Design Calculation of Pulley & Belt Drive

PULLEY & BELT CALCULATION – M.S.GURU

1. Selection of belt section : Select the type of a belt depending on the power to be transmitted. (see V-Belt Table in pg no:7.58 )

Example :

Motor power = 100 kW Motor speed(N1) = 1440 rpm Pump speed(N2) = 340 rpm

For power 100 Kw,D section is selected. (see V-Belt Table in pg no:7.58 )

2. Selection of pulley diameters (d and D) For power given, select small pulley diameter (d) from V-Belt Table in pg no: 7.58.then using the speed ratio, Calculate the larger pulley diameter (D). Speed ratio =

=

Larger pulley diameter, D = Speed ratio x d These pulley diameters should be rounded off to a standard diameter by using Table in (pg no:7.54). For Power 100 Kw, small pulley diameter d = 355 mm For power given,select small pulley diameter (d) from V-Belt Table in pg no: 7.58 Speed ratio =

=

= 4.235

Larger pulley diameter,, D = Speed ratio x d D = 4.235 x 355

D = 1503.53 mm Preferred larger pulley diameter, D = 1600 mm. Consulting Table, (pg no:7.54). 3. Selection of centre distance (C) : Considering Speed ratio, ratio can be obtained (see Table in pg no:7.61). Centre distance, C = D x (C/D) C max = 2(D+d) C min = 0.55(D+d) +T Centre distance,C = D x ( ) ( ratio can be obtained see Table in pg no:7.61). = 1600 x 0.95 C = 1520 mm C max = 2(D+d) = 2(1600+355) C max = 3910 mm C min = 0.55(D+d)+T (T value taken from table,see pg no: 7.58) = 0.55(1600+355)+19 = 1075.25+19 C min =1904.25 mm

4. Determination of nominal pitch length : Determine the length of the belt L ( Which is also known as nominal inside length ) by using formula. L = 2C + ( /2 ) (D + d ) + (D-d)2/4C For the Calculated nominal inside length and belt section,Consulting Table in 7.58,7.59 & 7.60),select the next standard pitch length. L = (2 x 1520) + (/2)(1600+355) +

= 3040 + (1.57 x 1955) + = 3040 + 3069.35 +254.938 L = 6364.28 mm

For this nominal inside length and D section,Consulting Table,pg no:7.58,7.59 @ 7.60, the next standard pitch length is selected as 6124 mm 5. Calculation of maximum power capacity : For selected section,calculate the maximum power capacity (in KW) of a V-belt using the formula given in Table in pg no: 7.62 Where Kw = Maximum power in kW AT 180° arc of contact for a belt of average length, S = Belt speed,



m/s,

De = Equivalent pitch diameter = dp x Fb, dp = Pitch diameter of the smaller pulley = d Fb = Pitch diameter factor to account for variation of arc of contact,from Table in pg no: 7.62 kW = ( 3.22 S-0.09 -

- 4.78 x 10-4 S2)

maximum power capacity (in KW) of a V-belt using the formula given in Table in pg no: 7.62 where =

S = Belt speed =





= 26.76 m/s de = dp x Fb dp = diameter of the smaller pulley Fb = Small diameter factor, for speed ratio of 4.235, From Table = 1.14 = 355 x 1.14 de = 404.7 Power, Kw = (3.22 x 26.760.09 -

- 4.78 x 10-4 x 26.762)

Power = 21.44 kW

6. Determination of number of belts(nb) : Determine the number of belts (nb) nb =

(see pg no : 7.70)

P = Motor power, kW, (i) Length correction factor (Fc) : For selected section,referring Table in pg no: 7.58,7.59 & 7.60) Length correction factor (Fc) can be obtained. (ii) Correction factor for arc of contact (Fd) : Arc of contact = 180° - (

) x 60°

For this arc of contact, consulting Table in pg no 7.68,correction factor for arc of contact is selected. (iii) Service factor (Fa) : Consulting Table in pg no 7.69 Select the service factor (Fa). Note: The details of driving units & driven machines under different duties are available in the data book,pg no 7.69 Kw = Rated power

nb =

(see pg no : 7.70)

FC= 1.00 referring Table in pg no: 7.58,7.59 & 7.60) Arc of contact,= 180° - (

) x 60°

For this arc of contact, consulting Table in pg no 7.68, correction factor for arc of contact is selected. = 180° - (

) x 60°

= 180° - (

) x 60°

= 130.85° For,130.85°,consulting Table,correction factor for arc of contact,Fd= 0.86 Service factor, consulting Table ,Fa= 1.3 nb =

nb = 7.05 nb = 7 Belts

7.Calculation of actual centre distance: C=A+√ Where

A = - (

)

B= C=A+√ A=

- (

)

A=

- (

)

= 1531 – 3.14(244.375) = 1531 – 767.337 A = 763.663 B= B= B= B = 193753.125 C = 763.663 + √ C = 763.663 + √ C = 763.663 + √ C = 763.663 + 624.041 C = 1387.704 mm C

8. Calculation of belt tensions (T1-T2) : Power transmitted per belt = (T1-T2)v

…… (i)

Determine mass per metre length( see Table in pg no : 7.58) Determine groove angle (2β) = 34° ( see Table in pg no : 7.70) Already found that arc of contact for smaller pulley, α =

radians

We know that the tension ratio for V-belts considering centrifugal tension, = e^µα/sin β = e^µα. βcosec

…….(ii)

Solving (i) & (ii), T1 & T2 can be obtained Power transmitted per belt = (T1-T2)v = (T1-T2)26.76,

S = v = Belt speed =



= =



=26.76 m/s

= (T1-T2) (T1-T2) = 467.12

……..(i)

Mass per metre length m = 0.596 kg/m ( see Table in pg no : 7.58) groove angle (2β) = 34° see Table in pg no : 7.70) Arc of contact, α = 180° - ( α = 180° - (

) x 60° ) x 60°

α = 180° - (

) x 60°

α = 130.85° α = 2.283 radians The tension ratio for V-belts considering centrifugal tension, = e^µα/sin β = e^µα. Cosec β = e^0.3x2.283. Cosec 17° = 10.53

T1-10.53T2 = -4067.79

……..(ii)

Solving (i) & (ii),we get T1 = 942.93 N T2 = 475.81 N 9. Calculation of stress induced: Consulting Table in pg no 7.58,cross-sectional area of selected section = Stress induced =

mm2

N/mm^2

Consulting Table in pg no 7.58,cross-sectional area of selected section = 475 mm2 Stress induced =

N/mm^2

Stress induced = Stress induced = 1.985 N/mm^2

10.Maximum power transmitted : P = (T1-T2)v v = S = 26.76 m/s

T1 = 942.93 N T2 = 475.81 N P = (T1-T2)v P = (942.93-472.81)26.76 P = (470.12)26.76 P = 12.58 kW

A belt is a loop of flexible material used to mechanically link two or more rotating shafts. Belts may be used as a source of motion, to transmit power efficiently, or to track relative movement. Belts are looped over pulleys. In a two pulley system, the belt can either drive the pulleys in the same direction, or the belt may be crossed, so that the direction of the shafts is opposite. As a source of motion, a conveyor belt is one application where the belt is adapted to continuously carry a load between two points.