Phan tfch dam tira dan chju vat the chuyen dong xet den bien dang dan hoi cua goi tina
Dynamic analysis of a simple beam subjected to a moving load considering the elastic support
Ngay nhan bai: 24/01/2015 Ngay su:a bai: 15/3/2015 Ngay chap nhan (Jang: 25/04/2015
Do Kien Quoc, Pham D a m Sdn Tung
TOM TAT
Bai bao phan tich dam tua dem chiu vat th6 chuyen dong - mo hinh mpt khoi lucmg xet den bien dang dan h6i ciia goi ti^a.
Phuong ttinh dao dpng cua he thfing bao g6m tiim va vat thi chuyen dpng dupc thiSt l|.p theo phuang phap phan tii huu han va dupc giai bang phuong phap tich phan true tiep dua tren thuat toan cila Newmarlc, Ma tran khoi lupng, ma tran dp ciing cua toan he thong va vecto luc mit tuong duong dupc tinh toan a moi buoc thoi gian. Anh hudng ciia van toe chuyen dpng, dp vong ciia dam, dp ciing ciia goi tua... duac khao sat chi tiet.
Tir khoa: Phan tich dgng, Tai trpng di dong, Goi tua dan hoi.
ABSTRACT
This article analyses a simple beam subjected to an oscillator moving along the beam considering of elastic support.
Equations of motion of system composed ofthe beam stioicture itself and the moving oscillator are formulated by using FEM and solved by means of the direct step-by-step-integration method based on Newmark algorithm. The overall mass, stiffoess and equivalent nodal force vector are calculated at each time interval. The influences of the moving velocity, stiffness of spring in ocsillator, deflection of beam, elastic support.., are investigated.
Keywords: Dynamic Analysis, Moving Load, Elastic Support,
PGS.TS. D6 Kien Quoc
Khoa Ky Thuat Xay Dung, Truong Dai Hpc Bach Khoa - Dai Hpc Qu6c Gia Tp.HCM
Email: dkquoc2004(gyahoo.com Dien thoai: 0903683422 ThS. Pham Dam Son Tung Khu QLGTDT s6 1 - So GTVT TP.HCM Email: [email protected] Dien thoai: 0938211535
I.Gidlthl^u
Viec nghien cu'u dao dong cCia ket cau chiu tai trong di dfing xuat phat tLf yeu can thUc tien, thuang gap trong cac linh viic nhU: xay du'ng, giao thong van tai, cO i<hf. . ThUc te chi ra rang, anh hu'Sng dong d o ti'nh chat di chuyen ciia t^i trong d^n phSn ung cua ket du la dang ke, phSt smh nhifng hien tu'Ong khong mong mudn nhU lam tang cac gia t n c h u y i n vj - n6i Iu'e, nguy hiem nhat la xuat hiin hien tu'Ong cong hUdng.
Ben canh 66, hi ket cau con dUOc dat tren he nin, mong cpc va lam vi^c dong thcfi vdii k^t du khi chiu t5i trong di dong. Hi thong goi tUa nay da du'dc de cap trong nhCtng nghl&n cClu ve phan u'ng dflng cCia dam v6i nhQng dieu kien bien khac nhau (giSn dOn, ngam) [1] [2] [3], Tuy nhien, nhCing nghi&n eilu ve 3nh hu'dng cCia sif dan hoi gdi tifa den ph5n ilng ddng cda d^m v3n cdn rat ft.
Vl vSy, vifc nghien cu'u phan u'ng dong ciia dam dudi tac ddng vat the dl ddng cd xet den bien dang cua gdi tUa de danh gia 3nh hu'dng cua mot thdng sd k f t cau va vat the di ddng den phan Ong cua toan hi la van 6i cd y nghTa thUc tien trong khoa hoc va dac biet la trong nganh giao thflng van t^i.
2. Co sfr 1]^ thuyet 2.1 Mid hinh ket cdu
Xet m d hinh vat the dl chuyen doc theo chieu dai dam L(m) vdi van tdc V(m/s) khdng ddi the hi&n 6 Hinh 1. Vat the dUOc mo hinh bdi he hai bac tif do g o m khdi lu'ong b3n than m ndi vdi banh xe khdi lUong m,, qua he gi^m chan Id xo vdi dd cUng Ko Trong thu'c ti, khdi luong banh xe m^ khong dang ke nhUng lam khdi lUcfng t i n h toSn rat nhieu nen trong bai bao nay bo qua m^. Ky hieu y la chuyen vi tuyet doi cua khdi luong m theo phu'ong thang dCtng, wo la chuyen vi theo phuong t h i n g diJng cua khdi luong m,. Gia thiet be mat dam bang phSng va ludn tiep xuc vdi banh xe rriv trong qua trinh di chuyen. Do d d , c h u y i n vi thang difng ciia banh xe (ivo) ludn bang dd vong ciia dam (iv) tai diem tiep xiic Vl t r i tong the cua vat the khi di chuyen tren dam la Xmft). Dam dUpc m d hinh la dam Euler-Bernouli tiet dien dong nhat khong doi. Oi xet den b i l n dang ciia gdi tUa, dam difoc tUa tren hai 16 xo 6 hai 5$u dam. Bo qua tfnh chSt dn eiia k i t cau.
Xm(t) ^
Hinhl Mo hinh ket
1
2.2 Md Mnh phan tCrhHru hgn
Dam du'oc chia thSnh n p h i n tCf dam cd chieu dai U va ndi vdi nhau tai dc nut n h u trin hinh 2. Vi tri cOa vat the so vdi dau ben trai phan ti!r damthCf"s"ISx('t)
lUt)
N| = 3 £ ^ - 2 t ^ , N | = ljl-5^ + e ' ) va 4 = — . K h i d d . p h J o n g trinh (1) v^ (2) duoc Viet gon lai la
M ' u ' + K ' u " = F ' ( t ) (3) trong do:
, , . _ [ M OIN'' ^ . r K O""" . , . mg\'''|
Hinh 2, MD hinh phan tiihijii han
Q5ni
phin doa a^n I
BA >un|| phin ma w^t »«o •>» Irtn
..u X i , JunB.«v.<,ia,
1
cmotaoii
Co the thay rang cac ma tran M', K" d (3) la cac ma tran theo thdi gian v) chiing bao gom thanh phSn khong thay ddi cCia bcin than k i t cau va thanh phan phu thudc theo t h d i gian do vat thi chuyen dong tren d^m tao ra. SU thay doi d | c t i n h ciHa he chm tSI trpng di dpng nay the hi^n tinh chat phi tuyen cua ket cau [5].
PhUdng trinh (3) la phuong trinh chu dao md tci dao ddng cua he thdng gom ket cau dam vh vSt the chuyen ddng tren nd va dUpc gioii bSng phuong phap tfch phan true t i l p dua trin thuat toan Newmark.
T h u a t t o 5 n g i 3 i nhusau
Mot chuong trinh may tfnh dupc lap trinh bSng ngon ngij' Matiab dupc v i l t theo luu do dUdc dung d l khSo sat vi du s6.
3. Ket qui so
Bai bao khao sat m d hinh dam gicin ddn cd chieu ddi L = 20m, mat dt m^t ngang cCia dam I - 0.048 m'. Vat lieu c^a dam cd cac dac trUng sau' mddun dan hoi 5 = 2 x ) 0 " N/m^, khdi luong t r ^ n mdt don vj chilu dai p = 1.5x IC kg/m. Dam chju vSt t h i chuyen dong c6 khdi lUdng m-=
SOtdnvi do cUng ciia l d x o / < ( j = 3 x T 0 ' N / m . Hai dau dam dUdc tUa tren hai 16 xo vdi 03 loai dfl cCfng dCing d l khSo s^t \h K n i = 1 0 0 ^ , Kn2 = 300-r- va K ^ j ^ S O O — . Ciing vdi so do dam tUa tren gdi cUng, bdn so do dam dUpc sijtdung de khdo sat
3.1 Khdo sdt hi sddgng chuyin vi tgi giQa nhip
Hinh 4 va Hinh 5 the hien dnh hUdng ciia van tdc ddi vdi he sd dong c h u y i n vi tai giifa nhip [DAFcv] d so do gdi ciJng vh sa 66 gdi dan hdi (K„=K^,) khi vat the c h u y i n ddng tren d i m . Cdc k i t qud cho thdy DAFcv d so dd goi dan hdi Idn hOn 1.2 d i n 1.3 lan so vdi DAFcv d sd do gdi cifng. K i t qud dUpc trinh bay d Bdng 1.
Hlnh 3: Thuat toan lap tiinh
Phuang trinh dao ddng cua ddm cd dang nhusau
M u + K u = N ^ P (1) vdi M, K la ma tran do cUng va ma tran khoi lUong cOa ddm dUoc
t h i l t Idp bdng cdch "chdng chat" cac ma trdn phan tif 14], P la ngoai lUc tdc dyng vdo dam, u vd « Id vectd chuyen vi, gia tdc ciia k i t cau ddm.
Phuong trinh dao ddng cua v§t t h i chuyen ddng Id
m y + K „ ( y - w ) - 0 (2) Lyc tuong tac glCTa vat the va dam duoc the hien bdi
P = m g - m y Khi vat the c h u y i n dong tren phan tif thU *s* tdi trong P dupc quy ddi ve cdc tai trong nut. Khi dd, vecto ngoai life la
F(t)=|o 0 0 0 ... f;(t) t;(t) mt) r.it) ._ 0 0 0 0}'^
trong dd, f f ( t ) , f | ( t ) , f | ( t ) , f^W Id cac thdnh phan lUc tac dung ISn hai nOtcila phan t i l ' s " . Vecto ham dang Ntai phan tif thif "s" la
> = {o 0 0 0 ... Nj N j N | Nl ... 0 0 0 oi Vdi Nf = l - B t ^ + 2 t ^ N | = l e { ^ 2 t ^ + ^ 3 ) ,
Hinh 4. He so dong chuyen vi tai giQa nhjp - so do goi ciftig
128 E
Hinh 5: Hk i6 dpng chuyen vi tai giifa i Mng 1: DAFcv Ifln nhal
- KJdo goi flan hdi vdi Kn=Kni
Gdi cAiq (a) Goi dJn hdi (K,=Kn,)[b) (b/a)
m/s V=20 1.2060 1.6064 1,332
m/
V=40 1,3995 1.8834 1.346
m/
V=60 16206 2 0430 1261
V=80 m/s
1.6668 19798 1.188 Hinh 6 vd Bdng 2 trinh bay d n h h u d n g cCia h^ sd Kn d d i v d i DAFcv k h i vat the chuyen d d n g tr&n d i m v d i V=60(m/s). Khi Kn c a n g Idn, s d d o g d i dSn hdt t i l n v l so 66 gdi citng, DAFcv g l d m d a n .
Hinh 6: Ht sd ddng chuyen vi giiia nhip vd Mng 2: DAfcv liin nhfl vfli V=60(m/sl
Kinh 8 He 50 dong momen tai giQa nhip-SCI fld goi dan hoi Kf^ICi 3.2 Khao sdt he so dgng momen tgi giifa nhip
Hinh 7 vd Hinh 8 the hien anh hiidng cua van tdc ddi vdi he sd ddng momen tai giCfa nhip IDAFmomen) 6 sd dd gdi ciJng va so dd gdi ddn hdi (Kn=Kn,) khi vat the chuyen ddng tren dam. Ket qud cho thay DAFmomen d so dd gdi dan hdi Idn hon khodng 12 l l n so vdi DAFcv d sd d d gdi cdng K i t qud dUdc trinh bay 6 Bang 3.
Bang 3 DAFmomen ldn nhat Gdi cifng (a) GOI dan hdi (K„=K„,) (b) (b)/(a)
V=20 m/s 1 1382 1.4183 1.246
V=40 m/s 1.3227 1.6407 1.24
V=60 m/s 13522 1 6114 1 192
V=80 m/s 1.3260 1.4817 1.117 Hinh 9 va Bang 4 trinh bay dnh hUdng cila h^ sd K„ 66i vdi DAFmomen khi vat the chuyen ddng tren dam vdi V=60(m/s}. Khi Kn cang Idn, so do gdi dan hdi t i l n ve so do gdi cifng, DAFmomen gidm dan, Od thi DAFmomen ciia sO dd gdi cilng tron hon nhieu so vdi sd dd gdi dan hoi.
Hinh 9: He sd dong momen tai gifla nhip vfli V=60(ni/5) Bdng 4: DAFmomen Idn nhat khi V=60(m/s)
Kni Kni Gdi Cling
DAFmomen 1.6114 1.6115 1.518- 1.3522
3.3 Khdo sdt luc cdt tai giOa nhip . ., * Hinh 10 va Hinh 11 the hien anh hUdng cOa vdn toe doi vdi lyc cdt tai giite nhip d so d d gdi cdng vd scf do gdi dan hoi (fC-/CiJ khi vat the chuyen dong tren dam. K i t qud eho tha'y lUc cSt tai giQa^nhip 6 so do goi dan hdi Idn hdn khoang 2.5 Idnsovdi l u c c d t d s d d o g o i cung vd ton tai m d t van tdc tdi han lam cho lUC cat nay la nhd nhat Ket qud dUdc trinh bay d Bang 5.
Unh 7:Hls6d6ng momen tai glOa nhip - sodSgSialng
12S
3.4 KhAo sdt chuyin vi ldn nhdt cua ddm theo KB vd V
Hinh 10 Luc ck tai gifla nhip - so do gdi cfltig
Hinh 11 Life c3t tai gifla nhip - so dd gdi dan hoi K,:=K,i B i n g 5: Lflc cat Idn nhat
Gdi cuHg (a) G d i d a n h 6 i ( K , = K , i ) ( b ) (b)/(a)
V = 2 0 m / s 154.1410 456,2151 2,96
V = 4 0 m / s 169,8641 4146817 2.441
V - 6 0 m / s 178.i091 331.8422 1861
V = 8 0 m / s 1642035 419 4825 2 555 Hinh 12 vd Bdng 6 trinh bay anh hudng cua hesd Kn ddi vdi lUccati giCfa nhip khi vdt the chuyen ddng tren dam vdi V=60(m/s). Khi Kn cang Idn, sd d d gdi ddn hdi t i l n ve so do gdi cdng, lUc cat nay giam dan. LUc cdt d giCa nhjp cCia so dd gdi dan hdi bien thien rat nhieu so vdi so dd g d i c d n g . T d n t a i m d t g i d tn/fncrlamcholUccatnaylaldnnhat.
Hinh 12: LiiccSt tai gifla nhip vfli V=60(m/s) Bang 6: Uc cSt Ifln nhat khi V=60(m/s)
Km KnJ Lflccat 331.8422 406.4857 K.1 403.0738
Gdi cuHq 1783091 Hinh 13 cho thay chuyen vi Idn nhat ciia dam ftV„ J theo he sd {KJ va van tdc c h u y i n ddng ciia vat the, W,„^ rat nhay khi van tdc chuyen ddng cilia vdt the dudi 30(m/s). Khi Kn cang Idn, so dd gdi dan hdi tien v l Sd dd gdi cdng, Wma< gidm dan.
Hinh 13: Chuyen vi Ifln nhat cua dam theo K, va V 3.5 Khdo sdt chuyen vi ldn nhdt cda ddm theo Kn vd Ko
Hinh 14- Cliuyen vi Idn nhat cua dam theo Ko va Kn
Hlnh 14 cho thdy chuyen vi Idn nhat ciia dam (Wn,<^.} theo h^ sd {Kn) va dd eUng cua nhip xe (KJ, Khi K„ cdng Idn, so dd gdi dan hdi tien ve so do gdi cifng, (WmJ giam dan. Vdi moi Kn chon trUdc, tdn tai mdt gia trj tdl han K ^ lam cho Wmai dat gia t n nhd nhat.
4. Ket luan
Bai bao trinh bay k i t qud nghi&n cUu ca bdn v l phdn Ung ddng ciia ddm tua ddn chiu vat the chuyen ddng - md hinh m d t khdi luong xet den bien dang dan hdi ciia gdi tUa. Phuong trinh dao ddng ciia toan hd thdng cung duac t h i l t ldp va gidi bdng phuong phdp tfch phdn true tiep dUa tren thudt toan Newmark de tien hanh khao sat mdt sd vi d u .
Viic xet den dnh hudng ciia bien dan hoi cUa gdi tUa da lam tang phdn ifng ciia dam, momen vd luc cat trong dam. Khdng chi tang v l do Idn ma cdn tang su b i l n thien ciia cac gia tri nay khi chju vat the c h u y i n ddng. S o eilng cua nhip xe [Ka] ddng vai t r d n h u mdt t h i l t bj gidm xdc hap t h u nang luong tU phdn Ung ciia ddm.Tdn tai mdt gia tri tdi han ciSa h& sd nen (/Cin) lam gid t n l u c cat tai giij'a nhjp la Idn nhat.
Tai lieu tham khao
[11 M AbuHilal and H.S.Zibdeh, "Vibration Analysis of Beams vjith General Boundaiy Conditions Traversed by a Moving Force," Sound and Vibration, vol. 229, no 2, pp. 377-388,2O0O.
[2] M Abu-Hilal and MMohsen,"Vibrationsof Beams with General Boundary Conditions Due to a Moving Harmonic Load," Soundond Vibration, voi 111, no 4, pp, 703-717,2000.
[3] Zibdeh H.S., and Hackwitz R., "Moving Loadson Beams Witti General Boundary Conditions,"
Jou/irfl'/6raflon,vol,195,no 1,pp.85-102,1996
[4] 05 Kien Quoc, Luong Van Hai, f)6ng lilc hoc ket cdu. HCM NXB flai hoc Quoc gia TPHCM. 2010 [5] i)o'iiinQwc,Hqi}yeplmng?hil6c,Cdcphiiangphdps6trongd6ngliichocketciu HCM:NXB
fiai hoc Qudc gia TPHCM. 2010