DIEN DAN KHOA HOC CONG NGH^
PHl/dNG PHAP NON CHOP TONG QUAT filNH Ll/ONG BIEN SUAT TRONG DAT N I N
I. PHUONG PHAP TINH
Phuong phdp ndn chdp Id mdt trong nhOng phuong phdp gidi tich cdn cd the gpi la phuong phdp ndn chdp gidi tich, dung eho viee djnh lupng dng suit tdi trpng thang ddng tdcdng thnh truyin xuIng ddt nIn. Ngodi nhdng nhdn t l chi phli khde, blnh dang d l mdng cpng trinh cung Id mpt nhdn t l cd dnh hudng d i n su phdn phli dng suit tdi trpng trong dlt nIn. Trudng hpp mdng hinh trdn hoac da gide d i u ed the tinh nhu mpt mdng trdn tuong duong thudng ed the sd dung chung mpt Idi gidi hay mpt cdng thdc tinh todn dng suit.
Cdn cdc trudng hpp cd hinh dang khdng trdn nhu mdng hinh vudng, blnh chd nhdt, hinh thang, da gide bit ky,... thudng thilt lap cdc cdng thdc rieng, cu the cho tdng trudng hpp nhdm dat dupe nhdng cdng thdc tien dung hen, phii hpp vdi timg hinh dang d l mdng cy the. Do dd viec djnh dang cdc viing dng suit vd biln dang cung vdn cdn thue hien cu the vdi tdng hinh dang d l mdng vl mil hinh dang d l mdng khde nhau cd nhdng ddc trung rieng khdng gilng nhau. Tuy vay, cung vdn cd the thilt lap mpt cdng thdc tinh phdn phli dng suit chung cho t i t cd cdc hinh dang d l mdng vita sd dung cho d l mdng phang co blnh dang bit ky, vCra Idm CP sd d l suy ra cdng thdc tinh vdi cdc hinh dang d l mdng djnh hinh quen thupc nhu hinh trdn, hinh vudng, blnh chd nhdt, da gide diu, nen cdn gpi Id phueng phdp ndn chdp tong qudt. Ngodi ra, biln dang vd dng suit cd the gpi ngdn gpn Id biln suit.
II. VUNG BIEN SUATTRONG DAT N I N
Trong gidi ban biln dang ddn hdi tuyIn tinh, quan h$ gida dng suit vd biln dang biln thien theo djnh luat Hooke. l\/lpt s l chi tieu ddc trung cP ly cua d l t nIn dd xdc djnh theo thuc nghiem dupc coi Id khdng thay doi hay bit biln trong qud trinh dd vd trong todn bp VLing dlt nIn cdng trinh. Trong dd he s l Poat - xdng (Poisson) V cdn gpi Id he s l nd hdng hay he s l biln dang ngang cung dupc coi Id mpt bit biln:
V = ex / EZ = ey/ez
Suy ra ed V=XEX/X8Z = yey/ysz = ZEX/ZEZ = zsy/zsz He thdc diln xult dupc coi nhu t i n tai trong d l t nIn hoac tdng Idp d l t trong nIn cdng trinh, cd tinh ddng nhlt vd dang hudng dudi cdc loai hinh dang d l mdng cdng trinh nhu mdng trdn hoac da gide d i u , mdng vudng liodc mdng chd nhdt vd da gide bit ky.
Ddi vdi d l t rdi, vung biln suit se trung nhau.
Nhung trong nIn ddt dinh ed lue dinh c <> 0 thi vung .
DOAN DINH THOA Trung T a m G r o u p f o r m
dng suit Idn hon vd bao phu vung biln dang theo chilu true thang ddng OZ hudng xuIng theo dp sau nhu dd neu trong khi dng dung d l i vdi nIn mdng tron bode mdng da gide d i u . Ddi vdi mdng bode chan coc ed dang hinh vudng thi tinh todn tuong tu mdng chi?
nhdt chi viec thay canh a = b trong cdc cdng thilc dung cho mdng chd nhdt. D l i vdi mdng dang bit ky thi sd dung dang cdng thdc tong qudt de ti'nh toan.
Vdi tdi trpng thang ddng tren eac d l mdng kich cS nhd cd dien tich d l mdng Fd nhd hon bode bdng di?n tich bdn nen FB, ed the coi nhu trudng hpp tai trong tap trung, thi vung biln dang hoac vung dng suit cb dang blnh ndn trdn xoay. Thue t l tinh todn thudng vln phdi ke d i n kich thude d l mdng thue t l cd kich thude hdu ban, ed bdn kinh r vdi blnh trdn, hoac canh a v^
b vdi cdc hinh dang chd nhdt hoac vudng a = b,...
Cling vdi su khde nhau gida cdc canh, nen pham vi ViJng biln dang vd vCing Ung suit cd cdc dang hinh ndn cut hoac blnh chdp cut, cd the gpi chung la hinh ndn chdp cut hoac ndn chdp hai day.
1. Vung biln dang
a. Coi ring he s l nd ngang Poisson khdng dli v = const vd cdc he thdc biln dang Cauchy, cd ty s l X/Z tuong dng vdi he s l luong gide eua gdc Id X/Z = tga vd:
X/Z = tga = 1+eXtt, = 1+eXtt, [1]
Uo = eXgj.RO = veZttjRo [2]
Trong dd: eZa, = p .p/E, vdi he s l p =(1 -2v^) [3]
Hoac chpn xdp xl Htb = p/(2y) va So = eZtt,.Htb [4]
=>Ztt, = So/Hp = 2.YSO/P [5]
b. Ddi vdi hinh chd nhdt cd kich thude d l mong canh Id a theo chilu true tpa dp OY, biln d?ing Uo theo a Id Ua vd canh thd hai gpi Id canh b theo chldu true tpa dp OX, biln dang dinh theo b Id Ub, thi vung dnh hudng tai sdt ddy mdng ciia viing biln dang v^
vung dng suit theo cdc chilu kich thude se khdng gilng nhau U a o U b .
c. Biln dang theo chilu ede canh mdng chd nhdt a vd b:
Theo canh a ed: Ua= Uoy = EXtt,.a/2 = v.eZa,.a/2;
Gdc md rdng tai dinh ndn chdp cut theo canh a tinh theo cdng thdc
=> tgaa = 1+ey«,= 1+ vez^,
Theo canh b cd: Ub= Uox = eXtt,.b/2 = veZa,.b/2;
Gdc dinh chdp cyt theo canh b tinh theo cdng thiJc: i
=> tgab = 1 + eXtt,= 1 +veZft, ^
NGUdi XAY Dl/NG S6 THANG 9 2011
PHi/ONG PHAP NON
C H 6 PTONG QUAT
=> tga = tgaa = tgab
=> Az = a+2.Ua+2.Z.tga [6]
=>Bz = b+2.Ub+2.Z.tga [7]
d. Vdi suy ludn nhu vay se xdc djnh phuong trinh viing biln dang nhu sau:
Rz = ±[R1+Z.(1+vez] [8]
Vdi R1 = (Ro + Uo ); Ro = r hoac Ro = rtd bode Ro = a, Ro = b,... phu thupc hinh dang thuc t l cua d l mong tinh todn. Vung biln dang blnh ndn chdp hai ddy dli xdng qua true OZ hudng xuIng sdu trong dlt nIn qua tdm hinh dang mdng xung quanh glc tpa dp dat tai mat nIn sat ddy mdng.
2. Vung ling s u i t
a. Tdi trpng thang ddng tong cpng dd gdm trpng lupng bdn thdn d l mdng Id Np, dien tich d l mdng Fd.
b. o l t nIn cat hoac d l t rdi thi viing biln dang cung la vung dng suit, trijng nh§ku.
c. Dlt nIn Id loai d l t dinh thi vung dng suit bao trijm ngodi viing biln dang, Idn hon mpt chut vdi ty s l biln dang hay gdc md rpng td d l mdng xuIng phi'a sdu tinh theo ede he thdc Cauchy, cd dang:
X/Z = tga*= (x+lJx+Ax)/z =(z+Ux+Ax)/z =
= (1+Ux/z) + Ax/z
tga*= (1+exa,)+ Ax /z = tga + Ax /z
tga"= tga + Ax /z [9]
Nhu vay cd phuong trinh viing gidi han dng suit cd dang sau:
Rz = ± [R1+Z.(1+VEZ) + Ax)] [10]
Hinh dang viing biln dang cung nhu vung dng suit cd dang ndn chdp hai day, d l i xdng qua true OZ.
Nlu nIn cdt hay nIn d l t rdi, thi viing biln dang triing vdi viing dng suit. Cd the eho ring chpn xdp xl gin dung theo each gidi ndo thi phuong trinh viing dng suit cung co dang:
Rz = +(R1+Z.tga*)= ±[R1+Z.(1+vez+Ax] [11]
Phia chilu duong true hodnh thi phuong trinh cd dang:
Rz = R1+Z.tga" }
=>Rz = R1+Z.(tga +A x /Z) = Rl+Z.tga+Ax } [12]
- Phia chilu duong true hodnh thi phuong trinh eo dang:
Rz = -(R1+Z.tga*) }
=>Rz =-[R1+Z.( tga +A X /Z)] = -[R1+Z.tg a+Ax]}
Trj s l X cd the tinh theo mpt s l cdeh tinh todn gin dung. Nlu sd dung cdeh gdn dung dua theo he thdc Coulomb thi se cd:
tga*= tga + Ax /z = tga + tgcp [13]
cp Id gdc ma sdt trong ciJa d l t nIn.
=>Rz = ±(R1+Z.tga*) = ±[R1+Z.(tga+tgcp)] [14]
Cdeh tinh ndy don gidn hon, khdng phdi tinh gdn dung ddn, sai s l kit qua tinh (tng suit thien Idn, cd the tin cdy dupc, dd dupe sddung thd nghiem khd phu hpp.
3. Vung ddt nen chat di/di de mong
a. Viing biln dang ngay sdt d l mdng theo hai phia c6 biln dang tuong duong Uo. Di mdng thue cd kich thude tUPng dng Id Ro, d l mdng ed ke d i n vijng biln dang ngang Uo Id R1 = (Ro + Uo ). Khi dd dng suit
thing ddng Idn nhlt trung blnh dudi ddy mdng ed dang:
Pd = Np/Bd [15]
-l^dngtrdn: Bd = 7r.(R1^+Ro^+R1.Ro)/3 [16]
- Cdc dang hinh ddng khde:
Bd = [Bl +Bo+(Bo.B1)"]/3 [17]
Ky hieu Bo vd Bl Id dien tich d l mdng tuong dng vdi Ro vd R1.
b. Dua theo kit qud thuc nghiem, cho ring cd the xult hien viing dlt nIn bj nen chat ngay dudi d l mdng. Su phdn phli tdi trpng trong viing nen chat phdt sinh ndy tuong d l i kho xdc djnh, ma kham phd theo CO sd ly thuylt lai cdng phdc tap hon. Viing nem ddt nen chat sat d l mdng hay dudi sat chan cpc nlu de mdng hoac chan d l cpc phang ndm ngang co the xdc djnh bdi ty s l biln dang ngang va dung hoac nda gdc dinh p* nhin td dinh nem ngupe dudi day mong theo dang:
Ro/Zn = tgp*= Ro.(l-ex)/(1-EZ)/Z;
=>Ro/Zn = tgp*= [RO.(I-EX)] / [(1-EZ).RO]=
= (1-extb)/(1-eztb) [18]
=>Zn = RO / tgp* = Ro.(l-ez) / (1-ex) =
=Ro.(1 - eztb) /(I - veztb) } [19]
Hoac chpn g i n dung
Zn= Ro.(l-EZ) =Ro.(1-.5z/E) } Cd the chpn gin dung de tinh Zn theo cdng thdc ddn xult td cong thdc tren:
Trong trudng hpp don gidn nhlt coi nem nen chat Idn nhlt cd dang hinh ndn ngupe md d l mdng cong trinh la ddy, chilu cao la Zn = Znmax Idn nhdt, nIn biln dang ddn hIi tai gidi han tuyIn tinh va phi tuyln.
Khi dd cd the chpn gdn diing de tinh Znmax theo cong thdc d i n xult tdcdng thdc tren:
• => Zn = Znmax= Ro.(1 -p.p/E) } [20]
Hoac Znmax = Ro.(1 -ez) = Ro.(1 -So/Htb)} [21 ]
Vdi Htb = p/y/2 . } He s l p = (1-2v^); ^ - He s l dp luc ngang.
c. Trong cdc trudng hpp khdng c l djnh nem dlt, ty s l biln dang hay gdc dinh thay doi thi nem co dang phi tuyln rlt phdc tap, nhilu biln dpng, ehl co xu hudng blnh ndn chop biln dpng xung quanh gid trj trung blnh.
III. PHAN P H 6 | l/NG SUAT
Trong viing nem dlt nen chat, dng suit tdi trpng thang ddng phdn phli theo dang hinh ndn hoac chop hai ddy Tai Zn cd su tdi phdn b l dng suit chuyen td dang phdn b l hinh ndn chdp cut sang dang hinh ndn hoac chop mpt ddy mpt dinh tai tit cd cdc dp sdu Idn hon Zn. ling suit Idn nhlt zmax tai mdi dp sdu Z ndm tren true OZ vd dng suit nhd nhlt azmin = 0 tai giao tuyln cua ddy ndn hoac chdp dng suit vdi mat phdng ndm ngang Zx = Z = Const. Theo cdc dudng sinh tren b l mat xung quanh cdc ndn chdp iing suit coi nhu dng suit phdn b l tuyln tinh theo cdc hudng ngang, tren dinh cd azmax, phia dudi cd azmin = 0.
1. Trong vung nem d l t nen chat
Tai cdc viing sdt d l mdng trong vung xult hien
NGUdi
XAY DUNG S 6 THANG9-2011
PHUONG PHAP
N 6 N C H 6 P T 6 N GQUAT..
nem ddt nen chat ed dp sdu nhd hon dinh ndn nem nen chat Z < Zn, cdng thdc tinh todn dng suit dudi dang tong qudt theo nem nen tuyln tinh cho ode loai hinh dang d l mdng cd dang nhu sau:
az max = 3.T<lp l[Fz + Fn + yfFz^Fn)] [22]
Np = Pd.Bd hay Pd = Np/Bd nhu ede cdng thdc [15], [16], [17] da d l cap.
a.Vdi mdng bit ky thi tinh theo cdng thdc tdng qudt, cdn cd the tdeh rieng ede loai d l mdng cu the gdm mong trdn bode tuong duong mdng trdn, mdng vudng, mdng chd nhdt, cd dang cdng thite eu the hon.
b. Mdng chd nhdt:
Kich thude d l mdng Fd = axb. Canh a vd canh b Id kich thude eac canh hinh chd nhdt d l mdng. Ki'ch thude ddy Idn chop dng suit Fz = Az.Bz.
Vdi Az = (Ao+2.Z.tga*) ; Bz = (Bo + 2.Z.tga*) [23]
Ddy nhd chdp ung suit viing nem nen Fn = A.B Trong do A= (a - 2.Z.tgP) ; B = (b - 2.Z.tgP) [24]
Ao = a + 2.Ua = a.(1+V£z) = a.(1+V£Ztb) } [25]
Bo = b + 2.Ub = b.(1+V£z) =b.(1+VEZtb) } Thay ede gid trj tinh todn hinh dang vao cdng thdc tinh toan tong quat [22], bode sd dung cdng thdc eo dang rieng cu the hon nOra:
ozmax = 3.Np/[A.B + Az.Bz + (A.B.Az.Bz)''^ [26]
c. 0§ mdng vudng:
D& mdng vudng tinh nhu d l mdng chd nhdt cd hai canh bdng nhau a = b.
d. o l mdng hinh trdn:
Viing nem dlt nen chat sdt d l mdng cd dang ndn trdn xoay ngupe dinh dudi vd ddy phia tren sdt d l mong, kieu hinh phlu. Phan phli dng suit trong dlt nIn dudi d l mdng trdn ed dang hinh ndn cut hay ndn hai day trong viing nem dlt chju nen vd cd hinh ndn trdn xoay dudi viing nem nen.
LTng suit Idn nhlt tai cac dp sdu tinh todn Z < Zn dudi mdng trdn hoac tuong duong trdn dupe tinh theo cdng thdc tong qudt [22] hoac theo cdng thdc rieng:
azmax = 3.Np/[7t(R2z+ R2n + Rz.Rn)] [27]
Trong dd:
Rz = R1 + Z.tga*= R1 + Z.(tga + Dx /z ) } [28]
R1 = (Ro + Uo ); Rn = (Ro - Z.tgp*) } Cdc s l hang Ro, Ax, Uo ti'nh nhu da neu tren.
Cdc mdng dang da gide d i u thi cd t h i coi nhu mdng trdn cd ban kinh tuong duong Ro = rtd =
yJF I n . vd tinh todn cdng thdc tuong tu mdng trdn.
f> Cdc trudng hpp sd dung nem dlt biln thien thi viee thilt lap cdng thdc vd tinh todn phdc tap hon trudng hpp e l djnh viing nem ddt nen chat. Khi nghien cdu su biln thien dd thi sd dung ede cdng thdc td [18]
d i n [28]. Cd the gap nhilu b i t thudng dj nghiem khdng ludng trude dupe hit. VI vay cdn nghien cdu nhilu hon nda hoac loai bd ban nhdng kit qua b i t thudng vd chl dung nem d l t nen chat dang e l djnh de gidi bdi todn don gidn hon vd ludn ed nghiem chde ehdn nhu dd trinh bdy d tren.
2. Tai cac vung c6 do sau Idn hern
Tai cdc vung ddt nIn dudi vung nen chat ed dd sdu idn hon dinh nem ndn chop ngupe Z > Zn, urng suit Idn nhlt trong d l t nIn cd the tinh theo cdng thdc tong qudt vdi cdc loai hinh dang de mdng:
CTzmax = 3.Np/Fz [29]
a> Cdc dang d l mdng b i t ky ed dien tich d l mong Fd, tai dp sdu Z cd dien tich day chop dng suit Id Fz thi tinh theo cdng thdc tong qudt.
b> o l mdng chd nhdt hoac vudng tinh theo cong thdc tong qudt hoac eu the hon nda vdi cdng thifc
rieng ed ke d i n cdc canh d l mdng theo dang sau:
azmax = 3.Np/(Az.Bz) [30]
c> o l mdng trdn ed ke d i n ddc trung cu the theo dang:
CTzmax = 3.Np/(7i.Rz^) [31]
d> Cdc mdng dang da gide d i u thi cung c6 thi diing dang mdng trdn tudng duong cd ban ki'nh tuong duong Ro = rtd = 4FIn , vd tinh todn tuong tir mdng trdn.
IV. VAN KET
Phuong phdp ndn chop thilt lap md hinh phan phli dng suit do tdi trpng thang ddng td cdng trinh truyin xuIng d l t n I n cd vung biln dang va dng suit dang hinh ndn vdi tai trpng tap trung hay vdi dl mdng trdn, vd blnh chop vdi cdc dang d l mong khac. De tien dung ddt ten Id md blnh ndn chop hoac phupng phdp ndn chdp. Mpt cdeh tong qudt c5 thi thilt lap dupc cdng thdc ti'nh dng s u i t Idn nhlt omax tai mdi mat cdt ngang ed dp sdu Z tfnh todn, ndm tren true OZ cd chilu duong hudng xuIng dudi va di qua trpng tdm blnh hpc eua dk mdng cdng trinh. ling suit nhd nhlt Tzmin tai mdi mat cdt ngang cd dp sau Z tinh todn nam tai dudng bien giao tuyln giufa mat phdng ndm ngang vd mat ndn chop gidi ban hay dudng sinh cua ndn chop vdi gid trj triet tieu amin = 0,0. Phan phli dng suit tai mdi mat cdt ngang tai dp sdu Z tinh todn trong vung nem d l t nen chat ngay dudi d l mdng Z<Zn cd dang ndn chop cut hay hai ddy Idn nhd, vd cd dang ndn chdp tai cdc dp sau dudi nem Z>Zn. Trj s l dng suit Idn nhdt omax tai dinh ndn chop cat true OZ, tinh theo cdng thdc ting qudt hay cdng thdc rieng ciJa mdi blnh dang de mdng tai mdi dp sdu tinh toan Z. Trong qua trinh tinh todn thd nghiem cho t h l y k i t qua dng suit vd phdn phli dng suit tuong d l i hpp ly, vd phuong phdp ndn chdp cung Id mpt phuong phap gidi tich dai s l kha chat che. Vide gidi bdi todn biln dang lun cung da dung Idi gidi giai tich rdt thuan ti$n. Qud trinli xdy dung phuong phdp giai tich dd thilt lap md thilt l$p ndn chdp de giai quylt bdi todn dng suit vd biln dang do tdi trpng thing ddng gay ra trqpg gidi h^n ddn hdi tuyln tinh, cung da vd dang dupe nghidn cdu phdt trien phuong phdp trong cdc tr^ng thdi gi^
han khde cung nhu cac dang khde nhau cua nSn mdng cdng trinh.Q
