K H O A H O C - C 6 N G NGHE
-S6 9/2015
Dao dpng cua 6 to hai cau theo mo hinh he khong gian 7 bac ttf do
T o m tat: Bii bio khao sit dao dgng cua d td hai ciu cd h$ thong treo phg thugc theo md hinh hi khdng gian tuyin tinh, 7 bac ttf do. Hg phtfdng trinh vi phin dao dgng vi phtfdng phip tinh dtftfc di xuit trong bii bio cho phip khao sit day dO dip tfng dgng Itfc hgc cua d td.
Jii khoa: Daa dgng, d td hai du, khdng gian tuyin tinh.
Abstract: This paper considers the response of a two-axled automobile with dependent suspensions in the model of a seven-DOF spatial vibration system. The system of differential equations of vibration and the method proposed for solving allow to investigate fully the dynamic response of the automobile.
Keywords: Oscillator, automotive seals, linear space.
1 . Dgt v ^ n d e
Bdi toan khdo sat dao ddng cua d td trong gud trinh c h u y i n d d n g tren dUdng khdng bdng phang nhdn dUpc sp quan tdm eua nhieu tdc gid lrong vd ngodi nUdc. Do tinh chdt phQe lap cija vide khdo sdt, tinh todn, cac m d hinh khdng gian ve dao ddng eija d t d mac du da dQpc de cap nhung cdn kha khiem t d n . Md hinh he dao ddng khdng gian 8 bde l y do c u a d td hai cdu da dUdc khdo sdt lrong [3], d dd tfnh c d n cua cae Idp xe dQpc bd qua, cdn than xe dQpc lap md hinh lheo hai khdi IQdng dao ddng trong cdc mdt phang vudng gde vdi trge dpc xe Trong [51 da sQ dung md hinh he khdng gian 7 bde tg do de khao sat dao ddng eija d td hai eau he t h i n g treo dde Idp vdi gidm chdn kieu khf ndn.
Bdi bao nay se khdo sat dao ddng cua d td hai cdu ed h$ thdng treo phg thudc theo md hinh he khdng gian 7 bde tQ do.
2. M d h i n b k h d o s a t d a o d p n g 2.1. Mgt si gii thiit dugc ip dgng Bai bdo si!f d u n g cdc gid t h i l t sau:
- Cdc b a n h xe ludn tidp xdc vdi mat d u d n g ; - T r o n g m d hinh dao d d n g , than xe, eau trudc va c d u sau dQpc xem la vgt rdn tuyet ddi;
- Coi c d n eija ldp xe Id ddn n h d l tuyen tinh;
- e u d n g tdc d y n g eua iQc Id xo vd iQc giam c h d n thude cung mdt c y m treo trung nhau tren mdt dQdng thSng t h i n g dQng.
• PGS. T S . VO CONG H A M
• PGS. T S . VU Q u d c T P y
• T h S . NCS. N G U Y I N D I N H DUNG Hgc vi$n Ky thuit Quin stf
2.2. Md hinh khao sit
Hinh 2.1 b i l u d i l n mdt d td hai edu vdi ba phan IQ khdi lupng: Thdn xe 1, cau trudc 2 vd cdu sau 3. Lidn k i t giQa edu trudc va edu sau vdi lhdn xe dupc thue hidn thdng qua cdc c g m Id xo, gidm chdn thudc hd thdng treo cija d Id.
Hmh 2.1:
Md hinh khao sit dao dgng cua d td
Md hinh sQ dgng hai he tpa dd, he ed djnh OXYZ gdn vdi mat dudng; hd dpng O'X'Y'Z' gdn vdi than xe. Cdc ky hieu S, S,, S^ - TQdng Qng Id khdi tam cua thdn xe, cdu trUde va eau sau; a,, a^ - Khoang each tQ khdi tdm lhdn xe den tryc c l u Irudc va true edu sau; 2b - Khodng cdch trung binh glQa hai vet banh xe phdi va trai ( e h i l u rdng ed sd); 2c - Khoang cdch giQa nhfp phdi vd trai; M^, Mj.,, Mp2 - TUdng Qng la khdi iupng phdn treo, phan khdng treo edu trude va cdu sau; J^^, J^.^ - TUdng Qng la md mem qudn tfnh khdi IUdng eua thdn xe ddi vdi true Q'X vd OY'; J^,, J^.^ - TUdng Qng la m d - men quan tinh khdi lupng cilia cdu trQdc vd cdu sau vdl trye song song vdl true QX; C^.^.,, K , C^c^, K.rc2 - TUdng Qng Id he s l dan hdi vd he so cdn cua he treo eau irude va cdu sau; C^p,, \c:\' '-'LC2' \C%' TQdng Qng la hd sd ddn h i i va hd so cdn cQa Top trUde va lop sau.
Zp,, Zpj, Z[,3, Zjj^ - Dp cao cilia bidn dgng mat dudng lai diem cac tdp xe IUdng Qng liep xuc vdi mat dudng. Khi bidn dgng mat dudng b i l l trudc thi dd cao ZQ. se chi phg thude vdo vj lri eua xe {tpa dd X cOa khd'i tam S) hay:
^=z^{x),k = l*4 (1) Trong cdc thi dy tfnh lodn, ham z^^ dUde eho
trude, tpa dp x d U p c xdc djnh qua vdn tdc ehuyen ddng ^ v a lhdi gian e h u y i n ddng f cCia xe.
2.3. Cic thdng si md ta dao dgng Dao dOng cua thdn xe vd cdc cdu xe se dupe khao s d i vdi cac thanh phan nhU sau:
- Thdn xe ed 3 c h u y i n ddng (3 bac IQ do) Id:
Dich e h u y i n t h i n g dQng cCia khdi ldm z^; djeh
5 7
KHOA H O C - C O N G NGHE
ehuyen gde cp^ (gde ldc dpe) quanh trge OY; djeh ehuyen gde *Fg ( gde ldc ngang) quanh tryc OX.
- Cac edu xe (khdi lupng khdng dupe treo) ed 4 chuyen ddng Id: Djeh c h u y i n thang dQng cua khdi tam cdu trUdc v d sau z , z^^ v d dich c h u y i n goc "Pf,,, T^2 cilia cau trUdc va sau quanh ede tryc song song vdi OX.
Quy Udc cac c h u y i n vi z^ z^,,, z^^, lay tQ v[
trf cdn bang ITnh, theo dd khdng can d l cap den trpng IQdng ban thdn cae khoi tupng dao ddng.
3. H^ phUc^g trinb vi pban dao dpng 3.1. Chuyen vj cua cic diem gin Id xo, giam chin tren md hinh dao dgng
Vdi eac bac tp do neu tren, c d the thanh lap vectd eac tpa dd suy rdng:
Trdn Hinh 2.1 bieu didn vj trf cQa 12 diem gdn td xo vd gidm chdn, dupc ddnh so tQ 1 den 8 cho cdc d i l m lidn k i t thpc vd bdn diem lien k i t danh nghTa V, 2', 3', 4'. C h u y i n vj t h i n g dUng cua cac d i l m gan 16 xo v d gidm chan dupe ky hidu IQdng Qng Id: _ _ _
j , , _ , . r 3 , _ j , i , , j g . . ^ - , Z g , Z j , r , . r j j Z j (3) c a e e h u y i n vj t h i n g dQng dupe bieu didn qua hd tpa dp suy rong nhu sau:
:^^;n+W'ci.2j=2ci-*'*'a- ^ =^+*'*'(--
^' • " '^ {4
3.2. Thiit lgp h$ phUdng trinh vi phin dao dgng cua cd h^
Ap dgng dinh lugt 2 New^ton cho 3 khdi lUpng dao ddng ta thie't l | p dupc he phUdng lrinh vi phdn dao ddng cija cd hd, gdm 7 phUdng lrinh vi phdn cdp hai nhQ sau:
' " . ^ = -'tr,U,-;;)-Cr,U.-^|-M--.--^:)-
-Q,(l.-r,ia-A.(.---^)a4.Q,(--r,io,+ ( g )
*Cnt.-.=;)f-K„(.-.-.-,)r-C^fe,-ri)c4- ( 7 )
* C „ ( i , - i i K K „ ( : , - ^ , 5 - C „ ( . - , - 0 - ( 8 ) J'a*==«n<--,---.V+'~„U-i;y-*„(--,--,>c-
^ Thay cdc bieu thQc vao hd phUdng trinh lren rdi thgc hien mdt s l phep bien doi ddn gidn la ed l h i dua he phQdng trinh vi phdn dao ddng d trdn
ve dgng ma tran, trong d d khdng cd mat c h u y i n vj t h i n g dQng c u a cae diem g i n Id xo va gidm chdn, chi cd mat cdc tpa dp suy rdng:
{M]'gHC]g+[K]q = F (12) Trong do: Ma Iran khdi lupng [M\, ma trdn cdn
[Q, ma trgn dd cQng [K] v a vectd iQc kfch thich f hodn toan xdc djnh lheo cae thdng s d hinh hpc, ddng IQc hpe cua xe va bidn dang mat dudng.
4. Dao dpng ci!ia 3 to trong hai trUdng hpp kich thi'cb tieu bieu
Khdo sat dao ddng eOa d td ddn d i n viec giSl hd phUdng trinh vi phdn dao ddjig luyen tinh theo d i l u kien dau cho Irudc. Bdi v i e l sijf d g n g phUdng phap s d (phUdng phdp Runge-Kulta) d l gidi h i phUdng trinh vi phdn dao ddrig bang cdch ISp ChUdng trinh tfnh trong phdn m e m MATLAB.
Theo d d , dau tien cdn bien doi h§ ve h§
phUdng lrinh vi phdn cap m p t tUdng Qng blng each dat^=[?,?]'^. Khi d d , he phQdng trinh vi phdn cdp mdt nhgn dupc c d d g n g :
p^[q,g]'=H(p.t) (13) Trong dd, H(,p,r) - Vectd dai d i ^ n cho ve phdi cua 14 phQdng lrinh vi phdn c a p mpt.
Theo (13) ed the tha'y 7 thdnh phan dau cua H(.p. I) la?. 7 thanh phan c d n lai Id bleu thQc cila
? dQpc rut ra iQ he phQdng lrinh H- bdng each chia cdc ve phdi eho hd sd cCia gia ldc suy rdng tUdng Qng.
DQdi ddy trinh bay kel qud tinh todn trong hai trudng hdp lidu bieu ve kfch lhich. Gid trj cfla cdc thdng s d hinh hoc va d d n g lpc hpe dupc la'y lheo xe GAZ-66 [4]:
Me=2200 k g ; M^,=660 kg; M.,=580kg;
JBx=756kg.m2; Jg^=2750kg.m2; J =1780kg.ni':
Jc2=1170kg.m2; K^,=246000 N/m;K^=196000N/m;
C^,^1500kg/s; C =1500kg/s; K, =800000N/m;
K^=800000N/m; CL,=62000kg/s; C^=62000kg/s;
a,=1,563m; a =1,737m; 2b=1,8m; 2c=1,2m.
Trudng hgp 1: Xe ehuyen ddng vdi van tic khdng doi V lren mdt dudng c d bidn dang hinh sin (Hinh 4.1).
Bien d d h va budc sdng L cQa bidn dgng m$t dudng cho hai banh b e n trdi dupe ky hi#u Id (/L, Lj), cho hai bdnh bdn phdi Id (bp, L^).
PhUdng trinh md ta c h i l u eao map m l tgi dilm tie'p xue cCia bdn bdnh xe vdi mat p h I n g danh nghTa cua dudng (trge Ox tren Hinh 2.1) lrong trudng hop nay Id:
Hinh 4.1: Biin dgng mit dudng hinh sin Trong ede edng thQc (14), x = Vt la tpa dd theo phUdng xeilia khdi tdm S cilia than d Id tgi thdi dilm t Gia lrj cy the cilia cdc thdng s d v l bidn dgng m§t dudng dUde sQ d u n g de tinh todn Id:
58
K H O A H O C - C 6 N G N G H l
V = 20km/h: h^ = 15mm; h^ = 25mm;
Lp = 5,0m; L^ = 5.0m. _ . ...
C d c d i l u kidn ddu la: ? L =°- ^ L " °
^ Hinh 4.2 Id cac do thi
bieu d i d n ba t h d n h phan dao ddng t h i n g dQng eua t h d n xe {z„ = zJfj), eau trudc (z„, = z„,{()) v d c d u s a u ( z , ^ = z,3(0V
-So 9/2015
Hinh 4.2: Dao dgng thing dUng cda thin xe vi hai cau khi bien dgng mgt dudng hinh sin (z^
- nit liin, z^ - net dtft, z „ - cham chim) Cae do thj Hinh 4.2 cho thay, vdi d g n g kich thfeh dang x e i , ba thanh phan dao ddng thang dQng d giai dogn sau qud dp eung ed dang hinh sin va hodn todn phu hpp vdi ly t h u y l t . D o lhi md td eac thanh phan dao ddng gde (khdng trinh bdy d d d y ) e u n g ed dgng tUdng t p .
Trudng hgp 2-. Xe dang chgy vdi v a n t i c khdng d l i \ / t r d n d u d n g bang p h I n g thi d g n g phdi mdt m d dat trdn vdt banh xe ben phdi, ndi hai banh xe s d 1 v d s d 3 di qua. Sau d d mot khodng bdng d t h e o h u d n g c h u y e n d d n g , xe Igi d y n g phdi mdt m d dat khae tren vet bdnh xe ben irdi, ndi bdnh xe s l 2 v d banh xe s d 4 di qua.
S Q thay d l i chleu eao c u a hai md ddt theo phQdng ehuyen ddng dupe md Id theo ham sin vdi gid trj Idn nhat Id h^ va h^ , ehieu dai IUdng Qng cua hai m d dat theo phQdng ehuyen d d n g Id Lp va L^ (Hinh 4.3). Ngoai hai m d da't d d , mat dQdng Igi hoan t o a n b a n g p h a n g . Trong trudng hdp ridng, khi xe ehl d u n g phdi mpt m d dat tren mdt vet bdnh xe, c h i c d n lay h = 0 hoac h.^ = 0.
z
0
' '~~J? \
=#F1
Hinh 4.3: Bleu diin biin dfng va vi tri tUdng ddi cua hai md dat trin hai vit binh xe
C a c gia tri s 6 duijc sCf d g n g de tinh toan la:
l/=251<m/h, h|,= 3 5 m m , h , = 2 5 m m L = 65cm, 1-,= 80cm, d= 5m C a c d i e u kien d a u : s L - ' . s L " " , Hinh 4.4 la i i thi mo t4 si.1 thay doi theo thdi gian c d a hai thanh phSn dao dgng goc c u a than xe<p„=<p,(l).>r, = 'i'B(''-
Hinh 4.4: D6 thi md ta hai thinh phin dao dgng gde cua than xe ((pg - nit liin, "Vg - nit dtft)
Cdc dd thi eho thdy, dgng kich thieh dang xet cd dnh hudng Idn hdn ddi vdi thdnh phan dao ddng gde ngang xe, d i l u ndy Id phu hpp vdi thpc t l . Do cd mat cda cdc gidm chdn nen hai dao ddng gde d lrdn d l u Id dao ddng tai dan. 5 thanh phan dao ddng cdn lgi eug cd he cung dupe xdc djnh mdt each ddng thdi vdi hai Ihdnh phan dao dpng gde neu trdn vd cd t h i dupe bieu didn lheo cdch IUdng t u .
5. K e t l u a n
Bai bao da xdy dung md hinh dao ddng dgng khdng gian, 7 bgc tu do eua d td hai c d u , cd he thong treo phu thudc, ehiu kfch thfeh ddng hpc khi ehuyen ddng t h i n g trdn mdt dUdng vdi van tde khdng doi, da thie't ldp hd phUdng trinh vi phdn dao ddng cija cd hd vd ehl ra phQdng phdp gidi hd phQdng trinh vi phan dao ddng bang phQdng phdp s d nhdm tim ddp Qng dao ddng cua cd hd theo tat cd cdc bde tQ do. Bai bao cung md Id hai dgng kich ihfcb ddng hpe khd tieu bieu IhQdng dupe dp dyng lrong tfnh toan dao ddng cua d td, da xdy dung ChQdng lrinh tinh todn va gidi thieu m d i vdi ket qua tieu bleu d dang do thi nham minh hpa linh kha thi cua phUdng phdp. Vdi hd phUdng trinh vi phdn dao ddng nhgn dUpc, cd the xac djnh cae tan sd dao ddng ridng va cac dang dao ddng ridng cua d Id bdng phUdng phdp giai tich. ChUdng trinh Unh da xay dung dupc cho phep khao sdi ky ludng hdn dao ddng cua d id khi e h u y i n ddng trdn mat dudng khdng bang p h l n g . •
Tai lipu t h a m k h d o
[1]. Vu Cdng Ham, Tran Van Binh (2014), Ly thuyet dao dong, NXB. Quan dpi nhan dan, Hd Ndi.
[2]. Vij Cdng Ham, Tran Quang Dung (2007), Dao ddng CO hoc, Hoc vien Ky thuat Quan sU.
[3]. Vu DQc Lap (2011), Dao dong 6 td. NXB.
Quan doi nhan ddn, Ha Npi.
[4]. Vu DUc Lap (2004), Si tay tra ctfu tinh nang ky thuat d to, Hpc vien Ky thuat Quan sp.
[5]. CHEN Ke, ZHANG Ming, TONG Xuefeng, Vibration characteristic analysis of vehicle air suspension based on fuzzy control. The second International Conference on Electronic & Mechanical Engineering and Infonnation Technology (EMEIT-2012), pp. 2196- 2199, Published by Atlantis Press, Paris, France.
N g a y n h | n b a i : 20/6/2015 N g a y c h a p n h g n d a n g : 31/7/2015 N g u d i p h d n b i ^ n : T S . V u Minh DQc
PGS. TS. Vu DQc Ldp
