.2
Calculate the equivalent bending load of rope from the above problem. Use
the following proportions: Dw = 0.067 Dr; Ds = 30Dr; Am = 0.4Dr^2 where Dr
= rope diameter, Dw = wire diameter, Ds = sheave diameter and Am =
metallic cross section area.
SOLUTION :
Dw
FB = D
Ds
0.067 Dr
( 30 x 104 psi ) ( )(0.4 Dr 3 ¿ = 26800 lbs
Dw
Find the static factor of safety of the wire rope in the above problem if its
breaking strength is 31.6 tons.
SOLOTION :
FU −F
N= B
FT
(32.6)(2000)−26800
= = 4.967
7329
Find the fatigue approach factor of safety of the wire rope in the above
problem if its ultimate strength is 200 ksi.
SOLUTION :
P
()(Dr)(Ds)(Sn)
N = Su
2 Ft
(0.0015)(30 x 1)(1)(200000)
= = 0.61
2(7329)
For a mine hoist cage weigh 5900 lb, the cars 2100 lb and the load of coal in
the car is 2800 lb one car loaded at a time on the hoist. The maximum depth
is 1500 ft. It take 6 seconds to accelerate the cage to 3285 fpm. Decide on a
grade of wire, kind and size of rope on the basis of life 2×10 ^5 cycles and N
=1.3 against fatigue failure.
SOLUTION :
W H = 5900 + 2100 + 2800
= 10 800 lb
ft min
V 3285( )( ) ft
A= = min 60 sec = 9.125 2
t sec
6 sec
F = ma
wL+ wH
F t=(wL −wH )= (a)
32.2
a
F t=(wL+wH )( +1)
g
Using 6 x 19 wire rope for general purpose
From table AT 28
lb
W = 1.6 Dr 2
ft
Ds = 45 Dr in.
Ft = ( 1.6 Dr 2 (1500)+(10800)¿ ¿ + 1 )
=3080 Dr 2 + 13860.56 lb
2 NFt
DrDs =
( p/ Su)Su
From Figure 17,30 for 6 x 19 wire rope with 2 x 105 cycles
P
=0.0028
Su
2(1.4)(3080+13860)
Dr(45Dr) =
0.0028 Su
36037
Dr 2 =
0.126 Su − 8008.312
For PLOW steel
2 36037
Dr = = Dr = 1.33 in.
0.126(22500)−8008.312
A traction elevator carries a maximum weight of 50 KN has an acceleration of
1 m/sec ^2, the 4 cables pass over the upper sheave 4 times and the lower
sheave twice. Compute the maximum weight of counterweight to prevent
slipping om the driving sheave, f= 0.07
SOLUTION :
w
F = ma = a
g
a
F - W E=W E
g
1
=(1+ ¿
9.81
= 500N
a
W cw (1− )=F 2
g
F1 f ∅
=e
F2
o
∅ =(2+3)180
o π
= 900 ( o
rad )
180
55.1
F 2= 0..07 (5 π )
e
= 18.35 N
18.35
w CW =
1
1−
9.81
W CW =20.43 N
The wire rope of mine hoist handles a total load of 12 kips. It lift coal at
maximum depth 1500 ft and take 6 seconds to accelerate the load cage to
200 fpm. The diameter of drum if 5 ft. Determine the Su.
SOLUTION :
Ds = 60 in.
P/Su = 0.0028
Ds = 450r
Dr = 1.33 in
1 2 5.56
Ft = ( 2.4 (1 ¿ + 12¿ ( +1)=18.477 kips
4 32.2
2 NFt
Su =
DrDs( p /Su)
Su = 228.68 ksi
A traction elevator carries a maximum weight of 36 KN has an acceleration of
1 m/sec ^2, the 4 cables pass over the upper sheave 4 times and the lower
sheave twice. Compute the maximum weight of counterweight to prevent
slipping om the driving sheave, f= 0.07
SOLUTION :
w
F = ma = a
g
a
Cw - F = cw
g
Find F :
F1 f ∅
=e
F2
o o
∅ =5(180 )=900
1
F 1 −36000=336000
9.81
= 39668 N
39669 N 0.07 (5 π )
=e
F2
= 13210 N
F2
CW= a
1−
g
CW = 14 710 N
A wire rope lift a load of 10 kips at a maximum speed 1200 fpm attained in 10
seconds starting from rest. The rope has a metallic cross sectional area of 0.4
in². Wire diameter of 0.067 in modulus elasticity of 12 × 10⁶ psi and ultimate
load of 76 kips. The sheave has a diameter of 3 ft. Neglecting weight of rope
and determine the factor of safety.
SOLUTION :
F U − Fb 36000− 8933
N= N= = 8.2
Ft 10 621
EDw
Fb = Sb Am = Am
Ds
= 8933 lb
1200/60 −0
a= = 2 ft/sec 2
10
a
Ft = W( 1 + ¿
g
2
= 10000 ( 1 + )
32.2
=10 621 lb
