Wire ropes

2
Wire ropes

- is made up of a number of strands laid helically about a metallic or non-metallic core.

- is an intricate device made up of a number of precise moving parts.

The moving parts of wire rope are designed and manufactured to maintain a definite
relationship with one another. This relationship ensures that the wire rope has the flexibility and
strength crucial to professional and safe hoisting operations.
Wire rope is stronger, lasts longer, and is much more resistant to abrasion than fiber line.
These factors, gave advantages for wire rope used for hoisting tasks that are too heavy for fiber
line to handle. Many of the movable components on hoisting devices and attachments are moved
by wire rope.

Figure 2-1: Lays

of Wire rope

(strands are laid to the right)

(strands are laid to the left)

(strands are wires are laid to the right)

(strands and wires are laid to the left)

( wires are laid in alternative directions,

Wire rope materials: but strands are laid to the right)

1. iron
o iron wire rope is largely used for low strength applications (elevator ropes, stationary
guy ropes, ropes not used for hoisting, etc.)

2. phosphor bronze
o used occasionally for elevator governor cable rope and other marine applications.

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 Wire ropes

3. Traction steel
o used primarily as a hoist rope for passenger and freight elevators.

4. Galvanized carbon steel

o used for small cords.

5. Polyvinyl plastic core

o common used where applications is exposed to moisture.

6. Plow steel
7. Improved Plow Steel (IPS)
8. Bridge rope steel

Classification of Wire ropes

1. 6 x 7 – used where resistance to wear is high, as in dragging operation.

2. 6 x 19 – most common and widely used class of wire rope for general hoisting
applications.
3. 6 x 37 - is flexible, making it suitable for cranes and similar equipment where sheaves are
smaller than usual.
4. 8 x 19 – due to its flexibility, used for high speed operation with reverse bends.

Wire rope is commonly designed by two figures, the first indicating the number of strands and
the second, the number of wire ropes per strands.
Example: 6x19 – having 6 strands with 19 wires in each strand

Figure 2-2: Strands and

Wire rope construction

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 Wire ropes

Figure 2-3: Wire

Core construction

Lays of Wire rope

The term lay refers to the direction of the twist of the wires in a strand and the direction
that the strands are laid in the rope ( see Fig. 2-1).
In some instances, both the wires in the strand and the strands in the rope are laid in the
same direction; and in other instances, the wires are laid in one direction and the strands are laid
in the opposite direction, depending on the intended use of the rope.
Most manufacturers specify the types and lays of wire rope to be used on their piece of
equipment. Be sure and consult the operator’s manual for proper application.

Type lays used in Wire ropes:

1. Right Regular Lay: the wires in the strands are laid to the left, while the strands are laid to
the right to form the wire rope.

2. Left Regular Lay: the wires in the strands are laid to the right, while the strands are laid to
the left to form the wire rope. In this lay, each step of fabrication is exactly opposite from
the right regular lay.

3. Right Lang Lay: the wires in the strands and the strands in the rope are laid in the same
direction; in this instance, the lay is to the right.

4. Left Lang Lay: the wires in the strands and the strands in the rope are also laid in the
same direction; in this instance, the lay is to the left (rather than to the right as in the right
lang lay).

5. Reverse Lay: the wires in one strand are laid to the right, the wires in the nearby strand
are laid to the left, and the wires in the next strand are laid to the right, and so forth, with
alternate directions from one strand to the other. Then all strands are laid to the right.

Factors affecting wire ropes selection

1. Tensile Strength
Tensile strength is the strength necessary to withstand a certain maximum load applied to the
rope. It includes a reserve of strength measured in a so-called factor of safety.

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 Wire ropes

2. Crushing Strength
Crushing strength is the strength necessary to resist the compressive and squeezing forces that
distort the cross section of a wire rope, as it runs over sheaves, rollers, and hoist drums when under a
heavy load. Regular lay rope distorts less in these situations than lang lay.

3. Fatigue Resistance
Fatigue resistance is the ability to withstand the constant bending and flexing of wire rope that
runs continuously on sheaves and hoist drums. Fatigue resistance is important when the wire rope must be
run at high speeds. Such constant and rapid bending of the rope can break individual wires in the strands.
Lang lay ropes are best for service requiring high fatigue resistance. Ropes with smaller wires around the
outside of their strands also have greater fatigue resistance, since these strands are more flexible.

4. Abrasion Resistance
Abrasion resistance is the ability to withstand the gradual wearing away of the outer metal, as the
rope runs across sheaves and hoist drums. The rate of abrasion depends mainly on the load carried by the
rope and the running speed. Generally, abrasion resistance in a rope depends on the type of metal that the
rope is made of and the size of the individual outer wires. Wire rope made of the harder steels, such as
improved plow steel, has considerable resistance to abrasion. Ropes that have larger wires forming the
outside of their strands are more resistant to wear than ropes having smaller wires that wear away more
quickly.

5. Corrosion Resistance
Corrosion resistance is the ability to withstand the dissolution of the wire metal that results from
chemical attack by moisture in the atmosphere or elsewhere in the working environment. Ropes that are
put to static work, such as guy wires, maybe protected from corrosive elements by paint or other special
dressings. Wire rope may also be galvanized for corrosion protection. Most wire ropes used in crane
operations must rely on their lubricating dressing to double as corrosion preventive.

Wire rope equations:

Safe working load, SWL

o is used to define the load which can be applied that allows the rope to provide
efficient service and also prolong the life of the rope.

SWL = D x D x 8 , in tons eq. (2-1)

where: D – wire rope diameter, inch

Sample problem:

1. The wire rope is 1/2 inch in diameter. Compute the SWL of the rope, kgs?

 Using eq. (2-1), with rope diameter of ½ -inch, SWL in tons is,

 SWL = (0.5)(0.5)(8) = 2 tons or 2000 kgs

2. Compute the SWL of a 2 – inch wire rope in tons?

 Using eq. (2-1), the SWL is equal to 32 tons

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 Wire ropes

Note: as rule of thumb, for old rope, worn out rope, or rope that is poor condition reduce the
value of SWL to at least 50% depending on the condition of the rope.

3. A six-year-old wire rope frequently used on a gate-lifting device located in a warm, humid
freshwater environment is inspected. The inspection reveals that the rope is near failing from
corrosion as there are many large corrosion pits and rusty areas. The inspection also indicates
fatigue is a problem as many broken wires are present. There is no indication of abrasion.
Look for the possible solutions to recover the purpose of the rope. Other data are as follows:
a. Drum/sheave arrangement = 10 part arrangement
b. Sheave pitch diameter = 45 inches.
c. Sheave bearings = Plain.
d. Maximum load (gate weight, friction draw down, etc) = 132 tons
e. Rope = 1 ½ -in. 6 x 30 flattened strand, right regular lay, improved plow steel
(uncoated), Independent Wire Rope Core (IWRC)
f. Dynamic rope tension = 15.6% of the load for a lift mechanism
g. Nominal strength of existing rope = 119 tons
h. D/d ratio = 30:1 for bending over sheaves
i. Rope efficiency = 95%

Dynamic rope tension = 0.156 (132 tons) = 20.59 tons eq. (2-2)

Factor of safety when new = eq. (2-3)

 Using eq. (2-3), its factor of safety is,

 FS = 5.5
 from above condition, we may thought of the replacement of the rope of similar
construction but different in material used. (stainless steel instead of IPS)
 “The stainless steels are many times more corrosion resistant than the carbon.”
 It is expected of 10 – 15% loss in strength if stainless steel will be used. At 15%
strength reduction on nominal strength,
 FS = = 4.66 (stainless steel), compared to 5.5 FS for carbon, which is
lower than required
 from above situation, broken wires are sign of fatigue on the materials, therefore
changing from carbon to stainless would not reduce fatigue.
 Given the same load, fatigue could be reduced by changing the rope construction to
lang lay.
 “Lang lay rope has better resistance to fatigue while it is equal in strength.”

Rigging/Hoisting Wire rope

Sheave size would affects wire rope strength. Severe
bending
is
a major
cause
of

short
rope life.
By contrast, the
larger
the
sheave
diameter is the
less
wear
on
the
rope
and
the greater
its
strength
efficiency.

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 Wire ropes

For hoisting blocks pitch diameter ratio is 16:1 (ANSI
B30.5 (Crawler, 
Locomotive and
Truck

Cranes) and
ANSI
B30.15
 (Mobile
Hydraulic
Cranes).
ANSI
B30.5
is requiredby
OSHA
regulations
as
printed
in

Federal
Register
on
June
24
and
27, 
1974)and,
For overhead cranes pitch diameter ratio is 24:1 (This P.D.
ratio is required
for
running
sheaves with
6
x
37
rope
by
CMAA Specification
No.
70 (Electric
Overhead
Traveling
Cranes).


P.D. ratio = eq. (2-4)

Mechanical Advantage, MA

MA = eq. (2-5)

MA = (N) x (E) eq. (2-6)

Efficiency = E = eq. (2-7)

where: W – load to be lifted

P – lead line to pull
N – no. of line parts supporting loads
E – rope efficiency
K – bearing constant (1.045 for bronze bushing: 1.02 for roller bearings)
S – total no. of sheaves in travelling block/top block/ boom point

FIGURE 2-4 Wire ropes

in block applications

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 Wire ropes

Example: Determine the lead line pull required to lift a 75 - ton block.

 Load (W) = 75 tons

 Line Parts (N) = 10
 With roller bearing sheaves, K = 1.02
 Using table (2-1), 10 parts of line, Mechanical Advantage = 9
 Using eq. (2-5), converting 75 tons to lbs.
 P = 16,666 lbs

Table 2-1 Mechanical

Advantage vs. Number of
Line parts supporting load

Example: Determine the load to be lifted of the rope if P is 10 tons; N = 12

 For roller bearing sheaves,

 Mechanical Advantage of 10.6
 W = P X M = 10 X 10.6 = 106 tons

Equivalent bending stress

Sb = eq. (2-8)

where: Fb – equivalent bending load due to curvature of the sheave/drum

Am – cross-sectional of metal

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 Wire ropes

Sb = eq. (2-9)

where: E – modulus of elasticity of material ( i.e. E for steel = 30 x 106 psi)

dw – wire rope diameter
ds – sheave or drum diameter

N= eq. (2-10)

N= eq. (2-11)

where: N – factor of safety

Fu – breaking strength of rope
Ft – tensile force due to load (acceleration forces etc.)

Design Factors for some common applications

Nmin = guy supports rope, 3.5

= miscellaneous hoisting rope, 5
= haulage ropes, cranes, derricks, 6
= small hoists, 7
= hot ladle crane, 8

Suggested N for wire ropes, based on ultimate strength:

Service Factor of safety (N)

Elevators 8 – 12
Mine hoists 2.5 – 5
Cranes, motor driven 4–6
Cranes, hand powered 3-5
Derricks 3-5

Deformation of wire rope

δ= = , in eq. (2-12)

where: S – bending stress

L – length of wire
Am – mean area wire rope
Er – modulus of elasticity wire rope materials (Table AT – 28), p. 605 (Faires, 1965)

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 Wire ropes

Pressure of the rope on the sheave, P

P= eq. (2-13)

P/Su = or eq. (2-14)

DrDs = eq. (2-15)

⁄

where: P/Su – fatigue failure ratio, ≤ 0.0015

Loads and efficiency of wire ropes

Example of determining the loads on various ropes of a hoist and its efficiency,

Figure 2-5 Raising

the Load, W

To represent its tensions,

T2 = CT1

T3 = CT2 = C2T1

T4 = CT3 = C3T1

T5 = CT4 = C4T1

F = CT5 = C5T1

where: C – rope size factor, Table 2-2, p.19

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 Wire ropes

In order to raise a load resistance, W

W = T1 + T2 + T3 + T4 + T5

W = T1( )
W = F( )

( )
F=

For lowering the load:

T1 = CT2

T2 = CT3

T3 = CT4

T4 = CT5

T5 = C(F)

F= ( )

Table 2-2 Values of

C for wire ropes

Dr C
3/8 1.090
7/16 1.083
½ 1.076
9/16 1.070
5/8 1.064
¾ 1.054
7/8 1.046
1 1.040

Sample problem:

1. A hoist is arranged below, in which a hydraulic cylinder applies a force W in order to

raise the load P = 1500 lb. Determine the diameter of the rope, dr required, assuming 6 x
19 steel rope with N = 3, sheave is 40(dr) and a coefficient C = 1.10. Neglect weight of
the rope. (Su = 280 ksi)

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 Wire ropes

F T2 T1

 T1+T2 = P =1500 lb; T2 = CT1; F = CT2, where, C = 1.10.

 T1 = 714.285 lb; T2 = 785. 714 lb; F = 864. 285 lb
 W ≈ 1650 lb (tensile load due to force)
 Applying eq. (2-15), where Ds = 40 Dr
 Dr = 0.767 ≈ ¾ in. (rope diameter)

2. A motor coupled to a sheave is used to lift a maximum load of 3000 lb from the depth of
1000 ft. The rope travels 20 fps attained in 5 sec. Determine the rope diameter of 6 x 19
rope, IPS sheave design factor of 1.3, and elongation of rope. (Su = 240 ksi)

 W = 3000 lb = 3 kips
 weight of the rope = wL = 1.6 dr2 lb/ft (1000 ft) = 1.6(dr2) kips, Table AT-28, p. 605,
Faires
 force on the rope to lift a load = Ft, summing all forces acting on the system,
 ∑ ∑F = Ft – W – wL = Ft – 3 – 1.6 dr2, m(a) = ,a= = = 4 ft/s2

 Ft – 3 – 1.6 dr2 = ( )(4)

2
 Ft = 3.373 + 1.799 dr
 Su = 240 ksi; N = 1.3; P/Su = 0.0015
 Using eq. (2-15), Ds = 24.361 + 12. 993 Dr; Ds = 45(Dr), (Table AT-28)
 45 Dr = 24.361 + 12.993 Dr, Dr = 0.76 in ≈ ¾ in.
 Bending stress of the rope, use eq. (2 - 9) Sb = 17, 866.67 psi; dw = 0.067 dr; ds = 45 dr;
Er = 12 x 106 psi; Am = 0.4dr2 (Table AT – 28)
 Elongation of the rope, use eq. (2-12), δ = 13.33 in

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 Wire ropes

Name:_______________________________________________Date:__________Score:_____

Problem Exercises: (to be submitted on May 1, 2013)

PROBLEM
A hoist is used to raise a load Q by a force P applied through a hydraulic cylinder. The ratio of
rope tension is equal to C > 1. Determine the value of P.

SOLUTION

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 Wire ropes

PROBLEM
A hoist is arranged as shown in Fig., in which a hydraulic cylinder applies a force P in order to
raise the load Q = 7500 lb. Determine the diameter D of the rope required, assuming 6 x 19 steel
rope, a factor of safety of 3, sheaves 45D in diameter, and C = 1.10. Weight of rope is negligible.

Su = 280 ksi

7500 lb

SOLUTION

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 Wire ropes

PROBLEM

Find the SWL of a 2 inches rope both for new and old rope.

SOLUTION

PROBLEM

The wire rope is 1/2 inch in diameter. Compute the SWL for the rope.

SOLUTION

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 Wire ropes

PROBLEM

A ¾ inch diameter rope is required to fill a drum having the following dimensions: B = 24 in, D
= 18 in, H = 30 in, and K = 0.74. Find the length of the rope.

SOLUTION

L = (A+d)*A*B*K eq. (2-16)

A = (H-d-2Y)/2 eq. (2-17)

where: A - depth of rope space on drum, in

B - width between drum flanges, in
D - diameter of drum barrel, in
H - diameter of drum flanges, in
K - wire rope factor
Y - depth no filled on drum

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 Wire ropes

PROBLEM

Find the bending stress and equivalent bending load due to the bending of a 6 x 19 wire rope 0.5
inch in diameter around a 24 inches pitch sheave diameter. Ans. Sb = 15.7 ksi; Pb = 1590 lbs

SOLUTION

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 Wire ropes

Assignment

Find machinery or device/machines that have wire rope application. Have a picture of it and
identify the details of the machine element. Submit this requirement(s) on ___________.
May 1, 2013

Notes

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 Wire ropes

DESIGN PROBLEM # 2

Name: Rating:
Course/Yr: Date:

WIRE ROPE DESIGN

PROBLEM

The pulley system shown below is used to raise a load of 10 000 lb with a maximum acceleration
of 10 ft/s2. Determine the diameter of the rope, assuming a factor of safety equal to 5, a
coefficient of 1.10, and sheave diameters of 20d, 30d, 40d, and 50d. Su = 280 ksi

Ft
a = 10 ft/s2

W=

SOLUTION

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 Wire ropes

DESIGN PROBLEM # 2 continuation

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