TRN CNG NGH HNG DN GII BI TP L THUYT N HI V C HC KT CU (TI LIU HC TP DNH CHO SINH VIN KHOA NG TU V CNG TRNH NI) THNH PH H CH MINH6 2009 I HC GIAO THNG VN TI TP H CH MINH 3 Trang ny trng 4Chng 1 LTHUYT N HI Tm tt Phng trnh cn bng: = +++= +++= +++000Zy x zYz x yXz y xyzzx zyz yx yxzxyx (1.1) trong X, Y, Z lc khi. Phng trnh bin dng: + =+ =+ ====xwzuywzvxvyuzwyvxuzxyzxyzyx(1.2) iu kin tnghp (lin tc): =+ =+ =+z x z xz y y zy x x yxz x zyzzyxy yx 222222222222222 v += += ++= z y x z y xz y x y z xz y x x z yxyxzyzzxyxzyz yxyxzyzx 222222(1.3) Cng thc chuyn ca tensor ng sut. Nu k hiuma trn cc cosin gc gia hai h trc l [c],tensorng sut im trong h ta Oxyz l [], tensor ng sut trong h ta mi [] tnhtheo cng thc: [ ] [ ][ ][ ]Tc c =*(1.4) 5vi[ ] [ ]==z yz xzyz y xyxz xy xzz zy zxyz yy yxxz xy xxc c cc c cc c cc ;* * ** * ** * * ng sut chnhxc nh tphng trnh: = + += + += + + 0 ) (0 ) (0 ) (m l km l km l kz yz xzzy y xyzx yx x (1.5) hoc di dng ma trn: } 0 { =mlkz zy zxyz y yxxz xy x (1.6) trong tng bnh phng cc cosin bng n vk2+ l2+ m2= 1. Li giih phng trnh: 3- 2J1 +J2 J3=0.(1.7) trong J1 = x+ y+ z J2 = yz + zx + xy - yx2 - zx2 -xy2 (1.8) J3= xyz+ 2xyyzxz -xy2z-yz2x-zx2y(1.9) Cc i lng J1,J2,J3 c gi bt bin th nht, bt bin th hai, v bt bin th ba ca tenso ng sut. Trng hp ng sutphng, trong h ta xOy ng sut chnh tnh theo cng thc: 2 2412 , 1) (2xy y xy x + += (1.10) Hng trc ng sut chnh tnh tcng thc: y xxyntg =22 (1.11) ng sut ct ln nht: 22 1min max, = (1.12) xyy xstg = 2 2 (1.13) Vng trn Mohr xy dngt phng trnh: 62222 2 = + +y x y x (1.14) nh lut Hooke p dng cho vt liu ng hng vi m un n hi E, h s Poisson . ( ) [ ]( ) [ ]( ) [ ]+ =+ =+ =y x z zz x y yz y x xEEE 111v===zx zxyz yzxy xyGGG 111(1.15) trong ) 1 ( 2 +=EG (1.16) Nu k hiu:z y xe + + = c th vit: ( )( )( )( )( )( )++ +=++ +=++ +=z zy yx xEeEEeEEeE 1 2 1 11 2 1 11 2 1 1 (1.17) + =+ =+ =z zy yx xG eG eG e 222(1.18) trong ( )( ) 2 1 1 +=E mang tn gi hng s Lam. Hm ng sut Airy (x,y) : 4(x,y) =0. ; ; ;22222y x y xxy y x = = = V d 1: Thnh lp hm ng sut cho dm di L, hnh 1.1, mt ct ngang hnh ch nht cnh ng 2c, chiu rng b,chu tc ng ti phn b u q = const. iu kin bin nh sau: a)Ti x = 0: x=0;xy= 0. b)Ti x = L: q Hnh 1.1 7=== 2210qL ybdybdyqL bdyccxccxccxy c)Ti y = c: 0 ; = =xy ybq d)Ti y = -c: y=0;xy= 0. Nhng nhn xt ban u: -iu kin u tin caa) trong trng hp c thkhng th tha mn.-T tnh cht i xng ca mt ct ngang vbqy = ti y = c vy=0 ti y = -c, c th rt ra y s l hm l ca y. -Hmxcng l hm l ca y. Hm Airy nnvit di dng: = Axy+Bx2+ Cx2y + Dy3 +Exy3 +Fx2y3 +Gy5 C ththy rng:4(x,y) =24Fy+ 120Gy = 0. T phng trnh cuisuy ra F = -5G. ng sut tnh theo cng thc sau: 3 22220 30 6 6 Gy y Gx Exy Dyxx+ + = = 32210 2 2 Gy Cy Byy+ + = = ) 30 3 2 (2 22Gxy Ey Cx Ay xxy + + = = T cng thc tnhxyc th vit: Tha mn iu kin xy= 0tix= 0:A + 3Ey2= 0, t A = E = 0. Tho mn xy= 0 ti y = cc th thy: 0=-(2Cx- 30Gc2x),hay l C = 15Gc2. Gii phng trnh xc nhy, tha mn iu kin bin cho php xc nh B, G: 83 3 320 2 10 30 2 Gc B Gc Gc Bbq+ = + = 3 3 320 2 10 30 2 0 Gc B Gc Gc B = + =T c th nhn c: 340;4 bcqGbqB = =Bit rng momen qun tnh mt ctngang tnh bng 332bc I = , biu thcca B v G s c dng: IqGIqcB60;63 = =Hng C tnh theo G s l:C = 15Gc2=- (qc2)/(4I) T phng trnh xc nh xc th vit: 3 2 3 23 26 20 30 6 6 yIqy xIqDy Gy y Gx Exy Dyx + = + + = Thay biu thccui vo iu kinbin tix = L c th thy: 2 3 2213 26 qL ybdy yIqy xIqDycc= ++ T c th vit: D =Iqc302 Trng ng sutc dng sau: ( )( )( ) = + = + =2 23 2 33 2 223 2632 510y c xIqy y c cIqyIqy c xIqxyyx V d 2:Phng trnh chuyn vdm trnh by ti hnh 1.2c dng: ( )( ) [ ] = =x L y x LxEJPy xy x LxEJPyy x u2 2 22 23 36) , (3 66) , (v 9YX Hnh1.2 Xc nh chuyn vim ti trcy = 0 v xc nh trng ng sut. Chuyn v theo phng thng ng ti y = 0: ( ) ( ) x LEJPxx LxEJPx = = 3636) 0 , (23 2vGc xoaydm tnh theo cng thc =yuxxyv21 , mang dng sau: ( ) ( ) ( )2 2 2 2 2 23 3 663 3 663 3 66 21y x LxEJPy x LxEJPy x LxEJPxy = =Ti y = 0 gc xoay s l: ( )23 66) 0 , ( x LxEJPxxy = Bin dng trong dm tnh theo: ( ) ( )y x LEJPyy x LEJPxuy x == == v;( ) ( ) 0 3 3 663 3 662 2 2 2= + =+=y x LxEJPy x LxEJPyuxxy v Trng ng sut tnh theo cch sau: == + = = =00 ) ( ) (1) ( ) ( ) (1222xyyxy x LEJPy x LEJP Ey x LJPy x LEJPy x LEJP E 10Vd3:Sdngphngtrnhcnbngtronglthuytnhithnhlphmngsuty,xy dm, tit din dm hnh ch nht2c x b, trong b chiu rng dm, 2c chiu cao dm, chu ti trng phn b ucng q(x) = const. ng sut x tnh timt ct bt k ca dm, ti v tr x, tnh theo cng thc: yJx Mx) (= (a) trong M = 221qx (b) Hnh1.3 Hnh 1.3 Nu b qua trng lng bn thn, lc khi dm s khng c nhc ti. T phng trnh cn bngy : = ++= ++00Byy xyBxxyxfy xfy x c th vit:(c) xyJqyxy =(d) Tin hnh tch phn phng trnh o hm ringnys nhn c: ) (2' 2x f xyJqxy= = (e) rng, trng hp khng c ng sut ct ti mp trnv mp di ca dm, xy=0 ti y = c vy = -c, hm f(x) s phi l: xJqcx f2) (2= (f) Ty c th vit: ) (2' 2 2y c xJqxy = (g) T phng trnh th hai ca (c )v FBy= 0 c th vit: ) (22 2y cJqyy = Sau tch phn c th nhn c: ) ( ) 3 (62 2x F y c yJqy+ = (h) iu kin bintiy = c: bqy = . Momen qun tnhqua trc trung ha mang gi trJ = 12) 2 (3c b. T y xc nh F(x) = Jqc33q11Hm y gi c th vit: ( )3 2 33 26y y c cJqy + = V d 4: Dng phng trnh tiu kintng hp (lin tc)xc nh chuyn v trong mt phng xOy dmcng xn nu ti v d 2.Mt ct dm trong trng hp ny l hnhch nht, dy, cng EJ, h s Poisson . YX Hnh1.4 Momen un dm tnh theo cng thc: M = -P(L x) 0 < x < L(a) Cc hm ng sut tnh theo cng thc quen thuc sau: ) ( x L yJPyJMx = = (b) y= 0; xy= 0. T nh lut Hooke c th vitcc phng trnh bin dng: ( ) ) (1x L yEJPEy x x = = ( ) ) (1x L yEJPEx y y = = 0) 1 ( 2=+=xy xyE(c) Quan h bin dng - chuyn vcho php vit: ) ( x L yEJPxux = =) ( x L yEJPyy = = v(d) 12Tin hnh tch phn hai phng trnh o hm ringdng (d) c th nhn c: ) ( ) 2 (2y f x L xyEJPu + =) ( ) (22x F x L yEJP+ =vHm f(y) l hm ch ca y, hm F(x) ch ca x.Sau tch phn, tin hnh thay vo hm bin dng gcyuxxy+=vchng ta c th vit: yy fx L xEJPxx FyEJPyuxxy+ ++ =+=) () 2 (2) (22v Thay biu thc cui vo (c )s nhn c phng trnh: 22) () 2 (2) (yEJPyy fx L xEJPxx F = +(e) Phng trnh (e) chtha mn khi c hai vl const, vdc hai bng C1. = = +1212) () 2 (2) (C yEJPyy fC x L xEJPxx F Gii h phng trnh ny c th vit: + =+ + =3 132 126) () 3 (6) (C y CEJPyy fC x C x L xEJPx F(f) Hm u v v gi y c dng: + + =+ =2 1223 13) 3 (6) (66) 2 (2C x C x L xEJPx LEJPyC y C yEJPx L xyEJPuv(g) Tha mn iu kin bin sau y:ti x=y= 0:u=v=xy= 0, cc hng s phi lC1 = C2 = C3 = 0. T c th vit: = =) 3 (6) () 2 (22 23x L xEJPx L yyeEJPx L xyEJPu2EJP- v 13 V d 5:Cho trc thp trn ng knh16mm, chulc ko dc trc P = 40kN. Lc P gy ng sutcttimtctab,gitrcabng60%ngsutphptimtab.Xcnhgc nghing mt ab. Li gii: Hnh 1.5 ngsut php tnh ti tit din trc thp trn: MPadP2004 / 16 .400004 /2 20= = = ng suttnh ti mt ct xin ab: == 2 sin2cos020 T iu kin ra =0,6 hay l0sincos=0,6 0 cos2c th vit: 6 , 0cossin= = tgT c th xc nh= 31. V d 6:Trng thi ng sut ti im P biu din bng tensor ng sut: MPaij=35 0 70 10 77 7 14 . Xc nh ng sut php v ng sut tip ti mt qua im, song song vi mt miu t bng phng trnh 2x-y+3z= 9. Li gii: Cosin php tuynmt 2x-y+3z- 9 = 0tnhnh sau: 14=+ +==+ + ==+ +=1433 ) 1 ( 231413 ) 1 ( 2) 1 (1423 ) 1 ( 222 2 22 2 22 2 2mlk(a) ng sut php tnh theo cng thc: mk lm kl m l kzx yz xy z y x 2 2 22 2 2+ + + + + = (b) trong , t tensor ng sutc cx = 14,y= 10, z= 35;xy=7,zx=-7,yz = 0.Kt qu ng sut php, tnh theo (b) s l = 19,21 MPa. ng sut tip tnh theo cng thc: ( ) ( ) ( )2 2 2 2 2 + + + + + + + + = m l k m l k m l kz yz zx zx y xy zx xy x(c) Sau khi thay cc gi tr ng sut v k, l, m vo v phi phng trnh (c), ng sut tip ctnh nh sau: =14,95MPa. V d 7:Trng thi ng sut ti im P, ghi trong h ta Oxyznh sau: MPaij =5 2 22 4 62 6 8Tnh trng thi ng sut ny trong h ta Oxyz, qua hai bc:ln u trc Oz xoay gc = 45, sau h trc va hnh thnh xoayquay trc Ox gc = 30. Li gii: Sau ln xoay quanh trc Oz, hta mi c mi lin h vi h ta Oxyz theo quan h: =zyxzyx1 0 00 cos sin0 sin cos""" , vi = 45 Lnxoay htrc sau th hin bng quan h: ="""cos sin 0sin cos 00 0 1'''zyxzyx , vi = 30 T : 15[ ]{ } X Czyxzyxx= = cos sin cos sin sinsin cos cos cos sin0 sin cos''', CngthctnhchuynngsutthtaOxyzsanghtaOxyzcdng: [ ]Tx ij x ijC C ] [' =Sau khi thay= 45,= 30 cc thnh phnma trn [Cx] tnh nh sau: [ ] =23424221464622220xCCc thnh phn ng sutim ang xt trong h ta Oxyz s l: MPax4220 ) 2 ( 20222 222226 2 0 522. 422. 82 2'= ++ + + ++ = MPa x xx xy x20 , 54602122246021222462246226210 ) 5 (4622446228' '=+++++ ++ = MPa x xx xx z34202322242023222422242226230 ) 5 (4222442228' ' =+ ++++ ++ = MPay8 , 42146) 2 ( 22146) 2 ( 246466 221546. 446. 822 2' = + ++ + ++ = 16MPaxz y71 , 24221462324221234624642464262321) 5 (4622446428' '=+ +++ + + + = MPaz2 , 84342) 2 ( 243422 242426 22323542. 442. 82 2' = + ++ + + = Kt qu tnh nh sau: MPaij =2 , 8 7 , 2 37 , 2 8 , 4 2 , 53 2 , 5 4' V d 8:Xc nh trc chnh v ng sut chnh phn tchiutc ng ng sut sau:x= 500 kG/cm2, y=300 kG/cm2, xy=100 kG/cm2. Li gii: Cng thc tnh ng sut chnh: 222 , 12 2xyy x y x + ++=Thay cc gi tr cho vo biu thc trn s nhn c: 4 , 541 4 , 141 400 1002300 5002300 500221= + = + +++= kG/cm2; 6 , 258 4 , 141 400 1002300 5002300 500221= = + ++= kG/cm2; Gc nghing trc chnh so vi trc Ox, Oy tnh theo cng thc: y xxytg =22Trng hp ny tg2= -1v do vy2 =-45; = -22 V d 9: Bit trc gi trbin dng im trong mt phng 2D sau y: x=0,002; y =-0,001;xy=0,003. Xc nh hng chnh v bin dng chnh. Li gii: 17Gc xoay hng chnh tnh theo cng thc: 2) 001 , 0 ( ) 002 , 0 () 003 , 0 ( 222 ===y xxytg T :2=63,4 v243,4 =31,7 v(31,7 + 90) Bin dng chnh: +=+++= 2 sin 2 cos2 22 sin 2 cos2 2''xyy x y xyxyy x y xx Sau thay th bng s cng thc cuic dng: x=0,00385 vy= -0,00285. V d 9: B cm bindng rectangular rosette, bacm bin b tr trong nhnh vng trn, gcgia chng 45, hnh1.6, ghi nhnbin dng im o nh sau:x= 200; 45= 900; y = 1000 Xc nh gi trv hng ng sut chnh, gi tr ng sut ct ln nht ti im o.Bit rng E = 200 GPa, = 0,285. xAHCx4545 Hnh1.6 Li gii: Bin dng gc tnh t cng thc: 600 1000 200 900 2 245= = =y x xy ng sut ti im tnh tquan h bin dng ng sut: ( ) [ ]2 6 6292/ 10 . 6 , 105 10 . 1000 285 , 0 200) 285 , 0 ( 110 . 2001m NEy x x= += += ( ) [ ]2 6 6292/ 10 . 1 , 230 10 . 200 285 , 0 1000) 285 , 0 ( 110 . 2001m NEx y y= += += 18ng sut tip: 2 6 69/ 10 . 7 , 46 10 . 600) 285 , 0 1 ( 210 . 200) 1 ( 2m NExy=+=+=Gc xoay hng chnh tnh theo cng thc: 5 , 1) 10 . 1000 ( ) 10 . 200 () 10 . 600 ( 2226 66 === y xxytg T :=-18,4 v71,6 ng sut phptnh theo cng thc: Vi =-18,4MPa snxyy x y x0 , 90 2 cos2 2= + +++= C th vit: 2 = 90,0 MPa. Trng hp = 71,6 tnh c1 = 245,7 MPa. ng sut tip ln nht, tnh cho trng hp trng thi ng sut phng, 3= 0: MPa 8 , 12220 7 , 24523 1max=== V d10: Tm ua-ra dy t = 2mm c np bng 4 np cng ti bn mp. Cc np ni vi nhau bng khp xoay.Ti v tr C t lc P = 25 kN, hnh 1.7 Xc nh: Thay i gc cc gc tm, Chuyn v c theo chiu ng, ng sut chnh trong tm, Thay i chiu di AC v BD. Bit rngE = 7.104 MPa; = 0,34. Li gii: Ti trng P phi cn bng lc ct cnh BC, bt tmchu ct thun ty.P=.A=.t.l T c thtnh: MPa Pa 50 10 . 525 , 0 . 10 . 210 . 5 , 2734= = = 19 Hnh 1.7 G( )MPa Pa4 101010 . 6 , 2 10 . 6 , 234 , 0 1 210 . 7= =+=Bin dng gc tnh bng cng thc radG310 . 92 , 1= = mm m lC48 , 0 10 . 8 , 4 .4= = = ng sut chnh: 1= = 50 MPa; 2= -=-50 MPa; Thay i chiu di on AB v BD: ( ) mmEll lACAC AC338 , 02 1 1= = = mm lBD338 , 0 = 20Bi tp 1.Xc nh bin dng trong lng vt th tha mn phng trnh chuyn v: u = A1x2 + B1y2 +C1z2 v = A2x2 + B2y2 +C2z2 w = A3x2 + B3y2 +C3z2 Ai,Bi,Ci, i= 1, 2, 3 l const Bin dng ny c tha mn iu kin tng thch hay khng? 2. Bin dng o c biu din bng cc hm sau: x=A(x2+z2); y=0; z=Az2 yz =0;zx =2Axz;xy =0; Xc nhchuyn v tng ng. 3.Trong v d 3 chng ta khng xt n nh hng lc ct. Bi ton ang nu ti hnh 1.2 ny c xem xt y hn, tnh n nh hng lc ct.Mt ct ngang dm hnh chnht, cnh ng 2c.Chuyn v dm c miu t bng hm u v v dng sau: ( ) ( ) ( ) [ ]( ) ( ) ( ) [ ]+ + =+ + + =x c x L x x L yEJPy xy y c x L xyEJPyy x u2 2 23 21 3 3 36) , (2 1 3 2 36) , ( v Xc nh xyxyca dm. Xy dng hmv(x, 0)v(x, 0). 4.Dm ngn chu nn, chu ng sut php 100MPa,ng sut tip40 MPa. Xc nh gc nghing mt tnh ton, so vi trc dm.Tnh ng sut php v ngsut ct ln nht. 5. Bit rngtm thp hnh vung chu ng sut nh sau:x= 150MPa, y= 50MPa.Tnhng sut php v ng suttip trn mt ct ab nghing gc - so vi Ox.6. Tm hnh chnhtkch thc300x100 mm dy t = 10mm, chu tc ng ng sut:x= 120MPa, y= 60MPa.Tnh thay i kch thctmdo bin dng. M un n hi vt liu 2.105

MPa,h s Poisson = 0,25. 7. Phn thnh vungchu ng sut:x= -200MPa, y= 100MPa,xy=-120 MPa. Xc nh hng trc chnh, ng sut chnh. 8. Trng thi ng sut phng ti im biu th trong h ta xOynh sau: MPa9 44 3 Xc nh gi trcc thnh phn ng sut ca im trong h ta xOy xoay theo chiu kim ng h 45. Gii bng hai cch: (1) s dng cng thcchuynv (2) s dng vng trn Mohr. 9. Trng thi ng sut xc nh nh sau:x =14 MPa, y=- 10MPa,xy= 5MPa.Xc nh ng sut chnh, trc chnh. 2110. Trng thi ng sut xc nh nh sau:x =14 MPa, y=- 10MPa, xy= - 5MPa.Xc nh ng sut chnh, trc chnh. 11. Trng thi ng sut xc nh nh sau:x =- 14 MPa, y=+ 10MPa,xy=- 5MPa.Xc nh ng sut chnh, trc chnh. 12. ng sut ti imtrong lng vt thmang gi tr th hin ti tensor ng sut: 0 2 12 0 11 1 3 Xc nh trc chnh, ng sut chnh v gi tr ln nht ng sut ct.13. Khi su mt lm tnhm chu tc ng cc ng sut sau:x= -40MPa, y= 100MPa,z

= 60MPa. Xc nh: ng sut ct ln nht. ng sut bt din oct. ng sut bt din oct. Thnng n v u0 do bin dng khi vt liu. Mun n hi vt liu E = 7.104MPa, h s Poisson = 0,35. Hng dn: ng sut chnh: 1=100 MPa;2 =60MPa; 3=-40 a. 23 1max =( ) ( ) ( )21 323 222 131 + + =oct ; 33 2 1 + +=oct ( )3 2 12 1 + +=EV ( ) [ ]3 51 2 1 3 2 2 0/ 10 . 1 , 1 221m JEu = + + + + = 14.ngsuttiimtronglngvtthmanggitrthhintitensorng sut:37 18 618 37 66 6 21 MPa, m un n hi vt liu E = 200GPa, h s Poisson = 0,3. Xc nh bin dngchnh.Xc nh bin dng gc.15.Kt qu o bin dngim vt th trongtrng thi ng sut phnga n kt qu: x= -90,y= -30,xy=120. Bit rng E = 209 GPa, = 0,29, lp tensor ng sut v tensor bin dngti im kho st ny. 2216. Bit rng tensor bin dng ti im c dng: 4105 4 24 0 12 1 3 , E = 70GPa, = 0,33. Vit tensor ngsut cho im ang kho st. 17. Bit trc cc gi trca cc thnh phntensor bin dngghi trong h ta xOy, xc nhcc thnh phn tensor bin dngca cng trng thi trong h ta rOs, nh biu din ti hnh . a/ x= 200,y =400 , xy= 400 , = 30.Tmr,s,rs b/ x= -400,y =0, xy= 300 , = -30.Tmr,s,rs c/ x= 0,y =0, xy= 300 ,=45.Tmr,s,rs d/ x=1,2.10-3,y =0,8.10-3, xy=-0,8.10-3Hnh 1.8 = 120.Tmr,s,rs d/ x=0,2.10-3,y =0,1.10-3, xy=0,05.10-3

= 45.Tmr,s,rs e/ x=1,2.10-3,y =0,8.10-3, xy=-0,8.10-3

= 120.Tmr,s,rs 18. Trng thi ng sut phng ghi li nh sau : x= -90,y =-30, xy=120.Bit rngE = 209GPa, = 0,29,xc nh cc thnh phn cn li catensor bin dng v tensor ng sut. a)s dng phng trnh chuyn gii bi ton b)s dng vng trn Mohr x l bi ton. 19.nh gi bn kt cu chu tc ngng sut trong khng gian 3D sau y: x= 90MPa, y= 70MPa,z= -30MPa.Bit rng ng sut gii hn cr= 120MPa. Hng dn: Tiu chun bn Tresca: max( 1- 2,2 - 3,3 - 1) =Y Tiu chun bn von Mises(l thuyt Maxell-Huber-Hencky-von Mises) c dng:eq= Y.Trong ng sut tng ng vit nh sau: ( ) ( ) ( )221 323 222 1 + + =eq

Trng hp3=0 tiu chun ny mang dng: Y = + 22 2 121 Li gii: xysr23Theo tiu chun Tresca: cr eqY