Chapt 5 Stress dueto bending：5章受力弯曲

1、chapt 5. stress due to bending 25.1 introductionstress distribution & the resultantwhat about bending? 3 geometry of deform. strainstress111limlim00ocdosdsdkcurvaturess radius of curvature-(1) 45.2 geometry of deformationconsidering a straight beam with the sym. cross-section of homogeneous mate

2、rial under pure bending 5sym. of abed & bcfe plane surfaces long. axis remain plane & plane ad,be,cf have a common point of intersection oassume no-deform. in y-dir. ad=a1d1 , be=b1d1 6for linear elements on the plane of sym. / long. axisshorten ”neutral axis.” () neutral surface or planeelo

3、ngate 7ysddyassyssypqpqqpyqpsrssrxxlimit the )( )(111111-(2)(1) 8assume (2) holds for every line element / long. axis (2) independent of materialfor transverse dirs.yxzy 9ex 5.1) a steel bar of rectangular section under a constant moment. if , min. permissible & angle change?3ys01029.204.05.375.

4、1) (15.37105.372751013max3maxradmmsmconstconstdsdmmmyx 105.3 stress distribution & equil. requirements for a line element / the long. axis,using hookes law like a tensile bar comp. for y tension for y requirements areyeexxmdaymdazmdafaxzaxyaxx00 11wherely)(identical 0section cross .0 andcentroid

5、 the throughpassmust 0asection -cross theofmoment 10 zydasymzydaedazyedazmaxisx ydastydaedaeydafaaaaxyaaaaxx(2)(2) 12zxzxzzaaaxzimyeimyydsdeiiwheremdayedayyedaymalso eimasection cross theof intertia) (ofmoment ) (2nd : day ,za22(2)-(3) 13ex 5.2)4 point bending of pure alumina beam .if the fracture s

6、tress =240mpa ,p at fracture? 14bmdpam .max(3)n.)mmm(mmpanmmmm).mm)(.(mpaabh pmpabhpabhhpabhdyybdzdyydayiiypaim-xxhyxxazzzyhhhhbb552151011010681212406fracturesit ,240when 612)2(12 where 232622max fractureat max232max3222x222222 15ex 5.3)(a)(b)location of centroid & iz ? 16(a)(2tdtwtarea,0get toa

7、yda(total area) c =ac +2ac must hold)(2)()2()2122)(2(2tdtwttdttdwtct(dt)t(d)twt(dtdtwtcwdt 1722222222)(asidasydasidasysdadaydasyisyydayizzzz letyzyzs“parallel axis theorem”centroid 18iz=(iz)+(iz)=(ic)i+ai(c-c)2+(ic)+a(c-c)22322)(21)(12)(2)2(12tdctdttdtctdwtwt(b) similarly,34812543tdidczc & iz pr

8、ogram “moment of inertia” 19fig 5.14 see appendix c,d,e for i 205.4 stress in symmetric elastic beams with variable bending momenteven for non-pure bending ,(1)(3) are still good approx.for slender & sym. beams 21ex 5.4)1000psisuch that determine1000 ,6max llbphb 22)4maxplmftinftinlbinlbpbhlbhpl

9、bhplhbhiwhereihpliyplzzzx12121100033661000232 2381212 244 232max23max3max(3) 23ex 5.5)p=7kn , l=6mmax. tensile stress?1th2tbmmtmmt75. 775. 521 24)if neglecting weightmpamknmmmmmmmnknmmbtthbt)t(h-twhere iihmimymknplmzzz194/1094.111010164.21004210164.2 22212122)2( 42243347max4722232321maxmaxmaxmax 25m

10、pamkn.thtbhbmknlalwwherewlm71.1910210164. 210036237. 02370 )2()2(277 297max2132maxdue to weight 26ex 5.6)max tensile & compressive stress?5,4 ,2bakipspab 27634. 2266172232277212332 c )sfdlbftpam800042000maxfor c & iac=aici+aiiciiaici+aiiciiai+aiior program “moment of inertia”mbdaba bappb2pa

11、28i=ii+iii=ici+(c-ci)2ai+icii+(c-cii)2aiipsi.maxpsiftinininlbftimy.-c yxx221761011234628000 62.34-cy ie. . bottomat comp. .4397126 .101654. 4800046547ie. surface. at top le tensimax.6 .101644.18167.5774.21472154. 11272223346. 12122342323 at x=a+b check () 29ex 5.7)stress? tensile.6,3,213000maxcbaftl

12、bw 30) neglecting the weightuse bending of beams to get sfd & bmdfig. 5.20(a) & (b) 315.20(c) fig (seeft 43. 9xat lbft10333. 1m5maxfor c & ic centerpsi550.25ftin12in6 .375in6ft10333. 1iym)(in6 .37512) 112() 18(12128i452hyzmaxmaxx433 32ex 5.8) weight theofeffect theeinvestigatstress comp.

13、 & tensilemax,mm7050t ifm5b,m10am/kn10q0ab 33) t=50mm using moment of inertia2448105,10604. 2,175mmammimmcmknaqwmknmmmmmmknalvlengthweight/85.13/85. 3101105/77023243 34using bending of beams(see fig. 5.23 & 24) m=-173.1knm at x=10m & m=97.38knm at x=3.75m 35mpampa.xmpampaxatmax44.6510604

14、.2)175(38.9705.2810604.2)175250(38.97 753at 3.11610604.2)175()1.173(86.4910604.2)175250(1.173 10 comp & tensile.8max8min8min8maxif t=70mmmax. compressivemax. tensile 36ex 5.9)? allowablemax 24ksi52.7 ,3with pipe steel hollow a6,8allowi0prrba a=b= 37) fbdrppbpabapmr6)(0)(4pipe hollowfor 4sinshaft

15、 solidfor 4404222izzazzrrirrdr dridayiiforab 38section modulusfor given load, we can select the cross section on the basis of simy smyi s msmaxmaxmaxmaxmax(3)dependent on geometry only 39ex 5.10)beam.-h a selectksilbpa201200 40)192009 .1412)2/1814181200(9141412,202090012)2/1815181200(2813151096.1212

16、10201812001812002max32max2max3332maxmaxin.s w ksipsiswlmin.swinftinlbinksiksiftlbmsftlbplmweightxma with select weinstead weightthe gconsiderin but with c.1 table usingnot good!ok 415.6 shear stress distribution in symmetric beams with variable bending momentvariable m shear force shear stress on th

17、e section cf) normal stress given by (3)vdxdm 42moment & force equil. fig 5.32 43 fig 5.33 44zyxaazyxxyxzazazazyxxaxyxxxaxiyv(x)qfdxdffydayqletydaidxdmxfdxdfydaixvydaixmxxmydaixmydaixxmfdafda110flow shear: & 1lim)()()()(011111 for a1for total a 45xyzyxbivq-(4)in a case of a rectangular beamv

18、dydzyhivdaavbhvivhyhiv yhbydybydydzydaqbbhhhybbhyzaxyzyxyxyzxya 222212/12/2/2/112220max21221242also2323842 42(4) 3/2at the neutral surfacezyxyxyxyxivqbfxbf b widththe across ddistribute uniformaly is if 46ex 5.12)p=1500lb b shear stress?(neglect the weight)1 47)psiinininlbbhdhdpbbhdhdbppv)d(hdbdhhby

19、bdyydaqwherebivqxyzxyyxh/dhh/y31000)3)(2() 13)(1500(6 )(612)(2222212bhi 3232322223z2221“v ” 48ex 5.13)p? allowable max.ft/lb45psi180psi12003aa oak 49lbinftinlbftinftftlbininpsilbinpsiftftftinlbininpsiftwlbhlipbhwlplbhmbhhmwlplmmaaalx 6610121)1236154510021200(62)12436604510021200(62)4366045310061200(

20、62)43(684612)2(84223322222222max3maxmax22max)bh15 50lbplblbinftftinftlbpsiinwlbhpbhwlpbhhbwlpbhybydaqbbhqwlpwlpvvallowableaahhx where also661066105 .1428712126106451801063412161064518010634344)(38)(68212)22(222232232max2202203max0max6 51ex 5.14)?maxmax 52)maxmaxmaxmax32max23max416232312866122)( beam

21、, slendar a forlhahplapapbbhbhpahplbhplbhhplslenderness 1/10 53circular cross- sectionassuming uniform distribution through bavrvrrrvbivqbivqyzxy34342432 :2430maxa but you must consider the free surface-effect!3232322sin 32342 whererdrrddrrryda q orrrrq 5422020200440330max34 )( 2 )(4)(32 iiiiiirrrrrravrrrrrrvbivq& for a thin hollow shafttbydaq2, 55ex 5.15)pxppxpbpbxpxslddd max0maxmaxmax 0)(407