NASA NACA-TM-0999-1941 Stress analysis of circular frames《圆形框架的应力分析》.pdf

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1、4.!, .distance of shear flow from center., ., .,., ;:.,.,:.” ,!.distance of shear flow from neutral fiber.,e, . . . , ,., , .:,., : . ,.j.,x distance of s”hear.:center from “centerM moment*llBerechnung der Be.a,fisprchung.kreisftlrmiger Ringspante. 11Luftfahrtforscnung , vol. 1”8,no= 4“,April 22, 19

2、41, pp.122-127.II . -. .Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-2BN2x8larvid10RcantheXAGA Technical Mernorandnrn No. 99? ;:, ;- .bending moment .,normal force,. .,.transverse force . .statically undetermined quantity”load and coeffi.cient, re

3、spect ively. , .,.:1-1;.THE FRAME IQU.ILIB.RIUM,. ;, ; ,”- . ,. .,For the.”ayTli:cAtioh of transverse. forces in a circ-,.Rshell with large ratio , circuli:rf.rarnesare Pr”o.-.ed. They are in equilibrium with the concentratedds and the shear forces from the shell. Eech loadinghe divided into the tra

4、nsverse force passing throughelastic centroid of the shell and the moment (fig. 1).ingsheala).WithTheuncle,r fltransverse force produces, as a result O”fbendr transverse force, a sinusoidally distributedow in the shell that reaches o:the frame” (fig.,?-s = :s.:denoting the mfunctionrepresents theThe

5、 distafor sinusoidalxomentsheI =ITR3Sof inertia of thearTsfl?owJ?_fiRva.nce of tva.riatiheon,.she(froR/Tsdu“(-) =g2riatar cm thchnical;Memo r,andum No-:.99%. 3, :The. shear” fl.o.whp;pli(ed-at$the f:ram as result of amoment is”con stant.”and ,mount.s:t.o(fig:.:1.6):, ,.,The, distance of the shear ce

6、nter of a circular half forconstant variation from the center: is.: :.- -, . . ,.x= ;R 1:57 R, .!, ,“The resulting shear,flow (fig. lc) follows from;Ts. - LoadHCa$e A”“ .Localized-Radial Force Acting on the Frame,.,. : ,. , i.”, ,.,:,. ,. .Ordinarily the circular frame is threefold staticallyundeter

7、mined. but. in this. In.st:anceand in the subsequentload cases the solution canbe considerably simplified bycleverly chosen sectionalization . At point O of theProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-4 MAGA Te6hnical .Membrload in figure2, th

8、e statically undetermined quantity isX3 = o. The.signs for carrying: out the anazysts are givenin figure 4. The depth of the sectional area of the frameis introduced by means of the ratio . The subsequent re-sults are valid for (fig. 3).=The elasticity equations red .The displacement quantities gene

9、rally follow atEI 6ik =J Bi Bk duDetermination of the bending moment curve B. in thestatically determined principal system referred to neutralfiber (fig. 4). The tangentially applied shear force ele-mentsets up s.tpoint T in teframe the bending momentdBo= - TS du eThe distance e follows fom the geom

10、etric relatione = R -r(sin CPo sin V + cos To COSQ)Then the bending moment B. at Q is:,9 QJ,!.BoG-S sin Q. do + sin J sin2 To dCPo +n l-l 09Ei.z+”r-:,iNo. :99 5; :,-.“)+-RBa=r (l- cos T) as a result of X2 = 1. : . ,. :For reasons of symmetry the “integration can te lim-ited to a half frame. .The,fol

11、lowing loads and factorsare obtained. , ,.fi :, . ,. :,”EI tjlo = r(,B1 % du= Pr - Rl/oEI “=o = Pr2” R(% -:2 “ “ “: EI il. = EI 623 ra+. ., :,.:. , .Solution of the elasticity eg.uations”gives the magnitudeof.the. st,a$,i,cally,undeter,min,edquantities, .r . , .:,4 : . :. . . . . ,. , ,.Xl = - =., .

12、X241-r $(-:-”iJ”“whence the ultimate bending moment !. =. . . . . . . .T( Pr. .g :in+=i-2TT ); C,os.?p:-.l. (1)., ,.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-. . . . . ,.,. , .,., , . ,6 IJACA,?ec,hnical Memorandum. No. .999The final normal for

13、ce follows from.N= No +. XINl + X2N2 .,The normal force No in the statically determinedprincipal system at po$nt cp %s obtained by”splittingthe shear force element in the tangential component fol-lowed by integration from O to Q (fig. ”4“),dNo = - TS du COS ( - o)The norme.1 force distribution in th

14、e statically undeter-mined system then is ,.and the transverse force variation(2)(3)Figures 5 to 7 “show bending moments, normal force,and transverse force plotted against the frame circumfer-ence. The ratio. serves as parameter forR $= 1,r=le2, =o.8. . .R R .,Load Case 3Localized Moment Acting Alon

15、g a,Diameter of th;e Frame (fii 8)For this load the frame is simply statically unde-termined at point O. The elasticity equation readsProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA Technical Memorandum No. 999, 7630 + X3833 = o.Bending moments,

16、 normal force-, and -transverse -force in.thestatically determined principal system are obtained as forcase A. The loads and factors are:The statically undetermined quantity follows atThe final bending moment isthe final normal force is . N=-Qo= Cos CP- :,1furthermoreand .I N=% ( 2RCp Cos q)+ r sin

17、V - )1sin V. )1(7)(8j(9)B and Q are plotted in figures 13 and 14. N hs thesame aspct as Q in load case A (fig. 7).Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-. . . . .,NACA Technical M“emorandumNo”O 99g”.: 9Localized Tangential Force Acting Alon

18、Then :,-. ., ,(B=%” 2 sin q) m-cpcoscp-sinq -q (lo)2R.L-(11)Figure 16 illustrates the bending moment distribution; NN in (Fig. lT. )By division of the ,forae P we get. , ,.,., .,., . . ,.,Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,- . -, ,.,.,NAC

19、A Technical Memorandum No. 999Load Case A: 0.9 1?= 1800 kg acting radiallyc: 0.4 F = 800, kg acting tangentiallyB: .40 mkg,.The resulting bending moment curve is fonnd, numerically by superposition of the results from equations(1), (7), and (4), or by graphical superpositin of thebending moment curv

20、es from the basic losds illustrated infigures 5, 13, and 9. The same method applies to thenormal and the transverse force. Of gretest interest isthe knowledge of the lngitudinal stresses from the bend-ing moments and normal forces.BENDING MOMENTS (mkg) IN FRAME FROMTHE NUMERICAL SOLUTIONi It90 “i80

21、180 270 360A “ “”-:7.2 I 53,9 -11. -141.5*53.9” -i7 .2-3.0 0 0 3.0 0B., , ,. ,: ,2,+ : 20”0c a71 , 20.0 -:2.7: 0 ,Result -“37;2 53.6 -1”.:”“-121.5 5 - “” “l?or the stress analysis of the rivets or welds be-tween circular frame and shell the shear flow distributionis employed. It is computed by the m

22、ethod indicated insection II and has for the particular frame loading theaspect shown in figure 19. The maximum shear flow amountsto 26 kg/cm.v. SCOPE OF VALIDITY1. The solutions hold for circular frames withsmall sectional depth compared to curvature radius r.In this case the curved member acts sim

23、ilar to a straightmember. Hence the stress distribution was assumed linearand the cross sections presumed to remain plane. The ef-fect of the longitudinal and transverse forces on the dis-placement factors was disregarded.2. The bending stiffness of the shell plate comparedwith that of the circular

24、frame wa,s presumed to be small.3. The departure of the frame contour from the cir-cular shape due to elastic strain was discounted.Translation by J. Vanier,National Advisory Committeefor Aeronautics,Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-3?

25、3cIF Figure 1.-P 2W L. -,.,aEquilibriumof frame and divisionintobasic load cases.;.fof ratio r/R by equal Fiyme 4.- Identificationof Bo, I?o,and SiB. cocoaProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-, , , -. -lEACATechnical Memorandum Ho. 999 Fig

26、s. 5,6,7,8afo f K /“ i /405 I . , ,.0 i /I 97 Wo0 1 2703# oh/ 10aosL*-423 1Figure 5.- Bending moments underradial load.oFigure 6.- Normal forces underradial load.(.M rB,= 2n Rslnv +NO= moment loWIlg.Figure 7.- Transverse forces under radialload, concurrent normal forcesunder tangential load (distanc

27、e r).Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-. . . . HAOA Technioal Memorandum No. 999 rigs. 9,10,11,3.2.a30Kwafo0 0 0-am-azo(230Figure 9.- Bending momente under Pigure 10.- IJormalforces undermoment loading.6?30A I I I I I I I I I I I 1075/

28、Ii-u-a-amoment loading.Figure 12.- Load case C;tangential loading(distance r). Figure 11.- Transverse forces undermoment loading.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA Technical Memorandum No. 999 Figs. 13,14,15,16f%06 / K f ao4 , q,. -

29、 - -, / “ o0Figure 14. - Transverse forces undertangential loading(distance r).ao81Ii I I I I I I IFigure 13.- Bending moments undertangential loading(distance r).Figure 15.- Load case D;tangential loading(distance r).Figure 16.- Bending moments undertangential loading(distance r).:3,*!.Provided by

30、IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-IUCA Technical Memorandum No. 999 rigs. 17,18,19.,. , -Figure 18.-Bendingmomentsandnormalforces.Figure 17.- Sample frloading./P. 4970 Kg600- 60 n/ (kg - mkg 300 - 30 “ j I / /N -8 I I 78000 -o I/ ,90 h/z 700 / 360f

31、 300 -30 I “./l / / ! ,/ ! !/ . , 600 -60 /-900 90 1. f ,/! %20-750 Rigure19.-Magnitude “”andvariationof shearflow atframe.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-IProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-