chuong 4 chuẨn- công nghệ chế tạo máy
DESCRIPTION
cong nghệTRANSCRIPT
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BI GING CNG NGH CH TO MY
H ni, 1/2015
TS. Trng c Phc
B mn Cng ngh ch to my
Vin C kh - HBKHN
T1
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Cng ngh Ch to my
Chng 4. Chun
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I. Khi nim
Khi gia cng chi tit cn c mt v tr xc nh so vi my hoc g. Chi tit c cc b mt:
B mt nh v: 1,2
B mt kp cht: 3
B mt gia cng: 4,5
B mt t do: 6,7,8
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B mn Cng ngh Ch to my Vin C kh BKHN
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II. nh ngha
Chun l tp hp nhng b mt, nhng ng, nhng im m ngi ta cn c vo xc nh v tr ca cc b mt, ng hoc im khc ca bn thn chi tit hoc ca chi tit khc
Nh vy, chun c th l b mt, ng hoc im
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III. Phn loi chun
1. Chun thit k
2. Chun cng ngh
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1. Chun thit k
L nhng b mt, ng hoc im c dng thit k chi tit
Chun thit k c th l: chun thc hoc chun o
Chun thc (mt A) Chun o (im O)
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B mn Cng ngh Ch to my Vin C kh BKHN
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2. Chun cng ngh
Chun cng ngh chia ra:
Chun gia cng
Chun lp rp
Chun o lng (chun kim tra)
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2.1. Chun gia cng
Nu g t t ng t kch thc H th mt A lm hai nhim v: mt t v mt nh v (a)
Nu g chi tit theo ng vch du B th mt A lm nhim v mt t (b), cn chun nh v l ng vch du B
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2.1.1. Chun th
L chun cha c gia cng
Chun th cn phi bng phng, khng c g gh, khng c h hng
Ch trong trng hp phi a vo xng dng gia cng s b th chun th mi l nhng b mt qua gia cng. Nhng trng hp ny thng gp trong ch to my hng nng. , cc chi tit rn ln chuyn n qua tin th vi mc ch pht hin ph phm khi to phi, vn chuyn v gia cng d dng.
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B mn Cng ngh Ch to my Vin C kh BKHN
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2.1.2. Chun tinh
L chun qua gia cng. Chun tinh chia ra:
Chun tinh chnh (chun dng khi gia cng v lp rp-l A trn hnh a)
Chun tinh ph (chun chi c dng khi gia cng m khng dng khi lp rp- mt 1 v l 2 trn hnh b, mt l tm trn hnh c)
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(a) (b)
(c)
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2.2. Chun lp rp
L chun c dng xc nh v tr tng quan ca cc chi tit khc nhau ca mt b phn my trong qu trnh lp rp
Cc bnh rng, bcchun lp rp l l v mt u
Chun lp rp c th trng vi mt t lp rp cng c khi khng
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B mn Cng ngh Ch to my Vin C kh BKHN
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2.3. Chun o lng
Chun o lng (chun kim tra) l chun cn c vo o hay kim tra kch thc, hnh dng hnh hc hoc v tr tng quan
Trong thc t c khi chun thit k, chun gia cng, chun lp rp v chun o lng trng nhau hoc khng trng nhau
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VD: khi kim tra ng tm ca cc bc trc, thng dng 2 l tm lm chun, chun ny l chun kim tra
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IV. Qu trnh g t chi tit
Mc ch ca qu trnh g t???
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IV. Qu trnh g t chi tit
1. Qu trnh nh v
2. Qu trnh kp cht
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1. Qu trnh nh v
Xc nh v tr chnh xc ca chi tit: hnh a: chi tit c nh v bng mt B t kch thc H, gc KT l bn my hoc b mt nh v. G t trn mm cp 3 chu t nh tm: sau khi a chi tit ln mm cp, vn cho cc chu tin vo, sao cho tm chi tit trng vi tm my. Lc ny qu trnh nh v kt thc.
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B mn Cng ngh Ch to my Vin C kh BKHN
B
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2. Qu trnh kp cht
L qu trnh c nh v tr ca chi tit sau khi nh v chng li tc dng ca ngoi lc (ch yu l lc ct) trong qu trnh gia cng lm x dch chi tit sau khi nh v
Qu trnh kp cht xy ra sau qu trnh nh . v, ch khi no qu trnh nh v kt thc th mi bt u qu trnh kp cht. Khng bao gi lm ngc li.
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B mn Cng ngh Ch to my Vin C kh BKHN
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V. Cc phng php g t chi tit
1. Phng php r g
2. Phng php t ng t kch thc
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1. Phng php r g
C hai PP r g: r trc tip trn my v r theo ng vch du. Theo PP ny th cng nhn dng mt vi dng c nh bn r, mi r, ng h so hoc ng knh quang hc (trn my doa ta ) xc nh v tr ca chi tit so vi my v dng c ct
PP r g c dng trong SX nh hay n chic hoc trong nhng trng hp mt phi qu th khng th dng g c
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1a. u im ca PP r g
C th t CX cao (dng ng h chnh xc v tay ngh cng nhn)
Tn dng mt s phi khng chnh xc phn b u lng d
Loi tr nh hng ca mn dao
Khng cn g phc tp
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B mn Cng ngh Ch to my Vin C kh BKHN
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1b. Nhc im ca PP r g
Tn thi gian cho r g
Bc th phi cao
Khi r theo ng vch du s c sai s v ng vch du c kch thc
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B mn Cng ngh Ch to my Vin C kh BKHN
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2. Phng php t ng t kch thc
Theo phng php ny th dng c ct c v tr tng quan c nh so vi chi tit (v tr iu chnh). V tr ny nh cc c cu nh v ca g m bo
Khi gia cng theo phng php T t kch thc th my v dao c iu chnh trc
Phng php c s dng trong SX ln
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B mn Cng ngh Ch to my Vin C kh BKHN
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2a. /im ca PP t ng t k/thc
m bo CX gia cng, t ph thuc vo tay ngh ca cng nhn
Sau khi g, ct mt ln l t kch thc
Nng sut gia cng cao
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B mn Cng ngh Ch to my Vin C kh BKHN
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2b. N/im ca PP T t kch thc
Chi ph tin v thi gian cho vic iu chnh
Chi ph cho vic ch to phi chnh xc
Chi ph cho vic ch to g
Nu dao mn nhanh s nh hng n kch thc gia cng
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B mn Cng ngh Ch to my Vin C kh BKHN
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VI. Nguyn tc 6 im khi nh v chi tit
Mt vt rn trong khng gian c 6 bc t do khi ta t vo h ta -cc (h ta khng gian 3 chiu). 6 bc t do l: 3 bc xoay xung quanh 0X, 0Y v 0Z; 3 bc t do tnh tin dc trc 0X, 0Y v 0Z
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VI. Nguyn tc 6 im khi nh v chi tit
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1. Mt s qui nh v s bc t do
Mt mt phng hn ch 3 bc t do
Khi V di hn ch 4 bc t do
Khi V ngn hn ch 2 bc t do
Cht tr di hn ch 4 bc t do
Cht tr ngn hn ch 2 bc t do
Cht trm hn ch 1 bc t do
Hai mi tm hn ch 5 bc t do (g trn my tin hoc my mi)
Mm cp 3 chu t nh tm hn ch 4 hoc 2 bc t do ty tng trng hp
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1. Mt s qui nh v s bc t do
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Mt s qui nh v s bc t do
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Mt s qui nh v s bc t do
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Mt s qui nh v s bc t do
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Mt s qui nh v s bc t do
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B mn Cng ngh Ch to my Vin C kh BKHN
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2. Siu nh v
Trong trng hp mt bc t do c hn ch qu mt ln th s sinh ra siu nh v
a) Mt phng v cht tr di (3+4=7). Nh vy l siu nh v
b) Nu ta c tnh cho mt phng tip xc vi mt t thi cht nh v s b b cong
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B mn Cng ngh Ch to my Vin C kh BKHN
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VII. Nguyn tc chn chun
Mc ch chn chun:
m bo cht lng gia cng
Nng cao nng sut v h gi thnh
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Chn chun cho chi tit dng hp
-Trng hp c c ta ly A lm chun gia cng l 0, sau ly 0 lm chun gia cng A, cui cng ly A lm chun gia cng B
-Trng hp l c rng: ly l lm chun gia cng A ri sau ly A lm chun gia cng B v 0
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B mn Cng ngh Ch to my Vin C kh BKHN
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2. Chn chun cho v ng c in
Khi g/c v ng c in phi m bo dy ca thnh u n. Nu ly l A lm chun th g/c mt y C, sau ly C lm chun g/c l A s m bo ng tm gia l A v mt B. Khi khng c g th ly du l A m bo thnh l gia A v B c b dy u n. Lm nh vy chnh l ly l A lm chun nh v
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B mn Cng ngh Ch to my Vin C kh BKHN
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3. Nguyn tc chn chun th (1)
Nu chi tit c mt b mt khng g/c th chn b mt lm chun th. V d: Hnh a: ly A lm chun th g/c cc b mt B,C,D. Hnh b: ly mt trong khng cn g/c lm chun th g/c mt ngoi, nh vy m bo ng tm gia mt ngoi v mt trong
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B mn Cng ngh Ch to my Vin C kh BKHN
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3. Nguyn tc chn chun th (2)
Nu chi tit c mt s b mt khng cn g/c th ly b mt khng g/c no c yu cu CX cao v v tr tng quan i vi b mt g/c lm chun th. V d: Cc mt A v B u khng cn gia cng. ta chn mt A lm chun th g/c l. Nh vy, s m bo ng tm gia l v mt ngoi
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B mn Cng ngh Ch to my Vin C kh BKHN
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Chi tit c nhiu b mt cn g/c thi chn b mt no c lng d u v nh lm chun th. V d: g/c bng my tin: ly B lm chun g/c mt A, sau ly A lm chun g/c mt B v B khi c nm na khun di nn c lng d u v nh hn.
3. Nguyn tc chn chun th (3)
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B mn Cng ngh Ch to my Vin C kh BKHN
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3. Nguyn tc chn chun th (4)
Chun th phi tng i bng phng, khng c g gh, u rt, u ngt
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3. Nguyn tc chn chun th (5)
Chun th chi c dng mt ln. V d: khi gia cng trc bc, nu ly mt 2 lm chun th g/c mt 3, sau li ly mt 2 lm chun g/c mt 1, nh vy s khng m bo ng tm gia 1 v 3
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B mn Cng ngh Ch to my Vin C kh BKHN
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4. Nguyn tc chn chun tinh (1)
Nn chn chun tinh l chun tinh chnh cho v tr ca chi tit khi gia cng v khi lp rp trng nhau. V d: i vi bnh rng tr ta chn l A lm chun tinh chnh v khi g/c v lp rp u dng l A
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B mn Cng ngh Ch to my Vin C kh BKHN
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4. Nguyn tc chn chun tinh (2)
Chn chun nh v trng vi gc kch thc cho sai s chun bng 0
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B mn Cng ngh Ch to my Vin C kh BKHN
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4. Nguyn tc chn chun tinh (3)
Chn chun sao cho khi g/c chi tit khng b bin dng do lc ct v lc kep. Mt chun phi c din tch nh v
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4. Nguyn tc chn chun tinh (4)
Chn chun sao cho kt cu ca g n gin v thun tin khi s dng
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B mn Cng ngh Ch to my Vin C kh BKHN
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4. Nguyn tc chn chun tinh (5)
C gng chn chun tinh thng nht (chun c dng nhiu ln trong qu trnh gia cng)
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B mn Cng ngh Ch to my Vin C kh BKHN
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VIII. Cch tnh sai s g t
Sai s g t l tng vc-t ca 3 thnh phn:
Sai s chun
Sai s kp cht
Sai s g
Gi tr ca sai s g t:
_
gd dgc k
2 2 2
gd c k dg
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Sai s chun (1)
Sai s chun c: l lng bin ng ca kch thc thit k theo phng kch thc thc hin. Sai s chun xut hin khi chun nh v khng trng vi gc kch thc
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1. Sai s chun (2)
a) Gia cng mt N: chun nh v l K, kch thc A c gc l K v vy c(A) = 0 (Hnh a)
b) Sai s chun ca kch thc B: c(B) = H (Hnh b)
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1. Sai s chun khi /v trn khi V (3)
Tnh sai s chun ca H1,H2,H3:
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1. Sai s chun ca H1 (4)
Sai s chun ca H1:
Sau khi bin i ta c:
1max max min min
1 1C H
D D D DBB CB CB (C0 0B)
2 22sin 2sin
2 2
1C H
1( 1)
2sin
2
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Sai s chun ca H2 (5)
Sai s chun ca H2:
Sau khi bin i ta c:
2 1 1 1 1C H
max max min min
AA CA CA (CO 0A) (C0 0 A )
D D D D
2 22sin 21sin
2 2
2C H
1( 1)
2sin
2
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Sai s chun ca H3 (6)
Sai s chun ca H3:
Vy:
3max min
1 1C H
D D00 C0 C0
2sin 2sin2 2
3C H
2sin2
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Sai s chun ca H3 (7)
Chi tit c nh v trn hai mi tm
C(a)=0 v kch thc A c nh c iu chnh sn. Sai s chun C(b) 0 v kch thc b khng c nh. C l kch thc t mt gia cng ti mi tm (C=constant). Sai s chun C(D) 0 v c th ly bng 0,25D (D khng chu nh hng ca hai l tm, nhng nu hai l tm lch nhau s gy ra sai s chun)
Hnh b: Sai s chun ca cc kch thc a v b u bng 0 v dng mi tm ty ng (mm)
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Sai s chun ca H3 (8)
Sai s chun ca kch thc b c tnh nh sau:
max min max minC
d d d d(b) ( )tg(90 ) ( )ctg
2 2 2 2
Do sai s chun ca kch thc b:
d dC(b) tg(90 ) ct g
2 2 2 2
d- dung sai ca
kch thc l tm
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1. Tnh s/s chun bng lp chui KT (9)
PP lp chui kch thc tnh sai s chun (PP cc i-cc tiu): chui kch thc xut pht t b mt g/c qua cc khu c nh, bin i ri li quay v b mt gia cng. Nh vy, s hnh thnh chui kch thc khp kn
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1. Tnh s/s chun H1 bng lp CKT(10)
Ta c phng trnh sau:
a b + c - d = 0
d = a b + c
d = a D/(2sin/2) + D/2
Vi phn 2 v ta c:
1
,,
,
c(H )
D D D 11
2 22sin sin
2 2
11
2sin
2
d
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Tnh s/s chun H2 bng lp CKT(11)
Ta c:
m n p + q = 0
q = m n p =
= m D/(2sin/2) D/2. Sau khi bin i ta c:
2C H
1( 1)
2sin
2
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B mn Cng ngh Ch to my Vin C kh BKHN
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1. Tnh s/s chun H3 bng lp CKT(12)
Ta c phng trnh:
x = z + y = D/(2sin/2)+z
z = x - D/(2sin/2)+z
Vi phn z ta c:
c(H3) = /2sin/2
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B mn Cng ngh Ch to my Vin C kh BKHN
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2. Sai s kp cht k (1)
Sai s kp cht k l lng bin ng ca kch thc do lc kp gy ra
k = (ymax ymin).cos
ymax , ymin bin dng ln nht v nh nht do lc kp gy ra; - gc gia phng ca kch thc v phng ca lc kp
Sai s kp cht bng 0 khi lc kp vung gc vi kch thc
Sai s kp cht ln nht khi lc kp v kch thc // vi nhau (trng phng)
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3. Sai s g dg
Sai s g bao gm:
Sai s ch to g ct
Sai s mn g m
Sai s iu chnh g dc
m = 0,18 . .Vi N- s c/tit g trn /g
dc = 5~10 m
Sai s ch to: ct
Ta cn xc nh sai s ch to ct t yu cu k thut ca g
N
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4. Sai s g t gd
Sai s g t:
T :
Ghi ch: tr s di cn phi > 0. Trong trng hp khng c ta tm cch gim cc thnh phn v th 2 trong ngoc n. Chn sai s g t g=(1/3~1/5)nc
2 2 2 2 2 2 2 2
gd c k dg c k ct m dc
2 2 2 2 2ct gd c k m dc
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B mn Cng ngh Ch to my Vin C kh BKHN
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5. iu kin k thut ca g
Khi c sai s ch to ct ta cn c vo gi tr ca n t yu cu k thut ca g
V d: ct = 0,03mm. Khi ta ghi cc iu kin k thut ca g nh sau:
khng song song gia b mt phin t v y g 0,03mm
khng vung gc gia tm bc dn v y g 0,03mm
khng trng tm gia hai khi V 0,03mm
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B mn Cng ngh Ch to my Vin C kh BKHN