The Mechangers
CHAPTER
17
Design of Machine Tool Gear Box
17.1
INTRODUCTION: MACHINE TOOL SYSTEM
The various elements of a machine tool are assembled together so as to provide maximum rigidity to the system; however. this assembly has revolving. sliding and fixed or stationary components. Generally the drives of a machine tool are covered and hidden. but operated by controls which are accessible to the operator. Variouselements of a machine tool are made integral or fabricated and assembled together to make a system quite homogeneous in appearance and operation. 17.1.1
DrIves and Regulatlon of MotIon on Metal-cuttIng MachInes
Metal-cutting machines receive working motions (speed and feed) from electric motors. which usually have constant revolutions per minute. In order to fulfil different operations. it is necessary to find out various numbers of spindle revolutions as wel! as different values of feed. For these purposes. speed and feed boxes
which work by either stepped or unstepped principle of regulation are used. 17.1.2
VarIous Motions of Machine Tool System
To drive the various components of the machine tool system, we can have four methods: (i) Mechanical (iii) Hydraulic
(ii) Electrical (iv) Pneumatic.
The choice of a particular method of drive will depend upon many factors such as cost. operating speeds and feeds. power to weight ratio. rigidity. reliability. maintenance costs. intended use. sophistication. and control. 316
The Mechangers 318
TEXTBOOK Of PRODUCllON ENGINEERING
Generally, today the user of machine tools demands beuer quality, improved performance, and higher operating speeds, and this has led to a design system quite complex in nature-consisting of anyone of the drive methods stated above-s-of the machine tool system. The field of machine dynamics, particularly that of the load bearing components of a system such as base, bed, table, saddle, columns and spindle support, are important for technological gains, as these components are made up of iron castings, but their design as steel weldments offers functional and economic advantages. The functional advantage is the possibility of using higber speeds and feeds, and the economic advantage is the low power to weight ratio and thus lower cost and ease of handling. There are two types of motion in a machine tool systems: (i) the main motions, viz. cutting speed and feed, and (ii) the subsidiary motions such as fixing of workpiece, tool seuing, machine control, ere, Sometimes the primary motion is only in one axis as in the case of a broaching macbine, but more often such motions are required in two or more than two planes. Insuch situations, we prefer to have an individual drive rather than a common drive. The line shaft drive is most obsolete. as there is not much control of speed in such a system of speed regulation. The hydraulic or pneumatic speed regulation devices are used where an infinite number of speeds. within a range of maximum and minimum speed, are required; however, stepless regulation can also be achieved by a DC motor with a resistance control or mechanical drives using pressure variations, but for a very limited range. Hydraulic speed regulation is good for straight line motions, e.g. broaching, grinding, milling and shaping machines. 17.2
FUNDAMENTALS OF MECHANICAL REGULATION
The ideal regulation of speed is stepless drive; but it has ccnain limitations. due to which it cannot be incorporated in all types of machine tools. Some idea regarding this fact has already been given in previous sections. The cost of a gear box increases as the number of speeds increases, and this limitation hampers many users; therefore such machine tools are not commercially prospective. Inview of the above fact, it is customary to design (for a set of speeds) a gear box which has a optimum numbers of speeds, with minimum speed loss. Alternative approaches to this problem have posed various solutions, out of which we have to pickup the best one. Ifstepless regulation is not available, then the increment of speeds from a minimum level can be arranged in an AP. GP or HP series; it has been 'proved that tbe GP (Geometric Progression) gives minimum speed loss, and the GP series has other advantages too. As we know that Va
nn 1000
-=--=k=tan¢
do
which reveals that to get v/np a constant for the same diameter, we need to change n such that "o,tnt is equal to trd,tlOOO. lf we have a GP series such as
then this can be written as II, n¢,
nf ... nI/f'-', ... , n¢!'
If you multiply this series by ¢, you will bave II¢.IIf. nif •... , nl/l'. n¢!,+1
'
The Mechangers ,,-
-
...'"'iiI<
."".
C~'~ ...
','
,
""
Thus we get another orientation of the speeds, just shifting the first speed to the next higher one and so on, without affecting the structural change in case of a geometric series with a common rates 4J whicb is not possible for AP or HP series. Preferred numbers can be used in this series, which is not possible in other cases. At present, we shall consider only rotary motions and other types of motions later on. Let the RPM of the lathe spindle be "l; "l; "1; ... ; "4--1; "k; "«I; ... What is the law governing these numbers? For this purpose, let us examine tberadial diagram. Let us tum a sbaft which has a diameter d•. According to the theory of metal cutting, we know that the cutting speed is given as under. Ird"" rnImm . V = -1000 0111
V =
or
A
1r11
1000
d A
If n is constant, then this relationship will be a straight line which passes through the origin (see Fig. 17.1). K = tan
I/!
In our example. it is necessary to have the speed corresponding to the point A. But we must work either with the speed VA' or with the speed VB (see Fig. 17.1). It is necessary to work with the speed VB because it is near the speed VN
f
t
n= constant
II------r<----".t
" _d (b) Loss of culling speed with change of diameter
-d (a) Retation between diameter d and speed V
Fig. 17.1 Let flV be the relative loss of the cuuing speed, where 6V=
VA -VB VA
The maximum of this value win be with the diameter d.: Max
flV=
V' - V' A
H
flA
lfwe take this value as constant (as this is profitable for the exploitation of machine tools), then wesball get Max fl V = constant
V~ -VB = --"---"-
V~
V'
I-..../!.
V' A
The Mechangers
= 1 _ 1rd~ nk-l 1000 = 1_ n. - 1 1rd~ 1000 nk
n.
So the ratio
n.n.- 1 must be constant for
V to be constant.
Then Let
nk_'
=
=.!..
n._,
n. n._'9 "
So, if max ~Vis constant, then the segment A'B' must be constant too, for all of the values of rpm of the machine tool as shown in Fig. 17.1. Then the full radial diagram will be as shown in Fig. 17.2. This radial diagram is useful for determining the number of revolutions if d and V are known.
t
v A
t.V
no
~ B~HY~~~--~--~~--~~-
Ray diagram or speed spectrum for geometric series. It is the law of the geometrical progression with the common ratio .p. nk = nk_, ¢ where ¢ is a constant Therefore we have the number of speeds of a virtual gearbox as follows:
n, n2=n,.¢ n3 =n2· ¢= n,¢2
n. = n,l/f nx = n,¢'-' Let us call the ratio nm."lnmin as the range of regulation denoted by R. n n ~=R=.....:L
''min or so
or
nl
nx = n,¢'-' R = ¢'-' ¢=
x-rR
The Mechangers ,
• ~ , .'
'"_". "I"
OF
Even if the values "m", and "mi. are the same, the number of speeds will be more if the constant .p is smaller. When choosing the denominator of the geometrical progression, we take into consideration the following two factors: 1. Desirability of having a higher number of speeds, as then it is necessary to have smaller values of .p;
2. Aspiration to have a compact structure of the speed box, as then it is necessary to have higher values of .p. This gives us a lesser number of speeds. The values of and 2.
.p are standardized, and the most common ones are
1.06, 1.12, 1.25, 1.41, 1.58. 1.87.
17.3 DEVELOPMENT OF SERIES OF NUMBERS What requirements should the standard values of the constant of geometrical progression satisfy? This is an important point of debate. The denominator of geometrical progression should be chosen by taking into consideration the following factors: (a) The possibility of applying multi-speed asychronous or synchronous motors (e.g. with the number of revolutions as 3000, 1500, 750, etc.); (b) The decimal system of a series. The condition as expressed in the first factor may be expressed mathematically as follows: If the series of speeds has a member ".<), then it must have a member "y' such that
",= 211,1 11 =11 . ",£, y xl I"
or
",E,
nx3 = 'f'
where
",£,
I"
=2
'''x3
l'=2 ¢= £#2
or
(17.1)
The condition of the second factor may be expressed mathematically as follows: If the series has member then it should have a number such that ny' = lOnx'
II;.
1J '
y
= mt'l '1"
1J '
x
1011'="'£'11' y 'Y' x ¢E, = 10
1/>= E<.ho
(17.2)
Let us take logarithms of expressions. 17.1 and 17.2. then we have
E, log I/> = log 2 = 0.3 E,log
1/>= log 10 = 1.0
Dividing Eq. (17.3) by Eq. (17.4). we get
E, 3 -=~ 10
(I7.3) (17.4)
The Mechangers ,
~ ir
".
.
'"
Therefore, according to common understanding, if E2 = 40, 20, 10 or 5, EI would be 12,6,3 or 1.5. Hence, the standard values of ¢ from Eq. (17.2) would be ¢4(J= ~
= J¥2 = 1.06
ifi = 1.12 ¢to = JlflO = Vi = 1.26 tf>w = ~
¢S = ~
=
= J,rz = 1.58
However, the series of two numbers for the values of ¢was found to be insufficient; therefore. this series is to be supplemented with the following values of the series of number 2: ¢=..J2 = 1.41 ¢
= t/iO = 1.78
The advantages of standardisation are the following: (i) The decimal system of series is enough to erect a series from 10 to 100.All other numbers of this series may be obtained by multiplication or division of 10, 100, etc., and it is convenient for calculations. (ii) If we put down every second member of the series with ¢40= 1.06. then we get a series with ¢20= 1.12; if we repeat this after every third member. the series with ¢IOwill be obtained. (iii) The revolutions per minute of a synchronous motor can be fitted into this series very well. As mentioned earlier. R = Range of regulation = n... /nm;n. and if the number of speeds in the gear box of a machine tool is denoted by k, then
We have
¢,=k-Jf,i;
or
V;;
n.
I log ¢= -Iogk-I
nJ !!A.
or
log" k= 1 +-log¢'
or
k=I+logR log¢'
Let us assume that k = 2£1·3£2, as this is one of the requirements of design of a speed regulation system. The value of k could be any number, as EJ and E2 are whole numbers starting from zero. Thus, K = 2, 3. 4, 6. 8,9, 12, 16, 18, 24,27, 36, OUI of which the most widely used numbers are 3, 4, 6, 8,9, 12, 16, 11,24 and 36. Therefore, these are the steps of a stepped regulation or mechanical regulation system.
The Mechangers •
~;
.'
DESIGN Of MACtlNE TOOL ClEAR BOX 321
:: ,j"~
Example 17.1 Design a gear box having 6 speeds. i.e., 3 x 2 and 3 shafts. T Ts T 1
r+'
_2.
I"
L
II
-
X
r-r-
X-
X
T.
T.
'r•
$
2
r-r-
X
I-
flo
2i
T,
r- -
r-
Xf--
-
-
.......
III
n, ... ns
'-T To 10
•
Fig. 17.3 Design of gear box-kine malic layout diagram.
Solution 10 !his system of transmission of speeds at 6 steps, we require 10gears as shown is Fig. 17.3. We can proceed for analytical investigation of the kinematic chain of gears as follows: Series of number of revolutions: 1(
Lei us assume
-=el T2
T
.2. = e2 T, Ts . T6
-=e)
" -=e4 T,
T8
T9 -=es 1(0
Therefore "I
n) = e) . el'o
= el . e4"o
"5 = e2 . eS"O "4 = el
"6
.3 es"o
''1 = /l,¢ = ¢ = e2 ·e4/l0
and
"l
Hence Similarly. we can obtain
nl
e2=e,. if>
e) ·e4"o
_ ~
el
= e3
. Cs1to
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!!1.=¢3 = el·eS·~ 111 el·e.·~
es = e•.
= eS e.
if
The link between the number of revolutions and the denominator 4> is called the main set. The first gearing set is that which bas the power of denominator equal to the number of independent changes of speed in the main set.
11 T7
nl =--~ =el·e4~
T2 T8
~ = ¢ = es nl
es = e.¢
e4
In this case, the double block of the gears is the main group, and the treble block of gears is the first gearing group. The above analysis is called the method of investigating kinematic chains. In cases where the number of sets is more than 2, such analysis becomes difficult because of bulky calculations. Example 17.2 Show the graphical method of investigation of speed regulation by drawing ray diagrams. Solution
In general we know that nx
= n1tP'C-t
log II, = log 111+ (x - I) log 4> Therefore. from the above formula prepresenting a straight line equation. it is possible to draw line diagrams which are called ray diagrams. These are sbown in Fig. 17.4. Let us draw the ray diagram of the gear box having 6 speeds, i.e. 6 = 3 x 2. Shaft nos. 1st III
II Main group
Gearing group
n.
II Gearing group
n.
n.
n.
n,
n,
n,
n,
n,
n.
n, (a) Crossed type speed layoul Flg.17.4
III Main group
Symmetric
n, (b) Open type speed layout ray diagrams.
The Mechangers
The ray diagram represents the method of gear arrangement, and also provides the values of motor speed and its power for various kinematic pairs as well as arrangements of groups. The ray diagram does not provide the values of transmission ratios, and this diagram is drawn symmetrically. Motor
II
III
Motor
II
III
n,. "'"
9c
<, no
1:1 ~2
~
"'"
1:1
-,
<,
9c
M
"""" """""" 1:~""
1st method (a)
• •
M
"" Flg.17.5
M4
M3 M2 M
1:1
1:1
1£-..._
~:2
1:1 \"'" 1:4 \ 1~
\
,
1:1
~
n,
2nd method (b)
Asymmetricray diagrams.
Sometimes a ray diagram, besides providing this information, shows the values of the transmission ratios, and then it is drawn asymmetrically. These are shown in Figs. 17.5(a) and (b). Tbe step between the motor and the first shaft with nospeed is called the compensatory step; it provides the possibility of framing a standard series of rpm's. Ao obvious question arises: Which arrangement or diagram is better? The one in Fig. 17.5(a) or 17.5(b)? The first arrangement, shown in Fig. 17.5(a), is better, because the size of the gearbox system in this case is smaller, and faster speeds are available; thus less torque is transmitted. Hence, smaller shaft diameters and modules of gears are preferable. Example 17.3 Plot a ray diagram for a 12-speed gear box, which is a structure of 12 speeds, i.e., 12 = 2 x 3 x 2. Solution Refer to Fig. 17.6. If Po is the number of independent shaftings of tbe main group; P, is the number of independent shaftings of the 1st gearing group; P2 is tbe numberof independent shaftings of the 2nd gearing group; then the power value is given as follows: for the main grouP-Po; for the first gearing group-p,: for the second gearing group-po.P,: for the third gearing group-po'p,pz. The following limitations are accepted in machine tool design for separate transmission ratios of gear drives: In the kinematic chain of gear drive
where R is the range of regulation.
The Mechangers K OF PRODUCTION
~
1stgearing
Main group
group
2ndgearing group
Po
p,
4>
;'
l
Fig. 17.6 Ray diagram for 12-speed gearbox. In the chain of feed motion 1 - $e$2.8 2 Let us consider the limitationsof value .p by taking up anexample of 4-shaft mechanism in the kinematic chain of motions. From the structural diagram as given earlier (12 = 2·3·2) it is obvious that e = e . t/> main group '. = ') . ¢ Is: gearing group and therefore '7 = e6 . ¢ 2nd gearing group All values of t/> are suitable for the main group. For the first gearing group, R e/"3 1/1'; the maximum value of ¢ is as follows:
=
for the main motion, ¢ = ~
=
= 1.67;
for the cbain of feed motion, ¢ = ~ = 1.94; for the second gearing group, R = .-/e) = rf/'; and therefore the maximum value of ¢ is as follows: for the chain of feed motion, ¢ = ~ = 1.41; for the chain of feed motion, t/> = ~ = 1.93. It appears that displacement of gearing group will change nothing. Hence, the value of t/> is limited and depends on the number of shans and on the number of steps of revolution are obtained here. 17.4
RAY DIAGRAM FOR OVERLAPPING SPEEDS
Overlapping speeds are obtained by reducing the power by one or more degrees in one of the gearing groups. Let us take up a mechanism of 12 = 2·3·2speeds. the main group of which is the travel block of gears.
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The 'open structure' ray diagram is shown in Fig. 17.7. II
/
/'
<,
<,
/ v
IV
III
r< f<
/
t<'--
-,
k:::""'"
-, /
<
•
/ 2nd gearing group
"" n,o
K
-. K/ '2 "1
1
n'2
"
T
",
1st gearing group Main group
Fig. 17.7 'Open' structure ray diagram.
The ray diagram for overlapping speeds in 'cross structure' is shown in Fig. 17.8. 1st gearing
2nd gearing
groupmustbe
Maingroup ¢.z
groupmustbe ¢S
1,
~---4-----+--~~~ ~---4-----+~~~~
t::$~~3: ~~~+-~~~~*-~~ ~~~~n4
~----+-~~~~~~~ ~----+-+---+-~~~~ OVerlapping
Fig. 17.8 Ray diagram for overlapping speeds in 'cross' structure.
Let us decrease the power of the first gearing group by 1 unit. Thus, the overlapping of two stage speeds is obtained, and we have ¢= ~
= 1.41
¢= ~
= 1.56
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o Regulation with overlapping speeds should be applied in cases when (i) It is necessary to increase the value of !p, and (ii) It is necessary to avoid bigh speeds. 17.5 RAY DIAGRAMS FOR RETURN STEP OF SPEED A barrel of gears is incorporated in the kinematic chain. It is freely suspended and gives a stage of return speeds. Let 12 = 3·2·2 be the kinematic cbain with a stage of return. The structural diagram for this mechanism can be drawn up by anyone of the methods shown in Figures 17.9 and 17.10 below. The ovedappings are also oblained here.
r11-
-
X
X
'-+--+-+--+'x
-
X
t---
'--
X
X 1-----
'-r--
X
- x~====hxl==1 tIII
_
~
X
-
-
__
X
--v
'--
IV-----------+L+-+-----~~~-----
--
--
Fig. 17.9 Kinematic layout for versatile gear train.
The disadvantage in case of the arrangement shown in Fig. 17.IO(a) is that a return step is not available. This arrangement is quite often used. The disadvantage in case of the kinematic arrangement in Fig. 17.IO(b) is that visual control is difficult, but here a stage of return is observed, which is an advantage. II
IV
III
V
II
-,
/ // /// /" //
,/
\\ \\\ \:0<:
X/ XX X'\
0< ~ ~
<, IX'\. \ 1\\\
/~
/// //
\\ \ (a)
Flg.17.10
IV
III
/ (b)
Ray diagrams for kinematic layout of Fig. 17.9.
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17.6
KINEMATIC
ARRANGEMENT
FOR TWO OR MORE SPEEDS AT INPUT
SHAFT
While choosing a multispeed input, it is necessary to consider the following: (a) Economic aspect (b) Smaller size and simpler control. In this case, the value of ¢ is limited to 2. That is ¢m =2 The motor should be considered either the main group or as one of the groups. If treated as the main group, 411 in the 15tgearing is 41 P, in the 2nd gearing group is 41 N, ans so on. In the general case, it is qI. It means that ¢ = lfi. or qI = 2. Hence 41 = 1.06 = J~ ¢ = 1.12 = ~ ¢ = 1.26 = ifi ¢= 1.41 = J2 ¢= 2 The values ¢ = 1.58 and 41 = 1.78 are not suitable. The motor should be assumed as an elementary mechanism, and the structural diagram should be drawn for a 12-specdkinematic chain, such that 12 = 2·2·3, and as shown in Fig. 17.11,which shows the 'cross' and 'semi-open' structure layouts. II
III
/
r=:::tst gearing{ group
II
n" n"
II n,. III IIII X/I XX
I<:::::::"
n" n,. n.
,.-{ group
X'\\
"s
,., "s "s n,
'\.'\.'\.\ \\'\ '\.'\ \ n,
I-~M----:".j n" f---"I
L---_'_--~n,
Main 2ndgearing group group (a) 'Cross' structure layout
2ndgearing Main group group (b) 'Semi-open' structure layout
Fig. 17.11
17.7
DETERMINATION OF NUMBER OF TEETH ON GEARS OF STEPPED CONTROL MECHANISMS
There are three methods: (i) Method of least common multiple; (ti) Method of difference; (iii) Constructive method.
The Mechangers .'
17.7.1 Method of Least Common Multiple Let us assume that all gears have the same module. From Fig. 17.12, we have 2A m(TI+ Tz) m(TJ+ TJ = mTs where To is the total number of teeth,
=
=
To=TI +T2=TJ+T. el=2i.. Tz
ez=TJ T4
eland ez are thus known to us. If To is equal to a certain number. we can find To= TI+ Tz. Now, elTz= T,; therefore To= e,Tz+ r, = Tz(e,+ 1).
'r3
A
T. T.
Flg.17.12
Designing layout with least common multiple.
Thus, in the same way we can find To =e
_I
TZ
'
To=T,+- T, =T, (11+- ) el
el
By analogy. we can also find
Let
TJ = Toez 1 + ez a e) =b
_ TOeJ T4--1 +
c ez ="d
The Mechangers
where a, b, C and d are prime numbers. Then
a
1j =To
a
·K=Tox--·K
b( I+ ~)
(a
+
b)
where Tois the least common multiple of the values ab and cd: K is the set for obtaining a minimum number of teeth of the smallest gear. I b Tz =To--·K=Tox--·K I+~ a+b
b
Example 17.4 Let I
e=, 2 2
e2 =
then a= I
b=2
c=2
5'
d=5
=3}
a +b {c+d=7
LCMTo=2
and Let us take K = 6. Then T, = 42, T2= 84,TJ = 36, and T.= 90. If the modules of a pair of gears are different, then we have
= m(TJ+ T.)
2Ao = m(T,+ T2)
If it is necessary to calculate by one module, say m .. then
13 +T4=5..(1j
+Tz)=5..To "'2
tnz
Example 17.5 Let I
e=I 2 a=1
ez=b=2
1
tn,=2mm
2.5 c=2
a
1j =--·ToK a+b
d=5
tn2=3mm a+b=3
c+d=7
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a 12 =--·ToK
a+b c
m, 13 =----·ToK c+dlllz
m,
d T4=----·ToK c+dlllz where To= LCM of values a + b, c + d and m/"'z' In this example, the LCM is 3, 7, and 213 To= 21. I
2
T2 =-21·K=14K 3
1j =-21·K=7K 3
52
22
Let us take K
17.7.2
T4=--21·K=lOK 73
T3=--·21·K =4K 73
= 5. Then
Method of Difference
This method bas the disadvantage over the first method that it does not take into account the situation in which it is necessary that where Tmin~ 5 teeth T2 = T, +
sr
Is = T4+"'T T2 To
;5.. r2-
'-
X
,_
-
r-r-
X
X
T.
•
T._
'-
T•
_l_
T2m
2.5m
T Fig. 17.13
Kinematic layout with gear details.
Since
To = T, + T. = T2 + Ts
and
e=E. 'T,
Also
e2=
1j + t.T T4 - t.T
e=T2 2
75
The Mechangers DESIGN OF MACHINE TOOL GEAR lOX
33
since
°z =
+ 6T
1j
e, (1j + 6T)
7j _ 6T
e,
1j -
0,
·6T
T,ez -e,ez 6T= e,T, +0, ·6T 0, . 6T(1+ "2) Therefore, 1j e2-e, In the same way we can determine T2• T3• etc. The kinematic layout with gear details is shown in Fig. 17.13.The corresponding ray diagram is shown in Fig. 17.17.
or
Example 17.6 I 2
6T= 5
e2= -
5'~(1 Therefore,
1j=
+.!..)
5 2 =3·10=30 I 2 5-4 2 5 T2= T, + 6T= 35
1j 30·5 T4=-=--=75
e,
2
17.7.3 Constructive Method The shortcomings of the previous two methods are that the diameters of shafts, keyways, etc. are not taken into account. The constructive method consists of determining teeth numbers, while the above factors are also taken into consideration. Example 17.7 Design a gear box of a machine tool (turret) having 9 spindle speeds ranging from 90 to 1800 rpm. The gear box should be a compact one. Also (a) Represent the speeds graphically. (b) Draw the structural diagram. (c) Show the layout of the gear box. (d) Find Outthe numbers of teeth on various gears.
",= 90 = 190 "s = 400 n., = 850 "3
liz = 130 "4 = 280 "6 = 585 liS = 1240 "9 =
Solution
tp =
,{l8OO = ~20 .= = 1.455 V90
1800
9-
The ray diagram. gear box layout and structural diagram for Example 17.7are shown in Figures 17.14. 17.15 and 17.16 respectively.
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1800
1240
190 130 Input Intermediate Output Flg.17.14 Ray diagram for Example 17.7. 90 r
{!2
1
~
2
...
3
190'C:::::
'-
..
~
~
r;;-
fe rrs
'"90 ~57
'9
'"
30
~
'a
30
.. '"
'i2 I-
4s --
r,o
II
r-f;-j III
~ 57 90 Fig. 17.15 Gear box layout for Example 17.7. '-
~:----{" :~ 80/38
Flg.17.16
57/5
Structural diagram for Example 17.7.
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Let the motor rpm be 1400. Using belt and cone pulley arrangement from motor 10 the spindle and taking the reduction to be 7 or 5, i.e. no = 200 or no = 280 So we take no = 190 Now choose narrow diagram for compactness. For speeds n., ns and n6, i.e., 286, 402 and 585
1i
T.
= 3.0
~=~ ~=~
Let
..
Since the centre distance between the shafts carrying these two gears is to remain constant, TI
+ T. = T2+ Ts =T)+T6 = 120
T2 = 2.1
But
Ts
Similarly Now 2nd stage bas alltbe ratios
Ts = 38 T) = 67
as 3.1.
!J.... 1io
= 3.1
Since the centre distance is to remain constant, T, + TIO = T) + Til = TI2 = 120
Ts
7i1
= I (the speed
ratio)
Tg = Til = 57
or Since
T9
I
7i2 = 3.1
T9=30
7i2 = 90
Example 17.8 (a) A manufacturing concern takes up the demand of supplying turret lathes to its customers having 9 speeds powered by a 8 kW motor. The speed range is from 90 to 1500 rpm. Design a suitable gear box giving all details.
(b) After a few years, the customers demand that the working speed range may be increased to 2500 rpm, as lower speeds are rarely used. Suggest a workable alteration in the gears so as to meet this demand without changing the structure of the gear box.
Solution (a) Gear box for turret lathe: Datal. Spindle speeds: 9 2. Range: 90 to 1500 3. Capacity: 8 kW 4. Preferred Number: 1.06, 1.12, 1.26, 1.41, 1.58, 1.78, 2.
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The nearest preferred number in the list of 1.41, Outof the given listed numbers. The various speeds in descending order sball be: n, = 1500 "2 = 1062 n; =755 ". = 535 ns = 380 "6 =270 n8 = 135 ", = 191
n9=96 According to the considerations of slip and strength etc., the speed ratio allowable ar the pulleys are 5 to 7. So the primary speed is between 1440n 206 and 1440/5 288. Assuming motor rating to be 444 V at 1440 rpm and Hence we take 270 as the primary speed. For economy and compactoess, narrow ray diagram will be preferred. The ray diagram is shown in Fig. 17.17.
=
=
1500 1062 755 535
380 "0 = 270
270 191
~------r-~~~135 L-
Flg.17.17
-L
~OO
Ray diagram for Fig. 17.13 and Example 17.8(a).
Let the minimum number of teeth on the gear be 20. Then
2i. = I. 99
T2 =@1
T2 Now. as the centre distance remains constant, T) + T. = T, + T2= 60 T;
T4 = 1.41 60
T. =-=24.9=~ 2.41 Tl=~ To 2. =T)+T.=T,+T2=6O
T6
The gear arrangement is shown io Fig. 17.18.
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"rl,.
_,I-;.
----------------
-----
'~I.rr
~
.,
'.
" "
J
20
r-I---1
-
-
------
n
38
37
Flg.17.18
Gear arrangement for Example 17.8(a).
Now for speeds 1500, 1062.755
!i
=2.8
1jo
Therefore T,=~ The structural diagram is shown in Fig. t7.19. 35125
20156
8E~0~50120~ 30135 38/38 Flg.17.19 Structural diagram for Example 17.8(a).
(b) Now, if after a few years the company wants to rearrange the speed to have maximum speed of 2500 rpm approximately, the ray diagram shifts towards the high value of speeds. Then, using the same preferred number t500 x 1.41 = 2118 (i) 2118 x 1.41 = 2980 (ii) There is no restriction on minimum speeds as they are rarely used. Let us assume that the minimum number of gear teeth is 20. 1j = 3.95
72 T2 =1lQ] T,=IW T,+T2=99 =T)+T. =Tl+T6
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T)
=iTIl
T4=~ Ts =i§§]
T6 =ITII
The ray diagram is shown inFig. 17.20.
r-----.----.,,2980 f----+-=~.,j2118 f---+---r.~H'500 ~--~~~~~1002 1-------,,+**---1755
~..;;;"':~~380 ~~--+-~~~270
f---+--~191 Fig. 17.20 Ray diagram for Example 17.B(b).
Por second stage
!J... = 2.8
TIO =~
110 Tg
=1
~
=2.8
111 112
=
Ta Til
T1 = 56
=1lID
T,z=20
Ts=56
The structural diagram is shown in Fig. 17.21. 66/33
38/28
~7~0~5M0~ 73/28
20/50
Fig. 17.21 Structural diagram for Example 17.8(b).
T1 + T,o= Ta+ T'1 = t; + T'2 = 76 Ta
11,
= 2.8
Til
= 20
Ts
=~
T9 = 1
li2
, 'I
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17.8
PRACTICAL ASPECTS IN THE DESIGN OF DRIVES
The theoretical calculations of the speeds and the number of teeth have been duly provided in the preceding section; however, the actual speeds available on the machine tool generally vary by ±¢n in such a way that !fin is equal to zero. Not only this. it is limited to ± 10% (0-1). If deviation exceeds this limit, re-design should be done. After all speeds, the number of teeth are determined, the number of gears and layout of gear box. location of input. output shaft, etc. are taken up. The design calculation of a gear wheel is taken up after the forces acting on the spindle of a machine tool have been estimated. Researchers have given empirical relationships for estimating these forces, and these have been presented in previous chapters. The gear tooth must be strong enough to resist bending due to pitch line pressure, and also hard enough to resist the surface wear that takes place during rotations. In view of this, suitable gear material should be chosen. 17.9
MECHANICAL REGULATION OF DRIVES
If the kinematic chain between the input and output shafts is of mechanical type, we will get stepped regulation; however. stepless regulation can also be obtained. but it is not so commonly used. The methods of stepped regulation are as follows: (a) Belt and cone pulley drive; (b) Belt and puUey with back gear drive; (c) Gear box drives. For stepless regulation there arc various methods. as mentioned previously and as will be discussed in Chapter 18. 17.9.1
Belt and Cone Pulley Drive
The belt and cone pulley drive (see Fig. 17.22) is supposed to be the simplest and oldest mechanism for
transmitting power from one shaft to another. It has certain inherent disadvantages such as large size. small range of speeds. and that the torque transmitted by the driven shaft is proportional to the speed at which it is rotating. However. it has the advantage of being simple in design and cheap. Thus it is still found useful for various purposes.
._--FB}--,, ,,
._--t}P--Fig. 17.22 Belt and pulley drive.
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There are many types of belts as follows: (i) Flat belt; (ii) Wedge belt; (iii) Round belt (for low power). If a speed box is used there, the belt drive may be situated either in the place where the motion is transmitted to the speed box or in the place of transmitting the motion from the speed box to the spindle (see Figures 17.23 and 17.24). Two methods of transmission from speed box to spindle are shown in Figures 17.25 and 17.26. Speed box
Speed box
Ag. 17.23 Transmission from motor to gear box.
Ag.17.24
Transmission from gear box to spindle.
Wedge· belt drives are usually employed in the first variation and flat-belt drives in the second one.
~
,
,
Flat beH
T
,, ,, ,,
,, ,, ,, Fig. 17.25 Transmission from speed box to spindle-Method 1.
Speed box
Fig. 17.26 Transmission from speed box to splndle-Method 2.
A V-belted pulley is shown in Fig. 17.27.
Ag. 17.27 V-belteo pulley.
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DESIGN OF ".ACHJNE TOOL G.EAR BOX
339
The second variation is applied if the S))C(.'<1 box (situnted below) and the spindle tire separate units. In Ihis case the machine tool is more steady. and hence the possibility of vibrations is reduced. This scheme is used for machines of higher precision, It necessitates the application of Oat bells (as it is difficult to place
wedge belts here), There arc three rncrhods of regulating the stress of a ben in such drives. They are:
(i) By moving a motor (in advance of swivel): [ii} Belt-tightening pulleys; (iii) Special construction or a regulated pulley (one. of the 1\\'Oin Pig. 17,28).
Mere than 3 Sl015are usually nor provided. The bending of shafts can be cliuumucd by making the belt transmit the torque alone. This is done iirst of all for spindles by an)' One of the. following three mctbods:
(i) Unloading of the barre)
(Sc,'C
Fig. 17.28); unloading barrel
x Spindle
x Pulley Ofgeaf
Fig. 17.28 Spocial pulley or gear for power transmission. (ii) The spindle does not participate in uansmluing the force (see Fig. 17,29): and
T,'tO gears in mesn
Fig. 17.29 Use of internal gear of small diameter. (iii) Unloading of the. first shaft of the speed box.
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In de~i;igning belt and especially chain transmissions. great attention should be paid lor their location. Generally. they Can be tocared in two ways:
(i) Vertlcaltocatlon(see Fig. 17.30). (ii) Horizontal location (see fig. 17.31).
or
or
The horizontal tocaucn transmlssicn is better as the gravity force .f; changes the character belt stress. and the character of cs cilia lions arc mere favourable man in the vertical locations. It is clearly revealed in chain transmissions with large distance between centres. OUI horizontally located belts require larger dimensions or the chain or belt: therefore angular location of belts is preferable as shown in Fig. 17.32.
9
o Io I
9
+
Fig. 17.30 Vertical pulley bell transmisslon.
Fig. 17.31 Horizontal pulley bell Iransmission.
Endless belt drives
Fig. 17.32 Indined pulley belt Iransmission.
To improve the transmission system, one more shafl is added as shown in Pig. 17.29. and by this the oscilkuions of one chain are absorbedby the other,
17.9.2
Belt Pulley Drive with Back Gear
f\ typical example of such a mechanism is shown the ray diagram in Fig. 17.35.
in Fig. 17.33. wlth the structural diagram in Fig. 17.:l4 and
___n__ .""...".. • ~
•
It
•••
.. " " .. • " " "
_Q_x'
'" ----vFig.17.33
Gear drive wilh Slapped pulley.
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DESIGN OF ".ACHJNE TOOL G.EAR BOX
I-
341
I::+H+E)-
__
A
Counter shaft
II
IV
TI dId
'do
T, Fig. 17.34 Structural diagram for gear drive \vith stepped pulley.
II
III
IV
III
-c: <-II
Fig.17.35
'--v-'
t.~amgtOtsl)
~ ~
r----
r-----.
n, t
n,
1$1gCtlJiog gtOVJ)
Ray dfagram for gea( Clrive\\lith stopped plJlQy.
This type of system has the following advantages in cornparison to plain belt and pulley drive mechanisms:
Advantages (i) Smoothness of spindle rotation (H) Absence of vibration
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ENClINEER.lNG
(iii) Safety in use {iv} Compactness and greater number of speeds. Disadvantages (i) Cannot be auromarizcrt. (ii) Distribution 01'power is directly propcnicnal
to the diameter of the stepped pulley.
If T is the tension ~n('!H is the power on the spindle. then 1
HSI = TV. -
kVl
V~=---nlpln
1 k \V 60.102
V = ---2....£.nlpm , 1000
1 Hs.. = tv, ---
-(>0.102
Jrd2110
-
I
1000
H{I II
fls) = TV, -Hs = T
11{j
VI= --'--n11)Jll 1000
60.102
-
nd
k\V
kW
lTdllto
60.102.1000
17.9.3 GearBox Dtlves There arc runny IY1)CS 01'mechanical drives wlth gears as fo110\'1s: (i) Gear box with sliding gears (ii) Norton gc.ar box with idler gear (ii) Meander's gear mechanism (i'l) Gear box with clutches (v) Gear box wlrh drive key (vi} Reversing gear box. Gelfr box with sliding gctJrs f\ typical example of this type of mechanism is shown in f'i~. 17.36. The ~ear block shlft in splines On the shari with rounded cogs as shown in here. A keyed block of gears mesh alternmetv with different gears mourned on the other shaft.
T,
II~=111)-
~
T4
The gear wheels 7'1and Z" arc either made of one 8C'.trblank or arc pcrruancruly connected with each Other. There are 2. :\ aml sctdcm 4 gear wheels. Thesearc keyed together to form a Sliding gear.
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DESIGN OF ".ACHJNE TOOL G.EAR BOX
'-0-0-
343
=no-T, T,
n1
1I---.:t[}J.-ti:....-------n1 or
na
T,
n,=n,,T,
T''T'T, Fig.17.36 Mechanismvlilh gears. Advantages
These are: (i) Much higher values or power can be rmnsmiued: [ii] Such drives give constant power except for frictional losses; (iii) High efficiency;
(iv) Provide a large range of speeds: ('I) Can be automated. Disadvantages (i) Ditficuhy in opemnon; (ii) It is possible 10shirl onty when the machine is stopped. In order to 111Ukc. the. shirtings easier. cogs are made with roundings. as shown in Fig. 17.36. Norton gear box mechanism This mechanism is shown in Fig. 17.37(;) and (b). and is generally used for The various speeds Ih~1can be obtained arc shown in the same tlgurc.
T, r, e T,
,
.--.i.,-) ..
II
x
T,
T,
,
T, T
X X X
, / /
_.__ Throw-on pinion
2,
,/ (a) KinomatiC layOvl of Norton gc)Ar OOx,
low speeds. i.e.. for feed boxes.
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**11 (b) Feed box with tumbler (Norton's) gear.
Fig. 17,37 Norton gear box mechanism, The gear wheel is fixed on the shaft by means of the sliding key (see I";·ig..17.38). It nKI)' be clutched (through the idler) with
· To '1=T, ·
To
·
To
~=T e ~=T,
r, Fig. 17.38 Structural representation for speeds on second shaft.
The advantageof Ihis mechanism is cornpncrness,while the shonccming is low rigfdlty which leadsto failure to transmit large power. It is generally used for various feed boxes. The driving shaft gets either clockwise- Of counter-clockwise reversing, depending on the position the double cone-clutch (on the right Or On the left side).
or
n.~
The steppedpulley gets one of the three rpm's. Jl~
-
=
(/,
(12
· T. 1,=-
r,
· =T.
J~
-
'/~
· =To
l)
T, If the throw-out SIOI)is pushed in and if the back gear is stopped. the spindle of the machine tools gels only these3 rpms. when the srop is pulled out and the gearbox is engaged.the spindle gets 3 additional rpms.
_ eI) 21 2) ,.._--J
dz Z2 24
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DESIGN OF ".ACHJNE TOOL G.EAR BOX
345
This back-gear is called a single back-gear. There.are double and even treble back-gears. but they are seldom used.TIley also work by the preceding principle.
Meander's gear mechanism It is similar to the rumbler gear box rncchanlsm, but this rncrhnd of speed regulation is slightly bcucr than the previous one in having slightly higher speeds and being quite rigid in comparison to rumbler gear box mechanism. The arrangement is shown in ri~. 17.39.
11-
T,
T,
T,
X
xf-
-
r--
T,
T,
'i;
T,
T,
-:;-
-;=-
lT,
] [
III
2
4
3
II
n"
f--
T, T,
T,
r--
-
f-
T,
T,
5
W
II r I r
III
7
I
8
e
Fig. 17.39 Moonder's gear mechanism.
",=(~J'
"0
'r,)'
II~ = ( T~ "0
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ENClINEER.lNG
'"=(~)~ n, =(~r'1I0 "0
11uses gecmerricat progression with the denominator TI17·l• °fhi: advantages. the field of usage and the shortcomings arc the same as in Norton's mechanism. Mecbanisms with the chain of gears are used in various feed boxes. speed boxes and in other machine-
tool mechanisms. Gear box drive Ivith clutches
'lllis arraugcruent is used (sec Fig , 17.40) where speed is regulated by the usc of mechanical or rrtcuon clutches. Furthermore, the use or more than one clutch can increase the range uf speed regulation to a much higher value. A clutch is engaged Or disengaged by moving the key.
Fig. 17.40
"r
T,
7j
Tl
T~
Mull~la.(flsc Iriotion dutch.
=-/I()~ltl =-"<.1
This type is good for compactness and automatic speed regulation. but it is quite expensive and the
speed loss is quite high, Elementary mechanisms wuh clutches: Refer 10 Fig. 17.41. Instead of gear chuchcs, il is possible 10usc friction clutches which facilitatc shifting without stopping a machine tool. The mechanical efficiency in this case is 10\\'<.'rthan in the case of IIICehnnisI11Swith chain of gears. because the idle pair of the gear consumes a pal'! of the power \\,llhout giving nny useful work. These mechanisms are applied in both speed and feed
boxes. Gear box Ivith drive key A typical arr'.tngelllent is shown in Fig. 17.42. In this design. the gears to be marked arc placed in a key which slides in or out as shewn in the Pig. 17.42.
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DESIGN OF ".ACHJNE TOOL G.EAR BOX
347
Fig. 17.41 cone-type Iriclion dulch.
Fig. 17.42 Gear box vJithdrive key. T, = II~I-
II~
-
1~
These mechanisms ore used only in various feed boxes. The advantagesand shcnccmings arc the same as in Norton's and Meander's rncchanlsms. Speed reversing mechanism
Sometimes thi: direction of speed is to be reversed without wasting much time: for this purpose.an arrangement is used asshown in Figures 17.43and 17.44.
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o A
A
~
X 1-----1 X
,-
X
e
1-----1X 1-'
l-
II
II-
'B
1-"
e
I-
r
III III....:-.t-Ft-j::._",
'"
'c
c
'E {b) Using helical gesl$
Jew ciutcn e (a) Us,;ngspur gears
Fig. 17.43
e
'Y
~
!!....
(c) Usingtumblergears
Reversing mechanism.
~
V .y
X
I,
"
X A
(0)
Fig.17.44 Reversing mechanism with gears, REVIEW QUESTIONS I. Discuss the various motions of machine tool systc-ms. 2. wfuu are the fundamental principles or mechanical regulation'! 3, Distinguish between ;J my diagram and a speed diagram. 4. \Vhy are the speeds arranged in GP't 5. Describe the principle of stepped speed regulation as al)plied 10machine tools, 6. Discuss the meihods of stepped regulation.
7. Discuss the advantages and limhations of gear box drive. 8. Write snon notes On (i) Norton gear box mccbantsrn. (ii) Meander's gear meehanism.
