REG NACA-TR-636-1938 Approximate Stress Analysis of Multistringer Beams with Shear Deformation of the Flanges.pdf

1、REPORT NO. 636APPROXIMATE STRESS ANALYSIS OF MULTISTRI.NGER BEAMS WITH SHEARDEFORMATION OF THE FLANGESBy PAUL KUHNSUMMARYT7te problem ofskin-ehinger combinaiion8 used asaxially loaded panels or aa cocergfor box bemn8 is con-sidered from the point of riew of the practical stre8sanalyst. By a simple s

2、ubstitution the problem is reducedto the problem of the tingle-stringer structure, whichhas been treated in N. A. C. il. Report ATO.608. Themethod of making this 8ubstWion is e88entially 6mpiri-cal; in order to justify it, compar figure 1 (b)shows one used as the tension side of a beam. Thestress di

3、stribution in structures of this type is materi-ally influenced by the shear deformation of the plate.In aeronautical structures, where the plate often con-sists of a thin sheet that may be allowed to buckleinto a diagonal-tension field, it becomes necessary toconsider the effect of this shear defor

4、mation morecarefully than is customary in other types of structure.Reference 1 discusses in detail the fundamental prin-ciples and the simplifying assumptions tht permit amathematical approach to the solution of the problem.I.,.,.(wGuEX1.-SkIn5trIngerWmbfnathm89sknahudekmmts.It is shown that numeric

5、al solutions can be obtained ifthere is only a single central stringer (fig. 2). Athorough familiarity with the method of analyzingsingle-stringer structures aa given therein is prmup-posed. For muhistringer structures the mathematicsbecomes so complex that there is very slight possibilityof obtaini

6、ng suiliciently general solutions on the basisof the assumptions that were used for the singlestringer structures.Methods combining a desirable degree of accuracy “tith a reasonable degree of generality d, W _z,- m,.,m(4FIGGBE2-6i7M theother method would be to idealize and simplify thephysical conce

7、pt of the structure until the mathematicalrelations become manageable. The second method isused in this paper.The results obtained in reference 1 show that thehighest stresses occur fit the flange and that theydecrease from the ffnnge toward the center line of thestructure. The stress in the flange

8、and the closelyrelated strrm in the longitudinal adjacent to the flangeare therefore of pmamount interest to the analyst.In beams with cambered cover, which were nottreated in reference , the highest stress in the longi-tudinal may occur adjacent to the flange or it mayoccur at the center line of th

9、e bcmm. When it occursat the center line, the stress there also becomes amatter of concern to the analyst,It is quite obvious that, in general, the most impor-tant physical actions will take place around the flanges,partly because the loads are applied there and partlybecause the strwes reach a maxi

10、mum there as long asthere is no violation of the basic requirement that thecamber be very moderate. Consequently, nny sim-plification that may be made should affect as little aspossible the picture of the physical relations in theimmediate vicinity of the flanges.In conformance with thk requirement;

11、 the simplifica-tion necessary for obtaining a solution was achieved byusing a ctsubstitute structure” obtained b leaving tlwjhn.ge (and shear web) intact but replacing the kmgitu-diruds thut are actuully unijormly distributed over thewidth oj the sheet by a single longitudinal equivalent totlwm as

12、far as action on the jlange is concerned. Thissubstitution reduces the problem of the multistringmstructuie to that of the single-stringer structure, whichcan be analyzed as shown in referenco 1. ThG met.hmlof substituting (temporarily) a simplified structuro forthe actual one corrwponds in part to

13、tho method ofusing “phantom members” in hwsscs.The substitute structure is used only to calculato thestresses in the part that it has in common with theactual structure, namely, the flange and the skin ud-jacent to the flange. After this object has beenattained, the substitute structure is discorded

14、. Thestressesin the actual distributed longitudinnls are t.hwiobtained by using the method clescrilxxiin rofercnco 1for distributing “corrected forces.”It is clear that, in any given case, at least ono cquiwt-lent single longitudinal mists. WMhcr or not there isa general method for finding this equi

15、valent longitu-dinal, however, is a question thut could l.mansweredtheoretically only if aIl the exact matl.wnlatiml solu-tions were known. They me not Iinovn, and themethod of finding the equivalent Ionbtitudimdis therc-ore essentially empirical and must IN justified by twits.This requirement-is no

16、t such a serious drawback m itmay seem to be, because the basic simplifications usedmesuch that experimental verification is required in anyment.The method of the use of this factor in expression (1) tends tocounteract the loss of effectiveness caused by movingthe stringer from its original location

17、 to the center line.The sum of the individual substitute stringcre attachedrit the center line constitutes the singlo cquivalen tlongitudinalAs the stresses Urand UCLare unknown at tho outset,for a tit approximation, the ratio UJU=Lis obttiincdfrom equation (17) of the constant-stress solution given

18、in reference 1. With the stresses thus computed, asecond approximation might be made. In all cases in-vedigatid thus far, it was found that the second approx-tiation agreed with the first one within the limits oierimental accuracy. The use of the second approxi-mation is therefore considered unncccs

19、mry. (It mustProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-STRESS ANALYSIS OF NHJLTLSTRINGER BEAMS WITH SHEAR DEFORIIATION OF FLANGES 471be borne in mind that the method of finding the equ”va-lent longitudimd is essentidy empirical. Conse-quently,

20、there is no valid reason to believe that thesecond approximation must be better thm the fist one.)AXALYSIS OF AXIALLY LOADED PANELk an example of the analysis of m axially loadedpanel, the anaIyai.sof the compr=ion panel -withsevenstiffeners, described in reference 2, will be discussed indetail. The

21、 pertinent data on this panel are gken infigures 3 (a) and 3 (b).Estimate of effective areas and of eiYective shearstiffness.-The test restits are given in reference 2 for2P=2,000, 4,000, and 6,000 pounds. The analysis willbe made for 2P=4,CK10 or P=2,000 pounds. It willbecome apparent that the cond

22、itions at this load arethe same as for very smsU loads, so that the analysiswill be valid for any load between O and 4,000 pounds.The mean stress in the panel (reference 2) is2,000uM= =2,860 lb./sq. in.This stress is fairly close to the compressive bucklingstress of the sheet; the effective width of

23、 the sheet willtherefore be taken as eqmd to the actual width. Theeffective stringer area for the flange is therefore.4=0.180+2 XO.024=0.228 sq. in.and for the sum of the other stringers.4L=2.5X0.0S8+10 X0.024=0.460 sq. in.The force at the bottom of the edge stringer k approxi-matelyF,=2,860X0.228=6

24、52 lb.leaving 1,348 pounds to be transmitted by shear in thesheet to the other stringers. The average shear stressin the sheet nest to the edge stringer is thereforeT=48:O:24= 1,170 Ib./sq. in.The critical buckbng stress for 0.024-iich dural sheet,4 inches wide and assumed simply supported, is, acco

25、rd-iu to Tiioshe the sheet is now aamuned to carryonly shear.If there are at least two intermediate stringers be-tween the center stringer and the flange, the calculationof the substitute stringer may be simplified by using afornda derived on the assumption that there are in-finitely many intermedia

26、te stringem; that is, on theP“.ZC700A B cl!mmmmymmefricol abedf= O.oz%f1 t L1024L-$8 I.“i=Ei=ih (c)F=-L2 (d)FIGURE: (3)The area of an individual stringer is nowti=dyand the area of the substitute stringer that rcpkesit at the center line is, according to equRtion (l),dzlm=+dy cosh ZProvided by IHSNo

27、t for ResaleNo reproduction or networking permitted without license from IHS-,-,-472 REPORT NO. 636-NATIONAL ADVISORY COMMITTEE FOR AERONAUTICSThe total area of the substitute stringw located at thecenter line is thereforeA.=+ COSh K,I/dy=A._ (4)In the case under considerationKJ= d2X0.46QX 12 =(),70

28、Go,024x4t?xo.40 .J.so thatAu= O.460X-SO.500 sq. in.Figure 3 (d) shows the cross section of the substitutestructure.Analysis of substitute structure,-The substitutestructure of figure 3 (d) can be analyzed by applying theformulas given in appendix B. By formula (AI)K,= OAOX0.024 1(112 0.228+0 )K=o.07

29、15r tAir1FIamE 4.FreOadydiagramfar inthe case of tite shear stifhwss, the shear strain isdefined directly by the longitudinal strains. In thecambered cover with infinite shear stiffness, the stressvaries along the chord according to the straight-line lawof the ordinary bending theory; in the cambere

30、d coverwith finite shear stiffness, the shear strains are definedby the differences between the longitudinal strains andthe corresponding strains of the ordinary bendingtheory as indicated by equation (5c).These di.flerencesbetween the cambered and the flatcover may be interpreted as arising from th

31、e fact thatthe cambered cover has bending sti.flness of ita mvnbecause it has a “beam depth” equal to ita camber.70:30-.88:34.52:3Q -p- ) ,58 :42. ,/.54;48./.500 1.0 2.0 3.0YbFIGURE7.-oraph for IU3tfonoflmnlfantfwcaon Wver.In the single-stringer beam it was not diflicultto takecare of the effect of

32、thisbending stifbss mathematicallyby introducing the terms uFFand ODinto equation (5c).In the multistinger beam it is more convenient tointroduce a physical equivalent, namely, an atiarysystem of longitudinal stresses distributed over thecover in such a manner as to make the stress uniformin the lim

33、iting case of infinite shear stiffnms. In figure6 (c) the broken line shows the stresses given by theordinary bending theory, the full line shows the uni-form stress, denoted by au, and the crow-hatched areabetween the two lines indicates the auxiliary stressesnecessary to achieve the uniform stress

34、 distribution.The magnitude of the uniform stress is determined bythe condition that, when the auxiliary stresses act onthe flange A4F and on the longitudinal AL, they mustnot -change the bending moment acting at the section,i. e., they must have zero moment about the assumedcentroidd line of the lo

35、wer cover. The auxiliarystresseswill be denoted by a second subscript A placedafter the first subscript, which ckmotw tho stringm orflange where the stress is measured.With the auxiliary stre9sos asaumcd mtic, thomethod of finding the distribution of tho stresses alongthe chord is amdogoua to the me

36、thod uacd for Iltitpanels and will be shown in dotuil for a nmmwicalexample, From the stresses thus calculated, thoauxikmy stresses me subtracted to obtnin tho finalstressm.One step not necessary in tha anulysis of M panelsis required for cambered covers. h intlcatwl in fiiuro6 (d), it is necessary

37、to locate the resultant force FL*acting on the cover (exclive of the flnngo) when thoactual and the ausilinry stresses aro acting. Thovertical location Ah of this redt ant dcterminos thoeffective depth of the bemnh.=h.+Ah (6)when the combined stressesare acting. The exact cal-culation of Ah woukl re

38、quire a very tedious intogmtiouinvolving the stress distribution and the slmpe of thocover, which lms to be repented SOVCMI1tifncs for eachcross section with slightIy differing vnlucs of stressdistribution. For practical purposes, it will theroforobe advisable to simplify the problem, although there

39、will be a slight 10SSin accuracy, by finding tho latcrnllocation Lof the resultant tmd by Msuming that Ah i9determined by the intersection of the line y=yL cudthe straight line joining F nnd L, as indicated in figuro6 (d). Under the assumption of InOd(?Ifie Carnk, YLk giVSIl byf: .*y the length L of

40、 the beam is 108 inches. It is assumedthat the effective width of the sheet hns been estimntmland that the value AL=:0,85 sq, in. includes the effectivowidth.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-STRESS ANALYSIS OF MULTISTRINGER BEAMS WITH

41、SHEAR DEFORMATION OF FLANGES 475The next step is to estimate the efhctive shearmodtilus. If the presence of camber and of sheardeformation is neglected, the maximum shear stmsa inthe sheet will be given by formula (A-8) of appendixB asPit. 250X0.85=”=m=0.0115 X3 X1.65 =3,730 lb./sq. in.The buckling

42、shear st.rws of a long dural plate 0.0115inch thick and 1.80 inches wide isrtii=1,960 lb./sq. in.The mfium shear stress being only about twice thecriticil stres, the average shear stres9 is sufhcientlyclose to the critical to neglect diagonal-tension effectson shear stiffness find to set Q.= G or QJ

43、E=O.40.Equation (3) then gives3b=d=RE=0535Inserting this value in equation (4) givesALS=O.855=0.892 sq in.The cross section of the substitute beam is shown infigure 8 (b). SinCe the substitute beam is of uniformcross section, it can be solved analytictiy. Formula(A-14) giVC!S=o.40xo.o1159 (%+*)K= O.

44、0378 KL=4.08Formula (A12) then gives for z= 108 inch=U=4,830 lb./sq. in.This computation completes the fkst step and thesubstitute beam is discarded.The next step is to calculate the streseesin the actualbeam by the ordina bending theory:7,()(y3xo.94Urp= 9.65 =2,630 lb./sq. in.7,000x2.94rap= 9.65 =8

45、,220 lb./sq. in.Figure 8 (c) shows the chordwise distribution of thestressesaccording to the ordinary bending theory as vreIlas the auxilimy stresses. In crder i% ahow that theaudiary stresses indicated by me 8 (c) fultlll therequirements, a check on their total moment is made.1,955X3 XO.80 = 4,6908

46、37X3.4X0.189 = 538_81 X3.8X0.189 = 2021,399X4.2X0.189 =1,1112,517X4.6X0.189 =2,18S3,635X5.0X0.0945=1,717The moment is zero with a negligible error. The flangesfmss used for mdcuhding the chordwise.stress distribu-tion is thereforeCF*=CF+FA=4,830+ 1,955= 6,785 lb./sq. ti.The moment furnished by the f

47、lange isil.L.*=6,785 XO.80X3.00= 16,280 irdb.z-3a7wn)(c)FIGUBE8.-CambfTedQ3wrMam fm samphimels.TiGThe moment to be furnished by the cover longitudhaIsis therefore3L*=27,000- 16280=10,720 in.-lb.Assuming he=3.77 inches, the force FL* becomesFL*= 1=2,840 lbThe average stress is thereforeand the ratioW

48、ith this valuereference 1=Lane=2j840=3,345“ 3345=SV $ UF* 6,785 .493as abscissa, read from figure 18 of1%=1.94Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-476 REPORT NO. 636-NATIONAL ADVISORY C(XWIWITTEEFOR AERONAUTICS”and, with this value,yL/b=O.

49、613from figure 7 so thatti=2(l-0.613) =0.77 in.o 2 4 6 8 JOx./fPmnu 9.-SQe figure S (d) shows grnphicfilythe flmd stress distribution.TABLE ISiriuger ICenti lfne_l.-._-2_a_.-. _.L -Flange. tBP Yu0.0 0.2 .am.776: L 164Lb the folIowing ones, designated as N. A. C. A. beams2, 3, and 4, will be discussed in this pa-per.In all N. A. C. A. beams, measurements were made onthe tension side of the beam in