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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
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