1.1 Introduction to machine tool gearboxes | unit 1 design of machine tool gear box

MSD

UNIT 1

DESIGN OF MACHINE TOOL GEAR BOX

1.1 Introduction to machine tool gearboxes

A gearbox is a mechanical device utilized to increase the output torque or change the speed (RPM) of a motor. Its basic requirement is to

Increase the torque or speed

Convert single input speed to multiple output speed

Change the speed of prime mover to the speed of machine

Use small prime mover for higher torque application

1.2 Design considerations for multispeed gearbox

Optimus speed steps: The total speed range should be divided into optimum number of steps with suitable increment such as 10% to 20%.

No motor Stoppage: The motor drive should not be stopped to change the speed.

There should be only one set of gears engages to obtain the desired speed.

There should be minimum gears and other parts used to attain the required objective.

The gear control unit should be ergonomically designed and should be easily accessible and should not cause fatigue to operator.

No protruding parts: There should not be projecting or protruding parts from the gearbox; if possible it should be located inside the machine tool itself.

Direct switching: The gearbox should change to the required speed without passing through intermediate steps.

Maintainability: It should be easy to dismantle and replace the parts if required. It must be located in machine tool such that it will have easy access for maintenance.

1.3 Laws of Stepped regulation of speed

Stepped regulation: it is the regulation of speed with discrete values of spindle speeds obtained on machine tools.

Various laws are proposed to obtain the discrete speeds, those are

Geometric progression

Arithmetic progression

Harmonic progression

Geometric Progression

In this progression, the ratio of any two successive spindles is constant.

Φ = geometric progression ratio

n1,n2,n3,……,nz = spindle speed in rpm

z = total number of spindle speed steps

R = range ratio of spindle speeds

n1 = nmin

n2 = n1ϕ = nminϕ

n3 = n2ϕ = n1.ϕ.ϕ = n1.ϕ2 = nminϕ2

n4 = n3ϕ = n1.ϕ3 = nminϕ3

nz = nz-1.ϕ = n1ϕz-1 = nmin.ϕz-1

Arithmetic Progression

In this progression, the difference between any two successive spindles is constant

n1 = nmin

n2 = n1 + d = nmin + d

n3 = n2 + d = nmin + 2d

nz = nz-1 + d = nmin (z-1)d

In this progression the difference between reciprocal of any two successive spindle speeds is constant.

Out of these three laws geometric progression law has more advantages due to the following reasons

Design featured are better and economical compared to other laws

Loss of economic cutting speed in total speed range is constant

1.4 Kinematic Design of multispeed Gearbox

The steps involved are

Evaluate Minimum and Maximum spindle speed

Based on available input values of the diameters and maximum diameters of workpiece to be minimum machined corresponding to maximum and minimum cutting speeds.

Evaluate range ration of spindle speeds (R)

Range ratio is the ratio of maximum spindle speed to minimum spindle speed

Choose a particular geometric progression ratio(ϕ)

Evaluate total number of spindle speed steps(z)

Range ratio R = ϕz-1

Taking log on both sides

(z-1) log10ϕ = log10 R

(Z-1)= +1

This equation is used to find z and is rounded off to the nearest integer

Normally selected values of z are 2,3,4,6,9,12,16,18,24,17,32 & 36

Write all the spindle speed from nmin to nmax

As n1, n2,…..,nz-1,nz

Find the number of stages in the gear box (N)

Z = S1 S2 S3--------SN

S1 – No of stepped steps in stage 1

S2 - No of stepped steps in stage 2

S3 – No of stepped steps in stage 3

Sn – No of stepped steps in Nth stage

Based on the assumptions that, for minimum number of gears to require speed steps ‘z’ is essential to have all stages with same number of speed steps

s1 = s2 = ------= sN = s

Where S is number of speed steps per stage

Z = S. S. S……S(N times)

Z = sN

log10 Z = N log10 S

s could be either 2 or3

for non-integer number‘s’

z = 2N1.3N2

N1 – number of stages with two speed steps

N2 – number of stages with three speed steps

N = N1 +N2

Estimate and write all standard formulae and draw corresponding structure Diagrams

Structural formulae: It is an element of kinematic diagram and is defines as the mathematical expression giving the distribution if number of speed steps in each stage and the number of spindle speed by which two adjacent speed values of each stages are separated

z = s1(x1) s2(x2) s3(x3)……..sN(xN)

Draw the corresponding Diagrams

Select Optimum Structure diagram

After drawing all the structural diagrams, a single optimum structure diagram is selected satisfying important conditions as

sk . xk must be a factor of z

Transmission range (tf) =

Size of the shaft to be minimum

Number of gears on output shaft to be minimum

Speed ration in last stage to be maximum

In a stage z ≤ 3.

Input point must be closer to motor speed

Selection of motor speed (nm)

Motor speed is required to decide

Input shaft speed

The transmission mode from motor shaft to input shaft

if ni/p = nm (then the shaft are to be coupled directly)

if ni/p ≠ nm (then the shaft are to be connected by belt drive or gear box)

Draw optimum speed chart from optimum structure diagram

Draw kinematic(gearing) diagrams by evaluating number of teeth on various gears

1.5 Elements of Kinematic(gearing) Diagram

Structural Formulae

Structural Diagram

Ray diagram and speed charts

Structural formula

it is the mathematical expression giving the distribution of number of speed steps in each stage and the number of spindle speed steps by which two adjacent speed values of each stage are separated

n = various spindle speed

z = Total number of speed

s = Number of speed steps

N = Number of stages in gear box

x1,x2,…xN = Number of speed steps by which two adjacent speed values of particular stage are separated for first stage, second stage,…,nth stage

Total number of spindle speed steps are

z = s1 . s2 ……sN

x = 1(for difference between adjacent speed values equal to one spindle speed steps)

x = 2(for difference between adjacent speed values equal to two spindle speed steps)

z = s1(x1) . s2(x2)…….sN(xN)

x values

x = 1 (Stage 1)

x2 = s1 (stage 2)

x3 = s1 . s2 (stage 4)

xN = s1 . s2 . s3 ….. sN-1 (stage N)

Structure Diagram

The diagram describing kinematic structure of the gear box based on a particular structural formula s called as structure diagram

Steps in constructing structure diagram:

 Draw (N + 1) vertical lines, each line represents a shaft with first line as input shaft and last line as output shaft.

 Construct z number of horizontal lines at a spacing of (log10ϕ), each line representing spindle speed

__________

 Plot lines on s1 . s2 ……sN on the vertical lines 2 & 3, at the distance x1 x2 …xN respectively using structure formula

z = s1(x1) . s2(x2)…….sN(xN)

 Construct transmission line giving distribution from input to output shaft

 As

n2 = n1 . ϕ

Taking log on both sides

log10n2 = log10n1 + log10ϕ

In general

log10np = log10n(p-1) + log10ϕ

 Thus structure diagrams shows number of shafts in a gear box, number of gears on each shaft, order of changing transmission in a group and transmission range of individual group.

 Types of structure diagram

1.6 Basic rules for Optimum gear box design

The basic rules to be followed while designing the gear boxes are as follows

The transmission ratio ‘i’ in a gear box is limited by

¼ ≤ i ≤ 2

In other words

imin = ≥ ¼ and

imax = ≤ 2

For stable operation the speed ratio at any stage should not be greater than 8

≤ 8

In all stages except in the first stage

Nmax ≥ Ninput ≥ Nmin

The sum of teeth mating gears in a gear stage must be the same for module in a sliding gear

The minimum number of teeth and the smallest gear in drive should be greater than or equal to 17

The minimum difference between the number of teeth of adjacent gears must be 4

1.7 Kinematic layout

The kinematic arrangement of a multi – speed gear box is shown in figure.

From the figure, it is clear that the kinematic layout shows the arrangement of gears in a

gear box. The kinematic layout provides the following information’s required for gear box design.

The number of speeds available at the spindle, i.e., at the driven shaft.

The number of stages used to achieve the required spindle speeds.

The number of simple gear trains required to obtain the required spindle

speeds and their arrangement.

The overall working principle of the gear box.

The information required for structural formula and ray diagram.

1.8 Ray Diagram(or Speed Diagram)

The ray diagram is graphical representation of the drive arrangement in general from. In

other words, the ray diagram is a graphical representation of the structural formula, as shown

It provides the following data on the drive:

The number of stages (a stage is a set of gear trains arranged on two consecutive

shafts)

The number of speeds in each stage.

The order of kinematic arrangement of the stages.

The specific values of all the transmission ratios in the drive

The total number of speeds available at the spindle.

Procedure

In this diagram, shafts are shown by vertical equidistant and parallel lines.

The speeds are plotted vertical on a logarithmic scale with log  as a unit.

Transmission engaged at definite speeds of the driving and driven shafts are shown on the diagram by rays connecting the points on the shaft lines representing these speeds.

Figure shows the ray diagram for a 9 speed gear box, having the structural formula, z = 3 (1) . 3 (3)

Draw kinematic arrangement for 12 speed gear box

Case.1	z = 2 (1) . 2 (2) . 3(4)
Calculation of No. of shaft to draw kinetic arrangement:
	No. of shafts = No. stages + 1
		=3+1=4
Calculation of No. of gears to draw kinematic arrangement:
	Shaft 1	= p1 gears	shaft 1	= 2 gears
	Shaft 2	= p1+ p2 gears	shaft 2 = 2 + 2	= 4 gears
	Shaft 3	= p2 + p3 gears	shaft 3 = 2 + 3	= 5 gears
		.		.
		.		.
		.		.
	Shaft n = pn		shaft4	= 3 gears
		Total No. of gear		= 14 gears

Case.2	z = 2 (1) . 3 (2) . 2(6)
Calculation of No. of shaft to draw kinetic arrangement:
	No. of shafts = No. stages + 1
		=3+1=4
Calculation of No. of gears to draw kinematic arrangement:
	Shaft 1	= p1 gears	shaft 1	= 2 gears
	Shaft 2	= p1+ p2 gears	shaft 2 = 2 + 3	= 5 gears
	Shaft 3	= p2 + p3 gears	shaft 3 = 3 + 2	= 5 gears
		.		.
		.		.
		.		.
	Shaft n = pn		shaft4	= 2 gears
		Total No. of gear		= 14 gears

Case.3

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

Calculation of No. of shaft to draw kinetic arrangement:

No. of shafts = No. stages + 1

=3+1=4

Calculation of No. of gears to draw kinematic arrangement:

Shaft 1	= p1	gears	shaft 1	= 3 gears
Shaft 2	= p1+ p2 gears		shaft 2 = 3 + 2 = 5 gears
Shaft 3	= p2	+ p3 gears	shaft 3 = 2 + 2 = 4 gears
		.		.
		.		.
		.		.
Shaft n = pn			shaft4	= 2 gears
		Total No. of gear		= 14 gears

A gear box is to be designed to provide 12 output speeds ranging from 160 to 2000 r.p.m. The input speed of motor is 1600 r.p.m. Choosing a standard speed ratio, construct the speed diagram and the kinematic arrangement.

Given data: n = 12; Nmin = 160 r.p.m.; Nmax = 2000 r.p.m.; Ninput = 1600 r.p.m.

To find: Construction of the speed diagram and the kinematic arrangement.

Solution:

Selection of spindle speeds:

We know that = ϕn-1

or = ϕ12-1 or ϕ = 1.254

We can write, 1.12 x 1.12 = 1.254 …

So =1.12 satisfies the requirement. Therefore the spindle speeds from R 20 series, skipping one speed, are given by

160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600 and 2000 r.p.m.

Structural formula: For 12 speeds, the preferred structural formula

=3(1) 2(3) 2(6)

1st stage 2nd stage 3rd stage

Speed diagram (or Ray diagram):

Procedure:

Since there are 4shafts, draw 4 vertical equidistant lines to represent shafts.

Since there are 12 spindle speeds, draw 12 horizontal equidistant lines.

From the structural formula, it is clear that there are stages. In the third stage, i.e., in 2(6), 2 represents the number of speeds available in that stage and (6) represents the steps or intervals between these two speeds.

Locate the first point A on the lowest speed i.e., at 160 r.p.m. on the last shaft. After 6 steps above, locate the second point B at 630 r.p.m. These are the two output speeds.

Locate the input speed at any point on the preceding shaft (i.e., shaft 2), meeting the ratio requirements. We find, the input speed 400 r.p.m. at point C satisfies the ratio requirements.

In the second stage, there are two speeds. Lowest is at C, which is already located. Now locate point D on the 3rd shaft, above point C, in a three step interval. For these two output speeds in the second stage, the input should be from shaft 2. We find, the input speed 630 r.p.m. at point E on shaft 2 satisfies the ratio requirements.

In the first stage, there speeds. Lowest speed is at E, which is already located. Now locate points F and G on the shaft 2, above point E, in a single step interval.

Input speed can be located anywhere on shaft 1 meeting the ratio requirements. But in this problem, given that, input speed is at 1600 r.p.m.

In stage 2, we find input speed at E gives two output speeds at C and D. Similarly, input speeds at F and G, should give two output speeds. This can be achieved by drawing lines parallel to EC and ED, from points F and G, as shown in figure

Now for stage 3, to get the output speeds to all the input speeds in shaft 3, draw lines parallel to CA and CB. Thus we have located all the input and the output speeds. The completed ray diagram is shown in figure

Stage 3
and