The d
-,sofatr has been in connec 1
f,
I
< Complete interchangeability in the ection and materials hqdling by usi ded according according tb method is especially useful Its in complctte prottxtion
!
! !
r
,i i
f a axeztain-tqrgeafpiston skirt each be O M mm.
=
0.04)ildm
1,1 .
3;".
'
P*
clearance = largest bore-smallest piston
. ..
i
.
k g by selective bsrltsmdomxraeethat I, of t b d e r of 2.5 thou more--I slnd'gradd W s. Only the inside and ens. An incidental h n t a g e of this suitable for various purposes b different gnorrfrs. W+at means, three desirable classes are dri bewing, a snug b@ lfilee bearing, a d a preloeded &in@ for lspidb. 9.7. WLVED EXAMPLES : G.mpb 1. ~ j n dthe values of allowance, hole tolerancd and s w t tolerance fw t& bollwing .&hhm.yjoni qf mated parts according,do basic h ~ i qvNe~. e
-
4P,,r'
H Q :~47.5Q mm 37.52 mm
Shaft , : 37.47 mm
37.45' mm
',i
C0 I
tnr
.
. ,
'.I
I j
f
\
.
. .
3130 w8 s a d s& hi+,trl;s13 > r = 0 . 9 rmR. = &&3Rti,@ y&$&J Shaft tolerrtrrce = High- -Jilrait = 3?*4,7 - 37.45 = 37.52,-
m?Writ bM
,
'
O
Allowance.= ,Max,@-
Example 2. A 75 mm,&@, r is 0.075 rrun and the required ailowan the bear&$ bore with the basic,b&. Solution. Refer to Fig. 9.15 It is clear with the' kit'
E u a p k 3. A medium force f& on a 75
.
~
~ ri;;*: =
meQal & d m of hale
~
~
~
It is clear with the basic hobi& .. ,. LOW limit of hole = ''
'
"
B‘%w
"
.
&a
w.m +$&!
TD.T{
<,
&ad;,
ls;.2&8 inm = 732& '+ 0.225
: b ~ f ~
(b) We know that for hob;&&
Shaft : Hi& limit =
8
-e?-.
~xupl..&&&@$&-n'-ert?~i~&an
for the shafifW W# 80 mm.
i
=gy&&~.&
~ i &
High l h i t of sm xu -3s
thai-;
~
wwdik
?+I
@H+
, a l ,
(ikik)
*
and tolerances and hence the limits ofsize ~in@~~W&f/. The diameter steps are 50 mm and t!?
&v7 -
is mfIT3w%,&ndamentaldeviation
-sincethe shaft 'j'lies below the mro
*tab m 9ld qwq>q
-
= 59.37 0.03 = 59.94 mm The fit is shown in Fig. 9.17. Jt. is a clearance fit with &Or93 @ ; & J & ~ . . ~ Qqwance. I Example 5. In a limit system the jhllowirtg limi& we sgwfled to give a clearance fit between a shaji and a h d e : t ? dod f o firnil @if+ ,
-
Determine :(a) ~ a s i = s&e (b) sh@ and h d e to era&& (3 the shajl and hole limits (4the ..[. c y : , ! I @ $ : > $ , , -: maximum and minimum clearance. " . -,-I , *!L. Solution. . 4 l - l ' ~ - ~ T l ,, ~kic'siz= e 30 mmr (a) (b) SWtolerance = 0.018 1'0.0b~jGb.013 mm . m .~l-,ij If. i:!: r',: t , Hole 'tolerand 0.020 km .'.acir $\im* J x & rtk '.1 ,L r ! ~ " * * .-4 ~ i g hr i i t of s h d = 30 - o,@$-,',,. 7 , : . -,>sf$,,'> $.
,I:,?
.7
+
+
,;.a,8
:kt
-$<$
#'I
<
I
% i t ~ \ ,
i t !
3 \-
t,.
= 29.995 m%?::r:c Low limit of shaft:, 30 - 0.018
-
5
.It
29.982 ,mtq , .'
P
:,,
.
*,!,*.
! ' 7
High - limiti'bfhole ='30 + 0.620 5(,@20~,iitm Low limit of hole = 30 mm Maximum clearance = ~ i jimit ~ .di. h h& - low limit
~ i n ' i m cl& s
=
8
.,
~maul* .,
.:,{*I
df shad
<
,I .
Low limit of hole - High limit of shah
Examp@ & A hble k d shaft &e a basic site of 25 mm, a)td are to have a clearance $it with maximum lolearanee of.O.02 mn a d a mi~limumclearance of 0.01 mm. The hole tolerance is to be 1.5 timdp the sk@ &lev=+. &itermine :limitsfor both hole and shafi (a) using a hole bast3 system (6) using a shaji bash wgtp- . Solution. ~ e f a tog Fig. 9.18. . k l :' 7 ::,2 nc.r.;~s$mrCt I
Fig. J.P8
.,
If x is the shaft Wmuice Imd y is the hole tolerance, then
6
1'.
,
A Textbook of Production Engirieeting
..
x = 0.004 mm and y = 0.006 mm
(d) wok basis system. The lower deviation is z;ero, 71.9 2t;i I!! r r ~ ~ dy i f l i t * .? %($Ziti? >f > Low limit of hole 25 mm (Basic size) v, $$?[\ "\r 3 % 7 * Ju !+fin i b A - 8 . k (,I.High limit of hole = 25 + tolerance ZWD --. = 25.006 mm m a , , , Qt I%'\?, Upper (High) limit of shaft = low limit of hole - minimum clearance t
1
~t
:I.
wmi
Low limit of shaft = 24.99 - 0.004 = 24.986 m6-- ', :lb0k&. Shaft basis system. The upper deviation is zero. . . High limit of shaft = Basic size = 25 mm -. ,.oiwi& Low limit of shaft = 25 - 0.904 yi3439fpm in) L y limit gf hole = 25 + 0.01 =, 22i0a, mmGnd ;,<.) ~i~h"limP'of hole = 25.0 1 + 0.006 = 25.01 6 ,mm Example 7. A jil is designafed as 100 G,/ q, Find the diYqg1h2&~he hole and the shafl. The diameter steps are : 80 rnm and 120 mm. Sohtioa. Geometric mean diamete{ 6 l r - lM i , 30 tim11 wOJ =9sm
-": ,
(b)
,..~.GI\
.x's
*:
t
A3
9
.
*
.
Tolerance grade, i = 0.45 (D): io . ~ I - Dmicrow , ' ; I d to ' dmil 1171'1 - 0.45 t9qi .@,$@I x 98 = 2.079 + 0.098 p 2.177 microns tl . . tolerance = 16 i = 34.832 microns ,L ,t, Now, for hole GRi, Itant 1.3 11 ww I I I S )'I: ". ' = 0.035 mm (rationahzed) 9' -- Or,Q.fbC For shaft 4, tolerance = 54.425 microns 0.054 mm (rationalized) %&7n t i t m i l d+rU - 4 4 4 b~ ~ t i f t t i W O , ~ I
1
8
(.
>.
m
mz&~a -
We can get these values of t o l ~ m c e sdimfly from Table : Appendix IE.
F.D.~~w h& G,;; + 2.5 microns (T 9.2) = 25 (98)OJ4= 0,012 nyn (rqtionalized) fliV r-: +. 5 3 . &it3 :-l>%a~'$- : US^ ,US>Q Id ~DO". mai m m (7~ 9.2) ,,fib .un,w&ihl,fir shahm \Q.a"fg Y ~ ,RUIR?U ~T ~ '=15, L) h . U i > $7i :-yli.~ ?)SD 9 h A lffUQ w\s\im'A . tmm-aq 1 (gsp11,m i c t o ~ w \' t. ' Now
>!
:XI\\
Dimension See F 9,!3. ( i ) Hole - d--- - - L.L. of H&
-
m.2:
+ F.D. 100 + 0.012 mm = 100.012 mm 81,. C C , . M.L. .of Hole = L.L. + Tolerance ' f 4 \i/m = 10&012 + 0.035 = 100.047 m &wic-#ze
=
(ii) Shaft
U.L.or H.L. Sbft r Basic Size - ED. L.L. of Shaft
-
- O,Q,72
94928 mm i H.L. - Tohw;slaFe I
= 100
=
,, ,. ,
.;I 1..
. ,.:
Example 8. A 100 mm diameter journal and bearing assembly has 3 clea~.mcefir,with the iQiqwing - specifcatiotp , ;, ,!, , , , , +' I -,.; , , . I * I Tolerance on bewin$ = 4005 a m , ,, t
sivrl 30 irniii 1awo.f AClew~nce T * 0 2 mm Determine the si;m sf the bewing ad!he jo~fz#,op~(i)Hole Basis System (ii) Shafi Basis System. Take Unilateral System of tolerances.
Sobtion. ( a ) Hole-Basis System : Refer
v
:.
*.
,t
pig. 9.13,
-
2
limit of sd&r!hg =''&ifs Size - 100 dm'' ' Highest limit iif ~ k r i n ~ ' ' & ~ of L ;Bearing ~. + Tekrancd ' LOW&
'&"
"@
Refer to Fig. 9.2,
-f
100
+ b.005 = 100.005 mm
33n814f~~J I P - ~ : ~:qqU
=
f i s h 'to nrnr; ~~twc.,
Higher limit of Joumd k: W&'limit of bearing - allowance .mar ~14KW0.002 99.99811~11 i ~ t z f 2d t d 116d? f r i l Lower limit af d m a l = Migh.limit T @ b a h~sqql; ~ -5
43i11;.t410r -
(h) Shgt$+@kJ
-7
.= 99,k)98-0.004;', ?L gqiI X
W
~
~
,,~mid.@@ - = 99.W1m Fig- 9.13 Upper limjt sf.hucnal = Basic Size = 100 mm Lower limti&
rSI#s@,fiver .w
awn 2IQ.W
..: :,.
-
-
Z
= l~-p.oo4 '
:1
r ~ i ~&Z~ t &kwer a limit of Bearing i.A
Example 9. In an assembly of two matingpof.IOQ,mmh a ' size, the48 is Inter$erence and the inter$erence variesfrom 0.05 mm to 0.12 ,mm. . q e tolerance on the two mating parts is equal. Determine the sizes of the two mating part5 oh (a) Hole Basis Systein (b) Shaft Basis System. m Wf = x i . = 4 7 : 13 i Solution. 52rtsrylrfi I t r ~ r ; i [1. :vr, l (a) Hble Basis System : Refer Pig. 9. tsF :. < I .: qdlli
'
.jti1)4
nl.
:q?j
T
ftdz $m,Lower limit of h o b f i ~M j c siza-=JOO mrmn ,,: Now Refer to Fig. 9.5 (a), , .q, 6 , ; ,:,J i Maximum interference will be when the hole is at its lower limit and t e shaft.is at its upper
-
limit,
- - I
I
ii
:.
Upper limit of shaft
=
Lower limit of hole + Maximum interference
. . ' ::, Now it is clear that, Maximum interference - Minimum interference = Up-r limit of shaft - Lower limit of hole - (Lower limit of shaft - Uppet limit of bole3 3 m, = (Upper limit of $haft Lower limit of shaft) + (Upper limit of hale - Lower limit of hole ,,t~,d71, \, ' ,, :\-.,\I traitk\ b\r?f.rt Tolerance on shaft + Tofirance on hole "{' "' " ,. =,,.. , ,.. . ... ,, ... / \ I $ 'J1112 : = 2T .;*1:$2ff$*l=,>-.
c
t,
+
L~
c 7
-
""'"'
;
2T = 0.12 - 0.05 = 0.07 q9# . ,, . % ,..,.%a-,b)r Tolerance ~p qbft = Tolerance on hole = 0.933 nun Hi* l j i o f hole. r , Lower limit + Toleranqt. ,,I I,., r,;m ~ ~ J ooi I I = i 0 1 ~ 100.035 mm Lower limit of shaft = Upper limit - Tolerance
:.
,..
I
'-= 100.QSS mm. (b) Shaft Basis System : Upper limitdshaft = Lower 'limit of shaft *
i
NQCY
.
B d ~ c i z e=.10Ornia;rnr! .: Upper [limit - Tolerance 100 - 0.035 = 99.965 mm Lower limit of hole Upper limit of lhxinilrrm ~ " ' awl I'w; = 100 - 0.12 = w:$8 hutr. : Iyq'J ..- ., :. Upper liniit of hole = Lo& limit'%&kbhke i ' l s t i r / . ~ * = &.838'+ 0.035 = 99.915 mm. , a k ,j:je *'? Example 10. For a number of mterchangeable mathgpms (holes andshqfii), the average allowance is 0.04 ~ l w nand t k allowance must nd exceed 0.012 mmthe merage value. lRe bark*she 168.rrmr.l @ i e ~ &o#;hdc + $ ' ~ - t b b & d~ l f A shap. s Lktermine the sizes of holes and Skofts using Hole h i s - ~ 4 e m and Unilateral system of tolerances. Solution. Refer to Fig. 9.2 L.OO.mt : Maximum ahitmce --Mhu'mtarn allomot * ' l b h w c on shaft + Tolerance on hole
-
u'-
1
1
C'
1
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.n't'i>w. ,t e.
,'
.
,.
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,
a I .
.
tr..,
f Now
?'amce
&f\
=t
6.0011 m.4 j5r:\,,~': I : N ~i t . ..
': ),I .!.: ,.v:<-. ,. T o b r a q o n m l e = P.016, .~aic...s Lower limit of Hole = Basic Size = 100 mm Upper limit of Hole = Lower limit + tolerance
. t n
'
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3u.b
v.,
7"..p
5 . ~.i$ ,*.v.yi . Q,.;.-!T
;f\\\.\i:-l?,,;;
~hfSns.r.3 3,,\\
i~+.tl,~\.,
'c,~,,:;
.,;!n,:..,5;~ ,\c.,\;.
Min. allawake = Lowc2r%mii ofhde -'Upp&rh i t of shaft.
C :'
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.
'
~~~~ .!hiit B l m @ mrn
mrw,mJ~'~ &MU
nil
-
GG
A
w ~ S I E O ~ J R * x: kq A
99.972 0.008 lil srij ':c IWb3 & .,a&!$ Zl Jfltr8 99.964 mmi Wi fl R* slslfi~01 p i fk;c;i b9ntrfi r+ .(fS msr .nmrsll m b #KT .ntm (. . ., , wtt Lm riod &'*L enoivrvmib ' f{ 1$3 -, $-~~IP~QPLEMS s l.nnosarrR m r. bhr 1. Define "Interchangeability" and dimmsiirs IiapbFtancdQ?2. Define tolerance. .&I ,at ? ~ [ ! F I J3ris I L.T~ ; J J ~ . , L ~ ~ I i t , 1 l1~f3g 9% ibl if'~nwxsible€aobtain .c#v exmd-.~w~ ~ l a ~ OoJl t t .c.:.q r d C i nA .IS Ea. i s q m r b ) . the w % 3 . wl&'& edt&cs bf fit.& 10 ~ f l " t & ~ t o ~ ! l : ' . , .r. tbicl .-. .-.....vst: b~a~~mtnd 5. Define : aflowandi, clearanck htederenci: '-;"' - f # , !; 17, a,# ?,q. 6.' What is zero line? 7. Define : upper i&vh&t,%.&%~r &&e~#sv31i;f;r;!!4~ 1 t m : t "!fl 8. kxpran &jj'&;ibitoI* ' ; B i k ~ f i i ' 3 U .ST qql 3mmm 3 m 2-/;3 .LT 9. What is meant by H&,d&~*pl'&&q4'.~L 10. Explain and compare "Hole basis sy&~'sda'4biKdas ( ' # t r d ~ * ~ g & % ~ ~ ~ ~ ~. llker&.dto ~ . p i ~have imum toral c-.l FS #iWW$k ~W?&hi(tgtlili?Iautince &a69 Irihrm. Wermine the spccificaticrns of the parts k)i@ @Q&
[email protected] (&).for S r a t i W average inmcbngesBil#y. : 611 ti , > L a$;> 12. How will you write the Rt : .&4rq&$\* basic size is 40 rHm? & * d - . m *I* m amiwii+ arl f 04 i* ule &sieuria 0 mm H - h ~ & t h et o l k c ; grade I R ULO+ ti# ( ant f.O_ 0 : 1 s l 3 1 ~ 1 ~ lthdr b ) 4 ilf =
I
3
a
ah
_ I
!I
I
&
uun~ad
r
.
iti%iIk-
2
IS
&
'qg
P;&!F&,.&
t&ga'&
($13 the above fig - -.
,
,
., *',
-: Msiy
t@$qns&
.
(b :&A
9:
-7%
*@1&!4$
~
~
I
e@jp thq .liWofdze for hole
tJ.wl D
P
T
?-
7.1
15 A gear ring of85 mm diameter bore is died bn
$&A,,
ti1 s ~ u w @ 3
A
qy r v &;&--;1w ,a b and f ~ ; ~ n i b e ~'~uaIi& e to h c e g d i ~ 25i r $
-
.+. .,, , .itqa-l baabubiresulting ih'a Ha6 h ~ . l e &? I
tolerances and hence the limits of size for tht h b 'kt& he g& h e . S@$@ttf;t type of fit. 8fWj.&is 0.mm. The diamcter steps are 80 mm and 100 mm. 'Ihc 16. Calculate the fbndamental deviations and tdgwn6~1~ tlW: ,$Qr shaft and hole p i r - + & n a t t as 60 mm HimC The tol-ce unit*is g j ~ p l
#am
a
.
:I,
i = 0.45 + 0.00i D mi .scsn&~ 3!{~ The diameter steps are 59 m,@ BB mw! Tka: @I ldg~x@&m, for. m shaft is = + (IT7 - IT6). For quality 7, the multiplier is 16'and that for quality 6 it is 10. 17. Calculate the tolerances, limits and ~lltowuxcsfor a 25 'kid hole pair wigmted u H,ld,. The diameter steps arc 24 mrn snd 30 mm. The multipl(d.kt quality 8 .is 25. The FD for 'd' shaft is - 1 6 P U microns Name the type of fi& jh.9 18. Determine the types of fits p~odUCdaby the following maEg &holes anh
-4%
,,/
&
A Textbook &~ProductionEngineering 19. A fit is designated as : 60 mm H,2 h,; D b t e n r r ~ h hi mum' &?&anc!t and m9rnum clearance . of the tit. rn0.d = lo. A turned shak is to rotate in a reamed h * m a 8 ~ r d fit is HJc,. ~eterhinethe actual dimensions of the hole and the shaft. The ba#c size is 60 mm. The diameter steps are SO mm 8nd 80 mm. The fundamental deMIdA.w]t&7.sh& c is, - (95 a)mi )wts :fhlFsl$~y*~ml.* suiifl .t For quality 7, the muitiplicr is 16. 21. h idler gear is to rotate over a mm s k i , .The :-=fit 'LH#p Dea?rmik(t ttk actual b bPnt Q$ the i#k.T?w &mew, stcpg m 39: and 40 mm. The dimensions of the shfdt PIlQ t hndmental deviation for the hole H is ~ r *,?tQ3f;?r o $9 sh& 5. iq : - (95 + 0.8 D)microns. .,:I
s$@:@
' "'"'
C
I
The multipli~, , f. quality ~ 7 -is 16 a@ that for q q r s 22. Define and compare : Wdoq , w b .l y , st&@,~~1l . . I . 23. Give some common applicatians of J q i v e . . '
-
r
>
.
g .is12$. ,.
-2 ,
h
b~6r.W.d
a
"..&&:, &+
,B;II$W Q . ~ mn, ~ l
-*
~marmt9m'
.$
<'
e*~J1!b
* I
2 4 what y,*>*n;,*,pf:yy. Y*',
8
0
;*
*@y "twmdI&l ida#e @& :a ~ h & f f . ~ , m 76 mt'Itg,_~$w !:.* (6)Tha raigtive mor in the dimension is greater in &aft A. ,@lirfas#ti6rt.mtiii
-16
rr. .-
i i ~ fliwl
"1) (;,fat&&& &&&&js~" is g,+& (d)The rclatbe error in the dimension is greater in s W B. , . .(
?tmw ,
*w
$+,&
!,::
,,
.$I
'
:'
(Am.:a)
d-
-a) Tmsitien FI~J(@ lW&m~eSFit (c) c l e m e c fit (&>Noat (GATE 1993) ( h a : e) eL -y7. h e fit 0" ( %1'&&$l1&ab) ,< . : 2 is &ifid . as H, - s, The type of fit is
7
!.
(a)Clearance fit
(c) push fit (transition fif)
*a n *?da)
~n thq r p .k i i , o a
diq-7
&
-
L-1 >
,
&:,";;;
wning fit (yliding fit) (d)force fit [interference (G#TF 1%) A=- :dl , d 9% , , . I ~bwdw+g 21~
%
a l J o w m is aqual t ~ h i l a t s F a l , C w . m8.31~) akvanec isto UftiMWl Weran&.
b
!
*t. h
i
1
:I
d
PI
!
'
R
-0.001 4
,
.-
um
r t q /I
&f
' ro2
s~qqrsr& 4nr *L'.-
4.W t r m ~ m ,r
,\
xu-
% n ~z t i d
sion specified by 998) [Ans. : el
-9 *.@TI h@fmmb
TTi j +
a
.(kt3
YIO *ltr.jffia:~
.f l
WIJS~:~
+0,001 & vq@ ii 4.002 m% Determine.: T o 1 6 on Journal, Toleran and type of fit. (Ans : 0.001 m%&003 m.-0.001 n w Bepriflg : SO
.. .. .i t
d U 3 .&I
the tolerance. .a&% 39. A journal md b a i n g aSsOLrrbly has h e i l & # b g sizes%' 1'
k' ,bT
-,
I
Yc) . . efiowa~ceiJ Mebed& eFtd&iwe.
brsfi
I
.-
**
,Maximum clearance t WC
W mm, TmstioPL:fif ?B-
$5)
30. In an assembly of journal and bearing, the basis size is 55 mm. The lower deviation, and upper deviation for bearing is 0 micron and 4 microns respectively. the corresponding values h r journal are - 3 and -7 microns. Detmine the sizes of journal and bearing and the type of fit. H.L. = 54.997 mm (Aas : Journal : L.L. = 54.993 mm ' Bearing H-LA= "*Oo4 mm , clearance fit). L.L. = 55.000 mm -@in the d i h c e betww t o I - n q , d aljomnce~ Why do m~rm&n~pro&k produce $at& d t b such a wide range af tolerances? 33. Thejoumal and M n gwembljr has a basii size of 206 m.For the bearing : ED.; (Here, L.D.) = 0 micron, and Toterance = 46 micron. o f j d For the journal, the values are - 820 and 115 microns respectively. F a tfiG -4 and bearing, assuming unilateral system of tolerances. Alp, +tpmining the allowance and type of '" ,. ' '; ''tit. Which spkms of fithas,kn adopted?
-
-.,
.J15#" .: -
<.
. 4. ijl$
I'
I
h
b:
'J'fl5.? & ' L ~ P , . ~ f c 3
Minimum size of the hole = 3 & h '
<.I
<.
9
.--a gihwrtC . = 50 mm. Minimum size sfthe shaft = 50 - 0.01 1 = 49.989 mm +i@@q 01 gfribo 23 Maximum size of the hole = 50 - 0 . q = 49.974 w f:, --t t,,., Minimum size of the hole = 50 - 0.065 = 49.935 mm = Maximum size of hole Minimm size of Be'shaft Maximum Clearance 49.9% -W.989 0.01?! nun = M4nbum &e of hole W m u m siza of the shaa MinimumClm = 49.935 50~000= 0.065 llun Since bdfi the maximum andmhiiarrpnc-l an n w v e , the fit obtrdned will be "ineaEemnce fit" :, ,%,k3 36(! ;,,lii . 1 t P % 4 38, 36. Determine the type of fit : . + 0.05 9-3' ., Hale size : + 0'05 mm ; S ~ : 20 B 20 - 0.05 - 0.05 SI'[ ' : @huv size ofthehole = 20.05 * , ; , , Minimw size oPtheQole = 19.95 pun . ~ \ c > ; q > < 13; Id,(* ..J& ,-rJy ?I{: : ..j&#( b L : a i Maximum Qze of the shaft = 20.05 mm r :31 qd t t l d ~ki4d.;.:>/~*: ~ ~ *>I: fii .#ficil ,:h = 19.* * . ' k ; ( ~ ~Tm ~ ' d ~ ~ ~ ~ d i a t o f k d e - M i n & n r m ~ . o f t b M * Z @ 1935~0,101~~n .@5efReld- wUIm ~ a = 19.96 -f~.055-0.10 ~ m positive 9nd the Minimum Clearance is negative, the tit obtainJ
W. Maximum @6f
t.k shaQ
"
(,,
-- --
-
-
~6
sw
-
--
,
. . I
,
-
,
*---
A
*
f
1
i
X i
Gauges and Gauge Design
407
(iii) Combined Bore / Face Gauge: The position and parallelism of a bore in relation to a datum face can be checked by means of a combined bore/face gauge, Fig. 10.38. The pin which locates in the bore is in effect, 'Go' plug gauge, and the steps ground on the other pin are the 'Go' and 'Not Go' limits for the datum face to hole axis dimension. The length of the plug gauge needs to be sufficient to enable the length of the 'Go' step on the pin to check for parallelism. The tolerance on the hole must be less than the tolerance on the dimension to the face for the gauge to operate satisfactorily.
Fig, 10.38. A combined bore/face gauge. 10.11. SOLVED
EXAMPLES
Example I. A 25 mm H8--j7 fit is to be checked. The limits oj size Jar H8 hole are: High limit 25.033 mm, low limit 25.000 mm. The limits of size Jar 17 shafts are: High limit 24.980 mm. low limit 24.959 mm. Taking gauge maker's tolerance to be 10% oj the work tolerance. design plug gauge and gap gauge to check the fit.
= H.L - L.L. = 25.033 - 25.000 = 0.033 mm .. Gauge makers tolerance for plug gauge = 0.1 x 0.033 mm- = 0.0033 mm = 0.003 mm (rationalised) Gauge makers tolerance for gap gauge = 0.0021 mm = 0.002 mm (rationalised)
Solution.
Tolerance for hole
As the work tolerances are less than 0.09 mm, wear allowance may not be provided. (i) Plug Gauge Basic size of 'Go' plug gauge .'. In unilateral system,
= L.L. of the hole (MMC) = 25.000 mm
Dimensions of 'Go' plug gauge
=
That is, High limit of 'Go' plug gauge
=
=
25.000 + 0.003 25.003 mm 25.000 mm
=
25.033 mm
=
Low limit of 'Go' plug gauge Now, Basic size of 'Not Go' plug gauge
+ 0.003 25.00 mm - 0.000
+ 0.000 •• Dimensions of 'Not Go' plug gauge
= 25.033 mm
- 0.003 (Fig. 10.40 shows a sketch of combined 'Go' and 'Not Go' plug gauge.)
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Fig. 10.40. Plug Gauge (combined type)
.'?.%? ' a h
(ii)lCap Gauge
'Go' side = H.L.of shaft (MMC) = 24.980 mm.
:.
Dimensions of 'Go" gap gauge = 24.980
rnq
'Not Go' side = L.L. of shaft = 2 4 . 9 5 6 '
:.
+ 0.002 Dimensions of 'Not Go' gap gauge = 24.959 mm b*i@-w
(Fig. 10.41 shows a sketch of combinid 'Go' and 'Not Go' gap gauge) ,
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Fig. 10.41. Gap Gauge (combined type).
~ramptd2. !b&h of '73 2 b!@'km'&thew are 'hi Be ~6kkkctby the Ire& of a Go, Not Go snap gauges. Design the gauge, sketch it and show its Go size'& Not go Y& dimensions. Assume normal wear allowance and m e maker's tolerance. Soluti~n. Highlimitofshaft=7d.@2l'.I 1.,11r2t?r~~11~i Low limit of &Rri 74.98 mm Work tolerance = 75.02 - 74.98 0.04 mm . , :. Gauge makers tolerance (lO?!Y = 0.& mm Wear tolerance = 0.001 mm 'Go side' of snap gauge = H.L.of shaft; (MMC) = 75.02 mm / 'Not Go' side of snap gauge = 74.98 mm ~ i ~~ rr l Wear allowance is to be applied first to 'GO' ride, before gauge maker's tolarnw is applied. -,---.2/111r1
-
(Refer to Fig. 10.8). ! ' 'Go' side of snap gauge after considering the wear allowance = 75-02 - 0.002 = 75,018 mm :. Dirnensi~mof smp gauge are given RS : 8
.
1 G
I
,,,,:,(;k.,j l
i! ,,i
l
Unilateral System
+ 0.000 'Go' 75.018 mm - 0.004
Bilateral System
+ 0.002
,,....
.. 'Not Go' 74.98 mm rnrn FC1 I) 3rl,rk ,, r., 1-i ltci x ~ r m ~ ! ~ ~ 6.002
'go' 75.018 mm
-
- 0.002
k
z
'-"
Example 3. Find the 'Go' and 'Not Go' gauge dim&%; bTuafihg gauge m%g'8ilateral and Unilateral System and including wear.allowmce&r gauging 75 =t(Lp(.my. ~ t ~ r J 1 0 1 e s . High limit of hole = 75.05 I t * ,m . ' soljtion. 91o~i10 t I l ! i i i ~ 1o.3 Low limit of hole = 74.95 mm spue;> 0;) work tolerance = 75.05 - 74.95 = 0.1 mm si\\~x43ttt.,bw s t . Oh (11 Gauge maker's tolerance = 0.01 mm .s,,r~,@i m # . ~ 1 \ b , & - & ~ r d ! a o,im5 UOI US br1oq4~1m~ Ilia 3 ~ n a~ 2
'm'side of plug g a ~ g B Af..c.'ofBole 2~a.W' "' -
-
*I2
-
= ~74.9$'hid
6lie&r*$ .of w & $ ~ ~ i + m ~1ms%Sku y ~ qnrai f = 7$.%Z0+ 0.005 (Fig. 10.8) = 75.955
Dimension of plug gauge are given as Unilateral System
+ 0.010 'Go' 74.955 mm
- 0.000 Bilateral System tPs2&0-?Q5 :I 'Go' 74.9H' mm
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n o i ? + bmm 08 161 'sgusl) oi)' 'hp ~ - 0 0 0 m m.08 = %ot Go' 75.05 mm lot 'quai3 03'fa s~ - 0-010 mi-& traS.0
+
+ 0.005
m ~ W.B8 n W)O.O
-
'Not go' 75.05 mm -0.005 - 0.005 J , , . * . FFr fff dq mi - c r 01 Example 4. The rectangular hole shown in Fig. 10.43 @ a be checked The "mi* of size for the 'wt?& + 0.04
"
80 m m - 0.00
4.04
Design the suitabb gauges ?used on Taylor's m * -0.00 principle. W m -Fig.,10.42 Solution. According to ~aylor's principfe, there will ,- - -+ be one 'Go Gauge' of full form @a@length e q d to the ;'-' length of the hole. There will be' two 'Not Go Gauges;' of p&Z$rm to check the width and breaMh , *vT.=.* &- 5 of the hole. a
-.
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A Textbook of f k l ~ t i o n Enginewing
High limit of width of hole
=
60.04 mm
Low limit of width of hole High limit of breadth of hole Low limit of breadth of hole Tolerance on width of hole Toleran~eon breadth of hole
=
60.00 mm
mm = 80.00 mm = 0.04 mm = 0.05 mm
= 80.05
,.,
.
Gauge marker's t o h n c e (1 0%) : o.m*.-,4 "'!' Fot b r d t b of hole = 0.00i mm For width of hole
, !,
Go Gauge
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(i) 60 mm dimension : Go gauge will correspond to low limit of )vdth of h ~ i (MYC), e i.e., 60 mm. :. Basic size of 'Go Gauge' f i 60 pimnsion 'to h i e c; y
,w
= 60.00 rpm
Using unilateral system, the limits of siye for 'Go Gaugq' fa,@:mp dhension are : 1a.t:' . + 0.004 ,
60.00
mm - 0.000
, (io 80 mm dikemion :
Basier*
.
9
'4
.$+
;
of 'Go Gauge' for 80 mm dimension = 80.00
mm
:. L i d b of size of 'Go Gauge' for 80 mm dimension are :
+ 0.005 80.00
mm
(unilateral system)
- 0.000 The 'bb Gauge' is- shown in Fig. 10.43.
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Fig. 10.43.
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7,
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Not Go Gauges (Refer Fig. 10.7)
The 'Not Go Gaugesy will correspond to the mhimum met?! cuwjitim of hole i.e. high limits of dimensions. . ,, Basic she of 'NoT Go Gauge' for 50 mm dimwshn n r i 1~3.: I ?:;me :~rclr? c .-r = 60.04 rn : .-w :li . I Q I Z ~ : .arg: rFOC
:.
:.
Limits of size of 'Not Go Gauge' for 60 mm dimension are
Y*
- ... L., Limits of size of 'No Go 6'+&' P i 80 dimension . : . .are ,. .. ir: .. .i.c..,- -*. . . -- ... .-.....2 lo\ .: . I-.
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1. What is a gauge? O*i" + 81,807 &tp pi3 2. How dots a gauge differ from a rheasuring instrum& 3. --**! Sre - . 4 s o l & ~ ,baPu i w d n.sl;/. IGI-I.,' ,.,*. 6 , I l . .,..,v 4. What is the di&tonce between standard gauge (non-limit page) and a limit 4
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:H : # Q P , P ~ i ~ p w i i &iie ~ n ~d -r 0w0. *' ~ d W tdimnce? s 5Mdw,is ibopaed ia the BdsigB of gauges? Give the advantages and disadvantages of unillttellll and bilateral s y m of -fig. z, ww is wcar'dl~w-'? EBW ib itragtrlied in, the4dcsign ofgitiges? , State and explain Taylor's principle s f &it gauging. Discuss the .vllrim$ nar*erbksu&?d for g a w imw8bEture. What are the advantages of limit gauges'? ..,.,
.
we kbt limitations of Ilma gauges'? .c I - ' t! -::.ri . , ,. 13. How the gkgds skula bc c a d for btfore and bRh u3eq' 2 : " '
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14. What is
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d "n6r 'ga""eiids'of a set of manufacturing and inspection plug gauges to be used in checking a hole with s ' t c t I l C : l l q n b l 3 ~. -* tc' + 0.075 dsloit rs.irirrraa L-J! diameter specification of 25 mm Ai:t* hh:x.,{. a
15. Determine the specification &$ the 'gdlv a
0 02 ' << 16. A (haft I00 + m m diameter is to be checked using a 'go-not go' sna! F g e . Give its SP! C Y ~ $r dimensions. Anow for wear &d b g e mak&rlstbikrance. r %Yaitaij ?,i
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.
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a @he& the hPle 50m a: : : (50 H i , The depth of hole is 200 mm. Design the gauge and sketch it with dimensions. 18. Discuss briefly various a s p e a ttkr W i n g the td-B on limit gauges. 3.1, , :. +: 19. A hole and shaft system has the following dimensions : 60 mm HWc8 - p j r r r l ' J r,q 1;d lo? '3;uhl: 0 3 ) 13d-5r; >\!.I.. - ?:.'F,.~ The standard tolerance is given by ~.:0.() i = 0.45 (D)lr3+ O.@I D Where D = Diameter of geometric mean of stups. mm i=s t a n w tolerance, micron The multiplier for grade 8 is 25.'The fund&ental, d,yiation for shaft c for D > 40, is given by - (95 + 0.8 D) The diameter range lies between 50 to 80 mm. %ketch the fit and show on it the actual dimensions of hole and shaft. b e the class aP fit. Also, design the suitable gauges to check the hole and the shaft (AMIE 1974 S) [Ans. - - le : LL, = 60.000 mm, H.L.= 59.046 mm ' Shaft : H.L 9-59.854mm, LL.= 59.808 rnm, Clearance fit. . -. :. .'--- "---17. A limit gauge is required
-
I
x.
'
; 60
plug ~aude,GO'& ,
I
- - (j";io5;
* 0.005~mm '4
b d
Not Go side : 60.04~- 0.000
*.
0.005 mm CLlI
Snap Gauge, Go side 59.854 f
kg
Note. Unilateral system has been used. Wear allowance has been neglected; work taYqace being less than $09 mm;, ,, -,. 20. The minimum size of a hole is 25.00 mm. Its maximum size is 25.002 mm. When matching shaft ,8fwrldldis &the f b n M r Z a l Beviation is found to be - 0.02 mm. Shaft toler&W@ O H 3 m. D d g a &es for hde snd.'shft. TaPce the usual valiles. of gauge qabfs, t o k w e and wear Jlowyrcc. 21. Design 'GO' aryl 'W.GQ;, awls ,&LpIqg g ~ ~ t to g ~wqwm i %hd@ pf s b 28.4XW %O 14 mm adopting (a) Unilateral system (b)' Bilateral system. . l t' $ + I !
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22. A bore of mm dia x 43 n t t ~long ~ is to be &&A. Iksig@'dilZwd~~& IlImknsior a plug gauge for this, based on Taylor's principle design. vbf, r-YE ::ti'/.' .!'
23. A square peg having limits of 25.00 nun and 24.97 mm is to ke dPw:W. Design a gauge (gauges) for checking this, W op Tqyhr's principle ~pfw g e dyign. ,, .,, - . 24. Discuss the principle of Taylor's for the design of gauges for checking i , , ad X i2r [IS
, ( ~ j . ~ t4016 ~ .with ~ ~a c~rilylrical d :Not ,Go' g+ge. 1 ~ 3 i t t ~ ~ r '~ o , . ~ F~t ~ ~ . , ~ ~ (b) of rc&ngular hole rw xuri g y&;m&, or ~ a z usa 0 1 b @ ~ : : g21~!r?ourr,x:ln. (c) circular holes? i r J uL (d) circular shafts. rrtnt ?r ?c. itot~:i3r!1?3~jd (e) Non-circular holes and shafts. 'OC -
'
vwi-"
Gauges and Gauge Design 25. Sketch and discuss various types of Plug gauges
,.a. Sketch and discuss various types of
snap gauges. $7. Discuss the procedure df manufacturing Limit plug gauges. 28. Discuss the procedure of manufacturing Limit snap gauges. 29. What role do gauges play in the mass production system? f.
,4
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8
. . ---- - - -
31. m a r e Snap geugesw? 32. Sketch and discuss the ,mof following gauges : (a) Length gauges
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f33!31 I-. if'^ Ern s f l .b.~nrardo 1:~:. :$I.J tm-07 issrr~w~vs$ ''f yl~li. t v l w r r - 3 3 6 rt-,, , , l ) i ~ b ' - rnt 9 ~ t rsq r; 10 * (4F'@ pin w % n p-scr' ,ttoi51 to l?lZm3 l - & l ! { z i ~ 4 3 93L.13Jr . , 33. What are sorcw gaugeax?, How .am.aEbdy dad to umtrok.the oompk dirncnshw d $ak&ds? (c) Receiver gauge
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w o r ~ . ~ ~ & i t a ~ ~ ~ ~ k b . ~ ~ - f r ~ . r i ~ ~ d*. jkpi kge 914 G&uhte.'L' &em 49 1°-L58j,Max. did, = 23.42 nnn, min. [ b h b .* (D - dY2 an @In; 0 . M ~ m ] jAlfigq dia 1 3 . 4 Q t ~ . . ,
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t~.J! a: bshrc I .-. S.I. fi='va&us t n h k I+,I+'mWI& ~ n 10.30), . Special W= mamiets lir q $ k t i m m W ' & k r " ~ ~ rapidly ctuc to itsirL i k very wear (for ex~mple,a small diameter screw plug gauge used on a cast iron
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&4abpowder dbldhqs ggnrd with d i d grit abrasive, though very agrasllp. iqmvkl wcar 'life. *3I6tr& - 38. W r b the -UWdvantages of limit Gauging. Sol. Two limitatiorrs_oiLliaitG-Aw ,ksndkrseed & Art 10.8. The otha drawbacks . ntr;tr,;, trt nr-M nomi c ? f a e i ~ p - -, I , :. . . ., btjc Theit c a myfm ~ be IB &an ->! J B V ~ C (a) .Suitebbfbf Maw~pmitiQI of
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iimitititWs ~b , ~ e ' ' @ ~ , di ~ ~, . , ~~ ~ jw, , ; a~ : xd: R u ~ l ,~ p J g tur: ;!j (d) Particular sources bf error in themampnukt an Irrvs(rk$ R e & w m s in machine tQol q d better P t M (. , ' .& due, Q e k q m j x ) haw a h ~ the ~ d >(j
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