INTRODUCTION TO BELT DRIVE SELECTION

Belt drive 1

Belt drive 2

By the end of this chapter, students should be able to:

A. Know the applications of belt drives and chain drives

List three types of transmission belts and chains in common use.

Compare and contrast belt and chain drives.

State the different engineering applications of belt and chain drives.

Describe the different parts of a vee groove sheave and belt.

B. Design a vee belt drive system ( Using Mitsuboshi Belting)

Determine the design power, and select the proper cross-section.

Select a pair of standard sheaves and belt pitch length.

Determine the number of belts required for a particular problem.

Calculate the forces acting on the shaft due to belt tension.

1. Introduction

Belt and chain drives are used for the transmission of power when the centre distance between drives are relatively large and need not be very precise.

Belts are machine elements that use friction for the transfer of energy. Belts drives can absorb shock and vibration and relatively quiet.

Chain drives offer the advantages of positive drive (no slip) and therefore greater power capacity

Ribbed belt driving alternator, steering wheel pump & air-con compressor.

Types of chain:

4. Advantages of Belt Drives over Chain Drives(a) Absorb shock and vibration.

(b) Simple maintenance as it does not need any lubrication.

(c) Relative low cost, initial cost of drive is moderate and replacement cost is low.

(d) Design flexibility. It permits power transmission over non-parallel shafts ( For Flat Belt only).

5. Advantages of Chain Drives over Belt Drives

(a) High efficiency. It permits smooth and efficient flow of power. Efficiency can be as high as 98%.

(b) Positive Drive (no slip); and therefore greater power of capacity. Normally used in automobile valve timing and printing machines.

(c) Can be operated in oil-surrounded or high temperature environment, where belt drives would be unsuitable.

(d) A single chain can drive many shafts at different speeds.

(e) Can be driven on both sides.

V-Belts

V-belts :

• trapezoidal cross-section

• run on pair of grooved pulleys.

• variety of standardized sizes and transmission powers, (kilowatts)

• operate best at belt speeds between 8 and 30 m/s. For standard belts the ideal speed is approximately 23 m/s.

• For most drive applications the maximum satisfactory speed ratio is approximately 7:1.

A typical cross section of a V-belt is shown below:

Material and Surface Finish of V-pulley

Important facts:

1. “” at smaller pulley is faster

2. “V” is a constant

3. V= R

4. (rpm) = (rads-1)

5. “Driver” : pulley to power source (eg. Engine, motor)

6. “Driven” pulley to load (eg. Conveyor, pump, fan…)

7. 1 Hp = 746 W

8. Use SI units during calculations.

60

2

Selection of V-BeltsSelection of V-Belts

•Belt design => selection of suitable belt size to transmit a required power.

•Use Mitsuboshi catalogues.

•Based on the empirical formulae.

Belt section ?

Std pulley size ? (pitch dia.)

Std pitch length ?

How many belts ?

S.R

C Ca

Follow arrangement of tables &

charts given

Example One :

Design a V-belt drive to transmit the power from an A.C. squirrel cage motor rotating at 1440 rpm and rated at 11 kW (14.7 HP) to a fan rotating at 720 rpm. Centres are to be near to, but not more than 750 mm apart and the driven pulley is not to exceed 250 mm outside diameter. The drive is required to run a minimum of 18 hours a day.

D

d

Centre Distance

1440 rpm

720 rpm

Driver

- A.C. squirrel cage motor

- rated 11 kw

- 18 hrs/day

Driven

- fan

Belt section ?

Std pulley size ? (pitch dia.)

Std pitch length ?

How many belts ?

Follow arrangement of tables &

charts given

(a) To select Belt Section

Service Factor (Table A2-2) = 1.3

Design Power = Rated Power x Service Factor

= 1.3 x 14.7

= 19.11 Hp

Belt Section Selected (Fig A2-3) = B

Belt section ?

Std pulley size ? (pitch dia.)

Std pitch length ?

How many belts ?

S.RFollow arrangement of tables &

charts given

(b) To determine Standard Pulleys Sizes

Speed Ratio = 2

Standard pulley selected (Table A2-4)

Pitch Diameter of Larger Sheave, D = 240mm

Pitch Diameter of Smaller Sheave, d = 120 mm

Belt section ?

Std pulley size ? (pitch dia.)

Std pitch length ?

How many belts ?

C Ca

Follow arrangement of tables &

charts given

( c) To Select Standard ( c) To Select Standard Pitch Length of BeltPitch Length of Belt

L = 2C + 1.57 (D+d) + (D-d)2

c

4C

= 2070 mm

( Table A2-1)

Standard pitch length, L (Table A2-5)s = 2057 mm

( Belt no. 81)

Belt section ?

Std pulley size ? (pitch dia.)

Std pitch length ?

How many belts ?

Follow arrangement of tables &

charts given

Power Rating per belt = 3.81 + 0.71Using Rpm of 1425 & pitch dia. of 115 mm.

= 4.52 Hp

Arc of contact, (D-d)/C = 0.16a

Arc of Contact correction factor = 0.97

Table A2-11

Length Correction Factors

Belt length correction factor (Table A2-11) = 0.98

Actual power rating per belt = 0.98 x 0.97 x 4.52

= 4.3 Hp

No. of belts = Design power/Actual power rating per belt

= 19.11 / 4.3

= 4.4

Choose 5 belts

Any Questions

Belt tension (Ts) generates hub load (Fs)

Hub load generates bending moment (M = Fs x L)

Power transmitted thru’ rotating shaft generates a torque (T = P/)

22ET TM

J

rTE Torsional stress of shaft:

Torque transmitted

Bending moment

Equivalent torque

Min. Shaft diameter

P

T

LFM s

22ET TM

316

all

ETd

Ts, N,

If the overhang at the fan pulley in Example 1 is 120mm find the maximum standard shaft diameter of the fan. Use the allowable shear stress of 50MN/m2

for the shaft material

From Example 1,

• Ls = 2057mm

• D = 240mm

•d = 120mm

•B section belt

•Ca = 746mm

•no. of belts = 5

•Pd = 19.11 Hp

Torque Transmitted by the shafts (use rated power)

T = P/

x 1000) / (2x 720)

= 146 Nm

Bending Moment, M on the Shaft

Belt speed, V = x R

= (2x 720) x (0.240 / 2)

= 9.05 m/s

Arc of contact, (D-d)/Ca = 0.16

Any Questions