Belt Design

8/12/2019 Belt Design

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Chapter 11

Power Transmission Drives

In order to transmit power from one machine member to other member, either flexible or

non flexible power transmission devices are used. Flexible device means the deviceswhose center distance between driving and driven members can be changed conveniently.

Examples of such drives are belt drive, chain drive or rope drives. In non flexible, the

center distance is almost remains constant. Examples are gear drives, clutches, couplings,power screw etc.

Belt drive: Belt drive consists of endless belt fitted over two pulleys and powertransmission between two shafts occur due to friction between the belt and pulley. Belt

drives are generally used when exact velocity ratio is not the prime reuirement in the

design. !ue to flexibility, it permits the designer to decide the relative distances between

the two shafts where the driving and driven machinery are to be "ept.

Basic types of belt drive

#here are four basic types of belt drives used in power transmission. #hese are$a% Flat belt drive

$b% &'Belt drive

$c% (ibbed type$d% #oothed type

Flat belt drive: Flat drives are used when moderate amount of power is to be

transmitted for longer distance up to maximum of )'1* m at a peripheral speed of about+'* m-s and speed ratio of '1*.

Open belt drive:/hen the shafts are parallel and rotates in same direction, open beltdrives is used.

Cross belt drive:0hafts are arranged in parallel but rotates in opposite direction

Quarter twist drive:In this case shafts are at right angle and roted in some definite

direction.

V belt drive:& belt drives are used when higher amount of power is to be transmitted for

considerable shorter distance up to maximum of '+ m at a peripheral speed of about +'

* m-s and speed ratio of '1+.

Belt design

2et the speed and diameter of the driver pulley is 3 1and !1and for driven pulley 34and

!4respectively. 5s the length of the belt passes over one minute is same, we can write

4411 NDND = or1

4

4

1

N

N

D

D= $11.1%

/hile deriving euation 11.1 the thic"ness of the belt is not considered. If belt thic"ness tis ta"en in to account, euation $11.1% becomes


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4411 %$%$ NtDNtD +=+ or1

4

4

1

N

N

tD

tD=

++

$11.4%

Euation $11.1 and 11.4% are derived without considering the slip. If these is slip s in the

belts with respect each other, then euation 11.1 or 11.4 can be written as

1

4

4

1

1**1

N

Ns

D

D=

and

1

4

4

1

1**1

N

Ns

tD

tD=

++

$11.%

Design formulation

#he basic euation for power transmission by a flat belt drive is given by

( )vTTP 41= m-s or watt $11.%where #1and #4are belt tension in tight and slac" side respectively., & is the belt speed

and is given by

1***6*=

DNV

when ! in mm, 3 in (78 $11.+%

!esign power can be obtained if a service factor is multiplied in Euation $11.%

!esign power 9( )

1***

41 SVKTT , "/ $11.6%

:s is service factor which depends on environmental and type of loading condition. #he

general values of :s varies between 1.4 to 1... #able 11.1 illustrates the service factorsgenerally used for different conditions.

#able 11.1 0ervice factors :s

2oading condition 3ormal loading

condition or torue

3on uniform

loading conditionor torue

;niform e.g. centrifugal pumps, fans, lightmachine tools conveyers etc.

1.*'1.4 1.1 < 1.

2ight shoc" , e. g. heavy duty fans, blowers,

compressors, reciprocating pumps, heavy dutymachines etc.

1.1'1. 1.4 < 1.

8edium shoc" e.g. vacuum pumps etc 1.4 < 1. 1. < 1.6

=eavy shoc", rolling mills, stone crushermachinery, grinders, hammering machineryetc.

1. < 1.+ 1.+ < 1.)

/e have from belt tension


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eT

T=

4

1$11.%

where is the angle of contact of the smaller pulley if both pulley are of same material.

If pulley materials are different, then should be the smaller value among the two.

/riting Euation 11. in the following form and simplifying we can derive

114

1 = eT

T$11.)%

14

41 = eT

TT or ( 1441 =

eTTT $11.>%

0ubstituting the value of #4from euation $11. % in above euation

( )1141

= e

e

TTT or

=

e

eTTT 1141 $11.1*%

=ence !esign power 7 9 ss Ke

eVTVKTT

=

1%$ 141 $11.11%

#ension in the tight side of the belt is

11

=

e

e

VK

PT

S

$11.14%

Belt Cross section

If t is the belt thic"ness and w is the belt width, then the

5rea

TensionMaximum

Area

Forcet == 9

wt

T1 $11.1%

Euation 11.14 is based on hori?ontal belt position. But in many circumstances the belt

position may not be hori?ontal and there may be some inclination with the hori?ontalposition. In these circumstances, the belt tension differs and slightly increases. =ence to

account this a correction factor "nown as correction factor to belt inclination $ %C may

be introduced. #he values of this correction factor is given in #able 11.4

#able. 11.4 Correction factor to belt inclination $ %C

#ype of drive Inclination to hori?ontal

*'6** 6*')** )*'>**

@pen belt drive 1 *.> *.)Cross belt drive *.> *.) *.

#he allowable value of belt tension at tight depends on belt cross section and strength ofthe belt material. 5s the belt ma"es one revolution it goes through a complex cycle of

fatigue loading.


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tC

Twt

1= $11.1%

0ubstituting the value of #1from Euation 11.14 in euation 11.1

1

1

=

e

e

VK

P

C

wtSt

9

e

eC

TT

t

1

41

$11.1+%

#he euation derived above is valid for belt that runs slowly enough that the centrifugal

loading can be neglected. For greater power capacity most of the belt drives operate atrelatively high speed. #he centrifugal force acting on the belt creates a tension #Cof

rmTC4

= 9 4wtV $11.16%where m is the mass per unit belt length, & is the belt velocity and r is the pulley radius.

/hen the velocity is high enough $&14 m-s% the centrifugal tension #C should beadded to belt tension #1and #4. #he corresponding euation of the ratio of belt tensionbecomes,

e

TTTT

C

C =

4

1$11.1%

It should be noted that due to centrifugal action the angle of warp reduces and hence the

torue transmission varies with speed. Considering the centrifugal action the design

euation becomes

t

C

C

TTwt

+= 1 $11.1)%

1TTwtC Ct = or 14 TwtVwtC t = $11.1>%

4

1

VC

Twt

t = $11.4*%

/ith centrifugal action

114

1 = eTT

TT

C

C$11.41%

( )1%$ 441 = eTTTT C $11.44%

(eplacing #49

e

T1 in the right side of the euation 11.44 and simplifying

=

e

eTTTT C

1%$ 141 $11.4%

9 ( )

e

ewtVwtC t

14$11.4%

!esign euation becomes


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( )

e

eVC

TTwt

t

14

41

=

$11.4+%

Belt initial tension

/hen the belt is wound around the pulley, initial tension is induced in the belt. #he initialtension in the belt depends on the elastic characteristics of the belt material. #he tensionsinduced in the tight and slac" side of the belt are shown in Figure 11.+

TTTT Ci ++=1 9 DMTT tCi -++

$a%

TTTT Ci +=4 9 DMTT tCi -+

$b%

Figure 11.+ Forces and torue on the pulley

From the tensions shown in Figure 11.+, the summation between the #1and #4is%-$%-$41 DMTTDMTTTT tCitCi ++++=+ $11.46%

9 Ci TT 44 +

4

441 Ci

TTTT

++= $11.4%

In absence of centrifugal action, initial tension should be average of #1and #4.

4

41 TTTi+

= $11.4)%

!ividing Euation 11.4) by euation 11.4

0implifying and rearranging we

can write euation 11.4) as

{ }

+=

e

eTTTTTTT Ci

1%$4-%%$$ 14141

1

1

+

=

e

e

D

MT ti $11.4>%

{ }

+=

e

eTTTT

TT

TC

i 1%$4-%$ 14141


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7/27

F E!

#1 #4

8t

#1'#4

#1 ' #4 #1 #i

#C #4 B 5 F E ! C B

Figure Flat belt tensions

. Centrifugal stressA #he stresses due to centrifugal force Fc is given by

AFCC= 9 wtgwtV

4

9 gV

4

=ence the maximum stress in the belt is

g

V

D

Et

wt

T 41max

++=

7ower transmission#he transmitted power is given by

( )

1***

41 VTTP

= , "/

Contact angle or arc of contact

#he contact angle for open belt is given by

C

dDS

4sin4 1

=

C

dDb

4sin4

1 +=


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#he contact angle for cross belt is given by

C

dDbS

4sin4 1

++==

where ! 9 diameter of larger pulley

d 9 diameter of smaller pulley

C 9 center distance 9 angle of contact

!engt" of t"e belt

#he length of the belt can be obtained from summation of two arc lengths and twice thedistances between the beginning and end of contact .#he length of open belt system is

given by

( ) ( )

C

dDdDC!

4

%4

4+

++=

#he length of cross belt system is given by

( ) ( )

C

dDdD

CC 4

%

4

4+

+

+

+=

Belt material

&arieties of belt materials are available which are listed in #able 11. with their mean

properties. For a good belt following material properties are desirable.

$i% It should have sufficient strength in tension as the belt is subected to tensionduring power transmission

$ii% It should be flexible as it passes over a pulley in every rotation

$iii% It should have good frictional surface characteristics as the power capacity

depends on the coefficient of friction.$iv% It should be light weight as it moves with high velocity and weight increases

centrifugal tension in the belt.

0ome of the materials which are used as belt materials are discussed below.

1. 2eather A4. Fabrics and canvas

. (ubber

. Balata

8aterial 0pecifications 0i?e,

mm

8inimum

pulleydiameter,

mm

5llowable

tensionper unit

width at m-s,3-mm

0pecific weight

3-m

$1*+%

Coeffici

ent offriction

2eather 1 ply t9.+

9+.*

+

>*

+.4

+.)

*.*>+'*.114

*.*>+'*.114

*.*

*.*

4 ply .*).*

11+1+4

.4).+

*.*>+'*.114*.*>+'*.114

*.**.*


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>.* 4* 1*.+ *.*>+'*.114 .**

(ubber '1* ply 1.4

'4.*

6.+'1*.+ *.1 *.4

Balata *.1*> *.4

Canvas *.14* *.*

Belt si#es

#he standard belt thic"ness and widths are listed in #able 11.+

#able.

8aterial Belt thic"ness, mm Belt width, mm

0tandard values +

6.+

)1*

14

+ to 6

+* to 1*

>* to 4414+ to **

4+* to 6**

2eather belt to +.+ single ply.+ to 1* double ply

4* to **4* to **

/oven cotton .+ to ).+ * to 4+*

/oven woolen 6.> to 11.6 +* to +**

Coefficient of friction

#he coefficient of friction depends on the belt and pulley material and their

environmental conditions. For leather belting with cast iron pulley generally *. is usedwhere as for rubber belting slightly lower value of co'efficient of friction $ 9'.4+% is

used. If pulley are of plastic, further reduced values of coefficient of friction is used.

Pulley Design

#able 0tandard pulley si?es in mm

*$+%++, +6, 6, 1, )*, >*, 1**, 114, 14+, 1*, 16*, 1)*, 4**, 44, 4+*, 4)*, 1+, ++,**, +*, +**, +6*, 6*, 1*, )**, >**, 1***, 114*, 14+*, 1**, 16**, 1)** and 4***


10/27

$%ample &'!esign a flat belt drive for a fan running at 6* rpm which is driven by a 1*

"/, 1* rpm motor. #he belt drive is open type and space available for center distance

is 4 m approximately.

olution

!ata given7ower 79 1*.* "/

0peed of motor 319 1* rpm

0peed of fan , 349 6* rpm8aximum C 9 4*** mm

5ssumption made

Belt position A =ori?ontal

0lip s 9 4.* !iameter of larger pulley in meter may be obtained from empirical relation

max

1 ,,.11.1N

PtoD = 9

1*

1*,,.11.1 to 9 *.*>16 to *.11*) m

!19>1.6 mm to 11*.) mmFrom the standard pulley diameter of mild steel or cast iron $ D >*, 1**, 114, 14+, 1*,

D % !19 1** mm is selected.

!iameter of driver pulley !19 1** mm$ii% !iameter of driven pulley $!4%

&elocity ration &(9 1*-6* 9

%1**-1$1

4

sD

DVR

= 9

%1**-41$1** 4

=

D , !49 >4 mm

5vailable diameters are D. 1+, ++, **, +** D etc

=ence, select !49 ** mm

Corrected velocity ratio &( $corrected% 9 **-$1**x$1'*.*4%% 9 .*)

5s the induced or corrected velocity ratio is different than given one, hence the deviationis to be determined from which conclusion regarding acceptance of the diameters can be

made.

!eviation from the given velocity ratio 9 $.*)'%- 9 4 #he general permissible deviation should be within the limit of to + . In this case it

is within the permissible limit, so diameters are accepted.

$iii% Belt speed &

1***6*

11

= ND

V

91***6*

%1*%$1**$

9 .+ m-s

$v% Centre distance C

Empirically ( )41%4+.1$ DDtoC + or 4+., DC

( )**1**%4+.1$ + toC or %**$+.,CC $min%9 +* mm to 1*** mm or 1** mm

5s per given data C 4*** mm


11/27

#a"e C9 1** mm

$v% 2ength of the belt

( ) ( )

C

dDdDC!

4

%4

4+

++=

9( ) ( )

%1**$

1****

4

%1****%1**$4

4+

++

9 6*4 mm

#a"e 2* 9 6** mm$vi% 5ngle of lap

C

dDS

4sin4 1

= 9

%1**$4

1****sin4 1

9 1.)+*9 .* radian

$vii% Coefficient of friction

Co efficient of friction can be selected from the pulley and belt combination. Consideringa leather belting, = *. is considered.$viii% Calculation of belt tension

5s both of the pulley are made of same material, hence the governing factor will be the

condition at the smaller pulley. =ence ( ) at smaller pulley should be ta"en in forcecalculation. 5lso as the speed is less than 14 m.s hence centrifugal action may be

neglected. /e have then,

eT

T=

4

1 9 %*,.,,.*$ e 9 4.)+

#1 9 4.)+ #4 $a%

From power to be transmitted

( )

1***

41 VTTP

= ,( )

1***

+.1* 41

TT = #1 ' #49 146.46 3

$b%From $a% and $b% #19 441>.6+ 3 and #49 )>.1* 3

From design euation

( )

e

eC

TTwt

t

1

41

=9

( )%*,.,,.*$

%*,..,,.*$ 14.1

1*.),>,6+.441>

e

et

9t

)66.1>4

Belt si?e can be determined from the allowable strength of the belt material.

From #able 11. we have

For t9. mm tw 9 +.4 3-mm

.4.+

)66.1>4

=w 9 ).14> mm for 4 ply without velocity correction

with velocity correction factor

>+.*.4.+

)66.1>4

=w 9 )6.4> mm for two ply


12/27

/idth per ply 9 .1+ mm

t9. mm w9 +.** mm

0imilar calculations are made for t9+.* and shown in following table.

8aterial thic"ness 8inimu

m pulleydia

5llowable tension

per unit width at m-s, 3-mm

&elocity

correctionfactor

/idth ,

mm

2eather, 1 ply t9.

9+.*

+

>*

+.4

+.)

*.>+

*.>)+

+

+

5s both width are available in standard si?e, hence any one may be recommended.

Belt specifications are

&'( m lengt" of () mm wide % ('( mm t"ic*ness and + plies

$%ample +' !esign a flat belt drive to connect two hori?ontal shafts. #he nominal power

transmission is to be + "/ under moderate shoc" loading condition. #he velocity ratiois 4.+ with speed of driven pulley is 1)** rpm. #he distance between the two shafts is

m. !esign factor may be ta"en as 1.1

0olution

iven 79 + "/&(94.+

319 1)** rpm

349 1)**-4.+ 9 +* rpmC9*** mm

2oading condition A moderate shoc" :s91.!ecision A Belt materialA leatherIn the present problem, as the nominal power is given, hence it is reuired to determine

design power as follows.

!esign power 9 3ominal power x service factor x design factor( ) ( ) dSa"noDesign NK#$#$ = min

From #able :s 9 1., 3d9 1.1!esign power, "/ 9 + $1.% $1.1% 9 +*.*+ "/

$i% !iameter of larger pulley in meter may be obtained from empirical relation

max

1 ,,.11.1N

PtoD = 9

1)**

*+.+*,,.11.1 to 9 *.1) to *.441 m

!191) mm to 441 mmFrom the standard pulley diameter of mild steel or cast iron $ D >*, 1**, 114, 14+,

1*,16*,1)*, 4**, 44 etcD % !19 4** mm is selected.

!iameter of driver pulley !19 4** mm$ii% !iameter of driven pulley $!4%

%1**-1$1

4

sD

DVR

= 9

%1**-+.41$4**+.4 4

=

D !49 ).+ mm


13/27

5vailable diameters are D. 1+, ++, **, +** D etc

=ence, select !49 +** mm

Corrected velocity ratio &( $corrected% 9 +**-$4**x$1'*.*4+%% 9 4.+6!eviation from the given velocity ratio 9 $4.+6'4.+%-4.+9 4.+6

#he general permissible deviation should be within the limit of to + . In this case it

is within the permissible limit, so diameters are accepted.$iii% Belt speed &

1***6*

11

=

NDV

9

1***6*

%1)**%$4**$

9 1).)+ m-s

$vi% Centre distance C C9 *** mm

$v% 2ength of the belt

( ) ( )

C

dDdDC!

4

%4

4+

++=

9 ( ) ( )

%***$

4**+**

4

%4**+**%***$4

4+

++

91+1*4.) mm

#a"e 2* 9 1+1*+ mm

$vi% 5ngle of lap

C

dDS

4sin4 1

= 9

%***$4

4**+**sin4

1 9 1.++*9 .*>> radian

$v% centrifugal force

For leather 91*** "g-m

Fc9 &49>.)1

%1***$1).)+ 49 *.64 87a

Cross section of belt

( )

e

eVC

TTwt

t

14

41

=

9( )

e

eVCV

#$

t

1

1*

4

,

ta"ing C 91.* ,%*>>.,,.*$

%*>>.,,.*$ 1

e

e9 *.6*6

( ) %6*6.*$44.*%1$)+.1)1**+.+*

,

=

t

wt

9,64.*

,,.,)6

t

#a"ing allowable tensile strength of oa" tanned leather as t9 .* 87a

wt 9 1664.+ mm4

for t9+.* mm w94.++ mm. In single ply, it is not available.

For t9+.+ mm w9*4. mm. 5vailable in single ply with w9** mm

$%ample ,5 belt drive transmits 1+ "/ at a belt speed of 4*m-s approximately and

velocity ratio of .+. #he center distance is approximately 4.+ times the diameter of larger

pulley. #he stress in the belt should not exceed 4.+ 3-mm4. !ensity of belt material is*.> gm-cc. 0electing a cast iron pulley and leather belt, determine the belt dimensions.

0peed of the driver unit is 1+* rpm

olution


14/27

$i% !iameter of smaller pulley

1***6*

11

=

NDV

9

1***6*

%1+*$4* 1

= D

9 41).4 mm

From available standard diameters selected d9 !19 44 mm

!iameter of bigger pulley !49 44+**

1+* 9 ) mm

0tandard diameter is ! 9 !49 )** mm

Corrected velocity is%1***$6*

11NDV = 9

%1***$6*

%1+*%$44$94*.+4+ m-s $without slip%

-ote : One can calculate ta*ing slip of +., /

difference between velocities 9 1**$4*.+4+'4*%-4* 9 4.64+ which is less than

permissible limits. =ence both pulley diameters are acceptable.

$ii% Center distance $ iven% C9 4.+ $!4% 9 4.+x)**9 4*** mm

Empirically ( )41%4+.1$ DDtoC +

( ))**44%4+.1$ + toC . =ence C 1+6 to 4*)5s the value is within limits, hence, it is acceptable. @ne can consider slightly smaller

center distance for the same problem.

$iii%( ) ( )

C

dDdDC!

4

%4

4+

++=

9( ) ( )

%4***$

44)**

4

%44)**%4***$4

4+

++

0+6>.>6 mm0 )1)2 mm

$iv% 5ngle of twist

C

dDS

4sin4 1

= 9

%4***$4

44)**sin4

1 9 16. deg 9 4.)+ radian

5s the speed is more than 14 m-s, centrifugal tension should be considered.

8ass of belt per meter length

m9density x Cross sectional area 9 >*1***1***

tw 9 *.>x1*'wt "g-m

/e have#c9 m&49 *.>x1*'$wt% $4*.+4+%49 *.*)wt

eTT

TT

C

C =

4

19

%)+,.4$,.*

4

1

*).*

*).*e

wtT

wtT=

94.+

,+.4%*).*$*).* 41 wtTwtT = 0 4.+ #4'*.>6* wt

0implifying we get #1< 4.+ #4G *.++4 wt 9* $i%

/e have also


15/27

( )

1***

41 VTTP

= 9 1+ 9( )

1***

+4+.4*41 TT

#1 ' #49 *.)16 $ii%AT =1 0 %$wt 0 4.+$wt% $iii%

From $i% to $iii%

$%ample (5 4+ "/,1+* rpm motor drives a machine through a flat belt. 5ssuming

maximum belt tension limited to 1+ "3 and the coefficient of friction to be at least *.4,determine belt si?e.

$%ample )5 + "/, 1+* rpm motor drives a machine through a multiple flat belts. #heweight of the belt is 4.6 "g-m. #he pulley on the motor shaft has 1* mm pitch diameter

and the angle of warp is 16)*. 5ssuming maximum belt limited to ** 3 and coefficient

of friction between belt and pulley *.4*, $a% determine how many belts are reuired $b%Belt dimensions.

0olution

1***6*

11

=

NDV

9

( )( )1***6*

1+*1*

9 14.) m-s

eTT

TT

C

C =

4

19 1)*-16)4.* e 9 +)6.*e 9 1.>)

8ass of belt per meter length m9 4.6 3-m

/e have

#c9 m&4

9 4.6 $14.)%4

9 4.>) 35lso given maximum belt tension limited to ** 3. =ence =1T ** 3

>).14

1 =

C

C

TT

TT

>).1>).4

>).4**

4

=

Tgives =4T +>.4 3

7ower transmission per belt in "/ ,( )

1***

41 VTTP

= 9( )

1***

),.144.+>**9 1.+

"/

3o of belt reuired 9#$

#$

+.1

+9 .4 9

number of belt will be reuired.

8ass of the belt per unit length 9 5rea x density 9 tw

5ssuming a leather belt type for which ,-1*** m#g=6.41*** === wttwm

,1*6.4 =wt m

4

3ow ta"ing standard thic"ness*.+=t mm +4*=w mm (emar" A 0tandard belt is not available


16/27

*.)=t mm ,4+=w mm (emar" A 0tandard belt is not available*.1*=t mm 46*=w mm 3emar* : tandard belt is available

$%ample 1: (ecommend a flat belt drive for driving a centrifugal pump with ) "/ motor

operating continuously at 1+* rpm. #he pump speed should be 16++1* rpm and the

center distance may be from +* mm to 1*** mm. #he preferable distance may be >**mm.

$%ample 4

$%ample 5

Flat metal belt design for $%pected !ife

#he belts are of simple cross section or geometry, but their life determination is verydifficult because the oint that made to ma"e the loop is not accurately "nown. #he

general rule for determination of the fatigue life can not be applied to belt.

5pproximately belts are tested on two eual pulleys to determine the belt life. #he data

available for metal belt is given in table

6aterial 7ield

strengt"

6Pa

7oung

6odulus8

6Pa

Poisson9s

ratio

Dt Bending

stress

$;' ,2

Belt passes

*1 or *4

stainless

steel

14** 1> *.4> 64+

**

4**

+4

61*++

61**.+** . 1*6

*.16+ . 1*6

*.*)+ . 1*6

#he regression euation between stress and life shows the following relation4*).*

>*+1 = %N with correlation coefficient (H*.>where 3fis the number of belt passes.

Following step may be followed to select a metal flat belt for transmission of power.

0tep 1. InputsA 7ower reuired $ rated power%, Center distance if available, speed ratio or

speed of driver and driven parts0tep 4. 0elect service factor and calculate design power

0tep . !etermine pulley diameters and select standard pulleys

0tep . &erify the velocity ratio, and its acceptability, If it is not acceptable, select nextpair of pulley diameters and verify its acceptability.

0tep +. Chec" for any other diameter constraints if available.

0tep 6. Find the ratio of belt tensione

T

T =4

1or

e

TT

TT

C

C =

4

1according to the

velocity from friction and geometry

0tep . Find endurance strength4*).*

>*+1 = %% NS for *1 or *4 stainless steel

,

y%S

= for others


17/27

0tep ). Find allowable tension ( ) wt

D

EtST %a""

=

41 or

allowable strength ( )DEt

S%a"" 41

=

0tep >. CalculateD

MTTT t

441 ==

0tep 1*. CalculateD

MTTTT ta""a""

4

4 == or from step 6 and step >

0tep 11 Find initial belt tension4

4TTT a""i+

=

0tep 14. Find ( )

e

eVC

TTwt

a""14

41

=

0tep 1 Find w for standard value of thic"ness t

0tep 1 &erify the friction if it is assumed or selected from the standard value.

$%ample


18/27

( ),

1,1.4)1

)*

=wt9 .1>

4.*-1>.min =w 0 +2'5< mmVerification

)*1.4)4 == wTTT a"" 9 )*411.4) 9 +44.>1 341 TTT += 9 )* G +44.>1 9 6*4.>1 3

e

T

T=

4

19

>1.+44

>1.6*4

( )41-ln1

TT

= 9 ( )1+4>.1ln1

9 *.*+ which is less than the assumed value.

=ence it is accepted.

$%ample &2. 0elect a stainless steel belt for continuous operation for a mine hoist inwhich hoist runs at +* rpm and the 6 "/ driving motor runs at ++ rpm for 1* 6belt

passes.

$%ample &&'Find belt tensions during high and low power demand for a compressor

driven by * "/ internal combustion engine at *** rpm. #he service factor is 1.,

center distance ++* mm, pulleys diameters are ** mm and 1** mm. #he machineoperates at an input power of * "/ for 1+ of the time and at an input power of 4* "/

for remaining time.

V Belt selection

& belt drive is generally used where a great amount of power is to be tranmitted from one

pulley to another pulley. #his is possible because of increased coefficient of friction%sin-$

I = as compared to flat belt. !ue to increased value of coefficient of friction,

the drive permits to operate at a reduced value of angle of contact on smaller pulley andwith a low initial tension.

Belt selection means to find a standard belt that will closely match the reuired power tobe transmitted at reuired velocity ratio between input and output shafts for a reuired

center distance. #he standard cross section of a & belt is shown in figure 11.1*. #he

dimensions of the cross section are standardi?ed by the manufacturers and Bureau ofIndian standard. #here are five basic standards & belts and named as 5, B, C, !, and E

type & belt.

*

*

#


19/27

/

#able 11.11 !imensions of standard & belt cross section

Belt0ection

/idth/,

mm

#hic"ness#, mm

Juantity to beadded for length

conversion

8inimumsheave-pulley

diameter

"/ range, oneor more belts

5B

C

!E

11

44

4)

)11

1

1>4

+

)11

+1+

44+

4++*

*.4 to .+*.+ to 1)

11 to +

to 1)++ or higher

0pecification of & belt is written as followsA

Belt section'inside circumference length, mmExample ! *1 is a ! type section belt having an inside circumference length of *1

mm. #he pitch length is obtained by adding the uantity mentioned in table 11.11. #he

pitch length of !*1 is *1G) 9 *> mm $ 7itch length of the belt is defined as the

circumferential length of the belt at the pitch width i.e width at the neutral axis of the beltcross section.

#he included angle between sides is generally **and the angle of groove in the pulley is

generally made less than that of the belt so that the belt wedges to increase the effectivecoefficient of friction. #his angle on the pulley is generally "ept * to +*depending

upon the diameter of the pulley and type of belt and angle of contact.

#able 11.14 Inside circumference length of standard belt

Belt

0ection

Inside Circumference length, mm

5

B

C

!

66*,), ))>, 1*6, 116), 141>, 16, 1>, 1, 1+4+, 1+, 1646, 166,14, 1)*, 1>*+, 1>)1, 4*4, 41+>, 44)6, 4), 466, 4)+, *), 4+1

))>, >6+, 1*6, 116), 141>, 14>+, 16, 1>, 1, 1+4, 1+, 1646, 166,14, 1)*, 1>**, 1>)1, 4**, 4*+, 41+>, 44)6, 464, 46, 4+*, 4616,

466, 4)>, *), 4+1, 4, +, 6), *1, >, +4, >+, +,6*>6, 6)+), 64*14>+, 1+4, 14, 1>*+, 4*+, 41+>,

*), 4+1, 6+), *1, 11+, >,


20/27

E +4, >+, +,

Because of standardi?ation belt selection has become a step by step procedure whichshould be followed. #he procedure is explained below.

Belt selection procedure

tep &. #he information stated here are generally given in the problem statement. If not

given, select specific motor or engine reuired for the drive system and the reuired

speed ratio, center distance between input and output shafts along with the tolerablelimits of speed ratio and center distance.

tep +'0elect service factor from #able 11.1*. If none of the machines characteristics areeuivalent to the driver and driven euipment, use figure for section of design power

from the "nown value of fastest speed of the euipment.

#able 11.1* 0ervice factors

!riven 8achine !river #he euipments shown below arerepresentative only. 0elect a machine

which is very close to the given one.

5C 8otorsA 3ormal torue,suirell cage, synchronous,

split phase

!C motorsA 0hunt wound

EnginesA 8ultiple cylinderinternal combustion

5C 8otorsA =igh torue,=igh slip, (epultion

induction,0ingle phase 0eries

wound, slip ring

!C motorsA 0eries wound,compound wound

EnginesA 0ingle cylinder

internal combustion

2ine shaft, clutches

Intermi'

ttent

service

3ormal

service

Contin'

uous

service

Intermi'

ttent

service

3ormal

service

Conti'

nuous

service

'+

h-day

)'1*

h-day

1*'4

h-day

'+

h-day

)'1*

h-day

1*'4

h-day

5gitator for liuids Blowers, Exhausters,

centrifugal pumps K compressors, fan up

to 1* =7, light duty conveyers

1.* 1.1 1.4 1.1 1.4 1.

Belt conveyers for sand, grain etc,!ough 8ixer, Fans over 1* =7,

enerator, 2ine shaft, 2aundry 8achine,

8achine tools, punches, presses , shears,printing machines, positive displacement

rotary pumps, revolving and vibratory

screens

1.1 1.4 1. 1.4 1. 1.

Bric" machinery 1.4 1. 1. 1. 1.+ 1.6

Crushes, hoist 1. 1. 1.+ 1.+ 1.6 1.)


21/27

100

1000

10000

1 10 100 1000

Design power, kW

Speedofthefastershaft,rpm

Figure 11.1* Cross section selection chart

tep ,. @nce the design factor is selected, design power is obtained as!esign 7ower 9 (ated 7ower 0ervice factor

tep (. @btain the belt cross section from the figure 11.1* by finding the intersection of a

line representing the rpm of the fastest shaft, given along the ordinate and a linerepresenting the design power, given in "/ along the abscissa. #he area in the

intersection point indicates the cross section is to be used. 0ome times the intersection

point falls near a boundary. In this case two selection be made one for each and decisionmay be ta"en after design and analysis. For Example +* "/ $!esign 7ower% is to be

transmitted at 1*** rpm. #he intersection point lies on the boundary of C belt and ! belt.

5nalysis should be made for both the belts.

5

C

!

B

E


22/27

tep ). #he fifth step is to select standard pulley for this cross section from #able 11.14

to provide the desired speed ratio.

#able 11.11 0tandard pulley pitch diameters for 5, B, C, ! and E belt series

Belt section 7ulley pitch

diameter, mm

Belt speed , m-s

+ 1* 1+ 4* 4+


23/27


24/27

0tep . Calculate the belt length for the desired centre distance for the pulley selected

above. 0ome times the exact center distance is not "nown. In this circumstances, it should

be decided from the space available for the drive.

4 !1 !4

C

( )144

1DD

4-

( ) ( ){ }[ ] 4-1414441 +.*44

44

DDCD

D ++=

where

=

C

DD

4cos4

141

0implifying above euation we can write

( ) ( )

C

DDDDC

44

414

14

+++=

0tep ). 0elect a standard length from the table 11.1+ and calculate the modified center

distance from the following euation.

( ) ( ) ( )

++

+= 414

4

1414 444

1DDDDDDC

+

+

+=

4-1

414

41414

4

%$411

44+.*

DD

DDDDC

If the center distance calculated above is satisfactory, move to the next step. @therwise,

repeat the process till a satisfactory and acceptable belt length and center distance is


25/27

obtained. If the center distance obtained is not satisfactory after so many trial,

recommend to select other drive such as chain, rope etc.

0tep >. #he next step is to calculate the power per belt. #o determine this, find out the

belt length correction factor corresponding to standard length selected from #able 11.1+,

correction factor for arc of length and power rating.

#able 11.1+ Correction factor for belt length, :2

3ominal belt length, mm

5 belt B type C type ! type E type 2engthfactor

4>*>6+'11*

144*'1**

1+**'1)+

1>+*'44+*4**'4)**

***

11*144*'1+**

1+*'1)+

1>+*'44+

464+'***4**'6**

>+*'+**

)+

1)+4*4+'4**

464+'***

4**'>+*

*+*')++4+*'6***

6+*'+**

)4+*

4**6**'*+*

4+'+4+*

6***

6+*')4+*>***'1*+**

14***

1+**

)++4+*'6***

6+*'+**

)4+*'>+*

1*+**'14***1+**'1+***

16+**

*.)+*.>*

*.>+

1.**

1.*+1.1*

1.1+1.4*

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

80 90 100 110 120 130 140 150 160 170 180

Angle of contact, degree

Angleofcontactcorrectionfactor

(egression euation 11+.***>.*1**.4 4+ ++= K with >>>,.*4 =r

Figure Correction factor for angle of contact


26/27

7ower per belt is determined from the following euation.

7ower per belt9 (ated power length correction factor 5ngle of contact correctionFactor

( ) KK#$be"t#$ rated =

0tep 1*. #he number of belt reuired is obtained from

3o of belt 9beltper

power!esign

#$

#he number of belt so obtained should be rounded up to nearest integer.

$%ample &+. 0elect a & belt drive for 1+ "/, 1* rpm motor, which drives a centrifugal

pump running at a speed of +6 rpm for a service of )'1* hours per day. #he distance

between the driver and driven shaft is approximately 1.4 m.

olution

0ervice factor sK 91.1. !esign factor aN 9 1.* $ assumed as it is not stated in the

problem%

!esign 7ower 9 (ated 7ower 0ervice factor !esign factor( ) asDesign NK#$#$ = 9 1+1.11.* 9 16.+ "/#he highest speed of the motor is 1* rpm. 3ow corresponding to 1* rpm and 16.+

"/, B type of belt is selected.#he width and thic"ness of the belt are /9 1 mm and #911 mm, respectively.

Corresponding to B type belt, minimum pulley diameter is 1+ mm. #he other pulleyavailable are 1+, 1+,1++ , 16+, 1+ or higher. 0electing

!19 1+ mm!49 !1&( 9 1+4.+ 9 64.+ mm#he preferred pitch diameter available are ++, +, **, 4+ etc.

0electing ++ mm as pitch diameter the design velocity ratio becomes 4.> andpercentage difference is 4.*6 which is acceptable.

Belt pitch length ( ) ( )

C

DDDDC

44

414

14

+++=

( ) ( )14** 1+,++1+,++414**44

+++= 9 4>>1.>+ mm

#he standard pitch length for B type is 4)> or *> mm. First considering shorter belt

length center distance is re calculated. #he shorter distance is ta"en because thecalculation is made considering the maximum space available for the transmission drive.

#he modified center distance is


27/27

+

+

+=

4-1

414

41414

4

%$411

44+.*

DD

DDDDC

9

+

4-1

4

4

4

+**4)>

%41*$411

4

+**4)>4+.*

9 1*>.* mm

Considering the higher belt length, the centre distance is

+

=

4-1

4

4

4

+**,*>,

%41*$411

4

+**,*>,4+.*

C 9 11>.** mm

#his is more than the space available. 0o this belt length is reected.

#o calculate the "/ per belt the multiplying factors are reuired.

=

C

DD

4cos4

141 9

1*+*4

1+,++cos4 1 9 16).+*

1***6*

11

=

NDV

9

1***6*

%1*%$1+$

9 1*.> m-s

K 9 1.*+, K 9 *.>), and rated "/ 9 4.1

( ) KK#$be"t

#$rated = 94.11.*+ *.>) 9 4.16*>

3o of belt reuired 916*>.4

+.169 .6 9 ) belts are reuired.

Documents

Belt Design