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SV thc hin: Nguyn Danh Minh - Lp CK5-k5

M sinh vin: 0541010372

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LI NI U

Tnh ton thit k h dn ng c kh l yu cu khng th thiu i vi mt k

s ngnh c kh, nhm cung cp cc kin thc c s v my v kt cu my.

Thng qua n mn hc Chi tit my, mi sinh vin c h thng li cc kin

thc hc nhm tnh ton thit k chi tit my theo cc ch tiu ch yu v kh

nng lm vic; thit k kt cu chi tit my, v khung v b my; chn cp chnh

xc, lp ghp v phng php trnhby bn v, trong cung cp nhiu s liu mi

v phng php tnh, v dung sai lp ghp v cc s liu tra cu khc. Do khi

thit k n chi tit my phi tham kho cc gio trnh nh Chi tit my, Tnh

ton thit k h dn ng c kh, Dung sai v lp ghp,... tng bc gip sinh vin

lm quen vi cng vic thit k v ngh nghip sau ny ca mnh.

Trong : Tnh ton thit k h dn ng c kh T1, T2

Chi tit my tp 1, tp 2 him v ca em l thit k hp gim tc hai cp. Loi hp l: Hp gim tc bnh

rng tr. H c dn ng bng ng c in b truyn qua kh p ni v ihp gim

tc v ni trc n hi truyn ng nxch ti.

Ln u tin lm quen vi cng vic thit k, vi mt khi lng kin thc tng

hp ln, v c nhiu phn em cha nm vng, d tham kho cc ti liu song khi

thc hin n, trong tnh ton khng th trnh c nhng thiu st. Em mong

c s gp v gip ca cc thy c gio.

Em xin chn thnh cm n cc thy c gio, c bit l thy gio Hong Xun

Khoa hng dn tn tnh v cho em nhiu kin qu bu cho vic hon thnh

n mn hc ny...

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I. Tnh chn ng c, phn phi t s truyn v momen

xon trn cc trc

1. Chn ng c

a. Xc nh ch tiu cn thit chn ng c

iu kin lm vic ca ng c in: c b truyn xch

chu va p v cng sut lm vic:

Plv= = = 4,81 KW

Cng sut tng i nh, ta chn ng c xoay chiu bapha khng i v ng c khng ng b ngn mch c u

im sau: s dng ph bin trong cc ngnh cng nghip,

kt cu n gin, gi thnh tng i h, d bo qun,

lm vic tin cy, c th mc trc tip vi li in ba

pha khng cn bin i dng in.

b. Tnh ton cng sut cho ng c in v cc trc

Cng sut lm vic ca ng c in phi chu nhit

cao v nhit sinh ra khng vt mc cho php.

Cng sut lm vic ca xch ti l cng sut cn thit

Plv = 4,81 KW cho qu trnh lm vic phi t c.

Tra bng 2.3 ta c bng 1: hiu sut gia cc cp chi

tit truyn chuyn ng

Hiu sut Khp ni

1 ln 2 Bnh rng

3Xch ti

4Gi tr 0,96 0,99 0,96 0,90

Trong ; c mt khp ni, bn ln, ba cp bnh rng

v mt b xch ti

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Cng sut ln lt ca cc trc

Trc IV ni trc tip vi xch ti v c mt cp ln,

ta c:

P4==

= 4,86 KW

trc III ni vi trc IV qua b truyn xch v c mt

cp ln, ta c:

P3 =

=

= 5,45 KW

Trc III c 2 cp bnh rng tng ng nn ti mi bnh

rng nghing ta c:

P3n = P3/2 = 2,73 KW

Trc III ni vi trc II qua 2 cp bnh rng nghing v

c 1 cp ln, ta c:

P2n= [P3n/(3.3.2) =

= 2.99 KW

Ti cp bnh rng thng: P2t= 2.P2n= 5,98 KW

Trc I ni vi trc II qua cp bnh rng thng v c

mt cp ln, ta c:

P = [P2t/(3.2)] = = 6,29 KWTrc I ni vi ng c qua khp ni nn c cng sut

cn thit ca ng c l:

Pct= P1/1= = 6,55 KWc. Chn ng c in cng sut cn thit

c1. Ta xt hai trng hp sau:

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TH1: ti trng khng i cng sut tnh ton l cng

sut cn thit cho qu trnh lm vc trn trc cng

tc:

Ptd= Pct

TH2: ti trng thay i, ta xt ti ng c lc chy

qu ti, lc chy non ti nhng nhit ng c sinh

ra khng vt qu mc cho php, khi cng sut tc

dng tnh theo cng thc:

Ptd= Plv. ( ) (1)vi T2= 0,68.T1, t1= 2,8h = 0,35.tck, t2= 3,6h =

0,45.tck

suy ra: Plv= 4,81. = 3,59 KWhiu sut ca c qu trnh lm :

n =1.2. 3.4= 0,96.0,90 = 0,73;ta c: Pct= Ptd/ = = 4,92 KWc2. xc nh ng ng c lm vic, ta chn ng c

theo ch ngn hn, ta xt hai trng hp sau:

TH1: ti trng khng thay i . V t1+t2= 6,4h < tck=

8h => ti trng thay i

TH2: ng c lm vic ngn hn, ti trng thay di . T

s thi gian lm vic trong mt chu k :

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ts= (t1+t2)/tck= = 0,8 > 0,6 => ng c coi nhlm vic trong ch di hn vi ti trng khng i.

Kt lun: theo yu cu tnh ton thit k hp gim tc

hai cp ta chn cng sut cn thit ca ng c Pct=

6,55 KW

d. Chn ng c in

Cng sut nh mc Pm.nc> Pct= 6,55 KW ng thi

momen m my tha mn iu kin Tmm/T = 1,4 < Tmax/Tdml

t s truyn xon trn trc ng c v chn ng cK160S4 c Pm.c= 7,5.0,875 = 6,56 KW > Pct= 6,55 KWcn c vo phc lc P1.1, P1.2, P1.3 ta c bng 2:

Kiungc

Cngsut

KW

VntcVg/ph

Hiusut

%

Tk/Tmax Cos Tk/Tn

K160S4 7,5 1450 87,5 2,2 0,86 5,8

2. S vng quay ca xch ti

nxt= = = 41,27 vg/ph3. Phn phi t s truyn

theo hnh v ta c 3 t s truyn u1, u2, uxtng ng:

bnh rng tr rng thng, bnh rng tr rng nghing,

xch ti.

ta c: uc= u1.u2.ux= nc/nxt== 35,13ta chn Uh= u1.u2phi tha mn 3 ch tiu:

-Khi lng nh nht

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-Momen qun tnh nh nht

-Th tch bnh ln nhng trong du nh nht.

ta chn Uh= 10 vi hp gim tc phn i v tra bng

3.1 suy ra:

u1= 3,58; u2= 2,79 => ux= 3.52

4. Tc quay cc trc

C 4 trc v 3 t s truyn

truc I ni vi ng c nn nc= n1= 1450 vg/ph

trc II: n2= n2/u1== 405,03 vg/phtrc III: n3= n2/u2= = 145,17 vg/phtrc IV: n4= n3/nx= = 41,24 vg/ph5. Momen xon trn cc trc ng c :

Tc= =

= 43139,66 NmmT1= = 41427,24 NmmT2t= =

= 140999,43Nmm

T2n= T2t/2 = 70499,72 Nmm

T3n=

T4=

Lp bng 3: cng sut, s vng quay v momen xon

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Trc Trcng c

Trc I Trc II Trc II Trc III TrcThngs

Rngthng

Rngnghing

CngsutKW

6.55 6,29 5,98 2,99 2,73 4,8

Vngquayvg/ph

1450 1450 404,03 405,03 145,17 41,2

Momen

xonNmm

43139,66 41427,24 140999,43 70499,72 179592,89 1125

II. Tnh ton b truyn ngoi

Tnh ton b tuyn xch vi cc thng s sau:

Cng sut bnh dn PX= P3n= 5,45 KW

T s tuyn ux= 3,52S vng quay trc III l: n3= 145,17 vg/ph

Gc nghing ng ni tm ca b truyn ngoi: 900

c tnh lm vic ca xch: chu va p

1. Chn loi xch

V ti trng tng i nh, vn tc thp, ta dng xch

con ln theo bng 5.4 ta c: z1x= 25...23, ta chn z1x=25, do s rng a ln z2x= z1x.ux= 25,3.52 = 88

t s truyn thc: ux== = ux2. iu kin m bo ch tiu iu kin mn

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iu kin m bo ch tiu iu kin mn ca b truyn

xch tnh theo cng

thc: Pt= P.K.Kz.Kn < [P]

(1x)

trong : P = Px= 5,45 KW

h s dng rng: Kz= 25/Z1x= 25/25 = 1

h s vng quay Kn= n01/n1xvi n1x= 145,17 vg/ph

tra bng 5.5 ta chn n01= 200 vg/ph, do Kn= 1,38

K c tnh theo cng thc: K = Ko.Ka.Kc.Kbt.K.Kc(2x)

tra bng 5.6 ta c:

ng ni tm a xch lm vi ng nm ngang mt gc

> [] ta lyKo= 1,25

khong cch trc ly a = 60.P => Ka= 0,8

v tr ca trc iu chnh bng xch cng, ta ly Kc=

1,1

ti trng va p ly K= 1,4

lm vic 2 ca ly Kc= 1,25

mi trng lm vic c bi v ch bi trn II. Taly Kbt= 1,3

Thay cc thng s vo (2x) ta c:

K = 1,25.0,8.1,1.1,3.1,4.1,25 = 2,50

t (1x) ta c: Pt= 5,45.1.1,38.2,5 = 18,80 KW

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tra bng 5.5 v n01= 200vg/ph, ta chn b truyn xch 1

dy c bc xch

P = 31,75mm, tha mn iu kin mn Pt= 18,80 KW < [P]

= 19,3 KW

3. Khong cch trc s b:

a = 60.P = 60.31,75 = 1905mm;

s mt xch tnh theo cng thc: X = [(

.P]/(4.a.

)

=> X = 2.60+(25+88)/2+.31,75/(4..2.1905) =178,18

u tin ly s mt xch chn: X = 178.

khong cch trc tnh theo cng thc:

a = (P/4).{X-(Z1x+Z2x)/2+ => a = 1902,17 mm

trnh cng xch qu ln, khong cnh trc a tnh

c cn gim bt mt lng

a = (0,002...0,004).a = 3,80...7,61 mm

vy ta ly khong cch trc a = 1898mm

s ln va p xch trong mt giy:

i =

ln < 25 ln.4. Kim nghim xch v h s an ton

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Vi cc b truyn xch b qu ti khi m my hoc

thng xuyn chi ti trng va p trong qu trnh lm

vic, cn tin hnh kim nghim v qu ti theo h san ton: S = Q/(K.Ft+Fo+Fv) > [S]

(3x)

trong : Q l ti trng ph hng, tra bng 5.2 ta c:

Q = 88500 N v q = 3,8kg/m

K= 1,4 l h s ti trng ng

vi

= 1,4 v c tnh lm vic ca ti l chu va

p nn ta ly Kd= 1,7

vn tc xch v =

m/slc vng Ft= lc cng do lc ly tm gy ra: Fv= q = 14,01N

Folc cng do nhnh xch b ng sinh ra, tnh theo

cng thc Fo= 9,81.Kf.q.a (4x)

vi a = 1,898 m; q = 3,8 kg/m; Kf= 2 v t (4x) suy ra: Fo= 9,81.2.3,8.1,898 = 141,51 N

thay cc gi tr vo (3x) ta c:

S = 88500/(1,7.2838,54+141,51+14,01) = 17,77

tra bng 5.10 ta c: [S] = 8,5 < S = 17,17. Vy b

truyn tha mn h s an ton.

5. Xc nh ng knh xch

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bnh dn: d1x= =

= 253,32 mm

bnh b dn: d2x= ta c: r = 0,5025.d1-0,05

(5x)

tra bng 5.2 vi bc xch P = 31,75 mm , ta c: d1=

19,05; t (5x) ta c:

r = 0,5025.19,05-0,05 = 9,52 mm

ng knh chn xch

df1= d1x-2.r = 253,32-2.9,52 = 234,28 mm;

df2= d2x-2.r = 889,55-2.9,52 = 870,51 mm.

6. Kim nghim bn tip xc

Kim nghim bn tip xc ca xch tnh theo cng

thc:

[(6x)

Ta c: Ft= 2838,54 N; Kd= 1,4; E = 2,1.MPa l mun n hi vi bc xch P = 31,75 mm v xch c m = 1

dy, da vo 5.2 ta c: A = 262 Kr-h s k n rng a xch ph thuc vo Z

i vi bnh dn: Z1z= 25 => Kr1= 0,44; bnh b dn:

Z2x= 88 => Kr2= 0,20

lc va p trn 1m xch, tnh theo cng thc: Fvd=

13..nx.p3.m;

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vi n1x= 145,17 vg/ph; n2x= 41,24 vg/ph . Suy ra:

Fvd1= 13..145,17..1 = 6,04 NFvd2= 13..41,24..1 = 1,72 Nt (7x) ta c :

= 470,61 MPa=0,47.= 317,11 MPa

tra bng 5.11 ta chn thp 45, ti ci thin c rnb mt rng HR(45...60) v ng sut tip xc cho php

[ MPa .7. Xc nh lc tc dng ln trc

Lc sinh ra trn nhnh b ng: F2= Fo+Fv =

141,51+14,01 = 155,52 N

Lc sinh ra trn nhnh ch ng: F1= Ft+F2=

2838,54+155,52 = 2994,06 N

III. Tnh ton b truyn bnh rng

A: b truyn cp bnh rng thng trn trc I v II

cng sut P1= 6,29 KW; s vng quay u1= 3,58; n1=

1450 vg/ph.

1. Chn vt liu

Chn vt liu lm bnh rng nh l thp 45, ti ci

thin, c rn r1= 220HB,gii hn bn b2= 750MPa, gii hn chy ch1= 450 MPa

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chn vt liu lm bnh rng ln l thp 40 c rn r2= 200HB, ti ci thin, gii hn bn

b2= 700 MPa, gii

hn chy ch2= 400 MPa.2. Xc nh ng sut cho php

a. Ta c ng sut tip xc cho php [H1] v [H2] cxc nh theo cng thc

[H1] = (Hlim/SH).ZR1.zV1.KXH1.KHL1(1)

[H2] = (Hlim)/SH).ZR2.ZV2.KXH2.KHL2(2)

s chu k chu ti tng ng NHE2ca bnh rng ln

tnh theo cng thc:

NHE2= 60.c. n2.ti(3)

mi vng quay bnh rng ch n khp 1 ln nn c = 1. T(3) suy ra:

NHE2= 60.1.405,03.12500.( ) = 1,49.s chu k c s ca bnh ln NHO2= 30. =1,00..V NHE2 > NHO2 => KHL2 = 1

tng t: NHE1= NHE2.u1= 1.49.3.58 = 5,33.chu k.S chu k c s bnh nh NHO1= 30. 30. =1,26.. V NHE1 = NHO1 => KHL = 1gii bn tip xc tnh theo cng thc :

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Hlim1 oHlim1.KHL1; Hlim2 =0Hlim2.

gii hn bn mi ca bnh rng nh v bnh ln ta trabng 6.2 ta c:

0Hlim = 2.HB1+70 = 2.220+70 = 510 MPa;oHlim1 = 2.HB2+70 = 2.200+70 = 470 MPa;trong tnh ton sbta ly: ZR1.ZV1.KHL1 = 1; ZR2.ZV2.KHL2= 1.

T(1) v (2) suy ra:

[H1] = (510/1,1) = 463,64 MPa[H2] = 470/1,1 = 427,27 MPang sut cho php tnh theo hai cch sau:

[H]' = 0,5.([H1]+[H2]) = 0,5.(463,64+427,27) = 445,46MPa

[H]'' = 1,18[H1] = 1,18.427,27 = 504,18 MPa.Vy ta chn: [H] = 445,46 MPab. Ta c ng sut un cho php:

Cng thc tnh ng sut un cho php tnh theo cngthc:

[] = (Flim1/SF1).YR1.YS1.KXF1.KFC1.KFL1(4)

[] = (Flim2/SF2).YR2.YS2.KXF2.KFC2.KFL2(5)

trong tnh ton s b ta ly YR1.YS1.KXF1 = 1 v YR2.YS2.KXF2= 1

tra bng 6.2 ta c:SF1 = SF2 = SF = 1,75

chu k chu ti tng ng NFE2ca bnh ln tnh theocng thc:

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NFE2 = 60.c n2.ti(6)

do y l thp ti ci thin nn mF=6.T (6) suy ra:

NFE2 = 60.1.405,03.12500.( = 1,20.chu k=> NFE1 = u1.NFE2

NFE1 = 3,58.1,20= 4,30.chu ks chu k c s NFO1 = NFO2 = NFO = 4.v NFE1 > NFOv NFE2 > NFO=> KFL1 = KFL2 = 1

gii hn bn mi un tnh theo cng thc:OFlim1 = 1,8HB1 = 1,8.220 = 396MPa;OFlim2 = 1,8.HB2 =1,8.200 = 360 MPa

suy ra:

[F1] = 396/1,75 = 226,29 MPa; [F2] = 360/1,75 = 205,71MPa

c.Ta c ng sut qu ti cho php:

c1.Ta c ng sut tip xc qu ti cho php

bnh nh: [] = 2,8.450 = 1260 MPabnh ln: [] = 2,8.400 =1120 MPavy ta chn [] = 1120 MPac2.Ta c ng sut un qu ti cho php:

bnh nh: [F1]MAX = 0,8.ch1 = 0,8.450 = 360 MPabnh ln: [F2]MAX = 0,8.ch2= 0,8.400 = 320 MPa3. Tnh khong cch trc s b

aW1 = Ka.(u1+1)

(7)

trong : Ka l h s, tra bng 6.5, ta c: Ka= 49,5.Ybal h s chiu rng vnh rng , tra bng 6.6 ta chnYba= 0,3. T cng thc Ybd= 0,53.(u1+1). Yba; suy ra:

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Ybd= 0,53(3,58+1).0.3 = 0,73

tra bng 6.7 vi Ybd= 0,73 v t s truyn s 7, ta c

KHB = 1,02. T (7) suy ra:

aW1 = 49,5.(3,58+1).

= 132,20 mm

4. Xc nh cc thng s n khp

m un bnh rng xc nh theo cng thc: m =(0,01...0,02)aW => m= (0,01...0,02).132,20 = (1,32...2,64) mm.

tr s m un ly theo tiu chun, u tin dy 1, ta

chn m = 2

v l bnh rng thng nn = 0. S rng bnh rng nhxc nh theo cng thc:

Z1 =

=

= 29 => Z2 = u1.Z1=29.3,58 = 104t s truyn theo s rng bnh 1 v bnh 2 l:

u1' = Z2/Z1 = 104/29 = 3,59

sai lch t s truyn l thuyt so vi t s truyn

thc t :

sai lch qu nh do ta ly

u1 = 3,58 => khong cch trc chnh xc:

aW =

.m = .2 = 133 mm5. Kim nghim rng v bn tip xc

Ta c ng sut tip xc xut hin trn mt rng ca btruyn phi tha mn iu kin sau:

H = ZM.ZM.Z < [H](8)

trong : ZMtra bng 6.5, ta c: ZM = 274MPa1/3

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ZHl h s an ton k n hnh dng rng v bn mt

tip xc tnh theo cng thc: ZH =

. V

= 0

=> = 1,76Zh s trng khp tnh theo cng thc: Z (9)

vi l h s trng khp ngang tnh theo cng thc: 1/Z1+Z2)] = [1,88-3,2.(1/29+1/104) = 1,74thay vo (9) ta c: = 0,87T1 = 41427,24 Nmm, KHB = 1,02;=1 v y l bnh rngthng

KH = KHB.KHV.KHVl h s k n ti trng ng, xut hin trongvng n khp, tr s KHVtnh theo cng thc: KHV =

1+(VH.bW.dW1)/(2.T1.KHB

)

(10)ng knh vng ln bnh nh l : dW1 = Z1.m = 29.2 = 58mm;

bW = Yba.a = 0,3.133 = 39,9 mm, ta ly dW = 40 mm

vn tc vng: V = = = 4,25 m/sda vo bng 6.13 v V < 6 m/s, ta chn cp chnh xcng hc l cp 8.

da vo bng 6.14 ta chn cp chnh xc mc lm vic ml cp 8

ta c cng thc: VH =.g V.tra bng 6.15 ta c = 0,004 vi iu kin bnh rngthng c vt mp u rng

tra bng 6.16 ta c: g = 56

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do : VH = 0,004.56.4,25.= 5,80 m/st (10) ta c: KHV = 1 = 1,15t (8) suy ra: = 403,55MPa

ta xt: ZR.ZV.KXH

ZRh s k n nhm mt rng lm vic, vi cp chnhxc mc lm vic m l cp 8, khi gia cng vi nhm R

a=2,5...1,25 um, do Z

R= 0,95;

vi dw < 700mm => KHV = 1; ZVl h s nh hng vntc vng , khi rn mt rng HB < 350 MPa => ZV =0,85.V0,1=> Zv = 0,95.4,250,1= 0,98

suy ra ng sut tnh li l: [ZR.ZV.KXH =445,46.0,95.1.0,98 = 414,75 MPa >Vy vt liu chn tha mn iu kin bn tip xc.

6.Ki m rng v bn un

m bo bn un cho rng, ng sut sinh ra ti

chn rng khng vt mc cho php:

=2.T1..KF..F1/(bw.dw1.m) [](11)

.YF1/YF2(12)

trong : T1 = 41427,24 Nmm; m = 2mm; bw = 40mm ; dw1 = 56mm v

= 1/1,74 = 0,57 l h s k n trng khprng

= 1 i vi bnh rng thng

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ZV1 = 29 v ZV2= 104 tra bng 6.18 vi h s dch rng

x1 = x2 = 0;

ta c YF1 = 3,8; YF2 = 3,6.

KFl h s ti trng khi tnh v un: KF = KFB.KFV.vi Yba = 0,73 v t s truyn 7, ta tra bng 6.7 ta c

KFB = 1,02; l h s k n s phn b khng u titrng cho cc i rng, i vi rng thng .KFVl h s k n ti trng ng xut hin trong

vng n khp khi tnh v un :

KFV = 1+VF.bw.dw1/(2.T1.KFb.)(13)

vi VF = g.VTng t tra ng sut tip xc, ta c:= 0,011; g = 56

=> VF = 0,011.56.4,25.= 15,96 m/s

t (13) suy ra:

KFV = 1+ =1,44t (11) suy ra:

= 2.41427,24.1,02.1.1,44.0,57..3,81/(40.58.2) =56,81 MPa

t (12) =>= 56,81.3,6/3,8 = 53,82 MPata xt: YR.YS.KXF = YR1.YS1.KXF1 = YR2.YS2.KXF2

YR = 1 l h s k n nhm mt ln chn rng

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YS = 1,08-0,0695.lnm = 1,08-0,0695.ln2 = 1,03 l h s

k n nhy ca vt liu i vi tp trung ng sut

KXF = 1 v dw< 400 mm => YR.YS.KXF = 1.1,03.1 = 1,03

ng sut un cho php tnh li:

[] = 226,29.1,03 = 233,08 MPa > MPa[ = 205,71.1,03 = 211,88 MPa > = 53,82 MPa7. Kim nghim rng v qu ti

trnh bin dng hay dn gy lp b mt, ng suttip xc cc i khng vt mc cho php: = 1120 MPaTng t khi qu ti v un:

= 360 MPa

= 320 MPa

Vy vt liu chn t yu cu bn.

8. Cc thng s v kch thc b truyn

Lp bng 4:

STT Thng s khiu

Gi tr n v

1 Khong cch trc aw 133 mm

2 ng knh chia d1 m.Z1 = 58 mmd2 m.Z2 = 208 mm

3 ng knh nh rng da1

m.Z1+2.m =62

mm

da2 m.Z2+2.m =212

mm

4 ng knh chn rng df1 m.Z1-2,5.m = mm

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51

df2 m.Z2-2,5.m mm

5 Gc profin rng gc 200

6 M un m 2 mm7 Cp chnh xc ng hc Ch 88 Cp chnh xc lm vic Clv 89 H s dch chnh X1 0

X2 010 H s trng khp ngang 1,7411 Vn tc vng V 4,25 m/s12 Chu rng vnh rng bw 40 mm

13 T s truyn u1 3,5814

B. B truyn bnh rng nghing

trc II v III, c hai cp bnh rng nghing ging nhau

do ta ch xt mt cp bnh rng nghing vi cng

sut P2n = 2,99 KW v s vng quay n2 = 405,03 vg/ph

mo men xon T2n = 70499,72 Nmm v t s truyn u2 = 2,79

1. Chn vt liu

Chn vt liu lm bnh rng nh l thp 45x, ti ci

thin, c rn r1=240HB, giihn bn b2 = 850 MPa, gii hn chy ch1 = 650 MPachn vt liu lm bnh rng ln l thp 40x c rn

r2 = 220HB, ti ci thin, gii hn bn

b2 = 700 MPa,

gii hn chy ch2 = 550 MPa.2. Xc nh ng sut cho php

a.Ta c ng sut tip xc cho php [H1] v [H2] cxc nh theo cng thc:

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[H1] = (Hlim/SH).ZR1.zV1.KXH1.KHL1(1n)

[H2] = (Hlim)/SH).ZR2.ZV2.KXH2.KHL2(2n)

s chu k chu ti tng ng NHE2ca bnh rng ln

tnh theo cng thc:

NHE2 = 60.c. )3n1.ti(3n)

mi vng quay bnh rng ch n khp 1 ln nn c = 1. T

(3n) suy ra:

NHE2 = 60.1.145,17.12500.( ) = 5,35.s chuk c s ca bnh ln: NHO2 = 30. =1,26..V NHE2 > NHO2 => KHL2 = 1

tng t: NHE1 = NHE2.u2 = 2,79.5,35.chu k.S chu k c s bnh nh NHO1 = 30. 30. =1,55..V NHE1 =NHO1 => KHL = 1

gii bn tip xc tnh theo cng thc:

Hlim1 oHlim1.KHL1; Hlim2 =0Hlim2.KHL2gii hn bn mi ca bnh rng nh v bnh rng ln ta

tra bng 6.2 ta c:

0Hlim = 2.HB1+70 = 2.240+70 = 550 MPa;oHlim1 = 2.HB2+70 = 2.220+70 = 510 MPa;

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trong tnh ton sbta ly: ZR1.ZV1.KHL1 =1; ZR2.ZV2.KHL2= 1.

T(1) v (2) suy ra:

[H1] = (550/1,1) = 500,00 MPa[H2] = 510/1,1 = 463,64 MPang sut cho php tnh theo hai cch sau :

[H]' = 0,5.([H1]+[H2]) = 0,5.(500+463,64) = 481,82 MPa[H]'' = 1,18[H1] = 1,18.463,64 = 547,10 MPa.Vy ta chn ng sut cho php vi gi tr nh hn: [H]= 481,82 MPab.Ta c ng sut un cho php

Cng thc tnh ng sut un cho php tnh theo cngthc:

[] = (Flim1/SF1).YR1.YS1.KXF1.KFC1.KFL1(4n)

[

] = (

Flim2/SF2).YR2.YS2.KXF2.KFC2.KFL2

(5n)

trong tnh ton s b ta ly: YR1.YS1.KXF1 = 1 vYR2.YS2.KXF2 = 1

tra bng 6.2 ta c: SF1 = SF2 = SF = 1,75

chu k chu ti tng ng NFE2ca bnh ln tnh theocng thc:

NFE2 = 60.c n2.ti(6n)do y l thp ti ci thin nn mF = 6. T (6) suy ra:

NFE2 = 60.1.147,17.12500.( = 4,30.chu k=> NFE1 = u2.NFE2

NFE1 = 2,79.4,3.= 1,20.chu k

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s chu k c s NFO1 = NFO2 = NFO = 4.v NFE1 > NFOv NFE2 > NFO=> KFL1 = KFL2 = 1

gii hn bn mi un tnh theo cng thc :

OFlim1 = 1,8HB1 = 1,8.240 = 432 MPa;OFlim2 = 1,8.HB2 =1,8.220 = 396 MPa

suy ra:

[F1] = 432/1,75 = 246,86 MPa;[F2] = 396/1,75 = 226,29 MPa.c. Ta c ng sut qu ti cho php:

c1. Ta c ng sut tip xc qu ti cho php:

bnh nh: []MAX= 2,8.= 2,8.650 = 1820 MPabnh ln: []MAX= 2,8.= 2,8.550= 1540 MPaTa chn []MAX= 1540 MPac2. Ta c ng sut un qu ti cho php:

[F1]MAX = 0,8.ch1 = 0,8.650 = 520 MPa[F2]MAX = 0,8.ch2 = 0,8.550 = 440 MPa3. Tnh khong cch trc s baW1 = Ka.(u2+1)

(7n)

trong : Kal h s, tra bng 6.5, ta c: Ka= 43. Ybal h s chiu rng vnh rng, tra bng 6.6 ta chn Yba= 0,3. T cng thc Ybd= 0,53.(u2+1).Yba; suy ra:

Ybd= 0,53(2,79+1).0,3 = 0,60 v t s truyn s 7 =>ta c KHB = 1,07

T (7n) suy ra:

aW2 = 43.(2,79+1).

= 118,89 mm. Ta ly aw2 =

128 mm

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4. Xc nh cc thng s n khp :

M un php bnh rng xc nh theo cng thc: mn =

(0,01...0,02)aW => mn =(0,01...0,02).128 = (1,28...2,56) mm.

tr s m un php ly theo tiu chun, u tin dy 1,ta chn mn = 2

chn s b gc nghing , do s rng bnh rngnh xc nh theo cng thc:

Z1 =

=

= 33,26. Ta ly Z1 = 33 rng=> Z2 = u1.Z1 = 33.2,79 = 92,07 => ly Z2 = 93 rngt s truyn theo s rng bnh 1 v bnh 2 l:

u2' = Z2/Z1 = 93/33 = 2,82

sai lch t s truyn l thuyt so vi t s truynthc t:

sai lch qu nh do ta ly

u1 = 2,82 gc nghing ca rng l: = mn.(z1+z2)/(2.aw2 ) = 2.(33+93)/(2.128) = 0,98=>= 10,140. V theo tiu chun 0nn ta chn= 100ph hp vi tiu chunh s dch tm y l: y = aw2/mn-(Z1+Z2)/2 = 128/2-(33+93)/2 = 1,00 mm

h s Ky = = = 7,94,vi Zt = Z1+Z2 = 126 rng

tra bng 6.10 ta c Kx = 0,445; vy h s gim nhrng :

y = = 0,06mmtng h s dch chnh Xt = y+y = 1,00+0,05 = 1,06 mmh s dch chnh bnh 1:

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X1 =

=

= 0,29 mmX2= Xt - X1= 0,77 mm

cui cng gc n khp: =

= 0,93=> = 22,330gc prfin gc = 200gc prfin ca rng:

= arctg( = arctg(/) = 20,2805. Kim nghim rng v bn tip xc

Ta c ng sut tip xc xut hin trn mt rng ca btruyn phi tha mn iu kin sau:

H = ZM.ZM.Z < [H](8n)

trong : ZMtra bng 6.5, ta c: ZM = 274 MPa1/3

ZHl h s an ton k n hnh dng rng v bn mt

tip xc tnh theo cng thc: ZH = (9n )= .=>= arctg( ) =arctg(cos20,280.tg100) = 9,390

t (9n) suy ra ZH = = 1,67.vi l h s trng khp ngang tnh theo cng thc:

1/Z1+Z2)].= [1,88-3,2.(1/33+1/93)= 1,72l h s trng khp dc tnh theo cng thc: (10n)

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vi bw= Yba.aw2= 0,3.128 = 38,40 mm

t (10n) suy ra:

= 1,06 > 1. T suy ra:

tnh theo cng thc: ==

= 0,76momen xon T2 = 70499,72 Nmm

dw1= 2.aw2/(u2+1) = 2.128/(2,82+1) = 67,02 mm

KH h s ti trng khi tnh v tip xc, KH= KHB.KHV.(12n)

trong : KHB= 1,07

l h s k n s phn b khng iu ti trng nkhp

vn tc vng: V = = = 1,42 m/stra bng 6.13 v 6.14 ta chn cp chnh xc ng hc l

cp 9 v cp chnh xc mc lm vic m l cp 8 => =1,05

KHVl h s k n ti trng ng, xut hin trong

vng n khp, tr s KHVtnh theo cng thc: KHV =1+(VH.bW.dW1)/(2.T2.KHB )(13n)

ta c cng thc: VH =.g V.tra bng 6.15 ta c = 0,002 vi iu kin bnh rngthng c vt mp u rng

tra bng 6.16 ta c: g = 56

do : VH = 0,002.56.1,42 = 1,07 m/st (13n) ta c: KHV = 1 = 1,02t (8n) suy ra:

= 393,15 MPa

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ta xt: ZR.ZV.KXH

ZRh s k n nhm mt rng lm vic,vi cp chnh

xc mc lm vic m l cp 8, khi gia cng vi nhm Ra= 2,50...1,25 m, do ZR = 0,95;vi dw < 700mm tc l dw1, dw2 < 700 mm => KXH=1; ZVl h s nh hng vn tc vng, khi rn mt rng HB< 350 MPa => ZV = 0,85.V

0,1

=> Zv = 0,85.1,420,1= 0,88

suy ra ng sut cho php tnh li l :

[

.ZR.ZV.KXH= 481,82.0,95.1.0,88 = 402,80 MPa >

Ta ly rng vnh rng bw= 40mm.

Vy vt liu chn tha mn iu kin bn tip xc

6. Kim nghim rng v bn un

m bo bn un cho rng, ng sut sinh ra ti

chn rng khng vt mc cho php:

= 2.T2..KF..F1/(bw.dw1.m) [](14n)

.YF2/YF1(15n)

trong : T2 = 70499,72 Nmm; m = 2mm php; bw= 40mm ;

dw1= 67,02 mm v

=1/1,71 = 0,58 l h s k n trng khp rng

= 1-/1400= 1-100/1400 = 0,93 l h s k n nghing ca rngYF1vYF2h s dnh dng ca bnh 1 v 2 ph thuc vo

s rng tng ng Ztd1 v Ztd2l:

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Ztd1=

= = 34,55; Ztd2 = = = 97,37tra bng 6.18 v h s dch chnh ta c:

X1= 0,29 => YF1= 3,54; X2= 0,77 => YF2= 3,48

KFl h s ti trng khi tnh v un: KF= KFB.KFV.vi Ybd= 0,73 v t s truyn 7, ta tra bng 6.7 ta

c KFB= 1,17

l h s k n s phn b khng u ti trng chocc i rng, vi cp chnh xc mc lm vic m l cp8: V = 1,42m/s < 2,5 m/s v tra bng 6.14

ta c = 1,22KFVl h s k n ti trng ng xut hin trong

vng n khp khi tnh v un :

KFV= 1+VF.bw.dw1/(2.T2.KFB.)(16n)

vi VF= g.V(17n)

tng t tra ng sut tip xc ,ta c: = 0,006; g = 56=>VF= 0,006.56.1,29.= 3,21

t (15n) suy ra :

KFV= 1+

= 1,04t (14n) suy ra:

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= 2.70499,72.1,17.1,22.1,04.0,58.3,54.0,93./(40.67,02.2) = 74,54 MPa

t (15n) =>= 74,54.3,48/3,54 = 73,28 MPata xt: YR.YS.KXF= YR1.YS1.KXF1= YR2.YS2.KXF2

YR= 1 l h s k n nhm mt ln chn rng i

bnh rng thng

YS=1,08-0,0695.lnm = 1,08-0,0695.ln2 = 1,03 l h s k

n nhy ca vt liu i vi tp trung ng sut

KXF= 1 v dw< 400 mm => YR.YS.KXF= 1.1,03.1 = 1,03

ng sut un cho php tnh li:

[] = 246,86.1,03 = 254,27 MPa > MPa[= 226,29.1,03 = 233,08 MPa > = 73,28 MPa7. Kim nghim rng v qu ti

ta c: Kqt= Tmax/T = 1,4

trnh bin dng hay dn gy lp b mt, ng sut

tip xc cc i khng vt mc cho php: = 1540MPatng t khi qu ti v un:

= 520 MPa

= 440 MPaVy vt liu chn t yu cu bn qu ti.

8.Cc thng s v kch thc b truyn

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Lp bng 5:

STT

Thng s khiu

Gi tr nv

1 Khong cch trcchia

a 0,5.m.(Z1+Z2) =127,94

mm

2 Khong cch trc aw a. = 129,74 mm3 ng chia d1 m.Z1/ = 67,02 mm

d2 m.Z2/ = 188,87 mm4 ng knh ln dw1 2.aw/(u1+1) = 67,93 mm

dw2 dw1.u1= 191,56 mm

5 ng knh nhrng

da1 da1+2(1+X1-y).m =71,94

mm

da2 da2+2(1+X2-y).m =195,71

mm

6 ng knh yrng

df1 df1=d1-(2,5-2.x1)m =63,18

mm

df2 df2=d2-(2,5-2.x2)m =186,95

mm

7 ng knh c s db1 d1.=62,98 mmdb2 d2. = 177,48 mm8 Gc profin gc 2009 Gc profin rng 20,28010 Gc n khp 22,33011 Tng h s dch

chnhXt X1+X2= 1,06 mm

12 H s dch chnhbnh 1

X1 0,29 mm

13 H s dch chnhbnh 2 X2 0,77 mm

14 H s trng khpngang

1,7115 H s trng khp

dc 1,06

16 H s trng khp 0,76

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17 Cp chnh xc nghc

ccx 9

18 Cp chnh xc mclm vic m ccx 8

19 Chiu rng vnhrng

bw 40 mm

20 T s truyn U2 2,8221 Vn tc vng v 1,42 m/s22

IV tnh ton thit k trc trong hp gim tc

1. Chn vt liu:

Chn vt liu ch to trc l thp 45 ti ci thin, c

= 600MPa, ng sut xon cho php l: [] =12...30 MPa2. Xc nh s b ng knh trc:

Xc nh s b ng knh trc theo cng thc: di (1t)

Ta c:

trc I: P1= 6,29 KW; n1= 1450 vg/ph; T1= 41427,24

Nmm; u1= 3,58;

trc II: P2t= 5,98 KW; n2= 405,03 vg/ph; T2t=

140999,43 Nmm; u2= 2,82;

P2n= 2,99 KW; T2n= 70499,72 Nmm

trc III: P3= 2,73 KW; n3= 145,17 vg/ph; T3=

179592,72 Nmm;

trc IV: P4= 4,86 KW; n4= 41,24 vg/ph; T4= 1125436,47

Nmm;

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t (1t) ta c ng knh trc s b ti ch lp bnh

rng ;

d1 > (24,0...19,0) mm; d2t> (36.0...28,6) mm; d2n>

(28,6...22,7) mm;

d3n > (39,1...31) mm; d4x> (72,1...52,27) mm.

Ta chn ng knh trc theo tiu chun: d1= 24 mm; d2t

= 30mm;

d2n= 25mm; d3= d3n= 35mm; d4= d4x= 60mm.

3. Xc nh khong cch gia cc gi v im t lc:

a. Da vo ng knh trc, s dng bng 10.2 chn

chiu rng ln tng ng vi cc trc l: b1= 15mm;

b2= 19mm; b3= 21mm; b4= 33mm;

chiu rng vnh rng v may tnh phn bnh rng,

ta c:

l1t = bw1 = l2t= l2n = bw2 = l3n = l = 40 mm;b. Xc nh chiu di phn na ni trc, ta chn ni

trc a, bao gm hai a may , mi a lp trn on

cui ca trc bng mi ghp then

hai na a ni vi nhau bng mi ghp bu lng, khng

c khe h , phi chu momen xon Tk > 43139,66 Nmm. Tra

bng 16.4 ta chn thp 45 c Tk= 63000 Nmm,ng knh

ngoi D = 100mm, d = d1 = 20mm, chiu di khp ni

L = 100mm , c phn nm bn trc I di l1,= L/2 = 50 mm.

cn c vo bng 10.3 v 10.4 tnh khong cch lki

trn trc th k t gi O n chi tit quay th i v

kt hp hnh 1

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ta chn khong cch gia hai chi tit k1= 15 mm;

b dy ca hp gim tc l: t = bi+2.k2i;

trong b chiu rng ln v k2il khong cch t

mt nt n thnh trong ca hp .

trc II c b2 =19 mm; ta chn k22 = 10 mm => t = 39 mm

trc I c b1 = 15 => k21 = (t-b1)/2 = (39-15)/2 = 12 mm

trc III c b3 = 21 mm => k23 = (39-21)/2 = 9 mm.

trc I v III c chiu cao np v bu lng l hi,tac :

h1+b1/2 = h3+b3/2 => h1+7,5 = h3+10,5

v h1v h3>k21,tra bng 10.3 ta chn h1=18 mm,h3=15 mm

c. Xc nh chiu bnh ai ta tnh theo cng thc:

B

.

t phn II ta c: R = RX= 5,45KW; K = 2,50; KZ= 1; Kn

= 1,38 => B 9,74 mmta ly B =12 mm

d. Xc nh chiu di cc trc.

Hnh v 1: xc nh chiu di cc trc.

d1. Chiu di trc I:

lc11 = L/2+k3+h1+b1/2 = 50+15+18+15/2 = 90,50 mm;

l12 = l23 = 103,50 mm; l11= l21 = 207 mm.

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h k1 21 k

21l

l ll

L/2

c11

12

11

k3

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k22

h3

k1

k23

l

ll

l

ll

l

21

22

23

24

3234

30

lc33

k3

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d2. Chiu di trc II:

l22 = b2/2+k2+k1+l/2 =19/2+12+20+40/2 = 51,5 mm;

l23 = b2/2+k2+2.k1+3.l/2 = 19/2+10+2.12+60 = 103,5 mm;

l24 = b2/2+k2+ 3.k1+5.l/2 = 19/2+10+3.12+100 = 155,5 mm;

l21 = b2+2.k2+4.k1+3.l = 19+2.10+4.12+3.40 = 207 mm.

d3. Chiu di trc III:

l32 = l22 = 51,5 mm; l34 = l24 = 155,5 mm;

l31= l21 = 207 mm; lc33 = 46,5 mm.

4. Xc nh tr s lc v ng knh trc :

a. Xt trc I:

chn h trc ta oxyz , trc I c mt bnh rng,

2 cp ln v 1 khp ni, ta

thu v trng tm chi tit lp trn trc. Chn trc quay

ngc chiu kim ng h.

chiu di trc: l10 = l11+lc11 = 207+90,5 = 297,5 mm;

lc vng Ft1 = 2.=

= 1428,53 mm;

vi gc = 0 => Fa = Ft1.tg = 0; Fr1 = Ft1. =1428,53.tg200= 519,94 N

ta xt trng hp xu nht ca trc l: cng chiu . Khi :Fk= 6.= 6. = 1246,21 Nmm;M1= T1= 41427,24 Nmm.

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Hnh 2: momen xon v momen un.

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Fr 1

Ft1

Ft1

Fr1

Fk

Mz

Mz

n

Fk F t1

YA

Fr

Y

C BA

D

90.50

103.50103.50

17539.88Nmm

26906.90 Nmm

41427.24 Nmm

XX

AB

Momen

truc Y

Momen

truc

X

Momen

truc Z

B

1

112782.01N mm

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a1. Xc nh momen trn trc y:

phng trnh cn bng ti A v B l:

= 0 => 90,5.Fk+207.XB-103,5. Ft= 0XB = = = 169,42 NXA = Fk+Ft-XB = 1246,21+1428,53-169,42 = 2505,32 N.

Tr s momen ti mt s im c bit:

MD = 0;

MAY = 90,5.Fk = 90,5.1246,21 = 112782,01 Nmm;

MCY = 194.Fk-103,5.XA = 194.1246,21-103,5.2505,32 = -

17535,88 Nmm;

MBY = 0;

a2. Momen trn trc x:

t hnh 2 ta c:

YA = YB = = = 259,97 N;momen xon ti im C l:

MCX = -103,5.YA = -103,5.259,97 = -26906,90 Nmm;

a3. Momen trn trc z:

ng c in sinh ra momen ch ng, do khp ni

bn phn trc sinh ra

momen b ng nhm chng li chuyn ng cho nn MZ

ngc chiu vi n1 v

MZ = M1 = 41427,24 Nmm

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a4. Momen tng hp ti mt s im v ng knh thc

ti

MA = = 120149,90 Nmm;Mc = =

= 52419,89 Nmm;

tra bng 10.5, vi thp 45 c tr s ng sut cho php:

[] = 63 MPa = 63 N/mm2ng knh trc c xc nh bi cng thc:

di => ng knh trc ti mt s im c bit :

dA

= = 26,72 mm;

dC = 20,26 mm;

dD

= 18,73 mm.

Theo tiu chun ta chn:

dA = 30 mm; dC = 22 mm; dD = 19 mm;

b. Trc II c mt bnh rng thng v hai bnh rngnghing , ta c cc lc:

bnh 1: Ft1 = = = 1355,76 mm;vi ; Fa1 = 0;

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Fr1 = Ft1. = 1355,76.tg200= 493,46 N;

bnh 2: Ft2 = = = = 2075,66 N;Fr2 = Ft2./ = = 865,71 N;Fa2 = Ft1. 2075,66.tg 100= 365,99 N;

bnh 3 tng t :

Ft3 = Ft2; Fr2 = Fr3; Fa2 = Fa3;

bnh rng thng l bnh b ng nn T2ngc chiu n2,cn T3cng chiu n2v

bnh nghing ch ng .

a lc v tm trc s sinh ra momen v lc tng ng

ta c:

Mt1 = T2t.

Mt2= Mt3 =T2n;

Ma2 = |Ma3| = Fa2./2 = 365,99.67,93/2 = 12430,85 Nmm;lc dc trc Fa2+Fa3 = 0, do ti tm lc dc trc chi

c tc dng ko nn , khng

c tc dng un v xon trc.

Hnh 3: biu momen xon v momen un.

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r1

F FFr2 t2

F

r3

t3F

a2

Fa3

a3

Ft1

Ft3

Ft2

Fr2

Fr3M

Ma3

a2

M

Mt2

t3

Mt1

141807.31 Nmm

177057.07Nmm

31877.4Nmm

44308.32 Nmm

18648.4Nmm

70499.72Nmm

70499.72Nmm

6217.55Nmm

Ft1

F

A BC

D E

51,5

52

5251,5

r1FFa2

F

Momen

trc

Y

Momen

trc

X

Momen

trc

Z

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b1. Xc nh momen un trn trc y:

ta c: XA = XE = (Ft1+Ft2+Ft3)/2 = (1355,76+2075,66.2) =

2753,54 N

momen xon ti mt s im c bit:

MBY = -51,5.XA = -51,5.2753,54 = -141807,31 Nmm;

MCY =-103,5.XA+52.Ft1 = -103,5.2753,54+52.2075,66 = -

177057,07 Nmm;

MDY = -155,5.XA+104.Ft2+52.

Ft1 = -155,5.2753,54+104.2075,66+52.1355,76 = -

141807,31 Nmm;

MEY = 0;

b2. Xc nh momen un trc x:

ta c: YA = YE = (Fr2+Fr3-Fr1)/2 = (865,71.2-

493,46)/2618,18 N;

momen xon ti mt s im c bit:

MBXT = 51,5.YA=51,5.618,98 = 31877,47 Nmm;

MBXP = MBXT+Ma2 = 31877,47+12430,85 = 44308,32 Nmm;

MCX = 103,5.YA+Ma2-52.Fr2 = 103,5.618,98+12430,85-

52.865,71 = 31478,36 Nmm;

MDXT = 155,5.YA+Ma2-104.Fr2 = 155,5.618,98+12430,85-

104.865,71

= 18648,40 Nmm;

MDXP = MDXT-Ma3= 18648,40-12430,85 = 6217,55 Nmm;

ME = 0;

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b3. Xc nh momen xon trc Z:

ta c: Mt1+Mt2+Mt3= 0 v Mt1/2 = |Mt2| = |Mt3| =

70499,72 Nmm

b4. Momen tng hp ti mt s im v ng knh thc

ti v tr :

MB = = =

164446,80 Nmm;

MC = = = 193158,75

Nmm;

MD = = = 159459,36

Nmm.

Tng t trc I ta c:

dB > 29,66 mm; dC > 31,30 mm; dD > 29,36 mm;

=> dB = dD = 30 mm; dC = 32 mm;

c. Trc III c hai bnh rng v mt bnh ai .

bnh rng 1: Ft1=

= 2.179592,89/191,56 = 1875,06 N

Fr1= Ft1. =1875,06.tg22,330/cos100 = 782,04 N;

Fa1= Ft1.tg = 1875,06.tg100 =330,62 N.

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Bnh rng 2 tng t :

Ft11=Ft2; Fr1= Fr2; Fa1= Fa2.

Bnh xch: F1= 2994,06 N;

F2= 157,52 N;

Fn= F1+F2 = 3152,58 N.

momen trn bnh: Mt1= Mt2= T3= 179592,89 Nmm

Mn= (F1-F2)d1x/2(2994,06-158,54)253,32/2 = 359146,96

Nmm.Hnh3: biu momen un v momen xon

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359146,96 Nmm

179592,89 Nmm

71914,84 Nmm

60128,83 Nmm

13452,22 Nmm

146454,57 Nmm

Ma1

F

M

F

F

F

F

F

M

M1

2

n

t1

a1

t2

a2

r2

n

r1

Ma2

Fa1 Fa2

Ft2

Ft1 F

F

1

2

Fr2

Fr1

A

BC

D

E

Momen

trc

Y

Momen

trc

X

Momen

trc

Z

51.5104.0

51.5 46.5

F

40275,06 Nmm

c1. Momen trn trc y:

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phng trnh cn bng ti A v E l:

= 0 => 51,5.Ft1+155,5.Ft2-207.XD+253,5.Fn= 0=> XD = (51,5.Ft1+155,5.Ft2+253,5.Fn)/207= (207.1875,06+253,53149,58)/207 =5732,15 N;

XA = Ft1+Ft2+Fn-XD = 2.1875,06+3149,58-5732,15 = 1167,55

N.

momen un ti mt s im c bit:

MBY = 51,5.XA = 51,5.1167,55 = 60128,83 N;

MCY = 155,5.XA-104.Ft1= 155,5.1167,55-104.1875,06 = -

13452,22 Nmm;

MDY = 207.XA-155,5.Ft1-51,5.Ft2= 207.(XA-Ft1)

= 207.(1167,55-1875,06) = -146454,57 Nmm;

MEY= 0.

c2. Momen un trn trc x:

ta c: YA = YD = Fr1= Fr2= 782,04 N;

momen un ti mt s im c bit :

MBXT = -51,5.YA= -51,5.782,04 =-40275,06 Nmm;

MBXP= MBXT-Ma1= -40275,06-31666,78 = -71941,84 Nmm;

MCXT= -155,5.YA-Ma1+104.Fr1= -155,5.782,04-31666,78+104.782,04

= -71941,84 Nmm;

MCXP = -40275,06Nmm; MD = 0.

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c3. Momen tng hp ti mt s im v ng knh trc :

MB= = = 202594,94

Nmm;

MC = = =

366440,40Nmm;

MD= = 387860,13 Nmm;tng t ta c: dB > 31,80 mm; dC > 38,75 mm; dD> 39,49

mm

tra bng 10.5 ta c: dB= 34mm; dC= 40mm; dD= 40mm.

5. Tnh kim nghim trc v bn mi

kt cu trc va thit k phi m bo an ton ti cc

tit din nguy him phi tha mn iu kin sau:

(1)

vi vl h s an ton xt ring ng sut php vng sut tip ti tit din j:

(2) ;

(3);

a. Vi thp 45 c = 600 MPa; = 0,58.600 = 384 MPa. Tra bng 10.7 vi

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ta c = 0,05; = 0.

b. Cc trc hp gim tc quay u, ng sut thay itheo chu k i xng, do :

v vi khi trc quay mt chiu ng sut xon thay i theo chu

k mch ng do :

vi Wj v Woj l momen cn un v momen

cn xon ti tit din j ca trc c xc nh theo

cng thc :

Wj=

-

(4); Woj=

(5)

c. Xc nh h s an ton cc tit din nguy him ca

trc:

Da vo kt cu trc cc hnh 2, 3, 4 v biu momen

tng ng, c th thy tit din nguy him cn kim tra

v bn mi: trn trc I ti im A v C (hay A1 v

C1); trn trc II ta kim tra ti B v C (hay B2 v C2);

trn trc III ta kim tra ti C v D (hay C3 v D3);

kch thc then tra bng 9.1, tr s momen theo cng

thc (4) v (5) ng vi tit din trc ta c bng6:

Titdin

ngknhtrc(mm)

bxh t1 W(mm3)

A1 30 0 0 2650,72 5301,44C1 22 6x6 3,5 882,02 1927,38B2 30 8x7 4 2290,19 4940,90

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C2 32 10x8 5 2647,46 5864,45C3 40 12x8 5 5364,44 11647,62

D3 40 0 0 6283,19 12566,37

Bng 7:tr s momen un v momen xon

Titdin

MU (Nmm ) T (Nmm) A1 112782,01 42,55 41427,24 7,81C1 32118,98 36,42 41427,24 21,49B2 148568,,30 56,98 70499,72 14,27

C2 179833,52 67,93 70499,72 12,02C3 73188,73 13,64 359146,96 30,83D3 146454,57 23,31 359146,96 28,58d. Chn lp ghp: cc ln lp trn trc theo k6; lp

bnh rng, bnh dai, ni trc theo h6 kt hp vi lp

then.

e. Xc nh cc h s v i vi cc tit dinnguy him tnh theo cng thc:

*

+

(6); (7)trong : KXl h s tp trung ng sut do trng thi

b mt, ta chn gia cng trn

my tin ti cc tit din nguy him yu cu t: Ra=

2,5...0,63 um, do theo

bng 10.8, h s tp trung ng sut do trng thi b

mt KX = 1,06.

khng dng cc phng php tng bn b mt, do h s

tng bn Ky= 1.

v -l h s kch thc k n nh hng ca kchthc tit din trc n

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gii hn mi, tra bng 10.10 ta c bng8:

Dngchu ti

Vt liu ng knh22 30 32 40

Xon thp 45 0,87 0,81 0,80 0,78Un thp 45 0,90 0,88 0,87 0,85

tra bng 10.12 ta chn dao phay a vi = 600 MPa tac = 1,46;= 1,54Ta lp bng 9: kt qu tnh ton h s an ton i vi

cc tit din ca ba trc :

Titdin

d(mm)

SA1 30 1,66 1,77 1,72 1,96 3,69 10,97 3,50C1 22 1,62 1,77 1,68 1,83 4,41 9,76 4,02B2 30 1,66 1,9 1,72 1,96 2,72 13,73 2,67C2 32 1,68 1,93 1,74 1,99 2,28 16,05 2,26C3 40 1,72 1,97 1,78 2,03 11,12 6,14 5,38

D3 40 1,72 1,97 1,78 2,03 6,31 2,9 2,646. Kim nghim bn ca then:

Vi cc tit din trc dng mi ghp then bng cn tin

hnh kim nghim mi

ghp then v bn dp theo cng thc :

v

; kt qu tnh vi lt = (0,8...

0.9).lmBng 10: bn dp v bn ct

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Tit

din

ng

knh

Chiudi

thenlt

Chntr

slt

bxh t1 T(Nmm)

(MPa)

(MPa)

C1 22 32...36

35 6x6 3,5

41427,24

50,21 10,46

D1 20 40...45

40 6x6 3,5

41427,24

55,91 13,98

B2 30 32...36

35 8x7 4 70499,72

19,50 7,34

C2 32 32...36

35 10x8 5 70499,72

17,08 4,27

D2 30 32...36

35 8x7 4 70499,72

19,58 7,34

B3 34 32...36

35 10x8 5 359146,96

76,54 22,96

C3 40 32...36

35 12x8 5 359146,96

55,42 13,86

theo bng 9.5 vi ti trng tnh [] = 150 MPa v []= 60...90 MPa .

vy tt c then u t bn dp v bn ct.

V. Tnh ton trc IV nm ngoi hp gim tc

T phn trn ta c: d4 = 60mm v b4 = 33mm; d2x = 889,55

mm; dxt = 602,97 mm

chiu di may a xch tnh theo cng thc: lm =

(1,2...1,5).d4 = (72...90) mm.

Ta ly: lm = 75 mm .

c ba xch, mt cp ln v bn khong cch chi tit,

ta ly k1 =12 mm .

vy chiu di trc: l40 = 3.lm+2.b4+4.k1 = 3.75+2.33+4.12

= 339 mm;

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1. Xc nh chiu di trc

Khong cch t tm chi tit ngoi cng hay trng tm

ln ngoi cng ti cc trng tm chi tit khc trn

trc l :

l41 = b4/2+k1+lm/2 = 33/2+12+75/2 = 66 mm;

l42 = b4/2+2.k1+3.lm/2 = 33/2+2.12+3.75/2 =153 mm;

l43 = b4+2.lm+3.k1 = 33+2.75+3.12 = 219 mm; l44 =

3.b4/2+5.lm/2+4.k1 = 285 mm;

Hnh 4: biu momen xon v momen un.

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6687 66

66

AB C

DE

Fo

Fn

Fo

Fo

Fo

Fo F

o

Fn

Fn

X

X

A

D

557747.25Nmm

1115494.50Nmm

208069.71Nmm

23263.77Nmm

59394.06Nmm

truc

Y

Momen

truc

Z

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c. Xc nh lc v dimt

bnh xch 2 l bnh b ng cho nn M1cng chiu quay

n4

Fn= F1+F2= 2994,06+158,52 = 3152,58 N; M1= T4=

1115436,47 Nmm;

M = 2.Mo= 2.Fo.dxt/2 = Fo.dxt= 3700.602,97/2 =

1115494,50 Nmm;

sai s momen chuyn ng xch l:

M = = .100% = 0,89%sai s M qu nh, do sai lch s trong tnh ton tac th b qua.

c1. Xc nh momen un trn trc y:

phng trnh momen cn bng ti A v D l :

= 0 -66.Fo-153.Fo+219.XD-285.Fn = 0=> XD = Fo+285.Fn/219 = 1850+285.3152,58/219 = 5952,67 N;

XA = 2.Fo+Fn-XD = 3700+3152,58-5952,76 = 899,91 N,

momen un ti mt s im bit :

MBY = -66.XA = -66.899,91 = -59394,06 Nmm;

MCY= -153.XA+87.Fo= -153.899,06+87.1850 = 23263,77 Nmm;

MDY= -219.XA+153.Fo+66.Fo= -219.(Fo-XA) = 219.(1850-

899,91)

= 208069,71 Nmm;

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ME= 0.

c2. Momen xon trn trc Z ta c:

Mo= M/2 = 557797,25 Nmm;

c4. Momen xon tng hp v ng knh thc ti mt s

im :

MB = = 560900,75 Nmm;MC= = 1115737,06 Nmm;MD= = 1134733,88 Nmm;dB suy ra: dB= dC= dD= 60 mm;

d. Tnh kim nghim v bn mi

Tng t nh phn IV ta c:

Bng 11: momen trc ti mt s im:

Titdin

ng knhtrc(mm)

bxh (mm2) T(mm)

Wx (mm3) WO (mm

3)

B4 60 18x11 7 18256,30 39462,05C4 60 18x11 7 18256,30 39462,05E4 60 18x11 7 18256,30 39462,05

Bng 12: tr s momen v ng sut

Tit din MU(Nmm) T(Nmm) B4 59394,06 3,25 55770,25 1,41C4 23263,77 1,27 1115494,50 28,27E4 0 0 1115494,50 28,27

=0,78;=0,75.

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Bng 13: kt qu tnh ton h s an ton

Titdin

d

S

B4 60 1,87 2,05 1,93 2,11 43,05 129,07 40,84C4 60 1,87 2,05 1,93 2,11 110,15 6,44 6,43E4 60 1,87 2,05 1,93 2,11 0 6,44 6,44

Vy vt liu t bn .

e. Kim nghim bn ca then

T cng thc (6) v (7) ta ch cn kim tra ti then c

momen xon ln nht l ti

im E4, chiu di then lt= 1,35.d = 1,35.60 = 81 mm.

Suy ra : . Vy then t yu cu.VI. Chn

1. Ta chn bi

Vi ti trng tng i nh v ch c lc hng tm,

lc dc trc trn trc II v

III b trit tiu bi cp bnh rng nghing, ta dng

bi mt dy cho cc gi : A1, B1, A2, E2, A3, D3.

Cn c vo cc lc ti cc gi ,mi trc c hai cp

bi, ta ch xt mt cp bi

chu lc ln hn mi trc, do ta ch xt cc gi : A2, A3, D3, D4.

2. cc thng s ca bi

Vi kt cu trc ta c ng knh ti ch lp bi, tra

bng P.2.7 phc lc

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Bng 14: cc thng s ca bi

Gi ngknhtrc(mm)

khiu bi

ngknhtrong bi(mm)

ngknhngoi bi(mm)

Khnngtingchophp C(kN)

Khnngtitnhchophp Co(kN)

A1 30 406 30 90 37,20 27,20

A2 25 405 25 80 29,20 20,80D3 40 408 40 110 50,30 37,00D4 60 312 60 150 64,10 49,403. Tnh kim nghim kh nng ti ca

a. Xt kh nng ti ng ca bi

Bng 15: lc tng hp ti cc gi .

Gi

Phn lc trcX(N)

Phn lc trcY(N)

Lc hng tm tnghp Fri (N)

A1 2505,32 259,97 2518,77

A2 2753,54 618,98 2822,25D3 5732,15 782,04 5785,5D4 5952,67 0 5952,67vi Fa=0, ta c cng thc tnh ti trng ng:

Qi= X.V.Fri.kt.k (1)

trong : Fri-l ti trng hng tm;

V-l h s k dn vng quay, v vng

trong quay nn V = 1;

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kt-l h s k n nh hng ca nhit ,

vi t < 100oC => kt= 1;

k-l h s k n ti trng ng, vi my va

p va , rung ng ta ly

k= 1,4;

X-l h s ti trng hng tm, v Fa= 0,

ta ly X = 1;

t (1) ta c: Qi=1.1.Fri.1.1,4 = 1,4.Fri (20)

kh nng ti ng tnh theo cng thc: Cdi= Qi. (3)

m-ng cong mi khi th v bi, m = 3 i vi bi.

Li-tui th tnh bng triu vng vi :

Li= 60.ni.Lh/= 60.ni.12500/= 0,75.ni triu vngBng 16: tr s ti trng ng, s vng quay v kh

nng ti ng

Gi Tr s titrng ngQi (N)

S vngquay(triuvng)

Kh nngti ngCdi (kN)

Kh nngti ngcho phpCi (kN)

A1 3526,28 1087,50 36,26 37,20A2 3951,15 303,77 26,56 29,20D3 8099,35 108,88 38,68 50,30

D4 8333,74 30,93 26,16 64,10Vy bi t yu cu kh nng ti ng

b. Xt v kh nng ti tnh

ta c cng thc tnh kh nng ti tnh: Qoi= Xo.Fri

trong X l h s ti trng hng tm vi Xo= 0,6;

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Bng 17: so snh Qoi v Co

Gi Kh nng ti tnhQoi(kN)

kh nng ti tnhcho php Co (kN)

A1 1,51 27,20A2 1,69 20,80D3 3,47 37,00D4 3,47 49,40Vy bi t yu cu kh nng tnh

VII. Tnh chn kt cu hp

Ch tiu c bn ca v hp gim tc l cng vng caov khi lng nh, vt liu ph bin nht dng c

hp gim tc l gang xm GX15-30.

1. Chn b mt ghp np v thn hp

b mt ghp phi c gim trng lng v kch thc

hp gim tc, cng nh to iu kin bi trn tt cho

cc cp bnh rng bng phng php ngm du .

2. Xc nh cc kch thc c bn ca v hp gim tc

hnh dng ca np v thn ch yu c xc nh bi s

lng v kch thc ca rng, v tr mt ghp v s

phn b ca trc trong hp, ng thi cn ph thuc ch

tiu kinh t, bn v cng .

Khong cch tm ta ly theo khong cch trc ln nht,

khi a = aw= 133 mm, thn hp = 0,03.a + 3 > 6 mm =>= 0,03.133+ 3 = 6,99 mm > 6 mm. Vy ta ly = 7 mmBng 18: quan h gia cc kch thc cu to nn v hpgim tc

Tn gi Biu thc tnh ton Tr s quytrn hay lytrong gii

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hn

Chiu dy: thn

hp,

Np hp, 0,03.a+3 > 6 mm= 0,9.= 0,9.7

7 mm

6,3 mm

Gn tng cng:

chiu dy, e

chiu cao, h

dc

e = (0,8...1)=5,6...7

h < 58 mm

khong 20

7 mm

ng knh:

Bu lng nn, d1

Bu lng cnh , d2

Bu lng ghp bchnp v thn, d3

vt ghp np , d4

vt ghp np cathm, d5

d1 0,04.a+10 = 15,32d2 = (0,7...0,8).d1 =11,2...12,8

d3 = (0,6...0,7).d2 =9,6...10,8

d4 = (0,6...0,7)d2 =7,2...8,4

d5 = (0,5...0,6).d2 =6...7,2

16 mm

12 mm

10 mm

8 mm7 mm

Mt bch ghp npv thn:

Chiu dy bchthnh hp, S3

Chiu y bch nphp, S4

B rng bch npv thn, K3

S3 = (1,4...1,8).d3 =

14...18

S4 = (0,9...1).S3 =15,3...17

K3 = k2-(3...5)

17 mm

17 mm

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Kch thc gitrc:

ng knh ngoiv tm l vt: D,D2,D3.

Tm bu lng cnh: E2 v C (k lkhong cch t tm

bu lng ti mp )

B rng mt ghpbu lng cnh : K2

Chiu cao h

Xc nh khong cchkch thc np bng

cch tra bng 18.2, tac: D

D2

D3

E2 = 1,6.d2 = 1,6.12=19,2

C = D3/2 =125/2

R2= 1,3.d2 = 1,3.12 =15,6 mm

K2= E2+R2+(3...5)

=19,2+16+(3...5)

= 38,2...40,2

K3= K2-(3...5) = 35,2

Xc nh theo kt cu,ph thuc tm l bulng v kch thc mtta

80 mm

100 mm

125 mm

19,2 mm

62,5 mm

16 mm

40 mm

35,2 mm

Mt hp:

Chiu dy khikhng c phn liS1

Chiu dy khi cphn li Dd, S1, S2

B rng hp, k1

S1 = (1,3...1,5).d1 =20,8...24

Ddxc nh theo kchthc dao khot

S1= (1,4...1,7).d1 =22,4...27,2

S2= (1,0...1,1).d1 =

22 mm

23 mm

16 mm

48 mm

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v q 16...17,6

K1 = 3.d1 = 3.16 = 48

q k1+2.= 48+2.7 =62

ly q = 65 mm

65 mm

Khe h gia ccchi tit:

Gia bnh rng vthnh trong hp

Gia nh bnhrng ln vi yhp

Gia mt bn ccbnh rng vi nhau

(1...1,2)=7...8,4

(3...5).=(21...35) v ph thucloi hp gim tc,lng du bi trntrong hp

= 7

21 mm

35 mm

S lng bu lngnn Z

Z = (L+B)/(200...300)

L, B: chiu dy vchiu rng ca hp

a. Khe h gia b mt gia cng ca chi tit quay vb mt khng gia cng ca v hp phi ln hn tng sai

s v chnh xc ca v tr cc vch c cng nh

sng mp m b mt c. c chn y l gia bnhrng dn thnh trong ca v hp c chn theo iu

kin m bo kch thc np v phn li kp bu lng

cnh , do vy kt cu v hp s n gin hn.b. Khe h : t nh rng n y hp cn tha mncc yu cu sau:

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cn ln khi bnh rng quay cht bn v sn vt

mi mn lng xung y hp khng b khuy ng ln,

i vi hp gim tc bnh rng th = (3...5).cn to ra lng du bi trn cn thit khi bnh

rng c bi trn bng phng php ngm du, lng du

bi trn cn thit l (0,4...0,8) lt cho cng sut

truyn 1 KW.

c. B mt ghp np v thn hp c ghp bng bu lng.

Chiu dy mt bch S v S1c chn theo iu kin m

bo cng, ta lyS = S1 . B mt bch K3phi xit cht c th xoay

cha vn mt gc 600.B mt ghp np v thn cmi cao lp st. Khi lp gia hai b mt khng c

dng m lt m bo kiu lp gia vi hp. Chn R3

= K3/2.

S lng v ng knh bu lng d3 ghp bch np v

bch thn c chn theo iu kin ng sut phn b u

trn mt ghp.

Ta tra bng P3.4 ta chn u bu lng v ai c . Ta c

bng 19

d u bu lng ai c thng

S D h S D H

10 17 18.9 7 17 18.9 8

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22.0

22.0

7.0

8.0

10.0

18.9

Mt chn khng nn lm phng m nn lm hai dy li

song song nhm gim tiu hao vt liu, gim thi giangia cng v to kh nng lu thng khng kh qua y

hp thot nhit tt hn.

m bo an ton cho mt chn v v hp, nn lm

thm cc ng gn .

d. Gi trc trn v hp

gi trc cn phi cng khng nh hng n s

lm vic ca . ng knh ngoi ca gi trc (D3) cchn theo bng 18.2. Chiu di gi trc khng nhng ph

thc vo chiu dy thnh hp m cn ph thuc vo kt

cu b phn nh chiu rng np , chiu cao np ,

chiu rng vng chn du khi c bi trn bng m,

ng knh bu lng d2.

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ng knh gi trc cng l ng knh np , trong

c D l ng knh ln,ta tra bng 18-2 ta c bng 20:

kch thc np :

D 47 80 90 110

D2 60 100 110 130

D3 70 125 135 160

D4 37 75 85 100

d4 M6 M8 M8 M10

S lng (Z) 4 4 4 6

3. Mt s kt cu khc lin quan n cu to v hp

gim tc

a. Bu lng vng hoc mc vng

hnh v: Kch thc bu lng vng, mm

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d1

d2

h

fh1

l

d

c

r

r2

d3

b

r1

d4

nng v vn chuyn hp gim tc trn np v thn,

thng lp thm bu lng vng hoc vng mc.Kch thc bu lng vng c chn theo khi lng hp

gim tc. Vt liu bu lng l thp 20 hoc thp 25.

Ta c bng 20:

Rend

d1 d2 d 3 d 4 d 5 h h1 h 2 l

M8 36 20 8 20 13 18 6 5 18

f b c x r r 1 r 2

2 10 1,2 2,5 2 4 4

Trng lng nng c, ta chn b = 160 kG.

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Hp gim tc bnh rng tr hai cp, ta chn trng lng

Q = 180 kG ca hp c xc nh gn ng theo khong

cch trc a1a2= 150 250b. Cht nh v

m bo v tr tng i ca np v thn trc, sau

khi gia cng cng nh khi lp ghp,dng cht nh v.

Nh cc cht nh v, khi xit bu lng khng lm bin

dng ngoi ca , do loi tr vng trong cc nguyn

nhn lm chng hng.

ta chn cht nh v l l cht trbng 21: hnh dng v kch thc cht nh v hnh tr

d (mm) 6

c (mm) 1

l (mm) 60

l

c

d

c. Ca thm

kim tra,quan sat cc tit my trong hp khi lp

ghp v du vo hp, trn dnh hp c lm ca thm.

Ca thm c y bng np,trn np c nt thng hi.

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Kch thc ca thm chn theo bng 18.5, do ta c

bng 22:

A B A1 B1 C K R vt S lng

100 75 150 100 125 87 12 M8x22 4

Hnh v ca thm:

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C

A

B

B1

R

A

B

B1

R

K

C1

A1

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d. Nt thng hi.

Khi lm vic, nhit trong hp tng ln. gim

psut iu ha khng kh bn trong v bn ngoi hp,

ngi ta dng nt thng hi lp trn np ca thm hoc

v tr cao nht ca v hp.

H

Q

MK

L

I

A

R

B

G

E C

D

R

O

A B C D E G H I

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M27x2 15 30 15 45 36 32 6

K L M N O P Q R

4 10 8 22 6 32 18 36

S

32

e. Nt tho du

sau thi gian lm vic, bi trn cha trong hp b bn

hoc b bin cht, do cn phi thay du mi, thodu 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 cho trong bng 18.7, ta c bng 22: hnh dng

v kch thc nt tho du tr.

S

D

Do

d

L

b

m

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d b m f L C q D S D0

M22x2 15 10 3 29 2,5 19,8 32 22 25,4

h. Kim tra mc tho

ta c vn tc bnh rng thng v bnh rng nghing c V

12 m/s th c chiu cao mc du trong hp c kimtra qua cc thit b ch du bng que thm du.

bng 23: kch thc ch du mc phng

kch thc mt knh(mm)

D D1 l h

20 55 40 10 6

4. Bi trn hp gim tc

gim mt mt cng sut v ma st, gim mi mn rng,

m b thot nhit tt v phng cc tit my b hn

g cn phi bi trn lin tc

a. Ta chn phng php ngm du

b. Du bi trn hp gim tc

dng du cng nghip v c dng rng ri bi trn

nhiu loi my khc nhau .

chn nht bi trn ph thuc vo vn tc, vt liu

ch to bnh rng

bng 24: chn nht du 500

C (1000

C ) bi trnbnh rng

Vn tc vng (m/s) (MPa) Vt liu nht2,5-5 470-1000 thp 80/11

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Ta chn du cng nghip 30 c C = 27-33

5. Lp bnh rng ln trc v iu chnh s n khp

truyn momen xon t trc ln bnh rng v ngc li,

bnh rng tr khng di ng trong cc hp tc lp

ln trc chu ti va c s thay i v va p nh, ta

chn kiu lp H7/k6.

trong hp gim tc , cng thng dng bc cch v mt

mt ca vng nh v bnh rng trn trc. Khi s

dng phng php ny cn m bo s tip xc chnh xc

gia cc mt mt bnh rng, bc, vng , chiu dai bccn m bo chnh xc di hn on trc lp.

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TI LIU THAM KH O

[1]. TTTKHDCK- Tp 1-2-Trnh Cht_L Vn Uyn-Nh xut bn Gio Dc -nm 2007.

[2]. CHI TIT MY- Tp 1_2- Nguyn Trng HipNh xut bn Gio Dc-nm 2006.

[3]. DUNG SAI v LP GHPGS.TS-Ninh c Tn- Nh xut bn Gio Dcnm 2007.

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Mc lc

I. Tnh chn ng c, phn phi t s truyn v momenxon trn cc trc 2

1. Chn ng c

2

2. S vng quay ca xch ti

4

3. Phn phi t s truyn

4

4. Tc quay cc trc

5

5. Momen xon trn cc trc ng c

6

II. Tnh b truyn ngoi

6

1. Chn loi xch

6

2. iu kin m bo ch tiu iu kin mn

7

3. Khong cch trc s b

8

4. Kim nghim xch v h s an ton

8

5. X nh ng knh xch

9

6. Kim nghim bn tip xc

9

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7. Xc nh lc tc dng ln trc

10

III. Tnh ton b truyn bnh rng

10

A. B truyn cp bnh rng thng trn trc I v II

10

1. Chn vt liu

10

2. Xc nh ng sut cho php

113. Tnh khong cch trc s b

13


13

5. Kim nghim rng v bn tip xc

13

6. Kim nghim rng v bn un15

7. Kim nghim rng v bn qu ti

16


17

B. B truyn bnh rng nghing trn trc II v III

17

1. Chn vt liu

18

2. Xc nh ng sut cho php

18

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3. Tnh khong cch trc s b

20


20

5. Kim nghim rng v bn tip xc

21

6. Kim nghim rng v bn un

23

7. Kim nghim rng v bn qu ti


25

IV. Tnh ton thit k trc trong hp gim tc

26

1. Chn vt liu

26

2. Xc nh s b ng knh trc26

3. Xc nh khong cch gia cc gi v im t lc

27

4. Xc nh tr s lc v ng knh trc

31

5. Tnh kim nghim trc v bn mi

41

6. Kim nghim bn ca then

43

V. Tnh ton trc IV nm ngoi hp gim tc

44

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1. Xc nh chiu di trc

44

2. Xc nh lc v im t

46

3. Tnh kim nghim v bn mi

47

4. Kim nghim bn ca then

48

VI. Chn

481. Ta chn bi

48

2. Cc thng s bi

48

3. Kim nghim kh nng ti ca bi

48

VII. Tnh chn kt cu hp50

1. Chn b mt ghp np v thn hp

50

2. Xc nh cc kch thc c bn ca v hp gim tc

50

3. Mt s kt cu khc lin quan n cu to v hp

gim tc 55

4. Bi trn hp gim tc

61

5. Lp bnh rng ln trc v iu chnh s n khp

61

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