1, - Mechanical Geek€¦ · Complete interchangeability in the ... kg by selective ... A 100 mm diameter journal and bearing assembly has 3 clea~.mcefir, ...

The d -,so fatr has been in connec

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Complete interchangeability in the ection and materials hqdling by usi r

ded according according tb ,i . method is especially useful i ',i

! Its in complctte prottxtion C 0 !

I !

1, 1 .

f a axeztain-tqrgeafpiston skirt each be O M mm.

3;". '

tnr = 0.04)ildm P* clearance = largest bore-smallest piston

k g by selective bsrltsmdomxraeethat

I, of tb 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 d the values of allowance, hole tolerancd and s w t tolerance fw t& i . bollwing .&hhm.yjoni qf mated parts - according, do basic h ~ i e qvNe~.

4P,,r' H Q ~ : 47.5Q mm Shaft : 37.47 mm , I \ .

37.52 mm 37.45' mm . , f '.I j

= 37.52,- 3130 w8 s a d s& hi+, trl;s13 > r

= 0 . 9 rmR. = &&3Rti,@ y&$&J Shaft tolerrtrrce = High Jilrait - m?Writ bM , - -

= 3?*4,7 - 37.45

' O . ~ ~ = ~ ~ ~ ri;;*: Allowance. = ,Max,@- meQal & d m of hale

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

It is clear with the basic hobi& thai-; " i 8

. , . . LOW limit of hole = B‘%w " ~ i & =gy&&~.& &ad;, < , ?+I @H+

'' . -e?-. ls;.2&8 inm ~ , a l ,

High l h i t of &a = 732& '+ 0.225 sm xu -3s : b ~ f ~ w . m w w d i k (ikik) *

' ~xupl..&&&@$&-n'-ert?~i~&an and tolerances and hence the limits ofsize for the shafifW W# +$&! ~in@~~W&f/ . The diameter steps are 50 mm and 80 mm. TD.T{ t!?

(b) We know that for hob;&& &v-7 is mfIT3w%,&ndamental deviation

Shaft : Hi& limit = -sincethe shaft 'j' lies below the mro

*ta b 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&~. .~QI qwance. Example 5. In a limit system the jhllowirtg limi& we sgwfled to give a clearance fit

between a shaji and a hde : t - ? dod fo firnil @ i f + ,

Determine : (a) ~asi= s&e (b) sh@ and hde to era&& (3 the shajl and hole limits (4 the maximum and minimum clearance. "

, I : , ? ..[. c y : , ! I @ $ : > $ .$ , , -:

.7 + . + : k t , ; .a,8 - $ < $ - , - I , *!L. Solution.

(a) . 4 l - l ' ~ - ~ T l # ' I ,, ~ k i c ' s i z e = 30 mmr

(b) SW tolerance = 0.018 1'0.0b~jG b.013 mm - . <

m .~l-,ij If. i:!: r',: t , Hole 'tolerand 0.020 km

.'.acir $\im* J x & rtk '- I i t ! . 1 % i t ~ \ , ,L ~ i g h r i i t of s h d = 30 - o,@$ r ! ~ " * * .- 4 -,',,. 7 , : . t , . -,>sf$,,'> 3 \ -

= 29.995 m%?::r:c 5 . I t :,, . *,!,*. ., ~maul* Low limit of shaft:, 30 - 0.018 - 29.982 ,mtq ,

. ' P ! ' 7 8 . : , { * I ., High limiti'bf hole ='30 + 0.620 -

5(,@20~,iitm Low limit of hole = 30 mm

< , I. Maximum clearance = ~ i ~ h jimit .di. h& - low limit df shad

~ i n ' i m s cl& = 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~limum clearance of 0.01 mm. The hole tolerance is to be 1.5 timdp the sk@ &lev=+. &itermine : limits for both hole and shafi (a) using a hole bast3 system (6) using a shaji bash wgtp- .

Solution. ~ e f a g to Fig. 9.18. I . k l :' 7 ::,2 nc.r.;~s$mrCt

Fig. J.P8 . , 1' . , 6

If x is the shaft Wmuice Imd y is the hole tolerance, then

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. . 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 ~ ~ d f y i l i t *

3 t 1 > Low limit of hole ~t 25 mm (Basic size) v, $$?[\ "\r .? %($Ziti? >f

% 7 * J u !+fin ibA-8 . k (,I.- : I . High limit of hole = 25 + tolerance

= 25.006 mm ZWD --. ma, , , Qt I%'\?,

Upper (High) limit of shaft = low limit of hole - minimum clearance

Low limit of shaft = 24.99 - 0.004 = 24.986 m6-- '- -": ,, , . . ~ . G I \ .x's :* t 9 , :lb0k&. * (b) Shaft basis system. The upper deviation is zero. A3 wminiwr-,m$a

. . -. 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 - l i , 30 tim11 wOJ = 9 s m M

. . Tolerance grade, i = 0.45 (D): i o . ~ I - D , microw

' ; I d to dmil 1171'1 - - 0.45 t9qi .@,$@I x 98

'

= 2.079 + 0.098 p 2.177 microns I t l ( . . 8 1 m > . .

Now, for hole GRi, tolerance = 16 i = 34.832 microns , L , t, Itant 1.3 11 ww I I I S )'I: ". ' = 0.03 5 mm (rationahzed)

9' -- Or,Q.fbC For shaft 4, tolerance mz&~a = 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 , ~

We can get these values of to l~mces dimfly from Table : Appendix IE. Now F.D.~~w h& G,;; + 2.5 microns (T 9.2)

&it 3 :-l>%a~'$ - : US^ ,U S>Q fliV r-: +. 53. >! = 25 (98)OJ4 = 0,012 nyn (rqtionalized)

.un,w&ihl,fir shahm \Q.a"fg Y ~ ~ ~ a ~ ~ ,RUIR?U T I d DO". m i m m (7 9.2) ,,fib '=15, .;I . L) h . U i > $7 i :-yli.~ ?)SD 9 h A lffUQ w\s\im'A . tmm-aq 1 (gsp11, m i c t o ~ w \ ' :XI\\ t . ' m . 2 : 1. . ,.:

Dimension See F 9,!3. ( i ) Hole - d--- - - - L.L. of H& &wic-#ze + F.D.

81,. C C ,, = 100 + 0.012 mm = 100.012 mm .

4 i/m M.L. .of Hole = L.L. + Tolerance ' f \ = 10&012 + 0.035 = 100.047 m

( i i ) Shaft U.L. or H.L. Sbft r Basic Size - ED.

= 100 - O,Q,72 = 94928 mm ,, , . , L.L. of Shaft - H.L. - Tohw;slaFe i

I

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~nc e 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. v

, t *. : . (a) Hole-Basis System : Refer pig. 9.1 3,

- 2 - f

LOW& limit of sd&r!hg =''&ifs Size - 1 0 0 dm'' ' '&"

Highest limit iif ~ k r i n ~ ' ' & ~ L ; ~ . of Bearing + Tekrancd ' "@ 100 + b.005 = 100.005 mm

33n814f~~ J I P - ~ : ~ :qqU = f i s h 'to nrnr; ~~twc., Refer to Fig. 9.2, Higher limit of Joumd k: W&'limit of bearing - allowance

.mar ~14KW0.002 -5 99.99811~11 i~tzf2 d t d 116d? f r i l

Lower limit af d m a l = Migh.limit - T @ b a h ~ ~sqql;

43i11;.t410r - .= 99,k)98-0.004; ', ?L gqiI X W ~ ~

,,~m id.@@ - = 99.W1m

(h) Shgt$+@kJ rSI#[email protected] ,fiver Fig- 9.1 3 Upper limjt sf.hucnal = Basic Size = 100 mm Lower limti&<!purna1. = Uppcr , . limit - T~krance ! lw+G

= l ~ - p . o o 4 awn 2IQ.W - Z 1: '

- ..: : ,. 7 r ~ i ~ ~ t a &Z &kwer limit of Bearing i.A

Example 9. In an assembly of two matingp- of.IOQ,mm h a ' size, the48 is Inter$erence and the inter$erence varies from 0.05 mm to 0.12 ,mm. . qe 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 ' .jti1)4 nl. :q? j

(a) Hble Basis System : Refer Pig. 9. tsF : . < I .: T qdll i

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 - - I

Maximum interference will be when the hole is at its lower limit and t e shaft.is at its upper limit,

I i i

:. Upper limit of shaft = Lower limit of hole + Maximum interference

Now it is clear that, t, c . . ,>-. + L~ ' : :, c 7

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~,d 71, \, ' ,, : \-.,\I traitk\ b\r?f.rt Tolerance on shaft + Tofirance on hole "{' ""'"' "' " ,,.., ..,,= , ,.. . ... ,, . .. / \ I $ 'J1112 : = 2T

.;*1:$2ff$*l=-

2T = 0.12 - 0.05 = 0.07 q9# . ,,,,, . % ,. .,.%a -,b)r I

:. 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 I I ooi = i 0 1 ~ 100.035 mm Lower limit of shaft = Upper limit - Tolerance

'-= 100.QSS mm. (b) Shaft Basis System : i NQ CY .

Upper limitdshaft = B d ~ c i z e =.10Ornia;rnr! .: Lower 'limit of shaft Upper [limit - Tolerance

* 100 - 0.035 = 99.965 mm

Lower limit of hole - Upper limit of u'- lhxinilrrm ~ " ' awl I 'w; = 100 - 0.12 = w:$8 hutr. : Iyq'J

..- ., 1

:. Upper liniit of hole = Lo& limit'% &kbhke i ' l s t i r / . ~

* 1 C ' 1

= &.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 n and t k allowance must nd exceed * 0.012 mm- the merage value. lRe bark*she 168.rrmr.l @ i e ~ & o#;hdc + $ ' ~ - t b b & d ~ l f A s shap. 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

I ,' .n't'i>w. ,t

e. ?'amce &f\ =t 6.0011 m. 4 j5r:\,,~': I : N ~ i t . '.',? 3u.b v., 7". .p ~hfSns .r .3 , . . . . , . ': ),I .!.: ,.v:<-. ,. ' 5 . ~.i$ ,*.v.yi . Q,.;.-!T i ~+ . t l , ~ \ . , 3,,\\ 'c,~,,:; a T o b r a q o n m l e = P.016- , . ~ a i c . . . s .tn ; f \ \ \ . \ i : - l? , , ; ; . , ;!n,: . . ,5;~ ,\c.,\;. ' .'.S.;r I . tr..,

f Lower limit of Hole = Basic Size = 100 mm ':C . . ,, . '

Upper limit of Hole = Lower limit + tolerance

Now Min. allawake = Lowc2r%mii ofhde -'Upp&r h i t of shaft.

mrw,mJ~'~ &MU ~~~~ .!hiit nil B l m @ mrn GG A w ~SIEO~JR* x: kq A

= 99.972 - 0.008 lil srij ':c IWb3 & .,a&!$ Zl Jfl tr8 99.964 mmi I 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{ 3 1 - $3 -, $-~~IP~QPLEMS s l.nnosarrR m r. bhr 1. Define "Interchangeability" and dimmsiirs IiapbFtancdQ? -

2. Define tolerance. .&I ,at ?~[!FIJIL~ 3ris .T ; J J ~ . , L ~ ~ I i t , 1

l 1 ~ f 3 g a 9% ibl if'~nwxsible €a obtain .c#v exmd-.~w~ ~ l a ~ o l OJ 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&~to~! l :

' . _I t b i c l !I

, . r .

5. Define : aflowandi, clearanck ah htederenci: '-;"' .-. .-.....vst: b~a~~mtnd

. 6.' What is zero line? - f # , !; 17, I a,# ?,q - 7. Define : upper i&vh&t,%.&%~r & &&e~#sv31i;f;r;!!4~ 1 t m : t "!fl 8. kxpran uun~ad &jj'&;ibitoI* ' ; B i k ~f i i '3U r . .ST

9. What is meant by H&,d&~*pl'&&q4'.~L qql 3mmm 3m 2-/;3 .LT

10. Explain and compare "Hole basis sy&~'sda'4biKdas ( ' # t r d ~ * ~ g & % ~ ~ ~ ~ ~ ~ . p i ~ iti%iIk- . llker&.dto have

imum toral cl-. FS #iWW$k ~W?&hi(tgtlil i?Iautince &a69 Irihrm. Wermine - the spccificaticrns of the parts k)i @ @Q& [email protected] (&).for Sra t iW average

inmcbngesBil#y. I S : 611 2 ti , > L a$;>

12. How will you write the Rt : is 40 rHm?

.&4rq&$\* & basic size

i* 'qg ule &sieuria & * d - . m *I* m amiwii+ arl f 04

0 mm H - h ~ & t h e t o l k c ; grade IR ULO+ P;&!F&,.& t&ga'& ti# ( ant f.O_ 0: 1sl31~1~lb) thdr 4 ilf

($13 the above fig - -. , ., *', , -: Msiy

t@$qns& e@jp thq .liW of dze for hole

. 9: - 7 % *@1&!4$ ~ ~ I P T A ?- tJ.wl D ti1 s~uw@3 $&A,,

(b : &A qy r v &;&--;1w,a 7.1 e to h c e g d i ~ b 25i and f ~ ; ~ n i b e r ~'~uaIi& $ - I .,, , .itqa-l .+. 15 A gear ring of85 mm diameter bore is died bn baa bubiresulting ih'a Ha6 h ~. le? &

tolerances and hence the limits of size for tht h b 'kt& he g& h e . S@$@ttf;t type of fit. The diamcter steps are 80 mm and 100 mm. 'Ihc 8- fWj.&is 0.- mm.

16. Calculate the fbndamental deviations and tdgwn6~1~ #am 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 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 ~lltowuxcs for a 25 -4% '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 - 16PU microns Name the type of fi& jh.9 ,,/ &

18. Determine the types of fits p~odUCda by the following maEg &holes anh

A Textbook &~Production Engineering

19. A fit is designated as : 60 mm H, 2 h,; D b t e n r r ~ h h hi mum' &?&anct! and m9rnum clearance ' " ' " - of the tit. . C rn0.d s$@:@ =

'

lo. A turned shak is to rotate in a reamed h * m a 8 ~ r d fit is HJc,. ~eterhine the 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 'L H#p Dea?rmik(t ttk actual dimensions of the shfdt PIlQ tb bPnt Q$ the i#k. T?w &mew, stcpg m 39: and 40 mm. The hndmental deviation for the hole H is ~ r o *,?tQ3f;?r $9 sh& 5. iq

I 8 : - (95 + 0.8 D) microns.

.,:I 0 < ' . $

- The multipli~ , f ~ quality 7 is 16 a@ that for q q g . is12$. ,. , . r s -

22. Define and compare : Wdoq , w b l y , st&@,~~1l r . > . . . I . * I e*~J1!b

23. Give some common applicatians of J q i v e . . ' 2 4 what y,*>*n;,*,pf:yy. h a "..&&:, &+

Y*', - 2 , b ~ 6 r . W . d ,B;II$W Q . ~ ~ l mn, ~marmt9m' -16 *@y "twmdI&l ida#e @& :a ~ h & f f . ~ , m -* 76 mt'Itg,_~$w ;* !:.*

(6) Tha raigtive mor in the dimension is greater in &aft A. ,@lirfas#ti6rt.mtiii

(;, fat&&& && &&js~" is g,+& "1) rr. i i ~ f liwl *w .$I ?tmw $+,& ' :'

(d) The rclatbe error in the dimension is greater in s W B. , . - , ,, !,:: .( . (Am.: a)

d- -a) Tmsitien FI~J(@ lW&m~e SF it (c) c l e m e c fit (&>Noat (GATE 1993) ( h a : e) 7 eL

-y7. h e fit 0" ( %1'&&$l1&ab) ,< .: 2 is &ifid . as H, - s, The type of fit is &:,";;; (a) Clearance fit !. - & wning fit (yliding fit) (c) push fit (transition fif) (d) force fit [interference L - 1 , % (G#TF 1%) A=- : dl

*a ~n thq r p k i i o a . , diq-7 > , d 9% ,, , . I ~bwdw+g~ 21

n ! *?da) a lJowm is aqual t ~ h i l a t s F a l , C w . b .- k'

m8.31~) akvanec is- to UftiMWl Weran&. I -, ,bT brsfi Yc) efiowa~ce iJ Mebed& eFtd&iwe. d U 3 .&I . .

sion specified by the tolerance. .a&% 998) [Ans. : el

I 39. A journal md b a i n g aSsOLrrbly has h e i l & # b g sizes %' -9 h@fmmb * 1' PI u m r t q / I s~qqrsr& 4nr ,\ xu- a .@TI - TTi j + -

-0.001 *t. h d ! ' 4 , .- &f ' ro2 -.'L* % n ~ z t i d .(kt3 YIO *ltr.jffia:~ .f l

i : I R WIJS~:~ 4.W t r m ~ r ,m 1 +0,001 &ii vq@

Bepriflg : SO 4.002 m% ** Determine.: T o 1 6 on Journal, Toleran , Maximum clearance and type of fit. t W C $5)

. . i t (Ans : 0.001 m%&003 m. -0.001 n w W mm, TmstioPL:fif .. . ?B -

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 : Bearing H-LA = "*Oo4 mm , clearance fit). L.L. = 54.993 mm ' L.L. = 55.000 mm -@in the d i h c e betww toI-nq,d aljomnce~ -., .J15#" .: Why do m~rm&n~pro&k produce $at& d t b such a wide range af tolerances? - 33. The joumal and M n g wembljr has a basii size of 206 m. For the bearing : ED.; (Here, L.D.) = 0 micron, and Toterance = 46 micron. For the journal, the values are - 820 and 1 15 microns respectively. F a tfiG 4- o f j d and bearing, assuming unilateral system of tolerances. Alp, +tpmining the allowance and type of

'" ,. ' '; ''tit. Which spkms of fithas,kn adopted? < .

. . I ' 4 ijl$ I h b: 'J'fl5.? & ' L ~ P , .~fc3

Minimum size of the hole = 3 & h '

< . I 9 <.

W. Maximum @6f t.k shaQ = 50 mm. .--a gihwrtC . 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 Clearance = Maximum size of hole - Minimm size of Be'shaft - - 49.9% - W.989 - - 0.01 !? nun

M i n i m u m C l m = M4nbum &e of hole - W m u m siza of the shaa = 49.935 - 50~000 = - 0.065 llun

Since bdfi the maximum andmhiiarrpn cl- 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 : . Hale size : + 0'05 mm ; S ~ B : 20 20 - 0.05

+ 0.05 - 9-3' A ., - 0.05 ['IS ' :

@huv size ofthehole = 20.05 - , * , ; , , * Minimw size oPtheQole = 19.95 pun

. ~ \ c > ; q > < 13; Id,(* ..J& ,-rJy ?I{: : ..j&# ( b L : a i i X

Maximum Qze of the shaft = 20.05 mm r

* . ' k ; ( ~ ~ ~ ' d ~ Tm ~6 = 19.* - :31 qd t t l d ~ ~ ~ ki4d.;.:>/~*: *>I: fii .#ficil ,:h i

~ ~ ~ d i a t o f k d e - M i n & n r m ~ . o f t b M * Z @ . @ 5 - 1935~0,101~~n sw efReld- wUIm ~ a f ~ = 19.96 - ~.055-0.10 m positive 9nd the Minimum Clearance is negative, the tit obtain J

1 , . . I - * - - - f

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 whichlocates 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 gaugeneeds 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 thegauge 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: Highlimit 25.033 mm, low limit 25.000 mm. The limits of size Jar 17 shafts are: High limit 24.980mm. 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.

Solution. Tolerance for hole = 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)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 = L.L. of the hole (MMC) = 25.000 mm.'. In unilateral system,

+ 0.003Dimensions of 'Go' plug gauge = 25.00 mm

- 0.000That is,

High limit of 'Go' plug gauge = 25.000 + 0.003= 25.003 mm

Low limit of 'Go' plug gauge = 25.000 mmNow,Basic size of 'Not Go' plug gauge = 25.033 mm

+ 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.)

- . .,c* . . _A. . it.

, ', ,,,.: , -/ 4 . . 4 =

, - ; , I * - 4 , ft 8 -& iu .. I 8 , , I . 0 0 Bt3 I. 1 ^- r . - , . t + , ' , *t . + I -.-.---.-.-.-. - ....-.-.-.-.-.-.-.-.-.-. -.-.-.- .-.-.-.-.-.-....--.-.-.-.-.--.- iti

, @'

'!! +$,

h - l

r I . . ; , $

.'?.%? 'ah Fig. 10.40. Plug Gauge (combined type)

(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) /

, i . , c ; * ' < P '

Fig. 10.41. Gap Gauge (combined type).

~ramptd 2. !b&h of '73 2 b!@'km '&thew are 'hi Be ~6kkkct by 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- -,---.2/111r1 ' . 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 ~ r ~ r l

Wear allowance is to be applied first to 'GO' ride, before gauge maker's tolarnw is applied. (Refer to Fig. 10.8).

! ' 1 G I i! 'Go' side of snap gauge after considering the wear allowance

= 75-02 - 0.002 8 . = 75,018 mm ,,,,:, (;k.,j l ,,i l

:. Dirnensi~m of smp gauge are given RS :

Unilateral System + 0.000

'Go' 75.018 mm - 0.004

Bilateral System + 0.002

'go' 75.0 1 8 mm - 0.002

,,.... .. 'Not Go' 74.98 mm

rnrn F C 1 I) 3rl,rk ,,, r., 1-i ltci x ~ r m ~ ! ~ ~ - - 6.002 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 .

soljtion. High limit of hole = 75.05 I t * ,m . ' 91o~i 10 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.,b w s t . Oh (11

Gauge maker's tolerance = 0.01 mm .s,,r~, @i m # . ~ 1 \ b , & - & ~ r d ! a o 5 mi,, UOI US br1oq4~1m~ Ilia 3 ~ ~ 2 na

'm' side of plug ga~gBA f..c.'of Bole 2~a.W' "' *I2 - - -

= ~74.9$ 'hid 6lie&r*$ .of w & $ ~ ~ i + m ~ y ~ 1ms%Sku qnrai f

= 7$.%Z0+ 0.005 (Fig. 10.8) = 75.955 && I

QOl] t.3 - Dimension of plug gauge are given as . Unilateral System woiztrsr&h tn4n 0% fii l

+ 0.010 n o i ? + b mm 08 161 'sgusl) oi)' 'hp ~ - 0 0 0

'Go' 74.955 mm m m.08 = %ot Go' 75.05 mm - 0.000 w g mi-& lot 'quai3 03' fa s~ - 0-010

Bilateral System traS.0 +

tPs2&0-?Q5 :I m ~ n W.B8 + 0.005

'Go' 74.9H' mm W)O.O - 'Not go' 75.05 mm - 0.005 FFr f f f dq mi - c r 01

-0.005 J , , . * .

Example 4. The rectangular hole shown in Fig. 10.43 @ a be checked The "mi* of size for the 'wt?& "

+ 0.04

80 mm - 0.00

4.04 Design the suitabb gauges ?used on Taylor's * -0.00 m

principle. W m

Solution. According to ~aylor's principfe, there will a -- Fig., 10.42 -. ,- - - -+ be one 'Go Gauge' of full form @a @length e q d to the ;'- ' - . " J:,

length of the hole. There will be' two 'Not Go Gauges;' of p&Z$rm to check the width and breaMh of the hole. , *vT.=.* &- 5

! I

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 = 60.00 mm

High limit of breadth of hole = 80.05 mm

Low limit of breadth of hole = 80.00 mm

Tolerance on width of hole = 0.04 mm

Toleran~e on breadth of hole = 0.05 mm

, ,., ,! Gauge marker's t o h n c e (1 0%) : . , I 1: L L . ~ \ . t ~~!qt::? c P

"'!' Fot b r d t b of hole o.m*.-,4 %!,.hL \ . ,,\< , . I - 1 ) , ,, . , t , 1 , ~ ' 6 \

sk+d ?Q Yk,;~:b f ! > l ( i - i i t t ! I;< For width of hole = 0.00i mm

d3LJ fo ftnlll r I.;

Go Gauge ~r - ijr), ?T c qnsdoi AX J.,

(i) 60 mm dimension : Go gauge will correspond to low limit of )vdth of h ~ i e (MYC), i.e., 60 mm. :. Basic size of 'Go Gauge' f i 60 ,w pimnsion y 'to h i e c;

= 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 : ' 4 9 .$+ ;

Basier* 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.

:..i\lnr-..i

Fig. 10.43. i I E

Not Go Gauges (Refer Fig. 10.7) '4 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: r F O C

:. Limits of size of 'Not Go Gauge' for 60 mm dimension are zfia.fi 'bdq

- . . - . I-. L., .__A. - Limits of size of 'No Go 6'+& P i 80 Y* dimension . . are :

:6 ?-1. ., .. . : - . 4 . ' ir: .. .i.c ..,- -*. . . -- ... . .-..... 2 lo\ 'i~fq~f;:+i-ilj ~ l ? +awl: & + (p i -

1. What is a gauge? Oi*" + 81,807 &tp pi3 4 2. How dots a gauge differ from a rheasuring instrum&

-4. a! kl 3. --**! Sre - . 4 s o l & ~ ,baPu i w d n.sl;/. IGI-I.,' 1 1 ' .9joll( ,.,*. 6 , I l . .,. .,v

4. What is the di&tonce between standard gauge (non-limit page) and a limit e? airid31 ' " I , - .,.- ) I 2 ,.It ' I , T1 % .. ltadZ .+ Q* : H : # Q P , P ~ i ~ p w i i ~ n &iie ~d -r 0w0. *'

94: ira 6. What h mu@ ~ d W s tdimnce? 5Mdw ,is ibopaed ia the BdsigB of gauges?

7. Give the advantages and disadvantages of unillttellll and bilateral sym of -fig. '[I,. 8. ww is wcar'dl~w-'? EBW ib itragtrlied in the4dcsign ofgitiges? z,

, , 9. State and explain Taylor's principle s f &it gauging.

, , , . I& Discuss the .vllrim$ nar*erbks u&?d for gaw imw8bEture.

11. What are the advantages of limit gauges'? ..- - , . , ~ U P P 42, we kbt limitations of Ilma gauges'? .c I - ' t ! -::.ri . . , , . a ,

13. How the gkgds skula bc c a d for btfore and bRh u3eq' 2 : " ' 14. What is M W I I , , I 5 ? , + , , r , a d

15. Determine the specification &$ the 'gdlv a 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

0 02 ' << > . $

16. A (haft I00 + mm diameter is to be checked using a 'go-not go' sna! F g e . Give its ayr!! 3.

dimensions. Anow for wear &d b g e mak&rls tbikrance. r %Yaitaij C Y ~ $ r SP! ? , i < l i r ¶ - f . , f , -: , / , I l l , :

17. A limit gauge is required a @he& the hPle 50 :::ma (50 H i , The depth of hole is 200 mm. Design the gauge and sketch it with dimensions.

18. Discuss briefly various aspea 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 -pjrrrl 'J r,q 1;d lo? '3;uhl: 0 3 ) 13d- 5 r ; >\!.I. ?:.'F,.~

The standard tolerance is given by ~ . : 0 . ( ) - i = 0.45 (D)lr3 + O.@I D I

Where D = Diameter of geometric mean of stups. mm i = stanw 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 9-59.854 mm, LL. = 59.808 rnm, Clearance fit. . x. .: -- --- - - ' I . .'- " - - - - (j";io5;

plug ~aude, GO'& ; 60 * 0.005~ mm , bd '4

0.000 Not Go side : 60.04~ - 0.005 mm CLIl *.

Snap Gauge, Go side 59.854 f k g

Note. Unilateral system has been used. Wear allowance has been neglected; work taYqace being less than $09 mm;,

,, $ + I ! & -,. '! 20. The minimum size of a hole is 25.00 mm. Its maximum size is 25.002 mm. When matching

shaft ,8fwrldld is & the fbnMrZal 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, tokwe and wear Jlowyrcc.

21. Design 'GO' aryl 'W.GQ;, awls ,& L pIqg g ~ ~ t g ~ i to wqwm %hd@ pf sb 28.4XW * %O 14 mm adopting (a) Unilateral system (b)' Bilateral system.

8 , . l > t' >

22. A bore of :;::! mm dia x 43 nt 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 ~pf wge dyign. ,, ., ,, - .

24. Discuss the principle of Taylor's for the design of gauges for checking i ,, , a d

Xi 2r [IS , ( ~ j . ~ t ~ . ~ ~ d 4016 with a c~rilylrical :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 u sa 0 1 b @ ~ : : g 21~!r? ourr,x:ln.

(c) circular holes- ? i r J u L

(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? ,4

f . I*' 8

- 31. m a r e Snap geugesw? -- . . - - - - - . - -

32. Sketch and discuss the ,m of following gauges :

(a) Length gauges Y ~ C ) ~ - ~ ~ V U ~ . I H ~ 8 i . I! - - -.

,"!I I I t;t (c) Receiver gauge Ern s f l .b.~nrardo 1:~:. :$I.J tm-07 issrr~w~vs$ f33!31 I-. if'^ rsq r; 10 * . I . ''f yl~li . t v l w r r -336 rt-,, , , l ) i ~ b ' - rnt 9 ~ t

c (4 F'@ pin w % n p-scr' ,ttoi51 to l? lZm3 l - & l ! { z i ~ 4 3 93L.13Jr . , . i 33. What are sorcw gaugeax?, How .am.aEbdy dad to umtrok.the oompk dirncnshw d $ak&ds?

% f i r r . , 1 f -. hi To ml.1 F r : . I J - ~ > :

3' ~ 5 6 k ' h \ h * m g J . ~ r j r t r ~ 1 ofptsq/&b jO d%o. , .,, , , ,a luaa;l:

, . A w o r ~ . ~ ~ & i t a ~ ~ ~ ~ k b . ~ ~ - f r ~ . r i ~ ~ p ~ ~ ~ - i ) - i ~ ~ . )-s;~ 'I d*. jkpi kge 914 G&uhte.'L' &em 49 1°-L58j, Max. did, = 23.42 nnn, min. , I , - jAlfigq dia 7 1 3 . 4 Q t ~ . 7 . , [ b h b .* (D - dY2 an @In; 0 . M ~m]

1 ~ 5 1 37. w b on wearfar gar$eslakrgsni v. I . z ! f i z l w s i t~ J! a: bshrc yZ-,-- ._ . . - I 1 . I . -.

" b313: 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'&kr"~iL~ its r ik very rapidly ctuc to wear (for ex~mple, a small diameter screw plug gauge used on a cast iron \ *at), t), a :-

.-. (a) h s . (B& w& resistawe than c-stee~). .. ., &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_oiLliait G-Aw ,ksn dkrseed & Art 10.8. The otha drawbacks .. . ntr;tr,;, trt nr-M * nomi c ? f a e i ~ p - , - I , :. . . ., ->! J B V ~ C (a) .Suitebb fbf Maw~pmitiQI of btjc Theit c a ~ myfm be IB &an a : 1 1 ' a5 ~ i ~ ~ ~ o f d ,.> l b n 6 2 : , .: tl3OSg 12.)TT&Yl!f?l Sf*

~ f ! (.. )I (by fie>* of limit gaw h sizes I F 1 1 . :con&&&, s s&H7asTWght~ ffb$& i w r f l m e n w g ' 16 1 ~ 1 , r .- .,,,,,,, , fi,&'kiih MY iimitititWs Ep ~b bS

~ ~ * + ~ ~ ' ~ , ~ e ' ' @ ~ , d ~ ~ w ~ d u ~ g b ~ i ~ , . , ~ ~ j , , ; a : x : R ~ l , p J tur: ;!j !-33 a

(d) Particular sources bf error in themampnukt an ~~ Irrvs(rk$ Re&wms in machine

>(j <i c Y ~ 3 , tQol -4 q d better P t M (. , ' .& due, Q e k q m j x ) haw a h ~ ~ d the , d ~ s b a y . b , ~ l ; # . m w T + whmg -.- i n e n ulr . I L H L C , !I

I . .?:qr ~~$i't","~,~at$$s TCF~LT ;rf l i s8 2

c h d g plain bores, both intemal,and ex& somn thrrrds, s#hes and serrations , r t