Belts (Flexible Mechanical Elements)

Purpose: Power transmission at a long distance Ability to absorb shock make belts is preferable.

Types of Belts

Flat and round belts can vary in length V and Timing have standard length

Basic Working Principle

Two pulleys drive and driven Drive pulley is normally to motor, engine etc. This motor will cause the pulley to rotate. As a result of the rotation of the pulley, friction between pulley and belt will PULL the belt. The pulling effect causes one side of the belt to tight (side that is being pulled) and the other side to be slacks. Application

Conveyor belt, CVT (scotter, Honda Jazz and City) Flat Belt Geometry Open: same direction output Crossed different

Tight Side

Slack side

Relationship between tight side F1 and F2 is

fc2

c1 eFFFF =

where Fc: centrifugal force f : coefficient of friction of the belts

: angle of contact (radian) Note on the formula 9 Avoid slippage f

c2

c1 eFFFF

, mean fe is the limits

9 As increases fe increases

Calculating ef Refer to Table 17-2 Coefficient of friction depends on material is the wrap angle in RADIAN, which depends on the configuration of the open or closed and the pulley we examine. Refer to formula 17-1

+=

=

C2dDsin2

C2dDsin2

1D

1d

D = diameter of larger pulley d = diameter of small pulley (govern b Type of belt: Table 17-2) C = center distance Calculating F1

Allowable Tension per unit width at 3 m/s (103) N/m Therefore the maximum allowable tension: Allowable tension x w (width of the belt)

9 Allowable tension of F1 allowable tension. 9 To increase the tension, increase the width of the belt 9 The value of the F1 is at speed 3 m/s, other speed correction

factor must be included Allowable Tension: (F1)all = b Fa Cp Cv

b: width of the belt Fa: allowable tension (Table 17.2) Cp: Pulley correction factor (Table 17-4) CV: velocity correction factor if the V > 600 ft/min CV : From Fig 17-9 pp 1057 (leather only) Cv = 1 (for other belts)

Calculating centrifugal force FC

Fact of centrifugal force

22

c mrF = Weight of a meter (per unit length)

bt =

= weight density (Table 17-2) [kN/m3] b = width of the belt [m] t = thickness [m]

g/VF 2c = Where ]/[60/ smdnV =

Calculating F2 Manipulation of torque and power equation

Transmitted Torque: ( ) 2/dFFT 21 =

Transmitted Power H: VFFH )( 21 = Other related formulae

Initial Tension: c21i F2FFF +=

STEP IN ANALYZING BELT (pp 868) 1. Calculate ef

2. Setting the belt geometry : such as d (min pulley diameter d:

Table 17.2) and D and calculate Fc

3. Calculating the T

Two simultaneous equations

( )

d/T2)FF(2/dFFT

21

21

==

Eq 1

( )VFFH 21 = Eq 2

60/dnHd

V2HdT

=

=

In this case, H is H = Hnominal (input) x Ks (service factor Table 17-15)

60/dndKHT snom=

4. From the T, find (F1)a F2 = 2T/D

5. Calculate F2 from ]F)F[()F(F 2a1a12 = 6. Calculate Fi

7. Check the friction developed f which should f < f

8. Calculate the f.o.s = Ha/HnomKs

Example Q17-10

Two shafts 6m apart, with axes in the same horizontal plane, are to be connected with a flat belt in which the driving pulley, powered by a six pole squirrel cage induction motor with a 75-kW rating at 1140rpm, drives the second shaft at half its angular speed. The driven shafts drives light-shock machinery load. Select a flat belt. Initial Selection (the selection may differ from this one) Polyamide A-5 Diameter drive pulley: d = 340 mm n = 1140 rpm Diameter driven pulley: d = 680 mm n = 570 rpm

V-Belts 9 More than 1 belt is allowed. 9 Types of V-Belts: A, B, C, D and E (Table 17-9 pp 879)

Selection of belt is based on (min) sheave diameter and kW

range

9 For each type, length has been standardized by manufacturers (Table 17-10) :

A950 (Type A and circumference length = 950mm ),

E 4875 (Type E and circumference length = 4875mm)

9 Advisable speed in between 5m/s to 25m/s 9 Eliminate vibration: D < Center distance < 3(D+d) as

excessive vibration will shorten the belt life

Relationship between tight side F1 and F2 is

2

CC

5123.0

c2

c1

)2/sin(f

c2

c1

4.2VKF

eFFFF

eFFFF

=

=

=

Kc (from table 17-16) Procedure V-Belt Selection

1. Calculate the pith length LP

C4)dD()dD(57.1C2L2

P

+++= 2. Calculate the inside circumference

L= Lp- Lc 3. Based on the calculated L, choose the belt from Table 17-10

4. Calculate the Lp

LP=L + LC 5. Calculate center distance between pulley C

++

+= 22

PP )dD(2)dD(2L)dD(

2L25.0C

6. Verify the value C to statisfy D < C < 3(D+d)

7. Calculate the number of belt required N = Hd/Hall round up the value N: number of belts

Hd : designed power

Hall : allowable power per belt

Hall = K1K2Htab

Where

Htab : tabulated value (from table)

K1 : angle of wrap correction factor (Table 17-13)

K2 : belt length correction factor (Table 17-14)

The above procedure will determine the dimensional of the belting system such d,D, C, Lp. The next step is to calculate the related parameters Centrifugal force

2

4.2

= VKF CC eq 17-21 Power transmitted per belt

60/dnN/HF

NFVH

bd

bd

==

*recall that Hd = Hnom Ks nd

Solve for F1 and F2

5123.0c1

5123.0c1

21

eFFF

eFFF

FFF

+==

=

Example Q17-22 A 1.5kW ELECTRIC MOTOR RUNNING AT 1720 rpm IS TO DRIVE A BLOWER AT A SPEED OF 240 rpm. SELECT A V-BELT DRIVE FOR THIS APPLICATION ABD SPECIFY STANDARD V-BELTS, SHEAVE SIZES, AND THE RESULTING CENTER-TO CENTER DISTANCE. THE MOTOR SIZE LIMITS THE CENTER DISTANCE TO AT LEAST 550 mm.

Example 2 A machine commissioned to be utilized at a construction site will consist of a V-Belt pulley system of 290mm and 1500mm sheave diameters at about 1.9m apart, driven by a 40 kW engine at 400 rpm. The machine will operate under normal torque of lightest medium shock condition with 1.1 design factor. The engineer has to select from either B2125, C6750 or D9000 belt based on minimum sheave and kW range. Determine

a) The suitable belt type and reason for choice b) The actual center distance of the pulleys c) The number of belt(s) required, d) The forces on one belt e) The factor of safety, and f) The life of the belts in passes and hours

Example with solution Example Q17-10

Two shafts 6m apart, with axes in the same horizontal plane, are to be connected with a flat belt in which the driving pulley, powered by a six pole squirrel cage induction motor with a 75-kW rating at 1140rpm, drives the second shaft at half its angular speed. The driven shafts drives light-shock machinery load. Select a flat belt. Initial Selection (the selection may differ from this one) Polyamide A-5 Diameter drive pulley: d = 340 mm n = 1140 rpm Diameter driven pulley: d = 680 mm n = 570 rpm Step no 1:

Calculate fe

= C2dDsin2 1s

D = 680mm d=340 mm C = 6 m = 6000 mm rad08.3s =

f = 0.8 (from Table 17-2)

fe = 11.75 Step no 2: Calculate Fc

NF

bF

gVF

c

c

c

284881.9

)3.20(8.67 2

2

==

=

mNbxbxTablefrombt

sm

dnV

/8.67)104.6()106.10(

217

/3.2060

)1140)(340(60

33

==

=

==

=

Step No 3: Calculate the Torque

Nm

nnKHT dsnom

4.89860/)1140(2

)1.1)(3.1()10(7560/2..

3

==

=

Ks = 1. 3(Table 17-15: can be used as a guide) nd = 1.1 Step No 4:

N

dTFF a

7.528434.0/)4.898(2

2)( 21

==

=

Step No 5:

])[()( 2112 FFFF aa =

vpaa CCbFF =)( 1 Cp=0.72 Cv=1

NbbF a34560

)1)(72.0)(48000()( 1==

7.528734560])[()( 2112

==

bFFFF aa

Step No 5

75.1128487.528734560

284834560

)(

2

1

bbbb

eFFFF fc

ca

Solve for b b = 0.182m = 182mm Therefore bmin = 182, choose b = 200mm Then calculate f, F1, F2, Fi and T.

One of the solutions is

Belt : Polyamide A-5 d1 = 340mm D2 = 680mm and b = 200mm Fi, F1, F2, T .

Example Q17-22 A 1.5kW ELECTRIC MOTOR RUNNING AT 1720 rpm IS TO DRIVE A BLOWER AT A SPEED OF 240 rpm. SELECT A V-BELT DRIVE FOR THIS APPLICATION ABD SPECIFY STANDARD V-BELTS, SHEAVE SIZES, AND THE RESULTING CENTER-TO CENTER DISTANCE. THE MOTOR SIZE LIMITS THE CENTER DISTANCE TO AT LEAST 550 mm. Solution Assumption belt Type A, Minimum sheave = 75mm Therefore, diameter of drive pulley d = 75mm diameter of driven pulley D = 537.5mm

(to reduce input 1720 rpm to 240 rpm) 1) Calculate Lp

C4)dD()dD(57.1C2L2

P

+++= C = 550mm d = 75mm D = 537.5mm Lp = 2159 mm 2) L = Lp - Lc = 2159 32 = 2127 mm Lc = length correction factor 3) From Table 17-10 Choose A2250 2250 mm

4) Lp = 2250 + Lc = 90 + 32 = 2282 mm 5) Recalculate new C

++

+= 2

2

pp )dD(2)dD(2L)dD(

2L25.0C

C = 616.6mm

6) Verify the value C to statisfy D < C < 3(D+d) 537.5 < C < 3(612.5) OK

7) Number of belts

all

d

HHN =

Hall = K1.K2. Htab K1 = angle of wrap (Table 17-13)

845.0K

75.0CdD

1 ==

K2 = 1.1 (based on Lp = 2.282 m) Htab (From Table 17-12)

sm

dnV

/754.660

==

Interpolation on Table 17-12 Htab = 0.581

Hall = K1.K2. Htab = (0.845)(1.1)(0.581) = 0.54 kW

Number of belt =

belts

HnK

all

dsnom

454.0

)1.1)(1.1)(5.1(

5.1

==

=

Fc, F1 and F2

NVKF CC

44.44.2

754.6)561.0(

4.22

2

=

=

=

N

dnNHF bd

1.6760/)1720)(075.0(4/)1.1)(1.1)(1500(

60//

==

=

5123.0

5123.01

1.7644.4 eeFFF c+=

+=

rad13.3C2

dDsin2 1d

=

=

Solve for F1 and then F2

Note: Smaller sheave leads to more belts Large sheave to larger D and larger V V increase Htab increase *However, constraint of V-Belt V 5 m/s 25 m/s

Example 2 A machine commissioned to be utilized at a construction site will consist of a V-Belt pulley system of 290mm and 1500mm sheave diameters at about 1.9m apart, driven by a 40 kW engine at 400 rpm. The machine will operate under normal torque of lightest medium shock condition with 1.1 design factor. The engineer has to select from either B2125, C6750 or D9000 belt based on minimum sheave and kW range. Determine

a) The suitable belt type and reason for choice b) The actual center distance of the pulleys c) The number of belt(s) required, d) The forces on one belt

Solution FromTable179BBelt:toolowpowercapacityDBelt:toobigsheaveforminrequirementTherefore,thesuitableisC6750Lp=6750mm+72mm=6822mm

++

+= 2

2

pp )dD(2)dD(2L)dD(

2L25.0C

C=1909.3mm

Fromequation171rad497.2

C2dDsin2 1d

=

=

Designedpowercapacity:Hd=HnomKsnd=52.8kWThedesignedpoweraboveisthecapacitytobetransferredthroughthebelt.Thenextstepistocalculatethecapacityperbelt.

s/m074.660dnV ==

RefertoTable1712fortabulationofpowerperbelt.Perbeltcapacitylowerforsmallersheaved1=275mmforV=6.074Interpolation:H1=3.35kWPerbeltcapacitylowerforsmallersheaved1=300mmforV=6.074Interpolation:H1=3.54kWPerbeltcapacitylowerforsmallersheaved=290mmforV=6.074Htsb=3.464kWFindK1(byinterpolation)

s/m634.0C

dDV == InterpolationusingTable1713K1=0.903FromTable1714,usingLp=6.822m@CtypeBeltThecorrectionfactorK2=1.15Theallowablepowercapacityperbelt,Ha=K1K2HTAB=3.598kW

Numberofbelt: 7.14HHN

a

db ==

Thereforenumberofbeltsis15TocalculateF1,F2andFiFromTable1716Kc=1.716

7.14HHN

a

db ==

FromEq1722 N544.579NH

nd60F

b

d ==

FromEq1719 N544.579)5123.0exp(FFFF

dc2

c1 ==

FromEq1724 FFF 12 = Solving: F1=814N, F2=234.5N

Initialtension: N25.513F2FFF c21i =+=

Torque: Nm10261.1nH30T 3d ==