CHNG XII

H THNG PHANHI. CNG DNG, PHN LOI V YU CU. 1.1. Cng dng. H thng phanh dng gim tc ca t cho n khi dng hn hoc n mt tc cn thit no y. Ngoi ra h thng phanh cn dng gi t ng cc dc. i vi t h thng phanh l mt trong nhng cm quan trng nht, bi v n m bo cho t chy an ton tc cao, do c th nng cao c nng sut vn chuyn. H thng phanh gm c c cu phanh hm trc tip tc gc ca cc bnh xe hoc mt trc no y ca h thng truyn lc v truyn ng phanh dn ng cc c cu phanh. 1.2. Phn loi. Ty theo cch b tr c cu phanh cc bnh xe hoc trc ca h thng truyn lc m chia ra phanh bnh xe v phanh truyn lc. t c cu phanh chnh t bnh xe (phanh chn) cn c cu phanh tay thng t trc th cp ca hp s hoc hp phn phi (t 2 cu ch ng). Cng c khi c cu phanh phanh chnh v phanh tay ph hp lm mt v t i bnh xe, trong trng hp ny s lm truyn ng ring r. Theo b phn tin hnh phanh c c phanh cn chia ra phanh guc, phanh u di v phanh a. Phanh guc s dng rng ri trn t cn phanh a ngy nay ang c chiu hng p dng. Phanh di c s dng c cu phanh ph (phanh tay). Theo loi b phn quay, c cu phanh cn chia ra loi trng v a. Phanh a cn chia ra mt hoc nhiu a ty theo s lng a quay. C cu phanh cn chia ra loi cn bng v khng cn bng. C cu phanh cn bng khi tin hnh phanh khng sinh ra lc ph thm ln trc hay ln bi ca may bnh xe, cn c cu phanh khng cn bng th ngc li. Truyn ng phanh c loi c, thy, kh, in v lin hp. t du lch v t vn ti ti trng nh thng dng truyn ng phanh loi thy (phanh du). Truyn ng phanh bng kh (phanh hi) thng dng trn cc t n ti ti v trng ln v trn t hnh khch, ngoi ra cn dng trn t vn ti ti trng trung bnh c ng c izen cng nh trn cc t ko ko on xe. Truyn ng phanh bng in c dng cc on t. Truyn ng c ch dng phanh tay. 1.3. Yu cu. H thng phanh phi m bo cc yu cu sau: Qung ng phanh ngn nht khi phanh t ngt trong trng hp nguy him. Mun c qung ng phanh ngn nht th phi m bo gia tc chm dn cc i. Phanh m du trong bt k mi trng hp m bo s n nh ca t khi phanh.

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iu khin nh nhng, ngha l lc tc dng ln bn p hay n iu khin khng ln. Thi gian nhy cm b, ngha l truyn ng phanh c nhy cm ln. Phn b mmen phanh trn cc bnh xe phi theo quan h s dng hon ton trng lng bm khi phanh vi bt k cng no. Khng c hin tng t sit phanh khi t chuyn ng tnh tin hoc quay vng. C cu phanh thot nhit tt. Gi c t l thun gia lc trn bn p hoc n iu khin vi lc phanh trn bnh xe. C kh nng phanh khi ng trong thi gian di. II. KT CU CHUNG CA H THNG PHANH. H thng phanh trn t gm c phanh chnh (phanh bnh xe hay cn gi l phanh chn) v phanh ph (phanh truyn lc hay cn gi l phanh tay). S d phi lm c phanh chnh v phanh ph l m bo an ton khi t chuyn ng. Phanh chnh v phanh ph c th c c cu phanh v truyn ng phanh hon ton ring r hoc c th c chung c cu phanh (t bnh xe) nhng truyn ng phanh hon ton ring r. Truyn ng phanh ca phanh ph thng dng loi c. Phanh chnh thng dng truyn ng loi thy gi l phanh du hoc truyn ng loi kh gi l phanh kh. Khi dng phanh du th lc tc dng ln bn p phanh s ln hn so vi phanh kh, v lc ny l sinh ra p sut ca du trong bu cha du ca h thng phanh, cn phanh kh lc ny ch cn thng lc cn l xo m van phn phi ca h thng phanh. V vy phanh du ch nn dng t du lch, vn ti c nh v trung bnh v cc loi t ny mmen phanh cc bnh xe b, do lc trn bn p cng b. Ngoi ra phanh du thng gn gng hn phanh kh v n khng c cc bu cha kh kch thc ln v nhy khi phanh tt, cho nn b tr n d dng v s dng thch hp i vi cc t k trn. Phanh kh thng s dng trn t vn ti trung bnh v ln. Ngoi ra cc t loi ny cn dng h thng phanh thy kh. Dng h thng phanh ny l kt hp u im ca phanh kh v phanh du. S kt cu cc loi h thng phanh ca t c trnh by sau y: 2.1. Phanh du. phanh du lc tc dng t bn p n c cu phanh qua cht lng (cht lng c coi nh khng n hi khi p) cc ng ng.

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1

Qbl

2

3d

l'

D

4 6 7 5

Hnh 12.1: S h thng phanh du t S h thng phanh du (hnh 12.1) g c 2 phn chnh : truy n ng m phanh v c cu phanh. Truyn ng phanh b tr trn khung xe gm c: bn p 1, xilanh chnh c b cha du 2 to ra p sut cao, cc ng dn du 3 n u cc c c phanh. C cu phanh t bnh xe gm c: xilanh lm vic 4, m u phanh 5, l xo ko 6, trng phanh 7. Nguyn l lm vi ca h thng phanh du nh sau: khi ngi li tc dng c vo bn p 1 qua h thng n s y pttng nm trong xilanh 2, do du b p v sinh ra p su cao trong xilanh 2 v trong ng ng dn 3. Cht lng vi t p sut cao s tc dng ln b mt ca hai pittng xilanh 4. Hai pttng ny thng lc l xo 6 s y hai m phanh 5 p st vo trng phanh 7 v tin hnh phanh t v tr ng phanh 7 c gn lin vi moay bnh xe. Khi nh bn p ngha l lc ngng phanh, l xo 6 s ko hai m phanh 5 v v tr ban u, di tc dng ca l xo 6 cc pttng trong xilanh lm vic 4 s p du tr li xilanh chnh 2. S lm vic ca phanh du lm vic trn nguyn l ca thy lc tnh hc. Nu tc dng ln bn p phanh th p sut truyn n cc xilanh lm vic s nh nhau. Lc trn cc m phanh ph thuc vo ng knh pttng cc xilanh lm vic. Mun c mmen phanh bnh xe trc khc bnh xe sau ch cn lm ng knh pttng ca cc xilanh lm vic khc nhau. Lc tc dng ln cc m phanh ph thuc vo t s truyn ca truyn ng: i vi phanh du bng t s truyn ca phn truyn ng c kh nhn vi t s truyn ca phn truyn ng thy lc. Nu pittng xilanh lm vic c din tch gp i din tch ca pittng xilanh chnh th c tc dng ln pittng xilanh l lm vic s ln gp i. Nh th t s truyn s tng ln hai ln, nhng trong lc hnh trnh ca pittng lm vic s gim i hai ln, v vy m chng c quan h theo t l nghch vi nhau cho nn lm kh khn trong khi thi k truyn ng t phanh. c im quan trng ca h thng phanh du l cc bnh xe c phanh cng mt lc v p sut trong ng ng du ch bt u tng ln khi tt c cc m phanh p st vo cc tr phanh khng ph thuc vo n g knh xilanh lm ng vic v khe h gia trng phanh v m phanh. H thng phanh du c cc u im sau: Phanh ng thi cc bnh xe vi s phn b lc phanh gia cc bnh xe hoc gia cc m phanh theo yu cu. Hiu sut cao.

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nhy tt, kt cu n gin. C kh nng dng trn nhiu loi t khc nhau m ch cn thay i c cu phanh. Khuyt im ca h thng phanh du l: Khng th lm t s truyn ln c v th phanh du khng c cng ha ch dng cho t c trng lng ton b nh, lc tc dng ln bn p ln. Khi c ch no b h hng th c h thng phanh u khng lm vic c. Hiu sut truyn ng s gim nhit thp. 2.2. Phanh kh. Phanh kh s dng nng lng ca kh nn tin hnh phanh, ngi li khng cn mt nhiu lc iu khin phanh m ch cn thng l xo van phn phi iu khin vic cung cp kh nn hoc lm thot kh cc b phn lm vic. Nh th m phanh kh iu khin nh nhng hn. Nguyn l lm vic ca h thng phanh kh theo s (h12.2) nh sau: My nn kh 1 c dn ng bng ng c s bm kh nn qua bnh lng nc v du 2 n bnh cha kh nn 3. Ap sut ca kh nn trong bnh xc nh theo p k 8 t trong bung li. Khi cn phanh ngi li tc dng vo bn p 7, bn p s dn ng n van phn phi 4, lc kh nn s t bnh cha 3 qua van phn phi 4 n cc bu phanh 5 v 6. Mng ca bu phanh s b p v dn ng cam phanh 9 quay, do cc m phanh 10 c p vo trng phanh 11 tin hnh qu trnh phanh. 8 1 3 2

9

5

7

4

10

11

6

Hnh 12.2 : S lm vic ca h thng phanh kh t Trong trng hp ko rmoc (on xe) h thng phanh cn m bo chuyn ng an ton cho on xe. B tr h thng phanh t ko v rmoc c th theo s hnh 12.3 Cc s phn bit vi nhau theo s lng ng ng dn ni t ko vi rmoc ra loi 1 dng hoc 2 dng. Cc phn cn li s ging nhau theo hnh 12.3a, khng kh c nn bng my nn kh 1 r i truyn ti bnh lc 2 v b phn iu chnh p sut 3 n cc bnh cha kh nn 4. Khi trong cc bnh cha kh 4 c y lng d tr

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khng kh nn th b phn iu chnh 3 s ct khng cp kh t my nn vo bnh cha na. phng trng hp p sut c th tng t ngt ng dn kh, trong h thng c t van an ton 5. Khng kh nn c i t bnh cha n van phn phi 11. Khi cn phanh ngi li s tc dng ln bn p phanh qua h thng n n van phn phi 11 v m cho kh nn vo cc bung phanh 9, t s dn ng cam phanh p cc m phanh vo trng phanh tin hnh qu trnh phanh. phanh rmoc, trong h thng c trang b van phn phi 6 cho rmoc. Khi khng phanh khng kh nn c truyn qua van 6 ng dn v u ni 7 cung cp kh nn cho h thng rmoc. Khi phanh th khng kh nn c thot ra ngoi khi ng ng ni t ko v rmoc qua van 6. Do p sut ng ng ni b gim nn h thng phanh rmoc bt u lm vic.

a)

b)

Hnh 12.3: S lm vic ca h thng phanh kh c phanh rmoc

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Khi c khng kh nn c th phanh rmoc bng tay n 10, tay n ny s tc dng ln van phn phi 6 ca h thng phanh rmoc. Khi t lm vic khng ko rmoc th ng ng dn ca h thng phanh rmoc c tch ra khi ng ng ca h thng t bi van bt kn 8. h thng phanh kh hai dng (h.12.3b) phn cung cp kh (gm my nn kh 1, bnh lc 2, b phn iu chnh 3, cc bnh cha 4 v van an ton 5) ging nh h thng phanh kh mt dng, ch khc l van 11 iu khin c h thng phanh ca t v h thng phanh rmoc c ni vi nhau bi hai ng ng. Mt ng ng ni vi ng cung cp 12, ng ny thng xuyn c kh nn dn n h thng phanh rmoc. ng ng th hai ni vi ng c khng kh vo iu khin h thng phanh rmoc. Khc vi h thng phanh kh mt dng h thng phanh kh hai dng, khi phanh p sut trong ng ng iu khin tng ln, nh th m h thng phanh rmoc s bt u lm vic. So snh h thng phanh kh mt dng v hai dng c th rt ra kt lun sau: H thng phanh mt dng c th iu khin ring r h thng phanh t ko v rmoc, hay c th iu khin cng mt lc t y theo yu c s phanh hp l u on xe. iu ny m bo tnh n nh ca xe khi phanh. H thng phanh hai dng, khng kh nn cp cho t ko v phanh ca rmoc bng mt van chung. V th s c hin tng cp khng kh nn khng kp thi cho phanh rmoc nht l i vi xe c ko nhiu rmoc. H thng phanh hai dng c u im l thng xuyn cung cp khng kh cho h thng phanh rmoc, iu ny c ngha ln khi phanh thng xuyn hoc phanh lu di. Cc th nghim h thng phanh trong phng th nghi m v trn ng chng t h thng phanh mt dng u vit hn h thng phanh hai dng. V th cc xe hin nay ch yu dng h thng phanh kh mt dng. H thng phanh kh c u im l lc tc dng ln bn p rt b. V vy n c trang b cho t vn ti ti trng ln, c kh nng iu khin h thng phanh rmoc bng cch ni h thng phanh rmoc vi h thng phanh ca t ko. Dn ng phanh bng kh nn m bo ch phanh rmoc khc t ko, do phanh on xe c n nh, khi rmoc b tch khi t ko th rmoc s b phanh mt cch t ng. u im na ca h thng phanh kh l c kh nng c kh ha qu trnh iu khin t v c th s dng khng kh nn cho cc b phn lm vic nh h thng treo loi kh. Khuyt im ca h thng phanh kh l s lng cc cm kh nhiu, kch thc chng ln v gi thnh cao, nhy t, ngha l thi gian h thng phanh bt u lm vic k t khi ngi li bt u tc dng kh ln. 2.3. Phanh thy kh. Trn hnh 12.4 trnh bay s phanh thy kh. H thng p hanh thy kh gm c my nn kh 1 d n ng bng ng c t, bnh lc 2, bnh cha kh nn 3, xilanh lc, van v xilanh phanh chnh 4 (ba b phn ny kt hp lm mt cm), ng dn du 5, xilanh lm vic 6, m phanh 7, trng phanh 8, bn p iu khin 9. My nn kh 1 qua bnhc 2 s cung cp kh nn n bnh cha 3. Khi tc l dng ln bn p 9 van s m kh nn t bnh 3 n xilanh lc sinh lc p trn pittng ca xilanh chnh 4, du di p lc cao s truyn qua ng dn 5 n cc

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xilanh 6 do s d n ng n cc m phanh 7 v tin hnh qu trnh phanh. Cc ng dn kh h thng phanh ny ngn cho nn nhy ca h thng phanh tng ln. Phanh thy kh thng dng trn t ti ti trng trung bnh v ln. N phi hp c u im ca phanh kh v phanh du c th l lc tc dng ln bn p b, nhy cao, hiu sut ln v c th s dng c cu phanh nhiu loi khc nhau. Phanh thy kh s dng cha rng ri do phn truyn ng thy lc c nhng nhc im: nhit thp hiu sut gim, chm sc k thut phc tp nh kim tra mc du v thot khng kh khi truyn ngvv..

Hnh 12.4: S h thng phanh thy kh mt dng

A. TNH TON C CU PHANHI. XC NH MMEN PHANH CN SINH RA CC C CU PHANH. Mmen phanh sinh ra c cu phanh ca t phi m bo gim tc hoc dng t hon ton vi gia tc chm dn trong gii hn cho php. Ngoi ra cn phi m bo gi t ng dc cc i (mmen phanh sinh ra phanh tay). i vi t lc phanh cc i c th tc dng ln mt bnh xe cu trc khi phanh trn ng bng phng l (tham kho gio trnh L thuyt t):Pp 1 =

cu sau l:Pp2 =

G1 Gb m1 p = m1 p 2 2LG2 Ga m 2p = m 2p 2 2L

(12.1)

(12.2)

y: G G1, G2

- trng lng t khi ti y; - ti trng tng ng (phn lc ca t) tc dng ln cc bnh xe trc v sau trng thi tnh, trn b mt nm ngang; m 1p , m 2p - h s thay i ti trng tng ng ln cu trc v cu sau khi phanh a, b - khong cch tng ng t trng tm t n cu; L - chiu di c s ca t; - h s bm gia lp v ng ( = 0,7 0,8)

Cc h s m 1p , m 2p xc nh theo l thuyt t nh sau:

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m 1p = 1 +

j max h g gb j max h g

=1+

' h g b

(12.3) (12.4)

m 2p = 1 -

ga

=1-

' h g a

Trong : h g - chiu cao trng tm ca t; g- gia tc trng trng; j max - gia tc chm dn cc i khi phanh; - h s c trng cng phanh ( =

jmax ) g

t c cu phanh t trc tip tt c cc bnh xe (phanh chn). Do mmen phanh tnh ton cn sinh ra ca mi c cu phanh cu trc l:M p1 =

cu sau ( t hai cu) l:M p2 =

G1 G m 1p rbx = (b + ' h g ) rbx 2 2L

(12.5)

Trong : r bx bn knh lm vic trung bnh ca bnh xe. Khi tnh ton c th chn = 0,4 0,5 v = 0,7 0,8. ng v kt cu ca c cu phanh m xt th mmen phanh M p1 v M p2 phi bng: M p1 = M p1 + M p (12.7) M p2 = M p2 + M p2 (12.8) y: M p1 , M p1 m men phanh sinh ra m phanh trc v m phanh sau ca mi c cu phanh cu trc; M p2 , M p2 m men phanh sinh ra m phanh trc v m phanh sau ca mi c cu phanh cu sau. II. TNH TON C CU PHANH GUC. 2.1. Quy lut phn b p sut trn m phanh. Mun tnh ton c cu phanh guc chng ta cn phi bit quy lut phn b p sut trn m phanh. Tu theo s tha nhn quy lut phn b p sut trn m phanh, chng ta c nh ng cng thc tnh t on phanh gu khc nhau. Th c nghim chng t rng hao mn cc im khc nhau ca m phanh khng ging nhau, bi th tha nhn quy lut phn b p sut u trn m phanh l khng ph hp vi thc t. Chng minh sau y cng chng t iu .

G2 G m 2p rbx = (a - ' h g ) rbx 2 2L

(12.6)

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B A

P

A'

0

1

a)

b)

0

Hnh 12.5: S dch chuyn m phanh trong trng phanh tm quy lut phn b p sut trn m phanh chng ta tha nhn gi thit sau: p sut ti im no y trn m phanh t l thun vi bin dng hng knh ca im y khi phanh, ngha l coi nh m phanh tun theo nh lut Hc. iu ny tha nhn c trong phm vi bin dng thng rt nh ca m phanh. Khi phanh trng v phanh guc khng b bin dng m ch m phanh (tm m st) bin dng. S d nh vy l v trng v guc phanh lm bng nguyn liu cng hn m phanh nhi u, kt cu ca trng v guc phanh c ng gn tng cng cng vng. B mt lm vic ca m phanh p st vo b mt lm vic ca trng phanh khi phanh. Trn hnh 12.5a trnh by dch chuyn guc phanh trong trng phanh s quanh tm O1. Gi s rng trong qu trnh phanh khi m phanh va mi chm vo b mt lm vic ca trng phanh (thi im bt u b bin dng) guc phanh cn quay thm 1 gc na do m phanh b bin dng di tc dng ca lc P ng xilanh lm vic. Nu xt im A trn m phanh chng ta th im A ng vi thi im m y phanh va mi chm vo trng phanh. Trong qu trnh bin dng im A phi quanh quanh tm O 1 vi bn knh O 1 A v ti im A tng ng vi gc quay rt nh ca m phanh, ngha l O 1 A=O 1 A. T A h ng thng gc AB xung bn knh OA, o AB c trng cho bin dng hng knh ca m phanh ti n im A khi m phanh quay gc . Gc BA'A OAO1 = v c AB AO v AA AO 1 (coi nh rt nh) Xt tam gic vung ABA ta c:AB = AA'sin

nhng AA = O 1 A. ( tnh theo rad) cho nn: AB = O1 A.. sin Tam gic OO 1 A cho ta biu thc sau:

(12.9)

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OO1 O1 A = sin sin

hay l:O1 A = OO1 sin sin

(12.10)

Thay tr s O1 A t biu thc 12.10 vo 12.9 ta c: p sut q ti im A theo gi thit th nht s t l vi bin dng hng knh, do : q = k AB = k OO1 .sin (12.11) y: k h s t l, hay l cng ca m phanh. Trong cng thc (12.11) k v OO1 l hng s, cn s l gc quay chung cho tt c cc im ca m phanh quay quanh tm O 1 , cho nn n l hng s i vi cc im ca m phanh. Thay cc hng s bng mt tr s khng i K v coi im A l mt im bt k xc nh trn m phanh bi gc ( l gc thay ), cui cng ta c cng thc i tng qut xc nh p sut bt k im no trn m phanh nh sau: q = Ksin (12.12) y: K - h s t l (K = k OO1 . ); - gc xc nh v tr ca im cn tnh p sut trn m phanh. Cng thc (12.12) cho chng ta thy rng p sut phn b trn m phanh theo quy lut ng sin. p sut cc i ng vi lc = 900 ngha l ti im C (h.12.5b) (im C ca m phanh nm trn trc X X thng gc vi trc Y Y i qua cc tm O v O 1 ). p su cc tiu ng vi lc = 00 v = 1800, ti cc t im y p sut bng khng. Biu phn b p sut m phanh c ch r hnh 12.5b. p sut cc i im C s l: q max = K do cng thc (12.12) cn c th vit: (12.13) q = q max sin Do p sut phn b trn m phanh khng u (theo lut ng sin) cho nn cc im trn m phanh s hao mn khc nhau, phn gn im C s hao mn nhiu hn, cn cc u cui hao mn t hn. Thc t ra, cc u cui ca m phanh hu nh khng lm vic cng v th m gc m o ca m phanh trn mi guc phanh thng ly nh hn 120 o,i vi t hin nay gc o thng nm trong gii hn 900 1100. Quy lut phn b p sut ny lm phc tp cho vic tnh ton c cu phanh. V gc m o hin nay khng ln lm v guc phanh c th b bin dng khi phanh cho nn s chnh lch v phn b p sut trn m phanh trong phm vi nh th khng ln lm. V th trong tnh ton ban u khi chn s b cc kch thc, chng ta coi nh p su phn b u trn m phanh n gin cho tnh ton. t Khi guc phanh c cng ln v mun tnh chnh xc chng ta phi ly quy lut phn b theo ng sin. Sau y chng ta s tnh c cu phanh cho c hai trng hp phn b p sut u v theo ng sin. 2.2. Tnh ton c cu phanh.AB = OO1. sin

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Tnh ton c cu phanh nhm mc ch x c nh cc kch thc v cc thng s c bn ca c cu phanh khi phanh c th sinh ra mmen phanh m bo hm c t. Mmen ny t m mi c cu phanh cu trc v cu sau phi sinh ra c xc nh tng ng theo cng thc (12.5) v (12.6). Cc mmen trn c coi l mmen phanh tnh ton c cu phanh. 2.2.1. Xc nh gc v bn knh ca lc tng hp tc dng vung gc ln m phanh 2.2.1.1. Trng hp tha nhn p sut phn b u trn m phanh q = q 1 = const: Mmen phanh sinh ra trn ng phanh ph thu c vo kt cu ca c cu tr phanh. Trn hnh 12.6a trnh bys tnh ton c cu phanh vi hai guc phanh c im ta c nh ring r v mt pha. Nu truyn ng phanh l loi thy lc (phanh du) th lc p P ln cc guc phanh s bng nhau khi ng xilanh lm vic c ng knh nh nhau. Nu dng cam p ln cc guc phanh (truyn ng c loi c kh hoc loi kh) th lc p P 1 v P 2 ln cc guc phanh s khc nhau, trong khi dch chuyn ca cc m phanh s ging nhau. S d P 1 khc P 2 l v chiu lc ma st T 1 v T 2 trn cc m phanh khc nhau, Trong khi tr s ca chng bng nhau (T 1 = T 2 ) do dch chuyn ca hai m phanh nh nhau (lc T sinh ra do c lc N, m tr s ca lc N ph thuc vo bin dng ca m phanh, nu bin dng ny bng nhau th lc N 1 = N 2 , do T 1 = T 2 ). Chng ta s xt trng hp khi hai guc phanh c p mt lc P nh nhau.

y1 O'

P x1N 1y N1N1x

0

0

rt

a

T1

dN1ydT1

R1 U1 Ux

2

d

1

x1

dN1

dN 1x

r0O1

c

y1

a)

Hnh 12.6a: S tnh ton c cu phanh vi cc guc phanh c im ta c nh ring r v mt pha v lc p ln cc guc phanh bng nhau.

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Trn hnh 12.6a trc Y 1 Y 1 i qua hai tm O v O 1 v thng gc vi trc X 1 X 1 i qua im c p sut cc i. Khi phanh mi phn t ca m phanh b tc dng t pha trng phanh bi lc thng gc dN 1 v lc ma st dT 1 . Lc ma st: dT 1 = .dN 1 y: - h s ma st gia trng phanh v m phanh. Chng ta xt m phn t ca m phanh nm cch trc Y 1 Y 1 mt gc . t Phn t ny chon gc d. Lc thng gc dN 1 trn phn t s l: dN 1 = q 1 br t d (12.14) dT 1 = dN 1 = q 1 br t d (12.15) y: q1 p sut phn b trn m phanh trc (q1 = const theo gi thit); b chiu rng m phanh; r t bn knh trong ca trng phanh; d - gc m ca phn t m phanh ang xt. Khi p sut phn b u trn m phanh (h.12.7) th tng hp lc N 1 ca tt c cc lc dN 1 phi nm trn trc i xng OD ca m phanh, ngha l D l im gia ca cung EF.

y

F x ND

O

x

E

O

yHnh 12.6a 1 : Xc nh gc t ca lc N 1 khi p sut phn b u Gc to bi lc N 1 v trc X 1 X 1 s l: = 90o - DOO1 = 90o - ( DOE + EOO1 ) = 90o - ( 2 - 1 + 1 ) 2 + 1 = 90o - 2 2

(12.16)

y: 1 , 2 gc u v gc cui ca m phanh (h.12.7) Chiu lc dN 1 trn trc X 1 X 1 v Y 1 Y 1 ta c: dN 1X = q 1 br t sind

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dX 1Y = q 1 br t cosd Tch phn trong gii hn t gc 1 n 2 ta c:N 1X = dN 1X = q 1 brt sin d = q 1 brt ( cos 2 cos 1)1 1 2 2

(12.17) (12.18)

N 1Y = dN 1Y = q 1 brt cos d = q 1 brt ( sin 2 sin 1)1 1

2

2

Lc tng hp thng gc N 1 tc dng ln m phanh l:2 2 N 1 = N 1X + N 1Y = q 1 brt ( cos 2 cos 1) 2 + ( sin 2 sin 1) 2

(12.19)

Mmen phanh do mt phn t m phanh sinh ra l: dM pl = r t dT 1 = q 1 br t 2d Mmen phanh tc dng trn c m phanh trc l:M' pl = dM' pl = q 1 brt2 d1 1 2 2

(12.20) = q 1 br t ( 2 - 1 ) = q 1 br t 2 o y: o gc m ca m phanh. Lc thng gc tng hp N 1 s sinh ra lc ma st tng hp T 1 = N 1 . Lc T 1 c im t cch tm O mt on . M men phanh m phanh tnh theo cng thc (12.20) cn c th tnh theo cng thc sau:2

M pl = T 1 = N 1 T : = M' pl N1

(12.21)

(12.22)

Thay cng thc (12.19) v (12.20) vo (12.22) ta c: = = q 1 brt2 0 q 1 brt ( cos 2 cos 1) 2 + ( sin 2 sin 1) 2 o rt =

=

= 2 + 1 2 + 1 2 1 2 2 1 2 ( 2 sin sin sin ) + (2 cos ) 2 2 2 2 o rt o rt = 1 1 + 1 + 1 2 sin 2 4 sin 2 2 ( cos 2 2 ) + sin 2 2 2 2 2 2 o rt 2 sin o 2

n gin na ta c: =

(12.23)

Nu thay 'o =

o , cng thc (12.23) s c dng sau: 2

'o rt = sin 'o

(12.24)

y:

o na gc m ca m phanh.

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Cn ch rng gc o v o trong cng thc (12.23) v (12.24) tnh theo rad. Nu o = 90o = o th: = rt ; nu o = 120 = 2 2 2 2 th: 3 = 2 3 3 rt

2.2.1.2. Trng hp tha nhn p sut trn m phanh phn b theo quy lut ng sin q=qmaxsin. Khi phn b p sut t heo ng sin cc phn t lc dN 1 v dT 1 tc dng ln m phanh l: dN 1 = q max br t sin d dT 1 = q max br t sin d Chiu lc dN 1 ln trc X1 - X 1 ta c: dN 1x = q max br t sin2 d t N 1x = dN 1x1 1 2 2

(12.25) (12.26)

sin 2 = q max brt sin d = q max brt = 4 1 2 1

2

sin 2 2 sin 21 1 q max brt 2 1 + = 2 2 2 1 = q max brt (2 0 + sin 21 sin 2 2 ) 4 =

(12.27a)

Chiu lc dN 1 ln trc Y 1 Y 1 ta c: dN 1y = q max br t sin cos.d = q max brt sin 2.dN 1Y = dN 1Y1 12 2 1 1 = q max brt sin 2.d = q max brt sin 2.d 2 = 2 4 1 1

1 2

2 1 1 = q max brt ( cos 2) = q max brt ( cos 21 cos 2 2 ) 4 4 1

(12.27b)

Gc to bi lc N 1 vi trc X 1 -X 1 l:1 q max brt ( cos 21 cos 2 2 ) N 1Y tg = = 4 N 1X 1 q max brt (2 0 + sin 21 sin 2 2 ) 4 cos 21 cos 2 2 2 0 + sin 21 sin 2 2

n gin i ta c:tg =

(12.28)

M men phanh sinh ra trn phn t ca m phanh l : dM p1 =r t dT 1 =q max br t 2 sin d M men phanh sinh ra trn c m phanh trc l :M' p1 = dM' p1 = .q max brt1 1 2 2 1

sin d = q

max

brt2 ( cos 1 cos 2 )

(12.29)

Lc tng hp N

1

l:

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N1 = =

Bn knh xc nh theo cng thc: = M' pl T1 = M' pl N1

2 2 N1X + N1Y = 1 q max br1 (2 o + sin 21 sin 2 2 ) 2 + ( cos 21 cos 2 2 ) 2 4

(12.30)

Lp cc tr s M p1 v N 1 t cc cng thc (12.29), (12.30) vo v n gin i ta c:= = = 4rt ( cos 1 cos 2 ) (2 o + sin 21 sin 2 2 ) 2 + ( cos 21 cos 2 2 ) 2 4rt ( cos 1 cos 2 ) [2 o 2 cos ( 2 + 1) sin ( 2 1)]2 + [2 sin ( 2 + 1) sin ( 2 1)]2 4rt ( cos 1 cos 2 ) 4 + 4 cos ( 2 + 1) sin o 8o cos ( 2 + 1) sin o + 4 sin 2 ( 2 + 1) sin 2 o2 o 2 2

= = =

Cui cng ta c:=

2rt ( cos 1 cos 2 )

2 o + sin 2 o 2 o cos ( 2 + 1) sin o

(12.31)

Cc cng th (12.16), (12.23) cho ta tnh ton gc v bn knh trong c trng hp p sut phn b u, trong trng hp p sut phn b theo ng sin chng ta dng cng th c (12.28) v (12.31) tnh. T cng thc trn thy rng gc v bn knh ch ph thuc vo cc thng s kch thc ca c cu phanh ( 1 , 2 , r t ) m khng ph thuc vo tr s ca p sut. Nu m phanh trc v m phanh sau hon ton i xng vi trc ng (ngha l cc thng s kch thc u bng nhau) th gc v bn knh ca m trc v m sau u nh nhau mc du p sut trn hai m phanh phn b theo cng quy lut (phn b u hoc theo ng sin), nhng vi tr s khc nhau. Khi b tr m phanh nh trn hnh 12.6b th p sut m phanh trc s ln hn m phanh sau v lc T1 m phanh trc tng cng cho s phanh, cn lc T2 m phanh sau li gim s phanh (h.12.6b), nhng gc v bn knh hai m phanh c tr s nh nhau. 2.2.2. Tnh ton lc cn thit tc dng ln guc phanh P1 v P2: Trong thc t khi tnh to n c c phanh, chng ta cn xc nh lc P i tc u dng ln guc phanh (h.12.6b) m bo tng s mmen phanh sinh ra guc phanh trc (M pl hoc M p2 ) v guc phanh sau (M p1 hoc M p2 ) bng mmen phanh tnh ton (M p1 hoc M p2 ) ca mi c cu phanh. Mmen phanh tnh ton M p1 v M p2 c xc nh trc theo cng thc (12.5) hoc (12.6). Sau y chng ta s xt quan h gia lc P i v mmen phanh M p1 v M p1 (gi s rng chng ta xt c cu phanh cu trc). Khi thit k c cu phanh chng ta chn trc qui lut phn b p sut trn m phanh trn c s chn trc cc thng s kt cu ( 1 , 2 , r t ) chng ta tnh c gc v bn knh , ngha l xc nh c hng v im t lc N 1 .

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U2

R2

y2 O' O'' P1 P2

y1

P2 P1

x1 N1

Ox2

r0

R2 N2

x2T2

U1

T1

R1 U1

x1 rt

c)

R1

U2

y1

y2

b)

Hnh 12.6b: S tnh ton c cu phanh vi cc guc phanh c im ta c nh ring r v mt pha v lc p ln cc guc phanh bng nhau Lc R 1 l lc tng hp ca N 1 , T 1 v R1 to vi N1 gc . Gc xc nh nh sau:tg = T1 = N1

r

r

(12.32)

Chn = 0,3 chng ta s xc nh c gc ngha l xc nh c hng r ca R1 . Gc m phanh trc v m phanh sau u bng nhau v cng mt h s ma st nh nhau. Mmen phanh ca c cu phanh l: M p1 = M p1 + M p1 = R 1 r o + R 2 r o = (R 1 + R 2 )r o (12.33) y: R 1 , R 2 lc tng hp m phanh trc v sau: ro - bn knh, xem hnh 12.6b. Bn knh r o xc nh theo cng thc. tg (12.34) ro = sin = = 2 1 + tg 1+ 2 Tr s M 1 tnh theo cng thc (12.5), r o xc nh theo cng thc (12.34) t chng ta xc nh tng s lc R 1 + R 2 theo cng thc sau:R1 + R 2 = M 1 ro

(12.35)

Mun xc nh ring r lc R 1 v R 2 chng ta dng phng php ha bng cch v a gic lc ca guc phanh trc v sau. Trn mi guc phanh c ba lc tc dng P 1 , R 1 , U 1 hoc P 2 , R 2 , U 2 (trng hp dn ng bng thy lc th lc P hai guc phanh bng nhau nu ng xilanh lm vic cng mt ng knh). Guc phanh trc v sau nm v tr cn bng cho nn ba lc tc dng phi gp nhau ti tm O hoc O (h.12.6b). Hng lc P 1 v R 1 bit (tr s ca chng cha bit), ko di chng cho gp nhau O, ni O vi O 1 chng ta c hng lc U 1 . Cng lm nh vy i vi guc phanh sau chng ta tm c hng lc U 2 . Sau xy dng a gic lc cho guc phanh trc v guc phanh sau vi cng

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mt t l nht nh (v lc P i hai guc phanh bng nhau: P 1 = P 2 = P, cho nn c th ly P lm mt n v chng hn, iu ny khng nht thit, ch yu l m bo t l ca hai a gic lc hai guc phanh nh nhau). Trn c s cc a gic lc v c chng ta tm c t s gia lc R 1 v R 2 (R1 R ). Bit c t s 1 R2 R2

c v bit c tng s R 1 + R 2 theo cng th (12.35) chng ta c th xc nh c tng tr s ring r R 1 v R 2 . C R 1 , R 2 chng ta s xc nh c tr s ca cc lc P, U 1 , U 2 . Bit c lc P chng ta c c s tnh ton truyn ng phanh. Ngoi ra lc P, U 1 v U 2 to iu kin cho chng ta tnh ton sc bn cc chi tit ca c cu phanh. Lc P m chng ta xc nh theo phng php nu trn s m bo cho c cu phanh sinh ra mmen phanh yu cu M p1 cu trc hoc M p2 cu sau. Nu guc phanh b p bng cam th lc P 1 v P 2 tc dng ln hai guc phanh s khc nhau. Trong tr ng hp ny khi cam quay, hai guc phanh s dch chuyn nh nhau. N thi gian u khe h gia m phanh v trng phanh guc u phanh trc c khc guc phanh sau i na th qua mt thi gian chy r p sut tc dng ln hai m phanh s bng nhau do dch chuyn ca hai guc phanh nh nhau. V p su hai m phanh bng nhau cho nn lc R 1 = R 2 . Nh v khi t y guc phanh b p bng cam quay chng ta c th xc nh ngay lc R 1 v R 2 . R1 = R2 =M p1 2ro

(12.36)

Bit c tr s lc R 1 v R 2 , da vo cc a gic lc ca guc phanh trc v sau v theo phng php trn chng ta tm c tr s lc P 1 , P 2 , U 1 v U 2 . Trn kia chng ta dng phng php ha xc nh lc P. C th dng phng php gii tch xc nh quan h gia lc P v mmen phanh nh sau: Xt cn bng guc phanh trc i vi tm O ta c (h.12.6a): U x c Pa = R 1 r o = M p1 (12.37) y: U x hnh chiu ca lc U 1 trn trc X 1 X 1 (h.12.6a); c, a cc kch thc, xem trn hnh 12.6a. T biu thc (12.37) rt ra Ux =M pl + Pa c

(12.38)

Chiu cc lc tc dng ln guc phanh trc trn trc X 1 X 1 ta c: Pcos o + U x N 1 cos - T 1 sin = 0 Thay tr s ca U x t cng thc (12.38) v thay N 1 = thc (12.39) chng ta c biu thc sau:P cos o + M' pl c + M' p1 Pa M' pl cos sin = 0 c

M pl

(12.39) , T 1=M pl

vo biu

(12.40)

Gii phng trnh (12.40) i vi P ta c biu thc sau:M' P1 = P(c cos o + a) c( cos + sin )

(12.41)

Tng t nh vy, nu xt cn bng guc phanh sau ta c:

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M''P1 =

P(c cos o + a) c( cos sin ) +

(12.42)

Cng thc (12.41) v (12.42) dng cho trng hp guc phanh dn ng bng cht lng. Khi guc phanh dn ng bng cam th lc R 1 = R 2 , do M p1 = M p1 . T c th rt ra biu thc sau: M p1 = 2M p1 = 2M p1 ==2 P1(c cos o + a) P2 (c cos o + a) =2 c( cos + sin ) c( cos sin ) +

(12.43)

y: P 1 , P 2 lc tc dng t cam quay ln guc phanh trc v sau, hai lc ny c tr s khc nhau. T s cc lc P 1 v P 2 xc nh nh sau:P1 c( cos + sin ) = P2 c( cos sin ) +

(12.44)

Hnh 12.7: S c cu phanh t cng ha Trn hnh 12.7 trnh by c cu phanh t cng ha. c cu phanh ny hiu qu phanh c tng ln nh dng lc ma st gia m phanh trc v trng phanh. Hai gu phanh c ni vi nhau bng thanh trung gian 1. Nh vy, c guc phanh sau c p vo trng phanh khng nhng bng lc P m cn bng lc U 2 c tr s bng lc U 1 . Coi nh guc phanh v trng phanh hon ton cng chng ta c th xc nh tr s v r 0 theo phng trnh (12.31) v (12.34). N u lc P v U 1 song song th l c R 1 cn bng cc lc trn cng phi song song v ng thi li tip tuyn vi vng trn bn knh r o . Chng ta s c cc phng trnh sau: R 1 = P + U 1 ; M p1 = R 1 r o iu kin cn bng guc phanh sau, khi U 2 = U 1 s l: R2 = P + U1 + U3 Do mmen phanh guc phanh sau: (12.45)

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c

b

a

M p1 = R 2 .r 0 M p1 = (R 1 + U 3 )r 0

(12.46)

So snh cng th c (12.46) vi (12.45) chng ta thy trong trng hp ny mmen phanh guc phanh sau ln hn guc phanh trc. iu kin cn bng mmen ca tt c cc lc tc dng ln guc phanh trc i vi im t lc U 1 l: P (a + c) = R 1 (c r 0 ) ; R 1 = P U1 = R1 P = P T y M p1 = Pa+c r0 c r0 a + r0 c r0 a+c c r0

(12.47)

iu kin cn bng mmen ca tt c cc lc tc dng ln guc phanh sau i vi im ta A (h.12.8) s l: P(a b) + R 2 (b r o ) = U 1 (b + c)R2 = P (a + ro )(b + c) (a b)(c ro ) (c ro )(b ro )

Bin i i ta c: M p1 = P(a + c)( b + ro ) ro (c ro )( b ro )

(12.48)

Cng thc (12.48) v (12.47) cho chng ta thy rng c cu phanh t cng ha khi c lc P tc dng, guc phanh sau s sinh ra mmen phanh M p1 ln hn nhiu so vi guc phanh trc. Nu gc 1 v 2 ca m phanh trc khc vi m phanh sau th v r o ca hai guc phanh cng s khc nhau. c cu phanh t cng ha trnh by trn hnh 12.8 hiu qu phanh (mmen phanh) khi t tin v li u nh nhau. 2.3. Phanh m du v n nh ca t khi phanh (hin tng t sit). Phanh m du v n nh ca t khi phanh ph thuc vo s phn b u lc phanh bnh xe phi v tri khi cc bnh xe khng b gi cng, vo s n nh ca mmen phanh M p i vi c cu phanh c, khi h s ma st thay i trong gii hn c th ca n (thng t 0,28 n 0,30) v vo kh nng b sit ca ca c cu phanh. Nu mmen phanh cc bnh xe phi v tri sai lch so vi mmen phanh tnh ton khong 10 15%, khi h s thay i th n nh ca t khi phanh (khi phanh khng b lch hng) vn m bo d dng c bng cch gi bnh li. Trong qu trnh phanh c th xut hin hin tng t sit. Hin tng t sit xy ra khi m phanh b p st vo trng phanh ch bng lc ma st m khng cn tc ng lc P ca truyn ng ln guc phanh. Trong trng hp nh vy, mmen phanh M p ng v phng din l thuyt m ni s tin ti v tn. i vi guc phanh trc (h.12.6a) hin tng t tit s xy ra khi c iu kin sau theo cng thc (12.41): c(cos + sin) - = 0

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Ngha l khi:= c cos c sin

(12.49

Bng cch chng minh n gin c th thy rng khi xy ra hin tng t sit lc tng hp R 1 s i qua tm quay O 1 ca guc phanh. Nu xt cng thc (12.42) dng cho guc phanh sau, chng ta thy rng mu s ca n khng th bng s khng c bi v lun lun m bo > csin v lc tng hp R 2 khng th i qua tm quay O 2 ca guc phanh sau c (h.12.6b). V th guc phanh sau khi lm vic khng thun chiu quay th khng bao gi sinh ra hin tng t sit. guc phanh t cng ha hin tng t sit s xy ra khi c = r o hoc b = r o (theo cng th 12.47 v 12.48) ngha l khi lc tng hp R 1 i qua thanh p c trung gian hoc khi lc tng hp R 2 i qua im ta A (h.12.8). Hin tng t sit s xy ra khi:= b 2 b2

v =

c 2 c2

(12.50)

C cu phanh t cng ha c mmen phanh t n nh hn khi h s ma st thay i v c kh nng b t sit nhiu hn so vi c cu phanh m guc c cc im ta c nh ring r. Cng v th m hin nay c cu phanh t cng ha khng dng trn t du lch. Khi thit k c cu phanh phi ch chn cc thng s kch thc th no trnh xy ra hin tng t sit, c nh th phanh mi c th m du v n nh c.

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