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V-belt and Rope Drives n 727V-Belt and Rope Drives Drive over Flat Belt Drive.. The power is transmitted by the *wedging * The wedging action of the V-belt in the groove of the pulley re

V-belt and Rope Drives n 727

V-Belt and Rope Drives

Drive over Flat Belt Drive.

5 Ratio of Driving Tensions

10 Sheave for Fibre Ropes.

11 Ratio of Driving Tensions

for Fibre Rope.

17 Properties of Wire Ropes.

18 Diameter of Wire and

Area of Wire Rope.

19 Factor of Safety for Wire

Ropes.

20 Wire Rope Sheaves and

Drums.

21 Wire Rope Fasteners.

22 Stresses in Wire Ropes.

23 Procedure for Designing a

We have already discussed that a V-belt is mostly used

in factories and workshops where a great amount of power

is to be transmitted from one pulley to another when thetwo pulleys are very near to each other

The V-belts are made of fabric and cords moulded in

rubber and covered with fabric and rubber as shown in Fig

20.1 (a) These belts are moulded to a trapezoidal shape

and are made endless These are particularly suitable for

short drives The included angle for the V-belt is usually

from 30° to 40° The power is transmitted by the *wedging

* The wedging action of the V-belt in the groove of the pulley

results in higher forces of friction A little consideration will show that the wedging action and the transmitted torque will be more if the groove angle of the pulley is small But a small groove angle will require more force to pull the belt out of the groove which will result in loss of power and excessive belt wear due to friction and heat Hence the selected groove angle is a compromise between the two Usually the groove angles of 32° to 38° are used.

action between the belt and the V-groove in the pulley or sheave A clearance must be provided at the bottom of the groove as shown in Fig 20.1 (b), in order to prevent touching of the bottom as it becomes narrower from wear The V-belt drive may be inclined at any angle with tight side either at top or bottom In order to increase the power output, several V-belts may be operated side by side It may be noted that in multiple V-belt drive, all the belts should stretch at the same rate so that the load

is equally divided between them When one of the set of belts break, the entire set should be replaced

at the same time If only one belt is replaced, the new unworn and unstretched belt will be more tightlystretched and will move with different velocity

Fig 20.1 V-Belt and V-grooved pulley.

20.2

20.2 Types of V-belts and PulleysTypes of V-belts and Pulleys

According to Indian Standards (IS: 2494 – 1974), the V-belts are made in five types i.e A, B, C,

D and E The dimensions for standard V-belts are shown in Table 20.1 The pulleys for V-belts may

be made of cast iron or pressed steel in order to reduce weight The dimensions for the standard

V-grooved pulley according to IS: 2494 – 1974, are shown in Table 20.2.

Table 20.1 Dimensions of standard V-belts according to IS: 2494 – 1974

Type of belt Power ranges Minimum pitch Top width (b) Thickness (t) Weight per

20.3 Standard Pitch Lengths of V-beltsStandard Pitch Lengths of V-belts

According to IS: 2494-1974, the V-belts are designated by its type and nominal inside length

For example, a V-belt of type A and inside length 914 mm is designated as A 914–IS: 2494 The standard inside lengths of V-belts in mm

According to IS: 2494-1974, the

pitch length is defined as the

circumferential length of the belt at the

pitch width (i.e the width at the neutral

axis) of the belt The value of the pitch

width remains constant for each type of belt

irrespective of the groove angle

The pitch lengths are obtained by

adding to inside length: 36 mm for type A, 43 mm for type B, 56 mm for type C, 79 mm for type D and

92 mm for type E The following table shows the standard pitch lengths for the various types of belt.

Table 20.3 Standard pitch lengths of V-belts according to IS: 2494-1974

Type of belt Standard pitch lengths of V-belts in mm

Note: The V-belts are also manufactured in non-standard pitch lengths (i.e in oversize and undersize) The

standard pitch length belt is designated by grade number 50 The oversize belts are designated by a grade

Material handler.

number more than 50, while the undersize belts are designated by a grade number less than 50 It may be noted

that one unit of a grade number represents 2.5 mm in length from nominal pitch length For example, a V-belt marked A – 914 – 50 denotes a standard belt of inside length 914 mm and a pitch length 950 mm A belt marked

A – 914 – 52 denotes an oversize belt by an amount of (52 – 50) = 2 units of grade number Since one unit of

grade number represents 2.5 mm, therefore the pitch length of this belt will be 950 + 2 × 2.5 = 955 mm.

Similarly, a belt marked A – 914 – 48 denotes an undersize belt, whose pitch length will be 950 – 2 × 2.5 = 945 mm.

20.4

20.4 Advantages and Disadvantages of V-belt Drive over Flat Belt DriveAdvantages and Disadvantages of V-belt Drive over Flat Belt Drive

Following are the advantages and disadvantages of the V-belt drive over flat belt drive :

Advantages

1. The V-belt drive gives compactness due to the small distance between centres of pulleys.

2. The drive is positive, because the slip between the belt and the pulley groove is negligible

3. Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.

4. It provides longer life, 3 to 5 years

5. It can be easily installed and removed

6. The operation of the belt and pulley is quiet

7. The belts have the ability to cushion the shock when machines are started

8. The high velocity ratio (maximum 10) may be obtained

9. The wedging action of the belt in the groove gives high value of limiting *ratio of tensions

Therefore the power transmitted by V-belts is more than flat belts for the same coefficient of

friction, arc of contact and allowable tension in the belts

10. The V-belt may be operated in either direction, with tight side of the belt at the top or

bottom The centre line may be horizontal, vertical or inclined

Disadvantages

1. The V-belt drive can not be used with large centre distances, because of larger weight per

unit length

2. The V-belts are not so durable as flat belts.

3. The construction of pulleys for V-belts is more complicated than pulleys of flat belts

4. Since the V-belts are subjected to certain amount of creep, therefore these are not suitable

for constant speed applications such as synchronous machines and timing devices

5. The belt life is greatly influenced with temperature changes, improper belt tension andmismatching of belt lengths

6. The centrifugal tension prevents the use of V-belts at speeds below 5 m / s and above 50 m / s.

20.5

20.5 Ratio of Driving Tensions for V-beltRatio of Driving Tensions for V-belt

A V-belt with a grooved pulley is shown in Fig 20.2.

Let R1 = Normal reactions between belts and

sides of the groove

R = Total reaction in the plane of the

groove

∀ = Coefficient of friction between the

belt and sides of the groove

Resolving the reactions vertically to the groove, we have

R = R1 sin ! + R1 sin ! = 2R1 sin !

Fig 20.2 V-belt with pulley.

* The ratio of tensions in V-belt drive is cosec ! times the flat belt drive.

Consider a small portion of the belt, as in Art 18.19, subtending an angle %& at the centre, the

tension on one side will be T and on the other side (T + %T) Now proceeding in the same way as in

Art 18.19, we get the frictional resistance equal to ∀ R cosec ! against ∀.R Thus the relation between

T1 and T2 for the V-belt drive will be

2.3 log (T1/ T2) = ∀∋& cosec !

20.6

20.6 V-flat DrivesV-flat Drives

In many cases, particularly, when a flat belt is

replaced by V-belt, it is economical to use flat-faced

pulley, instead of large grooved pulley, as shown in

Fig 20.3 The cost of cutting the grooves is thereby

eliminated Such a drive is known as V-flat drive.

Example 20.1 A compressor, requiring 90 kW, is to run at about 250 r.p.m The drive is by V-belts from an electric motor running at 750 r.p.m The diameter of the pulley on the compressor shaft must not be greater than 1 metre while the centre distance between the pulleys is limited to 1.75 metre The belt speed should not exceed 1600 m / min.

Determine the number of V-belts required to transmit the power if each belt has a sectional area of 375 mm 2 , density 1000 kg / m 3 and an allowable tensile stress of 2.5 MPa The groove angle of the pulleys is 35° The coefficient of friction between the belt and the pulley is 0.25 Calculate also the length required of each belt.

cross-Solution. Given : P = 90 kW = 90 × 103 W ; N2 = 250 r.p.m ; N1 = 750 r.p.m ; d2 = 1 m ;

x = 1.75 m ; v = 1600 m/min = 26.67 m/s ; a = 375 mm2 = 375 × 10– 6 m2; ( = 1000 kg / m3; ) = 2.5MPa = 2.5 N/mm2; 2! = 35° or ! = 17.5° ; ∀ = 0.25

First of all, let us find the diameter of pulley on the motor shaft (d1) We know that

0.33 m750

d N d N

∃

V-belt pulley Flat pulley

Fig 20.3. V-flat drive.

5-tine clamps of a material handlesr

For an open belt drive, as shown in Fig 20.4,

We know that mass of the belt per metre length,

m = Area × length × density = 375 × 10–6 × 1 × 1000 = 0.375 kg / m

+ Centrifugal tension,

TC = m.v2 = 0.375 (26.67)2 = 267 Nand maximum tension in the belt,

T = ) × a = 2.5 × 375 = 937.5 N + Tension in the tight side of the belt,

T1 = T – TC = 937.5 – 267 = 670.5 NLet T2 = Tension in the slack side of the belt

We know that

1 2

2.3 log T

T

− / 0

1 2 = ∀∋& cosec ! = 0.25 × 2.76 × cosec 17.5°

Power transmitted per belt #16.085# Ans.

Length of each belt

We know that radius of pulley on motor shaft,

r1 = d1/ 2 = 0.33 / 2 = 0.165 mand radius of pulley on compressor shaft,

Solution. Given : n = 2 ; 2 ! = 30° or ! = 15° ; a = 750 mm2 = 750 × 10–6 m2; ∀ = 0.12 ; ( = 1.2Mg/m3 = 1200 kg/m3; ) = 7 MPa = 7 × 106 N/m2 ; d = 300 mm = 0.3 m ; N = 1500 r.p.m.

We know that mass of the belt per metre length,

m = Area × length × density = 750 × 10–6 × 1 × 1200 = 0.9 kg/m

and maximum tension, T = ) × a = 7 × 106 × 750 × 10–6 = 5250 N

We know that tension in the tight side of the belt,

T1 = T – TC = 5250 – 500 = 4750 NLet T2 = Tension in the slack side of the belt

Since the pulleys are of the same size, therefore angle of lap (&) = 180° = , rad

We know that

1 2

2.3 log T

T

− / 0

1 2 = ∀∋& cosec ! = 0.12 × , × cosec 15° = 0.377 × 3.8637 = 1.457

1 2 =

1.457

0.63352.3 # or 1

We know that for maximum power, centrifugal tension,

d

d or

1 1 2

2

800 mm = 0.8 m375

d N

We know that the mass of the belt per metre length,

m = Area × length × density = 400 × 10–6 × 1 × 1100 = 0.44 kg / mand velocity of the belt,

T = ) × a = 2.1 × 400 = 840 N + Tension in the tight side of the belt,

1 2 = ∀.& cosec ! = 0.28 × 2.64 cosec 20° = 0.74 × 2.9238 = 2.164

1 Number of belts required

We know that the power transmitted per belt

= (T1 – T2) v = (731.4 – 83.9) 15.71 = 10 172 W = 10.172 kW + Number of belts required

2 Diameter of driven pulley shaft

Let D = Diameter of driven pulley shaft.

We know that torque transmitted by the driven pulley shaft,

T =

3

3 2

P N

Since the driven pulley is overhung and the distance of the centre from the nearest bearing is

200 mm, therefore bending moment on the shaft due to the pull on the belt,

Solution Given : P = 60 kW ; N1 = 750 r.p.m ; N2 = 300 r.p.m ; d2 = 1500 mm ; x = 1650 mm ; Overload factor = 1.5 ; a = 350 mm2 = 350 × 10–6 m2; ( = 1000 kg/m3; ) = 2 MPa = 2 N/mm2;

∀ = 0.28 ; 4 = 40 MPa = 40 N/mm2

1 Design of the belt drive

First of all, let us find the diameter (d1) of the motor pulley We know that

1 2

N

2 1

d

2 2 1

1

600 mm = 0.6 m750

d N

We know that the angle of contact,

& = 180° – 2∗ = 180 – 2 × 15.83 = 148.34°

= 148.34 × , / 180 = 2.6 radLet T1 = Tension in the tight side of the belt, and

T2 = Tension in the slack side of the belt

Assume the groove angle of the pulley, 2 ! = 35° or ! = 17.5° We know that

1 2

2.3 log T

T

− / 0

1 2 = ∀∋& cosec ! = 0.28 × 2.6 × cosec 17.5° = 2.42

1 2 = 2.42 / 2.3 = 1.0526 or

1 2

and mass of the belt per metre length,

m = Area × length × density = 350 × 10–6 × 1 × 1000 = 0.35 kg / m

+ Centrifugal tension in the belt,

TC = m.v2 = 0.35 (23.66)2 = 196 Nand maximum tension in the belt,

T = Stress × area = ) × a = 2 × 350 = 700 N + Tension in the tight side of the belt,

T1 = T – TC = 700 – 196 = 504 N

11.28 11.28

T

We know that the power transmitted per belt

= (T1 – T2) v = (504 – 44.7) 23.66 = 10 867 W = 10.867 kW

Since the over load factor is 1.5, therefore the belt is to be designed for 1.5 × 60 = 90 kW

+ Number of belts required

Power transmitted per belt #10.867 # Ans.

Since the V-belt is to be designed for 90 kW, therefore from Table 20.1, we find that a ‘D’ type

of belt should be used

We know that the pitch length of the belt,

+ Pitch length of the belt,

Let D = Diameter of the shaft.

We know that the torque transmitted by the driven or compressor pulley shaft,

T =

3 2

3 Design of the pulley

The dimensions for the standard V-grooved pulley (Refer Fig 20.1) are shown in Table 20.2, from which we find that for ‘D’ type belt

w = 27 mm, d = 28 mm, a = 8.1 mm, c = 19.9 mm, f = 24 mm, and e = 37 mm.

We know that face width of the pulley,

B = (n – 1) e + 2 f = (9 – 1) 37 + 2 × 24 = 344 mm Ans.

4 Design for key

The standard dimensions of key for a shaft of 80 mm diameter are

Width of key = 25 mm Ans.

and thickness of key = 14 mm Ans.

Example 20.5 A V-belt is driven on a flat pulley and a V-pulley The drive transmits 20 kW from a 250 mm diameter V-pulley operating at 1800 r.p.m to a 900 mm diameter flat pulley The centre distance is 1 m, the angle of groove 40° and ∀ = 0.2 If density of belting is 1110 kg / m 3 and allowable stress is 2.1 MPa for belt material, what will be the number of belts required if C-size V-belts having 230 mm 2 cross-sectional area are used.

Solution. Given : P = 20 kW ; d1 = 250 mm = 0.25 m ; N1 = 1800 r.p.m ; d2 = 900 mm = 0.9 m ;

x = 1 m = 1000 mm ; 2 ! = 40° or ! = 20° ; ∀ = 0.2 ; ( = 1110 kg/m3; ) = 2.1 MPa = 2.1 N/mm2;

a = 230 mm2 = 230 × 10–6 m2

Fig 20.5 shows a V-flat drive First of all, let us find the angle of contact for both the pulleys.

From the geometry of the Fig 20.5, we find that

&2 = 180° + 2∗ = 180° + 2 × 18.96 = 217.92°

= 217.92 × , / 180 = 3.8 rad

We have already discussed that when the pulleys have different angle of contact (&), then thedesign will refer to a pulley for which ∀∋& is small

We know that for a smaller or V-pulley,

∀∋& = ∀∋&1 cosec ! = 0.2 × 2.48 × cosec 20° = 1.45and for larger or flat pulley,

∀∋& = ∀∋&2 = 0.2 × 3.8 = 0.76Since (∀∋&) for the larger or flat pulley is small, therefore the design is based on the larger or

Mass of the belt per metre length,

m = Area × length × density = a × l × (

= 230 × 10–6 × 1 × 1100 = 0.253 kg / m

+ Centrifugal tension,

TC = m.v2 = 0.253 (23.56)2 = 140.4 NLet T1 = Tension in the tight side of the belt, and

T2 = Tension in the slack side of the belt

We know that maximum tension in the belt,

1 2 = ∀∋&2 = 0.2 × 3.8 = 0.76

1 2

log T

T

− / 0

1 2 = 0.76 / 2.3 = 0.3304 or

1 2

We know that number of belts required

Power transmitted per belt # 4.302 # A Ans.

20.7

20.7 Rope DrivesRope Drives

The rope drives are widely used where a large amount of power is to be transmitted, from onepulley to another, over a considerable distance It may be noted that the use of flat belts is limited forthe transmission of moderate power from one pulley to another when the two pulleys are not morethan 8 metres apart If large amounts of power are to be transmitted, by the flat belt, then it wouldresult in excessive belt cross-section

The ropes drives use the following two types of ropes :

1 Fibre ropes, and 2 *Wire ropes

The fibre ropes operate successfully when the pulleys are about 60 metres apart, while the wireropes are used when the pulleys are upto 150 metres apart

20.8

20.8 Fibre RopesFibre Ropes

The ropes for transmitting power are usually made from fibrous materials such as hemp, manilaand cotton Since the hemp and manila fibres are rough, therefore the ropes made from these fibresare not very flexible and possesses poor mechanical properties The hemp ropes have less strength ascompared to manila ropes When the hemp and manila ropes are bent over the sheave, there is somesliding of the fibres, causing the rope to wear and chafe internally In order to minimise this defect, therope fibres are lubricated with a tar, tallow or graphite The lubrication also makes the rope moistureproof The hemp ropes are suitable only for hand operated hoisting machinery and as tie ropes forlifting tackle, hooks etc

The cotton ropes are very soft and smooth The lubrication of cotton ropes is not necessary But

if it is done, it reduces the external wear between the rope and the grooves of its sheaves It may benoted that the manila ropes are more durable and stronger than cotton ropes The cotton ropes arecostlier than manila ropes

* Wire ropes are discussed in Art 20.12.

Notes : 1. The diameter of manila and cotton ropes usually ranges from 38 mm to 50 mm The size of the rope

is usually designated by its circumference or ‘girth’.

2. The ultimate tensile breaking load of the fibre ropes varies greatly For manila ropes, the average

value of the ultimate tensile breaking load may be taken as 500 d2 kN and for cotton ropes, it may be taken as

350 d2 kN, where d is the diameter of rope in mm.

20.9

20.9 Advantages of Fibre Rope DrivesAdvantages of Fibre Rope Drives

The fibre rope drives have the following advantages :

1. They give smooth, steady and quiet service

2. They are little affected by out door conditions

3. The shafts may be out of strict alignment

4. The power may be taken off in any direction and in fractional parts of the whole amount

5. They give high mechanical efficiency

20.10

20.10 Sheave for Fibre RopesSheave for Fibre Ropes

The fibre ropes are usually circular in cross-section as shown in Fig 20.6 (a) The sheave for the fibre ropes, is shown in Fig 20.6 (b) The groove angle of the pulley for rope drives is usually 45°.

Fig 20.6. Rope and sheave.

The grooves in the pulleys are made narrow at the bottom and the rope is pinched between the edges

of the V-groove to increase the holding power of the rope on the pulley The grooves should be

finished smooth to avoid chafing of the rope The diameter

of the sheaves should be large to reduce the wear on the

rope due to internal friction and bending stresses The

proper size of sheave wheels is 40 d and the minimum

size is 36 d, where d is the diameter of rope in cm.

Note : The number of grooves should not be more than 24.

20.11

20.11 Ratio of Driving Tensions for FibreRatio of Driving Tensions for Fibre

Rope

A fibre rope with a grooved pulley is shown in

Fig 20.5 (a) The fibre ropes are designed in the similar

way as V-belts We have discussed in Art 20.5, that the

ratio of driving tensions is

1 2

2.3 log T

T

− / 0

1 2 = ∀∋&5cosec !

where ∀, & and ! have usual meanings

Rope drives

Example 20.6 A pulley used to transmit power by means of ropes has a diameter of 3.6 metres and has 15 grooves of 45° angle The angle of contact is 170° and the coefficient of friction between the ropes and the groove sides is 0.28 The maximum possible tension in the ropes is 960 N and the mass of the rope is 1.5 kg per metre length Determine the speed of the pulley in r.p.m and the power transmitted if the condition of maximum power prevail.

Solution Given : d = 3.6 m ; n = 15 ; 2 ! = 45° or ! = 22.5° ; & = 170° = 170 × , / 180

= 2.967 rad ; ∀ = 0.28 ; T = 960 N ; m = 1.5 kg / m

Speed of the pulley

Let N = Speed of the pulley in r.p.m.

We know that for maximum power, speed of the pulley,

Power transmitted

We know that for maximum power, centrifugal tension,

TC = T / 3 = 960 / 3 = 320 N + Tension in the tight side of the rope,

T1 = T – TC = 960 – 320 = 640 NLet T2 = Tension in the slack side of the rope

We know that

1 2

2.3 log T

T

− / 0

1 2 =

2.170.9435

Example 20.7. A rope pulley with 10 ropes and a peripheral speed of 1500 m / min transmits

115 kW The angle of lap for each rope is 180° and the angle of groove is 45° The coefficient of friction between the rope and pulley is 0.2 Assuming the rope to be just on the point of slipping, find the tension in the tight and slack sides of the rope The mass of each rope is 0.6 kg per metre length.

Solution. Given : n = 10 ; v = 1500 m/min = 25 m/s ; P = 115 kW = 115 × 103 W ; & = 180°

= , rad ; 2 ! = 45° or ! = 22.5° ; ∀ = 0.2 ; m = 0.6 kg / m

Let T1 = Tension in the tight side of the rope, and

T2 = Tension in the slack side of the rope

We know that total power transmitted (P),

115 × 103 = (T1 – T2) v × n = (T1 – T2) 25 × 10 = 250 (T1 – T2)

We also know that

1 2

2.3 log T

T

− / 0

T # (Taking antilog of 0.714) (ii)

From equations (i) and (ii), we find that

Solution. Given : P = 600 kW ; d = 4 m ; N = 90 r.p.m ; & = 160° = 160 × ,/180 = 2.8 rad ;

T1 = T – TC = 2400 – 533 = 1867 NLet T2 = Tension in the slack side of the rope

We know that

1 2

2.3 log T

T

− / 0

1 2 = ∀∋& cosec ! = 0.28 × 2.8 × cosec 22.5° = 0.784 × 2.6131 = 2.0487

1 2

log T

T

− / 0