dme ppt for all mechanical engineering student

1. DESIGN OF MACHINE ELEMENTS
BELT AND CHAIN DRIVES
GUIDED BY:- Prof. Alok kumar dwivedi
NAME ENROLLMENT NO:-
VRAJ TRIVEDI 180053119057
CHIRANSHU UPRAITY 180053119058
SUMIT VADGAMA 180053119059
MITESHKUMAR VANKAR 180053119060

2. Introduction
• Beltsandchainsarethe major typesof flexiblepower transmission
elements
• Beltsoperate on sheavesor pulleys
• Whereaschainsoperate on toothed wheelscalledsprockets
• Often usedastorque increaser(speedreducer)
• (Geardriveshaveahigher torque capability but not flexibleor cheap)

3. TYPESOFBELT
• Flat belt is the simplest type, oftenmade from leather or rubber-coatedfabric
• Thesheavesurface is also flat and smooth, and the driving force istherefore
limited by the pure friction between the belt and the sheave

4. TYPESOFBELT
• Cogbelt, applied to standardV-grooved sheaves
• The cogsgive the belt greater flexibility and higher efficiency
compared with standardbelts
• They can operate on smaller sheavediameters
• A widely used type of belt is theV-belt drive
• The V-shape causesthe belt to wedge tightly into the groove,
increasingfriction and allowing high torques to be transmitted
before slipping occurs.

5. TYPESOFBELT
• Other types
V-band Poly-rib belt Double angle V-belt

6. V -BELTDRIVES
 The typical arrangement of the elements of aV- belt drive is
shown.
 The linear speed of the pitch line of both sheaves is the same as
and equal to the belt speed,then
• vb =R1ω1 =R2ω2

8. V -BELTDRIVES

9. V -BELTDRIVES

10. • Standard Belt CrossSections
V - BELTDRIVES
V - BELTDRIVES

11. V - BELTDRIVES
• Standard Belt CrossSections
X

12. V -BELTDRIVES
• Thebasicdata required for drive selection are thefollowing:
1. The rated power of the driving motor or other prime mover
2. Theservice factor basedon the type ofdriver and driven load
3. Thecenter distance
4. Thepower rating for one belt asafunction of the sizeandspeed of the smallersheave
5. Thebelt length
6. Thesizeof the driving and drivensheaves
7. Thecorrection factor for beltlength
8. The correction factor for the angle of wrap on the smaller sheave
9. Thenumber of belts
10. Theinitial tension on thebelt

13. V -BELTDRIVES
• Example
• Design a V-belt drive that has the input sheave on the shaft of an electric
motor (normal torque) rated at 50.0 hp at 1160-rpm, full- load speed. The
drive is to a bucket elevator in a potash plant that is to be used 12 hours (h)
daily at approximately 675 rpm.
• Solution
• Given
• Pin=50 hp
• N1=1160 rpm N2=675 rpm

15. • Step 1.Calculate
• design power (PD)
• PD=Pinxsf
• =50 x 1.3
• =65 hp

16. • Step 2. Select belt type.
• for 65 hp at 1160-rpm input speed.
• 5V belt is recommended

17. • Step 3.Calculatethe nominal speedratio(SR) SR=N1/N2
• SR=1160/ 675
• =1.72
• Step 4.Determine sheavesizes
• Choosebelt speedof 4000ft/min
• vb =π D1N1
 D2=22.7in

18. • Step 5. Find “standard” sheavesize
The two trials in boldface in Table 7-3 give only about 1 % variation from the desired
output speed of 675 rpm, and the speed of a bucket elevator is not critical. Because no
space limitations were given, let's choose the larger size.

19. • Step 6a.Find
• rated power(PR)
• For small sheave Ø12.4 in and 1160 rpm
• PR= 26.4 hp
Step 6b.Find“power added” to ratedpower
For SR = 1.72
and N1 = 1160 rpm
Power added
= 1.15 hp
∴ PR
=26.4 hp +1.15 hp
=27.55 hp

21. • Step 7.Findestimated centerdistance
• D2 < C < 3 (D2+D1)
• 21.2 < C < 3 (21.1+12.4)
• 21.2 < C < 100.5
• In the interest of conserving space,let's tryC=24.0 in.
• Step 8.Calculatebelt length
L=101.4in

22. • Step 9. Select standard belt length
•Lcalc = 101.4 in
• The nearest standard length is 100 in

23. • Step 10. Calculate actual center distance
• B=4L– 6.28 (D2+D1)
• B=4(100) – 6.28 (21.1+12.4) =189.6
C=23.3in

24. • Step 11. Calculate wrap angle, small sheave
θ1 =1580

25. • Cθ=0.94 CL=0.96

26. • Step 13. Calculate corrected power
• Corrected power =CθCLPR
• =(0.94)(0.96)(27.55 hp) =24.86 hp
• Step 14. Calculate Belts needed
=
2.61 belts (Use3
belts.)

27. ChainDRIVES
 Achain isapower transmission element madeasaseriesof pin- connectedlinks
 Thedesignprovides for flexibility while enabling thechain to transmit large tensileforces
 When transmitting power between rotating shafts, thechain engagesmating toothed
wheels, calledsprockets

28. Types of Chains

29. • Roller Chain Construction (Most commontype)

30. • General recommendations for designing chaindrives:
1. The minimum number of teeth in asprocket should be 17 unless
the drive is operating at avery low speed, under100 rpm.
2. The maximum speed ratio should be 7.0, although higher ratios are
feasible. Two or more stages of reduction can be used to achieve
higher ratios.
3. The center distance between the sprocket axes should be
approximately 30 to 50 pitches (30 to 50 times the pitch of the chain).
4. The larger sprocket should normally have no more than 120 teeth.

33. • General recommendations for designing chaindrives:
9. The arc of contact, θ1, of the chain on the smaller sprocket should be greater
than 120°
• θ1 =180° - 2 sin-1 [(D2 –D1)/2C]
10. For reference, the arc of contact, θ2, on the larger sprocket is
• θ2 =180° +2 sin-1 [(D2 –D1)/2C]

34. • Lubrication
• Itisessential that adequate lubrication be provided for chain drives.
• Thereare numerous moving parts within the chain, alongwith the interaction between the chain
and thesprocket teeth.
• Must bedefined the lubricant properties and the method of lubrication.

35. • Lubrication
• Lubrication methods
• (American ChainAssociation,Naples,FL)

36. • Lubrication
• Lubrication methods
• (American ChainAssociation, Naples, FL)

37. • Example
• Design achain drive for aheavily loaded coal conveyor to be driven
by agasoline engine through amechanical drive. The input speed will
be 900 rpm, and the desired output speed is 230 to 240 rpm. The
conveyor requires 15.0 hp.
• Solution
• Given
• Power transmitted, P=15 hp Speedof
motor, N1=900rpm
• Output speed range, N2=230 to 240rpm

38. • Step 1.
• Calculate design power (PD)
• PD=Pxsf
• =15 x 1.4
• =21 hp
TABLE Servicefactors for chaindrives
Typeof driver

39. • Step2.CalculateSpeedRatio(SR)SR=N1/ N2
• Usingthe middle of the requiredrangeof output speeds,
• SR = 900 / 235 = 3.83
• Step 3.
• Refer to the tables for powercapacity (Tables7-5, 7-6, and 7-7), and select the
chainpitch.
• For asingle strand, the no. 60 chain with p =3/4 inseemsbest.
• A17-tooth sprocket is rated at 21.96 hp at 900 rpmby interpolation.
• At this speed, type Blubrication (oil bath) isrequired.

43. • Step4.Computethe required number of teeth on the large sprocket:
• Z2=Z1xSR
• Z2=17x3.83 =65.11
• Let's usethe integer: 65teeth.
• Step 5.Computethe actual expectedoutputspeed N2 =N1(Z1/ Z2)
• =900 (17/65)
• =235.3 rpm

44. • Step 6.Compute the pitch diameters of thesprockets
• D1=4.082 in
• D2=15.524 in

45. • Step 7.Specify the nominal centerdistance.
• Let's usethe middle of the recommendedrange, C=40 pitches.
• Step 8. Compute the required chainlength
• L=122.5 pitches, let’s use122pitches

46. • Step 9.Compute the actual theoretical centerdistance
• C=39.75 pitches
• C=39.75 x 0.75 in =29.8 in

47. • Step 10. Compute the angle of wrap of the chain
• θ1 =180° - 2 sin-1 [(D2 –D1)/2C]
• θ1 =180° - 2 sin-1 [(15.524 – 4.082)/2(29.8)]
• θ1 =157.86°
• Becausethis is greater than 120°, it isacceptable.
• θ2 =180° +2 sin-1 [(D2 –D1)/2C]
• θ2 =180° +2 sin-1 [(15.524 –4.082)/2(29.8)]
• θ2 =202.14°

48. • SummaryofDesign
• Pitch: No.60
• chain. 3/4-in pitch
• Length:122pitches
• =122(0.75)
• =91.50 in
• Centerdistance:
• C= 29.825 in(maximum)
• Sprockets:Single-strand, no. 60. 3/4-in pitch Small:17teeth. D=4.082in
• Large:65teeth, D = 15.524in
• TypeBlubrication isrequired. Thelarge sprocket candip into anoilbath.

49. THANK YOU