buoi 3

TRNG I HC KINH T K THUT CNG NGHIP

Cng ngh ch to my

(Manufacturing Engineering and

Technology)

Kim tra bi c:

Cu 1: V nh hng ca nhm b mt ti tnh chng

mn ca CTM, quan nim no sau y khng ng :

A. nhm ban u ca cp chi tit i tip khc nhau th

tc mn ban u khc nhau.

B. Qu trnh mn ca cp chi tit tip xc tng khi nhm

b mt chi tit thp.

C. Chiu cao, hnh dng ca cc nhp nh t vi trn b mt

c cng chiu vi vt gia cng.

D. Qu trnh mn ban u, mn bnh thng, mn kch lit

ca cp chi tit tip xc ph thuc vo nhm b mt

2

Kim tra bi c:

Cu 2: Quan nim no sau y khng ng :

A. sng b mt l chu k khng bng phng ca b mt

chi tit my c quan st trong phm vi t 1 100 mm.

B. Tnh cht c l lp b mt chi tit l ch tiu quan trng

ca chnh xc gia cng.

C. ng sut d ca b mt chi tit my c quan st

trong phm vi t 0,01 1 mm.

D. nhm b mt l sai s ca b mt thc quan st

trong phm vi nh 1mm2.

3

Kim tra bi c:

Cu 3: nh hng ca cht lng lp b mt ti

chnh xc cc mi lp ghp nh sau:

A. chnh xc cc mi ghp c kh cng cao th chiu

cao cc nhp nh t vi cng ln

B. Chiu cao nhp nh t vi Rz tng th bn mi ghp c

di (lp ghp cht) gim.

C. nhm ca cc b mt lp ghp c nh hng ti

bn, n nh cc mi ghp c kh.

D. Cht lng lp b mt vt liu ph thuc vo chnh

xc mi ghp trong kt cu c kh

Chn sai

4

Kim tra bi c:

Cu 4: Kh nng t nhn bng b mt ca phng

php tin cn c vo tr s Rz (m), c cho nh sau:

A. Tin tinh vi dao hp kim cng : Rz = 2,5 10

B. Tin bn tinh : Rz =10 40

C. Tin bc v (tin ph) : Rz =1 2,5

D. Tin th: Rz = 40 100

Chn sai

5

2.2.1. Khi nim

6

2.2 chnh xc gia cng

CX GC l ch tiu kh t v gy tn km nht k c trong

QT xc lp cng nh trong QT ch to

L mc ging nhau gia chi tit l tng trn bn v

thit k v chi tit thc c gia cng

7Thc t, dng gi tr sai lch nh gi CX GC.

Sai lch GC

SLch 1 C/tit SLch1 lot CT

SL Kch thuc SL BM CTit Tng sai s

Kc

h thuc

V tr

t.quan

H.d

ng h

.hc

Sng b

m

t

Nhm

b m

t

T.c

c l lp B

M

SS

h thng

SS

ngu n

hi

n

8 Sai s h thng khng i:

- Sai s l thuyt ca PP ct.

- Sai s ch to ca my, g, dng c v.v

- Do s bin dng nhit ca chi tit

Sai s h thng thay i:

- Dng c ct b mn theo thi gian

- Bin dng nhit ca my, dao, g

Tnh cht ca sai s GC:

9 Sai s ngu nhin:

- Tnh cht ca VL GC khng u

- Lng d GC khng ng u

- V tr ca phi trong g thay i Sai s g t

- S thay i do ng sut d

- Do mi dao v g dao nhiu ln

- Do thay i nhiu my GC mt chi tit

- Do dao ng nhit ca ch ct

Tnh cht ca sai s GC:

12

3

4

5

2.2.2. Nguyn nhn gy sai s GC.

7

6

NN do chnh xc cua MCC

NN do dng c ct.

NN do BD n hi cua HTCN.

NN do g t chi tit.

NN do bin dang nhit v ng sut d.

NN do rung ng cua HTCN.

NN do dng c o v PP o.

10

1. NN do chnh xc cua MCC

chnh xc ch tao MCC.

11

MCC thng c nhng sai s:

o hng knh ca trc

chnh.

o ca l cn trc chnh.

o mt u ca trc

chnh.

Cc sai s ca cc b phn

khc nh sng trt, bn

my

nS

n

S

a b

12

chnh xc ch tao MCC.

a) Tm trc chnh khng song2 sng trt trong m.p ngang

b) Tm trc chnh khng song2 sng trt trong m.p ng

nn

a b

S S

A

A A A

13

chnh xc ch tao MCC.

a) M.p GC khng song2 m.p y chi tit

b) Trc chnh khng vung gc vi bn my theo phng dc

mon cua MCC.

o Khng ln vi may co

tc mon chm (tr

bng may, ban trt..).

o Sng trt thng

mon nhanh hn vi chu

lc ln hn, do o lam

dao b h.

14

2.Nguyn nhn do dng c ct.

Do ch tao.

C nh hng ln n CX GC, c bit l dng c

nh hnh v nh kch thc.

15

Do mon.

u

D

Dt

16

III

III

U (m)

a

L(Km)Lb 2LHL

HU

U2

17

Do mon.

o Giai on ban u I:

dao mn nhanh

o Giai on hai II: dao

mn bnh thng

o Giai on III: dao

mn kich litQuan h mon dao U v chiu di ct L

Do g t dao.

Vic g t dao trn

my khng chnh xc

cung gy ra sai s GC

18

3. NN do BD n hi cua HTCN.

Bin dang n hi cua HTCN.

19


V d: Tin trc

R

R+z

y

R

zyx

Trong :

X - lng chuyn v ca dao theo

phng chy dao x.

Y - lng chuyn v ca dao theo

phng php tuyn vi b mt

GC

Z - lng chuyn v ca dao theo

phng tip tuyn vi b mt GC

R y

y

PJ

y (KG/mm)

J

PyR

y

20


Gi =1/J l mm deo ca HTCN, i mm

deo ca chi tit chu lc th i trong h thng. Ta c:

= 1+ 2 +... + n

yPyR .

21

Trong thc t, cng vng ca HTCN c xc

nh theo CT:

CTDM JJJJ

1111


CTDM

Hoc mm deo c xc nh theo CT:

Trong : JM, JD, JCT, M, D, CT l cng vng;

mm deo ca my, dao v chi tit GC.

22

yy

l

l/2

Pyy

us

d

ut

yus= Py /2Jus.

yut= Py /2Jut.

yct = Py /Jct.

yd = Py /Jd

Anh hng cua BDDH n chnh xc GC.


)(2

1usutdct

yyyyy

)

11(

4

111.

usutdct

yJJJJ

PRy

23

A

A1

No

S chuyn v ca mykhi tng v gim lc


Sai s in dp.

Do phi c sai s HDHH nn sau mi ln ct se gy sai

s cng loi trn c/tit GC, tuy nhin tr s se gim i

nhiu ln. Hin tng sai s v HDHH ca phi gy nn

sai s cng loi trn chi tit GC gi l sai s in dp.

ph

ct

idK

Trong :

ct: sai s ca chi tit sau mi ln ct.

24

ph: sai s ca phi.

Nu qu trnh GC gm n ln ct th sai s in dp

tng cng xc nh theo CT:

n

i

idiidKK

1

n

i

idiphoiidphoictKK

1

Nhn xet:

Nh vy, sai s GC sau n ln ct c xc nh

theo CT:

Trong :

n - s ln ct hoc s bc GC.

25

Kidi - h s in dp ln ct th i.

Gia cng chi tit nh hnh ve

Nguyn nhn gy ra sai s in dp.

D

D

D

y

y

ph

min

ctmin

min

min

t

ct

ma x

ph

max

max

maxt

D

26

V tr s:

4. NN do g t chi tit.

G t chi tit l nguyn nhn gy sai s GC.

Sai s do g t chi tit c xc nh theo CT:

dgkcgd

222

dgkcgd

27

5. NN do b.dang nhit v SD.

a. Bin dng nhit ca my v g.

b. Bin dng nhit ca dao.

c. Bin dng nhit ca chi tit GC.

d. nh hng ca ng sut d.

28

6. NN do rung ng cua HTCN.

gim nh hng ca rung ng n CX GC, c

th s dng cc bin php:

S dng my v cc kt cu my c cng vng cao.

Chn s g t hp l nng cao cng vng

ca HTCN.

Trnh GC cc b mt khng lin tc.

Gim cc rung ng t bn ngoi ti...

29

7. NN do dng c o v phng php o.

30

2.2.3. PP am bao c/xc GC trn my cng c.

2.2.3.1. Phng php o do ct th.

1

2

3

n

S

31

u im:

C th t CX GC mt cch ch ng:

+ Khi mun t CX cao: s dng thit b o c

CX v th c tay ngh cao.

+ Nu CX GC khng yu cu cao: s dng cc

thit b o thng thng v th c tay ngh thp.

C th loi tr nh hng sai s do mn dao

n CX GC.

Khng cn g phc tp.

32

Nhc im:

CX GC ph thuc tay ngh CN v b gii hn

bi chiu dy lp ct b nht.

Do tp trung cao nn th chng mt mi do

d gy ph phm.

NS rt thp, gi thnh cao.

33

2.2.3.2. Phng php chinh sn dao.

CX c m bo

nh DCC c v tr tng

quan c/xc so vi c/tit

GC hay c/tit GC c mt

v tr tng quan c/xc

so vi DCC

V tr ny c m bo nh nh v ca g.

QT c thc hin cho c lot chi tit GC.

34

w

s

w

n

L

2.2.3.3. PP khao st CX GC.

1. PP thng k kinh nghim.

L CX t c mt cch kinh t nht trong iu

kin SX bnh thng.

a. Khi nim v CX bnh qun kinh t.

/kin SX bnh thng c nhng c im:

o Thit b GC hon chnh.

o Trang b cng ngh t yu cu v cht lng.

35

o S dng bc th trung bnh.

o Ch ct, nh mc k thut theo tiu chun.

u im:

CX GC t c n nh, ko ph thuc tay ngh CNv ko ph thuc vo chiu dy ct b nht.

Ch ct mt ln l t k/thc nn NS cao.

36

i hi g phc tp do ch hiu qu khi snlng GC ln.

Phi cn m bo mt c/xc nht nh.

mn ca dao se nh hng rt ln ti CX GC

Nhc im:

Pham vi s dng:

Thng c s dng trong SX lot ln hng khi.

C s ca PP l CX bnh qun kinh t. T thc t

SX c rt thnh kinh nghim, cc s liu c

thng k v c a vo cc bng trong s tay CN

CTM.

b. Phng php thng k kinh nghim.

37

u im: S dng n gin, nhanh chng.

Nhc im: khng st vi thc t, ngi cn b

phi c trnh nht nh th mi tra cu v x l cc

kt qu trong bng mt cch hp l.

2. Phng php thng k xc sut.

GC th lot 60 100 c/tit, kho st v tm quy

lut xut hin sai s GC. Coi quy lut chnh l quy

lut xut hin sai s trn lot i tr vi /kin mi

yu t CN c gi nguyn nh khi th.

38

u im: kt qu st vi hin trng.

Nhc im: tn km cc chi ph trong vic GC

th. Mun tin cy cao th s c/tit th phi nhiu

do cng tn km.

a. Kho st CX GC bng ng cong phn b l lun.

Ni dung PP:

39

Thc t thng s dng PP thng k x/sut sau:

Gia cng th lot 60 100 chi tit.

XD ng cong phn b thc ca lot th.

Nhn dng ng cong v tin hnh XD ng

cong phn b l lun.

T ng cong phn b thc, tin hnh nhn

dng ng cong. Nu t CX GC bng chnh

sn dao th sai s GC thng phn b theo quy

lut chun. Nu dng ca ng cong phn b

thc ging vi dng ng cong ca quy lut

chun th ta s dng cc phng trnh ca quy

lut chun XD ng cong phn b l lun.

Cch xy dng ng cong phn b l lun:

40

ng cong phn b thc:

KThuoc

m

n

1

23

4

5

6

7

8

9

1011

L

L

min

max

L

41

Xc nh trung tm phn b theo CT:

Xc dnh chiu rng khong phn tn 6 vi

c xc nh theo CT:

n

L

L

n

i

i

tb

1

21

2)(

n

LLn

i

tbi

42

Cch xy dng ng cong phn b thc :

u im: Qu trnh tnh ton n gin.

Nhc im:

Do khng quan tm ti sai s h thng bin i nn khi

c thnh phn sai s dng ng cong se khc i

Do khng quan tm n th t GC nn khng th xc

nh thi im xut hin ph phm v thi im iu

chnh li dao.

43

Pham vi s dng: Khi GC tinh, my tt, cng vng

ca HTCN cao, dao t mn. Sai s GC ch yu l sai s

ngu nhin nn s dng PP ny cho tin cy cao.

Tm im un v im cc i ca ng cong.

Ve th

6

L LLmin tb max

44

Coi quy lut xut

hin sai s trn lot

th cung l quy lut

trn lot i tr. Mun

x/nh sai s ch cn

t trng dung sai

vo trng phn tn.

b. Kho st CX gia cng bng biu im.

Ni dung PP:

GC th lot chi tit 60 100. GC c c/tit no tin

hnh o ngay k/thc c/tit . Kt qu c biu

din bng mt im trn biu . T biu se x/nh

c chiu rng ca phn b v trung tm phn b.

45

Mun xc nh sai s GC ch vic t trng dung

sai vo trng phn b.

V d: Sai lch GC l , thi im xut hin sai s

GC l nk, v thi im iu chnh dao l nk.

1

2

3

4

n

n-1

Th t gia cng

Kch th-c

nk

46

u im:

Bit c thi im no xut hin ph phm v

n thi im no phi iu chnh li dao.

Qu trnh kho st tng i n gin v khng

cn quan tm n tnh cht xut hin sai s GC

47

Nhc im:

QT thc hin phi cn thn trnh ghi nhm ln.

Pham vi s dng:

PP GC m sai s h thng chim tnh tri (ch yu

l do mn dao) th PP ny cho tin cy rt cao.

3. Phng php tnh ton phn tch.

Cn c vo iu kin c th ca tng NC xc

nh cc nguyn nhn gy sai s GC, quy lut xut

hin cc loi sai s GC, tr s ca cc sai s , sau

tin hnh tng hp c sai s tng cng.

Sai s chia 3 loi:

+ Sai s h thng c nh.

+ Sai s h thng bin i.

+ Sai s ngu nhin.

48

Sai s h thng c nh v bin i c tng

hp theo PP i s. Sai s ngu nhin c tng

hp theo PP xc sut.

Tng cc sai s h thng c nh l mt sai s h

thng c nh v c xc nh theo CT:

p

i

iAA1

Trong : Ai - Sai s h thng c nh th i.

49

Tng cc sai s h thng bin i l mt sai s

h thng bin i v c xc nh theo CT:

BJ(t) - Sai s h thng bin i th j.

q

j

j tBtB1

)()(

Tng cc sai s ngu nhin l mt sai s ngu

nhin v phng sai ca n c xc nh theo CT:

n

z

z

1

2

50

Nu gi 1, 2... n l cc sai s ngu nhin. K1,

K2...Kn l cc h s th sai s tng cng c xc

nh theo CT:

Nu phn b chun th K = 1.

n

z

zzKK1

2)(

51

Tng hp ba loi sai s trn bng PP th ta se

nhn c sai s tng cng

Thi giant t t

A

A

A

B

B

B

C

C

C

A

B

3

0 1 k

(t1)

0

0

0

1

1

1

k

k

k

L0

Kch th-c

52

Nh vy trong khong thi gian t t0 tk th

trung tm phn b se di chuyn trn ng

A0A1Ak v kch thc ca lot chi tit se nm

trong vng gii hn B0B1Bk C0C1Ck v sai s

tng cng se l ng cong y rng nh

bng c chiu rng khong phn tn:

= B(t) + 6

Vi phng sai c xc nh theo CT:

22 b

53

Trong : b - phng sai ca B(t) v c xc

nh theo CT:

kt

tok

b BdtBtt

0

222 1

Vi l gi tr trung bnh ca B(t) v c xc

nh theo CT:

B

kt

tok

dtBtt

B

0

.1

54

u im: Qu trnh tnh ton c c s khoa hc.

Nhc im:

Vn cha lng ht cc yu t ngu nhin c th

xy ra nn kt qu tnh cha st vi thc t

Khi lng tnh ton ln nn trong SX t dng.

55

2.2.4. iu chinh my.

2.2.4.1. Khi nim.

L cng vic cn thit ca tng NC nhm m bo

CX gia cng.

Trong sn xut n chic, lot nh CX t c

bng PP o d ct th.

G t g, DCC vo v tr c li nht cho iu

kin ct gt.

Trong SX lot ln hng khi, t CX GC bng

chnh sn dao, th iu chnh my c nhim v:

56

Xc nh ch lm vic ca my.

m bo v tr tng quan c/xc gia DCC v

chi tit GC. y l vn phc tp nht v n

quyt nh n CX GC.

Thc t thng s dng cc PP sau:

iu chinh tinh.

iu chinh ng: iu chnh ng c 2 pp.

o iu chnh theo c/tit ct th bng Kalp lm vic.

o iu chnh theo c/tit ct th bng dng c o.

57

2.2.4.2. iu chinh tinh.

Ban cht: c th hin ro ch l QT tnh ton

kch thc Lttc c coi nh l c lc ct tc ng

vo h thng nhng lc ct trng thi tinh.

o Tnh ton kch thc iu chnh tnh ton Lttc .

Ni dung:

o G t DCC theo ng Lttc a c xc nh.

58

Xc nh Lttc

o Kch thc Lttc c xc nh theo CT:

Lctc k/thc iu chnh c/tit c tnh theo

k/thc cng ngh cn t c ca NC.

Lttc = Lct

c bs

Trong :

bs - lng b sung.

Du - khi GC mt ngoi, du + khi GC mt trong.

59

Xc nh Lctc

o Vic xc nh Lctc phi cn c vo cc /kin cng

ngh c th nh: PP t CX GC, iu chnh...

o V d:

s

n

L

K

mdctt

bs

n

K

n

60

Nu t CX GC bng PP o d ct th th

k/thc Lctc c tnh theo k/thc cng ngh K.

Nu t CX GC bng chnh sn dao th Lctc c

tnh ton theo kch thc cng ngh m. m c xc

nh khi gii chui kch thc m = n + K. Nu:

C iu chnh theo k/thc gii hn ln

nht th: Lctc = mmax

61

C iu chnh theo k/thc gii hn nh

nht th: Lctc = mmin

Nu mun iu chnh trung tm phn b trng trung

tm dung sai th:

2

minmax mmLctdc

Nu mun iu chnh trung tm phn b nm trn v

tr c li nht trn trng dung sai th:

Lctc = . mtb

Trong : - h s k n lch gia trung tm

phn b v trung tm dung sai.

62

Xc nh bs

bs = 1 + 2 + 3 + . . .

Trong :

- 1: lng b sung do BDH ca HTCN.

J

1

yP

y

- 2: lng b sung do nhp nh t vi b mt.

2 = Rz

63

V d: my tin trng thi hon chnh th 3 =

(0.02 - 0.04 )mm

- 3: lng b sung do khe h hng knh cc

c trc, c bit l c trc chnh. Tu theo cc loi

my cng c th m 3 c tra trong cc s tay

Cng ngh CTM.

64

u im: Vic tnh ton kch thc iu chnh Lttc

l c c s khoa hc.

Nhc im: Sai s GC ln v sai s ca lng b

sung bs ln v c bit l khng lng trc ht

cc yu t ngu nhin xy ra trong qu trnh ct. V

vy khng nn dng PP ny l PP iu chnh duy

nht m nn kt hp vi cc PP iu chnh khc

trong iu chnh tinh l bc iu chnh ban u.

65

2.4.3. iu chinh c/tit ct th bng Kalp lm vic.

Sau khi iu chnh my t yu cu, tin hnh ct

th mt vi chi tit. Nu DS chi tit ct th nm trong

phm vi DS cho php (c kim tra bng Kalp lm

vic) th vic iu chnh coi nh xong.

Ni dung:

u im: Qu trnh iu chnh n gin.

66

Nhc im: tin cy thp (vi khng bit c

s chi tit ct th nm v tri no trn trng phn

b, v cng khng bit trng phn b nm v tri

no trn trng dung sai)

6 6

Ph phm

Ph phm

67

2.4.4. iu chinh theo c/tit ct th bng dng c

o van nng.

G t dng c ct theo kch thc Lc sau ct

th m chi tit. Nu kch thc trung bnh cng m chi

tit th nm trong phm vi dung sai iu chnh c th

qu trnh iu chnh coi nh hon thnh.

Ni dung:

Vn l cn xc nh Lc v c.

68

Xc nh Lc v c

(Xt trng hp khi khng k n sai s h thng

bin i)a/ Quan h gia trng phn b v trng dung

sai u iu kin xc nh c ( > 12).

dc

M N

69

b/ Quan h gia trng phn b v trng dung

sai khng u iu kin xc nh c ( < 12).

- C s l da trn nh l: Nu c mt lot c/tit m

k/thc ca n phn b theo lut chun vi phng sai

l v nu phn lot c/tit thnh nhiu nhm mi

nhm c m chi tit thi kt/ thc trung binh ca cc

nhm a phn cng phn b theo quy lut chun vi

phng sai l:

m

1

70

- Gi s cc chi tit ct th nm trn v tr xu nht ca

trng phn tn v trng phn tn li nm trn v tr

xu nht ca trng dung sai, khi gia cng xong vn

khng c ph phm. iu chng t rng nu dung

sai ca chi tit th nm trong khong MN th khi gia

cng xong se khng c ph phm. MN chnh l khong

dung sai iu chnh:

c = MN = - 12

71

M1

- p dng nh l xc sut vo vic x/nh dung sai iu

chnh nh sau: chia lot c/tit GC thnh n nhm, mi

nhm c m c/tit th kch thc trung bnh ca cc nhm

se phn b theo quy lut chun vi phng sai l:

72

- T hnh ve ta thy khong MN c chn lm c .

m

1

11

M N

dc

73

mMNc

11666 1

Hay :

mc

11

11

Trong : = /6 l h s an ton.

74

Ta thy c ph thuc vo dung sai ch to c/tit ,

vo h s an ton , vo s chi tit th m. Nu m

tng, c se tng do d iu chnh nhng se lm

tng thi gian v chi ph ct th. V vy thng s chi

tit ct th m xc nh theo cng thc:

2

6

6

m

Thng m = 2 8 chi tit.

75

m bo khng c ph phm th trung tm

phn b phi trng vi trung tm dung sai > 12.

- Nu tnh c dung sai iu chnh th iu kin khng

c ph phm l:

c

m

116

V trung tm phn b trng vi trung tm dung sai nn

k/ thc iu chnh c xc nh theo CT:

76

2

minmaxLL

Ldc

. iu chinh t ng.

C th m bo

CX GC bng s

dng b iu chnh

c lin h ngc.

Cho php gim s

ph phm n mc

bng 0, ko di chu

k iu chnh li.

77

2.2. chnh xc gia cng

2.2.1. Khi nim

2.2.2. NN gy ra sai lch gia cng

2.2.3. PP am bao CX GC trn my cng c

2.2.4. iu chnh my

Tng kt bi

78

Nhim v nh:

Tm tt bng s ND chng 1 v cc phn

chng 2 sau:

A. n tp:

79

2.1 Cht lng b mt GC

2.2 chnh xc GC

Nhim v nh:

2.3. Chun

2.3.1. nh nghia v phn loi chun

2.3.2. Qu trnh g t chi tit

2.3.3. Nguyn tc nh v 6 im

2.3.4. Tnh sai s g t

2.3.5. Nguyn tc chn chun

B. Xem trc ti liu:

80

buoi 3

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