Flexible Mechanical Elements

INTRODUCTION

• Belts, ropes, chains, and other similarelastic or flexible machine elements areused in conveying systems and in thetransmission of power overtransmission of power overcomparatively long distances.

• In often happens that these elementscan be used as a replacement for gears,shafts, bearings, and other relativelyrigid power transmission devices.

Flexible mechanical elements

Chapter outline

1.Belts

2.Flat and round belt drives

3.Flat metal belts3.Flat metal belts

4.V belts

5.Timing belts

Belt

Belt

• Belts are the cheapest utility for powertransmission between shafts that may not beparallel.

• They run smoothly and with little noise, andcushion motor and bearings against load

• They run smoothly and with little noise, andcushion motor and bearings against loadchanges, albeit with less strength than gearsor chains. However, improvements in beltengineering allow use of belts in systems thatonly formerly allowed chains or gears.

Belt• Belt drive, moreover, is simple, inexpensive, and

does not require parallel shafts.

• It helps protect the car from overload, anddamping it from noise and vibration.

• Load fluctuations are shock-absorbed (cushioned).

• They need no lubrication and minimal maintenance.

• They have high efficiency (90-98%, usually 95%),

• high tolerance for misalignment, and areinexpensive if the shafts are far apart.

• Clutch action is activated by releasing belt tension.

� Cheap� Allows misalignment (parallel shafts)� Protects from overload� Absorbs noise and vibrations� Cushion load fluctuations� Needs little maintenance� High efficiency (90-98%, usually 95%),

Advantages

Disadvantages� Speed ratio is not constant (slip & stretch)� Heat accumulation� Speed limited – 2000 m/min,� Power limited - 700 kW� Endless belts needs special attention to install

Disadvantages

Belt

Belt types

Flat-belt geometry

Open belt

2sin2

2sin2

1

1

d

D

C

dDC

dD

πθ

πθ

−−=

−+=

−

−

Crossed belt

[ ] )(2

1)(4

22/122

dD dDdDCL

C

θθ ++−−=

[ ] θ

πθ

)(2

1)(4

2sin2

2/122

1

dDdDCL

C

dD

+++−=

−+= −

22

22

2222

0)(

022

)(

,

)(

ωθ

θωθθ

θθθ

θθθωωθ

fmrfFdF

dfmrfFd

fdSfFdfdNdF

dFFFfdNF

dSFddNdSdNd

Fd

dFFF

massbeltthemspeedbeltV

dFdmVdmrrmrddS

t

r

c

−=−

−=

−==

=++−−=

−==>=++−+−=

======

∑

∑

ωθ

fmrfFd

−=−

g

VF

mdnVf

fFFFF

mrFfFF

FF

mrF

mrF

mrfmrFFF

mrfmrFF

sidelooseatstartstconsAmrFAmrfAF

c

c

cc

c

2

121

22

2

122

2

221

222221

22222

222

22

sec/)exp(

1)exp()(

)exp(

)exp()(

)exp()(

,tan)exp(

ω

πφ

φ

ωφωω

ωφωωθω

θωωθ

φθ

=

==>−−=−

==>=−−=

−−

+−==

+−=

===>−==>+=

= the belting equation

gFc =

Fi = initial tensionFc = hoop tension due to centrifugal forceΔF’ = tension due to the transmitted torque TD = diameter of the pulleyD = diameter of the pulley

VFFH )( 21 −=H = the transmitted power

The difference between F1 and F 2 is related to the pulley torque.

Subtracting

adding

Dividing

(1) provided, (2) Sustained(3) in the proper amount

For satisfactory flat-belt drive, the initial tensio n must be:

If Fi equals zero, then T equals zero: no initial tension, no torque transmitted.

(3) in the proper amount(4) Maintained by routine inspection.

From the above equation:

Similarly ,

Plot of initial tension Fi against belt tension F1 or F2, showing the intercept Fc, the equations of the curves, and where 2T/D is to be found.

• Manufacturers provide specifications for their belts that include allowable tension Fa (or stress σall) , (N/unit width)• Belt life is usually several years. The severity of flexing at the pulley and its effect on life is reflected in a pulley correcti on factor Cp. • Speed in excess of 600 ft/min and its effect on lif e is reflected in a velocity correction factor Cv . For polyamide and urethane belts use Cv = 1. • A service factor Ks is used for excursions of load from nominal, applied to the nominal power as Hd = HnomKsnd , where nd is the design factor for exigencies.

CCbFF =)( vpaa CCbFF =)( 1

b = belt width, mmFa = manufacturer’s

allowed tension, N/mmCp = pulley correction

factorCv = velocity correction

factor(F1)a = allowable largest

tension, N

The steps in analyzing a flat-belt drive

1. Find exp(fØ) from belt-drive geometry and friction.

2. From belt geometry and speed find Fc.

3. From find necessary torque.

4. From torque T find the necessary

(

5. Find F2 from

6. From Equation find necessary initial tension Fi.

7. Check the friction development, f’<f. Used Equation solved for f’:

8. Find the factor of safety from

Example 17-1

• A polyamide A-3 flat belt 150 mm wide is used to transmit11 kW under light shock conditions where Ks= 1.25, and afactor of safety equal to or greater than 1.1 isappropriate. The pulley rotational axes are parallel and inthe horizontal plane. The shafts are 2.4 m apart. The 150mm driving pulley rotates at 1750 rev/min in such a waymm driving pulley rotates at 1750 rev/min in such a waythat the loose is on top. The driven pulley is 450 mm indiameter. The factor of safety is unquantifiableexigencies.

a) Estimate the centrifugal tension Fc and the torque T.

b) Estimate the allowable F1, F2, Fi and allowable power Ha.

c) Estimate the factor of safety. Is it satisfactory?

a) Estimate the centrifugal tension Fc and the torque T.

rad

d

0165.3

)2400(2

)150450(sin2 1

=

−−= −πθ

b) Estimate the allowable F1, F2, Fi and allowable power Ha.

vpaa CCbFF =)( 1

Cv=1.0 (Polyamide &urethane belts)

(F ) =the allowable

N

CCbFF vpaa

1890

)0.1)(70.0)(18000)(15.0(

)( 1

==

=

(F1)

a=the allowable

largest belt tension

c) Estimate the factor of safety. Is it satisfactory?

rad

d

0165.3

)2400(2

)150450(sin2 1

=

−−= −πθ

f=0.8 f’<f=>0.314<0.8

no danger of

slipping

Belt tension scheme

where; d = dip, mL = center-to-center d = dip, mL = center-to-center distance, mω = weight per foot of the belt, N/mFi = initial tension, N.

A decision set for a flat belt

• Function: power, speed, durability, reduction, service factor, C

• Design factor: nd

• Initial tension maintenance.• Initial tension maintenance.

• Belt material

• Drive geometry, d, D.

• Belt thickness : t

• Belt width : b

Thin flat metal belts fabricated by laser welding and thin rolling technology made possible belts as thin as 0.002 in and as narrow as 0.026 in.

Flat metal belts

• High strength-to-weight ratio• Dimensional stability

Thin metal belts exhibit:

• Dimensional stability• Accurate timing• Usefulness to temperatures up to 700°F• Good electrical and thermal conduction properties

Flat metal belts

A thin flat metal belt with the tight tension F1 and the slack side tension F2 revealed. The relationship between F1 and F2 and the driving torque T is the same as in Equation

Flat metal belts

WhereE= Young’s modulus

The tensile stresses (σ)1 and (σ)2

imposed by the belt tension F1 and F2 are

E= Young’s modulusT = belt thicknessv = Poisson’s ratioD= pulley diameterσ b = bending stress

The largest tensile stress is

The smallest tensile stress is

Table: Belt Life for Stainless Steel Friction Drives

Steps for the selection of a metal flat belt

1. Find exp(fØ) from geometry and friction

2. Find endurance strength;

for 301, 302 stainless steel. Np is the number of belt

5. F2 =F1a – ΔF = ab – ΔF

6. Fi

7. bmin =Np is the number of belt passes.

3. Allowable tension

4. ΔF=

8. Choose

9. Check frictional development f ’ :

Example 17-3

• A friction-drive stainless steel metal belt runs over two 100-mm metal pulleys

(f = 0.35). The belt thickness is to 0.08 mm. For a life exceeding 1000000

• Solution

a) Select the belt if the torque is to be 3.4 N.m

From step 1, Ø = θd = π

therefore exp(0.35π)=1

From step 2, a life exceeding 1000000 belt passes with smooth torque (Ks = 1),

a) Select the belt if the torque is to be 3.4 N.m

b) Find the initial tension Fi.

From step 2,

From step 3, 4, 5, and 6

Example 17-3

)10)(08.0)(10(193

)1(

681.0

)4.3(22

353)10(97702)(

3)35.0exp()exp(

39

21

407.06

106

abtbDv

EtSF

NmD

TF

MPaS

f

fa

f

=

−−=

===∆

==

==

−−

−

πφ

247213281

21368281

281)019.0(1479614796

10)(08.0()1.0)(285.01(

)10)(08.0)(10(193)10(353

21

12

1

32

396

1

NFF

F

NFFF

NbF

F

a

a

a

a

=+=+

=

=−=∆−====

−−= −

−

[ ]

)19(

7.12

9.60069.0

13

3

14796

68

1)exp(

)exp(

14796

14796168049849)10(353

)10)(08.0()1.0)(285.01(

)10)(08.0)(10(193)10(353

min

min

61

32

396

1

mmb

mmb

mmmb

f

f

a

Fb

ba

bNF

bF

a

a

==

==−

=−

∆=

==−=

−−= −

−

φφ

)(35.00882.0'

0882.0213

281ln

1ln

1'

2472

213281

2

2

1

21

okff

F

Ff

NFF

F ai

<=><

===

=+=+

=

πφ

V Belts

1. To specify a V belt, give the belt-section letter , followed by the inside circumference in inches. For example, C60 is a C-section belt having an inside circumference of 60 in .

2. Calculations involving the belt length are usually based on the pitch length. For any given belt section, the pitch length is obtained by adding a quantity to the inside circumference. For example, a C60belt has a pitch length of 62.9 in.

3. The groove angle of a sheave is made somewhat smaller than the

Note that:

3. The groove angle of a sheave is made somewhat smaller than the belt-section angle. This causes the belt to wedge itself into the groove, thus increasing friction.

4. The optimum working speed of the V-belt should be between 5000 ft/min and 1000 ft/min.

should not be greater than 3 times the sum of the sheave d iameters(D +d) and no less than the diameter of the larger sheave(D) .

center-to-center distance(C):

The design power is given by:

The number of belts, Nb, is usually the next higher integer to H d/Ha.

The centrifugal tension Fc is given by:

when the belt is used under other conditions, the ta bulated value Htabis adjusted as follows:

V belt tension

Bending induces flexural stresses in the belt ; the corresponding belt tension that induces the same maximum tensile stress is Fb1 at the driving sheave and Fb2 at the driven pulley. These equivalent tensions are added to F1 as:

the Gates Rubber Company used the tension versus pass the Gates Rubber Company used the tension versus pass trade-off in the form:

The Miner rule is used to sum damage incurred by the two tension peaks:

The lifetime t in hours is given by:

If NP > 109, report that N P = 109 and t > NP Lp/(720V)The analysis of a V-belt drive can consist of the f ollowing steps:

PROPERTIES

V-BELTS are oil and heat resistant and manufactured from material to prevent formation of static electricity.

1) OIL RESISTANT V-BELTS are resistant to damage from influence of mineral oil, splash and fatty components. This property provides longer belt life.

2) HEAT RESISTANT 2) HEAT RESISTANT To prevent aging and decomposition of the belts in high temperature, V-BELTS have obtained heat resistant properties.

3) ANTI STATIC V-BELTS are of electrical conductivity which prevents the builtup o static electricity. This property is of greatest importance when working with inflammable materials.

C

dDD 2

sin2 1 −+= −πθ

Timing belts

Timing belts

Sophisticated characteristics at low speeds and high torques open up new possibilities of use in sectors where only chains had been a possible solution so far. Features and Benefits of the Industrial Belt : Exceptional power transmission capability. · Versatile. · Clean · Quite running. · No Maintenance.

#1. Design a friction metal flat-belt drive to connect a 1-hp, four-pole squirrel-cage motor turning at 1750 rev/min to a shaft 15 in away, running at half speed. The circumstances are such that a service factor of 1.2 and a design factor of 1.05 are appropriate. The life goal is 106 belt passes, f = 0.35, and the environmental considerations require a stainless steel belt.

• Function: Hnom = 1 hp , n = 1750 rev/min , V R = 2 , C =˙ 15 in , Ks = 1.2 ,Np = 106 belt passes.• Design factor: nd = 1.05• Belt material and properties: 301/302 stainless steel

Given data

d• Belt material and properties: 301/302 stainless steelTable 17-8: Sy = 175 000 psi, E = 28 Mpsi, ν = 0.285• Drive geometry: d = 2 in, D = 4 in• Belt thickness: t = 0.003 in

• Belt width b• Belt loop periphery

Design variables:

Solutions:

The transmitting torque T and the design power :

#1. Design a friction metal flat-belt drive to connect a 1-hp, four-pole squirrel-cage motor turning at 1750 rev/min to a shaft 15 in away, running at half speed. The circumstances are such that a service factor of 1.2 and a design factor of 1.05 are appropriate. The life goal is 106 belt passes, f = 0.35, and the environmental considerations require a stainless steel belt.

• Function: Hnom = 1 hp , n = 1750 rev/min , V R = 2 , C =˙ 15 in , Ks = 1.2 ,Np = 106 belt passes.• Design factor: nd = 1.05• Belt material and properties: 301/302 stainless steel

Given data

d• Belt material and properties: 301/302 stainless steelTable 17-8: Sy = 175 000 psi, E = 28 Mpsi, ν = 0.285• Drive geometry: d = 2 in, D = 4 in• Belt thickness: t = 0.003 in

• Belt width b• Belt loop periphery

Design variables:

Solutions:

The transmitting torque T and the design power :

For full friction development,

Decision #1: b = 4.5 in

Existing friction

#3. A 60-hp four-cylinder internal combustion engine is used to drive a brick-making machine under a schedule of two shifts per day. The drive consists of two 26-in sheaves spaced about 12 ft apart, with a sheave speed of 400 rev/min. Select a V-belt arrangement. Find the factor of safety, and estimate the life in passes and hours.

Hnom = 60 hp, n = 400 rev/min, Ks = 1.4, d = D = 26 in on 12 ft centers.Design task: specify V-belt and number of strands (belts). Tentative decision: Use D360 belts.

Given data:

Lp = 360 + 3.3 = 363.3 inPitch length

Center to center distance C

Inside circumference Quantity to be added from table

Center to center distance C

Table 17-14: For D360, K2 = 1.10

Table 17-13: For θ= 180°, K 1 = 1

Table 17-12: Htab = 16.94 hp by interpolation b/n 2000 and 3000 ft/min

Number of belts, Nb

Nb = 5

At fully developed friction