Giáo trình: Phân tích ứng suất Lý Trường Thành

I( MaaAu Phdn tich Ung sudt 13 mQt mon khoa hoc thuQc nganh co hoc v~t rin bi~n dang N6 nghien ciru trang thai irng su~t, bi~n dang cua v~t th~ can bfulg du&i tac dung cua cac tac dQng ben ngoai, rna v~t li~u cua v~t th~ chua vuot qua gioi han dan hAL Tuy co sg gi3ng ve muc dlch voi mon Suc bJn vdt li?u va men ClY h9C kit cdu nhung d3i tUQ11g nghien crru thl khac N~u nhir mon Suc bJn vdt li?u va mon Ca hoc kit cdu lAy d3i nrongnghlen ciru chi la cac v~t th~ dang thanh, thl dAi ttrong nghien ciru cua mon Phdn tfch Ung sud! la cac v~fth~ dang thanh, dang t~m, vo vadang kh3i, tire la lien quan t&i vi~c gilli bai toan 1, hoJc chleu Men Phdn tlch fmg sudt cling khac voi men Suc bJn w;it li?u va ClY h9C kit cdu ve tfnh chJt chS cua nghiem Loi giai cua StfC bJn Vt;lt li?u VaCCZh9C kit cdu thl chu y~"t h\ gdn dilng vlkhong thoa man h~t cac dieu ki~n bien cua bai toano Khi nghien ctru cac bai toan sieu tinh, mon Suc bJn wJ/ li?u va ClY h9C kit cdu phai dua bd sung them cac dieu ki~n thuQc vao hlnh dang va lien k~t cua v~t th~ dS I~p them cac phuong trlnh bd sung, thi mon Phdn tichteng sudt chi cdn sir dung cac gill thi~t co lien quan tai tinh ch~t cua v~t thS giAng nhir cac mon Suc bJn w)t li?u va mon ClY hoc kit cdu, la cac gill thi~t: - V~t li~u lien tuc, d6ng ch~t va ding huang - V~t Ii~u dim h6i tuy~n tfnh va tuan theo dinh lu~t Hooke - Bi~n dang cua v~t la be Hai gia thi~t sau lam cho cac phirong trinh mo 13 trang thai chiu hrc cua v~t th~ la cac phuong trinh tuy~n tfnh va do co thS ap d\lng nguyen Iy ch6ng ch~t nghi~m (nguyen Iy ce)ng tac dl:mg-cua: ll!c)~ -' - - - Mon Phiin tfch img sud! khong nhling ghii dugc cac bai toan da gilli dugc mon Suc bdn vt;l! li?u va ClY h9C kit cdu rna giai dugc cll nhling bai toan rna mon Suc bJn vt;lt li?u va CO' h9C kit cau khong giai dugc nhu: Cac bai toan ve (mg su~t C\lC be), bai toan ve (mg su~t ~p trung, bai toan ti~p xuc v.v No co.~g cho kha nang danh gia dQ chfnh xac va gi&i h~n ap d\lng nghi~m bai toan giai theo cac pl1uangtrinh cua mon Suc bJn Vt;lt li?u va ClY h9Ckit cau Trong tili lieu trinh bay mon Phtin tfch Un!! sudt tren ca sa k~t hero hai mon hoc la Lv _thuyit dan h6i (mg d¥ng va Phuong philp phdn tir him h{ln dS giai mQt sA bai toan thuang gJp xay d\l'Og cong trinh, dJc bi~t la xay d\l'Dg cac cong trlnh Thuy Igi _ _ ," ._ .• _~._ •• _ _.• , _~~ ,., • •, • _.;, _.,.":._._ _,~ ••• ~, • ~ ,",,~,,;,_,,,,',_,.,.",_'~ ' , , ,.""~',,, • • , , ' , , _,._" "-" ' ,, e,"_ ._, _" , :.,;~~~ "., , •." ' _,,".' • " " " M _.· ,;' ·•• , ,,.-I· Tac gia xin chan cam an cac b~n dAng nghi~p Be) mon Suc b~n v~t Ii~u-Co hQc k~t c~u Truang D~i hQc Thuy Igi da cung c~p tai li~u va dong gop cac y ki~n quy bau dS hoan cuAn sach Do IAn ddu bien so~n nen khong tranh khoi cac khi~m khuy~t, mc gia mong nh~n dugc cac y ki~n d6ng gop cua cac b~n dQc d~ IAn xu~t ban sau dugc tAt hon Cac y ki~n dong gop xin gui v~ BQ mon Suc b~n v~t li~u-Ca hQc k~t c~u TruCrng D~i hQc Thuy 19i TAc GIA ntem ve """ l Xet mQt v~t th~ dan h6i chiu tac dung cua h~ hrc can b~ng d~t h~ true toa dQ xyz nhtr tren hlnh 1-1 a Tach quanh di€m A v~t mQt phan t6 hinh hQp boi cac m~t phing song song voi cac m~t toa dQ Tren cac m~t cua phan t6 co phAn irng suit nhu tren hlnh 1-1 b dy y Y P O"x 'tzx O"z dx x x z z a) b) Trong cac phAn co phAn trng suit phap la ax, a y, a z va phAn trng su~t ti€p la 1:xy ,1:yx ,1:yz ,1:::y , ' 1:xz ,.1:;:x' Cac m~t c~t dugc goi ten b~ng phap tuyen ngoai cua no, chang han ta goi m~t x tuc Ia m~t co phap tuy~n ngoai co phirong song song voi true x Trong ky hi~u irng suat phap, chi s6 bi~u th] ten cua m~t cit co ung suat phap, cling la chi phuong cua irng su~t do; ky hi~u irng sutt ti€p, chi s6 thu' nhtt bi€u th] m~t co irng su~t ti~p, chi s6 thir biSu thi phuong cua irng sudt VI du (J' x Ia trng suAt phap tren m~t x, 1::cy hi trng Slitt ti~p tren m~.t x co phuong song song voi true y Ddu cua cac img sUdt duoc quy tree nhu sau: Niu phap tuyin ngoai ella m4U edt huang thea chieu duong ella true fog aQ, thi cae tmg suat co ddu duong chung cung huang theo PHA.N rtca UNG SUAT chieu duong cua h? true toa t/lj, tren cdc m(zt co phdp tuyin ngoai nguoc chilu vai chiJu h? true toa t/lj, thi cac Ung sudt co ddu duang chung eo hutmg nguoc vdi ehiJu cua h~ true toa a{i, cac irng su~t ve tren hlnh 1-1 d~u eo d~u duong Cac irng su~t khong nhtrng thay ddi thea tim;s di~m v~t (phu thuQe to~ dQ ella dj~m) rna phu thuQe vao phtrong cua m~t eit di quadi~m 'do (phu thuQe goc nghieng cua m~t eit di qua di~rn d6) N~u ky hi~u ehung cac (rng su~t la S thl ta eo th~ viSt: S = S(x, y, z, n) , voi n la cac eosin chi phirong cua rn~t eit di qua di~m d6 .~- 1.1.2 PhLPCYng trint: vi phalJ canpj!}g Navier Khao sat rnQt phan to dxdydz tach t11 rnQt v~t th~ cdn bAng Ngoai cac frng su~t tac dung tren cac m~t cua phan to, phan to eon e6 cac hrc th~ tich voi cac phdn hlnh ehiSu ella n6 len cac true toa dQ Iii X, Y, Z tac dung len phan to ntra N~u nhir tren m~t e6 toa dQ la x ta eo cac phdn (mgsudtJii:· O"x(x,y,z) 'Z"xy(x,y,z) 'Z"x:(x,y,z) ',' thl tren m~t eQtQ::t dQ la (x-dx) e6 cac phdn lrng su~t la: ~~+~~~ ~~+~~~- ~~+~~~ Dung khai tri~n Taylor va be qua cac vo cung be b~e eao ta dUQ'e: O"x(x + dx,y,z) =O"x (x,y, z) + dcr, =O"x(x,y,z) + a~x dx Lam hoan toan tuong til dOi voi cac phdn trng sudt khac duoc cac phdn (mg su~t tren cac m~t cua phan to cho tren hlnh 1-2 t xy + > "0 y x z ocr dx cr x +_X Ox at t i tzy + ::4: dz ) - xz ~~ ;f>tzx dz Oz ocr z dz cr + z Oz rdX~ Hinhl-2 Ot xy a;dx + Ot""xz dx ox Nhfingphwng CO' ban cua bai toen dan hOltuyentinh d~ng huting :han t6 can b~ng duoi tac dung cua cac thanh,ph~n trng suAt tren cac m~t va cac p~~n hrc the tich nen no thea man cac phtrong trinh can bang tinh hoc: LMx=O ~M LJ y =0 (1-1) LMz=O B~t d~u vai nh6m thir nhfrt cua (1-1), chang han voi phuong trinh oax ) (ot'yx ) ( ax + Ox dx dydz - a ~dydz + ryx +ay dy dzdx - + (,'" + a;; dz )dx4Y - LX = : t yx dzdx + ,,,,dxdy + Xdxdydz =0 Cac phuong trlnh hlnh chi~u thea phuongj/ va phuong z lam nrong nr, Sau rut gonta diroc: oax 0,]X ot'zx X - Ox +ay+az+ ot'XY oay ot'Zy + + +Y= Ox 8y oz (1-2) ot' 0, 00" E + E.+ = +Z =0 ox ay OZ ho?c vi~t duoi dang rna tr~n: cs + P = (1-3) Trang do: C la rna tr~n cac toan tfr vi phan: C = [ tx ~ 3z ] S la rna tr?n irng sutt: S=[:~ ~ ::] r; «; Pia vee to' hrc th~ tich: P = {X Y az zy (1-2, ), (1-3) lit cac phuong trinh can bimg hay lit h¢ phuong trinh vi phdn can brmg Navier Nh6m tlur cua (1-1) lit cac phirong trinh can b~ng momen 06i voi cac true Ch~ng han phuong trinh d~u tien lit: (,,, + = duoc: a;; dz)dxdydZ-(,,,, + Ok dyYXdydz=O PHAN riCH UNG SUAT Hay Lam nrong t1l ta dtroc: t:xy =t:yx (1-4) (1-4) bi~u di~n lud: adi tmg cua img sudt dip 1.1.3 (rng su§t tren m~t nghieng ~ Didu ki~n bien v~ Ivc o tren ta da xet phan t8 hlnh hQp tach til mQt v~t th~ can bAng va cac phuong trlnh can bAng (1-2) hoan toan thoa man d8i voi cac phan t8 Tuy nhien, n~u ta chia v~t th~ bAng cac m~t phang song song voi cac m~t phing toa dQ each nhtmg khoang vo cung nho thise chia v~t th€ vo s8 cac phan t8 hlnh hQpva inQt s8 cac phdn t5 tit ai~n Cac phan t3 ttr di~n ngoai cac m~t song song voi cac m~t toa dQ chju mc dung cua cac phAn img sudt, co m~t nghieng voi cac m~t toa dQ, tren co cac 11Ic b~ m~t tac dung nhu tren hlnh 1-3 Ta xet rnQt tfr di~n tach tit v~t th~ tren hlnh 1-3 Tfr di~n co cac m~t x, y, z va rn~t nghieng vcho tren hlnh 1-4 Gia sir luc b~ m~t toan phAn tren·m~t v cocac phAn hlnh chi~u len cac true toa dQ Ia Xv, Y», Z» Phap tuy~n ngoai v cua m~t nghieng hop voi cac true x, y, z cac gee a, (3, 'Y f)~t: e = coso = cos(v,x) m = cos (3 = cos(v,y) n = cos'Y = cos(v,z) e, m, n duoc goi la cac eosin chi phirong cua phap tuy~n v ·Y Phap tuyenv Ll!C be m~t P, c Hinhl-3 Hin~1-4 z ,"; ; Ky hieu dien tlch cua m~t ABC Ia dF, thl di~n dch c ~~ ~'~~19~t BOe, '~O~, AOB JAn hrot la dF e, dF.m, dF.n Phan t8 tlr di~n co th~ tfch rdt nho nen hi co th~b6- ciualh~nh phdn'l,!c ' th€ tich ; "~ t; '- •• "" ~ ' " :- ", r ,,~, ' , /T • • lnnltrfI I1JI'S"'''''''~ trtnn eel ban eua bai loan dan Phan t6 a trang thai din b~ng nen ta co: LX = LY = ~ Xv·dF-O'x·dF.l-1:Yx.dF.tn-1:zx.dF.n = ~ ~.dF -:-1:x~.dF.R - O'y.dF.m -1:zydF.n = z =0 ~ ; Z".dF -1:x;:.dF f -1:yz.dF.m - O';:dF.n =0 Rut gon ta duoc: (1-5) ho~c vi~t duoi dang rna tran: P, = S.L (1-5') Trong do: P, lit vecto irng su~t toan phdn tren m~t nghieng: I 2 t; ="Xv +Yv +Zv Y L la vecto eosin chi phtrong: L = {R m n S Ua rna tr~n trng sudt Ta co tha phan P, thanh phdn: ung sutt phap 0' v vuong goc voi m~t p~~ng v va trng su~t ti~p t-, n~m rn~t ph~ng v, Thanh phdn O'vdugc tinh b~ng cong thtrc: (1-6) N~u phan t6 tren duoc tach tai m¢t di€m tren bien v~t th€ thl X v'Yv ,Z v la cac ph~n tai phan b6 tren m~t nghieng, nen hie bi€u thirc (1-5) chlnh 1(\ tliiu ki?n bien vJ luc cua bai todn (cac di~u kien bien tinh hoc) nhieu bai toan dan h6i (nhu cac bai toan v~ thanh, t§m, vo, vv ) hAu nhtr khong th~ tim duoc nghiem thea man h8t moi di~u kien bien, d~c bi~t la tren bien co hrc t~p trung tac dung, Nham giam bot kho khan ta co th€ him "mern hoa" di~u ki~n bien v~ lire b~ng each ap dung d nguyen ly Saint-Venant, nguyen ly duoc phat bieu nhir sau: Khi mien t/(lt tai nho han moi kich thucc cua vdt thd, thi trang thai tmg sudt va bien dang tai nhii:ng tlidm xa nat a(lt 11!e thay tl6i rtit it ta thay h¢ hre btmg h¢ luc khac luang duong PHA.N TieH UNG SUAT _ H~ hrcnrong duong a day duoc hiSu la h~ 19c c6 cung vecto 19c chlnhva vecto momenchfnh Vi d\l co m~t cit ngang F la h~ng s6, chiu hai h~ 19c tuong duong nhir tren hlnh 1-5, trang thai irng suat.trong hai tnrong hQP chi khac a vung hai dAu gAn noi d~t I\fC, cang xa hai dAu thaeh, trang thai irng sudt cang gi6ng Ta se ap dung cu thS nguyen ly giiti mQt s6 bai toan a cac chuang sau Thanh co mat cAt ngang F' p-rt.l~'-'-'-'-'-'~;?:,fh:;n'~~ _._ ~r.- ,t r1:: _ .:.:.:l P :.:.' \:.:.: - F P =q - :.:.:.:.'/ , -.:.:- q q b) a) Hinh 1-5 1.1.5 Trflng thai ung su§t t{li mi}t (fj~m Nhir tren da n6i, img suat khong nhfmg phu thuQc vao tung diSm rna phu thuec vao phirong cua m~t cit di qua diSm d6 T~p hop tat cit cac ungsudt tac dQng tai mQt diSm theo moi phirong goi la trang thai ung sUdt tai diSm d6 Nhir v~y, dS nghien ctru trang thai trng sudt tai rnQt diSm M nao d6 v~t thS can bAng, ta phai bi~t h~t moi ung sudt tac dQng tai diSm M tren tat cit cac m~t di qua di~m d6~ Truce den, ta bi~u th] cac irng sudt tai diem nao d6 bAng cac irng suat tren cac m~t cua mQt phan t6 bao quanh diSm d6 Nhir v~y, trang thai tmg su,ft t~i mQt di~m la t~p hop tat cit cac ung suat tren cac m~t cua cac phan t6 bao quanh diSm d6 Nhir da chimg minh a 1.1.4 (cong thirc 1-6): N~u bi~t trng suat tren m~t cit vuong gee voi di qua mQt diem thi co th~ xac dinh dugc trng suat tren bat cu m~t cit nao di qua diSm d6 MQt m~t rna tren d6 khong co (mg suat ti~p dugcgQi la m{it chlnh tmg sudt Phap tuyen ngoai cua m~t chlnh dircc goi la phuong chinh va img -suat phap tren m~t chlnh dugc gQi la ung s-M.dL,,-hinb~~gY'Q'il~_.cU ng~YngJniDltd!J' q~Ja: Q:tJ~ ITI_Qtdi~m,llaOgQ!?~~LgibCUf!g!l!!Lg!!Q'C rnQt_ phan t6 hlnh hQp rna tat ca cac m~t cua n6 d~u la m~t chinh Phan t6 d6 dugc gQi la phiin t6 chinh Ky hi~u c~c frng suat chfnh cua phan t6 Ia 0'1, 0'2, 0'3 va v6'i quy uac la: O't ~ 0'2 ~ 0'3 (kS ca dau) Bay giG hay xac dinh cac (mg suat chfnh Xet phan t6 tu di~n nhu tren hlnh 1-4 Giit sir m~t ABC la m~t chinh va gQi i'rng suat chfnh tren m~t la 0' K , cac cosin chi phuang cua phuong chinh la K' m K ,n K e v 0' ~v Thay tro l~i Yv = O'K·mK Zv =O'KnK (1-5) dugc: + 'ryx·m K +'rzxn K =0 'rxyR K +(O'y -O'K).m K +'rzyn K =0 'rx:R.K + 'ryz.m K + (O'z =0 (O'x -0' K)l K ;~o - VK·-I.K - (j K )n K (1-7) NhCfngphurmg trinh ca ban cua bal toen dan hOI tuyen finh dAng huti'ng (1-7) la h~ phtrong trinh thuAn nh!t M?t khac eK, mK, nK khong th~ d6ng thai b~ng nen d~ phuong trinh co nghiem khong tAm thirong thi djnh thirc cua h~ phuong trinh phai b~ng 0: (O'x-O'K) D = 1:xy £)~t: t; (1-8) (O'y -O'K) I) = ax + O'y + a;: 12 = ax 1:yx 1:xy O'y -'ax 13 = 1:xy 1:xz + O'y 1:y;: 1:zy CJ;: + ax 1:;:x v; 0';: (1-9) 1:yx 1:zy O'y v; 1:y;: a;: 11,12,13 goi la cac bdt bi~n cua trang thai ung sudt (vi xoay phan t6 di thl cac d~i hrong khong thay d6i) Tir (I -8) va (1-9) ta ducc phuong trlnh b~c d6i vci img su!t chlnh: O"~ -1!O"i +12 0"K -1 =0 (1-10) Giai phuong trinh (1- 10) ta se ducc nghiern d6i v6'iO"K va ngiroi ta cling chimg minh ducc ba nghiem cua (1- 10) luon Iii thirc, d6 Iii cac irng sudt chinh O"J, 0"2, 0"3 Thay IAn hrot O"J, 0"2, 0"3 -vao (1-7) va k6t hop voi (*) gi3.i h~ se duQ'c phuong chinh ei+mi+ni=l (*) Nhtr v~y ta da xac dinh duoc cac irng su!t chlnh va phirong cua cluing Bay gio ta ti~n hanh xac dinh cac irng su~t tiep eire trio Cac img su~t tiep cue tri n~m mij,t phang chua mQt true chInh va nghieng mQt goc 45° so voi hai m~t chlnh tuong (mg Cac irng su~t tiep cue tr] ducc tinh bing cong thirc sau: - -~_ 0'1-0"2 '1: 12 2- '1: 23 = '1: =_3_1 0'2-0'3 2' (1-11) 0" -0" 31 Xet m9t v~t th~ dim h6i au lien k6t (v~t khong co chuyen vi cung) nhu tren hlnh 1-6 Khi bien dang, di~m M se b] djch chuyen toi vi trf MI, ta noi di~m M da thay d6i vi trl SI! thay dbi vi tri cua di~m v~t z 11 PHA.N TICB UNG SUAT NQi hrc tal cac tiBt di~n cua 'ddup~fu1 tir 3,-(D-B), d6 o Xoa' EJ 0, -1200 Yoa Xao 1200 =-1.3 ,0 Yao o -1200 o o 1200 o 0 [8]3' theo 4-34: 0'0 q£3 ql -+, -0,0003£ EJ -0,0009£ -q£ ° ° = +1,08q£ ql -1,08ql -ql NQi Igc ~i cac tiBt di~n ddu (thuc chAt la phan It}'c cac Ii~nk~t nut thea chiSu cua h~ true toa dQ C\lC bQ cua rn8i phdn tir) dugc thS hi~n tren hlnh 4-33 Cac phan 19c ghi tren la ddng tho; tai trong va cac chuyen vi nut ddu B gay nen, Tren co can blng giiia cac 19c diu va tai trong cua rn6i phdn ttr ve diroc biSu dd nQi hrc cua timg nhu hlnh :34 sa Vi diJ :Ve biSu dd nQi Igc cila khung chiu mc dung cua chuyen vi cU'Ong buc lien ket tira, Bi~t khungco.so dd, kfch thiroc, ti~t di~n nhir da cho hlnh vi d\l4-2 (hlnh 35) C(OO) b) A(O,O,O) B CD 8(1.23) "A e ,t e J ; /5 ~ ,; j I, 2£ /5 ·r-; , ; L c·~,mnh·4 j5 ~ Giai: '-' - , Cac s6 rna phdn tir, nut, chuyen vi vin chon nhu vi du tnroc DS xet djch chuyen cua g6i ttra a day ta thay dich chuyen g6i W-a blng cac 'Igc nut qui dai wang dtrong, Vec to luc nut {F} luc n~y la chuyen vi CU'Ong b(r~, ·lien kSt tua, dugc tang hop tic cac l1!c nut {PI!.} e cpa rn6i phdn tu co lien ket tga chuySn vi: mnh4 36 112 \ ;; ' o Gis! biJl {PAL =[L]: {PAL {p A} e nh~n dtroc b~ng phan hrc lien kSt ~ut chuyen vi g&i nra v6i dAu ngiroc lai: o {PAL =-{RA}l =~Ll ~£ , {PAL =-{RA}2 =~Ll -6 2£ {F} ={P:}I +{p:L -720 -2,4 -2,4£ , 720 2,4 {PAll =-{RA}, =0 =[~o ~ ~]Tl_~)~Ll +.[~~6 ~:: ~]Tl~~O )~Ll =l~~~:~)~~ 2£ 0 -2,4£ -0,4£ I! Va ta dugc h~ phirong trinh co ban [Kl {~} ={F} cua bai toan: 1-574' 56) 1969,08 574,56 -1,8£ 11.6.1 ) ~ 574,56 1645,92 -3,6; [ -1,8£ - 3,6£ 7£ 6.3 Ll2= -439,92 ~Ll -0,4£ Giai h~ tren ducc: , (.6.1 =.6.Bx, 11- ' 238 ) Llr Ll2 : Ll~ = ~,185 6., , 6.3 - CPB 0,214/ e i NQi 19c tai cac ti€t dien dAu cua m6i phAn til' ~~nh: {NL =[st.{8L +{NAL ,,;1 Cac rna tIi-n [8]1,[8]2,[8]3 da dugc tinh vi du tnrcc {N A} e NQi 19c tal cac ti€t di~n dAu chuyen vi cua lien ket tva gay nen, chinh ~ b~ng phan 19c cac lien ket dAu phAn tir ( hay b~ng {p A} ' e ddi dAu) eta tinh a trenD Cu6i cung, nQ!!gc tai cac ti~t di~n dAu cua m6i phAn til' tinh dtroc: {NL = XAB YAB MAE = XBA YBA MBA 113 PHA.Nl1CH UNGSU~T 1200 12 0-1200 0 6£ -12 6£ 6£ ' 4£2 o -6£ 2£2 EJ =7 -1200 o 1200 0 -12 -6£ 12 -6£ 6£ 2£2 - 6£ 4£2 ' {N}2 = = '0 -6 -4£ A+ -0,185 ETA -= £3 ETA -285,72 £3 5,069 -1,74£ -2£ -0,214/£ 2,8572 -5,069 " -3,32£ = 960 720 ; 960 - 720 -0,238 720 -1, 2, 3£ 2,4 1,~£ -0,185 0,211/£ 720 o -720 , 1,74£ -358,5,1 o -~,4 -1,74 EJ -0,238 XBC Yoc MB Xes Yes 1,8 - 2, -1,8 2,4£ 3£ ~960 - 720 960 1,8 -2,4 ~3£ -.1,8 ~,~, {NL = XDS 'YDS XSD Yeo 1200 -1200 EJ =£3 o -12pO A+ ~ 385,51 24t· 1,74 ETA "',7= 0 0 1200 -0,238 0 -0,185 A+ 0 221,52 0 EJA - -221,52 EJA -'£3 0, Dua vela k6t qua tren ta ve diroc b.~fu d6 nQi 19c cua khung nhir tren hinh 4-37 t 'J ~, @x~f 5,069 358,51" Hinh4-37 ' 114 Vi dlJ 4-4: Ve bi~u 06 nQi hrc cua dAm lien tuc cho tren hinh 4-38a Bi8t dAm co mat cit ngang khong o6i va la hlnh chfr nh~t (0,5xl)rn2 • Cho E = 2.104 kN/cm2 • • Giii: V&i dAm lien tuc chju u6n khong co 19c doc, An chuyen vi nut chi bao g6m chuyen vi th~ng vucng goc voi true dAm va gee xoay, Sau xem xet di~u ki~n bien, voi dAm aa cho chi ~onAn chuyen vi nut } ;F= {F; =M B} 6.= { 66 ==