Hgt Khai Trien3

Thit K H Dn ng Ti Ko

n chi tit my_Hp gim tc 2 cp Ging vin hng dn c Nam

Li ni u

Tnh ton thit k h dn ng c kh l ni dung khng th thiu trong chng trnh o to k s c kh . n mn hc Chi tit my l mn hc gip cho sinh vin c th h thng ho li cc kin thc ca cc mn hc nh: Chi tit my, Sc bn vt liu, Dung sai & lps gheps, V k thut .... ng thi gip sinh vin lm quen dn vi cng vic thit k v lm n chun b cho vic thit k n tt nghip sau ny.

Nhim v c giao l thit k h dn ng t ke gm c hp gim tc bnh rng v b truyn ai. H c dn ng bng ng c in thng qua b truynf ai ti hp gim tc v s truyn chuyn ng ti tang quay.Do ln u tin lm quen thit k vi khi lng kin thc tng hp cn c nhng mng cha nm vng cho nn d rt c gng tham kho cc ti liu v bi ging ca cc mn c lin quan song bi lm ca em khng th trnh c nhng sai st. Em rt mong c s hng dn v ch bo thm ca cc thy trong b mn em cng c v hiu su hn , nm vng hn v nhng kin thc hc hi c.

Cui cng em xin chn thnh cm n cc thy trong b mn, c bit l thy cs Nam trc tip hng dn, ch bo cho em hon thnh tt nhim v c giao .

Mt ln na em xin chn thnh cm n !H ni, ngythngnm 2007 Sinh vin thc hin

Nguynx V Binhf

Thit K H Dn ng Ti Ko

S liu cho trc - Lc ko bng ti F=8400 [ N ]

Vn tc ko cp v=0,7 m/s

ng knh tang D=340 mm

Thi hn phc v lh=19000 h

S ca lm vic : S ca =2

Gc nghing ng ni tm b truyn ngoi = 300 c tnh lm vic : va p vachng I: chn ng c in1.Chn ng c in

1.1.Xc nh cng sut t trn cc trc ng c

Trong :

Pct (kW) l cng sut trn trc my cng tc

(: l hiu sut ca HD

Ta c

F (N): lc ko bng ti

Hnh 1v (m/s): vn tc di ca bng ti

: H s ti trng tng ng, c tnh nh sau:

, ( l hiu sut ca cc b truyn v ca cc cp c trong h thng dn ng (c th chn theo bng 2.3 ti liu TTTKHDCK - T1)

(=(ot. (k. (ol3(br2(dTrong .

(ol Hiu sut ca ln

(ot Hiu sut trt

(kn Hiu sut khp ni

(ai Hiu sut b truyn ai

.(br Hiu sut b truyn bnh rng

Chn theo bng 2.3 ti liu TTTKHDCK - T1 ta c

(ol =0,99 ; (ot =0,98 ; (kn =1 ; (ai =0,95 ; (br =0,98

( = 0,98.1.0,993.0,982.0,95 = 0,87 Thay s vo ta c

Hnh 21.2.Xc nh s b s vng quay ng b ca ng c in

nsb = nct.usb

Trong

- nct : s vng quay ca trc my cng tc

i vi tang quay

v [m/s] l vn tc di ca bng ti

D [mm] l ng knh tang quay

T s truyn ca h thng

usb = usbH usbBtngTrong

usbH l t s truyn ca hp (chn theo bng 2.4 ti liu TTTKHDCK - T1)

usbBtng l t s truyn s b B truyn ngoi (chn theo bng 2.4 ti liu TTTKHDCK - T1)

Chn usbH = 10 Chn usbBtng =uai = 4 (tha mn nm trong khong t 3 n 5)

usb = 10.4=40nsb = 39,32.40 = 1572,8 (v/ph)

1.3.Chn ng c

Khi c c Pyc v s vng quay s b nsb, ta c th chn quy cch ng c theo bng P1.3(tr237_TK1)tha mn iu kin:

Cng sut:

S vng quay:

ng thi tha mn:

ng c c chn c cc thng s sau:

K hiu ng c in: 4A132S4Y3Cng sut danh ngha Pc =7,5 (kW)

S vng quay thc nc =1455 (v/ph)H s qu ti Tk/Tdn =2,0 >Tmm/T1=1,5Khi lng 77 kg

ng knh trc ng c 38mm (Bng P1.7_TK1_tr242) 2.phn phi t s truyn

2.1.xc nh t s truyn chunguc = ndc/nctnc(v/ph) l s vng quay ca ng c chn

nct(v/ph) l s vng quay ca trc my cng tc

uc =1455/39,32=37,002.2.Xc nh t s truyn ca b truyn trong hp

ung = uc/uh Da trn quan im v mi tng quan kch thc gia HGT v b truyn ngoi , ta chn t s truyn uai=3

uh=37,00/3 = 12,33 ung l t s truyn ca b truyn ngoi HGT (l b truyn ai)

uh l t s truyn ca HGT

Li c uh=u1.u2 vi u1,u2 ln lt l t s truyn ca hai cp bnh rng n khp

Chn phng n phn phi t s truyn uh theo phng php giI bI ton a mc tiuTheo bng 3.1(TK1_tr43) chn

Bng ni suy ta c

=> u1=4,398

=> u2=

Khi :

un=

3.tnh cc thng s trn cc trc

3.1.s vng quay trn cc trc

S vng quay ca ng c

nc=1455 [v/ph]

S vng quay ca trc 1

n1= nc/uai = 1455/3 =485[v/ph]

S vng quay ca trc 2

n2 =n1/u1 = 485/4,398=110,28 [v/ph]

S vng quay ca trc 3

n3=n2/u2=110,28/2,80=39,38[v/ph]

3.2.Cng sut trn cc trc

Cng sut trn trc 3

P3=Pct/ot.k=5,88/0,98.1=6,00Cng sut trn trc 2

P2 = Pct/(ol .(br =6,00 / 0,99.0,98 =6,18 [Kw] Cng sut trn trc 1

P1 = P2 / (ol .(br = 6,18/0,99.0,98 =6,37 [Kw]

Cng sut thc trn trc ng c

Pc = P1 / (ol .(ai = 6,37/0,99.0,95 =6,77 [Kw]

3.3.Mmen xon trn cc trc

T cng thc: Ti = 9,55.106.Pi/ni ta tnh c:

Mmen xon trn trc ng c

Tc = 9,55.106.Pc/nc = 9,55.106.6,77/1455 = 44435,39 (Nmm)

Mmen xon trn trc I

TI = 9,55.106.PI/nI = 9,55.106.6,37/485= 125429,897 (Nmm)

Mmen xon trn trc II

TII = 9,55.106.PII/nII = 9,55.106. 6,18 /110,28= 535174,10 (Nmm)

Mmen xon trn trc III

TIII=9,55.106.PIII/nIII=9,55.106.6,00/39,38=1455053,33 (Nmm)

Mmen xon trn trc cng tcTct = 9,55.106.Pct/nct = 9,55.106.5,88/39,38 =1425952,26 (Nmm)

Bng 1. Cc thng s trn cc trc

Trc

Thng sng c IIIIIITrc ct

P (KW)6,776,376,186,005,88

U 3,00 4,398 2,80 1

n(vg/p)1455485110,2839,3839,38

T (Nmm)44435,39125429,897535174,101455053,331425952,26

chng II: thit k cc b truyn

1.thit k b truyn ai thang

Cc thng s u vo:

+> Cng sut thc ca ng c : P/dc=6,77 kW

+> S vng quay trc ng c : ndc=1455 v/ph

+> T s truyn : u=3,00

+> Gc nghing ng ni tm 2 bnh ai : 30o1.1.Chn loi ai:

T hnh 4.1 (TKI_tr59) vi cng sut cn truyn 6,77 kW;

S vng quay ng c 1455 v/ph ta chn tit din ai Cc s liu kch thc ai tra bng 13.3(TLI_tr22) nh sau: b0 =14 mm b = 17 mm

h = 10,5 mm y0 = 4,0 mm

A1 = 138 mm2l0 = 2240 mmChiu di gii hn : 800 6300 mm

Khi lng 1 m ai : 0,18 kg/m

Hnh 3T bng 13.5 (TLI_tr23) cho d1min=125 mm

1.2. Xc nh cc kch thc v thng s b truyn :

1.2.1.Chn ng knh bnh ai nh :

Ta c : d1=(1,11,2)d1min =(1,11,2).125

=137,5150 (mm)

T bng 4.26 (TKI_tr67) chn d1=140 mm

Vn tc ai :

V= (m/s)

Tnh ng knh bnh ai ln:

d2 = u.d1.(1- )

vi : h s trt , chn = 0,02

d2 = 3.140.(1- 0,02) = 411,6 (mm) => chn d2 = 450 mm theo tiu chun

1.2.2.Chn khong cch trc :

Chn s b khong cch trc tho mn :

0,55.(d1+d2) +h a 2.(d1+d2)

0,55.(140+450) a 2.(140+450)

335 a 1180

Chn s b a = d2 = 450 mm

Theo cng thc tnh chiu di ai :

L=2.a +

L=2.450 + = 1880 (mm)T bng 4.13 (TKI_tr59) chn chiu di ai tiu chun : L = 1800 mm

Tnh li khong cch trc a

a =

a =

a = 407 (mm)

Gc m ca ai trn bnh ai nh

1 = 1800 -

1 = 1800 - = 136,590 > 1200 (tho mn)1.2.3. Nghim s vng chy ca ai trong mt giy

i = < 10 (tho mn)

1.3. Xc nh s ai z

S ai z c xc nh theo cng thc

z =

Trong :

P1 : Cng sut trn trc bnh ai ch ng (kW)

N chnh l cng sut thc ca ng c P1 = Pc = 6,77 kW[P0] : Cng sut cho php

T bng 4.19 (TKI_tr62) bng ni suy ta c :

Vi V=10 m/s

x = 2,56

Vi V= 15 m/s

y = 3,16

=>

=> [P0] = 2,64 (kW)

K : h s ti trng ng

T bng 4.7 (TKI_tr55) vi ti trng m my n 150% ti trng danh ngha , chn K=1,1.

C : h s xt n nh hng ca gc m 1T bng 4.15 (TKI_tr61) ta c

C=0,88Cl : h s k n nh hng ca chiu di ai

T bng 4.16 (TKI_tr61) vi

ta c Cl = 0,95

Cu : h s k n nh hng ca t s truyn

T bng 4.17 (TKI_tr61) ta c Cu = 1,14

Cz : h s xt n nh hng ca s phn b khng u ti trng cho cc dy ai

T bng 4.18 (TKI_tr61)

vi z = ta c Cz = 0,95

Thay s vo ta c

z=

Chn z=3 ztnh-zchn = 0,11 < 0,3 (tho mn)

Chiu rng bnh ai B :

T bng 4.21 (TKI_tr63) ta c vi tit din ai t= 19 mm

e= 12,5 mm

h0 = 4,2 mm

B = (3-1).19 + 2.12,5 = 63 (mm)

ng knh ngoi ca bnh ai

da = d + 2.h0 = 140 + 2.4,2 = 148,4 (mm)

1.4. Xc nh lc cng ban u v lc tc dng ln trc :

Hnh 4Lc cng trn 1 ai c xc nh bng cng thc

F0 =

Trong

Fv : lc cng do lc li tm sinh ra , xc nh bng cng thc

Fv = qm.V2qm : khi lng 1 m chiu di ai

T bng 4.22 (TKI_tr64) vi tit din ai qm = 0,178 kg/m

Fv = 0,178.10,662 = 20,23 (N)

F0 = (N)Lc tc dng ln trc

Fr = 2.F0.z.sin(1/2)

= 2.226,63.3.sin (136,59/2)

= 1263,37 (N)

Vi t s truyn thc ca b truyn ai u = 3,21 ta c

uh =

T bng 3.1 (TKI_tr43) bng ni suy ta c

u1 =4,2

u2 =

T ta c bng cc thng s ca ai

Bng 2 . Cc thng s ca ai Thng s Gi tr

ng knh bnh ai nh d1 (mm) 140

ng knh bnh ai ln d2 (mm) 450

Chiu rng bnh ai B (mm) 63

Chiu di ai l (mm) 1800

S ai 3

Tit din mt ai A1 (mm2) 138

Lc tc dng ln trc Fr (N) 1263,37

2. THIT K B TRUYN CP NHANH (BNH TR RNG THNG)

1.Chn vt liu.

Vt liu lm bnh rng p ng cc i hi sau:

- Vt liu lm bnh rng phi tho mn cc yu cu v bn b mt trnh hin tng trc mi, mi mn, dnh rng v bn un trong qu trnh lm vic. Cho nn vt liu lm bnh rng thng l thp c ch nhit luyn hp l hoc c lm bng gang hay cc vt liu khng kim loi khc.

- Theo yu cu ca bi th b truyn bnh rng thng phi truyn c cng sut ti a chnh l cng sut truyn ln nht ca trc I l 6,37 (kW) ng vi ch trung bnh cho nn vt liu lm bnh rng thuc nhm I c cng t HB ( 350.- m bo ch tiu kinh t ta phi chn vt liu v phng php gia cng hp l cho cp bnh rng c thi gian s dng khng c chnh lch nhau khng qu nhiu.

Cn c vo cc tiu chun v Bng 6.1 (Trang 92-Tp 1:Tnh ton thit k h dn ng c kh) ta xc nh s b vt liu lm cp bnh rng nh sau:

Bnh nh: Chn vt liu thp C45 v ch nhit luyn l tin hnh ti ci thin sau khi gia cng c cc thng s k thut ( cng,gii hn bn v gii hn bn chy) ln lt nh sau:

HB = 241 ( 285; (b1 = 850 MPa ; (ch 1 = 580 Mpa

Vy ta chn cng ca bnh rng 1 l HB1 = 245.Bnh ln: Chn vt liu thp C45 cng tin hnh ti ci thin sau khi gia cng c cc thng s v vt liu ( cng, gii hn bn v gii hn bn chy) ln lt nh sau:

HB = 192 ( 240; (b2 = 750 MPa ; (ch2 = 450 Mpa

Vy ta chn cng ca bnh rng 2 l: HB2 = 230.2. Xc nh ng sut tip xc [(H] v ng sut un [(F] cho php.

a. ng sut tip xc cho php c xc inh bi cng thc nh sau:

.

Trong : - SH l h s an ton.

- ZR l h s xt n nh hng ca nhm b mt.

- ZV l h s xt n nh hng ca vn tc vng.

- ZL l h s xt n nh hng ca bi trn.

- KxH l h s xt n nh hng ca kch thc bnh rng.

Chn s b ZR.ZV.KLKxH = 1 nn ta c

Do gii hn bn mi tip xc ng vi chu k chu ti NHE c xc nh nh sau:

.

Trong : -l gii hn bn mi tip xc ca b mt rng.

- KHL l h s xt n nh hng ca chu k lm vic.

Theo Bng 6.2 (TKI_tr94) ta c cng thc xc nh v SH nh sau:

= 2.HB + 70 (MPa) cn SH = 1,1.

Vy ta c gii hn bn mi tip xc ca bnh rng nh v bnh rng ln nh sau:

((H lim1 = 2.HB1 + 70 = 2.245 + 70 = 560 (MPa).

((H lim2 = 2.HB2 + 70 = 2.230 + 70 = 530 (MPa).

H s chu k lm vic ca bnh rng c xc nh nh sau:

KHL=

S chu k c s NHO c xc nh bi cng thc nh sau: NHO = 30.HB2,4.

(

S chu k thay i ng sut tng ng NHE c xc nh nh sau:

Trong :- c l s ln n khp trong mt vng quay. Nn ta c c =1.

- Ti l mmen xon ch i ca bnh rng ang xt.

- ni l s vng quay ch i ca bnh rng ang xt.

- ti l tng s gi lm vic ch i ca bnh rng ang xt.

Vy vi bnh ln (lp vi trcII) ta c:

Thay s vo cc gi tr tng ng ca cng thc ta c:

Ta li c :

Thay s vo ta s xc nh c ng sut cho php ca bnh rng nh sau:

(MPa).

(MPa)..

Do y l cp bnh tr rng thng n khp cho nn ng sut tip xc cho php xc nh nh sau:

(MPa).

b. ng sut un cho php c xc inh bi cng thc nh sau:

Trong : - [(Flim] l gii hn bn mi un ng vi chu k chu ti NFE.

- SF l h s an ton ly bng 1,75 do b mt c ti ci thin.

- YS = 1,08 0,16.lgm l h s xt n nh hng ca kch thc rng.

- YR (1 l h s xt n nh hng nhm mt ln chn rng.

- KxF l h s xt n nh hng ca kch thc bnh rng.

Chn s b YR.YS.KxF = 1 ( .

Do gii hn bn mi un ng vi chu k chu ti NFE c xc nh nh sau:

.

Trong : -l gii hn bn mi un ca b mt rng.

- KFL l h s xt n nh hng ca chu k lm vic.

Theo Bng 6.2 (Trang 94-Tp 1: Tnh ton thit h dn ng c kh) ta c cng thc xc nh v SF nh sau:= 1,8.HB v SF =1,75.

Vy ta c gii hn bn mi tip xc ca bnh rng nh v bnh rng ln nh sau:

((F lim1 = 1,8.HB1 = 1,8.245 = 441 (MPa).

((F lim2 = 1,8.HB2 = 1,8.230 = 414 (MPa).

H s chu k lm vic ca bnh rng c xc nh nh sau:

KFL=

M s chu k c s NFO =4.106 c xc nh cho mi loi thp.

Cn s chu k thay i ng sut tng ng NFE c xc nh nh sau:

Trong :- c l s ln n khp trong mt vng quay. Nn ta c c =1.

- Ti l mmen xon ch i ca bnh rng ang xt.

- ni l s vng quay ch i ca bnh rng ang xt.

- ti l tng s gi lm vic ch i ca bnh rng ang xt.

- mF l bc ca ng cong mi khi th v un y mF = 6.

Vy vi bnh rng ln (lp vi tr II) ta c:

Tin hnh thay cc gi tr bng s vo cng thc ta c.

Ta c :

Thay s vo ta s xc nh c ng sut cho php ca bnh rng nh sau:

(MPa).

(MPa).c.ng sut cho php khi qu ti :

+ ng sut tip xc khi qu ti :

[H1]max = 2,8.ch1 = 2,8.580 = 1624 (MPa)

[H2]max = 2,8. ch2 = 2,8.450 = 1260 (MPa)

+ng sut un cho php khi qu ti :

[F1]max = 2,2.HB1 = 2,2.245 = 539 (MPa)

[F2]max = 2,2.HB2 = 2,2.230 = 506 (MPa)3. Xc nh s b khong cch trc:

Cng thc xc nh khong cch trc a( ca b truyn bnh rng tr rng thng bng thp n khp ngoi nh sau:

a(1 ( 50 (u1 + 1)

Trong : - T1 l mmen xon trn trc bnh ch ng (l trc I)

- (d = b(/d(1 = 0,5.(a.(u+1) l h s chiu rng bnh rng.

- KH( l h s k n s phn b ti trng khng u trn chiu rng vnh rng khi tnh v tip xc.

- KHv l h s k nh hng ca ti trng ng.

- u1 l t s truyn ca cp bnh rng. y ta c:

- T1 = 125429,897 (N.mm); u1 = Unh = 4.2; (a = 0,3 v [(] = 481,8 (MPa)

-(d = 0,5.(a.(u+1) = 0,5.0,3.(4,2+1) = 0,78 Tra Bng 6.7 (Trang 98-Tp 1: Tnh ton thit k h dn ng c kh) ta xc nh c KH( = 1,115 (S 3).

- Chn s b KHv = 1.

Thay s vo cng thc ta s xc nh c khong cch gia 2 trc a(1:

a(1( 50.(4,2+1). (mm)

Vy ta chn s b a(1 = 205 (mm).

4. Xc nh cc thng s n khp

( Mun ca bnh rng tr rng thng (m) c xc inh nh sau:

m = (0,01 ( 0,02).a(1 = (0,01 ( 0,02).205 = 2,05 ( 4,1.

Theo dy tiu chun ho ta s chn m = 3 mm.

* S rng trn bnh ln v bnh nh ln lt l Z1v Z2 ta c :

Chn Z1 = 26 rng.

( Z2 = U1 Z1 = 4,2.26 = 109,2 (rng).

Vy Zt = Z1 + Z2 = 26 + 109 = 135 ;Nh vy t s truyn thc cp nhanh

u1, =

Ta c

(tho mn)

Tnh li khong cch trc a :

a= (mm)

chn a = 203 mm

* y ta phi tin hnh thm qu trnh dch bnh rng tng khong cch trc t a(1 =202,5(mm) ln a(2 = 203 (mm) m vn bo m qa trnh n khp.

H s dch tm :

y = - 0,5(z1 + z2) =

H s :

ky =

T bng 6.10a (TKI_tr101) bng ni suy ta c :

kx = 0,015

H s gim nh rng :

Tng h s dch chnh :

xt = y + y = 0,17 + 0,002 = 0,172

H s dch chnh bnh rng 1

x1 = (mm)

x2 = xt x1 = 0,172-0,034 = 0,138

Gc n khp :

cos =

ng knh vng ln bnh nh :

d1 = (mm)

Vn tc vng ca bnh rng :

v1 = (m/s)

T bng 6.13 (TKI_tr106) chn cp chnh xc ch to bnh rng l 9

Chiu rng bnh rng :

b = (mm)

ng knh vng chia bnh ln :

d2 = m.z2 = 3.109 = 327 (mm)

5. Kim nghim rng v bn tip xc.

Yu cu cn phi m bo iu kin (H ( [(H] Do (H = ;

Trong : - ZM : H s xt n nh hng c tnh vt liu;

- ZH : H s k n hnh dng b mt tip xc;

- Z( : H s k n s trng khp ca rng;

- KH : H s tp trung ti trng

- KHv : h s ti trng ng - b( : Chiu rng vnh rng.

- d(1 : ng knh vng chia ca bnh ch ng;

Ta bit c cc thng s nh sau:

- T1 = 125429,897 (N.mm).

- b( =60,9 mm;

- u1 = 4,19 v d(1 = 78,23 (mm).

- ZM = 275 MPa1/2 v bnh rng lm thp tra Bng 6.5 (TKI_tr96).

- ZH = H s trng khp (( = 1,88 3,2 . Z( =

KH = 1,115 Cn

Bng 6.15 (TKI_tr107) ( (H = 0,006.

Bng 6.16 (TKI_tr107) ( go = 73.Thay s vo ta xc nh c ng sut tip xc tc dng trn bn mt rng nh sau:

(H = (MPa).Tnh chnh xc ng sut tip xc cho php ca cp rng: [(H], = [(H]. ZRZVKxH.

Vi v = 1,986 m/s ( ZV = 1 (v v < 5m/s ).Vi cp chnh xc ng hc l 9 v chn mc chnh xc tip xc l 8. Khi nhm b mt l Ra = 1,25(2,5 (m ( ZR = 0,95 vi da< 700mm ( KxH = 1. Vy [(H] = 463.1.0,95.1 = 457,71 MPa.

Do (H = 423,38 < [(H], =457,71 nn bnh rng tho mn iu kin bn tip xc.

(tho mn)6. Kim nghim rng v bn un.

bo m bnh rng trong qu trnh lm vic khng b gy rng th ng sut un tc dng ln bnh rng (F phi nh thua gi tr ng sut un cho php [(F] hay: (F ( [(F].

; (F2 = (F1 . YF2 / YF1Trong : - T1 : Mmen xon tc dng trn trc ch ng- KF( : H s tp trung ti trng.

- KFv : H s ti trng ng.

- YF : H s dng rng.

- b( : Chiu rng vnh rng.

- d(1 : ng knh vng chia ca bnh ch ng;Do

Theo Bng 6.18 (TKI_tr109.).

Cn

Bng 6.15 (TKI_tr107) ( (F = 0,016.

Bng 6.16 (TKI_tr107) ( go = 73.

Bng 6.7 (TKI_tr98) ( KF( = 1,23.Vy ta c: (MPa).

( (F2 = (F1 . YF2 / YF1 = 99,31.3,49/3,68= 94,18 (MPa).

Do ng sut un thc t bnh rng c th chu c khi lm vic xc nh nh sau.

[(F1]= [(F1].YS .YxF.YR v [(F2]= [(F2].YS .YxF. YR.

Vi m = 3 mm ( YS = 1,08 0,0695.Ln(3) ( 1. Cn YR = 1 v KxF = 1:

( [(F1] = [(F1].1.1.1 = 252 MPa.

( [(F2] = [(F2].1.1.1 = 236,5 MPa.

Nhn thy rng c hai bnh rng u p ng c iu kin bn un v :

7. Kim nghim rng v qu ti. b truyn khi qu ti (xy khi m my hoc hm my... Lc momen xon tng t ngt) khng b bin dng d, gy gin lp b mt ca rng hoc bin dng d, ph hng tnh mt ln chn rng th ng sut tip xc cc i (Hmax v ng sut un cc i (F1max lun lun phi nh hn ng sut qu ti cho php [(H]max v [(F1]max.

* ng sut qu ti pht sinh khi chy my c xc nh nh sau:

(*)

Ta c h s qu ti Kqt = Tmax/ T = 1,5.

Thay s vo cng thc (*) ta c:

Kt lun: Vy cp bnh rng ta tnh ton c trn hon ton m bo c rng b truyn cp nhanh lm an ton.8.Lc tc dng ln trc :

Lc vng :

Ft = (N)

Lc hng tm :

Fr = Ft . tg = 3206,69.tg20,380 = 1191,28 (N)

Bng 3. Thng s c bn ca b truyn cp nhanh : Thng sKH Cng thc tnh Kt qu

Khong cch trc chia a a = 0,5.m.(z1 + z2)202,5 mm

Khong cch trc a( a( = a + (x1 + x2 - y)203 mm

ng knh chia d d1 = m. Z1 d2 = m.Z2d1 = 78 mmd2 = 327 mm

ng knh lndw dw =

dw1 = 78,23 mmdw2 = 327,78mm

ng knh nh rngdada1 = d1 + 2(1+ x1 -(y).m da2= d2 + 2(1+ x2 -(y).m84,192mm333,816mm

ng knh y rngdfdf1 = d1 - (2,5-2.x1).mdf2 = d2 - (2,5-2.x2).m70,70 mm320,33 mm

ng knh c sdbdb1 = d1. cos (db2 = d2. cos (73,296 mm307,28 mm

Gc prfin gc(200

Gc n khp(t((t( = arcos(acos(t/a()20,38(

Tng h s dch chnhxt0,172 mm

Gc prfin rng(t(t = arctg(tg(/cos)200

H s trng khp ngang

1,73 mm

Vi u1 = 4,19 ta c t s truyn ca cp chm

u2 =

3.Tnh ton b truyn cp chm (bnh tr rng nghing).3.1.Chn vt liu.

Theo bng 6.1[TK1_tr92] chn:

Bnh nh : Thp 45 ti ci thin t rn HB 241 ( 285 c:

(b3 = 850 MPa ;(ch3 = 580 MPa. Chn HB3 = 260 (HB)

Bnh ln : Thp 45, ti ci thin t rn HB 241...285 c:

(b4 = 850 MPa ;(ch4 = 580 MPa.Chn HB4 = 245 (HB)

3.2. Xc nh ng sut cho php.

3.2.1.ng sut tip xc cho php:

;

Chn s b ZRZVKxH = 1 (

SH : H s an ton khi tnh v tip xc, SH =1,1.

: ng sut tip xc cho php ng vi s chu k c s;

= 2.HB + 70 ( ((H lim3 = 590 MPa;

((H lim4 = 560 MPa;

KHL=

mH: Bc ca ng cong mi khi th v tip xc,vi mH = 6.

NHO: S chu k thay i ng sut c s khi th v tip xc.

NHO = 30. H

HHB : rn Brinen.

NHE: S chu k thay i ng sut tng ng.

c: S ln n khp trong mt vng quay.

Ti , ni, ti : Ln lt l mmen xon , s vng quay v tng s gi lm vic ch i ca bnh rng ang xt.

=> ly NHE=NHO tnh => KHL3 = KHL4=1 ([(H]3 = ; [(H]4=

V b truyn l b truyn bnh tr rng nghing nn theo 6.12[TK1] :

3.2.2.ng sut un cho php:

Chn s b:YR.YS.KXF =1 => [(F] =(((F lim/SF).KFC.KFL Tra bng 6.2[TK1_tr94]: ((F lim = 1,8.HB ; SF =1,75 ;

=> ((F lim3 = 1,8.260 = 468 MPa.

((F lim4 = 1,8.245 = 441 MPa.

KFC: h s xt n nh hng ca t ti.Vi ti trng mt pha => KFC=1

KFL: h s tui th.

KFL=

mF: Bc ca ng cong mi khi th v un, vi mF = 6.

NFO: S chu k thay i ng sut c s khi th v un.

NFO = 4.v vt liu l thp 45,

NEE: S chu k thay i ng sut tng ng.

c : S ln n khp trong mt vng quay.

Ti , ni, ti : Ln lt l mmen xon , s vng quay v tng s gi lm vic ch i ca bnh rng ang xt.

Ta c : NFE > NFO => tnh ton ly NFE =NFO => KFL3 = KFL4=1

Thay vo cng thc trn ta c:[(F3 ]=468.1.1/1,75 =267,43 MPa[(F4 ]= 441.1.1 / 1,75 = 252 MPa,3.2.3.ng sut cho php khi qu ti :ng sut tip xc cho php khi qu ti ca mi bnh rng:

Theo ct6.13[TK1]:

Bnh 3: [(H]3max=2,8 (ch3 =2,8.580 = 1624 MPa

Bnh 4: [(H]4max=2,8 (ch4 =2,8.580 = 1264 MPa

Vy ta chn [(H]max =1264 MPa

ng sut un cho php khi qu ti:

[(F3]max= 0,8(ch3 = 0,8.580= 464MPa;

[(F4]max = 0,8(ch4 = 0,8.580 = 464MPa;

3.3. Xc nh s b khong cch trc: Theo ct6.15a[TK1]: aw23 = Ka(u2+1)

Vi: T2: Mmen xon trn trc bnh ch ng ca cp chm, (Nmm) ;

T2= 535174,10(Nmm) Ka : h s ph thuc vo loi rng v vt liu cp bnh rng ; Theo bng 6.5[TK1_tr96],vi bnh rng nghing Ka =43

H s chiu rng vnh rng (ba = bw/aw1;

Theo bng 6.6[TK1_tr97] chn (ba =0,3

Tra bng 6.7[TK1_tr98] ( s 5) ta c KH(=1,028 ; [(H]= 522,5 MPa

Thay s ta nh c khong cch trc :

aw23= 43.(2,75 +1). (mm)

Chn aw23 = 225 (mm)3.4. Xc nh cc thng s n khp

( Mun : m

m = (0,01 ( 0,02). aw23 = (0,01 ( 0,02).225 = (2,25 ( 4,5). Theo bng 6.8 _ bng v gi tr mun tiu chun

Chn m = 3 (mm)

Chn s b =100,do cos=0,9848,theo 6.31[TK1] ta c :

S rng bnh nh :

Z3 = 2 aw23cos/ [m(u2 +1)] = 2.225.0,9848/[ 3(2,75+1)] = 39,38

Ly Z3=39 rng

S rng bnh ln: Z4 = u2 Z3 = 2,75.39 = 107,25 (rng)

Ly Z4=107 rng

Zt = Z3 + Z4 = 39+ 107 = 146Do t s truyn thc s l :u2=107/39 = 2,74

Cos=mZt/(2.aw23)=3.146/(2.225)=0,97

=> =13,260Vi bnh rng nghing ta khng dch chnh nn x3=x4=0

3.5. Kim nghim rng v bn tip xc.

Yu cu cn phi m bo (H [(H] Theo 6.33[TK1]:

(H = ZM ZH Z( Trong : - ZM : H s xt n nh hng c tnh vt liu; ZM = 275 MPa1/3 (tra bng 6.5) ;Theo 6.35[TK1] :

tgb=cost.tg=cos(20,56).tg(13,26)=0,22 ( b=12,440vi t=tw34=arctg(tg/cos)= arctg(tg20/cos13,26)=20,560do theo 6.34[TK1] : ZH = = = 1,74 (H s k n hnh dng b mt tip xc)

Theo 6.37[TK1], =bwsin/(m)=67,5.sin(13,26)/( .3)=1,64 >1

Do theo 6.38[TK1] : - Z( : H s k n s trng khp ca rng; Z( = = =0,76 - KH : H s ti trng khi tnh v tip xc; KH = KH(.KHVKH( Tra bng 6.7[TK1_tr98]: KH( = 1,028Vn tc vng bnh dn : v = (m/s)

Theo bng 6.13[TK1_tr106] .Chn cp chnh xc 9, tra bng 6.16[TK1_tr107] chn go= 73Theo cng thc 6.42 [TK1] :

Trong theo bng 6.15[TK1_tr107] => (H =0,002

Tra bng 6.14[TK1_tr107]: KH(=1,13

( KH = 1,028.1,13.1,006 = 1,17

Thay cc gi tr va tnh c vo ct6.33[TK1] : (H = 275.1,76. 0,76.= 481,73 (MPa)

Tnh chnh xc ng sut tip xc cho php : [(H] = [(H]. ZRZVKxH.

Vi v = 0,69 (m/s ) ( ZV = 1 (v v < 5 m/s ) . Cp chnh xc ng hc l 9, chn mc chnh xc tip xc l 8. Khi cn gia cng t nhm l Ra = 2,5...1,25 (m. Do ZR = 0,95, vi da< 700(mm). ( KxH = 1.

[(H] = 522,5.1.0,95.1 = 496,38 MPa , (H [(H] .

Rng tho mn v bn tip xc.

Xt t s :

(tho mn)3.6.Kim nghim rng v bn un.

Theo ct6.43,6.44[TK1] : (F3=2T3KFYF3Y(Y(/(bw34dw3m) ( [(F3]

(F4=(F3. YF4/ YF3Trong : Y( l h s k n s trng khp ca rng

Y(=1/ (=1/1,71=0,58 Y(= 1 -13,6/140=0,91S rng tng ng :

Zv3=Z3/cos3=39/0,973=42,73Zv4=Z4/cos3=107/0,973=117,23 Theo bng 6.18[TK1_tr109], c YF3= 3,685; YF4=3,6Theo bng 6.7(TKI_tr98), KF( = 1,07; KF(=1,37: theo bng 6.14(TKI_tr107) vi v< 2,5m/s v cp chnh xc 9.

Theo cng thc

Trong theo bng 6.15 [TK1_tr107], (F= 0,006, theo bng 6.16[TK1_tr107], g0= 73.Do theo cng thc

KFv=1+(Fbw34dw3/(2T3KF(KF() =1+2,73.67,5.120/(2.535174,1.1,07. 1,37)=1,01 ( KF=1,07.1,37.1,01=1,48Vy (F3= 2.535174,1.1,07.1,37.1,01.3,685.0,91.0,58/(67,5.120.3) = 126,84 MPa

(F4= (F3 YF4/ YF3 = 126,84.3,6/3,685= 123,91 MPatnh chnh xc ng sut tip xc cho php:

[(F3] =[(F3]. YR. Ys. KxF [(F4] =[(F4]. YR. Ys. KxFvi m = 3 (Ys= 1,08- 0,0695ln(3) = 1,00: YR=1: KxF=1(da< 400), do ng sut un cho php thc t l

[(F3] = 267,43.1.1,00.1= 267,43 MPa

[(F4] = 252.1.1,00.1=252 MPa

(F3, (F4 u nh hn cc gi tr cho php, vy bn un ca rng m bo .

3.7. Kim nghim rng v qu ti

trnh bin dng d hoc gy gin lp b mt, ng sut cc i khng c vt qu mt gi tr cho php

H s qu ti: Kqt=Tmax/T1=1,5 Theo 6.48[TK1] : (Hmax= (H= 496,38 = 607,94MPa < [(H]]max= 1264MPa;

(F3max=(F3Kqt= 126,84.1,5 = 190,26 MPa < [(F3]max= 464 MPa;

(F4max = (F4Kqt = 123,91.1,5 = 185,86MPa < [(F4]max = 464 MPa;

Vy rng bn v qu ti.3.8.Lc tc dng ln trc : Lc vng :

Ft = (N)

Lc hng tm :

Fr = (N)

Lc dc trc :

Fa = Ft.tg13,260 = 8919,57.tg13,260 = 2101,92 (N)Bng 4.Thng s c bn ca b truyn cp chm : Thng s KH Cng thc tnh Kt qu

Khong cch trc chiaaa = 0,5m(z1+z2)/cos225,77 mm

ng knh vng chiadd3 = m. z3/cos

d4 = m.z4/ cos120,62 (mm).330,93 (mm).

ng knh nh rngdada3 = d3 + 2m da4 = d4 + 2m 126,62 (mm).336,93 (mm).

ng knh y rngdfdf3 = d3 2,5mdf4 = d42,5m113,12(mm).323,43(mm).

Gc prfin gc(200

Gc prfin rng(t(t = arctg(tg(/cos)20,560

Gc n khp(t((t( = arcos(acos(t/a()20(

H s trng khp(( = [1,88-3,2(1/z3+1/z4)]cos1,71

CHNG V : THIT K TRC

I.Chn vt liu : Chn vt liu ch to cc trc l thp 45 c , ng sut un cho php

II.Tnh s ng knh trc : 1.Vi trc 1 :

d1 =

Vi T1 = 125429,897 Nmm : mmen xon trn trc I

: mmen xon cho php

Thay s ta c :

d1 = (mm)

d1 = 35 mm

T bng 10.2(TKI_tr189) chn s b chiu rng ln : b0 = 21 mm

2.Vi trc II :

d2 = (mm)

d2 = 55 mm

T bng 10.2(TKI_tr189) chn s b chiu rng ln b0 = 29 mm

3.Vi trc III :

d3 = (mm) chn d3 = 80 mm

T bng 10.2(TKI_tr189) chn s b chiu rng ln b0 = 39 mm

IV.Xc nh khong cch gia cc gi v im t lc : Chiu di moay bnh ai :

lm = (1,21,5)d1

= (1,21,5).35 = 4252,5 mm

lm = 65 mm

Chiu di moay bnh rng 1,2 c k n chiu rng bnh rng :

b = 60,9 mm => lm1,2 = 65 mm

Chiu di moay bnh rng 3,4 c k n chiu rng bnh rng :

b = 67,5 mm => lm3,4 = 80 mm

T bng 10.3(TKI_tr189)chn : - k1 : khong cch t mt mt ca ch tit quay n thnh trong ca hp hoc gia cc chi tit quay . k1 = 10 mm- k2 : khong cch t mt mt ca n thnh trong ca hp . k2 = 8 mm- k3 : khong cch t mt mt ca chi tit quay n np . k3 = 15 mm

- hn chiu cao np v u bulng . hn = 15 mm

T bng 10.4(TKI_tr191) ta c :

l22 = 0,5.(lm22 +b0) + k1 + k2 = 0,5.(65+29) + 10 + 8 = 65 mml23 = l22 + 0,5(lm22 + lm23) + k1 = 65 + 0,5.(65 +80) + 10 = 147,5 mm

l21 = lm22 + lm23 + 3.k1 + 2.k2 + b0 = 65 +80 +3.10 + 2.8 + 29 = 220 mm

Chiu di moay na khp ni :

Chn ni trc vng n hi

lmk = (1,4 2,5).d = (1,4 2,5).80 = 142 200 mm

Khong cch cng xn trn trc I :

lc12 = 0,5.(lm + b0) + k3 + hn = 0,5.(65 + 29) + 15 + 15 = 119,5 mm

Nh vy khong cch gia cc gi trc :

l11 = l21 = l31 = 220 mm

1.THIT K TRC I

1.1.Xc nh cc khong cch gia cc gi v im dt lc :

Ta c : l12 = l22 = 65 mm ; l11 = 220 mm

lc12 = 77 mmT chng 2 _ Thit k cc b truyn _ ta c : Lc tc dng ln trc ca b truyn ai : Fr = 1263,37 N

Vi gc nghing ca b truyn ngoi = 300

Frx = Fr . sin = 1263,37.sin300 = 631,69 N

Fry = Fr.cos = 1263,37.cos300 = 1094,11 N

Hnh 5

Hnh 6

Lc tc dng t b truyn cp nhanh_bnh tr rng thng ln trc : Lc vng : Ft = 3206,69 N => Ft1 = Ft2 = 3206,69 N Lc hng tm : Fr = 1191,28 N => Fr1 = Fr2 = 1191,28 N 1.2.Xc nh phn lc ti gi : Xt trc I :

Hnh 7

(1)

(2)

(2) => R2x = == -1406,48 (N)

(1) => R1x = - Ft1 + Fsin R2x = -3206,69+631,69 (-1406,48) = -1168,52 (N)

R2y =

R2y = (N)

R1y = Fr1 Fd.cos R2y = 1191,28 1094,11 (- 637,74) = 734,91 (N)1.3.V biu mmen :

Ta c :

Fr1.(l11 l12) = 1191,28.(220-65) = 184648,4 (Nmm)

Fy.lc12 = 1094,11.77 = 84246,47 (Nmm)

Ft1.(l11 l12) = 3206,69.(220 - 65) = 497036,95 (Nmm)

Fx.lc12 = 631,69.77 = 48640,13 (Nmm)

Ft1. = 3206,69. = 125429,68 (Nmm)

1.4.Tnh chnh xc ng knh trc : Mmen un tng ti v tr lp bnh rng 1 :

Mbr1 = (Nmm)Mmen tng ng ti ch lp bnh rng 1 :

Mt = (Nmm)

ng knh trc ti v tr lp bnh rng 1 :

dbr1 =

vi tra trong bng 10.5 (TKI_tr195) l 59,75 MPa

dbr1 =

chn dbr1 = 45 mm theo tiu chun

Momen un tng ti v tr lp ln :

Mol = (Nmm)

Mmen tng ng ti v tr lp ln :

Mt = (Nmm)

Hnh 8ng knh trc ti v tr lp ln :

dol = =

chn ng knh ch lp ln : dol = 40 mm

Mmen tng ng ti ch lp bnh ai :

Mt = (Nmm)


dbr1 = = (mm)chn ng knh ch lp bnh ai :

db = 32 mm

1.5.Kim nghim trc v bn mi :Chn lp ghp : Cc ln lp trn trc theo k6 , lp bnh rng , bnh dai , ni trc theo k6 , kt hp vi lp then.

1.5.1.Xt ti v tr lp bnh rng 1 : T bng 9.1 (TKI_tr173) chn :

b h = 14 9

t1 = 5,5

t2 = 3,8

W = (mm4)W0 = (mm4)

T bng 10.8 (TKI_tr194) chn Kx = 1,06

T bng 10.9 (TKI_tr197) chn Ky = 1

T bng 10.12 (TKI_tr199) chn

T bng 15.2 (TLI_tr56) tra h s kch thc

T bng 10.7 (TKI_tr197) ta c :

=>

T bng 10.11 (TKI_tr198) vi kiu lp chn

;

=2,12 (v Ky = 1)

=1,7

Mmen un tng ti v tr lp bnh rng 1 :

Mbr1 = 530227,08 Nmm

ng sut un :

ng sut xon :

Ta c :

=>

1.5.2.Xt ti tit din lp ln :Mmen un tng ti v tr lp ln : Mol = 90266,51 (Nmm)

ng sut un :

(MPa)

ng sut xon :

(MPa)

T bng 15.2 (TLI_tr56) :

=>

=>

Ta c :

EMBED Equation.DSMT4

=>

1.6.Chn then :

1.6.1.Ti v tr lp bnh rng 1 :Chn then bh=149

t1 = 5,5

lt = (0,80,9)lm = (0,80,9).65 = 52,5 58,5

chn lt = 56 mm

T = 125429,68 Nmm

1.6.2.Ti v tr lp bnh ai :

Chn then

b h = 10 8

t1 = 5

lt = (0,8 0,9).65 => lt = 55 mm

=>

2.THIT K TRC II2.1.Xc nh cc khong cch gia cc gi v im dt lc :

Ta c : l22 = 65 mm ; l23 = 147,5 mm ; l21 = 220 mmT chng 2 _ Thit k cc b truyn _ ta c : Lc tc dng t b truyn cp nhanh_bnh tr rng thng ln trc : Lc vng : Ft = 3206,69 N => Ft1 = Ft2 = 3206,69 N Lc hng tm : Fr = 1191,28 N => Fr1 = Fr2 = 1191,28 N Lc tc dng t b truyn cp chm_bnh tr rng nghing ln trc :

Lc vng : Ft3 = 8919,57 N

Lc hng tm : Fr3 = 3346,86 N

Lc dc trc : Fa3 = 2101,92 N2.2.Xc nh phn lc ti gi : Xt trc II :

Hnh 9

(1)

(2)

(2) => R2x = == 5198,66 (N)

(1) => R1x = Ft2 + Ft3 R2x = 3206,69 + 8919,57 5198,66 = 6927,6 (N)

R2y =

R2y = (N)

R1y = Fr3 Fr2 R2y = 3346,86 1191,28 (- 309,62) = 2465,2 (N)

2.3.V biu mmen :

Ta c :

Fr3.(l21 l23) = 3346,86.(220-147,5) = 242647,35 (Nmm)

Fa3. = 2101,92. = 126115,2 (Nmm)

Fr2.l22 = 1191,28.65 = 77433,2 (Nmm)

Ft3.(l21 l23) = 8919,57.(220-147,5) = 646668,83 (Nmm)

Ft3. = 8919,57. = 535174,2 (Nmm)

2.4.Tnh chnh xc ng knh trc : Mmen un tng ti v tr lp bnh rng 3 :

Mbr3 = (Nmm)

Mmen tng ng ti ch lp bnh rng 3 :

Mt = (Nmm)


dbr3 =

vi tra trong bng 10.5 (TKI_tr195) l 49,8 MPa

dbr3 =

chn dbr3 = 55 mm theo tiu chun

Momen un tng ti v tr lp bnh rng 2 :

Mbr2 = (Nmm)

Mmen tng ng ti v tr lp bnh rng 2 :

Mt = (Nmm)

Hnh 10ng knh trc ti v tr lp bnh rng 2 :

dbr2 = =

chn ng knh ch lp bnh rng 2 : dbr2 = 55 mm

=> ng knh ti v tr lp ln : dol = 50 mm2.5.Kim nghim trc v bn mi :Chn lp ghp : Cc ln lp trn trc theo k6 , lp bnh rng , bnh dai , ni trc theo k6 , kt hp vi lp then.

2.5.1.Xt ti v tr lp bnh rng 3 : T bng 9.1 (TKI_tr173) chn :

b h = 16 10t1 = 6t2 = 4,3W = (mm4)

W0 = (mm4)

T bng 10.8 (TKI_tr194) chn Kx = 1,06




=>


;

=2,26 (v Ky = 1)

=2,09Mmen un tng ti v tr lp bnh rng 3 :

Mbr3 = 690694,08 Nmm

ng sut un :

ng sut xon :

Ta c :

=>

2.5.2.Xt ti tit din lp bnh rng 2 :T bng 9.1 (TKI_tr173) chn :

b h = 16 10

t1 = 6

t2 = 4,3

W = (mm4)

W0 = (mm4)

T bng 10.8 (TKI_tr194) chn Kx = 1,06




=>


;

=2,26 (v Ky = 1)

=2,09

Mmen un tng ti v tr lp bnh rng 2 :

Mbr3 = 222353,29 Nmm

ng sut un :

ng sut xon :

Ta c :

=>

2.6.Chn then :


t1 = 6

lt = (0,80,9)lm = (0,80,9).65 = 52,5 58,5

chn lt = 56 mm

T = 535174,2 Nmm

2.6.2.Ti v tr lp bnh rng 3 :

Chn then

b h = 16 10t1 = 6lt = (0,8 0,9).80 => lt = 70 mm

=>

3.THIT K TRC III3.1.Xc nh cc khong cch gia cc gi v im dt lc :

Ta c : l32 = l23 = 147,5 mm ; l33 = l31 + lc33 = 220 + 119,5 = 339,5 mmT chng 2 _ Thit k cc b truyn _ ta c :

Lc tc dng t b truyn cp chm_bnh tr rng nghing ln trc :

Lc vng : Ft4 = 8919,57 N

Lc hng tm : Fr4 = 3346,86 N

Lc dc trc : Fa4 = 2101,92 N

Chn khp ni trc vng n hi theo mmen xon T3 = 1455,05 Nm

T bng 16.10a (TK2_tr69) c :

d = 63 mm

D0 = 200 mm

l1 = 48 mm

D = 260 mm

z = 8

D3 = 48 mm

l = 140 mm

nmax = 2300 mm

l2 = 48 mm

d1 = 110 mm

B = 8 mm

B1 = 70 mm

T bng 16.10a (TK2_tr69) c kch thc c bn ca vng n hi

d0 = 24 mm

l1 = 52 mm

d1 = M16

l2 = 24 mm

D2 = 32 mm

l3 = 44 mm

l = 95 mm

Lc t khp ni tc dng ln trc :

Fk = (N)3.2.Xc nh phn lc ti gi :

Hnh 11Xt trc III :

(1)

(2)(2) => R2x = =- =-5310,48 (N)(1) => R1x = -Ft4 + Fk R2x = -8919,57 + 4365,16 (- 5310,48)= 756,07 (N)

=> R2y =-

R2y =- (N)R1y = Fr4 R2y = - 3346,86 + 2679,38 = - 667,48 (N)3.3.V biu mmen :Ta c :Fr4.(l31 l32) = 3346,86.(220-147,5) = 242647,35 (Nmm)Fa4. = 2101,92. = 346816,8 (Nmm)Fk.lc33 = 4365,16.119,5 = 521636,62 (Nmm)Ft4.l32 = 8919,57.147,5 = 1315636,57 (Nmm)Ft4. = 8919,57. = 1455053,33 (Nmm)2.4.Tnh chnh xc ng knh trc : Mmen un tng ti v tr lp bnh rng 4 :Mbr4 = (Nmm)Mmen tng ng ti ch lp bnh rng 3 : Mt = (Nmm)ng knh trc ti v tr lp bnh rng 1 : dbr4 =

vi tra trong bng 10.5 (TKI_tr195) l 48,8 MPadbr4 =

chn dbr4 = 70 mm theo tiu chun Momen un tng ti v tr lp ln :Mol = (Nmm)Mmen tng ng ti v tr lp ln :Mt = (Nmm)ng knh trc ti v tr lp ln : dol = =

chn ng knh ch lp ln : dol = 65 mmMmen tng ng ti v tr lp khp ni :

Mt = .T = .1455053,33 = 1260113,15 (Nmm)

ng knh ti v tr lp khp ni :.

Hnh 12

dk = (mm)

chn dk = 60 mm theo tiu chun3.5.Kim nghim trc v bn mi :Chn lp ghp : Cc ln lp trn trc theo k6 , lp bnh rng , bnh dai , ni trc theo k6 , kt hp vi lp then.3.5.1.Xt ti v tr lp bnh rng 4 : T bng 9.1 (TKI_tr173) chn :b h = 20 12t1 = 7,5t2 = 4,9W = (mm4)W0 = (mm4)T bng 10.8 (TKI_tr194) chn Kx = 1,06T bng 10.9 (TKI_tr197) chn Ky = 1T bng 10.12 (TKI_tr199) chn


=>


;

=2,38 (v Ky = 1)

=2,09Mmen un tng ti v tr lp bnh rng 4 :Mbr4 = 1337825,59 Nmmng sut un :

ng sut xon :

Ta c :

=>

3.5.2.Xt ti tit din lp ln :T bng 10.8 (TKI_tr194) chn Kx = 1,06T bng 10.9 (TKI_tr197) chn Ky = 1T bng 10.12 (TKI_tr199) chn


=>


;

=2,34 (v Ky = 1)

=2,09Mmen un tng ti v tr lp ln:Mol= 521636,62 (Nmm)ng sut un :

(MPa)ng sut xon :

(MPa)Ta c :

EMBED Equation.DSMT4

=>

3.6.Chn then :


t1 = 7,5

lt = (0,80,9)lm

chn lt = 70 mm

T = 1455053,33 Nmm

chn 2 then. Khi : =0,75.131,48=98,61

2.6.2.Ti v tr lp khp ni :

Chn then

b h = 18 11t1 = 7lt = (0,8 0,9).140 => lt = 125 mm

=>

Chng IV : CHN LN. 1.Chn ln cho trc vo ca hp gim tc:(TrcI)1.1.Chn loi :Xt t s Fa/Fr : ta thy t s Fa/Fr = 0 v Fa = 0, tc l khng c lc dc trc nn ta chn loi l bi mt dy, c s b tr nh sau:

Hnh13 Da vo ng knh ngng trc d1A =d1B = 40 mm, tra bng P2.7[TKI_tr255] chn loi bi c nh c k hiu : 208 ng knh trong d = 40mm, ng knh ngoi D = 80 mm

Kh nng ti ng C = 25,6 kN, kh nng ti tnh Co = 18,1 kN;

B =18(mm) r = 2,0 (mm)

ng knh bi db = 12,7 (mm)1.2. Kh nng ti ng:

Phn lc tng trn hai :

Fr1=

Fr2=

Ta kim nghim chu ti ln hn, Fr 2 =1544,31 (N)

Theo CT11.3[TK1] vi Fa = 0 , ti trng qui c :

QII = X.V.Fr2.kt.k

i vi chu lc hng tm X= 1

V =1 khi vng trong quay

kt = 1 v (nhit t (( 100oC )

k = 1,3 ,ti trng va p va.( QII = 1.1.1544,31.1.1,3 = 2007,6 (N)

Do ti trng thay i nn ta c ti trng tng ng :

QE= = Q1

Trong , Q1=QII=2007,6 N , m=3 i vi bi.

( QE= 2007,6.= 1826,96 N

Theo ct11.1[TK1], Kh nng ti ng :

Vi L=60.n1.Lh/106=60.485.19000/106=552,9 triu vng

( Cd=QE.=1826,96. =14,995 kN1.3. Kim tra kh nng ti tnh ca .Ti trng tnh ton theo ct 11.19[TK1] vi Fa = 0 :

Qt = X0.Fr2 Vi X0 = 0,6 (tra bng 11.6[TLI_tr104])

Qt = 0,6.1544,31 = 926,59 (N) Theo ct11.20[TK1] th Qt = Fr2=1544,31 (N)

Chn Qt = 1,54431 kN kim tra v gi tr ln hn.

Ta thy : Qt = 1,54431 kN < C0 = 18,1 kN.

( kh nng ti tnh ca c m bo.

2 .Chn ln cho trc trung gian ca hp gim tc.(Trc II)

2.1.Chn loi

Xt t s Fa/Fr : Fa=Fat=2101,92 N hng t D ti C C : Fr1== = 7353,15 N

suy ra: Fa/Fr =2101,92/7353,15 =0,3 D : Fr2== =5207,87 N

suy ra: Fa/Fr =2101,92/5207,87 = 0,40

;

Vy ta dng bi d chn . Da vo ng knh ngng trc d = 45 mm,

tra bng P2.12 (TKI_tr263) chn loi bi -chn c trung hp 46310 c cc thng s

ng knh trong d = 50 mm, ng knh ngoi D =110 mm

Kh nng ti ng C = 56,03 kN, kh nng ti tnh Co =44,8 kN;

B =27(mm) , r =3,0 (mm) , r1 =1,5 (mm)

S b tr kiu O

Hnh14 Kim nghim kh nng ti : tin hnh kim nghim cho C v ny chu lc ln hn2.2. Kh nng ti ng:

Theo ct 11.3[TK1] , ti trng qui c :

Q =( X.V.Fr+Y.Fa).kt.k y lc dc trc Fa bao gm lc dc trc ngoi v lc dc trc Fs do lc hng tm sinh ra

Fs=e.Fr( bi chn), chn =260t bng 17.1 (TLI_tr101) e=0,68FsC=e.FrC =0,68. 7353,15=5000,14 NFsD=e.FrD =0,68. 5207,87=3541,35 N

= FsD+ Fa = 3541,35+2101,92=5643,27 > FsC suy ra FaC = 5643,27 N

= FsC - Fa =5000,14 2101,92 =2898,22 > FsD suy ra FaD = 2898,22 N

V =1 khi vng trong quay FaC/FrC=5643,27/7353,15=0,76 >e

T bng 17.1 (TLI_tr101) c X1 = 0,41 ; Y1 = 0,87Tra bng cc h s c:

kt = 1 v (nhit t (( 100oC )

k = 1,3;

QC = (0,41.1.7353,15+0,87.5643,27).1.1,3=10301,77N

Theo ct 11.1 Kh nng ti ng :

Tui th ca bi (m=3):

L=lh..60.n2/10=19000.60.110,28 /10=125,72 (triu vng)

Ti trng tng ng

QE= 10301,77.[]1/3= 9374,83 N

H s kh nng ti ng : Cd = 9374,83. = 46963,98 N=46,96 kN

Do Cd = 46,96 kN < C = 56,03 kN ( loi ln chn m bo kh nng ti ng.2.3. Kim nghim kh nng ti tnh.

Ti trng tnh ton theo ct11.19[TK1] :

Qt = X0.Fr+Y0.Fa Tra bng 11.6[1] : X0=0,5 , Y0=0,37

( Qt=0,5.7353,15+0,37.5643,27=5764,58 N

Theo 11.20[TK1] : Qt= Fr=7353,27 N

Chn Qt = 7353,27 N kim tra v gi tr ln hn. Ta thy Qt = 7,35 kN < C0 =44,8 kN.

( loi ln ny tho mn kh nng ti tnh.

3. Chn ln cho trc ra ca hp gim tc:(Trc III)3.1.Chn loi :-V trn u ra c lp ni trc vng n hi nn cn chn chiu ca Fkx ngc vi chiu dng khi tnh trc tc l cng chiu vi lc Ft4. Khi phn lc trong mt phng xoz :

R2x=(Fklc33-Ft4.(l31-l32))/l31=(4365,16.119,5-8919,57.(220-147,5))/220=-568,33 NR1x = -(R2x+Fk+Ft4)=-(-568,33+4365,16+8919,57)=-12716,4 (N)Xt t s Fa/Fr : Fa=Fat=2101,92 N hng t 1 ti 2 1 : Fr1== =12733,85 N

suy ra: Fa/Fr =2101,92/12733,85 = 0,17 2 : Fr2== =2738,99 Nsuy ra: Fa/Fr =2101,92/2738,99 = 0,77

;

Vy ta dng bi chn 1 dy . Da vo ng knh ngng trc d = 65 mm,

tra bng P2.12 chn loi chn c trung hp 46313 c cc thng s

ng knh trong d = 65 mm, ng knh ngoi D =140 mm

Kh nng ti ng C = 89,0 kN, kh nng ti tnh Co =76,4 kN;

B =33 (mm) S b tr

Hnh15 Kim nghim kh nng ti : tin hnh kim nghim cho 1 v ny chu lc ln hn.3.2.Kim nghim kh nng ti :

3.2.1. Kh nng ti ng:

Theo ct11.3[TK1],ti trng quy c :

QI = (X1.V.Fr1+Y1.Fa1).kt.kTrong :

V =1 khi vng trong quay

kt = 1 v (nhit t (( 100oC )

k = 1,3 , ti trng va p va.

Ta c :chn gc tip xc = 26.0 ta c e=0,68T ta c :

Fs1 = e.Fr1 = 0,68.12733,85 = 8659,02 (N)

Fs2 = e.Fr2 = 0,68.2738,99 = 1862,51 (N)

Ta c :

Fa1 = Fs2 Fat = 1862,51 2101,92 = - 239,41 < 0

Fa2 = Fs1 + Fat = 8659,02 + 2101,92 = 10760,94 (N)

X1 = 1 ; Y1 = 0Do : QI = 1. 1. 12733,85.1.1,3 = 16554,01 (N)

Do ti trng thay i nn ta c ti trng tng ng :

QE= = Q1

Trong , Q1= 16554,01 N , m=3 i vi bi.

( QE=16554,01.= 15064,51 N

Theo ct11.1[TK1], kh nng ti ng :

Tui th ca ln :

L = 60.n3.Lh/106 = 60.39,38.19000/ 106 = 44,89 triu vng

H s kh nng ti ng : Cd =6,29634. = 53,54 kN.

Do Cd =53,54 kN < C = 89,0 kN ( loi ln chn m bo kh nng ti ng.

3.2.2. Kim nghim kh nng ti tnh.

Ti trng tnh ton theo ct11.19[TK1], Qt = X0.Fr1+Y0.Fa1Vi X0 = 0,5; Y0=0,37 (tra bng 17.4 (TKI_tr104))

Qt = 0,5. 12733,85+0,37. 1862,51 = 7056,05 (N)

Theo ct11.20[1] th Qt = Fr1 = 12733,85(N)

( chn Qt=12,73 kN kim tra.

Ta thy Qt=12,73kN < C0=76,4 kN ( loi ln ny tho mn kh nng ti tnh.

chng V.Thit k v hp gim tc, bI trn

v Iu chnh n khp.

1.Tnh kt cu ca v hp:

Ch tiu ca v hp gim tc l cng cao v khi lng nh. Chn vt liu c hp gim tc l gang xm c k hiu GX 15-32.

Chn b mt ghp np v thn i qua tm trc .

Cc kch thc c bn c trnh by trong bng2.

2. Bi trn trong hp gim tc:

2.1.Bi trn bnh rng

Do vn tc vng ca cp nhanh v1=1,986 m/s 0,04.a+10 = 0,04.225 + 10 =19 ( d1 = M20d2 = 0,8.d1 = 0,8. 20 = M16

d3 = (0,8( 0,9).d2 ( d3 = M14d4 = (0,6 ( 0,7).d2 ( d4 = M10d5 =( 0,5 ( 0,6).d2 ( d5 = M8

Mt bch ghp np v thn:

Chiu dy bch thn hp, S3Chiu dy bch np hp, S4B rng bch np hp, K3S3 =(1,4 ( 1,8) d3 , chn S3 = 25mm

S4 = ( 0,9 ( 1) S3 = 25 mm

K3 = K2 ( 3(5 ) mm = 50 5 = 45 mm

Kch thc gi trc:

ng knh ngoi v tm l vt, D3, D2B rng mt ghp bulng cnh : K2

Tm l bulng cnh : E2

k l khong cch t tm bulng n mp l

Chiu cao hnh theo kch thc np

K2 = E2 + R2 + (3(5) mm = 25,6 + 20,8 + 4 = 50,4 mm

E2= 1,6.d2 = 1,6 . 16 = 25,6 mm.R2 = 1,3 . d2 = 1,3. 16 = 20,8 mm

k ( 1,2.d2 =19,2( k = 20 mm

h: ph thuc tm l bulng v kch thc mt ta

Mt hp:

Chiu dy: Khi khng c phn li S1

B rng mt hp, K1 v qS1 = (1,3 ( 1,5) d1 ( S1 = 30 mm

K1 ( 3.d1 ( 3.20 = 60 mm

q = K1 + 2( = 60 + 2.10 = 80 mm;

Khe h gia cc chi tit:

Gia bnh rng vi thnh trong hp

Gia nh bnh rng ln vi y hp

Gia mt bn cc bnh rng vi nhau. ( ( (1 ( 1,2) ( ( ( = 10 mm

(1 ( (3 ( 5) ( ( (1 = 30 mm

(2 ( ( = 10 mm

S lng bulng nn ZZ = ( L + B ) / ( 200 ( 300) ( 718,06 / 200(300 = 4,9(3,27 Z= 6

4. Kt cu bnh rng, np , cc lt ...Vt liu bnh rng bng thp, quy m sn xut nh, n chic. Dng phng php dp to phi. Bnh rng ln c ch to lm dng nan hoa gia kt hp vi c l gim khi lng bnh rng v d dng trong vn chuyn cng nh kp cht khi gia cng. Ti cc cnh rng c vt mp trnh tp trung ng sut. Cc kch thc c th chn nh sau :

= ( 2,5 ~ 4 ) m > 8 ~ 10 mm

h s nh dng cho bnh rng c kch thc ln

l = ( 0,8 ~ 1,8 ) d

h s nh dng i vi mi ghp cht, h s ln dng vi mi ghp di ng

D = ( 1,5 ~1,8 ) d trong h s nh dng vi bnh rng ch to bng thp v s dng lp ghp c di.

C = ( 0,2 ~ 0,3 )b

do = ( 12 ~ 25 ) mm , c 4 n 6 l

Np c ch to bng gang GX15-32. Trong HGT ny ta s dng 2 kiu np . Kiu 1 np c l thng cho trc xuyn qua, Mt np phnh ra to b dy khot rnh lp vng pht. Phn lp vo l hp c ch to vi dc nh d c, on g tip xc vi thnh l hp khng yu cu ln khong 3 ~ 4 mm dng nh tm np . Kiu np 2 tng t nh kiu 1 nhng khng c l xuyn thng qua. Mt np lm vo nhm gim bt kch thc np . Chiu dy bch np c 2 kiu trn ly bng 0,7 ~ 0,8 chiu dy thnh np .

5. Cc chi tit ph ca Hp gim tc

- Nt tho du : Sau mt thi vic, du bi trn cha trong hp b bn, hoc b bin cht, do cn phi thay du mi. tho du c, y hp c l tho du. Lc lm vic, l c bt kn bng nt tho du. Kt cu v kch thc nt tho du tra bng 18-7

Hnh 16 Que thm du : Dng kim tra mc du trong hp. C nhiu dng kim tra mc du, ta chn que thm du c kt cu nh hnh v

Hnh 17 Nt thng hi : Khi lm vic, nhit trong hp tng ln. gim p sut v iu ho khng kh bn trong v bn ngoi hp, ta s dng nt thng hi. Nt thng hi c lp trn np ca thm. Kt cu v kch thc nt thng hi chn theo bng 18-6 nh hnh v

Hnh 18 chng VI. Lp ghp v dung sai

1.Chn kiu lp u tin s dng h thng l v khi c th tit kim c chi ph gia cng nh gim bt c s lng dng c ct v dng c kim tra khi gia cng l .

Kiu lp phi hp trn bn v : lp np ln ln v H7/d11 ;lp bc chn gia bnh rng v ln H7/h6 ; lp bng rng ln trc H7/k6

thun tin khi lp ln ta chn kiu lp ln ln trc k6 ; kiu lp ln ln v hp H7, cho c ba cp .

Sai lch gii hn ca kch thc then theo chiu rng b - h9

Sai lch gii hn ca rnh then trn trc , ghp trung gian - N9

2 . Dung sai

lp ghp vng trong ln trc v vng ngoi ln v ,ngi ta s dng cc min dung sai tiu chun ca trc v l theo TCVN 2245-7 phi hp vi cc min dung sai ca cc vng . Bng6.Thng k cc kiu lp v tr s ca sai lch gii hn ca cc kiu lp

V trKiu lpSai lch gii hn

ca trc(m)Sai lch gii hnca l(m)

TrcI bnh rng45H7/k6

Trc II-bnh rng50H7/k6

55H7/k6

TrcIII-bnh rng70H7/k6

Bc chn-trcI30D11/k6

Bc chn-trc II 45D11/k6

Bc chn -trcIII65D11/k6

L hp ln

Trc I62H7

L hp ln

Trc II 85H7

L hp ln

Trc III100H7

Trc I - ln 30k6

Trc II - ln45k6

Trc III - ln65k6

Bnh ai-trcI28H7/h6

Khp ni-trcIII63H7/h6

L hp np

Trc I62H7/d11

L hp np

Trc II85H7/d11

L hp np

Trc III100H7/d11

Then trc Ib=14,b=10N9/h9h9 N9

Then trcIIb=16N9/h9h9 N9

Then trc III

b=20,b=18N9/h9h9

N9

Ti liu tham kho.

[TK1] - Trnh Cht, L Vn Uyn Tnh ton thit k h dn ng c kh, tp 1 Nh xut bn Gio Dc. H Ni 2001.

[TK2] - Trnh Cht, L Vn Uyn Tnh ton thit k h dn ng c kh, tp 2 Nh xut bn Gio Dc. H Ni 2001.

[TL2] Ninh c Tn Dung sai v lp ghp. Nxb Gio Dc. H Ni 2004.

[TL1] - Nguyn Trng HipChi tit my, tp 1, 2. NxbGioDc.H Ni 1994.[TK]- Nguyn Trng Hip,Nguyn Vn Lm-Thit k chi tit my .Nxb Gio Dc.H N 2003 Ht

tmm

t2

t1

tck

T2

T1

Tmm

PAGE 32Nguyn V Bnh Lp C in t 2_K49

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Documents

Hgt Khai Trien3