Home | | Design of Transmission Systems | Solved Problems: Design of Gear Boxes

# Solved Problems: Design of Gear Boxes

Mechanical - Design of Transmission Systems - Design of Gear Boxes

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

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

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

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Mechanical : Design of Transmission Systems : Design of Gear Boxes : Solved Problems: Design of Gear Boxes |