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# Design of Gear Boxes

When the spindle speed are arranged in geometric progression then the ratio between the two adjacent speeds is known as step ratio or progression ratio.

DESIGN OF GEAR BOXES

Standard progression

When the spindle speed are arranged in geometric progression then the ratio between the two adjacent speeds is known as step ratio or progression ratio.

Step ratio Basic series : Step ratio (φ)

R5    : 1.58

R10  : 1.26

R20  : 1.12

R40  : 1.06

R80  : 1.03

Preferred basic series

Basic series          Preferred number

R5    (φ=1.6)       1.00, 1.60, 2.50, 4.00, 6.30,10.00

R10  (φ=1.26)     1.00, 1.25, 1.60, 2.00, 2.50, 3.15, 4.00, 5.00, 6.30, 8.00, 10.00

R20  (φ=1.12)     1.00, 1.06, 1.25, 1.18, 1.60, 1.25, 2.00, 2.24, 2.50, 2.80, 3.15, 3.55, 4.00, 4.50, 5.00, 5.60, 6.30, 7.10, 8.00, 9.00, 10.00

R40  (φ=1.06)     1.00, 1.06, 1.18, 1.25, 1.32, 1.18, 1.40,1.60, 1.70, 1.25,1.80, 1.90, 2.00, 2.10, 2.24, 2.36, 2.50, 2.65, 2.80, 3.00, 3.15, 3.35, 3.55, 4.00, 4.25, 4.50, 4.75, 5.00, 5.30, 5.60, 6.00, 6.30, 6.70, 7.10, 7.50, 8.00, 8.50, 9.00, 9.50, 10.00

Design of gear box

1.     Selection of Spindle speed:

Find the standard step ratio by using the relations, 2.      Structural formula

It can be selected based on the number of speed:

Number of speed Structural formula

3(1)   2(3)

6       2(1)   3(2)

2(1)   2(2) 2(4)

8       4(1)   2(4)

3(1)   3(3)

9

3(1)   2(3) 2(6)

12     2(1)   3(2) 2(6)

2(1)   2(2) 3(4)

3(1)   3(3) 2(5)

14     4(1)   2(4) 2(6)

15     3(1)   3(3) 2(6)

4(1)   2(4) 2(8)

16     2(1)   4(2) 2(8)

2(1)   2(2) 4(4)

3(1)   3(3) 2(9)

18     3(1)   2(3) 3(6)

2(1)   3(2) 3(6)

3.  Ray Diagram

The ray diagram is the graphical representation of the drive arrangement in general from. In other words, The ray diagram is the graphical representation of the structural formula.

The basic rules to be followed while designing the gear box as

ü Transmission ration (i): ü For stable operation the speed ratio at any stage should not be greater than 8.

Nmax / Nmin  ≤ 8

4.  Kinematic Layout:

The kinematic arrangement shows the arrangement of gears in a gear box.

Formula for kinematic arrangement,

n = p1(X1) . p2(X2)

5.     Calculation of number of teeth.

In each stage first pair,

Assume, driver Zmin  ≥ 17,

Assume Z = 20 (driver)

SOLVED PROBLEMS

1.     Design a 4- speed gear box for a machine. The speed vary approximately from 200 to 400 rpm. The input shaft speed is 600 rpm.

Sol:

ɸ  = ( Rn) 1/z-1

Rn= 450/200 =2.25; Z=4

ɸ =2.25 0.334 =1.31

the nearest standard value of ɸ is 1.25

the speed of shafts are 180,224,280 and 355 rpm or 224,280,355,450 rpm Ta + Tb = Tc + Td

N a / Nb = Ta / Tb  =  355/600

Nd / Nc = Tc / Td  =  450/600

Assuming minimum number of teeth on the smaller gear as 18( Ta =18)

18/ Tb  =  355/600

Tb = 30

Ta + Tb = 48 = Tc + Td

Tc / Td  =  450/600

Tc  = 20 and Td = 28

Considering the transmission between the intermediate and output shafts

Te+ Tf = Tg + Th

Nf / Ne = Te/ Tf  = 224/355

N a / Nb = Ta / Tb  = 355/355

Assuming minimum number of teeth on the smaller gear as 20( Te =20)

20/ Tf  =  224/355

Tb = 32

Te+ Tf  = 52 =  Tg + Th

Tg / Th  =  450/600

Tg = Th  = 26

2. Design a gear box for a drilling machine to give speed variation between 100 and 560 rpm in six steps. The input shaft speed is 560 rpm. The intermediate shaft is to have three speeds.

Sol:

The progression ratio,

ɸ = ( Rn) 1/z-1

Rn= 560/100 =5.6; Z=6

ɸ =5.6 0.2 =1.411

the nearest standard value of ɸ is 1.4

the speed of shafts are 100,140,200,280,400 and 560 rpm.

Ta + Tb = Tc + Td = Te+ Tf

Nb/ Na = Ta/ Tb = 280/560

Nd / Nc= Tc/ Td= 400/560

Nf / Ne = Te/ Tf  = 560/560

Assuming minimum number of teeth on the smaller gear as 20( Ta =20)

20/ Tb = 280/560

Tb = 40

Tc/ Td= 400/560

Ta + Tb = 60 = Tc + Td

Tc  = 25 and Td = 35

Te/ Tf= 560/560 and  Te + Tf = 60

Te = Tf =30

Considering the transmission between the intermediate and output shafts

Tg + Th= Ti+ Tj

Nh/ Ng = Tg/ Th = 100/280

Nj / Ni= Ti/ Tj= 280/280

Assuming minimum number of teeth on the smaller gear as 20( Tg =20) 20/ Th = 100/280

Th = 56

Ti/ Tj= 280/280

Ti + Tj = 76 = Tg + Th

Ti = Tj=38

2.     Design a nine speed gear box for a grinding machine with a minimum speed of 100 rpm and a maximum speed of 700 rpm. The motor speed is 1400 rpm. Determine the speed at which the input shaft is to be driven.

Sol:

The progression ratio,

ɸ = ( Rn) 1/z-1

Rn= 700/100 =7; Z=9

ɸ =7 0.125 =1.275

The nearest standard value of ɸ is 1.25

The speeds of shafts are 112,140,180,224,280,355,450,560 and 710 rpm

Ta + Tb = Tc + Td = Te+ Tf

Nb/ Na = Ta/ Tb = 224/560

Nd / Nc= Tc/ Td= 280/560

Nf / Ne = Te/ Tf  = 355/560

Assuming minimum number of teeth on the smaller gear as 24( Ta =24)

24/ Tb = 224/560

Tb = 60

Tc/ Td= 280/560

Ta + Tb = 84 = Tc + Td

Tc  = 28 and Td = 56

Te/ Tf= 355/560 and  Te + Tf = 84

Te = 33 and Tf =51

Considering the transmission between the intermediate and output shafts

Tg + Th= Ti+ Tj= Tk+ Tl

Nh/ Ng = Tg/ Th =112/224

Nj / Ni= Ti/ Tj= 224/224

Nl/ Nk= Tk/ Tl= 450/224

Assuming minimum number of teeth on the smaller gear as 20( Tg =20) 20/ Th = 112/224

Th = 40

Ti/ Tj= 224/224

Ti + Tj = 60 = Tg + Th

Ti = Tj=30

Tk/ Tl= 450/224

Tk+ Tj = 60

Tk=40 and Tl=20

4. Design an all geared speed gear box for a radial machine, with the following specifications:

Maximum size of the drill to be used = 50 mm

Minimum size of the drill to be used = 10mm

Maximum cutting speed = 40m/mt

Minimum cutting speed = 6 m /mt

Number of speeds = 12

Sol:

Max. Speed N max = 100 * Vmax / π Dmin = 1000*40 / π * 10= 1280 rpm = nz

Min. Speed N min = 100 * Vmin/ π Dmax = 1000*6 / π * 50= 38 rpm = n1

The progression ratio,

ɸ = ( Rn)

Rn= 1280/38 =33.68; Z=12

ɸ =33.68 1/12-1 =1.38

The nearest standard value of ɸ is 1.4

The speeds of shafts are 31.5,45,63,90,125,180,250,355,500,710,100 and 1400 rpm

Ta + Tb = Tc + Td = Te+ Tf

Nb/ Na = Ta/ Tb = 500/1400

Nd / Nc= Tc/ Td= 710/1400

Nf / Ne = Te/ Tf  = 1000/1400

Assuming minimum number of teeth on the smaller gear as18( Ta =18)

18/ Tb = 500/1400

Tb = 50

Tc/ Td= 710/1400

Ta + Tb = 68 = Tc + Td

Tc  = 23 and Td = 45

Te/ Tf= 1000/1400 and  Te + Tf = 68

Te = 28 and Tf =40

Considering the transmission between the second and third shafts

Tg + Th= Ti+ Tj

Nh/ Ng = Tg/ Th =125/500

Nj / Ni= Ti/ Tj= 355/500

Assuming minimum number of teeth on the smaller gear as 18( Tg =18) 18/ Th = 125/500

Th = 72

Ti/ Tj= 355/500

Ti + Tj = 90= Tg + Th

Ti = 37and  Tj=53

Considering the transmission between the third and output shafts

Tk + Tl= Tm+ Tn

Nl/ Nk = Tk/ Tl=31.5/125

Nn/ Nm= Tm/ Tn= 250/125

Assuming minimum number of teeth on the smaller gear as 18( Tk =18) 18/ Tl = 31.5/125

Tl = 72

Tm/ Tn= 250/125

Tm + Tn = 90= Tk+ Tl

Tm= 60 and  Tn=30

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