Part 5 MT Drives and Mechanisms 1

Islamic University of Technology (IUT) الجامعة اإلسالمية للتكنولوجيا

Université Islamique de Technologie

Dr. Mohammad Ahsan Habib Assistant Professor

E-mail: [email protected] Department of Mechanical and Chemical Engineering (MCE)

Room: 405, Level 4, First Academic Building

MCE 4627 Machine Tools

Part 5 Machine Tools Drives and Mechanisms: 1

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© Dr. Mohammad Ahsan Habib

References:

Machine Tool Design N. K. Mehta

Elements of Machine Tools M. Anwarul Azim

Machine Tools Hand Book P. H. Joshi

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Machine Tool (MT) Components Fri, 19 Jul, 2013

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Machine Tool

Drive System

Main Drive

Spindle Drive (SGB)

Feed Drive (FGB) Bearing System

Structure bed or stand

Clamping and Locating System

Control System

Coolant System



Classification of MT drive elements Fri, 19 Jul, 2013

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MT Drive Elements

Mechanical

Frictional Elements

Formed Elements (Gear etc.) Electrical and

Electronic

Hydraulic

Pneumatic



Recommendations for the design of a drive system Depending on the type of the machine tool and the motions required etc. the

requirement of a drive system can vary. But the following points should be kept in view, while designing a drive system.

1) It should be possible to produce variable cutting speed to match the economic cutting conditions.

2) There should not be any marked variation in the speed, if the cutting condition suddenly changes.

3) The cutting speed should be reversible (rotated in both directions). 4) The transient time for starting, reversing or stopping the drive should be a minimum. 5) The efficiency (mechanical) should be as high as possible. 6) The lost energy should not heat up the drive systems and/or the machine tool beyond

a reasonable limit. 7) Availability of the technology and of the material of the system 8) Minimum cost of production. 9) Dimensional constraints. The machine tool has designed dimensions. The drive

system has to fit within these dimensions. 10) Aesthetic aspects.

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Determination of the maximum and the minimum cutting and feed speeds Maximum and minimum spindle speeds can be calculated using the following

expressions: 1000 1000 / / The Speed Range ratio is defined as . ;

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For straight line cutting motion (as in slotting, shaping etc.) Rn depends only upon Rv. But for rotary cutting motions (as in drilling, turning, milling), Rd must also be taken into consideration.

Cutting speed range varies widely depending on the operating conditions. For example, while machining metals, the economic cutting speed for a tool life of 240 min. may vary from 18m/min to 1800 m/min. Thus the value of Rv is 1800/18 =100. If the diameter range varies from dmax = 200 mm to dmin = 10 mm (for example, on a center lathe), Rd becomes 20. It is quite difficult and uneconomic to design a gearbox for a speed range ratio, Rn = 100x20 = 2000. Consequently the operating range of the machine will have to be limited to reduce the value of Rn within an economically feasible limit. Once this is done, the values of speed range ratio (Rn) and the extreme cutting speeds (for example, nmin and nmax) can be set.

Figure 1(a) shows the (economic) cutting speed for the minimum manufacturing cost. This economic cutting speed can be replaced , without loosing significant economy, by a range of cutting speeds Vu and Vl (Vu = upper limit and Vl = the lower limit of the range). Figure 1(b) shows the cutting speed as a function of diameter (for example, diameter of a job on a lathe, diameter of a drill in a drilling machine) for different rotational speeds.(n1, n2 … … … n5).

If the machine is to operate economically for all the diameters (d1 to d6) and if a particular speed (say. n2) cuts at surface speed, Vu at a certain diameter (d2). the next speed (n1) should also theoretically cut at surface speed, Vl at the same diameter (d2). This logic holds good for all other speeds. Thus,

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Determination of the maximum and the minimum cutting and feed speeds



1000 1000 1000 1000 1000 1000 ……………………1000 1000

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Figure 1: (a) Economic cutting speed (b) Saw diagram …………… The stepped speeds constitute a Geometric Progression (GP) series with the progression ratio of Two other methods for the distribution of the spindle speeds are: a) Arithmetic Progression (AP) b) Logarithmic Progression (LP).



Analysis of the geometric progression (GP) series

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If V is the limiting cutting speed for a given set of cutting conditions and economy and Vj is the corresponding actual cutting speed, because the speed, n corresponding to the cutting speed. V is not available (Figure 2).

The loss of cutting speed (∆ ) and the relative speed loss (SL) are given in equation: ∆ ∆ 100 % 100 %

Therefore, the maximum relative loss of cutting speed SLmax occurs when n approaches nj +1 (i.e.

the limiting cutting speed approaches Vj+1). ∆ 1 1 1 100%

Figure 2: Analysis of speed loss in a

stepped drive



Layout of speeds in GP series: In a G. P. series, . . 1 We know, ,

Example: If n1 = 10 rpm, nz = 450 rpm and z =12 45010 1.4 The calculated speeds are: 10, 14, 20, 28, 40, 56, 80, 112, 160, 224, 315, 450.

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Layout of speeds in AP series: In a A. P. series,

Example: If n1 = 10 rpm, nz = 450 rpm and z =12 1 450 1012 1 40

The calculated speeds are: 10, 50, 90, 130, 170, 210, 210, 250, 290, 330, 370, 410, 450

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Layout of speeds in LP series: We know, _ _ Kronenberg proposed that the depth of cut at each step can instead be taken proportional

to the square root of the diameter cut, when d1 is the starting diameter i.e. 2 2 2 2

…………………………………… 2 Thus … 2 … 2 … ∑

The value of k is determined by trial and error method, so that a compatible final value of dm is obtained. This leads to LP series, which is shown below from an example.

Example: n1 = 10 rpm; nz = 450 rpm; z = 12. Calculate the L.P. series, if the economic cutting speed is 20m/min.

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Solution: The biggest diameter, 636

The smallest diameter, 14

Let 1, then 2 636 2 1 636 585 2 586 2 1 586 537 193 Let 1.9, then (if you take k=2 then d12 will negative which is not possible) 2 636 2 2 636 540 2 535 2 2 535 452 3.75 For the given cutting conditions the value of d12 should be 14. By interpolating the value of k for a compatible value of d12 = 14 mm k = 1.8 Now using k=1.8 find all the 12 diameters value and then use to find n The corresponding speeds are 10, 11.65, 13.76, 16.5. 20.2, 25.4, 32.8. 44.2, 63, 98,

177 and 450 rpm.

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Figure 3: Semi-log plotting of d12 against the value of k



Kinematic advantages of GP Series 1. From a GP series having the common ratio , if terms are removed in such a manner

that only every term remains, the remaining terms form a new GP series having the common ratio . If from a series

, , , , … … … … … Four terms are removed so that every 5th term remains, the new series become

, , … … … … … 2. By multiplying the terms of a GP series by a factor , the whole series is shifted by y

terms. By multiplying the series in (1) by , the new series becomes , , , , … … … … …

Which is same as the original one, but the starting term is shifted by 3 terms to the right. 3. If the series in (1) is multiplied by a constant term C, a new GP series is obtained having

the same common ratio as the original one, and its every term is C times greater. , , , , … … … … …

, , , , … … … … … 4. The standard values of are such that they are certain root values of 2 and 10.

Consequently a certain term is double the value of another term in the series, whereas another term is 10 times of this term.

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5. If any two terms of a GP series, which contains a term with the value of 1, are multiplied by each other, the product is again a term of the same GP series.

6. Because the speeds are uniformly distributed over the saw diagram (log - log scale): The max. relative losses of (a) the cutting speed, (b) the formative capacity and (c) the

production rate are constant over the whole range of the speed layout. Average speed loss is independent of the speed. The speed loss can be calculated with respect to the mean speed bet between two steps (Figure 4)

6. Average speed loss is

The relative average speed loss is:

7. A GP series of spindle speeds can be easily obtained by arranging basic speed

changing units in series.

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Figure 4



Standard Values of Geometric Progression Ratio and Guidelines for Selecting a Proper Value

The standard values of are established from the following two main considerations: In machine tool drives two speed motors are often used; the ratio of the two speeds is generally equal to 2 (e.g., motors have rpm values of 3000 and 1500, or 1500 and 750, etc.). If the spindle rpm values constitute a geometric progression for the lower rpm of the motor, then according to the property of geometric progression, the spindle speeds should increase two times when the motor speed is switched to the higher one. This means that if there is a spindle rpm value , then after a certain number of steps there must occur a spindle rpm value 2 . . , The geometric progression (decimal) should be developed by keeping the standards of preferred numbers and preferred series in mind. The geometric progression should then satisfy the condition . . , The standard values of are obtained from the above two equation that they must simultaneously satisfy

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The standard values of , their characteristics and the specific loss of economic cutting speed are given in Table 1.



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Table 1: Standard value of in machine tools

1.06 1.12 1.26 1.41 1.58 1.78 2.00

12 6 3 2 1.5 1.2 1.0

40 20 10 20/3 5 4 20/6 11 100 3 5 12 17 22 28 33

ISA Standard 1.06 1.12 1.27 1.4 1.6 - 2

ISA designation 40 20 20 2 20 3 20 4 20 6

The advantages of preferred numbers in the standard series are: 1. The series is built in dual as well as in decimal progression. 2. Loss in economy due to unnecessary variations in the design of the producer and

the demand of the market is reduced. 3. The use of standardized main dimension of motor and shaft according to the

preferred number system.



Selection of the progression ratio, of spindle speeds, strokes and feeds After the determination of n1 and

n2 (i.e. minimum and maximum speeds) for a machine tool, it is necessary to establish z, the number of speeds (which is again related to choosing of the progression ratio, ). The relation between z, and is given by : . ⇒

This is represented in a semi-log

plot in Figure 5

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It is evident that for a given , the number of steps, z increases rapidly with the reduction of the value of . In selecting the values of and Z it is thus necessary to make a compromise between the efforts to reduce the losses in cutting speeds by increasing the number of steps (thereby making the design complicated and costly) and the effort to reduce the cost of the machine tool by keeping its construction as simple as possible.

It is a good practice to select a number of speed steps, Z having the factors 2 and 3 so that. ,where p and q are zero or any integer number. Hence, z = 2, 3, 4, 6, 8, 9, 12, 16, 18,24, 27, 32, 36 etc. The most frequently used number of steps are z = 3,4, 6, 8, 12, 18 and 24.

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Table 2: Recommended values of in machine tools Machine Tool values of

Heavy-duty machine tools and automats 1.12

Large-to medium-size general-purpose machine tools and automats 1.26

Medium-size general-purpose machine tools 1.41

Medium-to small-size general-purpose machine tools 1.58



The number of basic speed changing units for a given numbers of speed steps If the first basic speed changing unit provides number of spindle speeds (Figure 6),

and the second unit number of spindle speeds, then the number of speed steps is .

Again, if there are r stages in the mechanism . . . …………… example in Figure 6; r = 2, 3 and 3 hence 3 3 9

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Figure 6

Kinematic diagram



The total number of gears required is ………… For a transmission having minimum number of gears for a given number of r steps, it should be …………… It shows that for minimum number of gears each basic speed changing unit should have equal number of gears, which amounts to 2 . Thus,

For example, if z = 27 and r = 3 27 3 and 2 3 3 3 18 If. however, is not a whole number, then suitable adjustment should be made to get the desired value of z. For example in a three stage 12 speed gearbox. 12 2.3 which is not practical. It may be hence adjusted as follows: 3 2 2 12

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Structural Diagrams and Their Analysis to Select the Best Possible Version Suppose a speed on one shaft yields two speed values on the next shaft, i.e., The

number of speed steps of the particular transmission group is p = 2. If the transmission is through gears, the transmission ratios that provide the two new speed values must lie in the following range:

The maximum reduction of speed is limited to four times to keep the radial dimensions of the speed box within reasonable limits, while the maximum increase of speed is restricted to two times due to limitations of the pitch line velocity. The transmission range of the group is given by,

Suppose there are z speed steps , , , ……… in a particular transmission group such that …………

Since the speeds on the last shaft of the speed box must constitute a geometric progression, the following relationship must be satisfied: .

Here X is known as the characteristic of the transmission group and it denotes the number of steps of the spindle rpm geometric progression by which two adjacent rpm values of the particular group are separated.

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A transmission group may thus be conveniently denoted as , where p is the number of speed steps in the transmission group and X its characteristic. The transmission range of the group can be calculated from the expression:

where is the number of speed steps in the group and its characteristic.

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Design of Feed box Fri, 26 Jul, 2013

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Ray Diagram and Speed Chart Fri, 26 Jul, 2013

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Documents

Part 5 MT Drives and Mechanisms 1