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

f

c2

c1 eFF

FF

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

: angle of contact (radian) Note on the formula

Avoid slippage f

c2

c1 eFF

FF

, mean fe is the limits

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

C2

dDsin2

C2

dDsin2

1

D

1

d

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)

Allowable tension of F1 ≤ allowable tension.

To increase the tension, increase the width of the belt 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 2

c

Where ]/[60/ smdnV

Calculating F2 Manipulation of torque and power equation

Transmitted Torque: 2/dFFT21

Transmitted Power H: VFFH )( 21

Other related formulae

Initial Tension: c

21

iF

2

FFF

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/dn

Hd

V2

HdT

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

60/dn

dKHT 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 More than 1 belt is allowed.

Types of V-Belts: A, B, C, D and E (Table 17-9 pp 899)

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

range

For each type, length has been standardized by manufacturers

(Table 17-10 pp 899)

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

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

Advisable speed in between 5m/s to 25m/s

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

4.2

VKF

eFF

FF

eFF

FF

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

1. Calculate the pitch length LP

C4

)dD()dD(57.1C2L

2

P

2. Calculate the inside circumference (pp 899)

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

2

2

PP )dD(2)dD(2

L)dD(2

L25.0C

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 17-12)

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/dn

N/HF

NFVH

bd

bd

*recall that Hd = Hnom Ks nd

Solve for F1 and F2

21 FFF

15123.0

5123.0

1

e

eFFF c

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

Procedure in the analysis of flat belt (refer to Section 17-2)

1) Calculate fe

C2

dDsin2 1

s

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

s

f = 0.8 (from Table 17-2)

fe = 11.75

2) Calculate Fc

= weight of a meter belt (N/m)

= density of the belt (N/m3)

NF

bF

g

VF

c

c

c

2848

81.9

)3.20(8.67 2

2

mNb

xbx

Tablefrombt

sm

dnV

/8.67

)104.6()106.10(

217

/3.20

60

)1140)(340(

60

33

3) Calculate the Torque

Nm

n

nKHT dsnom

4.898

60/)1140(2

)1.1)(3.1()10(75

60/2

..

3

Ks = 1. 3(Table 17-15: can be used as a guide) nd = 1.1

4) Calculate the 2T/d (= (F1)a – F2

N

d

TFF a

7.5284

34.0/)4.898(2

2)( 21

5) Solve for F2

])[()( 2112 FFFF aa

vpaa CCbFF )( 1

Cp=0.72

Cv=1

Nb

bF a

34560

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

7.528734560

])[()( 2112

b

FFFF aa

5)

75.1128487.528734560

284834560

)(

2

1

bb

bb

eFF

FF f

c

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 AND 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

Based on power input possible Belt type is A or B

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)

* Note: you have other soln such as different sheave diameter or even Belt B.

1) Calculate Lp

C4

)dD()dD(57.1C2L

2

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

* Note: if the C is out of range, you have to choose other belt size…

If the value is smaller than D, repeat step 1) by setting Lp = D

If the value is larger than 3(D + d), repeat step 1) by setting Lp = 3(D+d)

At the end of this stage, the final configuration of the belting is confirmed

Type A2250

Input sheave , d = 75mm Output sheave, D = 537.5

n = 1720 rpm n = 240 rpm

Center to center distance = 616.6 mm

7) Calculate the number of belts

Basic formula : Power design/ Allowable power by each belt

all

d

H

HN

Allowable power by each belt

* Table 17.12 is the allowable power for each belt based on the sheave diameter and

speed. Interpolation is required. If the velocity is not within the range, redesign is

required.

Allowable power for each belt

Hall = K1.K2. Htab

sm

dnV

/754.6

60

Interpolation on Table 17-12 Htab = 0.581

K1 = angle of wrap (Table 17-13)

* Assumption: VV is used

88.0

75.0

1

K

C

dD

K2 = 1.05 (based on Lp = 2.282 m)

Htab (From Table 17-12)

Hall = K1.K2. Htab = (0.88)(1.05)(0.581) = 0.537 kW

Design Power

Hd = Hnom Ks nd

Hnom: the required power (1.5 kW)

Number of belt =

belts

H

nK

all

dsnom

4

516.0

)1.1)(1.1)(5.1(

5.1

must be round number…

Analysis of the V Belt (please refer to Section 17-3 for the steps)

1 ) Calculate the Fc (Eq 17-21)

N

FC

44.4

4.2

754.6)561.0(

2

Kc table 17.16

2) Calculate DF (using eq 17-22). Note of the units used ( d in m , H in W)

N

F

1.67

60/)1720)(075.0(

4/)1.1)(1.1)(1500(

3) Calculate F1 using Eq 17-23

15123.0

5123.0

1

e

eFFF c

Wrap angle calculation in radian

rad

C

dDd

37.2

2sin2 1

F1 = 108.7 N

4) Solve for F2 and Fi using Eq 17-24

F2 = 32.82 N

Fi = 67.61 N 6) Calculate the nfs

snom

all

KH

NH

fsn

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

7) Durability of the belt

Due to belt being flexible, it create flexural stress and this flexural will increase the value of F1 and F2 to respective T1 and T2.

Belt A : K = 2999 b = 11.089

D

KFFFT

d

KFFFT

bb

bb

1212

1111

)(

)(

Kb = Table 17-16

T1 = ….. N (note Edition 9 please use the corrected value) T2 = …. N

Lp = 2250 for A 2250

If (Np > 109) 3600

PP LNt * report as Np = 109

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

From Table 17-9 B-Belt: too low power capacity D- Belt: too big sheave for min requirement Therefore, the suitable is C6750 Lp = 6750mm + 72mm = 6822mm

2

2

pp)dD(2)dD(

2L)dD(

2L25.0C

C = 1909.3 mm

From equation 17-1 rad497.2

C2

dDsin2 1

d

Designed power capacity: Hd = Hnom Ks nd = 52.8 kW

The designed power above is the capacity to be transferred through the belt. The next step is to calculate the capacity per belt.

s/m074.660

dnV

Refer to Table 17-12 for tabulation of power per belt.

Per belt capacity lower for smaller sheave d1= 275 mm for V=6.074 Interpolation: H1= 3.35 kW Per belt capacity lower for smaller sheave d1= 300 mm for V=6.074 Interpolation: H1= 3.54 kW Per belt capacity lower for smaller sheave d= 290 mm for V= 6.074 Htab = 3.464 kW Find K1 (by interpolation)

s/m634.0C

dDV

Interpolation using Table 17-13 K1 = 0.833 From Table 17-14, using Lp = 6.822 m @ C type Belt The correction factor K2 = 1.15

The allowable power capacity per belt, Ha= K1 K2 HTAB = 3.318 kW

Number of belt: 9.15a

db

H

HN

b) Therefore number of belts is 16 To calculate F1, F2 and Fi

From Table 17-16 Kc= 1.716

9.15a

db

H

HN

From Eq 17-22 NN

H

ndF

b

d 3.54360

From Eq 17-19 N544.579)5123.0exp(FF

FFd

c2

c1

From Eq 17-24 FFF 12

Solving: F1= 814N, F2= 234.5N

Initial tension: N25.513F2

FFF c

21i

Torque: Nm10261.1n

H30T 3d

Number of passes