metrology solved examples

8/2/2019 Metrology Solved Examples

1/17

The d -,soatr has been in connec1

II f , r= 0 . 9 rmR. = &&3Rti,@ y&$&J

Shaft tolerrtrrce = High Jilrait -m?WritM- -= 3?*4,7- 7.45' O . ~ ~ = ~ ~ ~i;;*:Allowance.=,Max,@- meQal & d m of hale

Example 2. A 75 mm,&@, ris 0.075 rrun and the required ailowanthe bear&$ bore with the basic,b&.

Solution. Refer to Fig. 9.15It is clear with the'kit'

E u a p k 3. A medium force f& on a 75


3/17

It is clear with the basic hobi& thai-; " i. ,. . 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 ~. m wwd i k (ikik) *' ~xupl..&&&@$&-n'-ert?~i~&an and tolerances and hence the limits ofsizefor the sha f i fWW# + ~in@~~W&f/.he diameter steps are 50 mm and

80 mm. TD.T{ t!?

(b)We know that for hob;&& &v7 is mfIT3w%,&ndamental eviation

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

*tab m 9ld qwq>q


4/17

= 59.37 - 0.03 = 59.94 mmThe fit is shown in Fig. 9.17. Jt. is a clearance fit with &Or93 @;&J&~ . . ~Q Iw a n c e .Example 5. In a limit system the jhllowirtg limi& we sgwfled to give a clearance fitbetween a shaji and a hde : t - ? dod fo firnil @ i f + ,

Determine : a) ~as i=&e (b) sh@ and hde to era&& (3 the shajl and hole limits(4hemaximum 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 olerance = 0.018 1'0.0b~jGb.013 mm -m .~l-,ij If. i:!: r',: t ,Hole 'tolerand 0.020 k m.'.acir $\im* J x & rtk '- I i t ! . 1 % i t ~ \ , ,L~ i g h i i t o f s h d = 30 - o,@$ r ! ~ " * * .-4-,',,. 7 , : . t, . -,>sf$,,'>\ -

= 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'bfhole ='30 + 0.6205(,@20~,iitmLow limit of hole = 30 mm

< , .Maximum clearance = ~ i ~ himit .di.h& - ow limit df shad

~ i n ' i m sl& = Low limit of hole - High limit o f shah

Examp@& A hble k d haft &e a basic site of 25 mm, a)td are to have a clearance$itwith maximum lolearaneeof.O.02mn a d a mi~limumlearance of 0.01 mm. The hole toleranceis to be 1.5 timdp thesk@ &lev=+. &itermine : imits or both hole and shafi (a)using a holebast3 system (6) using a shaji bash wgtp- .Solution. ~ e f a go Fig. 9.18.

I . k l :'7 ::,2 nc.r.;~s$mrCt

Fig. J.P8 . , 1' . ,6If x is the shaft Wmuice Imd y is the hole tolerance, then


5/17

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 ~ ~ d fi l i t *

3 t 1 Low limit of hole ~t 5 mm (Basic size)v, $$?[\ "\r .? %($Ziti? >f% 7 * Ju !+fin i b A - 8 . k ( ,I .- : I .High limit of hole = 25 + tolerance= 25.006 mm ZWD --.ma, , , t 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 \ . : t 9 , :lb0k&. *(b)Shaft basis system. The upper deviation is zero.A3wm. .- . High limit of shaft = Basic size = 25 mm ,.oiwi&Low limit of shaft = 25 - 0.904 yi3439fpm in)L y imit gf hole = 25 + 0.01 =,22i0a,mmGnd ; %a~'$ : US^ ,U>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, . .L) h . U i > $7 :-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 - -- - - - - L.L. of H& &wic-#ze + F.D.

81,. C C , = 100 + 0.012 mm = 100.012 mm. 4i/m.L. .of Hole = L.L. + Tolerance' f \ = 10&012 + 0.035 = 100.047 m( i i ) Shaft U.L. r H.L. Sbft r Basic Size - ED.

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


6/17

Example 8. A 100mm diameter ournal and bearing assembly has3 clea~.mcefir, ith theiQiqwing specifcatiotp ; ,!, , , , , + I -,. , , . I * I

Tolerance on bewin$ = 4005 a m ,,,

sivrl 30 irniii 1awo.f A C l e w ~ n c T * 0 2 mmDetermine the si;m sf the bewing ad!he jo~fz#,op~(i)Hole Basis System (ii) Shafi BasisSystem. Take Unilateral System o tolerances.Sobtion. v

,t *. : .(a) Hole-Basis System : Refer pig. 9.13, - 2 -fLOW& limit of sd&r!hg =''&ifs Size - 1 0 0 dm'' ' '& "Highest limit iif ~ k r i n ~ ' ' & ~ L ; ~ .f Bearing +Tekrancd '

"@ 100+ b.005 = 100.005 mm33n814f~~ I P - ~ : ~ qqU = f i s h to nrnr; ~~twc.,efer to Fig. 9.2,

Higher limit of Jo u m d k: W& ' l i m i t of bearing - llowance.mar ~14KW0.002

-5 99.99811~11~ t z f 2 t d 116d? f r i lLower limit afd m a l = Migh.limit -T @ b a h ~s qq l ;43i11;.t410r .= 99,k)98-0.004; , ?L gqiI X W ~ ~

,,~m d .@@= 99.W1m(h ) Shgt$+@kJ rSI#[email protected] Fig-9.13Upper limjt sf.hucnal = Basic Size = 100 mm

Lower limti&


7/17

: Upper limit of shaft = Lower limit of hole +Maximum interferenceNow it is clear that, t, c . . ,>-. + L~ ' : , c 7

Maximum interference -Minimum interference = Up-r limit of shaft - Lower limit ofhole - (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 1, \, ' ,, : -.,\I t ra i tk \ b\r?f.rt Tolerance on shaft + Tofirance on hole "{' ""'"' "' ", . , ,.. . ... ,, . . / \ I $ 'J1112 :2T .;*1:$2ff$*l=-

2T = 0.12 - 0.05 = 0.07q9#,,,,, . % ,..,.%a,b)r I:. Tolerance ~p qbft = Tolerance on hole = 0.933 nun,.. Hi* l j i o fhole.r , Lower limit + Toleranqt. ,,I I,.,r,;m ~ ~ J I Ioi = i 0 1 ~ 00.035 mm

Lower limit of shaft = Upper limit - Tolerance'-= 100.QSS mm.

(b ) Shaft Basis System : iNQ Y .Upper limitdshaft = B d ~ c i z e.10Ornia;rnr! .Lower 'limit of shaft Upper [limit- Tolerance* 100 - 0.035 = 99.965 mmLower limit of hole - Upper limit ofu'-hxinilrrm ~ " '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 ofmterchangeablemathgpms (holes andshqfii), the averageallowance is 0.04 ~ l w nnd t k allowance must nd exceed * 0.012mm- the merage value.lRe bark*she 168.rrmr.l@ i e ~ & #;hdc+ $ ' ~ - t b b & d l f A shap. Lktermine the sizesof holes and Skofts using Hole h i s -~ 4 e mnd Unilateral system of tolerances.

Solution. Refer to Fig. 9.2 LOOmt :Maximum ahitmce --Mhu'mtarn allomot * l b h w c on shaft + Tolerance on hole

I, .n't'i>w.te. ?'amce &f\ =t 6.0011m. j5r:\,,~': I : N ~ t . '.',? 3u.b v., 7". p ~h f S n s . r . 3

, . . .. , . ': ),I .!.: ,.v:


8/17

mrw,mJ~'~&MU ~~~~.!hiitilB l m @ mrnGG w S I EO~ JR * x : kqA= 99.972 - 0.008 lil srij ':c

IWb3 & .,a&!$ Z l Jfl tr8 99.964 mmi I Wi fl R* slslfi~ 1 p i fk;c;i b9ntrfi r+ .(fSmsr .nmrsllm b KT .ntm (. . , wtt Lm riod &'*L enoivrvmib 'f { 3 1 - $3 -, $-~~IP~QPLEMS.nnosarrR m bhr

1. Define "Interchangeability" and dimmsiirs IiapbFtancdQ?2. Define tolerance. .&Iat ?~ [ ! F I J I L ~ris .T ; J J ~ . , L ~ ~ I t , 1

l 1 ~ f 3 ga 9% ibl if'~nwxsiblea btain.c#vexmd-.~w~ ~ l a ~ o lJ t .c.:.q r d C i nA .ISEa. i s q m r b ) . the w % 3 . wl&'&dt&csf 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~#sv31fr!!4~ 1tm: "!fl8. kxpranuun~ad&jj'&;ibitoI* ' ; B i k ~f i i '3U. .ST9. What is meant by H&,d&~*pl'&&q4'.~Lql 3mmm 3m 2-/;3 .LT

10. Explain and compare "Hole basis sy&~'sda'4biKdas( ' # t r d ~ * ~ g & % ~ ~ ~ ~ ~ ~ . p i ~iti%iIk- . lker&.dtoaveimum toral c FS #iWW$k ~W?&hi(tgtlil?Iautince &a69 rihrm. Wermine- the spccificaticrns of the partsk)i@ @Q& [email protected](&).for S r a t i W average

inmcbngesBil#y. I S : 11 2 ti , > L a$;>12. How will you write the Rt :is 40 rHm? .&4rq&$\* & basic sizei* 'qg ule &sieuria & * d - . m ** m amiwii+ arl f 04

0 mm H - h ~ & t h e o l k c ; gradeIR ULO+P;&!F&,.& t&ga'&i# ( ant f.O_ 0 1 s l 3 1 ~ 1 ~ l b )hdr 4 ilf($13 the above fig - - . , ., *',, -: Msiyt@$qns& e@jphq .liWof ze 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.1e to h c e g d i ~ b5i and f ~ ; ~ n i b e r'~uaIi&- I .,, , .itqa-l .+.15 A gear ring of85 mm diameter bore is died bn baa ubiresulting ih'a Ha6h ~ . l e ?tolerances and hence the limits of size fo r tht h b kt& he g& h e . @$@ttf;t type of fit.The diamcter steps are 80mm and 100mm. 'Ihc 8- fWj.&is 0.- mm.

16. Calculate the fbndamental deviations and tdgwn6~1~amlW: ,$Qr shaft andhole p i r - + & n a t t as 60 mm HimCThe tol-ce unit*s g j ~p l:I,. i= 0.45 + 0.00i D mi .scsn&~ 3!{~The diameter steps are 59 m,@ BB mw!ka: @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 ~lltowuxcsor a 25 -4% 'kid hole pair wigmted uH,ld,. The diameter steps arc 24 mrn snd 30 mm. The multipl(d.kt quality 8 .is25. The FDfor 'd' shaft is - 1 6 P Umicrons Name he type of fi& jh.9 ,,/ &8. Determine the types of fits p~odUCdaby the followingmaEg&holes anh


9/17

A Textbook&~ProductionEngineering19. A fi t isdesignated as :60mm H, h,;D b t e n r r ~ hh hi mum' &?&anc andm9rnum clearance

' " ' " - of the tit. .rn0 .d s$@:@ =' lo. A turned shak is to rotate in a reamed h * m a 8 ~ r d fit is HJc,. ~eterhinehe actualdimensions of the hole and the shaft. The ba#c size is 60 mm. The diameter steps are SO mm8nd 80 mm. The fundamental deMIdA.w]t&7.sh&c is,

- 95 a) i )wts :fhlFsl$~y*~ml.* suiifl .tFor quality 7, the muitiplicr is 16.21. h idler gear is to rotate over a mm s k i , The :-=fit 'L #p Dea?rmik(tttk actualdimensions of the shfdtPIlQ tb bPnt Q$ the i#k. ?w&mew, stcpgm 39:and 40mm. Thehndmental deviation for the hole H is ~ r o,?tQ3f;?r $9 sh& 5. q

I 8 :- 95 + 0.8 D)microns. . 0 *n;,*,pf:yy. h a "..&&:,+

Y*', - 2 , b ~ 6 r . W . d ,B;II$W . ~ ~ ln,~marmt9m'-16 *@y "twmdI&l ida#e @& :a ~ h & f f . ~ , m*6 mt'Itg,_~$w* :.*(6)Tha raigtive mor in the dimension is greater in &aft A. ,@lirfas#ti6rt.mtiii(;,at&&& &&&&js~" is g,+& "1)rr. i ~ f iw l *w .$I?tmw $+& ' :'(d)The rclatbe error in the dimension is greater in s W B. , . - , ,, !,::( (Am.: )d- -a) Tmsitien FI~J(@W&m~e F t (c) c l emec fit (&>Noat (GATE 1993) ( h a : )

7 eL-y7. h e it 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) orce fit [interference

L - 1 , % G#TF 1%) A=- : l*a nthq r p k i i o a, diq-7 > , d 9% ,, , . I ~ b w d w + g ~1n! *?da) alJowm is aqual t ~ h i l a t s F a l , C w . b .- k'm8.31~)kvanec is- to UftiMWlWeran&. I -,bTbrsfi Yc) efiowa~ceJMebed& eFtd&iwe. d U 3 .&I.

sion specified bythe tolerance. .a&% 998) [Ans. : lI 39. A journal md b a i n g aSsOLrrbly has he i l & # b g sizes%-9@ fmmb*

1' PI u m t q / I s~qqrsr& 4nr ,\ xu- a .@TI - TTi + --0.001*t. d ! ' , .- &f ' o2 - % n ~ t i d .(kt3 YIO *ltr.jffia:~ .f li : R WIJS~:~4 .W t r m ~ rm1 +0,001 &iiq@Bepriflg : SO 4.002 m% **Determine.: T o 1 6 n Journal, Toleran ,Maximum clearanceand type of fit. t W C $5).. i t (Ans :0.001 m%&003 m.0.001 n w W m, TmstioPL:fif.. . ?B


10/17


11/17

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.)


12/17

-..,c* . . A.. t.

, ', ,,,.: , -/ 4 . . 4 =, - ; , I * - 4 , ft 8 - & i u.. I 8 , I . 0 0 Bt3 I. 1 ^- r . - , . t + , ' , *t .+ I -.-.---.-.-.-. - ...-.-.-.-.-.-.-.-.-.-. -.-.-.- .-.-.-.-.-.-....--.-.-.-.-.--.- iti

, @' +$,

- lr I . . ; , $.?.%? 'ahFig. 10.40. Plug Gauge (combined type)(ii)lCap Gauge 'Go' side = H.L. f 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 mmb*i@-w(Fig. 10.41 shows a sketch of combinid 'Go' and 'Not Go' gap gauge)

/ , i . , c ; * '


13/17

Unilateral System+ 0.000

'Go' 75.018 mm- 0.004

Bilateral System+ 0.002

'go' 75.018 mm- 0.002,,.... .. 'Not Go' 74.98 mmrnrn F C 1 I) 3rl,rk , r., 1-i l tci x ~ r m ~ ! ~ ~ -- 6.002

k z -"Example 3. Find the 'Go' and 'Not Go' gauge dim&%; bTuafihg gauge m%g'8ilateraland 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 = 75.05t * ,m' 91o~i 0 t I l ! i i i ~1o.3Low limit of hole = 74.95 mm spue;> 0;)work tolerance = 75.05 - 74.95 = 0.1 mm si\\~x43ttt.,b s t . Oh (1 1Gauge maker's tolerance = 0.01 mm.s, ,r~,@i m # . ~ 1 \ b , & - & ~ r d ! a o 5 UO I US br1oq4~1m~lia 3 ~ ~ 2a

'm'ide of plug g a ~ g B A..c.'ofBole2~a.W'' *I2 -- - = ~74.9$hid6lie&r*$ .ofw & $ ~ ~ i + m ~ y ~ms%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% i i l

+ 0.010 n o i ? + b m 08 161 'sgusl) oi)' 'hp ~ - 0 0 0'Go' 74.955 mm mm.08 = %ot Go' 75.05 mm

- 0.000 w g mi-& lot 'quai3 03'a s~ - 0-010Bilateral System traS.0 +

tPs2&0-?Q5 :I m ~ n .B8 + 0.005'Go' 74.9H'mm W)O.O - 'Not go' 75.05 mm- 0.005 FFr ff f dq mi - c r 01 -0.005J , , . * .

Example 4. The rectangular hole shown in Fig. 10.43 @ a be checkedThe "mi* of size for the 'wt?& "+ 0.0480 m m- 0.00 4.04Design the suitabb gauges ?used on Taylor's * -0.00 mprinciple. W m

Solution. According to ~aylor'srincipfe, there will a - - Fig.,10.42 -.,- - - -+e one 'Go Gauge' of full form@a@length e q dto the ;'- - . "J:,length of the hole. There will be'two 'NotGo Gauges;' of p&Z$rm to check the widthand breaMhof the hole. , *vT.=.* &-5 !I


14/17

A Textbook of f k l ~ t i o nnginewingHigh limit of width of hole = 60.04 mmLow limit of width of hole = 60.00 mmHigh limit of breadth of hole = 80.05 mmLow limit of breadth of hole = 80.00 mmTolerance on width of hole = 0.04 mmToleran~e n breadth of hole = 0.05 mm

,,.,, Gauge marker's t o h n c e (10%) : . , I 1: L L . ~ . t ~~!qt::?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 tt ! I;


15/17

Not Go Gauges (Refer Fig. 10.7)'4 7, ' The 'Not Go Gaugesywill correspond to the mhimum met?! cuwjitim of hole i.e. highlimits of dimensions.. ,,: Basic she of 'NoT Go Gauge' for 50 mm dimwshn n r i 1~3.: I ?:;me : ~ r c l r ? c .-

= 60.04 rn : .-w :li . I Q I Z ~ : .arg: FOC

: 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 80Y*imension. are ::6 ?-1. . .. . : - . 4 . ' ir: .. i. c .,- -*. . .-- .. . -..... lo \ 'i~fq~f;:+i-ilj l ?+awl: + ( p i -

1. What is a gauge? O + 81,807 &tp pi3 42. 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' .9joll(,.,*. 6 , I l . .,. ,v4. What is the di&tonce between standard gauge (non-limit page) and a limite?airid31 ' " I , - .,.- ) I 2 , .It ' I , T1 % ..ltadZ .+Q : H : # Q P , P ~ i ~ p w i i ~ niie ~d - 0w0. *'94: ira 6. Whath mu@ ~ d W sdimnce?5Mdwis 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 i b itragtrlied i n the4dcsignofgitiges? z, ,9. State and explain Taylor's principles f &it gauging., , , I& Discuss th e .vllrim$nar*erbksu&?d forgaw imw8bEture.

11. What are the advantages of limit gauges'? ..- - , . ,~ U P P 2, we kbt limitations of Ilma gauges'? .c - ' 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 d15. Determine the specification &$ the 'gdlv a d n6r 'ga""eiids'of a set of manufacturing andinspection 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


16/17

002 ' . $16. A (haft I00 + mm diameter is to be checked using a 'go-not go' sna! F g e . Give itsayr!! 3. dimensions. Anow for wear &d b g e mak&rls bikrance. r %Yaitai jC Y ~ $ r SP!? , i < l i r - f . , f , -:, , I l l , :17. A limit gauge is required a @he&the hPle 50 : (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 he 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- r ; >\!.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. mmi = sanw tolerance, micron

The multiplier for grade 8 is 25.'The fund&ental, d,yiation for shaft c for D > 40, s given by- 95 + 0.8 D)The diameter range lies between 50 to 80 mm. %ketchthe fit and show on it the actualdimensions of hole and shaft. b e he class aP fit. Also, design the suitable gauges to checkthe hole and the shaft (AMIE 1974 S)[Ans. - - le :LL, = 60.000 mm, H.L.= 59.046 mm

' Shaft : H.L -59.854mm, LL.= 59.808 rnm, Clearance fit.. . -- --- - - ' I. .'- " - - - - (j";io5;plug ~aude, O& ; 0 * 0.005~ m, bd '40.000Not Go side : 60.04~ 0.005 mm CL *.

Snap Gauge, Go side 59.854f k gNote. Unilateral system has been used. Wear allowance ha s 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 matchingshaft ,8fwrldldis& he fbnMrZal Beviation is found to be - 0.02 mm. Shafttoler&W@ O H 3 m. D d g a &es for hde snd.'shft. TaPce the usual valiles.of gaugeqabfs, o k w e and wear Jlowyrcc.

21. Design 'GO' aryl 'W.GQ;,wls,&LpIqg g ~ ~ t g ~ io wqwm%hd@f sb 28.4XW * %O 14 mmadopting (a)Unilateral system (b)' Bilateral system.8 , . l > t' >

22. A bore of : mm dia x 43 nt t~~ong is to be &&A. Iksig@'di lZwd~~&lImknsior a pluggauge 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 ~pfwge dyign. ,, .,,, - .

24. Discuss the principle of Taylor's for the design of gauges for checking i , , a dXi r IS , ( ~ j . ~ t ~ . ~ ~ d016with a c~rilylricalNot ,Go' g+ge. 1 ~ 3 i t t ~ ~ r ' o , . ~ ~ ~ ~ . , ~ ~t

(b) of rc&ngular hole rw xuri g y&;m&, or ~ a z u a 0 1 b @ ~ : : g 1~!r? urr,x:ln.(c) circular holes- ? i r J u(d) circular shafts. rrtnt ?r ?c. itot~:i3r!1?3~jd(e) Non-circular holes and shafts. 'OC ' vwi-"


17/17

Gauges and Gauge Design25. Sketch and discuss various types of Plug gauges,.a.ketch 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 nap geugesw? --. . - - - - - . - -32. Sketch and discuss the ,mf following gauges :

(a ) Length gauges Y ~ C ) ~ - ~ ~ V U ~ . I H ~i . !- - -.,"!I I I t;t (c) Receiver gauge Ern s f l .b.~nrardo 1 : ~ : . :$I.J tm-07 issrr~w~vs$33!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@ pin w % n p-scr' , t to i51 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?

% f i r r . , f - . hi To ml.1 F r : . I J - ~ > :3 ~ 5 6 k ' h \ h * m g. ~ r j r t r ~ 1 ofptsq/&bO d%o. , . , a luaa;l:, . A w o r ~ . ~ ~ & i t a ~ ~ ~ ~ k b . ~ ~ - f r ~ . r i ~ ~) - s ; ~ 'I d*. jkpikge914 G&uhte.'L' &em 49 -L58j,Max. did, = 23.42 nnn,min., I , - jAlfigqia 7 1 3 . 4 Q t ~ . . , [ b h .* (D - dY2 an @In; 0 . M ~ m ]1 ~ 5 1 37. w b n wearfar gar$eslakrgsni v. . z ! f i z l w s i t~ J! a: bshrcyZ-,-- ._ . - I 1 . I .-." 313: S.I. fi='va&us t n h k I+,I+'- mWI& n .0.30), Special W=mamiets lir q $ k t i m m W'&kr"~ iL~ts r i k very rapidly ctuc towear (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 rit abrasive, though veryagrasllp. iqmvkl wcar 'life. .-- *3I6tr&-38. W r b the -UWdvantagesof limit Gauging.Sol. Two limitatiorrs_oiLliaitG-Aw ,ksn krseed & Art 10.8. The otha drawbacks.. .ntr;tr,;, trt nr-M* nomi c ? f a e i ~ p - , I , :. . . .,

-> ! J B V ~ C (a) .Suitebbfbf Maw~pmitiQI of btjc Theit c a ~ yfmbe IB ana : 11 ' a5 ~ i ~ ~ ~ o f d.> l b n 6 2 : , . tl3OSg 12.)TT&Yl!f?l Sf*~ f ! )I (by fie>* of limitgaw h sizesI F 1 . :con&&&, s s&H7asTWght~fb$& i w r f l m e nwg16 1 ~ 1 ,.-.,,,,,,, fi,&'kiih MY iimitititWs Ep~b S~ ~ * + ~ ~ ' ~ , ~ e ' ' @ ~ , d ~ ~ w ~ d u ~ g b ~~ , . , ~ ~ j , , ; a : x : R ~ l , p Jtur: ;!j !-33 a(d) Particular sources bf error in themampnukt an~~ rrvs(rk$R e & w m s in machine>(j

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