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BELT DRIVES
Functions of belts
   Transmit motion between shafts that
    are located at a considerable distance
    from each other

   They are not used for exact fixed
    speed ratio
       slipping

   They are very flexible
     – distance or
     – the angle between the two shafts.
Types of Belts
Belt Pulleys
•Flat belts:
     Crowned pulleys

• Round and V- belts
    grooved pulleys or
    sheaves

• Timing belts
    toothed wheels or
    sprockets.
Layout of Flat belt drive :

Non-reversing Open Belt

Reversing Crossed Belt


 Reversing Open Belt


Quarter Twist Belt drive
V-Belts – Standard Sizes


                                                     1
                                                 1
                                        1            2
                                    1
                            7           4

         1        21        8
                  32
         2
                                            29
                                3
    5        13        17                   32
                                4
    16       32        32



         A        B         C           D        E
Selection of V-Belts

    Power to be transmitted

    Speed of the small or large pulley

    Speed ratio

    Field of application.
    Approximate distance between the
     centres of the two pulleys
Steps of V-belts Selection:
  Determine the service factor Based on
  application,
  Obtainthe design power from the
  equation:
 Design power = transmitted power x service factor

  Select   a suitable belt size from fig.(5. )
Steps of V-belts Selection:
   Find the diameter of the small pulley
    (d) from table (5. ).
   Find the diameter of the large pulley
    (D) from the equation:
    D= d x speed ratio
   Find the length of the belt using the
    equation:                      ( D d )2
               L   2C   1.57( D   d)
                                       4C
Steps of V-belts Selection:
    Obtain the standard length of the belt
     from table (5. )
    Calculate the exact centre distance
     from the equation
                   b   b2    32 ( D d )
               C
                            16
    Find the angle of lap (arc of contact),
     from the equation
    Angle of lap = 180  ( D d )60
                                   C
Steps of V-belts Selection:
       Find the capacity of one belt from the
        equation:
                                       YxS
        Capacity of one belt =
                                  0.91
                              XS          ZS 3
                                               de
       The values of X,Y and Z can obtained
        from table (5. )
       The equivalent small pulley diameter de
        can be obtained from the equation:
        de = diameter of small pulley x coefficient of small
                               pulley
Steps of V-belts Selection:

   Find the power transmitted by one
    belt from the equation:
    Power of one belt = belt capacity x length
    coefficient x coefficient of arc of contact



   The required number of belts can be
    obtained from the equation:
    No. of belts = Design power/ power of one belt
Example:
An engine lathe is driven by a squirrel cage electric
motor through a V-belt. The electric motor runs at
1500 rpm with a maximum power of 3 hp. If the
input speed to the engine lathe is 500 rpm and the
centre distance between the motor pulley and the
lathe pulley is 30 in. Select a suitable size, length
and number of belts if the lathe is expected to be
working for two shifts, 16 hours/day.
Solution:
                        Pitch diameter                                         AC Motor: High torque , High-slip
                                               AC Motor: torque, Standard Groove Dimensions
                                                                     Squirrel
From table (5.1), service factor for machine tools with Squirrel cage
Size of belt       Minimum                     Cage Groove
                                                      ,Synchronous,     Split
                                                                               Repulsion-Induction Single-phase
                                      Range                          W          D          X          S
                                                                               ,Series Wound, Slipping, DC Motor  E
  electric motor is = 1.2
                 recommended
                 Application                   Phase angle Motor :Shunt
                                                       ,DC
                                                                               : series wound Compound Wound
                                     2.6 to 5.4wound, 340Engines 0.494:Multi-
                                                                               Engine :Single- cylinder Internal 3/8
     A                  3                              Internal, Combustion 0.490        0.125       5/8
Design power = transmitted power x service factor
                                     Over 5.4 cylinder 380         0.504       Combustion, Line shafts :Clutches
Hour in daily service                4.6to 7.0         340         0.637
     B                 5.4                       3--5       8-10      16-24 0.580  3-5 0.1758-10 ¾ 16-24 ½
            = 3 x 1.2
Agitators for liquids, Blowers and
                                     Over 7.0          380         0.650
 exhausts , Centrifugal pumps and7.0 to 7.99              0
            = 3.6 hp                             1.0 34 1.1        0.879
                                                                       1.2         1.1         1.2          1.3
 compressor , Fan up9.0 10 hp and 8.0 to 12.0
     C                  to                             36 0        0.887      0.780      0.200        1       1 11/16
 machine tool, Light-duty conveyors 12.0
                                    Over               38 0        0.895
From figure (5.3) size A is selected
Belt conveyors for sand, grain, etc.
Dough mixers and Fan over 10 hp --12.99
                                    12
Then the recommended diameter of small pulley from table (5.2)
Generators and line-shafts, Laundry and
                                    13.0 -- 17.0
                                                       340         1.259
                                                          0
 printing machinery 13.0
     D                , Punches, presses         1.1 36 1.2        1.271
                                                                       1.3    1.0501.2 0.300 1.3 1 7/16 1.4 7/8
               d = 3 in.
 ,shears , Positive displacement rotary
                                    Over 17.0
                                                       380         1.283
 pumps, Revolving and vibrating screens
     speed ratio = 1500/500
Brick and textile machinery        18.0 to 24.0        360         1.527
BucketE elevators and exiters ,Piston
                      21.0                                0                   1.300      0.400      1 3/4       1 1/8
                                    Over 24.0          38          1.542
                 =3
 pumps and compressors ,Hammer-mills
 and paper-mill beaters , Conveyers and
                                                 1.2         1.3       1.4         1.4         1.5          1.6
Diameter of large pulley (D) = d x speed ratio
 pulverizers, Positive displacement
 blowers, Sawmill and wood-working
 machinery                      =3x3
Crushers ,mills and hoists
                                                 1.3         1.4       1.5         1.5         1.6          1.8
                                = 9 in.
Rubber calendars , extruders and mills
The length of the belt can be obtained from the equation:

                                    ( D d )2
                L    2C 1.57( D d )
                                       4C
                                    ( 9 3 )2
                L 2 x30 1.57( 9 3 )
                                      4 x30
                L = 79.14 in.
A       B       C                     A     B       C      D       E
      Standard                               Standard
                 Standard Pitch Lengths ,
     Designation                            Designation   Standard Pitch Lengths , Inches
                         Inches
From table (5.3) the standard length of the … …= 79.3 in
         26       27.3     ….      …….      belt  …
         31       32.3     ….       ….    107.9 … …
with a designation number A78.
         33       34.3     ….       ….          … …
         35       36.3             …..            …
The exact centre distance from the …
         38
         42
                  39.3
                  43.3
                                   equation: … …
                                   …..
                                    …             …
         46       47.3
                                   b…       b2   32(D     … 2
                                                           d)                          …
         48       49.3      C …                           …                            …
         51       52.3                           16       …   …                        …
         53       54.3              …                     …                            …
                  56.3              …                     …
                  61.3      b 4 L 6.28( D                 …   d)
                  63.3              …                     …
                  65.3              …                     …
                  67.3       b      4x 79.3 6.28(9 3)
                                    …                     …
                  69.3                                    …
       b = 241.84 72.3
                  76.3
                                    …                     …
                                                          …
                                                              …
                                                              …
                                                            2 …
                  79.3
                            …
                                    …
                               241.84
                                    …
                                                  241.84  …
                                                          …   …
                                                                  32(9 3) 2
                  81.3
                   ….
                        C                                 …   …         …
                   ….               …                16   …   …         …
                  86.3                                    …   …         …
       C = 30.08 in.
                  91.3                                    …   …         …
                  97.3      …                    …        …   …         …      …       …
The angle of lap (arc of contact), from the equation:
Angle of lap =
                      ( D d )60
                  180
                          C


 Angle of lap =       ( 9 3 )60
                  180
                         30.08

 Angle of lap = 168.03o
Speed Ratio     Small        Speed        Small              Speed      Small
  Range        Diameter    Regular Quality Belts
                             Ratio       Diameter            Ratio     Diameter
 The capacity of one belt from the equation:
                Factor       Range        Factor
                              Belt Cross Section            Range       Factor
                                                         1.341 - 1.429
 Capacity of one belt =
1.000 - 1.019
        FACTORS
1.020 - 1.032
                 1.00
                 1.01
                            1.110 - 1.142
                                    BXS
                                               1.05
                        A 1.143 - 1.178 0.91 YxS ZS 3D - 1.562 E
                                               C
                                               1.06
                                                         1.430
                                                                         1.10
                                                                         1.11
                                               de        1.563 - 1.814
1.033 - 1.055 Values of X , 1.179 - 1.222 be Used in H.P. Formula
                 1.02       Y and Z to         1.07                      1.12
                                                         1.815 - 2.948
1.056 - 1.081    1.03       1.223 - 1.274      1.08                      1.13
                                                          2.949 - and
  From table(5.4)
1.082 - 1.109 X  1.04 1.945 1.275 - 3.434
                                    1.430     6.372 13.616 19.914 1.14
                                               1.09
                                                              over
      X =2.684 3.801 9.830 26.899 93.899 177.74
         Y
         Z        0.0136 0.0234 0.0416 0.0848 0.1222
      Y =5.326
      Z =0.0136 Premium Quality Belts
  S = (3.142 x3 x 1500)/12000 = 1.18
                        Belt Cross Section
  de FACTORS =A
     = 3 x 1.14 3.42 B                  C      D       E
          Values one ,belt = to684Used in91 5.326 x1.18 0.0136 x1.1833
  Capacity of of X Y and Z 2. be x1.180. H.P. Formula
       X               2.684   4.737  8.792              3.42
                                                      18.788 24.478
       Y               5.326 13.962 38.819            137.70 263.04
  Capacity
       Z        of   one belt = 1.26 hp
                      0.0136 0.0234 0.0416            0.0848 0.1222
Find theStandard Belt Cross Section by one Contact fromCthedrive E
       Arc of Contacttransmitted Standard of belt Belt Cross Section
          power A Type ofCdrive Length
          Length              B
                                               Arc
                                                      A   B
                                                               Type of equation:
                                                                        D
              on                    V Designation on
                          V to VFactor to Flat              V to V Factor to Flat
                                                                       V
Power of one belt =belt …
        Designation
        Small sheaves
            26
                      Correction
                     0.81     …
                                     capacity x sheaves Correction … …
                                               Small length coefficient x
                                                      …          …
coefficient of arc of … …Factor
            31
            33       0.86     …
                               contact
                     0.84 Correction
                                    …
                                                             Correction Factor …
                                                                        …
                                                                        …        …
                                    …                                           …
Coefficient of arc of contact (from130
            35
             180
            38
                     0.87
                     0.88
                           1.00     …
                                       0.75          table (5.6)= 0.974.
                                                             0.86         0.86 …

Length coefficient0.95 … 0.77 (5.7)= 1.03.
             170
            42
             160
            46
                     0.90
                     0.92
                           0.98
                            (from table
                                    … 0.8
                                                   120
                                                   110
                                                      …
                                                      …
                                                             0.82
                                                             0.78
                                                                        … 0.82 …
                                                                          0.78 …
                                    …                 …                   0.74 …
Power of one belt = 1.26 0.82
             150
            48
            51
             140
                     0.93 0.92
                     0.94 0.89
                                       x 0.974 x901.03… 0.74
                                       0.84
                                                   100
                                                      …      0.69         0.96 …
                                    …                 …                         …
            53       0.93
                           = 1.264 hp
                                    …                 …
                                                      …
The required number of…belts can be obtained from the …
                                    …                 …
equation:                           …                 …
No. of belts = Design power/ power…of one belt
                                    …                 …   …
                                                      …   …
No. of belts = 3.6/ 1.264           …                 …   …

                      …
                       = 2.85……                       …
                                                      …
                                                          …
                                                          …      …
Take 3 belts …                      …                 …
                                                      …
                                                          …
                                                          …
                                                                 …
                                                                 …
                                                  …     …     …
                            …            …        …     …     …     …     …
Thank You
   Dr. Salah Gasim Ahmed
           MET 103         20

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MET 304 Belt drives

  • 2. Functions of belts  Transmit motion between shafts that are located at a considerable distance from each other  They are not used for exact fixed speed ratio slipping  They are very flexible – distance or – the angle between the two shafts.
  • 4. Belt Pulleys •Flat belts: Crowned pulleys • Round and V- belts grooved pulleys or sheaves • Timing belts toothed wheels or sprockets.
  • 5. Layout of Flat belt drive : Non-reversing Open Belt Reversing Crossed Belt Reversing Open Belt Quarter Twist Belt drive
  • 6. V-Belts – Standard Sizes 1 1 1 2 1 7 4 1 21 8 32 2 29 3 5 13 17 32 4 16 32 32 A B C D E
  • 7. Selection of V-Belts  Power to be transmitted  Speed of the small or large pulley  Speed ratio  Field of application.  Approximate distance between the centres of the two pulleys
  • 8. Steps of V-belts Selection:  Determine the service factor Based on application,  Obtainthe design power from the equation: Design power = transmitted power x service factor  Select a suitable belt size from fig.(5. )
  • 9. Steps of V-belts Selection:  Find the diameter of the small pulley (d) from table (5. ).  Find the diameter of the large pulley (D) from the equation: D= d x speed ratio  Find the length of the belt using the equation: ( D d )2 L 2C 1.57( D d) 4C
  • 10. Steps of V-belts Selection:  Obtain the standard length of the belt from table (5. )  Calculate the exact centre distance from the equation b b2 32 ( D d ) C 16  Find the angle of lap (arc of contact), from the equation  Angle of lap = 180 ( D d )60 C
  • 11. Steps of V-belts Selection:  Find the capacity of one belt from the equation: YxS Capacity of one belt = 0.91  XS ZS 3 de  The values of X,Y and Z can obtained from table (5. )  The equivalent small pulley diameter de can be obtained from the equation:  de = diameter of small pulley x coefficient of small pulley
  • 12. Steps of V-belts Selection:  Find the power transmitted by one belt from the equation: Power of one belt = belt capacity x length coefficient x coefficient of arc of contact  The required number of belts can be obtained from the equation: No. of belts = Design power/ power of one belt
  • 13. Example: An engine lathe is driven by a squirrel cage electric motor through a V-belt. The electric motor runs at 1500 rpm with a maximum power of 3 hp. If the input speed to the engine lathe is 500 rpm and the centre distance between the motor pulley and the lathe pulley is 30 in. Select a suitable size, length and number of belts if the lathe is expected to be working for two shifts, 16 hours/day.
  • 14. Solution: Pitch diameter AC Motor: High torque , High-slip AC Motor: torque, Standard Groove Dimensions Squirrel From table (5.1), service factor for machine tools with Squirrel cage Size of belt Minimum Cage Groove ,Synchronous, Split Repulsion-Induction Single-phase Range W D X S ,Series Wound, Slipping, DC Motor E electric motor is = 1.2 recommended Application Phase angle Motor :Shunt ,DC : series wound Compound Wound 2.6 to 5.4wound, 340Engines 0.494:Multi- Engine :Single- cylinder Internal 3/8 A 3 Internal, Combustion 0.490 0.125 5/8 Design power = transmitted power x service factor Over 5.4 cylinder 380 0.504 Combustion, Line shafts :Clutches Hour in daily service 4.6to 7.0 340 0.637 B 5.4 3--5 8-10 16-24 0.580 3-5 0.1758-10 ¾ 16-24 ½ = 3 x 1.2 Agitators for liquids, Blowers and Over 7.0 380 0.650 exhausts , Centrifugal pumps and7.0 to 7.99 0 = 3.6 hp 1.0 34 1.1 0.879 1.2 1.1 1.2 1.3 compressor , Fan up9.0 10 hp and 8.0 to 12.0 C to 36 0 0.887 0.780 0.200 1 1 11/16 machine tool, Light-duty conveyors 12.0 Over 38 0 0.895 From figure (5.3) size A is selected Belt conveyors for sand, grain, etc. Dough mixers and Fan over 10 hp --12.99 12 Then the recommended diameter of small pulley from table (5.2) Generators and line-shafts, Laundry and 13.0 -- 17.0 340 1.259 0 printing machinery 13.0 D , Punches, presses 1.1 36 1.2 1.271 1.3 1.0501.2 0.300 1.3 1 7/16 1.4 7/8 d = 3 in. ,shears , Positive displacement rotary Over 17.0 380 1.283 pumps, Revolving and vibrating screens speed ratio = 1500/500 Brick and textile machinery 18.0 to 24.0 360 1.527 BucketE elevators and exiters ,Piston 21.0 0 1.300 0.400 1 3/4 1 1/8 Over 24.0 38 1.542 =3 pumps and compressors ,Hammer-mills and paper-mill beaters , Conveyers and 1.2 1.3 1.4 1.4 1.5 1.6 Diameter of large pulley (D) = d x speed ratio pulverizers, Positive displacement blowers, Sawmill and wood-working machinery =3x3 Crushers ,mills and hoists 1.3 1.4 1.5 1.5 1.6 1.8 = 9 in. Rubber calendars , extruders and mills
  • 15. The length of the belt can be obtained from the equation: ( D d )2 L 2C 1.57( D d ) 4C ( 9 3 )2 L 2 x30 1.57( 9 3 ) 4 x30 L = 79.14 in.
  • 16. A B C A B C D E Standard Standard Standard Pitch Lengths , Designation Designation Standard Pitch Lengths , Inches Inches From table (5.3) the standard length of the … …= 79.3 in 26 27.3 …. ……. belt … 31 32.3 …. …. 107.9 … … with a designation number A78. 33 34.3 …. …. … … 35 36.3 ….. … The exact centre distance from the … 38 42 39.3 43.3 equation: … … ….. … … 46 47.3 b… b2 32(D … 2 d) … 48 49.3 C … … … 51 52.3 16 … … … 53 54.3 … … … 56.3 … … 61.3 b 4 L 6.28( D … d) 63.3 … … 65.3 … … 67.3 b 4x 79.3 6.28(9 3) … … 69.3 … b = 241.84 72.3 76.3 … … … … … 2 … 79.3 … … 241.84 … 241.84 … … … 32(9 3) 2 81.3 …. C … … … …. … 16 … … … 86.3 … … … C = 30.08 in. 91.3 … … … 97.3 … … … … … … …
  • 17. The angle of lap (arc of contact), from the equation: Angle of lap = ( D d )60 180 C Angle of lap = ( 9 3 )60 180 30.08 Angle of lap = 168.03o
  • 18. Speed Ratio Small Speed Small Speed Small Range Diameter Regular Quality Belts Ratio Diameter Ratio Diameter The capacity of one belt from the equation: Factor Range Factor Belt Cross Section Range Factor 1.341 - 1.429 Capacity of one belt = 1.000 - 1.019 FACTORS 1.020 - 1.032 1.00 1.01 1.110 - 1.142 BXS 1.05 A 1.143 - 1.178 0.91 YxS ZS 3D - 1.562 E C 1.06 1.430 1.10 1.11 de 1.563 - 1.814 1.033 - 1.055 Values of X , 1.179 - 1.222 be Used in H.P. Formula 1.02 Y and Z to 1.07 1.12 1.815 - 2.948 1.056 - 1.081 1.03 1.223 - 1.274 1.08 1.13 2.949 - and From table(5.4) 1.082 - 1.109 X 1.04 1.945 1.275 - 3.434 1.430 6.372 13.616 19.914 1.14 1.09 over X =2.684 3.801 9.830 26.899 93.899 177.74 Y Z 0.0136 0.0234 0.0416 0.0848 0.1222 Y =5.326 Z =0.0136 Premium Quality Belts S = (3.142 x3 x 1500)/12000 = 1.18 Belt Cross Section de FACTORS =A = 3 x 1.14 3.42 B C D E Values one ,belt = to684Used in91 5.326 x1.18 0.0136 x1.1833 Capacity of of X Y and Z 2. be x1.180. H.P. Formula X 2.684 4.737 8.792 3.42 18.788 24.478 Y 5.326 13.962 38.819 137.70 263.04 Capacity Z of one belt = 1.26 hp 0.0136 0.0234 0.0416 0.0848 0.1222
  • 19. Find theStandard Belt Cross Section by one Contact fromCthedrive E Arc of Contacttransmitted Standard of belt Belt Cross Section power A Type ofCdrive Length Length B Arc A B Type of equation: D on V Designation on V to VFactor to Flat V to V Factor to Flat V Power of one belt =belt … Designation Small sheaves 26 Correction 0.81 … capacity x sheaves Correction … … Small length coefficient x … … coefficient of arc of … …Factor 31 33 0.86 … contact 0.84 Correction … Correction Factor … … … … … … Coefficient of arc of contact (from130 35 180 38 0.87 0.88 1.00 … 0.75 table (5.6)= 0.974. 0.86 0.86 … Length coefficient0.95 … 0.77 (5.7)= 1.03. 170 42 160 46 0.90 0.92 0.98 (from table … 0.8 120 110 … … 0.82 0.78 … 0.82 … 0.78 … … … 0.74 … Power of one belt = 1.26 0.82 150 48 51 140 0.93 0.92 0.94 0.89 x 0.974 x901.03… 0.74 0.84 100 … 0.69 0.96 … … … … 53 0.93 = 1.264 hp … … … The required number of…belts can be obtained from the … … … equation: … … No. of belts = Design power/ power…of one belt … … … … … No. of belts = 3.6/ 1.264 … … … … = 2.85…… … … … … … Take 3 belts … … … … … … … … … … … … … … … … … …
  • 20. Thank You Dr. Salah Gasim Ahmed MET 103 20