Design of Machine Tool Gearboxes Driven by Double …ijaegt.com/wp-content/uploads/2014/12/IJAEGT-409317-pp-404-413-as… · Design of Machine Tool Gearboxes Driven by Double Speed

ISSN No: 2309-4893

International Journal of Advanced Engineering and Global Technology

I Vol-03, Issue-03, March 2015

404 www.ijaegt.com

Design of Machine Tool Gearboxes Driven by Double Speed

Electric Motors

Dr M. A. Asy Department of Production Engineering & Mech. Design

University of Menoufia Shebin El-kom, Menoufia, Egypt

[email protected]

Abstract: In Machine tool applications, the variety of

speeds are very important because of the cutting conditions.

Cutting conditions, specially cutting speed, and feed speed depend

mainly on the machined material properties and the surface finish

accuracy. The purpose of a gearbox system is to convert input

speed and torque into a different output speeds and torques. The

wide range of speeds with near steps requires a large number of

elements i.e. gears, shafts and bearings. This leads to large

gearbox size and losses. To bridge those problems double speed

electric motor as a prime mover for the gearbox is used. The 24

and 12 speeds were achieved through 12 and 6 structural

diagrams gearboxes. The structural formulae, structural

diagrams, and ray diagrams, as well as the number of teeth for all

gears, actual speeds, deviation between theoretical and actual

speeds, saving percentage ratio for the used elements and

utilization factors for these elements were considered in this

design. The proposed design achieves the required number of

speeds with small number of rotating elements, smaller

dimensions and higher overall efficiency at the least deviation

between theoretical and actual speeds. Keyword- Gearbox, Structural Diagram, Ray Diagram,

Transmission Ratio.

1- Introduction

The materials of the tool and work piece, the shape of the

tool, the type of machining process and the required quality of

the surfaces to be produced determine the optimum and most

economical speeds for the two machining movements- the

cutting and the feed movement. Single -purpose machines,

which are intended and designed for only single operation

often only need to be designed for a single cutting speed and

feed rate required for that operation. The designer of multi-

purpose machines has to provide a certain speed range which

covers the requirements of different operations, types and

shapes of work pieces and qualities of the surfaces that are to

be machined. The values of the required cutting speeds depend

upon technical (cutting properties of the tools, surface finish of

the machined surfaces) and economic considerations

(minimum tool life between-regrinds, grinding costs). The

greater variety of materials used for tools and work piece, the

wider speed range is the required for cutting speed range.

Literature survey clears that there is a great lack in the

published material about the design of multispeed machine

tool gearboxes [1-11]. For roughing operations this is often

measured by the rate of metal removal and during batch

production by the quantity of work pieces which can be

machined in a minimum time between regrinds [3][4][6].

Concept from optimization and decision theory can play an

important role in all steps of design stages. Gearboxes present

a very important group of machine members. They are utilized

in a great number of engineering fields. They which must

satisfy very rigorous technical requirements regarding

reliability, efficiency, precise manufacturing of products,

bearing, etc. Literature survey clears that there is a great lack

of the published material about the design of machine tool

multi-speed gearboxes especially those driven by double speed

electric motors. In this study a proposed design for those

gearboxes is presented. Minimization the teeth number leads to

reduction of the gears dimensions and consequently decreasing

the overall size of the gearbox. Structural formulae, structural

diagrams, and ray diagrams i.e. speed chart were constructed.

The number of teeth for all gears, deviation between

theoretical and actual speeds is estimated in more easier and

accurate than that given in reference [7]. Improvement of

gearboxes by reducing the number of rotating elements in 12,

16, 18, 24& 32 speeds gearboxes is studied for these systems

when they are driven by double speed electric motors. The

saved number and saving percentage ratio of each element in

these gearboxes are calculated. The utilization factors of the

rotating elements in the same gearboxes are also calculated.

Improvement by using double speed electric motors is

significant. Two case studies for achieving 24 and 12 speeds

through 12 and 6 structural diagrams gearboxes respectively

are considered in this paper. Improvement in utilization factor

in gears is 39, 33, 20, 28, 25 %, while it is 33, 25, 33, 25, 21%

in shafts and 33, 25, 33, 25, 18.5% in bearings for 12, 16, 18,

24 and 32 speeds gearboxes respectively.

2- Design Procedure:

The proposed design procedure to achieve gearbox of

minimum deviation between theoretical and actual speeds as

well as maximum saving in rotating elements and maximum

utilization factor of these elements can be summarize in the

following statements:

Structural Diagram

mailto:[email protected]

ISSN No: 2309-4893



405 www.ijaegt.com

The structural diagram can be achieved according to the

structural formulae. It gives information about the number of

shafts and the number of gears on each shaft. By analyzing the

gearbox speed number to its coefficients i.e. the transmission

stages, the characteristics of each stage and the number of

transmission ratio in each stage can be calculated [1] & [4].

Ray Diagram (Speed Chart) [1] & [4]

Ray diagram is the diagram from which the transmission ratio

and the rpm values of each gearbox shaft can be determined.

The speed chart construction is necessary to determine the

transmission ratio. Therefore, the following remarks must be

considered.

- The horizontal ray in the speed chart means that there is no

speed change i.e. the transmission ratio i=1

- The upward inclination ray represents speed increasing, i.e.

i>1

- The downward inclination ray means speed reduction, i.e.

i<1.

From the speed chart the minimum transmission ratio, i.e.

maximum speed reduction can be calculated.

In the following, the steps of designing a two stages gearbox

(3*2) driven by a double speed electric motor to deliver twelve

speeds were carried out.

Formula Required: [3], [4], [6], [7] and [8].

1- Rn= N1/Nz= fz-1 (1)

Where:

Rn= (N1/ Nz) i.e. Range ratio, N1= Nmax, Nz= Nmin, Z= No. of

gearbox output speeds, and f= Progression ratio.

2- f= Rn(1/z-1) (2)

3- Z= log (Rn. f/ log f (3)

Structural formula Z= P1 (x1). P2(x2) (4)

Where: P1= No. of speed steps (transmission ratios) in the first

transmission stage, and x1 its characteristics

Im= f (pm-1)xm (5)

Where: Pm= No. of speed steps in mth stage and xm= its

characteristics, Im= transmission range of mth group.

5- Transmission ratio restriction (Ig≤ 8)

The transmission range of group

6- Ig = Imax/ Imin = 2/ (1/4) =8 (6)

Where Imax ≤ 2 & Imin ≥ ¼

minimum tot shaft size (Σdmin).

7- x1≤ x2 ≤ x3 ≤ ….. ≤ x u-1 ≤ xu (7)

UT = No. of stages.

3-Improving Gearbox System by Reducing its Elements

Number

Saving a number of power transmitting elements in gearboxes

driven by double speed motors added a considerable

improvement. This improvement because of decreasing the

overall size of the system and consequently reduce its cost. In

the following we consider some gearboxes with different

numbers of speed output and calculate the number of each

main element used in each system for both cases of driving i.e.

single or double speed motors. Also, we compare between the

number of these elements for the same gearbox capacity, and

we calculate the saved number of each element, saving

percentage ratio for each element and the minimum percentage

ratio of the overall saving.

Assume:

Nsaved = Ns - Nd (8)

Nsaved = No. of saved elements.

Ns = No. of elements used in case of single speed motor

Nd = No. of elements used in case of double speed motor.

Nsaved %= (Ns - Nd)/ Ns*100 % (9)

4- Utilization Factor Analysis

Reducing the number of power transmitting elements in

gearboxes driven by double speed motors leads to the increase

in the utilization of the rest of the elements. In the following,

the utilization factors for the main elements in the gearbox i.e.

shafts, gears and bearings, for different speed capacities are

calculated.

Utilization Factor for the Main Elements of Gearboxes Driven

by Single Speed Motors:

Cu-ssh = (Z/ No. of shafts), (i)

Cu-sg = (Z/ No. of gears), (ii) (10)

Cu-sb = (Z/ No. of bearings) (iii)

Utilization Factor For the Main Elements of Gearboxes Driven

by Double Speed Motors:

Cu-dsh = (Z/ No. of shafts), (i)

Cu-dg = (Z/ No. of gears), (ii) (11)

Cu-db = (Z/ No. of bearings) (iii)

Where:

Cu-ssh and Cu-dsh = Utilization Factor for shafts if the gearbox is

driven by single or double speed motor respectively.

Cu-sg and Cu-dg = Utilization Factor for gears if the gearbox is

driven by single or double speed motor respectively.

Cu-sb and Cu-db = Utilization Factor for bearings if the gearbox

is driven by single or double speed motor respectively.

Percentage Ratio of Increasing Utilization Factor Cu for Each

Element in Different Gearboxes Speed Capacities Driven by

Double Speed Motors

For shafts: Cu - sh= [(Cu-ssh - Cudsh)/ Cussh]* 100 (i)

For gears: Cu-g= [(Cu-sg - Cudg)/ Cusg]*100 (ii) (12)

For bearings: Cu= [(Cu-sb - Cudb)/ Cusb]* 100 (iii)

Where: Cus = Utilization Factor in case of using single speed

motors,

Cud = Utilization Factor in case of using double speed motors.

ISSN No: 2309-4893



406 www.ijaegt.com

In the following through two case studies a complete solution

to calculate the minimum number of teeth for all gears in the

gearboxes, driven by double speed motors, according to the

previous procedure, and calculate the actual produced speeds.

The actual speeds are compared with theoretical speeds to

calculate the deviation which is found in the permissible limits.

Case Study (1):

For a gearbox driven by double speed electric motor to deliver

12 speeds, it is required to calculate the number of teeth for all

the gears, actual speeds carried out from the gearbox, the

deviation occurs between the actual and theoretical speeds

considering the following data: Nmin= 80 rpm, Nmax = 1000

rpm, Um (ratio between the motor speeds)= 2, f = 1.26 Nm=

Motor speeds =1440/720 rpm

Solution:

Step 1: Structural Formulae [9].

The delivered speeds (Z) which can be achieved by a gearbox

driven by a double speed electric motor may defined as:

Z= [number of motor speeds (2)] * [number of gearbox speeds

(6)]

From data: Z = 12 speed, then

12 = 2*[6]

By analyzing the number of gearbox speeds into its

coefficients i.e. transmission stages UT= 2, one can obtain the

following two arrangements:

12= 2 * [3*2] (I)

12= 2 * [2*3] (II) (13)

(Here the transmission stages i.e. UT=2 i.e. No. of coefficients)

Structural formulae for the first arrangement:

a) Z= 2 * [P1(x1)* P2(x2)] (i)

Z= 2* [3(1)* 2(3)]

b) Z= 2 * [P1(x2)* P2(x1)] (ii) (14)

Z= 2* [3(2) * 2(1)]

Here x1= 1 & x2= characteristics of x1= P1=2

Step 2: Structure Diagrams

To construct the structural diagram the following steps must be

considered:

1) Draw UT+1 vertical line at convenient distance where UT=

No. of stages (for above formula UT=2; two transmission

group).

Note: The first vertical line represent the transmission from

motor shaft and the rest represent the transmission groups

of speed box.

2) Draw array of horizontal lines equal to the number of

speed steps Z of speed intersecting the vertical lines

distance of (log f) from each other.

Step 3: The Best Structure Diagrams:

The best structure diagram should ensure the following:

1- The nmin values of the intermediate shafts are maximum

and nmax values are minimum.

2- The rays should be narrower towards the starting point.

3- The number of gear pairs between the last two shafts in

the gearbox must be minimum.

I) Transmission Ratio Restriction

Ig ≤ Imax/ Imin = 2/(1/0.25) =8 & Im= f (pm-1)xm (15)

By analysing structure diagrams of two options shown in Figs.

(1-a & 1-b) i.e. equations (14-i & 14-ii).

Apply stage restriction or transmission ratio restriction

condition:

Option 1:

Between shaft I & II

m=1, Ig= Imax/Imin = f(p1-1)x1= f2 (16)

Between shaft II & III

m=2, Ig= Imax/Imin = f(p2-1)x2 = f3 (17)

Hence the maximum transmission range for option (1) is f 3.

Option 2

Between shaft I & II

m=1, Ig= Imax/Imin = f 2 (18)

Between shaft II & III

m=2, Ig= Imax/Imin = f1 (19)

Hence the maximum transmission range for Fig. (1- b) is f2

Now, Analyze for condition, ig ≤ 8

For Figs. (1- a& 1- b), f2 < 8 = (1.26)2 ≤ 8 (condition is

satisfied)

Apply Minimum Shaft Size Criterion i.e Σdmin i.e x1<x2<x3

For Fig. (1-a) x1<x2 (satisfy)

For Fig. (1-b) x1> x2 (Not satisfy)

Then, Fig. (1- a) satisfies both conditions. Also, maximum

speed on intermediate shaft is minimum and minimum speed is

maximum.

Indications of Structural Diagram Forms

The layout of rays on the gearbox shafts in the structural

diagram has an indication for the selection of the best version.

For example option (1) has a narrower ray towards the starting

point i.e. input point. Hence option (1) is considered as the

best version of structural diagram

Step 4: Ray Diagram Construction

To construct the ray diagram of a gearbox driven by a double

speed electric motor, the following considerations must be

considered

i- The first shaft has two speeds i.e. one according to the high

speed of the motor (represented by solid circle) and the other

according to the low one (represented by an empty circle

filled with sign x).

ii- Each shaft has a number of speeds equal to double the

speeds on the same shaft if a single speed electric motor is

used. Halve of these speeds obtained from the high speed of

the motor and the other obtained from the low one.

iii- The number of high and low speeds on each shaft is equal.

ISSN No: 2309-4893



407 www.ijaegt.com

iv- The higher speeds on each shaft is attached to the high

speed of the motor, and followed by the speeds attached to the

low speed of the motor.

(a)

6 = 3(1) * 2(3).

(b)

6= 3(2) * 2(1).

Fig. 1 Structural Diagrams of the Two Structural Formulae for the First

Arrangement.

v- The change from high speeds to low speeds and vice versa

occurs after a number (k) determined by the following

relation:

Um = f k (20)

Where:

Um = the ratio between the motor speeds, f = Progression

ratio,

k = the number after which the speeds change from high to

low and vice versa.

TABLE (1)

Comparison between Speed Charts for 6 Speeds Gearbox Driven by a Double

Electric Motor with Different Values of Transmission Ratios between Second

and Third Shafts.

Item Option (1)

Fig. 2-a

Option (2)

Fig. 2-b

Gear ratio ( ray restriction) f f

Stage restriction (Rs) f f

Summation of node should be minimum 5+9+13 = 28 6+9+13=27

Gearbox Speeds and Transmission Ratios Analysis

Gearbox speeds and transmission ratios analysis, when it is

driven by a double speed electric motor, can be derived as

follow [8]:

Z = motor speeds * [gearbox speeds]

12 = 2 * [6]

12 = 2 * [3*2] (21)

No. of gears = (3+2) *2 = 10 gears (22)

Rays (transmission ratios) for the first stage steps in the speed

chart shown in Fig. (2-a)

i1 = z1/z2 = 1/f (i)

i2=z3/z4=1/fii

i3 = z5/z6fii

Assuming fj/gi the ratio between the teeth number of the driver

to driven gear for pair number j according to the transmission

ratio ij, and (fj+gj) is the summation of numerator and

denominator of that ratio.

TABLE 2

Conversion of Fraction into Form of Numerator and Denominator.

ij f fj/gj (fj+gj)

i1 = z1/z2

i2 = z3/z4

i3 = z5/z6

a b

Fig. 2 Speed Charts for 6 Speeds Gearbox Driven by a Double Speed Electric

Motor with Two Options of Transmission Ratios between Second and Third

Shafts.

The maximum power involves the minimum size gear i.e.

passes through the minimum transmission ratio between the

first and second shafts.

Hence, imin = 4/10 = fmin/g (24)

Here fmin.

where fmin./(fmin +g) = 4/(4+10) = 4/14 = 2/7 (25)

Assuming the minimum number of teeth for machine tools

gearboxes equals 17 teeth, the summation of teeth for each pair

of gears between the first and second shafts (CI) is calculated

as follows:

CI= (fmin + g )* 17/ fmin (26)

ISSN No: 2309-4893



408 www.ijaegt.com

CI= 7*17/2 = 60 teeth

The gear teeth calculations assumed a constant module for the

first stage gears:

Z1= CI * f1/(f1+g1) = 60 * 6/16 = 23 teeth (i)

Z2= CI * g1/(f1+g1) = 60 * 10/16 = 37 teeth (ii)

Z3= CI * f2/(f2+g2) = 60 * 6/15 = 20 teeth (iii)

Z4= CI * g2/(f2+g2) = 60 * 10/15 = 40 teeth (iv) (27)

Z5= CI * f3/(f3+g3) = 60 * 4/14 = 17 teeth (v)

Z6= CI * g3/(f3+g3) = 60 * 10/14 = 43 teeth (vi)

In the same arrangement for the second stage

Ray i4= z7/z8= f = (1.262) = 1.6 = 16/10 (i)

i5 = z9/z10 = f = 1/(1.264) = 0.4 = 4/10 (ii) (28)

TABLE 3


Ij 1/ f fi/gi (fi+gi)

i4= z7/z8 16/10 8/5 13

i5 = z9/z10 0.4 2/5 7

The maximum power involves the minimum size gear i.e.

passes through the minimum transmission ratio between the

second and third shafts.

imin = 2/5 = f5/g5 while fmin =2

Assuming the minimum number of teeth for machine tool

gearboxes equals 17 teeth, the summation of teeth for each pair

of gears between the second and third shafts (CII) is calculated

as follows:

CII = (fmin. +gi)* 17 / fmin = 7*17/2 = 60 teeth (29)

The gear teeth calculations assumed constant module for the

second stage gears:

Z7= CII * f4/(f4+g4) = 60*8/13 = 37 teeth (i)

Z8 = CII * g4/(f4+g4) = 60* 5/13 = 23 teeth (ii)

Z9 = CII * f5/(f5+g5) = 60*2/7 = 17 teeth (iii) (30)

Z10 = CII * g5/(f5+g5) = 60 * 5/7 = 43 teeth (iv)

Theoretical Speeds:

From machine tool design references theoretical speeds when

the progress ratio f= 1.26 can be listed as following:

1000, 800, 630, 500, 400, 315, 250, 200, 160, 125, 100, and

80 rpm

Actual Speeds:

N1 = 1000 * 23/37 * 37/23 = 1000 rpm

N2 = 1000 * 20/40 * 37/23 = 804 rpm

N3 = 1000 * 17/43 * 37/23 = 635 rpm

N4 = 500 * 23/37 * 37/23 = 500 rpm

N5 = 500 * 20 /40 * 37/23 = 402 rpm

N6 = 500 * 17/43 * 37/23 = 317 rpm (31)

N7 = 1000 * 23/37 * 17/43 = 246 rpm

N8 = 1000 * 20/40 * 17/43 = 198 rpm

N9 = 1000 * 17/43 * 17/43 = 156 rpm

N10 = 500 * 23/37 * 17/43 = 122 rpm

N11 = 500 * 20/40 * 17/43 = 99 rpm

N12 = 500 *17/ 43 * 17/43 = 78 rpm

TABLE (4)

Theoretical and Actual Speeds and Deviation Calculations. [i=1…Z]

Fig. 3 Kinematic Diagram for Six Speeds Gearbox for Single Speed Motor

and 12 Speeds for Double Speed Electric Motor.

Case Study (2)

For a gearbox driven by a double speed electric motor to

deliver 24 speeds, it is required to calculate the number of

gears teeth, actual speeds delivered from the gearbox, the

deviation occurs between the actual and theoretical speeds

considering the following data:

Nz = 5 rpm, N1 =1000 rpm, Z = 24 speed, UT=3, f1.12 &

Um = 1500/750 rpm.

Solution:

ISSN No: 2309-4893



409 www.ijaegt.com

Step (1) Transmission Ratio Stages:

The number of gearbox speeds has to be analyzed to its

coefficients:

Z = [motor speeds (2)] * [gearbox speeds (12)]

24 = 2*[12] = 2* [3*2*2]

According to the sequence of these coefficients the following

arrangements can be carried out:

= 2* [3*2*2] (I)

= 2* [2*3*2] (II) (32)

= 2* [2*2*3] (III)

Transmission speed stages (UT = 3)

For each arrangement, we can obtain six structural formulae,

each one is used to draw a corresponding structural diagram.

For 1st arrangement:

Structural formulae can be written in the following forms: 24 = 2*[3*2*2] (I)

= 2*[P1(x1)* P2(x2) *P3(x3)] = 2*[3(1)* 2(3)*2(6)] (i)

= 2*[P1(x1)* P2(x3) *P3(x2)] = 2*[3(1)* 2(6)*2(3)] (ii)

= 2*[P1(x2)* P2(x1) *P3(x3)] = 2*[3(2)* 2(1)*2(6)] (iii) (33)

= 2*[P1(x2)* P2(x3) *P3(x1)] = 2*[3(2)* 2(6)*2(1)] (iv)

= 2*[P1(x3)* P2(x1) *P3(x2)] = 2*[3(4)* 2(1)*2(2)] (v)

= 2*[P1(x3)* P2(x2) *P3(x1)] = 2*[3(4)* 2(2)*2(1)] (vi)

With the same sequence, we can obtain six structural formulae

to draw six structural diagrams for each one of the other two

arrangements. So, for the present system, we can obtain

eighteen structural formulae for eighteen structural diagrams.

Step (2) Structural Diagrams Construction:

Considering the first arrangement, one can draw the structural

diagrams, according the previous sex formulae, as shown in

Fig. (4).

(i) (ii) (iii )

2 *3(1) * 2(3) * (6)

Imax/Imin = 8= f6

2 *3(1) * 2(6)* 3)

Imax/Imin = 8= f6

2*3(2)*2(1)* 6)

Imax/Imin = 8= f6

(vi)

2 *3(4) * 2(2) 2(1)

Imax/Imin = 8= f8

(v)

2 *3(4) * 2(1) * (2)

Imax/Imin = 8= f8

(iv)

2 *3(2)* 2(6)*(1)

Imax/Imin = 8= f6 Fig. 4 Structural Diagrams According to the Structural Formulae of the First

Arrangement.

(Apply min. shaft size criterion i.e. Σdmin i.e.

x1 ≤ x2 ≤ x3

It is noticed that this condition is satisfied for Fig. (4- a) only,

and all other Figs. are not satisfied. So, Fig. (4-a) Satisfies

both conditions i.e. transmission ratio restriction & min. shaft

size criterion. Table (5)

Compare between Speed Charts Given in Figure 5

Item Option (1) Fig. 5-a Option (2) Fig. 5-b

Gear ratio ( ray

restriction) f f

Stage restriction (Rs) f f

Summation of node

should be minimum

9+14+17+25 = 65 8+12+17+25=62

So, Fig. (5- b) is the best speed chart. For that figure:

In the first stage between shaft I&II

No. of rays =3

i1 = z1/z2= 1/ f = 1/1.122 = 0.8 = 8/10 (i)

i2 = z3/z4=fii

i3 = z5/z6= f= o.63 = 63/100 (iii)

ISSN No: 2309-4893



410 www.ijaegt.com

a b

Fig. 5 Twenty Four Gearbox Speed Charts driven by Double Speed Electric

Motor.

TABLE 6


ij 1/ f fj/gj (fj+gj)

i1 = z1/z2 1/f2 =1/1.26 4/5 9

i2 = z3/z4 1/f3 = 1/1.4 7/10 17

i3 = z5/z6 1/f4 = 0.63 63/100 163

The maximum power involves the minimum size gear.

Hence, imin. = 63/100 = f/g

Here fmin. = 63 where fmin./ (fmin.+g) = 63/163

Assuming the minimum number of teeth for machine tools

gearboxes equals 17 teeth, Assuming the minimum number of

teeth for machine tool gearboxes equals 17 teeth, the

summation of teeth for each pair of gears between the first and

second shafts (CI) is calculated as follows:

CI = (fmin+g)* 17/ fmin

CI = 163* 17/63 = 44 teeth

The gear teeth calculations assuming constant module for the

first stage gears:

Z1 = 44 * 4/9 =19 teeth (i)

Z2 = 44 *5/9 = 25 teeth (ii)

Z3 = 44 * 7/17 = 18 teeth (iii)

Z4 = 44 *10/17 = 26 teeth (iv) (35)

Z5 = 44 * 63/163 = 17 teeth (v)

Z6 = 44 * 100/163 = 27 teeth (vi)

For the second stage:

i4= z7/z8 = 1/ f = 1/1.122 = 1/1.25 = 4/5 = 0.8 (36)

i5 = z9/z10 = 1/ f (37) TABLE 7

Conversion of Fraction into Form of Numerator and Denominator

ij 1/ f fj/gj (fj+gj)

i4= z7/z8 1/1.122 4/5 9

i5 = z9/z10 1/1.125 57/100 157

The maximum power involves min. size gear.

Hence, imin= 57/100 = f/g

Here fmin. = 57

Where (fmin+g) = 157

Assuming the minimum number of teeth equals 17,

CII = (fmin+g) * 17/ fmin

CII = 157*17 /57= 47 teeth


second stage gears:

Z7 = 47 * 4/9 =21 teeth (i)

Z8 = 47 *5/9 = 26 teeth (ii)

Z9 = 47 * 57/100 = 17 teeth (iii) (38)

Z10 = 47 * 57/100 = 30 teeth (iv)

For the third stage in this arrangement:

i6 = f= 1.124= 1.6 = 8/5 = 1.6 (39)

TABLE 8


Ij 1/f fj/gj (fj+gj)

i6 = z11/z12 1.57= 157/100 157/100 257

i7 = z13/z14 0.4 2/5 7

The maximum power involves min. size gear.

Hence, imin= 2/5 = f/g

Here fmin. = 2

Where (fmin+g) = 7

Assuming the minimum number of teeth equals 17, this leads

to:

CIII = (fmin+g) *17/ fmin

CIII= 7*17/2= 60 teeth


third stage gears:

Z11 = 60* 157/257 =37 teeth (i)

Z12 = 60 *100/257 = 23 teeth (ii)

Z13 = 60* 2/7 =17 teeth (iii) (40)

Z14 = 60* 5/7 =43 teeth (iv)

ISSN No: 2309-4893



411 www.ijaegt.com

Fig. 6 Kinematic Diagram of 12 Speeds Gearbox for Single Speed Motor and

24 Speeds for Double Speed Electric Motor.

Results:

The present study clears that; the solution technique has a

good suitability to calculate the minimum number of teeth for

gears used in machine tool gearboxes driven by double speed

electric motors. The output speeds carried out by these systems

have allowable deviations from the theoretical speeds. Also,

the using of double speed electric motor as a prime mover

leads to reduce the number of main elements in the gearboxes

(shafts, gears and bearings) which consequently reduce the

gearbox size, and increasing the utilization factor for the rest

elements with significant percentage ratios. The utilization

factor for each element increases as well as increasing the

system speeds. Improvement in utilization factor of rotating

elements for gearboxes driven by double speed electric motors

compared with those driven by single speed electric motor.

Improvement in utilization factor in gears is 39, 33, 20, 28, 25

%, while it is 33, 25, 33, 25, 21% in shafts and 33. 25. 33, 25,

18.5% in bearings for 12, 16, 18, 24 and 32 speeds gearboxes

respectively. The calculations of saved number of gearbox

elements (Ns) for various gearboxes driven by single or double

speed electric motors and the percentage saving ratio for each

element are listed in table (9), and plotted in Figs. (7, 8,

9&10). The utilization factor for the same elements in the

same gearboxes also listed in table (10) and plotted in Figs.

(11, 12& 13).

TABLE 9

Saved Number of Gearbox Elements (Ns) for Various Gearboxes Driven by

Single or Double Speed Motors and the Percentage Saving Ratio for each

Element.

Fig. 7 Shows the Relation between the Number of Speeds and Number of

Used and Saved Gears in Case of Using Single or Double Speed Motors.

Fig. 8 Shows the Relation between the Number of Speeds and Number of

Used and the Saved Shafts in Case of Using Single or Double Speed Motors.

ISSN No: 2309-4893



412 www.ijaegt.com

Fig. 9 Shows the Relation between the number of Speeds and Number of

Used and the Saved Bearings in Case of Using Single or Double Speed

Motors.

Fig. 10 Shows the Relation between the Number of Speeds and the Saving

Ratio for Gears, Shafts, and Bearings.

Table (10) shows Cu for each element in different gearboxes,

and rate of Cu increasing in each case. Figs. (10- 12) show the

relation between gearboxes speeds and Cu for different

elements, in the case of single and double speed motors.

TABLE 10

Utilization Factors for Gearbox Components (Cu) for Gearboxes Driven by

Single and Double Speed Motors and the Percentage Improvement for each

Element.

Fig. 11 The Relation between the Number of Speeds and Gear Utilization

Factor for Gearboxes Driven by Single and Double Speed Motors

Fig. 12 The Relation between the Number of Speeds and Shaft Utilization


Fig. 13 The Relation between the Number of Speeds and Bearing Utilization


Conclusions:

1. The proposed design of gearboxes driven by double speed

electric motors achieves the required number of speeds with

small number of rotating elements, smaller dimensions and

higher overall efficiency at the least deviation between

theoretical and actual speeds.

2. Gearboxes driven by double speed electric motors give

output speeds in the allowable limits of permissible deviation.

3. Lower cost as a result of reducing the number of the

elements.

ISSN No: 2309-4893



413 www.ijaegt.com

4. Gearboxes driven by double speed motors insure high

utilization factors reaches to 39% in gear, 33% in shafts and

24% in bearings.

5. The increase of utilization factor values is significant

when using gearboxes driven by double speed motors.

References

1. Acherkan N., Push V., Ignatyev N., Kudinov V.,

“Machine Tool Design”, Vol. 3. Mir Publishers, 1969.

2. Shigley, J. E. “Mechanical Engineering Design”, 7th

Edition, São Paulo: Artmed Publisher, 2004.

3. Mehta N. K., “Machine Tool Design and Numerical

Control”, 2nd Edition, Tata McGraw-Hill Publ. Co. Ltd., New

Delhi.

4. Joshi P. H., “Machine Tools Handbook Design and

Operation”, Tata McGraw Hill Publ. Co. Ltd., New Delhi,

2007.

5. Chernov N., Machine Tools, Mir Publishers, Moscow,

1984.

6. Koenigsberge F., “Design Principles of Metal-cutting

Machine Tools”, Pergamon Press Book. 1964.

7. Pravin S. Ghawade, Tushar B.Kathoke, Satish B.

Chawale, Ravikant V. Paropate and Ranjan K.Waghchore,

“An Appropriate Stepwise Solution for Design of Speed Box –

a Practical Approach”, Current Trends in Technology and

Science ISSN: 2279- 0535. Volume: 3, Issue: 4 (June-July)

2014.

8. Nasser A., "Machine Tool Design (Power Transmission)",

Faculty of Engineering, Menoufia University, Egypt, 1979.

9. Faisal.S. Hussain, "Optimization of Gear Reduction Unit

through Ray Diagram", International Journal of Advanced

Engineering and Global Technology Vol-1, Issue-4, November

2013.

10. Božidar Rosić, Aleksandar Marinkovic, Aleksandar

Vencl, "Optimum Design of Multispeed Gearboxes and

Modelling of Transmission Components", International

Scientific Conference Heavy Machinery-HM'05 Kraljevo, 28.

June - 3. July 2005.

11. Deb, Kalyanmoy and Jain, Sachin, “Multi- Speed gearbox

design using multi objective evolutionary algorithms”, Journal

of mechanical design” Journals of mechanical Design, Vol.

125, pp 609- 619, September 2003.

Documents

Design of Machine Tool Gearboxes Driven by Double …ijaegt.com/wp-content/uploads/2014/12/IJAEGT-409317-pp-404-413-as… · Design of Machine Tool Gearboxes Driven by Double Speed