Chuong 4 Botruyenxich

Chi tiet may Chng IV

41

CHNG 4 B TRUYN XCH

4.1. KHI NIM CHUNG 4.1.1. Nguyn l

F1

F2

T2

n2

z2

T1

n1

z1

F2

F1

B truyn xch thc hin truyn ng t bnh xch ch ng sang bnh xch b ng nh vo s n khp gia rng trn a xch v cc mc xch.

4.1.2. Phn loi Theo cng dng chung, xch c chia lm ba loi:

- Xch ko. - Xch ti. - Xch truyn ng.

Trong chng trnh chng ta ch tm hiu v xch truyn ng m thi. Xch truyn ng c chia lm 3 loi chnh: xch ng, xch con ln v xch rng.

4.1.3. u nhc im v phm vi s dng u im:

- Khng c hin tng trt nh b truyn ai, c th lm vic khi c qu ti t ngt, hiu sut cao.

- Khng i hi phi cng xch, nn lc tc dng ln trc v nh hn - Kch thc b truyn nh hn b truyn ai nu cng cng sut - Gc m khng c ngha nh b truyn ai nn c th truyn cho nhiu

bnh xch b dn Nhc im:

- Bn l xch b mn nn gy ti trng ng, n. - C t s truyn tc thi thay i, vn tc tc thi ca xch v bnh b dn

thay i. - Phi bi trn thng xuyn v phi c bnh iu chnh xch. - Mau b mn trong mi trng c nhiu bi hoc bi trn khng tt.

Phm vi s dng

- Truyn cng sut v chuyn ng gia trc c khong cch xa, cho nhiu trc ng thi trong trng hp n < 500v/p

- Cng sut truyn thng thng < 100 kW - T s truyn < 6, hiu sut 0,95..0,97


42

4.1.4. Kt cu xch truyn ng a. Xch con ln

pc

b0

1 2 3 4 5 6

Thng thng s mt xch l s chn. Cc kch thc c bn c th tra trong

bng (4.1)[1]

b Xch ng Cu to ging xch con ln nhng khng c con ln 6, nn mn nhanh. Nhng

khi lng v gi thnh thp.

c. Xch rng pc

Gm nhiu m xch xp xen k nhau, cc m xch ni vi nhau bng khp bn l. Cc m xch n khp vi bnh xch bi hai mt phng u m xch. Xch rng lm vic m, t n, kh nng ti cao xch khi lm vic khng b trt khi a xch theo phng dc trc ta dng cc m dn hng khng c rng t gia xch. Trn a xch c phay rnh.

* a xch Kt cu bnh xch nh hnh


43

- Kch thc bnh xch nh, c th dng phi dp. Khi kch thc bnh xch ln, c th ghp ri may v a xch bng mi ghp hn hay bu lng. - Bin dng v kch thc rng xch ph thuc v loi v kch thc xch

4.2. THNG S HNH HC B TRUYN XCH 4.2.1. Bc xch pc

- L thng s c bn b truyn xch. - Bc xch cng ln th kh nng ti cng cao. ng thi ti trng ng, va p v ting n cng tng theo, nht l khi lm vic vi vn tc cao - tng kh nng ti c th tng s dy xch (xch ng con ln) hoc tng chiu rng xch (xch rng) - Bc xch c chn theo tiu chun

4.2.2. S rng a xch - Thng thng Z1 < Z2, nu s rng nh th xch mau b mn (v gc xoay bn l ln) v ti trng ng cng nh va p. - Do , ta hn ch s rng nh nht. Thng thng, khi v>= 2m/s. th zmin >= 19, khi v


44

)50...30(2/)dd(a 2a1amin (4.2) + Khi u > 3:

10

)u9(

2

)dd(a 2a1amin

- S b, trong thc t c th chn a = (30..50)pc (4.3)

- Chiu di xch xc nh theo cng thc:

a4

)dd(

2

dda2L

22121

- S mt xch X :

cp

LX

a

p)/2z(z

2

)z(z

p

2a c2

1221

c

(4.4)

Nn chn X chn - Sau khi tnh s mt xch. Cn xc nh li chnh xc khong cch trc a:

2

21

2

2121c

2

zz8

2

)zz(X

2

)zz(Xp25,0a

(4.5)

Thng gim a mt khong a = (0,0020,004)a to chng cho b truyn xch.

4.3. NG HC TRUYN NG XCH 4.3.1. Vn tc v t s truyn trung bnh

- Vn tc trung bnh b truyn xch:

60000

p.z.nv c

(4.6)

n s vng quay ca a xch z s rng ca a xch pc bc xch


45

- T s truyn trung bnh u = n1 / n2 = z2 / z1 (4.7)

Thng thng u


46

4.4. NG LC HC B TRUYN XCH 4.4.1. Lc tc dng trong b truyn xch

- Lc vng Ft lin h vi lc trn nhnh cng F1 v chng F2:

t21 FFF (4.11)

- Lc cng do ly tm:

2mv vqF (4.12)

qm Khi lng mt mt xch, kg/m v vn tc vng , m/s - Lc cng ban u do trong lng nhnh xxh t do:

gaqKF mf0 (4.13)

a chiu di on xch t do bng khong cch trc g gia tc trng trng Kf h s ph thuc vng xch Kf = 6 khi xch nm ngang Kf = 3 khi xch nm nghing < 40

0 so vi phng ngang

Kf = 1 khi xch thng ng

- v02 FFF . C th ly gn ng: F1 = Ft (4.14) - Lc tc dng ln trc

tmr FKF (4.15)

Km h s trng lng xch +Km = 1,15 Khi xch nm ngang hoc nghing < 40

0

+Km = 1 Khi xch thng ng hoc nghing >= 400

4.4.2. Ti trng ng - S thay i c chu k ca vn tc nh trn l nguyn nhn thay i t s truyn tc thi, l nguyn nhn gy ti trng ng ph trong b truyn. - Vn tc v1 gy va p gia bn l xch v rng bnh xch.

4.4.3. ng nng va p - Ti thi im bn l B vo n khp, v1 v v1

gp nhau gy nn va p. Tn

tht ng nng :


47

2vmv5,0E

m = 103qmpc khi lng xch tham gia va p.

vv vn tc va p.

)z/360(sinpnq5,0E 10232

m c1 (4.16)

- trnh va dp, ngi ta chn bc xch theo bng (4.2)[1] 4.5. TNH TON B TRUYN XCH

4.5.1. Cc dng hng v ch tiu tnh

FF

pc

d0

1 2

Cc dng hng ch yu trong b truyn xch bao gm: + Mn bn l + Mn rng a xch + V con ln

4.5.2. Tnh ton xch con ln theo bn mn - Tui th xch theo bn mn ph thuc vo khong cch trc,s rng z1, p sut trong bn l p, iu kin bi trn Trong , p sut sinh ra trong bn l l nh hng quan trng nht i vi tui th - Tnh ton bn mn theo p:

K

K]p[]p[

A

Fp x0

t (4.17)

Ft lc vng, N; 1

1t

d

2TF ;

1

1

6

1n

P10.55,9T ;

c11

pzd

1000

1

60000

npz

K

K.A]p[

1000

v.FP

K

KA]p[F

1c1x0

t1

x0t

Nhn c t v mu cho z01, n01

K

K

n

n

z

z

1000.60000

p.z.n.A]p[P x

01

1

01

1c01010t


48

t

- 1000.60000

p.z.n.A]p[]P[ c01010

-1

01n

n

nK

-1

01z

z

zK

Ta c:

K

K

K

1

K

1]P[P x

nz

1

]P[K

KKKPP

x

nz1t

A = d0b0 din tch bn l xch 1 dy. C th ly gn ng: A=0,28pc2, mm

2

[p0] p sut cho php, tra bng (4.3)[1] Kx h s xt n s dy xch. Nu x = 1;2;3;4 th tng ng Kx = 1;1,7;2,5;3 K h s iu kin s dng xch

K = Kr Ka K0 Kdc Kb Klv (4.18)

( Cc h s ny tham kho trong tai liu [1]) [P] cng sut cho php b truyn mt dy c bc xch pc Kz = 25/z1; Kn = n01 / n1.

Tra bng 4.4[1], tm c pc 4.5.3. Kim nghim xch theo s ln va p trong mt giy

]i[X15

nz

Xp60

pnz4

L

v4i 11

c

c11 (4.24)

X s mt xch tnh theo (4.4) Z1,n1 s rng v s vng quay a xch b dn [i] s ln va p cho php. Tra bng (4.5)[1] - Kim tra theo h s an ton:

6...5]S[FFF

QS

021

(4.25)

Q ti trng ph hu cho php ca xch, bng 4.1[1] [S] h s an ton cho php. Tra bng (4.6)[1]


49

4.6. TRNH T THIT K B TRUYN XCH 1. Chn loi xch 2. Chn s rng xch theo cng thc z1 = 29 2u v z2 = uz1 3. Tnh cng sut tnh ton Pt, chn bc xch pc 4. nh khong cch trc a, tnh cc ng knh a xch.. Tnh s mt

xch X theo (4.4) nn chn X l s chn v kim tra s ln va p xch trong mt giy

5. Tnh lc tc dng ln b truyn (4.15)

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Chuong 4 Botruyenxich