ME-804: MACHINE DESIGN, UNIT 1, DESIGN OF BELT, POPE AND GAIN DRIVES INTRODUCTION:
Belt, hope chain and gear drives etc are all means of power transmission from one shaft (driving shaft)
to the other driven shaft). Following are the factors considered for selection of the transmission
mechanism:
a.) Distance between the two shafts
b.) The magnitude of the torque to be transmitted
c.) Availability of time
d.) Relative position of the shafts
e.) Velocity – ratio
Belt drives: This is a type of friction drive which uses a belt & is generally employed for transmitting
power over a large distance (up to 25 meters) between two parallel shafts. It consists of an endless belt
wrapped tightly over the pulleys mounted either side (i.e. on driving and driver shafts).
The power transmission takes place due to the friction between pulleys and
belt. The flexibility & tension of the belt facilitates absorption of vibrations & shocks and in turn ensures
high transmission efficiency. Belt drives are capable of operating satisfactorily in open environments
where they are exposed to dirt, dust and hazardous condition.
The tension on either side of the belt is same at rest. However during transmission, difference
in tension starts to develop i.e. tension at one side increases while that on the other side decreases..
This difference in tension is responsible for power transmission in belt drives.
It should be noted that since belt drive is a type of friction drive, slip between belt and pulleys in an
inherent characteristic of this drive, and thus, the velocity (or speed) ratio to these drives is never
constant. Besides, the problem of creep is also prominent in belt-drives.
Factors affecting the selection of the belt drive ;
1. Centre distance between the shafts
2. Power to be transmitted
3. Speed of driving and driven shaft
4. Speed reduction ratio
5. Positive drive requirements
6. Availability of space
7. Service conditions
8. Shaft relationship
Classification of belt drives: Belt drives can be classified as follows:
A. On the basis of peripheral speed :
i) Light drives : used for small power transmission (up to 12m/sec)
ii) Medium drives : used for medium power transmission (up to 24m/sec)
iii)Heavy drives : used for large power transmission (above 24/sec)
B. On the basis of the type of belt ;
i) flat belt drive
ii) V-belt drive
Iii) Round (circular) belt drive
iv)Timing belt drive
i) Flat belt drive: It uses a flat belt which is in the form of a thin flat band, designed to run on a
cylindrical pulley. It is used for moderate range power transmission when the pulleys are
repeated by not more than 8meters.
ii) V-belt drives: It used V-belt which has a trapezoidal cross – section and runs in a pulley groove of
V-shape. It in used for high range power transmission between the closely spaced pulleys.
iii) Round (or circular) belt or rope drives: They use circular belts made to fibrous material such as
hemp. The application of this drive for power transmission is very limited. However, they are
used for high range power transmission between the two pulleys separated by a large
distance (more than 8m).
iv) Timing (or toothed belt drive): It is a positive drive which has teeth corresponding to the teeth
provided on the pulley. They are used for power transmission inside automotive engines to
control the valve timing for following reasons:
a) Constant speed operation
b) No slippage
c) Require least tension of all the belts
d) High efficiency
e) High load bearing capacity
Belt – materials: The materials selection for belts and ropes depends upon the operating conditions
and service requirements.
The material used must have following characteristics:
a) Strength
b) Flexibility
c) Pliability
d) Durability
e) High co-efficient of friction
f) Light weight
Following are the commonly used materials:
1.) Leather: It is one of the most commonly used materials for manufacturing of flat belts. They are
known for their flexibility and load bearing capacity. They are capable of withstanding fluctuating
loads as well as high over loads. They made of leather strips of thickness between 3 to 6 mm&
length between 1to1.5 meters. Belt of different lengths and thickness are formed by joining
these individual strips. On the basis of thickness, flat belts can be classified as followers:
a.) Single ply: They use only one leather strip & thus, have thickness between 3to6 mm.
b.) Double ply : They are formed by joining (cementing) two strips of leather (one over the
other) and thus, have the thickness between (6mm to 13mm)
c.) Triple ply: They are formed by joining three strips leather (one over the other) & thus, have
thickness between 9mmto20mm.
Leather may be oak tanned, mineral salt tanned or vegetable compound, tanned. In order to keep
leather soft and flexible, it is cleaned or treated with some cleansing compound.
It should be noted that the modulus of elasticity of leather belts varies between 140to320 Mpa and
its ultimate strength can be up to 28MPa. These belts have a density of 1000Kg/m3.
2.) Fabric or cotton belts : these belts are made by folding canvas or cotton duck into three or more
layers & stitching together. These belts can also be made by weaving the material into strips of
desired width and thickness. In order to make the fabric or cotton belts water – proof and
prevent any damage to the fabric, they are impregnated with linseed oil.
These belts are cheap and provide a tough service. They are
particularly very useful for farm machinery because of their high durability in dry and dusty
environment. The density of these belts range between 1150to1250Kg/m3.
3.) Rubber belt : these belts are formed in a similar way as cotton belts except that between every
pair offloaded canvas or cotton duck layer, these is a layer of rubber. Besides, a thin layer of
rubber is also present on the faces of the belt assembly.
These belts are very flexible and are resistant to
moisture; acids etc and thus, are generally preferred for outdoor service. Besides, they have an
advantage over the other belts that it is very simple to form an endless assembly of these belts.
They are vulnerable to heat and oil and thus, get readily destroyed in their presence.
4.) Balata belts : The method of manufacturing these belts is same as that of rubber belts except
that in these belts the rubber is replaced by compound “Balata” for impregnating as well as
covering the layers of canvas or cotton duck.
Balata belts are water proof & acid proof and are stronger than
the rubber belts. These belts are not affected by animal oils or alkalis. However, their operating
temperature is kept maximum up to 500 c beyond which these belts start to become soft and
sticky. These belts have a densely of 1110Kg/m3.
5.) Steel belts: Thin steel belts (thickness approximately 1mm) have the quality to run very high
speeds since the centrifugal force acting on the belt are very small. However, these belts require
perfect alignment of pulleys. Besides the co-efficient of friction of these belts is also low.
Belt fastenings: The strips of flat belts are fastened together to form a continuous loop of belts
(endless belts). Following are the types of fasteners generally used for this purpose:
a.) Cementing
b.) Lacing
c.) Metal hinges (or hooks )
a.) Cementing: the joints formed by cementing are considered to be strongest (with efficiency
up to 100%) and thus, are more prepared than any other joint.
b.) Lacing: Laced joints are formed by punching the belt ends and lacing the two belts together
by a raw hide strip to form a joint. The efficiency of raw hide laced joint is between 40 to 60%.
Sometimes, metals are also used to lace the two belts together. The
joint formed in this fashion is made like a staple connection. The efficiency of these joints
ranges between 80to90 percent.
c.) Metal hinges: These joints are made by securing a row of wire loops in the ends & locking
them with a steel wire pin. This fastening is generally used for narrow belts. However, heavy
belts of rubber or canvas are fastened by means of belts and clamps (formed by two steel
plates).
Types of flat belt derive: Following are the major flat belt drives used to transmit power from
one pulley to the other.
a.) Open belt drive: In the type, shafts are arranged parallel to each other and are made to
rotate in the same direction. In this type, the driver wheel- along its direction of rotation
pulls the belts from one side & supplies it to the driven on the other side. The tension of the
pulled side is more & thus, is called the tight side. Whereas, the tension of the supply side is
less and thus, is called as slack side.
It should be noted to that when the centre distance of the driving & driven wheels is large,
the lower side of the belt should be the tight one.
b.) Crossed or twist belt drive : In this type, the shafts are arranged parallel to each other but
unlike the open belts drive, are rotated in opposite direction. In this type, the driving wheelalong is direction of rotation- pulls the belts form one side &supplies it on the other to the
driven wheel. The tension on the pull side is more and therefore, it is known as tight side
whereas, the tension of the supply side is less and thus , it is known as slack side.
It should be noted that in order to avoid excessive wear and tear, due to rubbing, at the
point where the belts cross each other, the centre distance between the shafts is kept at a
distance of about 20b (where, b=width of the belt) and the speed of the belts is kept within
15m/sec.
c.) Quarter turn belt drive : Also known as “right angle belt drive”, in this type, the shafts are
arranged perpendicular to each other and are made to rotate in either clock wise or anti –
clock wise direction. In order to prevent the belt from leaving the pulley, the face of the
pulley is kept sufficiently wide (i.e. width >1.4b).
However, when a reversible motion is desired or when
the pulley cannot be arranged in right angle, a guide pulley is used in between the driven &
the driver pulley. The drive thus formed is known as “quarter turn drive with a guide pulley.”
d.) Stepped pulley drive : Also known as Cone pulley drive, in this type, the driving shaft runs
at a constant speed and the speed of the driven shaft is varied by shifting the belt from one
step to the other.
e.) Jockey pulleys drive (belt drive with an idler pulley): In this type, an idler pulley is used with
the driver &driven pulley for power transmission. Idler pulleys are employed with the shafts
arranged in parallel for any of the following reasons:
i.) When it is impossible to employ a simple open belt drive due to two small angle of
contact between the belt and the smaller pulley.
ii.) To obtain a high velocity ratio
iii.) To obtain the required belt tension, which cannot be obtained by any other means.
f.) Fast & was pulley drive : This drive is used when the driver shaft is to be rotated too often
or stopped without effecting the operation of the driver shaft. In this arrangement, two
pulleys are provided on the driver shaft : A pulley keyed to the driven shaft, known as “fast
pulley,” which rotates with same angular velocity as that of the driven shaft and a pulley
known as slow pulley, which is set to run freely over two driver shaft and thus, is capable of
transmitting any power.
Thus, the driver shaft suns at a uniform velocity and the rotation of the
driven shaft is stopped by pushing the belt from the fast pulley to the slow pulley by means
of a shifting lever.
g.) Compound belt drive: This drive is used when the power from the drive shaft is to be
transmitted to the driver shaft through a number of pulleys.
Laws of belting: The law of belting states that in order to prevent the belt from running off
the pulley, the centre line of the bet as it approaches any pulley must lie in a plane
perpendicular to the axis of that pulley or must be coplanar with the pulley.
Belt speed : Any increase in the speed of the belt results in increase of the centrifugal force
acting on the belt, which tries to pull the belt away from the pulley and thereby, leads to
decrease in the power transmitted by the belt. Therefore, in order to ensure an efficient
transmission of the power, the belt speed is generally kept within the range of 15to30
m/sec.
Co-efficient of friction between belt and pulley: Co-efficient of friction is an imperative
factor, which has to be taken into account to ensure on efficient power transmission.
Following are the factors, which affect the co-efficient of friction between belt and pulleys: a.) Pulley and belt material
b.) Angular velocity of pulley and belt
c.) Slip
Velocity ratio of open Belt drive: Velocity ratio can be defined as the ratio of the velocity of
the follower ( or driven) to the velocity of the driver.
Let, 𝑑1 & 𝑑2 be the diameters of driver & driver pulley respectively N1& 𝑁2 be the angular
velocity (in r.p.m) of driver &driven pulley respectively
Now, length of the belt passing over the driver in one minute
= circumference of driver × number of revolutions per minute
= 2  𝑟1 𝑁1
=  𝑑1 𝑁1
Similarly, length of belt passing over the driver in one minute
=  𝑑2 𝑁2
We know that, length of the belt passing over driver in one minute = length of the belt
passing over the driver in one minute
i.e.  𝑑1 𝑁1 =  𝑑2 𝑁2
𝑑
Thus, 𝑑1 =
2
𝑁2
𝑁1
= velocity ratio
It should be noted that when the belt thickness is considered, the mean diameter of rotation
will be equal to the sun of diameter of the driver (or the driven) & the thickness of the belt
(d+t) and the velocity ratio would be expressed as follows:
𝑁1
𝑁2
=
𝑑1 +𝑡
𝑑2 +𝑡
Alternatively, the velocity of all open belt drive can be calculated as follows :
Angular velocity of the driver (1) =
2 𝑁1
60
Angular velocity of the driven (2) =
2 𝑁2
60
Now, when there is no slip between the belt & the pulleys and a negligible belt thickness, the
peripheral velocity of the driver & driven pulley will be the same and would be equal to the
velocity of the belt (V)
Peripheral velocity of the driver = 1 r1
Peripheral velocity of the driven = 2 r2
for no slip : 1 r1 = 2 r2
=
2 𝑁1
60
× 𝑟1 =
2 𝑁2
60
× 𝑟2
=  𝑑1 𝑁1 =  𝑑2 𝑁2
=
𝑑1
𝑑2
=
𝑁2
𝑁1
= velocity ratio
Velocity ratio for compound drive: Velocity of the compound drive is given by the following expression:
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑑𝑟𝑖𝑣𝑒𝑛 𝑝𝑢𝑙𝑙𝑒𝑦
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑑𝑟𝑖𝑣𝑒𝑟 𝑝𝑢𝑙𝑙𝑒𝑦
=
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑑𝑟𝑖𝑣𝑒𝑟𝑠
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑑𝑟𝑖𝑣𝑒𝑛𝑠
Slip of the belt : It can be defined as the difference between the linear speed of the pulley rive & the
belt. It is expressed in percentage (%).
The belt rotates along with the driver pulley and intern relates the driver pulley because of the friction
between the belt & the pulleys. However, the frictional grip is sometimes (generally at high velocities)
insufficient to ensure the same velocity of the driver and the belt and thus, driver will have a relative
motion with the belt. Similarly, sometimes the belt and the driven pulley will have a relative motion with
each other.
The slopping of the belt results in a reduced velocity ratio of the drive. The phenomenon of slip
increases with increase in the drive load.
Let,
𝑁1 & 𝑁2 = Speed (in rpm) of driver & driven pulley respectively
𝑠1 & 𝑠2 (in %) = slip between the driver & belt and slip between the
belt and the driven pulley respectively
V= velocity of the belt passing over the driver pulley/ second
Peripheral velocity of the belt = 1 r1
Where, 1 = angular velocity of the driver =
r1 = radius of the driver pulley =
 1 r1 =
2 𝑁1
60
𝑑1
2
×
=
2 𝑁1
60
𝑑1
2
𝑑1 𝑁1
60
Loss of peripheral velocity due to slip between the driver & the belt =
𝑑1 𝑁1
60
×
𝑠
100
Therefore, the total peripheral velocity of belt on considering slip between the belt & the
driver pulley:
𝑑1 𝑁1
60
=
𝑑1 𝑁1
60
-
𝑑1 𝑁1
(1 −
𝑠1
100
×
60
𝑠1
)
100
Now, this loss in peripheral velocity of belt would also result in loss of peripheral velocity of
the driver pulley and thus, the velocity of driven pulley (or the velocity of belt passing over
the driven pulley per second) can be expressed as:
Peripheral velocity of belt – loss of belt velocity due to slip
Where, velocity of belt
And loss of velocity =
𝑑1 𝑁1
=
60
𝑑1 𝑁1
60
(1 −
𝑠1
)
100
𝑠
𝑠
1
2
(1 − 100
) (1 − 100
)
 Peripheral velocity of the belt passing over the driven pulley per second:
V=
𝑑1 𝑁1
60
𝑠
𝑠
1
2
(1 − 100
) (1 − 100
)
Where, V= peripheral velocity of the belt over the driven pulley per second
= 2 r2 = =
2 𝑁2
60
×
𝑑2
2
=
𝑑2 𝑁2
60
Therefore,
𝑑2 𝑁2
60
=
𝑑1 𝑁1
60
𝑠
𝑠
1
2
(1 − 100
) (1 − 100
)
𝑠
𝑠
𝑠 𝑠
2
2
1 2
= 𝑑2 𝑁2 = 𝑑1 𝑁1 (1 − 100
− 100
+ 100
)
𝑠 𝑠
1 2
= 𝑑1 𝑁1 [1 − ( 100
)] (𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔
𝑠
𝑠1 𝑠2
100
)
𝑑2 𝑁2 = 𝑑1 𝑁1 (1 − 100) (where, s = total slip = s1 + s2)
𝑁2
=
𝑁1
𝑑1 +𝑡
𝑑2 +𝑡
(1 −
𝑠
)
100
Creep of the belt : It can be defined as the relative motion of the due to the elasticity of the
belt material. It is expressed in %
In a belt drive, belt is subjected to tensions during power transmission between the driver
and the driven pulley. The tension in the belt on the tight side is always greater than its
tension on the slack side and thus, any belt portion with a unit length expands (or extends)
when it passes from the slack side to the tight side. Thus, the belt undergoes a change in
length between the driver & driven pulley and thereby, there develops a relative motion
between the belt & the pulley surface.
This relative motion results in reduction to the speed of the driven pulley.
In practice, the amount of creep is about 1% and the combined effect of slip &creep should
not exceed 3%.
Let,
E= young’s modulus of the belt material
1 = stresses on the tight side of the belt
2 = stresses on the tight side of the belt
Now,
Velocity ratio of an open belt drive, considering creep is expressed as:
𝑁2
𝑁1
=
𝑑1
𝑑2
𝐸+√𝜎2
)
𝐸+√𝜎1
(
Miscellaneous transmission losses : Apart from slip & creep. Following are some
miscellaneous losses in transmission through open belt drive:
1. Resistance of air to the movement of belt and pulleys
2. Friction in bearings of the rotational members.
3. Bending of belt.
Length of an open belt drive: Let two-pulley i.e. larger and smaller pulley with center A&B
respectively rotate in clock wise direction.
diagram
Let,
r1 = ratios of large pulley
r2 = ratios of the small pulley
𝑥 =centre distance between the driver & driven pulley
From the diagram,
Belt repeaters, form larger pulley at D & meets it back at M and from the smaller pulley it
separates at F & meets it back at C.
Join AD,AE and BC, BF
Let us assume, a point (‘G’) on AD and point ‘H’ on AE such that BG is parallel to CD & perpendicular
to AD and BH is parallel to EF and perpendicular to AD
Let ∠G B A = 𝛼

Thus, ∠G A B = (900 -𝛼) or (2 − 𝛼 ) radians
And, ∠OAD = 𝛼 and similarly, ∠OBC=𝛼
Now, since,
AD=AG+GD
r1 = AG+ r2
AG= r1- r2
In  ABG,
AB 2 = AG2 + GB2
𝑥 2 = (r1- r2)2 + GB2
 GB = √𝑥 2 − (r1 − r2 ) = DC ------ (i)
And ,
Sin 𝛼 =
AG
AB
=
𝑟2−𝑟1
𝑥
And, since 𝛼 is very small, sin 𝛼 = 𝛼
𝛼 =
𝑟1−𝑟2
------- (ii)
𝑥
Total length of the belt (L) = 𝐴𝑟𝑐 𝐸𝑋𝐷 + 𝐷𝐶 + 𝐴𝑟𝑐 𝐶𝑌𝐹 + 𝐸𝐹
= 2 (𝐴𝑟𝑐𝑋𝐷 + 𝐴𝑟𝑐 𝐶𝑌 + 𝐶𝐷)
-----(III)
[Since, EF=DC, Arc XD = Arc XE & Arc CY = Are YF]
Where,


Are XD = r1 (2 ∓ 𝛼 ) & Are CY = r2 (2 − 𝛼 )
Therefore, the total length of the belt:
𝜋
𝜋
2 [𝑟1 ( 2 + 𝛼) + √𝑥 2 − (𝑟1 − 𝑟2 )2 + 𝑟2 ( 2 − 𝛼)]
𝜋
= 2 [ 2 (𝑟1 + 𝑟2 ) + 𝑥 (𝑟1 − 𝑟2 ) + √ 𝑥 2 − (𝑟1 − 𝑟2 )2 ] --- (iv)
The expression (iv) gives the enact length of the open belt drive alternatively, an approximate value of
the total length of the belt can be calculated as follows:
GB = DC = √ 𝑥 2 − (𝑟1 − 𝑟2 )2 ---- form (1)
= 𝑥 √1 −
(𝑟1 −𝑟2 )2
𝑥
(𝑟1 −𝑟2 )2 1/2
= 𝑥 [1 −
]
𝑥
---(V)
Using Binomial expression in equation (v)
1 𝑟1 −𝑟2 2
)
𝑥
DC = x [1 − 2 (
+ − − −] = 𝑥 −
(𝑟1 −𝑟2 )2
2𝑥
On putting this value in eq. (iii) the total length of the belt :
𝜋
2[𝑟1 ( 2 + 𝛼) + 𝑥 −
(𝑟1 −𝑟2 )2
2𝑥
𝜋
+ 𝑟2 ( 2 − 𝛼) ]
𝜋
=2[ 2 (𝑟1 + 𝑟2 ) + 𝛼(𝑟1 − 𝑟2 ) + 𝑥 −
(𝑟1 −𝑟2 )
2𝑥
=[𝜋(𝑟1 + 𝑟2 ) + 2𝛼(𝑟1 − 𝑟2 ) + 2𝑥 −
Now, 𝛼 =
(𝑟1 −𝑟2 )2
𝑥
]
(𝑟1 −𝑟2 )2
𝑥
]
-----from (II)
 Length of the belt:
[𝜋(𝑟1 + 𝑟2 ) + 2
= 𝜋(𝑟1 + 𝑟2 ) + 2
=[𝜋(𝑟1 + 𝑟2 ) +
(𝑟1 −𝑟2 )
𝑥
(𝑟1 −𝑟2 )2
𝑥
(𝑟1 −𝑟2 )2
𝑥
(𝑟1 − 𝑟2 ) + 2𝑥 −
+ 2𝑥 −
(𝑟1 −𝑟2 )2
𝑥
]
(𝑟1 −𝑟2 )2
𝑥
+ 2𝑥] ------(vi)
The expression (vi) gives the approximate length of the open belt drive
Length of cross-belt drive : Let the two pulleys i.e. larger and smaller pulley with centers A&B
respectively rotate in opposite directions
Drawing ???????????????/
Let, 𝑥 = distance between the centers of two pulleys
𝑟1 & 𝑟2 = radius of larger and smaller pulley respectively
L= total length of the cross belt
From the diagram:
Belt separates from the larger pulley at E and meets it back at D and from the smaller pulley it separates
at ‘C’ & meets it back at ‘F’
Join AE, AD, BC & BF.
Now, let us extend AE to an imaginary point ‘G’ & AD to an imaginary point ‘H’ such that BG is parallel to
FE & perpendicular to AG and BH is parallel to C& perpendicular to AH
Let, ∠G B A = 𝛼

Thus, ∠ G A B = (900 -𝛼) or (2 − 𝛼 ) radians
And ∠OAD = 𝛼 and similarly, ∠OBC=𝛼
Now, since,
AG = AE+EG
AG = 𝑟1 + 𝑟2
And In  ABN,
AB 2 = AG2 + GB2
2 = (r1+r2)2 + GB2
= GB2 = 𝑥 2 - (r1+ r2)2
GB = √𝑥 2 − (r1 + r2 )2 = EF ------ (i)
AG
And, Sin 𝛼 = AB =
𝛼=
𝑟1 +𝑟2
𝑥
𝑟1+𝑟2
𝑥
Since is very small; sin 𝛼 = 𝛼 ------ (ii)
Total length of the belt (L) = Arc EXD + DC + Arc CYF + EF
= 2(Arc EX + Arc CY + CD)
Since EF = DC, Arc XD = Arc XE and Arc CY = Arc YF
Where,


Arc XD = r1 (2 ∓ 𝛼 ) & Arc CY = 𝑟2 (2 − 𝛼 )------------- (iii)
Therefore total length of the belt:
𝜋
𝜋
2[𝑟1 ( 2 + 𝛼) + 𝑟2 ( 2 ∓ 𝛼) + √ 𝑥 2 − (𝑟1 ∓ 𝑟2 )2 ]
𝜋
=2[ 2 (𝑟1 + 𝑟2 ) + 𝛼 (𝑟1 ∓ 𝑟2 ) + √ 𝑥 2 − (𝑟1 + 𝑟2 )2 ]-------------- (iv)
The expression (iv) gives the enact length of the cross – belt drive.
Alternatively, an approximate value of the cross – belt length can be calculated as follows:
From equation (1) :
GB = EF = √ 𝑥 2 − (𝑟1 + 𝑟2 )2
𝑟1 +𝑟2 2
)
𝑥
= 𝑥 √1 − (
= 𝑥 [1 −
(𝑟1 +𝑟2 )2 1/2
]
𝑥
---(V)
On applying the binomial expression in (v)
1 𝑟1 +𝑟2 2
)
𝑥
EF = 𝑥 [1 − 2 (
+ − − −] = 𝑥 −
(𝑟1 +𝑟2 )2
2𝑥
On putting this value in eq. (iii) the total length of the belt will be:
𝜋
2
2[𝑟1 ( + 𝛼) + 𝑥 −
(𝑟1 +𝑟2 )2
2𝑥
𝜋
2
+ 𝑟2 ( + 𝛼) ]
𝜋
2
=2[ (𝑟1 + 𝑟2 ) + 𝛼(𝑟1 + 𝑟2 ) + 𝑥 −
(𝑟1 +𝑟2 )2
2𝑥
=[𝜋(𝑟1 + 𝑟2 ) + 2𝛼(𝑟1 + 𝑟2 ) + 2𝑥 −
(𝑟1 +𝑟2 )2
𝑥
]
]
Now,
𝛼=
(𝑟1 −𝑟2 )2
𝑥
-----from (II)
Total length of the belt :
[𝜋(𝑟1 + 𝑟2 ) + 2
= 𝜋(𝑟1 + 𝑟2 ) + 2
=[𝜋(𝑟1 + 𝑟2 ) +
(𝑟1 +𝑟2 )
𝑥
(𝑟1 +𝑟2 )2
𝑥
(𝑟1 +𝑟2 )2
𝑥
(𝑟1 + 𝑟2 ) + 2𝑥 −
+ 2𝑥 −
(𝑟1 +𝑟2 )2
𝑥
]
(𝑟1 +𝑟2 )2
𝑥
+ 2𝑥] ------ (vi)
The expression (vi) gives the approximate length of the cross - belt drive
Note :
1. In open belt drive : The angle of arc of contact :
a.) At larger pulley : 𝜋 + 2𝛼
b.) At smaller pulley : 𝜋 − 2𝛼
2. In cross belt drive : The angle of are of contact on either of the pulley : 𝜋 + 2𝛼
Ratio of belt tensions: The value of ratio of belt tension plays a significant role in power transmission
since the power transmission capacity of a belt drive depends upon the difference in tensions of tight &
slack side of belt.
Let, T1 = tension on the tight side of the belt.
T2 = tension on the slack side of the belt.
 = angle of contact of belt with the pulley.
µ= co – efficient of friction between the belt and the pulley.
Drawing ????????????????????????/
Consider the driven pulley rotating in clock – wise direction. The ratio of belt tension can be found by
considering an elementary Strip ‘EF’ of the belt subsuming an angle ”θ” at the center of the pulley.
Following are the force which keep the elementary strip ‘EF’ in equilibrium:
1. Tension (T+T) at E, acting tangentially on the tight side.
2. Tension (T) at F, acting tangentially on the stack side.
3. Normal reaction (RN) acting racially outward at ‘G’
4. Frictional force (F=µR) acting tangentially at point G (perpendicular to reaction force (RN) and
opposite to the diction of rotation of pulley.
Now,
∠EBG=∠GBF = ⁄2
Therefore, ∠TGF = ⁄2
On resolving all the forces tangentially:


+ (T+) sin 2
2
 

Where, sin = [𝑠𝑖𝑛𝑐𝑒, 𝑎𝑛𝑔𝑙𝑒
2
2
2


R= T + (T+T)
2
2



R= T 2 + T 2 + T 2
 T
R= T+ 2
R= T sin
𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 ]
R= T [ 𝑠𝑖𝑛𝑐𝑒 T is very small & 𝑡ℎ𝑢𝑠,
On resolving all the force vertically:


F(=𝑅𝑁 ) = (T+T) cos 2 - T cos 2
F (=𝑅𝑁 ) = T+T –T [ 𝑠𝑖𝑛𝑐𝑒
𝑅𝑁 =T

2
 T
can be
2
𝑖𝑠 𝑠𝑜 𝑠𝑚𝑎𝑙𝑙 𝑡ℎ𝑎𝑡 𝑐𝑜𝑠
neglected ] ----(i)

𝑐𝑎𝑛
2
𝑏𝑒 𝑐𝑜𝑢𝑠𝑖𝑑𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 1 ]
𝛿𝑇
RN= 𝜇 ------------- (ii)
Now, on equating the values of RN for both vertical as well as the horizontal direction:
𝛿𝑇
= 𝑇𝛿𝜃
𝜇
𝛿𝑇
= 𝜇𝛿𝜃
𝑇
On integrating the equation (iii) between 𝑇1 & 𝑇2 and 0andθ:
𝑇 T
1 𝑇
?????? ∫𝑇 2
𝜃
= 𝜇 ∫0 𝛿𝜃 ? ? ? ? ? ? ?
𝑇
Loge (𝑇1 ) = 𝜇𝜃
2
𝑇1 𝜇𝜃
=𝑒
𝑇2
This equation (eq.iv) can be used to express the ratio of tight & stack side tension in terms of co-efficient
of friction and angle of contact.
Angle of contact :
A.) For open belt drive: Angle of contact at smaller pulley in open belt drive is smaller in
comparison to that of the larger pulley.
𝑟 −𝑟
We know that, for open belt drive: sin 𝛼 = 1 2
𝑥
Where, 𝑟1 = radius of larger pulley
𝑟2 = radius of smaller pulley
𝑥 = distance between the two pulleys
Now, from the construction, angle of contact can be expressed as:
Angle of contact () = (1800 -2 𝛼)
B.) For cross belt drive ; angle of contact at smaller pulley in cross belt drive is larger in comparison
to the open belt drive &can be expressed as :
Angle of contact () = (1800 -2 𝛼)
The value of 𝛼 can be calculated from the following expression:
𝑟 +𝑟
Sin 𝛼 = 1 𝑥 2
Power transmitted by the belt :
Let, 𝑇1 = tension in the tight side of the belt
𝑇2 = tension in the slack side of the belt
V= velocity of the belt (in m/sec)
We know that, the tension on the tight side of the belt is always greater than the slack side and thus, the
effective turning force at the circumference of the driver pulley is the difference between the tensions
on both sides (𝑇1 − 𝑇2)
Now, work done = force × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
= (𝑇1 − 𝑇2 ) V Nm/sec or (𝑇1 − 𝑇2 ) V W
And, torque transmitted on the driving pulley = (𝑇1 − 𝑇2 ) 𝑟1
Similarly, torque transmitted on the driven pulley = (𝑇1 − 𝑇2 ) 𝑟2
Centrifugal tension: Centrifugal force acting on the belt is a consequence of its mass. An inertial force
due to mass acts on the belt and tends to lift the belt off the pulley. The inertial force also causes an
increase in tension of both slack as well as the tight side.
It should be noted that at lower belt speed the centrifugal
tension is negligible. However at high speeds (i.e. speeds more than 10m/sec), it effect is considerable &
thus, taken into account.
Consider a small element AB of the belt of length (r ) which is in equilibrium under the action of
following forces:
???????????????????????????????????????????????/
A.) Centrifugal force (Fc)
B.) Centrifugal tension (Tc)
Let, m = mass of belt per unit length
r= radius of the pulley
V= peripheral velocity of the belt
= angel of contact by the elementary portion ‘AB’ with the pulley.
Now,
Length of the elementary portion ‘AB’ = r
& mass of the elementary portion ‘AB’ = r × 𝑚
Centrifugal force = mass × accleration
𝑉2
 Centrifugal force acting on the elementary position ‘AB’ = m r  × 𝑟 = m 𝑉 2 
In order to keep the belt is equilibrium, the centrifugal tension (Tc) acts in opposite directions at A & B.
On resolving the forces (centrifugal force & centrifugal tension) in vertical direction, we get:


Fc = Tc sin ( 2 ) + Tc sin ( 2 )




Fc = Tc 2 + Tc 2 [since sin ( 2 ) is very small and considered equal to( 2 )]

Fc = 2 Tc 2

m v2  = 2 Tc 2
Tc =mv2
This expression gives the expression for centrifugal tension in terms of mass & peripheral and thus, it is
clear that centrifugal tension is independent of the tension of the belt. Therefore,
Total tension of the tight side of the belt (considering centrifugal tension): 𝑇𝑇1 = 𝑇1 + 𝑇𝐶
Similarly, tension of the stack side of the belt (considering centrifugal tension): 𝑇𝑇2 1 = 𝑇1 + 𝑇𝐶
Maximum tension in the belt: Maximum tension in the belt is always equal to the total tension on the
tight side of the belt(𝑇𝑇1 ). This can be seen as follows:
Let
: Maximum safe stress
𝑏 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑒𝑙𝑡
t= thickness of the belt
Maximum tension in the belt: Maximum stress × cross-sectional area of the belt
.b.t
And, this tension is equal to the tension of the tight side, i.e.
T=T1 (when centrifugal force is neglected)
= T1 +Tc (considering centrifugal force)
Maximum power transmission:
Power transmission of any belt is given by the relation:
P= (𝑇1 − 𝑇2 ) V ------ (i)
Where, 𝑇1 = tension on the tight side
𝑇2 = tension on the slack side
V = peripheral velocity of belt
And, the ratio of driving tensions of two belts is expressed as:
𝑇1
= 𝑒 µ ------ (ii)
𝑇
2
𝑇
𝑇2 = µ1 ----- (iii)
𝑒
On putting the value of 𝑇2 from (iii) in (i), we have:
𝑇
1
1
P= (𝑇2 − 𝑒 µ1 ) V = 𝑇1 (1 − 𝑒 µ )V = 𝑇1 KV [𝑤ℎ𝑒𝑟𝑒, 𝐾 = (1 − 𝑒 µ )] -- (iv)
Now, considering centrifugal force, the total tension on the tight side (or the maximum tension in the
belt) = 𝑇𝑇1 (=T) =T1 +Tc or 𝑇1 = T- Tc
On substituting this value of 𝑇1 in (iv) we have,
P= (T- Tc) VK
[𝑠𝑖𝑛𝑐𝑒 Tc = m𝑣 2 ] ----(v)
= (T-m𝑣 2 ) V.K =(TV-m𝑣 3 ) K
For the condition of maximum power transmission, the equation (V) should be differentiated with
respect to the velocity ‘v’ & equated with zero.
Thus,
𝜕𝑝 𝜕(𝑇𝑣−𝑚𝑣 3 )𝐾
=
𝜕𝑣
𝜕𝑉
=0
= (T-3m𝑣 2 )K=0
= (T-3m𝑉 2 )=0
Or T=3m𝑉 2
---- (vi)
T=3Tc
Therefore, for maximum power transmission, the maximum allowable belt tension must be equal to
three times the centrifugal tension.
Also, for maximum power transmission, the velocity of the belt can be expressed as:
𝑉
---- From eq. (vi)
𝑇
max=√
3𝑚
Inifial belt tension : It can be defined as the tension experienced by the belt even when the pulleys are
stationary. When an endless belt is wound over the two pulleys, the friction between the belt and the
pulleys ensures the traction of belt and thereby, the pulley. In order to have a better grip between belt
& the pulleys, a small tension is induced in the belt. This tension is called the initial tension of the belt
(T0).
When power is supplied to the driver pulley, it tends to rotate and pull the belt from one side &
release it on the other. Thus, the tension on one of the side is increased while on the other side it is
decreased. The side with increased tension is termed as the tight side (T1) & side with reduced tension is
termed as slack side (T2).
Let us assume a perfectly elastic belt whose length is always constant.
𝛼 = co=efficient of increase of length of the belt per unit force
Increase in tension of the tight side due to rotation = T1-T0
Therefore, increase in length of the tight side portion of the belt = 𝛼 (T1-T0) and, decrease in length of
the length of the slack side portion of the belt = 𝛼 (T0-T2)
Now, for any belt of constant length:
Increase in length of belt on tight side = decrease in length on the slack side
Therefore,
𝛼 (T1-T0) = 𝛼 (T0-T2)
=T1+T2 = 2T0
T +T
or T0= 1 2 2
And when centrifugal tension is also considered:
(𝑇 +𝑇 )+(𝑇 +𝑇 )
𝑇 +𝑇 +2𝑇
2
𝑐
T0= 1 𝑐
= 1 2 𝑐
2
2
However, in practice the material of the belt is never perfectly elastic and the following expression
suggested by C.G. Barth is followed:
2√𝑇0 = √T1 + √T2
Belt – pulleys: Belt pulleys are the circular members of the belt (or rope) type transmission system
which are used to transmit the power from one shaft to other by means of a belt.
In belt (or rope) a drive, the velocity ratio is inversely proportional to the diameters of
𝑉
the driving & driver pulley i.e. 𝑉1 =
1
𝑑2
𝑑1
and therefore, an appropriate diameter of the pulley should be
selected in order to ensure the required velocity ratio.
Requirements of Belt – pulleys: Following characteristics are expected in any belt drive:
1. Good shock absorbing ability
2. Good heat conduction ability
3. Maximum resistance due to air.
4. Smooth surface for minimum wear
5. Resistance to corrosion
6. Light weight
7. High strength
8. High co-efficient of friction of face.
Materials used for pulleys: Following materials are generally used for manufacturing the pulleys used in
power – transmission:
a.) Cast iron
b.) Cast steel or pressed steel
c.) Wood pulley
d.) Wood rim with metallic centre
e.) Paper rim with metallic centre
Types of pulleys: Following are the generally used pulley types:
1. Split pulley
2. Solid pulley
3. Fast and loose pulleys.
1. Split pulley: In order to ensure a convenient mounting of pulley on shafts, a common practice is
to cast them in two halves which are joined by the means of bolts and links after mounting on
the shaft.
2. Solid pulley: They are the pulleys which are cast in a single piece. The mounting of such pulleys
is usually not convenient and absence of joints generally leads to problems in inclination of the
pulley. These pulleys are usually small & suitable for low speed applications.
3. Fast & loose pulleys: As the name suggests they are the pulleys having a fast pulley and a loose
pulley. The fast pulley is keyed to the machine shaft (drive shaft) and the loose pulley is set free
over the machine shaft.
This arrangement is used on shafts to ensure the operation of machine at will i.e. the machine
transmits the power only when it is desired and can be stopped without interfering with the
operation of the other machines operating on the same shaft.
When power is to be transmitted, the belt is made to
run over the fast pulley. However, when it is not required to transmit any power, the belt shifts
from the fast pulley to the loose pulley.
Cast iron pulleys: For its low price cast iron is one of the most widely used material for
manufacturing of the pulleys. The casting must be of a close grain structure and free from
blowholes, contraction cracks, porosity etc.
The rims of these pulleys are supported by central boss (or hub) through arms
or spokes. The arms are generally kept curved in order to increase their stress withstanding
capacity.
Crowning of pulleys: The face of the pulleys is never kept flat, it is given a convex curvature in
order to ensure the running of the belt at the centre of pulley width & there by prevent the belt
from slipping off the pulley during operation.
Design of cast iron pulley: Following steps are involved in designing of a cast iron pulley:
1. Calculation of pulley dimensions
2. Calculation of hub dimensions
3. Calculation of arm dimensions `
1. Calculation of pulley dimensions : Following are the parameters to be considered for designing a
pulley :
a.) Width of the pulley (B) : It can be calculated from the expression :
B = 1.25 b
Where, b= width of the belt
b.) Diameter of pulley (D) : It can be calculated from the expression :
t = v2
Where, t = centrifugal stress
 = density of pulley material
(For cast iron = 7200 Kg/m3)
𝜋𝐷𝑁
60
V= velocity of the rim =
[where, N=speed (in rpm), D=diameter of pulley]
c.) Thickness of the rim (t) : It can be calculated from following expression :
𝐷
𝐷
+2mm ≤ 𝑡 ≤ 200 + 3𝑚𝑚 − − − −for single belt drive
200
𝐷
𝐷
And, 200 2𝑚𝑚 ≤ t≤ 200 + 6𝑚𝑚 − − − − for double belt drive
2. Calculation of hub diversions : Following are the parameters to be considered for designing hub :
a.) Diameter of the hub (d1) : It can be calculated from following expression :
d1 = 1.5d+25mm
Where, d = diameter of the shaft
b.) Length of the hub (L) : It can be calculated from following expression :
𝜋
L=2 × 𝑑
2
It should be noted that the length of the hub ranges between 3 B and B.
3. Calculation of arm diversions : Following are the parameters to be considered for designing the
arms of the pulley :
a.) Cross – section of the arm : The cross – section of the arms is usually elliptical with the
major axis (a1) equal to twice the size of the minor axis (b1).
The cross – section of arm is obtained by considering the arm as cantilever (fixed at the hubend and carrying the load of the rim on the other end).
The length of the cantilever beam can be assumed to be equal to the radius of the pulley.
Besides, it is also assumed that power at any time is transmitted from the hub to the rim
through only half of the total number of arms.
Let, T = torque transmitted
R= pulley radius
n= number of arms
Tangential load per arm (WT) =
𝑇
𝑅×
𝑛
2
=
2𝑇
𝑅𝑛
Minimum bending moment on the arm (at hub end):
2𝑇
M= 𝑅×𝑛 × 𝑅 =
𝜋
2𝑇
𝑛
Section modules (‘Z’) = 32 b1 × (a1)2
Cross section of the arm can be obtained from the relation:
b (or t) = 𝑀⁄𝑍
b.) Number of arms : They can be : 4 for pulley diameter of (200to600mm)
6 for pulley diameter of (600to1500mm)
c.) Taper of arms: arms are tapered between 1⁄48 to 1⁄32 from hub to rim.
Steel pulleys: These pulleys are made from pressed steel sheets with rim made in two
halves and hub & arms pressed separately. These pulleys are known for their strength,
durability & light weight and are capable of running at very high speeds. Usually, steels
pulleys are of split type with rim halves, arms and hub riveted (or bolted) together. The
assembly thus formed is mounted on the shaft without the help of any key (since, the
clapping action of the hub is sufficient to hold the pulley to the shaft). However, for high
service requirements, keys are also used.
These pulleys are generally equipped with interchangeable bushings in
order to facilitate the used of shafts of different sizes with them.
Minimum length of the hub =
𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑓𝑎𝑐𝑒
2
Thickness of the rim = 5mm (for all sizes of pulley).
Advantages of steel pulleys:
1.) They are lighter in weight
2.) They possess high strength and are more durable
3.) With the belt of leather, they offer co-efficient of friction which is equal to or more than
cast iron pulleys.
Wooden pulleys : they are made from maple segments glued together under high pressure. A protective
coating of shellac or varnish is provided over these pulleys in order to prevent them from moisture and
thereby, warping. Following are the general characteristics of wooden pulleys:
1. High co-efficient of friction (higher than cast iron or steel)
2. Light weight
3. These pulleys are suitable for use in applications where arc of contact between the belt and the
pulleys is very small
4. They can be of both: split and solid type.
Paper pulleys: They are made by compressing paper fiber and are formed with metal at the centre.
These belts are usually preferred for applications with small centre distance.
V-BELT DRIVES: V-belts are made in trapezoidal section and are generally made of cotton fabric &
cords, moulded in rubber and covered with fabric and rubber. These belts are made endless and
therefore, the strength is same across the complete belt. These belts can satisfactorily operate with
slack side top or slack side bottom ad with any operating angle but cannot be used for large distance
drives. Following are the major compound sections in a V-Belt:
Drawing????????????????????????????????????
1. Cover made up of rubberized fiber
2. One layer of load carrying cords made of rayon, nylon, Dacron, steel and glass fibre.
3. Flexible section comprising of several layers to carry the tension during bending when belt runs
around the pulley.
4. The bottom rubber of compression section subjected to compression during bending. This
section is located below the pitch line and serves the following two function :
a.) To support the tensile section
b.) Transmitting the wedging pressure from tensile section to the sheave groove through side
walls.
5. The cushion section which serves to :
a.) Keep the cords bonded between the top tensile & bottom compression section
b.) Absorb shock during varying load conditions.
ADVANTAGES:
Following are the characteristic features of V-belt drive:
1. Large power transmission capacity
2. Shorter centre distance
3. Negligible slip and thus, a more positive drive
4. Facilitates use of multiple belts on the same pulley to increase the power transmission
5. The installation and uninstallation is simple.
6. It can be used in any position (i.e. the centre line between the shaft axis of the two pulley can be
inclined at any angle with respect to the horizontal )
7. They are durable
8. V-belts are make endless (i.e. with no joint) and thus, the drive is smooth
9. High velocity ratios of the orders of 10 are obtainable by use of V-belts.
10. Space requirements are minimum owing to the drives compact construction.
11. They have quieter operation and require minimum maintenance.
12. They are capable of transmitting higher tensional moment with lesser width and tension than
flat belts.
13. These drives provide shock absorption between driver and driven shaft
14. They are capable of powering multiple driver shafts through a single driver without using belt
tighteners
DISAPVANTAGES: Following are the major limitations of V-belt drive :
1. The efficiency of V-belt drive is low as compared to the flat drive.
2. They cannot be used with large centre distances.
3. The life of V-belt drives is comparatively less as they deteriorate faster under the effect of wear
and tear.
4. The design of V-belt drive pulleys is complicated than the normal pulley.
5. The construction of V-belt is complicated.
6. These belts are subjected to certain creep and thus, cannot be used in devices like timing belts
where synchronous speeds are required.
7. V-belts are subjected to very high centrifugal tension at high velocities i.e. velocity more than
50m/sec. besides, at low velocities (i.e. V<5m/sec), these drives become uneconomical.
8. Improper tensioning or mismatching of the belts can significantly affect the durability of the
drive.
9. In a multiple belt v-belt drive breaking of one belt results in replacement of all the belts.
10. The belt life gets significantly reduced at operating temperatures above 850 c or below 150c.
Ratio of driving tension of v-belt: We know that, in v-belt drive a trapezoidal section belt is
designed to run in a v-shaped groove pulley and thus, because of the wedging action of the belt with
v-groove, the normal reactions are perpendicular to the face of the groove.
Let,
R=total reaction between the belts and side of the grooves
µ=co-efficient of friction between belt & sides of the grooves
R1= normal reaction between the belts and sides of the grooves.
Drawing ???????????????????????????????????
On resolving the reactions vertically to the groove:
𝑅
R= R1 sinα+ R1 sinα=2 R1 sinα or R1=2sinα
Now, frictional force = 2 µR1
µR
µR
=2 2sinα = sinα =µ. R. cosec α
Consider an elementary strip (AB) of the v-belt subtending an angle so at the centre of the pulley. The
strip ‘AB’ will be in equilibrium under the following forces:
1.
2.
3.
4.
Tension (T+𝛿𝜃) at point ‘A’ on the tight side.
Tension (T) at point ‘B’ on the slack side.
Normal reaction (R) acting upward between the elementary length of the belt and the pulley.
Frictional force (F=2 µR) acting tangentially to the pulley rim, which resists slipping of the
elementary strip on the pulley.
For equilibrium, the algebraic rum of forces acting in horizontal and vertical direction must be zero.
Resolving the forces in horizontal direction :
𝛿𝜃
2
F=(T+ 𝛿𝑇) cos
- T cos
𝛿𝜃
2
2 µR1 = T + 𝛿𝑇-T [𝑠𝑖𝑛𝑐𝑒  𝑖𝑠 𝑣𝑒𝑟 𝑠𝑚𝑎𝑙𝑙 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ cos
𝛿𝜃
2
≈ 1]
𝛿𝑇
2 µR1 = 𝛿𝑇 or R1 = 2 µ ---- (i)
Resolving the forces in vertical direction:
𝛿𝜃
𝛿𝜃
R= (T+ 𝛿𝑇) sin 2 + Tsin 2
𝛿𝜃
2
2R1 sinα = T + (T+ 𝛿𝑇)
2R1 sinα = 2T.
𝛿𝜃
2
+
𝛿𝜃
2
[𝑠𝑖𝑛𝑐𝑒 𝛿𝜃𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ 𝑠𝑖𝑛
𝛿𝜃
2
≈ 𝛿𝜃⁄2]
𝛿𝑇𝛿𝜃
2
2R1 sinα = T 𝛿𝜃 [𝑠𝑖𝑛𝑐𝑒
𝛿𝑇𝛿𝜃
2
𝑇𝛿𝜃
𝑖𝑠 𝑛𝑒𝑔𝑙𝑖𝑔𝑏𝑙𝑒 ] or R1=
----------- (ii)
2sinα
On equating the value of R1 in (i) &(ii), we have :
𝛿𝑇 𝑇𝛿𝜃
=
2 µ 2sinα
𝛿𝑇 2µ𝛿𝜃 R𝛿𝜃
=
=
T 2sinα sinα
Integrating the equations between T1&T2 and 0 &𝜃, we have:
𝑇 𝛿𝑇
2 𝑇
??????? ∫𝑇 1
𝑇
=
𝜃
µ
∫ 
sinα 0
µ𝜃
Log. (𝑇1 ) = sinα
2
𝑇1
𝑇2
µ𝜃
= 𝑒 sinα
µ
From the above expression it is clear that the co-efficient of friction increases of ( ⁄sinα) and thus, ratio
of the tension of the v-belt is greater than that of the flat belt. Hence, the power transmission in this
belt drive is more.
Types of v-belts: Following are the major types of v-belts ad their characteristics
Type
A
Power range
(KW)
0.7-3.5
Minimum pitch
diameter (mm)
75
Thickness (t)
Width(b)
Wight per meter length
8
3
1.06
B
C
D
E
2-15
7.5 – 75
20-150
30-350
125
200
355
500
11
14
19
23
17
22
32
38
1.89
3.43
5.96
-
V-BELT PULLEYS: following are the materials generally used for v-belt pulleys:
1. Cast iron
2. Formed and pressed steel
3. Molded or machined plastic
4. Die-cast aluminum
V-belts (continued ) : In v-belts, the included angle (2α) , i.e. the angle between the belts sides is
generally 400 and the angle of the groove is kept between 300 to 380. The angel of the groove is kept less
than the included angle in order to increase the effective co-efficient of friction acting between the belt
and the pulley.
Sheave design: The sheaves used in v-belts drives are generally made of cast iron, fabricated steel or
aluminum alloys. Cast iron pulleys are preferred for speed below 35m/sec and for higher speeds
aluminum pulleys are preferred. For automotive & agricultural purposes, formed steel sheaves are
preferred. However, for special applications hard anodized aluminum alloy sheave are preferred.
The successful application of v-belts depends upon the
prefect modeling between v-belts ad pulley grooves (since v-belt run in the pulley grooves). Thus, the
pulley grooves should be such that the belt does not touch the bottom of the groove. Also, the top
section of the belt should also not protrude outside.
When the belt is fixed on the pulley, the
top section is subjected to tension while its lower section is subjected to compression and consequently,
the angle between the side walls of the belt reduces. Smaller the sheave diameter, greater the reduction
in the angle and thereby , reduction in the life of belt.
Following is the table of various v-belts sheave dimensions:
Drawing ??????????????????????????????
Type
W
D
A
C
T
E
A
B
C
D
E
11
14
19
27
32
12
15
20
28
33
3.3
4.2
57
8.1
9.6
8.7
10.8
14.3
19.9
23.4
10
12.5
17
24
29
15
19
25.5
37
44.5
Face width of the pulley (B) = (n-1) e+2f. Where, n = number of belts.
No. of
sheave
grooves
(𝑥)
6
9
14
14
20
Groove
angle
(2𝑥)
32,34,38
32,34,38
34,36,38
34,36,38
---
Design procedure of belts:
4. For flat belt drives: The belt drives are designed on the basis of the power transmission
requirements. Power transmission depends on the following factors:
a. Net belt tension
b. Angle of contact
c. Centre distance
d. Co-efficient of friction
The design power can be calculated by multiplying the power to be transmitted and the service factor.
Design power = power to be transmitted × service factor.
Where,
Service factor is : a.) 1.35 for only, wet or dusty environment
b.) 1.20 for jerky loads
c.) 1.90 for shock and revered loads.
1.) Design of belt section: The design of the belt section is based on the max. tension of the belt (i.e. the
tension on the tight side of the belt). And, the tension on the tight as well as the slack side can be
determined by following expression:
a.) power transmitted (P) = (T1-T2) V
b.) ratio of belt tension =
T1
T2
= 𝑒 µ𝜃
Where,
µ=co-efficient of friction
𝜃 =angel of contact on the smaller pulley
𝜋
𝜃 =(180-2α) 180 rad
Where, α = sin−1
𝑟1 −𝑟2
𝑥
(for open belt drive)
𝑥 = centre distance
α = sin−1
𝑟1 ∓𝑟2
𝑥
(for cross belt drive)
It should be noted that centrifugal tension (Tc) is taken into account wherever velocity of the belt (V) is
greater than 10m/sec and thus, power transmitted (considering centrifugal tension) = (𝑇𝑇1 − 𝑇𝑇2 ) V and
𝑇𝑇1−Tc
ratio of belt tension = 𝑇
𝑇2− Tc
= 𝑒 µ𝜃
Where, 𝑇𝑇1 − = 𝑇1 + Tc and 𝑇𝑇2 = 𝑇2+ Tc
And Tc =
𝑚𝑉 2
𝑟
Now, the thickness of the belt section cae be determined from the equation :
𝑇
1
b×t = 𝜎×𝑐
(for V<10m/sec)
1
& b×t =(𝜎−
𝑇
(for V> 10m/sec i.e. considering centrifugal force)
 𝑉 2 )×𝑐1
Where, b= width of the belt width
t=thickness of belt
𝜎= maximum allowable stress in belt
C1= correction factor, whose value depends upon the angle of centre line of drive with
horizontal.
Value correction factor, c1 is different for different angle as should in the following table :
Types of drive
Inactivation to horizontal
Open belt drive
Cross belt drive
Quarter turn drive
0-600
1.0
0.9
0.8
600-800
0.9
0.8
0.7
800-900
0.8
0.7
0.6
Following are the generally used values for width & thickness of the belt:
a.) Width ( b):
25,32,40,50,60,71,80,90,100,112,125,140,160,180,200,224,250,280,315,355,400,450,500,560,
and 600mm
b.) Thickness (t) : 5mm for width (b) of 25 to 63 mm
6.5mm for width (b) of 50 to 140 mm
8.0mm for width (b) of 90 to 224 mm
10mm for width (b) of 125 to 400 mm
12.0mm for width (b) of 250 to 600 mm
𝜋
Length of the belt (L) : 2 (𝑑2 + 𝑑1 ) + 2𝑐 +
𝜋
2
(𝑑2 + 𝑑1 ) + 2𝑐 +
(𝑑2 −𝑑1 )2
4𝑐
(𝑑2 +𝑑1 )2
4𝑐
(for open belt drive)
(For cross belt drive)
B. For V-belt drives: The expression used for v-belt drives are similar to that of the belt drives except
that the pitch diameter instead of the outer diameter is used in the expressions.
i) The design power can be determined by multiplying the rated power with the service factor.
Where, service factor represents the severity of the load transmitted.
Following are the generally applicable service factors:
Driven machine
Service factor
Agitators, padded propeller
1.0-1.2
Bakery machinery
1.0-1.2
Conveyors
1.0-1.5
Mills (flour, cereals etc)
1.0-1.4
Generators and exciters 1.2
Paper machinery
1.2
Compressors, screens
1.2-1.4
Brick and clay machinery
1.2-1.6
Textile machinery
1.2-1.8
Fan, pumps etc
1.2-2.0
Mills, crushing machinery
1.4-1.6
Line shafts
1.4 – 2.0
ii) cross – section of belt : The cross section of the belt can be selected from the following table :
Type of belt
Power range
(in KW)
Minimum
pitch dia. (D)
Top width
(b)
Thickness (t)
A
B
C
D
E
0.7-3.5
2-15
7.5-75
20-150
30-350
75
125
200
355
500
13
17
22
32
38
8
11
14
19
23
𝑁
iii) Velocity ratio = 𝑁1
2
𝑁
µ𝜃
Where, ratio of tension =𝑁1 = 𝑒 𝑠𝑖𝑛𝛼
2
Weight per
meter
length (in
Newton’s)
1.06
1.89
3.43
5.96
--
Area of
cross.
Section (in
mm2)
81
140
230
475
695
iv) For an economical drive for smaller pulley pitch diameter giving 20-25 m/sec belt speed is selected.
The value of the pitch diameter for the smaller value can also be calculated from the following
expression:
1⁄
3 mm
D1 = (38-42) (t1)
Where, t1 = torque at the driving shaft (in N-m)
v.) The diameter of the larger pulley (D2) can be calculated from the expression :
D2 = D1 (1-s) (VR)
Where, s=slip factor = 0.01 to –0.03 and VR is the velocity ratio
vi) Belt speed can be calculated from the expression:
V=
𝜋D1 𝑁1
60
=
𝜋D2 𝑁2
60
vii) Length of the belt can be calculated from the expression:
L=2c+1.57(D2 + D1) +
(D2 −D1 )2
4𝑐
Where, C= centre distance between the pulleys.
The centre distance can be calculated from the following relation:
C=P+√𝑝2 − 𝑞
𝐿
𝜋
Where, P=4 - 8 (D2 + D1) & q =
(D2 −D1 )
8
viii) Number of turns of the belt per second (𝑥) =
𝑉
𝐿
𝑉
𝐿
. In order to ensure desirability of the belt
≤10m/sec
ix) Angle of contact of the smaller pulley (𝜃):
D2 −D1
)
2𝑐
𝜃 = π-2𝑆𝑖𝑛−1 (
The value of 𝜃 should be greater than or equal 2.1 radians.
x.) Maximum tension in the belt:
P= (𝑇1 − 𝑇2 ) VW
xi) Number of belts:
𝑇
n ≥ 𝜎 𝐴𝑐1 𝑐 (ignoring centrifuge tension i.e. for V < 10m/sec)
𝑡
1 2
𝑇
n ≥ (𝜎 −𝜌𝑉 21)𝐴𝑐
𝑡
1 𝑐2
(considering centrifugal tension i.e. for V > 10m/sec)
Where, A= area of cross –section of the belt profile
C1 = arc of contact factor on the smaller pulley
C2= belt length correction factor
It should be noted that the number of belts should never exceed 8-12, otherwise the next larger belt
section should be used since with large number of belts the load distribution is difficult to manage owing
to the variation in length of belts and dimensions of the pulley grooves.
Rope drives: Under certain conditions, rope drives find a wide application for power transmission. For
example, Rope drives are widely used for the application in which a large amount of power is to be
transmitted over a considerable distance.
It should be noted that the use of belt drives for large distance (i.e. more than 8mm) power transmission
is not justified since it would require an excessive cross section which is not disable.
Types of rope drives : following are the two types of rope drives generally used :
a.) Fiber ropes : used for power transmission distance of up to 60mm
b.) Wire ropes : used for power transmission distance of up to 150m
a.) Fiber ropes: These ropes are used for both power transmission and load lifting & hoisting
applications. They are made of fibrous material such as :
i.) Cotton: Cotton ropes are known for their soft and smooth texture and thus, they never undergo any
internal damage or deterioration. However, lubrication is not necessary for these ropes; it could reduce
the internal wear between rope and grooves of ropes sheave.
ii.) Hemp & manila rope: Hemp and manila fibers are rough and thus, the ropes made from these fibers
are not very flexible and possess poor mechanical properties. As compared to the manila ropes, hemp
ropes have less strength and are subjected to rapid abrasion, rapid damage from sharp objects,
atmospheric effects etc.
When hemp and manila ropes are bent over the sheaves, there is some sliding of its
rough fibers and thereby the rope wear and chafe internally. In order to minimize these problems, ropes
are impregnated with a lubricant such as tar, tallow, graphite etc. Lubricants, in addition to preventing
the wear at the same time also make the rope heavier and stiff.
Hemp ropes are suitable only for manual hoisting machinery and as the ropes for lofting tackle, hooks
etc.
iii.)Leather
iv.) Nylon
v.)Rubberized fabric
Advantages of power transmission through fiber rope drives:
1.
2.
3.
4.
5.
6.
High mechanical efficiency
The shafts may be out of alignment
Smooth, quit and steady operation
Low cost and economical service
little affect of the outdoor conditions.
Ease of operation with both short and long shaft centre and thus, the space requirements are
considerably less.
7. Power can be taken off in any direction and in fractional parts of the whole amount.
Classification of fiber ropes: According to the mode of money acuter and number of strands in each
rope, they can be classified as follows:
a.)Plain laid type (or hawser laid type) : This rope has three strands twisted together without a central
core. The direction of twisting is known as lay. Each strand is made of repeated yarns by twisting the
yarns together in left hand direction. These yarns are spun from fibers having a right hand twist
These ropes are formed by twisting three strands together in right hand
direction. This lay for the rope is designated as “Z” lay in which the strands run from left to right across
the top of the rope (as in the right hand screw thread).
It should be noted that in this rope if the strands coil in the same direction as the left hand screen
thread, the rope is said to be of the left hand or “s” lay. And, if the fibers in the strand coil in the same
direction as the strand, the rope is called to be of “parallel “or” long” lay. Moreover, if the fibers coil in
opposite dissection to that of the strand, the rope is said to be “regular” or “cross” or “opposite” lay.
Shroud laid: The lay in this ropes is of ‘Z’ type and the number of strands are “4”.
C.)Cable laid : The lay is this lay is of ‘Z’ for primary ropes & “s” for final ropes and the number of strands
is 9. This rope is formed by, twisting three primary ropes together without a central core – each primary
rope is plain laid and consist of three strands
Ratio of driving tension: The ratio of driving tension for fiber ropes is same as that of v-belts since they
are designed in the smaller way as v- belt. Following is the expression for driving tension of fiber ropes:
𝑇
2.3 log ( 𝑇1 ) =  cosec  or
2
𝑇1
𝑇2

= 𝑒 sin
Where, T1 = tension in the tight side (ion Newton’s)
T2 = tension i the slack side (in Newton’s)
=angle of contact (in radians)
=co-efficient of friction
2=groove angle
Velocity of fiber ropes: The velocity of these ropes varies between 15 to 30m/sec.
Factor of safety:
This factor multiplied to the power transmission desired in order to calculate design power. In fiber
ropes, the factor of safety is generally kept high so as to ensure durability. It can be expressed as:
F. S. =
𝑤𝑙𝑖𝑚𝑎𝑡𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑛𝑔ℎ𝑡
𝑛𝑜𝑟𝑎𝑚 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑜𝑝𝑒
It should be noted that “F.S.” for cotton ropes should be more than 30 and for manila ropes, it should be
taken as 35 or 36.
Sheave for fiber ropes: The fiber ropes are usually circular in cross – section and as they bend around
the pulleys, bending stresses are set up. These bending stresses can be reduced by increasing the sheave
size. If ‘d’ is the rope diameter , the sheave sage should atleast be 36d. Following parameters are
considered for redesigning a sheave for fibrous rope:
a. The form of the grooves & effective diameter should be same for all grooves of the same
sheave.
b. The groove angel of the pulley for rope drives is usually kept 450.
c. The diameter of the sheave should be large enough to reduce the wear on the rope due to
internal friction and bending stresses.
d. The surface should be accurately finished and well polished so as to ensure durability of the
rope.
e. The grooves in the pulleys are made narrow at the bottom and the ropes are pinched between
the v-belt edges so as to improve the grip between pulley and the rope.
f. The groove should be deep enough to prevent the rope from recalling the bottom.
g. Since the wedging action of the grooves is harmful for the ropes, the sheaves are grooved when
used only as idler or tension sheave.
h. The number of groove should not be more than 24.
i. The selected groove angle should be such that it provides great adhesion without slipping,
prevents undue wedging and offers least resistance to the rope movement through the groove.
Power transmitted: Power transmitted by one rope is expressed in the same way as the v-belt i.e.
Power transmitted = (𝑇1 − 𝑇2 ) V
However, for multiple ropes, power transmission is expressed as:
P= (𝑇1 − 𝑇2 )V.Z
Where, Z= number of ropes
P=(𝐷2 + 𝐷1 ) +
3
2
𝐷2
Where, 𝐷1 = diameter of smaller pulley
𝐷2 = diameter of larger pulley
Length of a rope : Length of one rope can be expressed by the same relation as of v-blet drives i.e. :
𝜋
e=2 (𝐷2 + 𝐷1 ) + (𝐷2 − 𝐷1 ) sin−1
(𝐷2 −𝐷1 )
2𝑐
+ 2𝑐 𝑐𝑜𝑠
where, D1 = dia of small pulley
D2 = dia of large pulley
2= groove angle
C= centre distance
Hoist fiber ropes : In hoisting operations, ropes are wound open drums and the sheaves are used for
changing the rope dissection. Following are the types of bribe ropes used for hoisting :
1.Cotton ropes : They are available in two types :
a.) plain (or however ) laid : In this type three strands are tueisted without the central core
b.) In this types, nine strands are used and the rope consisted 3 primary ropes in each of which
three strands are right handly laid. Primary ropes are laid left handly to from the find rope.
2.Hemp roes : In this type, three strands are tueisted together. They are available in following lades :
a.)Grade I : It is made of a good quality hemp & is used for lifting for locoing loads.
b.)Grade II : It is made of an average quality fiber.
3. Manila ropes : They are of following types :
a.) naweser or plain laid
b.)Cable laid
c.) should laid
now cenclature of fiber ropes : The ropes are designated in the following order :
Name – nominal diameter – standard
For example : For a cotton rope of a nominal diameter of 28mm, the nomenctature would be :
Cotton rope, 28 IT-3143
Pulley system : It can be defined as a combination of several movable and fined pulleys or sheaves.
Pulleys system can be of following types :
1. Gain in speed type
2. Gain in force type
Out of which, the tatter (i.e. the gain in force type ) is employed for hoisting purposes. Following are the
two gain in force type “arrangements used for hoisting applications :
a.)Rope running off fined pulley (fined pulley are the ones with fiend angles. They are also venom as
gvinding pulleys for they change the direction of the rope )
b.)Rope running off a moveable pulley (moveable pulley is the one with moveable angle to which load or
force is applied ).
Drawing ?????????????????????????????????
Design of pulley system : For rope passing over the pulley with one end carrying a weight “W” and an
effort “P” applied on the offer and, theoretically :
Effort (P) = weight (W)
Nowever, in practice, effort (P) is always greater than the weight (W).
P>W or P =CW.
Where, c is a constant whose value depends upon the following factors : a ) size of the
rope b.) co-efficient of friction C.) relative size of pulley and pin d.) type of gearing and lubricant used
with pulley angle. In general, for manila ropes, the Type equation here.value of “c” is taken as 1.4.
Efficiency of the pulley ??? The efficiency of the pulley can ??? be expressed as :
1
=𝐶
Drawing ??????????????????????????????
In a hoist rope system following situations are possidible :
i.)Raising the lead : let T1 T2 T3 T4 T5 & T6 be the tension in different
ropes. Then,
𝑇2 = 𝐶𝑇1
𝑇3 = 𝐶𝑇2 = 𝐶𝛼𝐶𝑇1 = 𝐶 2 𝑇1
𝑇4 = 𝐶𝑇3 = 𝐶𝛼𝐶 2 𝑇1 = 𝐶 3 𝑇1
𝑇5 = 𝐶𝑇4 = 𝐶𝛼𝐶 3 𝑇 = 𝐶 4 𝑇1
𝑇6 = 𝐶𝑇5 = 𝐶𝛼𝐶 4 𝑇 = 𝐶 5 𝑇1
P=𝐶𝑇6 = 𝐶𝛼𝐶 5 𝑇 = 𝐶 6 𝑇1
Now, for equilibrate of forces on the lower block :
Thus, of a rope passes ‘n’ times over the pulleys, the effort (P) required can be expressed as :
P=W.
𝐶 𝑛 (𝐶−1)
(𝐶 𝑛 −1)
If the friction is absent, 𝑃𝑖 =
𝑊
𝑛
(ideal condition )
Therefore, efficiency of the hoist can be expressed as :
𝑃
 overall = 𝑃𝑖 =
𝐶 𝑛 −1
𝑛𝐶 𝑛 (𝑐−1)
ii)wwering the load : Let T1 T2 T3 T4 T5 & T6 be the tension in different ropes then,
T1
𝑇2 = 𝐶𝑇1
𝑇3 = 𝐶𝑇2
𝑇4 = 𝐶𝑇3
𝑇5 = 𝐶𝑇4
𝑇6 = 𝐶𝑃
(𝑐−1)
And, therefore , P = 𝐶(𝑐 6 −1)  𝑊
For any rope passing ‘n’ times over the pulley, the effort 9P) required can be expressed as:
𝑐−1
P=𝐶(𝑐 𝑛 −1)  𝑊
Wire ropes : They are used for application in which long distance transmission of a large amount of
pouler is required. (Although, the wire ropes have largely been replaced by the advanced modes of
power transmission , they are widely used for following purposes :
a.)
b.)
c.)
d.)
e.)
Elevators
Cranes
Hauling devices
Suspension bridges
Mine hoist
f.)
g.)
h.)
i.)
Trasnways
Suspension bridges
Aerial conveyors
Oil well drilling
Materials used for wire ropes : Wire ropes are made of cold driven wires fro higher strength and
daradailty. Following are materials used for wire ropes for increasing their strength : wrought iron, cast
steel, plough steel, alloy steel, extra strong cast steel. Also, for contain purposes, wire ropes are made of
copper, brogue, alnmimimu alloys and stainless steels.
Advantages of wire ropes :
1.)
2.)
3.)
4.)
5.)
6.)
7.)
8.)
9.)
Algidity to withstand shock loads
Durable
Light is weight
High efficiency
Silent operation even at high speeds
Reliable
Low in cost
NO sudden failures
Should be noted that the strength of wire ropes increases with the decrease in length.
Classification of wire ropes : On the basis of the direction of tuiest of the individual wires as well as of
shavts relative to each other, wire ropes can be classified as follows:
1. Cross or regular lay ropes
2. Parallel or long lay ropes
3. Composite or reverse laid ropes
1.Cross or regular lay ropes : In this rope, the strands are tuusted into a rope in the direction opposite
to the direction of the tuist of the wires in the strands. These ropes are most widely used.
2.parallel or long lay ropes : In this rope, the strands are tuusted into a rope in the same direction as the
direction of tuuist of the wires in the strands . these ropes are more flemible, resist wear more
effectively and offer a belter bearing suefacel. Also these ropes are harder to splice and tuuist easily
when loaded but have a tendavcy to spin. These ropes find wide application in hoist & lifts and also as
haulage
ropes.
3.)composite or reverse laid ropes : In this rope, the wires in two adjacent strands are tueisted in
opposite direction.
It should be noted that the direction of the lay can either be right handed or belt handed and
accordingly the ropes are termed as “right lay ropes “or “left lay ropes”.
Wire – rope designation : wire ropes are designated in terus of the number of strands and number of
wires in each strand.
Following are standard designations used in wire ropes :
a.)6× 𝟕 : It represents a wire rope having ??? strands and seven wires in each strand. It is also known as
“standard coarse laid rope. It is usually used as haulage ropes in munes, tramways and on power
transmission.
b.) 6× 𝟏𝟗: It represents a wire rope with ?? strands and nineteen wires in each strand. It is also known
as “standard hoisting rope”. It is usually used for hoisting purposes in mines, cranes, tramways, dredges,
quarries, well drilling, elevators, derricks, or docks etc,
c.)6× 𝟑𝟕 ∶ It represents a wire rope with ?? strands and thirty. Seven wires in each strand. It is also
known as “extra flexible hoisting rope. It is used in steel mill ladles, cranes, high speed elevators,
d.)𝟖 × 𝟏𝟗 ∶ It represents a wire rope with?? Eight strands and nineteen wires in each strand. Its is also
known as “Extra flailed hoisting rope:.
Properties of wire ropes : The linear deadly, breading strength etc depends upon the purpose for which
the rope is to be used. Following are the properties of various types of are ropes :
(It should be noted that the symbol ‘d’ the following tables represents the diameter of the wire rope in
mm)
Class
Type of
rope
Nominal diameter
(in mm)
Average
Without
(N/m)
For
haulage
purpose
in mines
For lifts
elevators
and hoists
6× 7
8,9,10,11,12,13,14,15,16,18,19,20,21,22,24,25,26,
27,28,29,31,35,
13,14,16,18,19,20,21,22,24,25,26,28,29,30,32,35,
36,38
6,8,10,12,14,16,18,20,22,25,
0.0347d2
6× 19
6× 19
8× 19
For oil
well and
oil well
drilling
6× 7
6× 19
6× 37
8× 19
18× 7
8,10,12,14,16,18,20,22,25,
10,11,13,14,16,19,22,25,
13,14,16,19,22,25,29,32,35,38,
13,14,16,19,22,25,26,32,35,38,
13,14,16,19,22,25,29
13,14,16,19,22,25
TENSILESTRENGHT (N)
TENSILESTRENGHT OF
WIRE
1600 MPa
1800
MPa
2
530d
600d2
0.0363d2
530d2
595d2
0.0383d2
1100-1250
MPa
12501400
MPa
435d2
445d2
2
0.034d
0.037d2
0.037d2
0.037d2
0.0338d2
0.04d2
385d2
355d2
16001800
MPa
550d2
510d2
490d2
-
18002000
Mpa
610d2
570d2
530d2
530d2
575d2
20002250M
pa
630d2
6000d2
-
For
cranes,
excavator
etc
6× 19
6× 37
For
shipping
purposes
6× 7
6× 19
6× 37
8,9,10,11,12,13,14,16,18,20,22,24,26,28,32,36,38,
40
8,9,10,11,12,13,14,16,18,20,22,24,26,28,32,36,38,
40,44,48,52,56
8,10,12,14,16,18,20,22,24,26,28,32,36,40,
8,9,10,11,12,13,14,16,18,20,22,24,26,28,32,36,38,
40,44,48,52,56
8,9,10,11,12,13,14,16,18,20,22,24,26,28,32,36,38,
40,48,52,46
16001750
MPa
17501900
Mpa
0.0375d
540d2
590d2
0.038d2
510d2
550d2
1400-1600N/mm2
490d2
480d2
510d2
Wire – diameter and effective cross –section of ropes : The diameter of rope and the diameter of the
wire can be expressed by the formula :
d = 1.5 𝑑𝑤√𝑍
where, d=diameter of rope
𝑑𝑤 = diameter of a wire
Z= number of wires (in a rope)
Type of wire rope
6× 𝟖
Wire diameter
0.106d2
(𝑑𝑤 )
Area of wire rope 0.38d2
(A)
6× 𝟏𝟗
6× 𝟑𝟕
8× 𝟏𝟗
0.063d
0.045d
0.050d
0.38d2
0.38d2
0.35d2
Permissible operating load : The operating loads are chosen such that the rope should have a
reasonable life and ti should not fail as a result of wear and fatigue. Following are the values of factor of
safety for wire ropes based on the wtiamde strength for different applications :
1
2
3
4
5
6
7
8
Application of
wire rope
Factor of
safety
Rope speed
(m/sec)
gups
hot ladle cranes
hadulage ropes
mine hoists:
deplhs
3.5
8
6
1.0
1.0
10
Upto 150m
300-600m
600-900m
Above-900
8
7
6
5
3-5
8
6-8
10-12.5
12.5-15
15
10
1.0
Motor
driven
Hand driven
4-6
1.0
3-5
4.2
7
0.5
1.0
1.0
8-12
8
1.5-6.25
1.0
derricks
slings
cranes :
track cables
small electric
and air hoist
9
10 elevators
11 hot ladle cranes
Wire rope sheaves and drums : In order to reduce the bending stresses ion the ropes while bending
around the sheaves or pulleys, the diameter of the sheave should be as large as possible following are
the values of sheave diameter for different type of ropes :
Type of rope
6× 7
6× 19
RECOMMENDED SHENE DIAMETER (D)
MINIMUM SHENE
PREPERPED SHEME
DIAMETER
DIAMETER
42d
72d
30d
45d
60d
100d
20d
30d
6× 37
18d
27d
8× 19
21d
31d
APPCICATIONS
Mines, laularge transmu
hoisting ropes
Cargocranes,mine hoists
Derricks, dirdges,
elevators tramuays well
drilling
Chanes, high speed
elevator sand small
sheaves
Extra ?? hoisting rope
It should be noted that large sheave diameter give a better and ?????? service and thus, should be used
wherever the space allows.
The sheave groove has a great influence on the life as well as the service of the rope.
1.griives with too small size leads to wedging of the rope into the groove and thus, the normal rotation
is prevented
2.griives with a bigger size than the rope is not able to provide sufficient support to the rope and thus,
the rope tends to change its circular shape to flat and thereby, the effects of fatigue increase .
Sheaves have a great influence on the life and service of the rope. They are usually mounled on the fined
angle on auti friction or browse bearings.
Material : rope sheave used for light and medium service are usually made of coition. However the
heavy duty sheave ar emade of steel castings. It should be noted that small sheaves are usually cast in
one piece without ribbing & on the other hand large sheaves are provided with ribs and hoes (or crossshaped spokes.)
Wire rope fasteners : following are the different types of fasteners generally used :
a.)Wire rope socket with yime : they offer on efficiency of 100%
b.)Three bolt wire cloups : They offer an efficiency of 75%
c.)Thimble with four or five wiretiuks : They offer an efficiency of 90%
d.)Special offset thimble with chips : They offer an efficiency of 90%
e.)Regular thimble with chips: They offer an efficiency of 85%
Selection of wire rope : Under loading, the various stresses induced in the wire rope are as follows:
1.derect stress due to anial load hoisted and the weight of the rope.
2.Bending stresses (when he rope winds around the sheave or drum )
3.Stresses during starting (or stopping)
4.Stresses due to change in rope speed.
It should be noted that the selection of rope should be done very carefully and in consideration of all the
operational stresses involved. Stresses induced in a wire rope are compel especially when it is under
bending. This is due to the fact that during bending stresses such as tension, bending, tuisting mutual
compression, rubbing of wires and strands are involved. besides , and additional tension acts on the
outer fibers of wires since, during bending, the iner wires contact and the other wires expand.
1.dorect stress due to anial load hoisting and weight of the rope : direct stress can be determined from
the expression :
𝜎𝛼=𝑊+𝑤
𝐴
Where,
W=lifted weight
W=Rope weight
𝜎𝛼 = direct stress
And
A= cross – sectional area of rope
2.Bending stress (when the rope winds around the sheave or drum ): following is the expression to
calculate the bending stress :
𝜎𝑏 =
𝐸𝑟 .𝑑𝑤
𝐷
Where,
𝑑𝑤 = wire diameter (in mm)
D= diameter of sheave or drum (in mm)
𝜎𝛼 = ?? ending stress
𝐸𝑟 = modulus of elastiaty of wire rope (in mm)
=84× 103 𝑁 𝑚𝑚2 for steel ropes
=77× 103 𝑁 𝑚𝑚2 for wrought ?? ropes
Equivalent bending load on the rope (Wb) :
Wb = 𝜎𝑏 × 𝐴 =
𝐸𝑟 ×𝑑𝑤 ×𝐴
𝐷
𝑁
Where, A= net cross sectional area of the rope (in mm).
It should be noted that when bending stress (𝜎𝑏 ) for each wire is given, the load on the whole rope due
to bending sam be expressed by the following relation :
𝜋
Wb = 4 (𝑑𝑤 )2 × 𝑛 × 𝜎𝑏
Where, n= number of weras in the rope
4.stresses during starting (or stopping) : An additional load in induced into to accelerate the load
supported by the load and the weight of the rope in mine hoists, elevators etc. this additional load can
be expressed by the equation :
𝑎
Wa =(𝑊 + 𝑤) 𝑔
Where, Wa = additional load
W=weight of the load to be hoisted
W=weight of the rope
a=acceleration of the rope and load =𝑣⁄60𝑡, where, V=speed t=since
generally, rope has a slack which has to be overcame before the rope gets taut and starts enerting a pull
on the load. Consequently, a considerable impact load applies on the load. This impact load can be
found by the following intact load relation :
Wst = (𝑊 + 𝑤) [1 + √1 +
𝑉𝑟 2 ×𝐸𝑟
]
𝜎𝑑 ×𝑙×𝑔
Where, vr = velocity of rope when it it tanut = √2𝑎 × ℎ. Where, a = accele –ration of the rope and load
h=slackness of rope
?? = cenght of the rope
g=accede to givinly
Wst= starting impact load.
𝜎𝑑 = direct stress
𝐸𝑟 = modulus of elasticity of the wire rope
Now, when stackness (a) = o
vr = 0 and thus, impact load during starting :
(𝑊+𝑤)
Wst=2(𝑊 + 𝑤)and corresponding stress (𝜎𝑠𝑡 ) = 2
𝐴
4.Stress due to change is speed : The additional load on the rope due to change in spped can be
obtained from the same relation is in the case of starting, which can be expressed as :
𝛼=
(𝑉2 −𝑉1 )
𝑡
where, (𝑉2 − 𝑉1 ) is the change is speed (in m/sec) & t is time (in sec)
A sudden stop of the hoist of drum while courering of the load , leads to selling up of sever stresses in
the rope . This is due to the fort that the k???? energy of moving mass S suddenly made yero. This kintic
energy is absorbed by the rope and the find the resulting stresses induced in the rope ki8netec energy is
equated to the resilience of the rope.
During stopping if the load moves down a certain distance, the corresponding change
of potential energy must be added to the kinetic energy. It should be noted that ‘t’ is alos necessary to
add the work of sthetcking the rope during stopping which may be obtained from the impact stresses.
5. Effective stress : Effective stress is the total stress acting on the rope on any particular condition.
Effective stress is the sum of the direct stress (𝜎𝑑 ) and bending stress (𝜎𝑏 ) during normal conditions.
during normal condition effective stress = 𝜎2 +𝜎𝑏
during starting, effective stress =𝜎𝑠𝑡 +𝜎𝑏
during acceleration of load , effective stress = 𝜎𝑠𝑡 +𝜎𝑏+ 𝜎𝑎
design of wire ropes : It should be noted that for design of any wire rope, the sum of all stresses should
be less than the estimate strength divided by the factor of safety.
Following are the steps involved in designing of the wire rope :
1.A suitable rope for the given application is chosen from the previous tables.
2.Design load is calculated (by assuming a factor of safety of 2 to 2.5 times than the factor of safety in
the table).
3.Calculate design diameter of wire rope (d) by equating the design load with the tenside strength of the
rope .
4. Calculate, diameter of wire (dw) and area (A) of the rope.
5.Calculate all the stresses in the rope
6.Calculate the effective stress (or load) during different condition such as starting stopping normal
working acceleration etc.
7.Now calculate the actual factor of safety.
8.Compare the actual factor of safety wit the factor of safety in the table.
The design is considered safe ,if actual faction of safety is in permissible limits.
Chain drives : A chain drive typically contains a chain running over the driving and driven shocked.
Chain drives and used to eusurea positive drive since unlike rope or belt drives chain drives operate
without any slip. This is due to the fact that during operation, the special protected on the shrocket ???
into the corresponding recesses in the chain links.
The chains used in chain drives are composed of number of rigid
links which are hinged together by pin joints which provide necessary flexibility for wrapping around the
drives and drives sprockets.
The chain drives are generally employed for application wit short centre distance (for
eg. Motareyeles, be – cyeles etc) for velocity upto 25m/sec and power transmission of 110km. however,
chain drives can also be designed to operate with larger centre distances, higher velocities and higher
power transmission.
Advantages :
1.The size is comparatively smaller &conpact than flat belt drive .
2.The load enerted on shafts is comparatively lower than belt drive.
3.It can be employed for centre distance as high as 8m.
4.A perfect velocity ratio (or a positive drive) is achied due to no slip
5.Under normal working conditions, efficiency as high as 99% can be achieved.
6. capable of transmitting power to multiple shafts by one chain only.
7.The power transmission capacity is comparatively higher than the belts.
8.Capaible of operating in adverse climatic conditions.
Disadvantages :
1.the alignment of shafts must be more accurate than belt drives.
2.cannot be used for application requiring praise times.
3.higher production cost
4.chain drive