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MODULE 2: DESIGN OF HOISTING MECHANISM

 INTRODUCTION TO CRANES
For lifting and conveying material within the unit of an industry, an overhead crane or a bridge crane
is used. An overhead crane consists of snatch block, trolley, trolley travelling mechanism and cross
travel mechanism to cover an entire area of unit. A hoist of a crane, travels along the bridge. An
overhead cranes are used for either manufacturing or maintenance applications. These cranes are
either Human operated or remote operated cranes.

 APPLICATIONS:
In manufacturing plants at every process the material is handled by crane till finished product leaves
a factory. For pouring Raw materials into a furnace, for rolling hot metal to specific thickness, for
tempering, annealing and for storing purpose an overhead crane is used. For lifting finished product
and loading in truck or train an overhead crane is used. Many industries including automobile uses an
overhead crane to handle the steel, raw material and finished product in the factory. Small cranes,
such as jib cranes handle lighter loads in a work area, such as CNC mill or saw. In the refinement plants
of metals like steel, copper, aluminum etc.
For regular maintenance in paper mills like removal of heavy press rolls and other equipment bridge
cranes are required. For installing drying drums and other massive equipment, the bridge cranes are
used.

 CONFIGURATIONS:
Based on applications, Overhead cranes are manufactured in a number of configurations. Some are
mentioned below.

1. EOT Crane (Electric Overhead Traveling Crane)

These cranes are electrically operated by a control pendant, radio/IR remote pendant or from an
operator cabin attached with the crane itself. In most factories EOT crane is used which is most
common type of overhead crane.

2. Rotary overhead crane

Rotary overhead crane has one end of the bridge mounted on a fixed pivot while the other end
carried on an annular track; the bridge traverses the circular area beneath. This crane is used for
longer reach and to eliminate lateral strains on the walls.

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 HISTORY:
In 1876 Sampson Moore in England designed and supplied the first electric overhead crane to hoist
guns at the Royal Arsenal in Woolwich, London. This crane was in service till 1980, and is now in a
museum in Birmingham, Alabama. Over the years, important innovations, such as the Weston load
brake and the wire rope hoist were implemented. The original hoist contained components mated
together in the built-up style hoist. They also provide for easier maintenance. Now many hoists are
package hoists, built as one unit in a single housing, generally designed for ten-year life, but the life
calculation is based on an industry standard. The true life calculation is based on load and hours used.

Figure 1 : Example of steam powered overhead crane from 1875, produced by Stuckenholz AG,
Wetter an der Ruhr, Germany. Design developed by Rudolf Bredt from an original installation at Crewe

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Figure 2 : Electrically operated overhead travelling crane

 INTRODUCTION:

Hoisting is the process of lifting and lowering some material or load or person from lower position to
higher position with the help of some device or mechanism.

 Hoisting Devices
A hoisting device is used for lifting or lowering a load by means of a drum or lift-wheel with the help of
rope or chain. It may be manually operated, electrically or pneumatically driven and may use chain, fibre
or wire rope as its lifting medium. Examples: Elevators, crane etc. The hoisting part of the EOT crane
consists of the following parts, 1. Hoist motor, 2.Gear box, 3. Drum, 4. Pulleys, 5. Wire rope, 6. Hook

A hoist motor is used as a driving system for the mechanism. The motor is coupled to a gearbox. The
gear box is coupled to the rope drum. The rope is wounded on the rope drum. The pulleys are arranged
with some rope falls. At the bottom of the pulley the hook is attached with the help of a thrust bearing.
Following figure show the schematic block diagram for the hoisting mechanism with four fall system.

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Figure 3 : Schematic view of the hoisting Mechanism

 Requirement of Hoisting Mechanism

1. Kind and properties of loads to be handled: For unit loads – their form, weight, convenient bearing
surface or parts by which they can be suspended, brittleness, temperature, etc. for bulk loads – lump
size, tendency to cake, volume weight, friability and the amount of crumbling liable to occur during
shipments, temperature, chemical properties, etc.
2. Required hourly capacity of unit: A practically unlimited hourly load moving capacity can be easily
obtained with certain types of devices as with some continuous-action conveyors. On the other hand,
there are devices such as power driven trucks or overhead travelling cranes following a definite cycle
of movements with a return idle run.
3. Direction and length of travel: Various types of devices can carry loads in a horizontal or vertical
direction or at an angle to the horizon. Some devices can easily negotiate track curves while others
move only rectilinearly, in one direction.

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4. Methods of stacking loads at the initial, final and intermediate points: Loading onto vehicles and
unloading at their destination differ considerably because some handling machines can be loaded
mechanically while others require special auxiliary fixture or manual power.
5. Characteristics of production processes involved in moving loads: This most important factor
essentially influences the choice of the type of transporting facility. As a rule, the movements of the
materials handling equipment are closely linked with and depend on manufacturing process;
sometimes these movements may even be directly involved in the performance of certain processing
operation.
6. Specific local conditions: It includes the size and shape of the area, type and design of the building,
ground relief, possible arrangement of the processing units, dust or humidity conditions in the
premises, temperatures etc.

 Type of Hoisting Mechanisms

a. Hoisting Machines
Hand Trolley Hoists, Portable power-operated hoists, travelling power-operated hoists,
Winches, Crane Trolleys
b. Cranes
Stationary Rotary Cranes, Cranes travelling on guide rails, Trackless Cranes, Bridge type Cranes,
Cable Cranes
c. Elevators
Portable air-operated hoists, Manually Propelled Stackers, Vertical skip elevators, Mast-type
elevators, Funiculars.

 Snatch Block Assembly

A snatch block is essentially a pulling block assembly which is used specifically to increase the load
pulling capacity of a winch. A Snatch Block is a pulley system to aid with winching. The unit is
essentially a pulley block assembly which opens to allow the easy connection of a looped rope or cable
rather than having the lengthy task of threading the cable. The block itself or 'sheave' is a wheel with
a grooved edge which carries the rope or cable. The side plates or 'cheeks' house the wheel assembly.
A snatch block can effectively allow double the line which in turn doubles the lifting capacity of the
snatch block and winch arrangement. A single snatch block or multiple snatch blocks can be deployed
in conjunction with a winch to maximize lifting load capacity. Figure shows the snatch block for four
fall system. It consist of two movable pulleys, pulley bearing , pulley axle, side plate, shackle plate,
crane Hook, cross piece, thrust bearing, washer, nut, spacer etc.

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Figure 4 : Snatch Block

 Pulleys

Pulleys are manufactured in fixed and movable designs. Pulleys with fixed axles are also guiding
because they change the direction of flexible hoisting appliance. A rope and pulley system that is, a
block and tackle is characterized by the use of a single continuous rope to transmit a tension force
around one or more pulleys to lift or move a load the rope may be a light line or a strong cable. This
system is included in the list of simple machines identified by Renaissance scientists.
If the rope and pulley system does not dissipate or store energy, then its mechanical advantage is the
number of parts of the rope that act on the load. This can be shown as follows.
Consider the set of pulleys that form the moving block and the parts of the rope that support this
block. If there are p of these parts of the rope supporting the load W, then a force balance on the
moving block shows that the tension in each of the parts of the rope must be W/p. This means the
input force on the rope is T=W/p. Thus, the block and tackle reduces the input force by the factor p.

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Different types of pulley systems are as follow

 Fixed: A fixed pulley has an axle mounted in bearings attached to a supporting structure. A
fixed pulley changes the direction of the force on a rope or belt that moves along its
circumference. Mechanical advantage is gained by combining a fixed pulley with a movable
pulley or another fixed pulley of a different diameter.
 Movable: A movable pulley has an axle in a movable block. A single movable pulley is
supported by two parts of the same rope and has a mechanical advantage of two.
 Compound: A combination of fixed and some movable pulleys forms a block and tackle. A
block and tackle can have several pulleys mounted on the fixed and moving axles, further
increasing the mechanical advantage.

Figure 5: (a) The gun tackle “ rove to advantage” has the rope attached to the moving pulley. The
tension in the rope is W/3 yielding an advantage of three. (b) The Luff tackle add a fixed pulley “ rove to
disadvantage.” The tension in the rope remains W/3 yielding an advantage of three.

Figure 6: Different pulley system with power weight relation and mechanical advantage

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Pulley Systems:
A pulley system is a combination of several moveable and fixed pulleys or sheaves. There are
system for gain in speed and far gain in force. Hoisting devices employ pulley far a gain in
force predominantly and only rarely, as for example, in hydraulic or pneumatic lifts, pulleys
far gain in speed. As independent lifting appliances pulley system are of secondary
importance; they are mainly used for power transmission in inches and cranes.

 Fixed Pulleys
 One end of rope passing around the pulley is loaded with weight and the other with pulling
force. The path of pulling force is equal to height to which load is raised. Disregarding the
resistance in the pulley, the pulling force equals weight.

 Movable Pulley
 These pulley have movable axles to which either a load or force (effort) is applied. Accordingly,
there are pulley for a gain in force and pulley for a gain speed. The efficiency of a movable
pulley is higher than of a fixed pulley.

Figure 7 : Movable Pulley

 Multiple Pulley Systems:

The following shortcomings due to direct suspension of loads from the rope end or from employing
simple pulleys for a gain in force in hoisting appliances can be pointed out.

1. The rope parts in one plane and this may cause the load to sway
2. Large diameter of ropes and pulleys
3. The load being lifted moves in a horizontal direction because a rope coiling on a drum moves
along its length.

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These shortcoming can be avoided, especially in the hoisting mechanisms of winches and cranes with an
electric drive, by using multiple pulley systems which raise the load in strictly vertical direction and keep
it more stable. These systems carry the load with twice as many parts as a similar pulley system.

Figure 8 : Pulley system with gain in force

In crane design, always multiple pulley system are used. Multiple pulley system is also known as multi-
fall system. It is done to get mechanical advantage and to reduce the load per strand of rope. It enables
us to use small cross-section of rope. This in turn reduces the size of pulley and cost of pulley. So that for
assembly we can go for light weight construction.

Following figure shows the types of multifall system.

Figure 9 :a) 2 fall system, b) 4 fall system with 3 bends, c) 4 fall system with 4 bends

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 Efficiency of Pulley system:

Consider a simple pulley system as shown below.

T1 T2 T3 T4 T5 T6

Let Q is a load to be lifted with efforts Z and T1, T2, T3, T4, T5, T6 are the tensions in the rope shown.

Now, T1 + T2 + T3 + T4 + T5 + T6 = Q

If ε be the transmission ratio, then , T5 = T6 X ε, Similarly,

T4= T5 X ε = T6 X ε2

T 3 = T 6 X ε3

T 2 = T 6 X ε4

T1 = T6 X ε5 , Hence

ε6 −1
T6 [ ε5 + ε4 + ε3 + ε2 + ε + 1 ] = Q, 𝑇6 [ ε−1 ] = 𝑄

ε−1 ε−1
𝑇6 = 𝑄 [ ] , 𝑍 = 𝑇1 𝑥 ε = 𝑇6 x ε6 ; 𝑍 = 𝑄. ε6 [ ε6 −1]
ε6 −1

𝑄
Now Ideal effort 𝑍0 = 6
;

𝑍0 𝑄 ε6 −1 ε6 −1
Efficiency of the pulley, ƞ = 𝑍
= 6 . 𝑄.ε6 [ε−1] = 6.ε6 [ε−1]

Hence general equation for the efficiency of n number of pulleys,

𝛆𝒏 −𝟏
ƞ=
𝒏.𝛆𝒏 .[𝛆−𝟏]

There are different fall system used for different loads which is shown in table 1.

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Table 1: fall system corresponding to different loads and efficiency of pulley system

Load to be Lifted No. of Fall System Efficiency of Pulley

Up to 40 KN 2 0.97
40-200 4 0.95
200-250 6 0.92
250-750 8 0.90
750-1000 10 0.85
1000-1200 12 0.8

Figure 10 : Pulley/Sheave

 SHEAVE MAINTENANCE:

Examine the sheave grooves for wear and proper diameter. To check the size, contour and amount of
wear, use a sheave gage. The gage should contact the groove for about 150o of arc. Inspect the fleet angle
for poor sheave alignment. The fleet angle is the side, or included, angle between a line drawn through
the middle of a sheave and a drum, perpendicular to the axis of each, and a line drawn from the
intersection of the drum and its flange to the base of the groove in the sheave. The intersection of the
drum and its flange represents the farthest position to which the rope can travel across the drum. There
are left and right angles, measured to the left or right of the centre line of the sheave, respectively. It is
important to maintain a proper fleet angle on installations where wire rope passes over a lead sheave and
onto a drum. A fleet angle larger than recommended limits can result in excessive rubbing of the rope
against the flanges of the sheave groove, or crushing and abrasion of the rope on the drum.

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This angle, for maximum efficiency and service, should not be more than 1-1/2° for a smooth drum,
nor more than 2° if the drum is grooved. The minimum angle which ensures that the rope will cross
back and start a second layer in a normal manner, without mechanical assistance, should be 0° 30
minutes. For smooth faced drums, this works out to a distance of 38 feet for each foot (76 feet for
two feet) of side travel from the center line of the sheaves to the flange of the drum. For a grooved
drum, the distance is 29 feet.

Figure 11 : Fleet angle

 SYNTHETIC SHEAVES:

When using synthetic sheaves or synthetic-lined steel sheaves, the inspector must carefully examine the
rope for diameter reduction or lengthening of lay, even if no visible damage is observed. Synthetic sheaves
greatly increase the contact area between the wire rope and sheave, by cushioning the rope. This
cushioning effect causes wire rope to wear internally (wire rope operating on steel sheaves will first wear
externally) before the damage is noted on the outer wires. This situation places the inspector at a great
disadvantage; therefore, he/she must be diligent in the detection of diameter reduction and lay
lengthening to prevent catastrophic failure from internal core damage.

 Bend

Bend is considered as a point where there is relative motion between the rope and the pulley and where
rope either moves over the pulley or leaves the pulley. A bend is considered as double bend where
direction of pulley changes and rope undergoes complete reversal of spaces. Life of rope is always
depends on number of bends. Hence select system with minimum number of bends so that life can be
increased. In order to reduce the stresses in the rope it would be necessary to increase the diameter of
𝑫𝒑
pulley. Hence as number of bends increases, 𝑫𝒓
ratio increases.

Individual wire in loaded bend Rope experiences a complex stresses consisting of tension, bending,
twisting stresses combined with material compression and rubbing of the wires and strands. As a result
the total stress can be determined analytically only to a certain degree of approximation. As they run over

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the pulleys and drums outer wire are subjected to abrasion which in turn reduces the total strength of the
rope. Experiments have shown that the life of rope is greatly affected by fatigue. It has been found that
each rope can withstand during its life only definite number of bends after which rapid disintegration sets
in. Depending on the number of bends life can be found from the ratio Dmin/d. Investigation have shown
that at the same Dmin/d ratio, the rope life is approximately inversely proportional to the number of bends.
One bend is assumed to mean the transition of the rope from its straight position to bent one or from a
bent position into a straight one. Reverse bending reduces the life approximately by one half or it is equal
to two single bends towards the same side. In determining the number of bends for multiple pulleys, the
compensating Pulley is not considered since it remains stationary when load is being raised or lowered.
To obtain the same rope life the effect of number of bends should be compensated for by an approximate
change in the ratio Dmin/d. Following figure show the bend measurement for two cases. 1. Bend toward
same side, 2. Reverse bend.

Example for Bend Measurement

 Compensating Pulley

Compensating pulley is located at the centre of the system and in normal course, compensating pulleys
do not rotate. If for lifting purpose, the hook assembly is pulled on one side then only the compensating
pulley will rotate and adjust the length of rope. Sometimes for higher heights, the weight of the ropes in
the suspension will cause an imbalance on the driving mechanisms. The ropes over the compensating
pulley help in maintaining appropriate balance. Compensating pulley diameter can be considered as 60
percent of the movable pulley diameter.

 Wire Rope

A hoisting device use chain, fibre or wire rope as its lifting medium. Wire rope consists of several strands
laid (or ‘twisted’) together like a helix. Each strand is likewise made of metal wires laid together like a
helix. Abrasion resistance increases with fewer, larger outside wires per strand and fatigue resistance
increase with more outside smaller wires per strand.

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 Advantage of wire rope over chain

Steel wire ropes are extensively used in hoisting machinery as flexible lifting appliances. As compare to
chains they have the following advantages:

 Lighter weight.
 Less susceptibility to damage from jerk.
 Silent operation even at high working speed.
 Greater reliability in operation.

 Requirements of rope
Depending on where they are used, wire ropes have to fulfil different requirements:

1. Running rope is bent over sheaves and drums. They are therefore stressed mainly by bending and
second by tension.
2. Stationary ropes, stay ropes have to carry tensile forces and therefore mainly loaded by static and
fluctuating tensile stresses. Ropes used for suspension are often called cables.
3. Track ropes have to act as rails for the rollers of cabins or others loads in aerial ropeways and
cable cranes. In contrast to running ropes, track ropes do not take on the curvature of the rollers.
Under the roller force, a so called free bending radius of the rope occurs. This radius increases
with the tensile force and decreases with the roller force.
4. Standard ropes are used to harness various kinds of goods. These slings are stressed by the tensile
forces but first of all by bending stresses when bent over the more or less sharp edges of the
goods.

Wire rope are manufactured from steel wire with an ultimate strength of σ = 1600 to 2000 N/mm2. In the
process of manufacturing the wire is subjected to special heat treatment which, combined with cold
drawing, imparts high mechanical properties to the wire.

A rope is a group of yarns, plies, or strands that are twisted or braided together into a larger and stronger
form. Ropes have tensile strength and so can be used for dragging and lifting, but are too flexible to
provide compressive strength. As a result, they cannot be used for pushing or similar compressive
applications. Rope is thicker and stronger than similarly constructed cord, line, string, and twine. Ropes
made from metal strands are called wire rope as shown in figure 12 .

Figure 12 : Rope construction

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 Construction of wire rope :

Rope may be constructed of any long, stringy, fibrous material, but generally is constructed of
certain natural or synthetic fibres. Material for steel wire ropes are generally high carbon steel. Most of
the wire ropes are made up of IPS (improved plaw steel) with ultimate tensile strength (σut) of 2000
N/mm2, Several other grades are also used, plaw steel (σut = 1600 N/mm2) and mild plaw steel (σut = 1200
N/mm2). Steel wires are manufactured by special machines. Initially separate wires are twisted to form
the strands and strands are again twisted to form rope.

construction of ropes is indicated by two numbers, 6×7, 6×19, 6×37 where 6 for No. of strands and 7,19,37
for No. of wires in each strands. More the number of wires in each strand, more the flexible the rope will
be. If number of wire is less the rope is stiffer.

6×7 > rope is made up of heavy wires and provides maximum resistance to wear and vibration.

6×19 > good compromise between flexibility and wear. It is most popular and widely used.

6×37 > Extra flexible are used where abrasion and wear are not very severe. Relatively sharp bends can
be tolerated.

The constructions of wire rope are shown in Figure 13 (a) and (b). The wire rope consists of a number of
strands, each strand comprising several steel wires. The number of wire in each strand is generally 7, 19
or 37, while the number of strands is usually six. The individual wires are first twisted into the strand and
then the strands are twisted around a fibre or steel core.

Figure 13: Wire Rope and its strands

The specifications of wire rope include two numbers, such as 6x7 or 6x19. The first number indicates the
number of strand in the wire rope, while the second gives the number of steel wires in each strand. The
popular constructions of steel wire ropes are as follows:

6x7 (6/1), 6x19 (12/6/1), 6x37 (18/12/6/1)

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The central portion of the rope is called core. There are three types of cores-fibre, wire and synthetic
materials. The fibre core consists of natural fibre like sisal, hemp, jute or cotton. The fibre core is flexible
and suitable for all conditions except when the rope is subjected to severe crushing.

The lay of the rope refers to the manner in which the wires are helically laid into strands and strand into
the rope. If the wires in the strand are twisted in the same directions as the strands, then the rope is called
a Lang’s lay rope. When the wire in the strand is twisted in directions opposite to that of strands, the rope
is said to be regular–lay or ordinary-lay. There are mainly three types of ropes parallel, composite and
crossed. The lays of wire rope is shown in the fig. (a), (b) and (c).

Figure 14 : (a) parallel, (b) cross, (c) Composite Lays of wire Rope

 Selection of steel wire ropes:

Extremely complex phenomena are involved in the operation of ropes which in some parts are
indeterminate. Individual wire in a loaded bent rope experience a complex stress consisting of tension,
bending and twisting stresses combined with mutual compression and rubbing of wires and strands. As a
result, the total stress can be determined analytically only to a certain degree of approximation. Besides
as they run over the pulleys and drums, the outer wire is subjected to abrasion which, in turn, reduces the
total strength of ropes. Ropes are selected based on bending and tensile stresses in the rope.

Experimentally it is seen that the life of wire rope is greatly affected by fatigue. It has been found that
each rope can withstand during its life only a definite number of bends after which rapid disintegrations
set in. one bend is assumed to mean the transition of the rope from its straight position into a bent one,
or from a bent position into a straight one. Ropes are used in various fashions. One can use different no
of rope falls. Increase in no of rope fall helps in decreasing the tension in the wire rope. Size of the rope is
depends on ultimate tensile strength of the material used, Maximum tension carried by rope fall, Number
of Bends, Dmin/d, types of rope, duty factor and factor of safety.

The relation between the diameter of rope and diameter of wire is given by,

drope = 1.5 𝑥 dw 𝑥 √𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠 where, dw − Diameter of wire, drope − Diameter of rope.

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 CALCULATION OF SIZE OF ROPE BASED ON DIRECT AND BENDING STRESSES

Let Tmax is maximum tension per fall

Q
Tmax =
ηf x ηp
Where, Q- Maximum load to lifted, ηf − No. of pulleys and ηp − Efficiency of pulley

Considering direct and Bending stresses,

Tmax 𝐸′dw 𝜎ut
+ ≤
A 𝐷𝑚𝑖𝑛 n
Where, A = cross sectional area ( Metal content)
 
A= 𝑥 𝑑2 wire X no. of wire = 0.4 𝑥 𝑥 𝑑2 rope
4 4

E’ =Modified Young modulus = 0.8 X106 N/mm2 ( Considering twisting which causes reduction in
bending stress)
Dw = wire diameter,
Dmin = minimum diameter of sheave,
𝜎ut = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑛𝑔ℎ
n = stress factor = n’ x duty factor , n’ = factor of safety.
drope = 1.5 𝑥 dw 𝑥√𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠

Rearranging the above equation,

Tmax 0.8 𝑥106 x dw 𝑥 𝑑𝑟𝑜𝑝𝑒 𝜎ut

+ =
A 𝐷𝑚𝑖𝑛 𝑥 𝑑𝑟𝑜𝑝𝑒 n

Tmax 0.8 𝑥106 x dw 𝑥 𝑑𝑟𝑜𝑝𝑒 𝜎ut

+ =
A 𝐷𝑚𝑖𝑛 𝑥 1.5 𝑥 dw 𝑥 √𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠 n

Tmax 𝜎ut 0.8 𝑥106 𝑑𝑟𝑜𝑝𝑒

= − 𝑥
A n 1.5 𝑥 √𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠 𝐷𝑚𝑖𝑛

Hence,
Tmax
𝐴=
𝜎ut 0.8 𝑥106 𝑑𝑟𝑜𝑝𝑒
− 𝑥 𝐷
n 1.5 𝑥 √𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠 𝑚𝑖𝑛

For 6 x 37 wire designation,

Tmax Tmax
𝐴= 𝜎ut 0.8 𝑥106 𝑑𝑟𝑜𝑝𝑒 = 𝜎ut 𝑑𝑟𝑜𝑝𝑒
n
−
1.5 𝑥 √6 𝑋 37
𝑥
𝐷𝑚𝑖𝑛 − 36000 𝑥
n 𝐷𝑚𝑖𝑛

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For 6 x 19 designation,
Tmax
A= 𝜎ut 𝑑𝑟𝑜𝑝𝑒
− 50000 𝑥
n 𝐷𝑚𝑖𝑛

Breaking strength of the wire rope is given by,

Tmax 𝑥 𝜎ut
𝑃 = 𝐴 𝑥 𝜎ut =
𝜎ut 0.8 𝑥106 𝑑𝑟𝑜𝑝𝑒
n − 𝑥 𝐷𝑚𝑖𝑛
1.5 𝑥 √𝑁𝑜. 𝑜𝑓 𝑤𝑖𝑟𝑒𝑠
( Breaking strength for different standard diameter of wire rope is given in PSG 9.4)

 Check for life:

𝑑𝑟𝑜𝑝𝑒
Usually failure is by fatigue and it depends on stress and ratio 𝐷𝑚𝑖𝑛
,

𝐷𝑚𝑖𝑛
No. of Cycles , Z = F1(𝜎ut ) and , Z = F2( )
𝑑𝑟𝑜𝑝𝑒
Z = 20000
𝐷𝑚𝑖𝑛 Z = 50000
For Z , stress  = F3( 𝑑𝑟𝑜𝑝𝑒
)

 Z = 150000
𝐷𝑚𝑖𝑛
= 𝑚. 𝜎ut . 𝐶. 𝐶1 . 𝐶2 + 8
𝑑𝑟𝑜𝑝𝑒

𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 8 𝑖𝑠 𝑎𝑑𝑑𝑒𝑑 𝑎𝑠 𝑋 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝐷𝑚𝑖𝑛

𝐶 − 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑎𝑛𝑑 𝑟𝑜𝑝𝑒 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑑𝑟𝑜𝑝𝑒

𝐶1 − 𝑅𝑜𝑝𝑒 𝑠𝑖𝑧𝑒 𝑓𝑎𝑐𝑡𝑜𝑟

𝐶2 − 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑓𝑖𝑛𝑖𝑠ℎ 𝑓𝑎𝑐𝑡𝑜𝑟
( REF: PSG 9.7 and PSG 9.8)

Value of Z are mentioned corresponding to values of m, m- a constant relating to life of rope

( Multiplication factor for m is 0.01 and for Z is 1000)

0.4 𝑥 𝑍
Life in No. of months , N = ,
a..𝑍2

where, Z = no. of cycles, a = No. of working cycles per month,

 = Endurance factor, 𝑍2 = No. of bends

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 Crosspiece:
Cross piece is secured in cross plate and casing with fastener. The main body is rectangular while the ends
are modified in cylindrical form called trunion. The trunion provides swinging effect; a provision is made
to house the thrust ball bearing which allows the loaded hook to turn easily in the handling load. The
recess for the seating rings is made to a depth from 3 to 10mm depending on the size of the bearing. A
cross piece is pivoted in the side plate or shackle plate made of steel. This enables the hook to be turned
in two mutually perpendicular directions. The cross piece is forged from the steel and turned trunion at
the ends are provided. The hole diameter should be little large than the hook shank diameter. Figure 15
and figure 16 shows the cross section of Cross Piece and types of cross piece.

(Note: For Cross piece design refer numerical)

Figure 15 : Cross-section of cross piece Figure 16 : Types of crosspiece

 Crosspiece design procedure

1. Material of crosspiece is selected. Then design stresses are determined.

2. Length of crosspiece can be determined based on axle length.
3. Trunion diameter can be calculated based on shear stress criteria.
4. Trunion has to be checked for bending and bearing failure.
5. Then width of crosspiece can be calculated from bearing, Nut, margin consideration and
sectional modulus can be found out.
6. Maximum Bending moment can be calculated considering UDL on cross piece and the
crosspiece has to be checked under bending failure.

 Shackle Plate:-

Shackle plate are the primary connecting link in all manner of rigging systems, from boats and ships to
industrial crane rigging, as they allow different rigging subsets to be connected or disconnected quickly.
A shackle plate is also the similarly shaped piece of metal used with a locking mechanism in padlocks. In
hoisting mechanism two shackle plates can be used along with side plates. Shackle plates are used for
securing crosspiece trunion and axle ends. It is subjected to tension, double shear and crushing failure.

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 Shackle plate design

1. The crosspiece is secured in the side plates which are strengthened with shackles or straps. As a
rule, only shackles are checked for strength neglecting the plates in view of their relatively small
thickness.
2. Material for shackle plate is determined.
3. Then width of shackle plate and height can be calculated based on empirical relation with the
other components dimensions.
4. Shackle plate has to be checked under three failures i.e. tensile, double shear and crushing failure.

(Note: For design refer numerical)

Figure 17 : Shackle Plate

 Hooks:-

A hook as shown in figure 18 , is a component consisting of a length of material that contains a portion
that is curved or indented, so that this portion can be used to hold another object. In a number of
uses, one end of the hook is pointed, so that this end can pierce another material, which is then held
by the curved or indented portion. Hook is made up of mild steel or high tensile steel. It is subjected
to different types of stresses at different cross sections.

Figure 18 : Hook

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 Crane hook manufacturing process.

Forging is an effective method of producing many useful shapes. The process is generally used to
produce discrete parts. The forged parts have good strength and toughness; they can be used reliably
for highly stressed and critical applications. Crane hooks are manufactured generally by open die
forging or press forging. Open die forging does not confine the flow of metal, the operator obtains the
desired shape of forging by manipulating the work material between blows. Some specially shaped
tools or a simple shaped die between the work piece and the hammer or anvil to assist in shaping the
required sections (round, concave, or convex), making holes, or performing cut – off operations may
be used. Press forging, which is mostly used for forging of large sections of metal, uses hydraulic press
to obtain slow and squeezing action instead of a series of blows as in drop forging. The continuous
action of the hydraulic press helps to obtain uniform deformation throughout the entire depth of the
work piece.
The design of the hook is based on curved beam theory. The inner fiber of the hook is subjected
to maximum stress which is combined effect of tensile stress and bending stress. Hence inner fiber is
more critical than outer fiber, to make the hook equally stronger at inner and outer fiber, more
material is required at inner side. Hence Crane Hook generally made with trapezoidal cross with bigger
side of trapezoid at inner side and smaller side of trapezoid at outer side.

 Crane hook types

Figure 19 : Crane hook types

Figure 19 shows different types of crane hook. Fig A is a Sling hook, the curve in the hook is designed to
allow the hook to hold a rope loop securely, it also allows the hook to be hooked over its own chain. These
can be made from 5 amp fuse wire as shown. Fig B is a common design with a large eye to take hemp
rope, there were often used in docks. These hooks were used on cranes and also attached to lengths of
chain or rope to make slings or 'spotter’. Fig C is a Liverpool hook, only used on cranes not on slings, the
projecting upper part is designed to prevent the hook catching on the sides of a ships hold when a load is
being lifted, it also serves to reduce the risk of the rope or chain sling jumping free if the load is jolted. Fig
D shows a Sling hook fitted with a hand loop, again only used on cranes, this allowed a man to release the
load without risk of getting his fingers trapped under the sling.

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 Hook material
Material used for hook is mild steel and high tensile steel.
For HTS material design stresses ,
Threaded part [𝝈t] = 100N/mm2 , Saddle part [𝝈t] = 180N/mm2
For mild steel design stresses can be used as ,
Threaded part [𝝈t] = 80N/mm2, Saddle part [𝝈t] = 150N/mm2
 Design procedure of hook
1. Material for hook is selected and stresses for shank and threaded part are determined.
2. Value of throat diameter C is decided based on the load to be lifted and the material of hook.
3. Then using data from PSG 9.11 and PSG 6.3 hook dimensions are calculated.
4. The design load acting on hook is determined.
5. The hook shank is checked for tensile stresses in the threaded portion.
6. The stresses (bending and shear stresses) at different sections are checked using different
design theories.
7. The most critical section at inner and outer fibre is checked for direct tensile and bending
stresses.
8. If stresses are beyond safe limit then either change the material of the hook or redesign it.

 Important sections of hook

There are mainly 4 important cross sections of hook where stresses are checked for safe hook design.
These section are shown in figure 20.
Section 1-1 is threaded part and it is checked for tensile failure.
Section 2-2 is critical section and checked for combined tensile and bending failure at inner and outer
fiber. Inner fiber is most critical as magnitude of total stress induced at inner fiber is maximum there.
Section 3-3 is critical stress region, subjected to tensile, shear and bending hence, Maximum Principal
Stress theory (MPST) and Maximum shear stress theory (MSST) are applied at this section to calculate
induced stresses.
Section 4-4 is subjected to direct shear stress, hence checked for shear failure

 Thrust bearing selection

Thrust bearing allows a loaded hook to turn easily in handling loads over 3 tonnes. it is mounted on
crosspieces and it supports the hook nuts. Selection of thrust bearing is done in following steps.

1. Bearing series is selected based on hook thread diameter.

2. The required dynamic capacity is calculated based on radial and axial load acting on hook.
3. A standard bearing is selected from the catalogue which has dynamic capacity more than required
dynamic capacity and inner diameter equal to hook thread diameter .

Fig. Different views of the crane hook

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 Rope drum
There are two types of construction of rope drums namely drum with helical grooves and plain cylindrical
drums without grooves. In most hoisting installation preference is given to grooved drums instead of plain
drums. The machine surface increases the bearing surface the drum and prevents friction between
adjacent turns of the rope. Consequently reduces the wear and improves life. The drums are usually made
up of grey cast iron FG 200 .Rarely welded drums are used. The drum is provided with helical grooves so
that rope winds up uniformly on the drum.

The rope drum should be made of seamless pipe machined & grooved accurately, to ensure proper seating
of wire rope in a proper layer. The drum should be fitted with two heavy duty Ball / Roller bearings of
reputed make for smooth operation & longer life. [ Note: for details design of drum refer Numerical ]

Figure 21 : Rope drum cross section

Figure 22 : Rope drum

 BEARINGS

A bearing is a machine element that constrains relative motion to only the desired motion, and reduces
friction between moving parts. The design of the bearing may, for example, provide for free linear
movement of the moving part or for free rotation around a fixed axis; or, it may prevent a motion by
controlling the vectors of normal forces that bear on the moving parts. Many bearings also facilitate the
desired motion as much as possible, such as by minimizing friction. Bearings are classified broadly
according to the type of operation, the motions allowed, or to the directions of the loads (forces) applied

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to the parts. For EOT crane design roller bearing and trust bearing is used as per the load lifted. In the
cross piece, for hook shank thrust ball bearing is used. For the pulley roller bearing or deep groove ball
bearing are suitable. There is large radial load (2Tmax) acting on the pulleys and very small or negligible
axial load because of rope run off horizontally. It is better to use two bearing for each pulley so that the
load will be shared and proper balancing will also maintained.

[ Note: for details design of Bearing refer Numerical ]

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NUMERICALS

1. Design a Hoisting Mechanism for lifting a load of 100KN.

 Solution:
(Design includes Rope, Sheave, Bearing , Axle, Hook, Thrust Bearing, Nut, Cross piece, Shackle Plate,
Drum, Hoisting motor, drum shaft, Bearing for drum shaft)

Note: PSG design data book is referred and suitable data is assumed.

STEP 1: SELECTION OF NO. OF FALLS AND PULLEY SYSTEM.

Fig: Four fall System

Fig: Block diagram for 4 fall Hoisting Mechanism

Selecting 4 fall system as load to be lifted = 100KN
Selecting pulley system as shown in above figure with No. of Bends = 3
Dmin
d
= 23 (for 3 bends) ……………………………………………………………. PSG 9.1

STEP 2: SELECTION OF ROPE

Selecting ordinary cross lay type of rope for hoisting, rope is suspended and not guided for economy and
avoid spinning, crossed lay type rope is selected.

Selecting 6x37 rope for greater flexibility.

Selecting Material for Rope as IPS (Improved Plaw Steel) with σut = 1800 N/𝑚𝑚2

STEP 3: FINDING MAXIMUM TENSION IN ROPE

Q
Tmax = F = where, Q – Load capacity, nf – No. of fall, ηp – Efficiency of pulley system.
nf x ηp

100
𝐹= Hence F = 26.315 KN
4 x 0.95

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STEP 4: ROPE SELECTION

Rope diameter can be found from either Cross sectional Area of Rope or Strength of the rope

Stress factor n = n’ x Duty factor

Where,
n’ (FOS) = 5 for Class II application …………… PSG 9.1
Duty Factor = 1.2 (Based on Strength and Life of crane of 20 years) …... PSG 9.2

Therefore, n = 5 x 1.2
n=6

For 6X37 type of rope, Rope strength P is given by ,

𝐹 . 𝜎𝑢
P=
𝜎𝑢 𝑑
𝑛 − 𝐷𝑚𝑖𝑛 3600
and area of cross section of the rope is given by ,
F
A=
σu d
n − 3600 (Dmin)
26.315 x 1000
A=
1800 1
6 − 3600 (23)
A = 183.44 mm2
𝜋
As there are voids, the useful Area = 0.4 x 4 𝑑2
𝜋 2
𝐴 = 183.44 = 0.4 𝑥 𝑑
4
𝑑 = 24.162 𝑚𝑚 ≅ 25 𝑚𝑚
∴ Selecting Standard Diameter = 25 mm ………………………………………………..PSG 9.4

∴ d = 25 mm and Dmin = 575 mm

STEP 5: ROPE LIFE CALCULATION

Using the equations from PSG 9.7

0.4𝑍 D
Life of rope in months is given by, 𝑁 = 𝑎.𝛽.𝑍2
and d
= m. σ. c. c1 . 𝑐2 + 8

Where:

N- Life of rope in months of 25 days,

Z- Number of repeated bends corresponding to constant m

a- Working cycle per months ( Table No.2, PSG 9.8)

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𝛽- Endurance Factor ( Table No.2, PSG 9.8)

𝑍2 - Number of repeated bends per cycle ( Table No.2, PSG 9.8)

m- constant value corresponds to Z ( Table No.1, PSG 9.8)

σ- tensile stress in rope in kgf/mm2

𝑐- Strength and rope construction factor ( Table No.4, PSG 9.8)

𝑐1 −Rope size factor ( Table No.3, PSG 9.8)

𝑐2 −Material and surface finish factor ( 0.63 to 1.15)

and Drum or Pulley diameter D ≥ e1.e2.d

Where,
e1 = Factor depends on types of hoisting device and its service condition (For power device medium
operating condition e1 = 25)
e2 = Factor depends on Rope Construction = 1 for 6x37 cross laid rope.
∴ D = e1.e2.d
D = 25 x 1 x 25 = 625 mm
Now,

F
σ=π
d2 ∗ 0.4
4
10F 10 x 2632 kgf⁄
σ= = = 13.41
πd2 π x 25 2 mm2
Assuming ,
C = 1.02 strength and rope construction factor for cross 6 X 37 type and 180 Kgf/mm2 strength ,PSG 9.8
C1 = 1.09 Rope size factor for d= 25mm ……………………………… PSG 9.8
C2 = 1 Material and surface finished factor (0.63 to 1.15) ……….…… PSG 9.7
𝐷 625
= 𝑚. 𝜎. 𝑐. 𝑐1 . 𝑐2 + 8, = 𝑚 x 13.41 x 1.09 x 1 x 1.02 + 8
𝑑 25

m = 1.1402 in Hundred
∴ m = 114.02 -------------------From PSG 9.8
Calculating Z By Interpolation,
Z 150 170
118 − 114.02 170 − 𝑥
= m 107 118
114.02−107 𝑥 −150
x = 162.76379
For m = 114.02 , Z = 162.763 x 103

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0.4Z
Life of rope in months, N =
a. β. Z2
Considering medium duty 16hrs/day
𝑎 = 3400 working cycles/month
𝛽 = 0.4 Endurance factor
𝑍2 = 3 no. of repeated bend/cycle
0.4 x 162.76 x 103
𝑁= 3400 x 0.4 x 3
, N = 15.95 months

STEP 6: SELECTION OF PULLEY/SHEAVE

For d=28mm, selecting standard pulley from PSG 9.10,

Where,
a= 80mm
Fig: Sheave cross section
b= 60mm
c= 45mm
h= 12mm
l= 20mm
Checking for fleet angle:
𝑏−𝑑 60 − 25
( 2 ) ( 2 )
tan 𝛼 = , α = tan−1
𝐷 𝐷
( 2𝑜 ) ( 2𝑜 )

Dimensions of sheave corresponding to next size of rope selection. Fleet angle α permissible will
increase due greater width of the groove and also it would give more space or width for bearing.
Do = Dmin + 2h = 575 + 2(45) ; Do = 665 mm
60 − 25
( )
α = tan −1 2
665
( )
2
α = 3.01° < 5° hence accepted.

STEP 7: SPEED OF SHEAVE (N),

𝑛
πDN = Velocity of sheave, V = Velocity of hoisting x 2
, where n- no. of fall
n 4
V x 6x
N= 2
, N= 2
= 6.11 rpm
πD π x 625 x 10−3

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STEP 8: SELECTION OF BEARING

Bearings are used between shaft and rope drum to permits relative motion between them. Generally
deep groove ball bearings are used because there is a negligible small axial force present. Selecting two
bearing per pulley so that the radial load acting on bearing is shared and small bearing can be used also
the balancing of load will be automatically taken care.
Now forces acting on each bearings are,
Fr = 26320 N
Fa = 0 N
Peq = (X.V.Fr + YFa) x S
Assuming , Radial load factor X = 1, Race rotation factor V = 1.2
and Service factor S = 1
Equivalent load, Peq = 31584 N
Assuming life of the bearing = 10,000 hours
Lh x N x 60
Life in millions of rev. is Lmr = 106

10000 x 6.11 x 60
Lmr = = 3.666 mr
106 Fig: Bearing
1
Dynamic Capacity , 𝐶 = 𝑃𝑒𝑞 x (𝐿𝑚𝑟 ) 𝑘 where K =1/3 for ball bearing
1
Dynamic capacity C = 31584 𝑥 3.663
C = 48673.677N, C = 4867.4kgf
∴ Selecting DGBB SKF 6215. ………….PSG 4.13
Having, d = 75 mm, D = 130 mm , B = 25 mm , C = 5200 kgf

STEP 9: DESIGN OF AXLE

Sheave axle material: C-30 steel with,

N N
[σb ] = 100 ; [τ] = 60 mm2
mm2

From bearing design, d = 75mm

From figure below, L1 = a + margin = 80 + 10 = 90 mm
L2 = a/2 + margin = 40 + 10 = 50 mm
L3 = casing thickness + shackle plate thickness = 4 + 12 = 16 Fig: Sheave axle, force and Bending Moment
mm diagram

L = support span = L1 + 2L2 + L3 = 90 + 100 + 16 = 206 mm

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Checking for Bending failure,

B.M max = (26320 x 2) 58 = 3053.12 KN.mm
M 3053.12 x 103 3053.12 x 103 N
(σb ) = = π 3 = π = 73.75 mm2 < [σb ]
Z d (75)3
32 32

∴ Safe in bending ,hence d = 75 mm

STEP 10: DESIGN OF HOOK

Hooks are made of mild steel or High tensile steel. After forging and machining operations hooks are
carefully annealed and cleaned from scale. The inner diameter of hooks should be sufficient to
accommodate two strands of chain or rope which carry the load. More often hooks have a trapezoidal
section, made wider on the inside. A trapezoidal section makes for better utilization of the material and
less complicated design. On the top, the hook ends in a round shank operating only in tension. The upper
part of die-forged hooks is threaded for suspension from crosspieces of load carrying devices.

Let material used for hook be mild steel,

For 25 tonnes load the dimensions of trapezoidal section point hooks

 Design load.
Assuming 10% extra load, [Wd] = W1 + (0.1W) = 100 + 10 = 110 KN

 Selection of Material.
Selecting Mild Steel , Assuming design stresses as,

[σt ] = 180 N⁄ 𝑓𝑜𝑟 𝑠ℎ𝑎𝑛𝑘 𝑎𝑛𝑑 𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑎 ℎ𝑜𝑜𝑘

mm2
[σt ] = 100 N⁄ 𝑓𝑜𝑟 𝑡ℎ𝑟𝑒𝑎𝑑𝑒𝑑 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑎 ℎ𝑜𝑜𝑘
mm2
For 12 tons , C = 134 ……. PSG 9.11

∴ C = 134

G = 70 mm, G1 = M68

A = 2.75 x C = 368.5 , B = 1.31 x C = 175.54

D = 1.44 x C = 192.96, H = 0.93 x C = 124.62

Z = 0.12 x C = 16.08, M = 0.6 x C = 80.4

bo = 2 x Z = 32.16

 Cross-sectional Area
1 1
Theoretical Area (A) = 2 (𝑏𝑖 + 𝑏𝑜 )H = 2
(80.4 + 32.16) x 124.62 , Ath = 7013.60 mm2

Actual Area a’ = 0.90 x 7013.61 = 6312.25 mm2

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 Locational Parameters
ri = C/2 = 67 mm

ro = ri + H = 67 + 124.62 = 191.62 mm
H bi +2bo
R i = ri + (
3 bi + bo
) …….PSG 6.3

1
(b )
RN = 2 x i x bo H
br − b r r
( i o H o i ) ln ( ro ) − ( bi − bo )
i

∴ R = 120.408 mm and RN = 110.45 mm

Fig – Cross Section of Hook
ho = H – hi = 81.17 mm

e = R – RN = 120.408 – 110.45 = 9.958 mm

hi = RN – ri = 110.45 – 67 = 43.45 mm

 Failures at Different cross section of the Hook:

1) At Section 1-1,

Tensile stress is given by;

[Wd ] 110 N
(σt ) = π 2 = π = 28.5829 mm2
< [σt ]
d (70)2
4 4

,Safe at section 1--1

2) At Section 2-2

(a) Inner fibre (tensile and bending stress) (2-2 is Most critical section) Fig – important Section of Hook

(𝜎𝑡 )𝑡𝑜𝑡𝑎𝑙 = (𝜎𝑡 )𝑖 + (𝜎𝑏 )𝑖

[Wd ] 110 x 1000
(σt ) = = = 17.426 N⁄
a′ 6312.25 mm2
Since, Mb = [Wd] x R = 13.24 x 106 N.mm

Mb x h i 13.24 x 106 x 43.45

(σb ) = =
a′ x e x ri 6312.25 x 9.958 x 67

(σb ) = 136.598 N⁄
mm2
(σt )total = 17.426 + 136.598 = 154.024 N⁄
mm2
(σt )total < [σt ] = 180 N⁄ ,
mm2
Safe at inner fibre that is at most critical section.

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(b) Outer Fibre (Compressive and bending stress )

(𝜎𝑐 )𝑡𝑜𝑡𝑎𝑙 = (𝜎𝑏2 ) − (𝜎𝑡1 )

Mb x ho W
(𝜎𝑐 )𝑡𝑜𝑡𝑎𝑙 = − d
′
a x e x ro 𝑎′

13.24 𝑥 106 𝑥 81.17

(𝜎𝑐 )𝑡𝑜𝑡𝑎𝑙 = − 17.426 = 71.7989 N⁄
6312.25 𝑥 9.958 𝑥 191.62 mm2

3) At Section 3-3 (tensile, bending and shear stress)

Section 3-3 is critical Section with regards to nature of stressed induced,

W1 = Wd cos 45, W2 = Wd cos 45, Mb = R. Wd cos 45

a) Tensile stress
Wd . cos 45
(σt1 ) = = 12.3223 N⁄
a′ mm2
b) Shear stress
Wd . cos 45 110 x 1000 cos 45
(τ) = = = 12.3223 N⁄
a ′ 6312.25 mm2
c) Bending stress
Mb x hi R cos 45 x Wd x hi 120.408x cos 45 x 110x1000 x 43.45
(σb ) = a′ x e x ri
= a′ x e x ri
= 6312.25 x 9.958 x 67
,

∴ (σb ) = 96.625 N⁄
mm2

Net stress at inner fibre at section 3-3

a) By Maximum Shear Stress Theory

σ +σ 2
12.32 + 96.62 2
(τ) = √( t b ) + τ2 = √( ) + 12.322 = 55.84 N⁄
2 2 mm2

(τ) < [τ] = 90 N⁄ ∴ 𝐒𝐚𝐟𝐞

mm2
b) By Maximum Principle Stress Theory

σb + σt σt + σb 2 12.32+ 96.62 12.32 + 96.62 2

(σt ) = (
2
) + √( ) + τ2 = ( ) + √( ) + 12.322
2 2 2

(σt ) = 110.32 N⁄ < [σt ] = 180 N⁄ :. Safe

mm2 mm2

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4) At Section 4-4 (shear stress)

[Wd ] 110 x 1000

(τ) = = = 17.426 N⁄ < [τ] ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑠ℎ𝑒𝑎𝑟 𝑓𝑎𝑖𝑙𝑢𝑟𝑒.
a′ 6312.25 mm2

STEP 11: DESIGN OF NUT FOR HOOK

Material Selection, Let C-25 with design stresses as,

[σt ]= 80 N/mm2 , [τ]=50 N/mm2, [σcr]=120 N/mm2

Proportions:

H= D = 68 mm; D1 = 2D = 136mm ----------PSG 9.11

load, WD= 110 KN

Failures: a) Shearing failure, b) Crushing of the thread

[Wd ] 110x103 110x103
(τ) = = = = 7.572 N/mm2 < [τ] hence safe in shear failure
As πxDxH πx68x68
Fig: Nut
WD WD
(σcr ) = =π
Acr x(d2o − d2i )x n
4
H 68
Where, n = no, of threads =
Pitch
= = 14 & do = 68mm ; di = 63mm
5

110x103
(σcr ) = π = 15.273 N/mm2 < [σcr] hence safe in crushing failure
x(682 −632 )x14
4

STEP 12: SELECTION OF BEARING FOR HOOK

Selecting single thrust ball bearing directly based on the dimensions and static load carrying capacity Co

Selecting Bearing No- 51314, series 513 ….. from PSG 9.11

From PSG 4.28, the value of C0 = 27700 kgf , d=70mm, d2= 72 mm , r = 2mm, D = 125mm, H= 40mm

Static load required , Pa = V x Fa x 1.2 = 1x 110 x 103 x1.2 = 13200 kgf < C0

Hence, Bearing is safe.

Fig: Thrust Bearing

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STEP 13: DESIGN OF CROSS PIECE

Cross piece is secured in cross plate and casing with fastener. The main body is rectangular while the
ends are modified in cylindrical form called trunion. The trunion provides swinging effect; a provision is
made to house the thrust bearing.

Material selected, Plain Carbon steel

Design stresses are, [σt ] = 100 N/mm2 , [σbr ] = k x[σt ] = 0.75x100 = 75 N/mm2 ,

[τ] = 60 N/mm2

( Constant K = 0.75 suggested by Rudenko for cross piece, K =0.1 to 1.5, depends on relative motion)

Calculate diameter of trunion of cross piece by considering

shear failure at trunion,

Wd ⁄2
[τ] = π = 60, d = 34mm
xd2
4
Hence diameter, d= 34mm , Assuming d= 50 mm

Let height of cross piece h = 1.5d = 75mm Fig: Cross Piece

M 55x103 x8 N
Checking trunion in Bending, (σb ) = = π = 35.85 mm2 < [σb ] , Therefore, safe in bending.
Z x503
32

Check in bearing failure,

W⁄ 55 x103 N N
2
(σbr ) = = = 68.75 < [σbr ] i. e 75 Therefore, safe in bearing failure.
dxL 50 x 16 mm2 mm2

B = Width of the cross piece = Size of the bearing or nut + clearance+2 x Flange thickness + End margin

Therefore, B = 136 + 4 + 2 x 12 + 46 = 170 mm

Figure show the force diagram and BMD for cross piece.
Wd l Wd D
BMmax = x − x
2 2 2 4
110x103 206 125
BMmax = x − 55 x 103 x
2 2 4
BMmax = 3946250 N. mm

Fig: Force diagram and BMD

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Figure shows the cross section for section modulus Z.

1
Z= (B − d)x H 2
6
1
Z = (170 − 50)x752 = 112500 mm3
6
Mmax 3946250
(σb ) = = = 35.08 N/mm2 < [σb ] ,
Z 112500

Therefore, safe in bending as the induced stress is less than the

design stress. Fig: the cross section for section modulus

STEP 14: SHACKLE PLATE

As the crosspiece is secured in the side plates which are strengthened with shackles or straps. Only
shackles are checked for strength neglecting the plates in view of their relatively small thickness.

Material : Plane Carbon steel C-20

[𝜎𝑡 ] = 100 𝑁/𝑚𝑚2 , [τ] = 60𝑁/𝑚𝑚2 , [𝜎𝑐𝑟 ] = 150 𝑁/𝑚𝑚2

d1= Axle size in shackle plate = 70mm, d2= Trunion size of cross piece = 50mm,

t = Thickness of plate = 16mm,

let h1 = d1 =70mm, h3 = d2 = 50mm and B = width of the plate = 3d1= 210mm,

h2= (Sheave diameter/2) +Height of cross piece above centre line + thrust bearing /nut height + margin

h2 =715/2 + 75/2 + 68 + 10 = 473mm = 480mm Approximately.

Now Checking for tensile failure at section 1-1,

𝑤⁄ 55𝑥103
(𝜎𝑡 ) = 2 =
(𝐵 − 𝑑1 )𝑡 (310 − 70)𝑥16
𝑁
(𝜎𝑡 ) = 14.32 < [𝜎𝑡 ] , ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑡𝑒𝑛𝑠𝑖𝑜𝑛
𝑚𝑚2
Checking for double shear stress at section 2 -2 ,
𝑤⁄
(𝜏) = 2 = 24.55 𝑁 < [𝜏] , ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑠ℎ𝑒𝑎𝑟
2𝑥ℎ1 𝑥𝑡 𝑚𝑚2
Similarly checking for crushing failure at section 3-3,
𝑤⁄ 3
(𝜎𝑐𝑟 ) = 2 = 55𝑥10 = 49.10 𝑁 < [𝜎 ]
𝑐𝑟
𝑑1 𝑥𝑡 70𝑥16 𝑚𝑚2
Fig: Shackle Plate
ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑐𝑟𝑢𝑠ℎ𝑖𝑛𝑔 𝑓𝑎𝑖𝑙𝑢𝑟𝑒.

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STEP 15: DESIGN OF DRUM

The rope drum should be made of seamless pipe machined & grooved accurately to ensure proper seating
of wire rope in a proper layer. The drum should be fitted with two heavy duty Ball / Roller bearings of
reputed make for smooth operation & longer life.

Compensating pulley diameter= 0.6 x diameter of movable pulley = 345 mm

Un-grooved length or unrolled length = 0.6 x Diameter of compensating pulley = 200 mm

From PSG 9.2 , length of drum for two rope is given as,
2𝑥𝐻𝑥𝑖
𝐿𝐷 = ( + 12 ) 𝑥 𝑆 + 𝑙1
𝜋𝑥𝐷
Where, H= Hoisting lift = 8 m
𝑛𝑝 4
i = Ratio of pulley system = 2
= 2 = 2;

l1 = Length of ungrooved length = 200 mm

D = Drum Diameter,

𝐷𝑚𝑖𝑛 = 575 𝑚𝑚, Therefore,

D = 575+25 = 600 mm Fig: Rope drum cross section

From PSG 9.9 and using interpolation method,

For d=25 mm; S=28 mm

2𝑥 8000 𝑥2
𝐿𝑑 = ( + 12) 𝑥 28 + 200 = 1011.34 𝑚𝑚
𝜋𝑥600
Approximately, 𝐿𝐷 = 1050 𝑚𝑚

Note- Keeping 50 mm out of 1050 mm on either side. Hence, mounting is at 950 mm. Depending on the
position of the load, the rope will be located at minimum distance of 200 mm and maximum distance of
950 mm. As the bending moment depends on the distance of the load from the support, the maximum
bending moment will occur when load is at distance L1.

selecting material for drum as C-40 with [σc ] = 140𝑀𝑃𝑎 also neglecting the weight of drum and rope
𝐿 𝐿1
B. M max = F x ( 2𝐷 − 2
) = 9.87 kN. m

Drum is subjected to direct compression over the area

(S x thickness)

Fmax 26.2 x 103

[σc ] = hence 140 = Fig: Force Diagram
Sxt 28 x t
t = 6.71mm , approx. t = 8mm

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Thickness is calculated and depth of groove C1 can be calculated based on diameter of 25 mm,

depth of groove, C1 = 6mm and t = 8mm below the groove, …..PSG 9.9

Hence actual thickness of the plate is 6 + 8 = 14 mm.

Now actual compressive stress is given by,

Fmax 26.32x103
(σc ) = = = 67.13 N/mm2
Sxt 28 x14
and stress due to bending is given by,

Mmax 9.87 x 106

(σb ) = = = 2.921 N/mm2
Z π 5754 − 5474
x( )
32 575
It is observed that compressive stress due to direct compression is the predominant stress, hence for
quicker analysis directly design based on the direct compressive stress.

Total compressive stress, σctotal = σc + σb = 67.13 + 2.9213 = 70.05 N/mm2

Do
Drum is subjected to torsion also, The maximum torque acting on the rope drum = 2 x Fmax x 2
600
T = 2 x 26.32 x 103 x 2
= 15.792 KN. m

Torsional shear stress induced is given by,

T T 15.792 x 106
(τ) = = = = 2.33 N/mm2
J D4 − D4i π 5754 − 5474
π x ( 16 x D ) 16 x( 575
)

Principal Stresses,
σctotal σctotal 2
σ1 = + √( ) + τ2
2 2

70.05 70.05 2 N 𝑁
σ1 = + √( ) + 2.332 = 70.127 2
< [140 ], ℎ𝑒𝑛𝑐𝑒 𝒔𝒂𝒇𝒆
2 2 mm 𝑚𝑚2

STEP 16: SELECTION OF MOTOR

Given , Hoisting Speed = 8 m/min

Output Power = Design load * Hoisting Speed

8
𝑊𝑑 𝑥𝑉 = 110𝑥103 𝑥 = 14666.6 𝑤𝑎𝑡𝑡, Considering transmission efficiency as 0.85
60

Output Power 14666.6

𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = = = 17.254 𝑘𝑤
transmission efficiency 0.85
Selecting Standard Motor of 18 KW ……………………………………………. PSG 5.124

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STEP 17: DESIGN OF DRUM SHAFT

Drum shaft is subjected to Torque and Bending Moment. SFD and BMD is as shown in figure.

BMmax = 26.32 x103 x 50 = 1.316 KN. m

𝐷𝑑𝑟𝑢𝑚
Torque T = 2F x =
2
600
T = 52.64 x 103 x = 15.793 KN. m
2
Fig: Drum shaft
Equivalent torque, T = √M 2 + T 2 = 15.85 KN.m

From PSG 1.13 , Let material be 40cr1 with

N
Syt = 540 ; FOS = 4
mm2
N
[σt ] = 135 ; [τ] = 80 N/mm2
mm2
π π
Now, Teq = 16 xd3 x[τ] , 15.85 x106 = 16 x d3 x 80

𝐝 = 𝟏𝟎𝟎 𝐦𝐦

Selecting shaft with diameter as 100mm.

STEP 18: Design of bearing for drum shaft:

For drum shaft lets select cylindrical roller bearing, Fig : Force and BM Diagram

Forces acting on each bearing are, Fr = 26.32KN, Fa = 0

Assuming life in hours, Lhr = 10,000 hrs

Linear speed of the drum = 2x linear speed of the load = 2 x 6 = 12 m/min

V = πxDxN , 12 = π x 0.6 x N

N = 6.36619 rpm

Life in millions of revolutions is given by,

Lhr x N x 60 10000 x 6.37 x 60

Lmr = =
106 106

Lmr = 3.81971 mr

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Equivalent load acting on the bearing , Peq = (X. V. Fr + Y. Fa )X S

Assuming X=1, V = 1 and S = 1.2

Peq = 31.584 KN
1
Dynamic load carrying capacity is given by, C = P. (Lmr ) K =
3
52.64x (3.822)10 = 47.2119 KN

Where k = 10/3 for roller bearing,

Therefore, C = 4720 kg. f

Therefore, selecting from PSG 4.21,

standard bearing as SKF NU 2220 OR SKF 6220 for d=100mm

with d = 100 mm; D = 180 mm;

C0 = 19000 kg. f; C = 20000 kg. f

Fig : Cylindrical roller bearing

====================================================================================

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Numerical 2. Design of Hoisting Mechanism for the following specifications

Load to be lifted = 300KN
Hoisting Speed = 5m/min
Lift = 10m
Solution -

(Design includes Rope, Sheave, Bearing , Axle, Hook, Thrust Bearing, Nut, Cross piece, Shackle Plate,
Drum, Hoisting motor, drum shaft, Bearing for drum shaft)

Note: PSG design data book is referred and suitable data is assumed

The given load to be lifted = 300KN

Therefore, selecting 8 fall system & efficiency of pulley system as ղp = 90 %

Minimum no. of bends = 7

𝐷𝑚𝑖𝑛
𝑑
= 30 …PSG 9.1

For no. of bends = 7,

Selecting cross lay type of rope and 6x37 type rope for greater flexibility.

Selecting material for rope as IPS with σut =1800N/mm2

Fig: Snatch block for 8 fall system
 To find max tension in rope and diameter of rope

𝑄 300
Tmax = F = ղ = 8 x 0.9 F = 41.667 KN
𝑓 𝑥 ղ𝑝

Stress factor n = n’ x S =5 x 1.2 = 6

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Where n’ = factor of safety (for class-2 n’ =5) …PSG 9.1

S= duty factor based on strength (S =1.2) …PSG 9.2

𝐹
Cross Section Area, A = σ𝑢𝑡 𝑑 …PSG 9.1
−3600 x
𝑛 𝐷𝑚𝑖𝑛

41.667 x 1000
A = 1800 1
−3600 x
6 30

A = 231.48 mm2

𝜋
Actual Area = 0.4 x x d2 = 231.48
4

Therefore, d = 27.14 mm

By taking standard value of rope from PSG 9.4 , d = 29 mm

hence Dmin = 30 X d = 870 mm

 Life of the rope

By Empirical Formula, D ≥ e1 x e2 x d

Where, e1= 25 (for medium power hoisting mechanism)

e2 =1 (for cross laid), d = 29 mm

Therefore, D = 25 x 29 x 1 = 725 mm

Selecting Dmin = 870mm, D = 870 +29 ≈ 900mm

10𝐹 10 x 41.667 x 103 N

Now, σ = = =14.7 kgf/mm2
𝜋𝑑 2 𝜋 x 292

𝐷
= m σ C1 C2 C + 8 …PSG 9.7
𝑑

C1 = 1.09 for d=29mm rope size factor …PSG 9.8

C2 = 1 (varies from 0.63 to 1.15) material and surface finish factor …PSG 9.7

C = 1.02 strength and rope construction factor …PSG 9.8

725
29
= m x 14.7 x 1.09 x 1 x 1.02 + 8

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m = 1.2642 , m = 126.42 (in hundred)

From PSG 9.8,

Z 170 X 190

M 118 126.42 129

Therefore, x = 154.69 (by interpolation)

Z in thousands, Z = 154.69 x 103

0.4 𝑍
N = 𝛼𝛽𝑍2 …PSG 9.7

From PSG 9.8, Table 2, Considering medium duty 16hrs/day

𝑎 = 3400 working cycles/month
𝛽 = 0.4 Endurance factor
𝑍2 = 3 no. of repeated bend/cycle
0.4 x 154.69 x 1000
N= 3400 x 3 x 0.4
= 15.656 months

Life of Rope in months is 15. 66 months

 Selection of pulley / sheave

Selecting std. sheave dimensions for wire dia.

d = 34.5mm from PSG 9.10

Where, a = 90mm, b = 70mm

C =15mm, h = 55mm, L = 22mm

Checking fleet angle α

𝑏−𝑑
tan α = 𝐷𝑜 2x 0.5

Do = Dmin + 2h = 870 + 2 x 55 = 980 mm Fig – Pulley Cross Section and fleet angle

Therefore α = 2.26o <5o Hence Acceptable .

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 Selection of bearing (Deep Groove Ball Bearing)

Selecting two bearing per pulley so that the radial load acting on bearing is shared and small bearing can
be used also the balancing of load will be automatically taken care.
Now forces acting on each bearings are,
Fr = 41667 N , Fa = 0 N

Peq = (X.Fr.V + Y Fa ) S

Assuming , Radial load factor X = 1, Race rotation factor V = 1.2

and Service factor S = 1
Equivalent load,

Peq = 50000.4 N
Fig. Deep Groove Ball Bearing
To find rpm of Sheave :

𝜋𝐷𝑁 = velocity of sheave = velocity of hoisting x n/2

𝑛 8
𝑉𝑥 5x
2 2
N= = = 7.07 rpm
𝜋𝐷 𝜋 x 0.9

Assuming life of the bearing = 10000 hrs.

10000 x 7.07 x 60
Lmr = = 4.24 mr
106
1
Dynamic Capacity , 𝐶 = 𝑃𝑒𝑞 x (𝐿𝑚𝑟 ) 𝑘 where K =1/3 for ball bearing
1
Dynamic capacity C = 50000 𝑥 4.243 = 80926.72 N = 8092.67kgf
From PSG 4.13 , Selecting DGBB SKF 6219. ………….PSG 4.13

Having, d = 95 mm, D = 170 mm , B = 32 mm , C = 8500 kgf

 Design of Axle

(Assuming two pulleys on axle are in between the shackle plate and two pulleys are at two ends of axle)

Let, Design stresses for the axle material be [ σt ] = 100 N/mm2, [ τ ] = 60 N/mm2

L1 = a + margin = 90 + 10 = 100 mm

L2 = a/2 + margin = 45 + 10 = 55 mm

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t = casing thickness + Shackle plate thickness = 4 +12 mm = 16 mm

Support Span, L = L1 + 2L2 + t

Support Span, L = 100 + 110 + 16 = 226 mm

Total length of Axle = L + t +2L2 + Margin

From Force and Bending Moment diagram,

Fig – Axle Span Length
B.Mmax = 41.667 x 103 x 2 x 63

B.Mmax = 5250.042 KN-mm

𝐵.𝑀𝑚𝑎𝑥 5250.042 x 1000

[ σb ] = = π 3 = 100
𝑍 d
32

Therefore d3 = 534764.88

d = 81.16 mm

Let, d = 95 mm (considering DGBB SKF 6219 )

and d1 = 85 mm

Fig – Force Diagram and Bending Moment Diagram

 Design of hook

Design load.
Assuming extra load, [Wd] = 320 KN

Selection of Material.
Selecting Mild Steel , Assuming design stresses as,

[σt ] = 200 N⁄ 𝑓𝑜𝑟 𝑠ℎ𝑎𝑛𝑘 𝑎𝑛𝑑 𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑎 ℎ𝑜𝑜𝑘

mm2
[σt ] = 100 N⁄ 𝑓𝑜𝑟 𝑡ℎ𝑟𝑒𝑎𝑑𝑒𝑑 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑎 ℎ𝑜𝑜𝑘
mm2
From PSG 9.11 for 32 tonne load,

C = 207 mm

G = 110 mm

G1 = M100 Fig – Sectional Diagram Hook

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A = 2.75C = 569.25 mm , .

B = 1.31C = 271.17 mm

D =1.44C = 298.08 mm , H = 0.93C = 192.51 mm

Z = 0.12C = 24.84 mm , M = 0.6C =124.2 mm

bo = 2Z = 49.68 mm

Theoretical Area of cross section,

1 1
(A) = (𝑏 + 𝑏𝑜 )H = (124.2 + 49.68) x 192.51,
2 𝑖 2

Ath = 16736.81mm2 , Actual area a’ = 0.9 Ath = 15063.13mm2 Fig – Cross Section of Hook

Location parameters …PSG6.2/6.4

ri = 0.5C = 103.5 mm

ro = ri + H = 296.01 mm

From PSG 6.3,

𝐻 𝑏𝑖+2𝑏𝑜
R = ri + [ ] = 186mm
3 𝑏𝑖+𝑏𝑜
Fig– Beam Diagram
0.5(𝑏𝑖+𝑏𝑜)𝐻 16736.81
RN = 𝑏 𝑟𝑜 −𝑏𝑜 𝑟𝑖 𝑟𝑜 = = 170.63mm
( 𝑖 ) ln( )−74.51 164.26 x 1.05082-74.51
𝐻 𝑟𝑖

hi = RN – ri = 170.35 – 103.5 = 66.86 mm

ho = H – hi =192.51 - 66.86 = 125.65 mm

e = R – RN = 15.37 mm

Checking for different Failures :

At section 1-1 (tension only)

𝑊𝑑 320000
(σt ) = 𝜋 =𝜋 = 40.74 N/mm2 < [σt] Hence Safe
𝑥𝑑𝑥𝑑 𝑥 100 𝑥 100
4 4

At Section 2-2
Fig – important Section of Hook
(a) At inner fibre (tension and bending)

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(σt)total = ( σt1 ) + ( σb1 )

[𝑊𝑑] 320000
(σt1) = 𝑎′
= 15063.13 = 21.25 N/mm2

Mb x hi 320000 x 186 x 66.86

(σb1) = a' x e x ri = 15063.13 x15.37 x 103.5 = 166.07 N/mm2

(σt)total = 21.25 + 166.07= 187.32N/mm2

(σt)total < [σt] Hence Safe

(b) At outer fibre [ Compressive stress only ]

(σc)total = ( σb2 ) + ( σt1 )

𝑀𝑏 ℎ𝑜 𝑊𝑑 320000 x 186 x 125.65 320000

(σc)total = [ ] – [ 𝑎′ ] = 15063.13 x15.37 x 296.01 – 15063.13 = 109.12 – 21.25 = 87.87 MPa,
𝑎 ′ 𝑒 𝑟𝑜

(σc)total = 87.87 N/mm2 < [σc] Hence Safe

At Section 3-3 (tension, bending and shear)

Tensile Stress induced :

𝑊𝑑 𝑐𝑜𝑠45 320000 cos 45

(σt1) = 𝑎′
= 15063.13
= 15.02 N/mm2

Shear Stress induced:

𝑊𝑑 𝑐𝑜𝑠45 320000 sin 45

(τ) = 𝑎′
= 15063.13 = 15.02 N/mm2
Fig . Hook section 3-3
Bending Stress induced:

𝑀𝑏 ℎ𝑖 𝑊𝑑 𝑐𝑜𝑠45 𝑥 𝑅 𝑥 ℎ𝑖 320000 xcos45 x 186 x 66.86

(σb) = = = = 117.43 N/mm2
𝑎 ′ 𝑒 𝑟𝑖 𝑎 ′ 𝑒 𝑟𝑖 15063.13 x15.37 x 103.5

 Net stress at inner fibre at section 3-3 –by MSST

𝜎𝑡+𝜎𝑏 2
15.02+117.43 2
(τ) =√( 2 ) + (τ)2 = √( 2
) + 15.022

(τ) = 67.9N/mm2 < [τ] = 90N/mm2 Therefore, Safe

By MPST

(σb+ σt ) 𝜎𝑡+𝜎𝑏 2 15.02+117.43 15.02+117.43 2

(σ1) = [ 2
] +√( 2
) + (τ)2 = 2
+ √( 2
) + 15.022

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(σ1) = 134.125 N/mm2 < [σt ] Therefore, Safe

 Design of nut

Material Selection, Let C-25 with design stresses as,

[σt ]= 80 N/mm2 , [τ]=50 N/mm2, [σcr]=120 N/mm2

Proportions – Cylindrical Nut,

H = D = 82 mm, D1 = 2D = 164 mm
Fig: Nut
Failures - (a) Shearing of thread

(b) Crushing of thread

[𝑊𝑑] [𝑊𝑑] 320000

Shear stress induced (τ) = 𝐴𝑠
= πDH
= π x 82 x 82 = 15.14 N/mm2 < 50 N/mm2 Therefore, Safe

[𝑊𝑑] [𝑊𝑑]
Crushing stress induced (σcr ) = 𝐴
= 0.25𝜋(𝑑𝑜2 −𝑑𝑖2 )𝑛

𝐻 82
where , n = 𝑝𝑖𝑡𝑐ℎ = 6
= 13.66 = 14, do= 82mm, di = 76mm

320 𝑥 1000
(σcr) = 0.25𝜋(822 −762 )14 = 330.7 N/mm2 < 120N/mm2 Therefore, Safe

 Selection of bearing for hook (Thrust Ball Bearing)

Selecting bearing based on the dimensions …PSG 9.11

Series = 513

Bearing no. = 51319

d H Co D

88 mm 49 mm 40800 kgf 150 mm

Static load , Peq = Y. Fa. S = 320 x 103 x 1.2 = 384000 N

Co= 38400 kgf < [Co = 40800 kgf ] Bearing is safe

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Design of Mechanical System / MEC801 / Semester VIII / Mechanical Engineering / Mumbai University.

 Design of cross piece

Cross piece is secured in cross plate and casing with fastener. The main body is rectangular while the
ends are modified in cylindrical form called trunion. The trunion provides swinging effect; a provision is
made to house the thrust bearing.
Material – Plain carbon steel

[σt] = [σb] = 100 N/mm2 , [τ] = 60 N/mm2

[σbr] = k [σt]

( Constant K = 0.75 suggested by Rudenko for cross piece,

K =0.1 to 1.5, depends on relative motion)

[σbr] = 75 N/mm2
Fig – Crosspiece for hooks
d1 = shank dia. + clearance

d1 = 100 +2 = 102 mm

D = outer dia. of bearing, D = 150 mm

L1 = 210 mm , L2 = 16 mm, H = 1.5d

Considering shear failure for trunion:

[Wd]/2 165 x 103

[τ] = π , 60 = π
dxd dxd
4 4

Trunion diameter d = 59.17 mm , let d = 60 mm , Height of cross piece H = 1.5 x 60 = 90 mm

Checking trunion in bending failure:

𝑀 165 x103 x 8
(σb) = , (σb) = π = 62.24 N/mm2 < [σb]
𝑍 x 603
32

Checking under bearing failure:

𝑊𝑑 /2 165 𝑥 1000
(σbr) = 𝑑 𝑥 𝐿2 = 60 𝑥 16
= 171.87 N/mm2 > [σbr] hence fail in bearing failure.

Now changing d to 80mm and thickness of shackle plate to 20mm, required bearing stress for material is
165 𝑥 1000
80 𝑥 20
= 103.125 MPa.

Hence changing material such that [σbr] is 110MPa. Hence H = 1.5d = 120mm

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Design of Mechanical System / MEC801 / Semester VIII / Mechanical Engineering / Mumbai University.

Width of cross piece B = size of bearing + clearance +2(flange thickness) + margin

B = 164 + 4 + 2(12) + 8 = 200 mm

𝑀
Checking cross piece for bending failure, (σb) = 𝑍
where,

𝑊𝑑 𝑙 𝐷
BMmax = [ − ]
2 2 4

BMmax = 165 x 103 [226/2-150/4]

BMmax = 12.45 x 106 N-mm Fig: Force and BMD

Section modulus (Z)

Z = (1/6) (B – d1) H2 = (1/6) (200 –102 ) 1202 =235.2 x 103 mm3

𝑀 12450 x 1000
(σb) = 𝑍
= = 52.93 N/mm2 < [σb] Hence Safe
235200

 Shackle plate design Fig – Cross Section of Cross Piece

As the crosspiece is secured in the side plates which are strengthened with shackles or straps. Only
shackle plates are checked for strength neglecting the side plates in view of their relatively small
thickness.

Material : Plane Carbon steel C-20

[𝜎𝑡 ] = 100 𝑁/𝑚𝑚2 ,

[τ] = 60𝑁/𝑚𝑚2 ,

[𝜎𝑐𝑟 ] = 150 𝑁/𝑚𝑚2

d1= Axle size in shackle plate = 70mm,

d2= Trunion size of cross piece = 50mm,

t = Thickness of plate = 20mm,

Fig– Cross Section of Shackle Plate
let h1 = d1 =82mm, h3 = d2 = 80mm

and B = width of the plate = 3d1= 246mm,

h2= (Sheave diameter/2) +Height of cross piece above centre line + thrust bearing /nut height + margin

h2 =900/2 + 120/2 + 68 + 20 = 598mm

h2= 600mm Approximately.

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 Checking for tensile failure

0.5𝑊𝑑 165000
(σt) = (𝐵−𝑑1)𝑡 = (246−82)𝑥 20
= 50.30 N/mm2 < [σt] hence safe

 Checking for Double shear

0.5𝑊𝑑 165000
(τ) = 2 𝑥 ℎ1 𝑥 𝑡 =2 𝑥 82 𝑥20 = 50.30 N/mm2 < [σt] hence safe

 Checking for Crushing

0.5Wd 165000
(σcr) = d1 x t
= 82 x 20
= 100.61N/mm2 < [σcr] hence safe

 Drum design

The rope drum should be made of seamless pipe machined & grooved accurately to ensure proper seating
of wire rope in a proper layer. The drum should be fitted with two heavy duty Ball / Roller bearings of
reputed make for smooth operation & longer life.

Compensating pulley diameter= 0.6 x diameter of movable pulley = 0.6 x 870 = 540 mm

Un-grooved length or unrolled length = 0.6 x Diameter of compensating pulley = l1= 0.6 x 540 = 324 mm
2𝐻𝑖
Length of drum , L= ( + 12)S+ l1 …….…PSG 9.2
𝜋𝐷

Where, H = hoisting lift

i = velocity ratio = no. of falls x 0.5 = 0.5 x nf

D = Dmin + d = 870 + 29 = 899 mm, Let D = 900 mm

H = 10 m = 10000 mm, i = 4, S= 33 …....PSG 9.9

Fig. - Rope drum cross section
2 x 10000 x 4
L= ( π x 900 ) x 33 + 330 = 1233.59 mm, Let L = 1300 mm

Keeping 50 mm out of 1300 mm on either side. Hence, mounting is at 1200 mm. Depending on the position
of the load, the rope will be located at minimum distance of 324 mm and maximum distance of 1200 mm.
As the bending moment depends on the distance of the load from the support, the maximum bending
moment will occur when load is at distance L1.

selecting material for drum as C-40 with [σc ] = 140𝑀𝑃𝑎 also neglecting the weight of drum and rope

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Drum is subjected to direct compressive stress over the area ( s x thickness )

σc = 𝐹𝑚𝑎𝑥 /(𝑆 𝑥 𝑡)

41.66 x 1000 41666

[σc] = 33 t
, 140 = 33 t

Therefore, t = 9mm

For d = 29 mm, C1 = 9 ……..PSG 9.9 ,

Therefore total thickness = t + C1 = 9 + 9 = 18mm

41666
( σc ) = 33 𝑥 18
= 70.13 N/mm2 and

𝐿𝐷 𝐿1 1200 324
B. M max = F x ( − ) = 41667 x ( − ) = 18.25 kN. m
2 2 2 2

Mmax 18.25 x 106

(σb ) = = = 3.285 N/mm2
Z π 9004 − 8824
x( )
32 900
It is observed that compressive stress due to direct compression is the predominant stress, hence for
quicker analysis directly design based on the direct compressive stress.

Total compressive stress, (σc)total = (σc) + (σb) = 70.13 + 3.285 = 73.41 N/mm2

Do
Drum is subjected to torsion also, The maximum torque acting on the rope drum = 2 x Fmax x 2
900
T = 2 x 41.67 x 103 x 2
= 37.5 KN. m

T T 37.5 x 106
(τ) = = = = 3.37 N/mm2
J π D4 − D4i π 9004 − 8824
x( )
16 x ( D ) 16 900

Principal Stress,
σctotal σctotal 2
σ1 = + √( ) + τ2
2 2

73.41 73.41 2
σ1 = + √( ) + 3.372
2 2

N 𝑁
σ1 = 73.56 2
< [140 ], ℎ𝑒𝑛𝑐𝑒 𝒔𝒂𝒇𝒆
mm 𝑚𝑚2

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Design of Mechanical System / MEC801 / Semester VIII / Mechanical Engineering / Mumbai University.

 Selection of motor

Given , Hoisting Speed = 5 m/min

Output Power = Design load * Hoisting Speed

5
𝑊𝑑 𝑥𝑉 = 330𝑥103 𝑥 60 = 27.5 𝐾𝑤,

Considering transmission efficiency as 0.85

Output Power
𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 =
transmission efficiency
27500
𝐼𝑃 = = 32.35 𝑘𝑤
0.85
Selecting std. motor with 32.5KW power

 Design of drum shaft

Drum shaft is subjected to Torque and Bending Moment.

Assuming bearing span is 1300mm,

B.Mmax = M = 83.32x 1000 x (650-600)

M= 4.166 x 106 N-mm

Torque T = 37.5 x 106 N-mm

Equivalent torque = √𝑀2 + 𝑇 2 = 37.73 x 106 N-mm

Let shaft material be 40Cr1 and FOS = 4

Design stresses are [τ] = 80 N/mm2 ,

Fig – Loading Diagram and BMD

[σt]= 135 N/mm2

π
Teq = xd3 x[τ] ,
16

π
37.73 x106 = x d3 x 80
16

Therefore, d = 133.92 mm,

Let d = 140 mm

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 Design of bearing (cylindrical roller bearing)

For drum shaft lets select cylindrical roller bearing,

Forces acting on each bearing are, Fr = 41.67KN, Fa = 0

Assuming life in hours, Lhr = 10,000 hrs

Linear speed of the drum = 2x linear speed of the load

V= 2 x 5 = 10 m/min

V = πxDxN , 10 = π x 0.9 x N

N = 3.54 rpm

Life in millions of revolutions is given by,

Lhr x N x 60 10000 x 3.54 x 60

Lmr = =
106 106

Lmr = 2.122 mr

Equivalent load acting on the bearing ,

Peq = (X. V. Fr + Y. Fa )X S Fig : Cylindrical roller bearing

Assuming X=1, V = 1 and S = 1.2

Peq = 50 KN

Dynamic load carrying capacity is given by,

1 3
C = P. (Lmr ) K = 50 x (2.122)10 = 62.66 KN

Where k = 10/3 for roller bearing,

Therefore required Dynamic load carrying capacity, C = 6266 kg. f

Selecting NU2228 …PSG 4.21

With d = 140 mm, D= 250 mm, Co = 38000 kgf , C = 36500 kgf

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Design of Mechanical System / MEC801 / Semester VIII / Mechanical Engineering / Mumbai University.

SUMMARY

Parts Dimensions

Rope Diameter d=29 mm

Life of Rope N=15.66 months

Pulley a=90mm , b=70mm , c=15mm, h=55mm

L=22mm, Do = 980 mm

DGBB SKF 6219, with d = 95mm ,D = 170mm

B = 32mm , C = 8500 kgf
Axle Support span L = 226 mm, d = 95 mm , d1 = 85 mm

Hook C= 207mm, ri=103.5 mm, ro = 296.01 mm, H =192.51mm, e = 15.37 mm

Nut H = 82 mm ,D1 = 164mm, di = 76 mm

Thrust Ball Bearing Bearing No. 51319

d = 88mm, H = 49 mm, Co = 40800 kgf , D = 150 mm

Cross Piece d1 = 102 mm, D = 150 mm, Trunion diameter, d = 80 mm

L1 = 210 mm, L2 = 20 mm, B=200 mm, H = 120mm

Shackle Plate t = 20mm, B = 246 mm, h1= 82mm, h2 = 600 mm, h3= 80mm.

Drum design Dia. of compensating pulley = 540 mm

Un-grooved length = l1 = 324 mm

D = 900 mm, L = 1300 mm, total thickness = 18

Drum Shaft d = 140 mm

Motor Motor power =32.5KW

Cylindrical Roller Bearing D = 140 mm, D= 250 mm

Co = 38000 kgf , C = 36500 kgf

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Design of Mechanical System / MEC801 / Semester VIII / Mechanical Engineering / Mumbai University.

References
1. Rudenko - ‘Material Handling Equipment’ M.I.R. publishers, Moscow.
2. Bhandari V.B., “Design of Machine Elements”, Tata McGraw Hill Pub.Co.Ltd.
3. “Design Data Book ”, P.S.G. College of Technology, Coimbatore
4. Pradyumna keshari maharana “Computer aided analysis and design of hoisting mechanism of an
EOT crane” department of mechanical engineering , NIT Rourkela, may 2012.
5. Crane Training Handbook with GPR 8719.1B
6. en.wikipedia.org/wiki/Overhead crane
7. http://www.stren-flex.com/wire-rope-inspection-drum-and-sheave.aspx
8. https://rparmanik.wordpress.com/2008/07/26/design-of-hoisting-arrangement-of-eot-crane/
9. https://en.wikipedia.org/wiki/Hoist_(device)
10. https://www.teachengineering.org/view_lesson.php?url=collection/cub_/lessons/cub_simple/cub_
simple_lesson05.xml
11. http://www.akrutiengineers.com/product_images/block_assembly.jpg
12. http://dir.indiamart.com/impcat/rope-drum-hoist.html
13. http://practicalmaintenance.net/?p=626
14. http://nptel.ac.in/courses/105106049/lecnotes/mainch10.html
15. http://www.nwideuk.com/bearings/g-deep-groove-ball.asp
16. partdetails.aspx?PartID=10-02082013-91896&class=NSK&clsid F_NSK. 010

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