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MW“ b IH‘ {III M M PRACTICAL TREATMENT OF STRUCTURAL TENSION FIELD WEBB Thasis for the Degree of CIVIL ENGINEER WCHIGAN STATE COLLEGE Leo Vaughn Nothstine 1944 :42 g“ n—H :A L This is to certify that the thesis entitled Practical Treatment of Structural Field Webs. presented by Leo Vaughn Nothstine has been accepted towards fulfihnent of the requirements for the degree fiflfiLQiflLLngineer Chester L len Major professor Date June 8, 1945 ‘H; | V." “ PRACTICAL TREATMENT OF STRUCTURAL TENSION FIELD WEBS by Leo Vaughn‘flgthetine A THESIS Submitted to the Graduate School of Michigan State College of Agriculture and Applied Science in partial fulfilment of the requirements for the degree of CIVIL ENGINEER Department of Civil Engineering 1944 :THESz'ij Part1 Part II PartIII CONTENTS Introductory Chapter Discussion of Tension Field Theory Simplified Desig of Tension Field Webs Construction and Test of a Production Beaded Spar 17889 E Introductory In structures shieh are designed for minim weight, the problem of load- in: thin sheet material is confronted; Loading thin sheets efficiently becomes a problem of stability. It is not the purpose of this paper to develop nee equations of stability but to reduce the practical prelininary equations for use by the engineer. InPsrtIofthispaper, tensiontieldtheoryisdiscussedfronthe stand- point of develoment. his naterial [is considered to be in agreement withthelesdingengineersottoday. Itieinnonsnneroonplete, hutit is indicative of the intricacies of the developed theory. Sines this work falls beyond the scape of m gredueteengineers, it see-s expedient to sinpliiythenaterislanddenlopanethod ordooign whichsillbeedeqnate and useahle by the enaneer. The second phase of the paper deals with a simplified nethod for solving thestressesinthetensionfieldesiellasthedevelopnent orocsart for designing the rivet connection at theseb to flanges 01' the heel. An ennpleisnadeuptodeuonstnatethense ofthe deeignehartshiehvill familiarise the engineer with the simplified nethod. Thethirdpartofthepaperdeelsiiththetestoi’abeadedsehbeelo This testisinnowadevelop-ent ofthefirstpertsofthisreport, hutitisineorpontedtoshosresultsohteinedbythebeededsebinocn- perison to the web of riveted stiffener angles. NOTATION (unless otherwise specified.) A = Area a = Long side of a shear panel OK = Angle of web srinkles b = Short side of shear panel hr = (subscript) bearing c = conpression (subscript) critical (subscript) = Diameter 0 H II = Vertical stiffener spacing FIDO = Ioungs Iodulns = strain Total force = force per unit area = Effective depth of web = lonent of inertia = A function of the type of support given to the sheet edges INHD‘H"! = Moment = rivet spacing, or pitch distance = Static mt d flanges about the neutral ends .0013 =8hearflowinpomdsperinch H = Thickness of sheet d ll tension (subscript) tn tensile yield (subscript) ty = tensile ultimate (subscript) p. = Poisson's ratio V =‘l'otel Shear 1‘ Unit shear stress coordinate axis 19y 0" = Normal Stress (unit) PM]: DISCUSSION OF TENSIOI FIELD resort! PART I Ultimate strength of flat panels under shear. In the structural analysis of the wing beans of airplanes, the stress analyst is faced with several problems which, in general, are not faced by the civil engineer. The civil engineer endeavors to make the web sheet of beans thick enough so that they will carry the load before buckling takes place in the web. In this case buckling is considered one form of failure. The shearing stresses which cause this buckling determines the allowable mums which can be applied. The equation developed will then apply}2 was t2 Tor =.- K 12 (l :32) ('5 This equation is principally used to determine the critical shear qu4 stress allowable in a shear resistant been (under that of the shear strength of the material) due to the physical dimensions of the panels. The coefficient 'K' has a value which is dependent upon the ratio of the dimensions of the panels as well as on the rigidity of the stiffeners and flanges, or it may be considered as the end fixity of the panels. (See figure 9, Part II). The equation is derived by setting up the differential equation for a loaded plate and considering the edges as being simply supported. Then expressing the boundary conditions where the edges have no deflection, substitute in the first equation analyzing for the critical stress. For very thin sheets, the buckling stress given by this equation is ex- tremely low and, in the interests of efficient design with regard to weight, the aircraft engineer raises the question of how much addi- tional shear can be carried by such a buckled plate before some portion of the sheet has a total stress equal to the yield point of the material, which would cause permanent deformations, or the ultimate strength of the structure is reached. The aircraft engineer would also like to know the effect of this web buckling on the rest of the structure. For instance, the additional load on the flanges and on the web stiffeners if there are any. This problem has been the subject of considerable research the past few years. Curb rently, there is continuous study to make the designs more accurate and reliable. At the present time, it can be said that most of the tension field design methods used in our modern airplane are too conservative. In order to take advantage of tension field design, it is necessary to be able to compute the true margin of safety'instead of a number which is termed a conservative margin because of assumptions used and certain qualities of design not considered. The fact that the true conditions taking place in the web when in the tension field condition are generally not fully understood has caused ultra conservatism in production methods and repairs designed. To get a mathematical picture of the tension field theory it is necessary to consider a.beam.having a relatively thick web under the action of shear and bending. For such a section, the usual bending moment and shear equations of applied mechanics are valid. ' 0;=!_X_I and T=IQ I Ib 0; = Normal stress where '2 = bending moment at any value of x y' = distance from neutral axis to fiber considered I = moment of inertia about the neutral axis of cross section V 8 applied shearing force Q = static moment for which 7" is determined about the neutral axis b =3 thickness of the section at which 7‘ is to be determined For the web, br= t. This corresponds with the equation onpage 1. Considering an element of the web on the neutral axis, so that bending stresses are absent, take an element whose sides are parallel and pork pendicular to the shear force. The stress pattern will then be seen to be uniform shearing stresses on the four faces of the element, see figure 1. However the same result can be represented by an elemant con- sidered to be at 45° to the direction of the applied shear by pure normal stresses of equal magnitude in which (7's:= aht:= 7p was shown in figure l.also. For web plates which are thick, this stress distribution plus bending stresses, will hold up to stress values ap— proaching the yield point of the material. At that time plastic flow will enter the picture and the mathematics will no longer hold. The foregoing discussion takes on new significance in regard to principal stress patterns if we assume that the web plate in the beam is very thin. Considering the compressive stress against which thin plates have very little resistance, it is seen that the tendency will be for the web to F19. / ;\\">.\\\ I \\\ \ \x\\\ \‘ s; \\\//\\\ \\\\ \\\ 4 buckle in a direction perpendicular to the line of action of the compres- sive stress. The value of the applied shear will diminish as the web becomes thinner and thinner, the limiting case being for a sheet of zero thickness. In this case, the sheet buckles upon application of any shear load and can only resist shear by means of the tensile stresses at 45°, as shown in figure 2. It is obvious, with such a stress pattern, that the tensile stresses will tend to pull the two flanges together. Thus it can be seen that some means of vertical stiffening is required to counteract this tendency. The limiting case of a web having no compressive strength has been treated in detail by Wagner (reference 2) and beams approximating this are known as Wagner beams. It is interesting to point out at this point that the center portions of the wing on the Consolidated Liberator Bomber 3-24 is treated as a Wagner bean. Wagner assumes that the web buckles imdiately upon application of the shear load and that the only stresses resisting the shear forces are the tensile stresses set up in the web. These tensile stresses act at approxi- mately 45". Assuring infinitely rigid parallel span flanges and vertical stiffeners, the following equations for this limiting condition can be shown to apply. Keeping in mind the above assumptions and referring to figure 3, for notation the diagonal tension stress developed in the web 0' 1-. = ht sin 2 ' Total force in the compression flange Fc=_k___v_cotoc .6 h 2 Eq Total force in the tension flange Ft’ fiz-%°°’°0¢ Eq. 7 Axial force in the vertical stiffeners I'=-V%Tan Eq.8 where Va: applied shear load h1= effective depth of web t== web thickness d»: vertical stiffener spacing oL=-angle of web wrinkles, theoretically 45° for this case, but actually somewhat less. In summary of the pure tension field case, the following assumptions were made: 1) The sheet carries the entire shear load. 2) The flanges are pinned to the verticals and are pinsconnected at the fixed end of the bean.(in cantilever beam). 5) There is no gusset action at the connec- tions of the vertical.f1anges. 4) The sheet immediately goes into the wave state and completely supports the applied shear by means of the diagonal tension field set up when the load is applied. Accuracy of results by the derived equations are not too consistent. For instance the equation for determining the compressive stress in the web stiffeners gives results which have been feund to be from two to five times above the actual. The other equations give reasonable approximations for the first analysis. For web thickness over about .050 inch, the re- sults are conservative and become more so as the web thickness and the flange stiffness increases. For the above reason which causes designs to be too heavy it is desired to analyze the situation with perhaps revised assumptions to obtain a.mere accurate picture of the actual stress situation. \\\x w\\ \\ \\\ /// // ///////// / 1/42 F/g. 5 r \./\. /\ \ \i / AH, - \/ 0c F/g. 5 F19- Investigations of this problem have been made and are for the most part incomplete at this time. Especially as for the experimental data which will be some time in catching up. At the present time there is very little published work to be found on this subject. A revised set of assumptions will change the picture and divide the shear stress up between the following parts: 1) Shear carried by the flanges due to their small but finite shear stiffness. 2) The shear carried by the web before buckling, as a resistant member. 5) The diagonal tension field in the web, which carries additional shear of the web. In order to account for the shear carried by the flanges, the shear flow between the flanges is assumed unifom. This shear flow is the value conventionauy calculated for such a beam at the flange rivet line by the following equation. This method was suggested by Green of Consolidated Aircraft. q = ¥9 Eq. 9 whereq shearflowinwebinpoundsperinch Y total shear in pomds Q statical moment of the flange about the neutral axis I moment of inertia of the entire beam cross section when V is not greater than the buckling strength; and eqmls the moment of inertia of the flanges only when V is sheer resisted by a diagonal tension field. The shear resisted by the web plate only, would then be given by .1. Eq. 10 V' y I/eh where h = effective depth of the web, taken from the distance between the centroids of the flange rivets. At the buckling stress of the web, the shear by the web is equal to var r: fcrht ' Eq. 11 and the stress distribution on a unit element of the and vertical (Points of shear application) is as sham in Fig. 5, in which 0' cl = 0‘“ = 0:1. The value error is given by the equation 4 where I! is'obtained from a table. It is conservatively seemed that the shear panels in such beams correspond to panels simply supported on‘ all four edges. Above the critical buckling stress, the compressive stress remains con- stant at its critical value, and any additional shear is carried by an _ increase in the diagonal tension only. The stress pattern for this case is shown in Fig. 5 and the shear carried-by the tension field only is Vt = 0‘“: ht sinoccosot- Eq. 12 where at is the angle of plate buckling. The total average tensile stress is then Vt... =52 +011 diam: Eq- 15 Inalloftheabove ithasbeenassumed thatthe flanges havebeenin— finitely stiff in bending. But if it is considered that they may defers in the direction of the web, there will be a tendency to unload the sheet between verticals owing to this deflection. This condition is taken into account by a correction factor which is a function of the flange nonents of inertia and the bean dimensions. The correction factor is given by 5 the following equations. Mme W %0'tave. ligcoshwd-coswd “'14 ‘___ __‘_ .-.-. --—-—-—t-— sin on. sq. 15 and R = correction factor a. Sechler a Dunn, "Airplane Structural Analysis and Design"(1942) pp 258 0-1; ave. average sheet tensile stress given by equations 0-1., max. maximum tensile stress developed in the sheet 1c Moment of inertia of the compression flange about it's own neutral axis. Ir Moment of inertia of inertia of the tension flange about it's own neutral axis. The correction factor B can be found either from the above equations or by extrapolation from curves which are not developed in this paper. This factor is only applied to the sheet tensile stress and to that portion of the flange load arising from sheet tension. The and load in the verticals, since it is independent of the tensile stress distribution between the panels is not affected. Because ‘of the deformation of the flanges under the action of the tension field, the correction factor must be applied to 0‘} ave to obtain the maximal value. This rivet factor is given by c = V ' V . 16 r rivet spacing Eq The maximum value of the tensile stress is then .. 0t “tm‘(’§g*’°r)%; Eq. 17 fromwhich ._ ' m , mtz ~(fim- 01.)ch Eq 18 and Vt =(0’tm-I-g-¥)Cr Bht sindcosd- Eq. 19 r lhen the maximum tensile stress equals the tensile yield point of the material, O’m, the value of the web shear above buckling is given by vty == (O‘typ “3%)01- Rht sinot cos“ Eq. 20 9 and when it is equal to the ultimate tensile strength, 02 a , the cor- responding shear is vtu = (0.31158 - Ecgjcr Bht sin 4 cos a< Eq. 21 r The total shear carried by the beam for 0‘1; max = o’typ and (rt max = o’uts' respectively, is then I Vy = (Vcr t v‘ty) 'Q‘h’ qu 22 and vu = (vc, + vtu) .55 Eq. 25 The foregoing equations are all based on vertical web—stiffening mem- bers. ‘ The case of the stiffenere not being at 90° to the flanges will not be considered in this paper. It can be said that the same method is used with an additional correction factor introduced. PART II SIMPLIFIED DESIGN OF TENSION FIELD WEBS PART II Due to the shortage of trained personnel and the urgency of the produc- tion situation in our factories, short cuts for determining approximate design criteria are at a premium. Many aircraft factories are attempting to instruct a portion of their engineering layout men in the fundamentals of structures so as to help relieve the burdens of their structural en- gineers. It is with this in.mind that this treatment of tension field beams is considered as a valuable instrument for determining a prelimi- nary design. Formation of the tension field may be visualized by considering one panel of the been shown in figure 1. Under the shear lead, V, the panel tends to deform.as it shown in figure 2. Evidently the diagonal AD is lengthened and BC is shortened as is evidenced by the dotted lines. The web carries the load, V, as tension along one diagonal and compression along the other. . Since the web is of thin sheet it will buckle as an effect of the compressive component. It then is unable to resist.any more force in that direction. This leaves the tension component to carry the shear. After the web has buckled it can be pictured as having a washboardlap- pearance with the tensile stresses acting parallel tome wrinkles, as indicated in figure 5. If an element is considered from the web as in fugure 4a, the stresses r§.and'rh are equal. These represent the horizontal and vertical shear for the web in the unbuckled state. In figure 4b another element from the some web is cut at an angle of 45°. To observe the stresses acting on this element Ichr's circle is con- structed in figure 4c. From this it is apparent that these stresses are a r—1 HH 0 —9——e o —-—- A -—— A B C o +0—0——O 3+— -————£1 M A V Figure /' AI \\\ I \\\ A ‘ “ /3 I I \ c'< \\\ c “’0 Rqafe Z compression and tension stresses of equal magnitude. After buckling, the web can take no more compression. If the compres- sion that the web has taken is neglected a free body of an element may appear as shown in 5a. lohr's circle for this case is shown in figm‘e 5b. From this the stresses on the new rectangular element in figure So can be obtained. From the solution of Hohr's circle the values for the stresses may be found as follows: 0-1 = ftcoaz 0 Ego 1 a”: o'temzo Eq.2 ”1}: 01811106080 3,-1.5 111:-0’tsin0cos0 Eq.4 Solution for ft is required. Since the vertical shear stress is fv and the vertical shear is V the following can be written. v = r, 1-. d = t(sin 0 cos 0) t d = (rt/.2) t d sin .20 Eq. 5 where t = thickness web (1 = depth of web =Z.l __1_ (rt tdxsinZO Eq‘e Tests show that the tension field Irinkles occur at an angle near 45°. Then sin 20 = l and equation 6 becomes .1... . Wt=ztd qu7 This indicates that the 1ensile stress is twice shear stress found for a shear resistant beam. This equation is an easy means for finding the approximate stress in a tension field web. FZyur’e 14- *r FI’U" 5—5 . F), are 5‘- The next problem confronting the designer is the size and spacing of the rivets connecting the web to the flanges. By plotting charts from basic equations, this can be determined readily. To find the force act- ing on any one rivet, note that tension is now responsible for stressing the web flange rivets. Figure 6 shows three common rivet patterns. The rivet pitch is defined as the distance between two rivets in one of the rows. In the calculationssDural Alclad 248T will be used for the web. The rivets are Al7ST (type AD) for diameters up to and including 5/32.. For 5/16” and 1/4" 1781‘ (type D) are used. The allowable stress values for rivets and sheet are: Al7St (Type AD) = 27,000 lb. per in.2 Ult. shear str. 1731' (five D) Alclad 24 31' = 82,000 lb. per 111.2 (PM) Ult. bearing str- = 56,000 lb. per in.2 (Ft) Ult. tensile strength. lhen using the design chart, corrections must be made for variation in 50,000 lb. per 111.2 Ult. shear str. rivet pattern. Assume one row as in figure 6a for the develOpment of the charts. Force per rivet = tension in the sheet. Fr 3 0.1; (P/fi ) t = 2 37—d- (p/ 12' ) t = fies/d) p Eq. 7 Shear on Rivet type AD Force on the rivet = Ult. single shear strength of rivet. r1. = 1r (02/4) 1?, Rewriting and substituting, V/d = 14,970 (ll/p) D Eq. 8 If the rivets are in doubleahear they will carry twice this leading. Notice that this equation is sheer per inch. a! (C) Frigur'c. 4 . Shear on type D rivets V/d = 16,630 (D/p) D Eq. 9 Bearing on the web or force on the bearing area is equal to the Ultimate bearing strength of.24st alclad. Fr=Dthr v/d = 58,000 (n/p) t Eq. 10 In any of the rivet patterns in figure 6 the full tensile load must be carried across a diagonal strip containing one rivet hole. Therefore, (the force on the sheet is equal to the ultimate tensile strength of the sheet) Fr:= lip/2 ) éDJyt Ft substituting from the above equations the following is obtained. We a 28,000 [1 - am» 1'2’] t Eq. 11 When equations (8-11) are plotted, a design chart for the web flange rivets of the tension field web is obtained. This has been done in figure 8. Equations 8 and 9 are plotted on the left side of the verb tical n/p axis. These lines will all pass thru the origin. The broken lines on this side of the chart represent rivet pitches. They were eb- tained by selecting a pitch and computing a.D/p for the various rivet sizes. The lower lines on the right hand side represent bearing on the web. They were plotted from equation 10, and also pass thru the origin. The upper side represents tension in the web. This was plotted from equation 1].. Solving equations 10 and 11 simultaneously will give a constant D/p value of .287. This is shown as a horizontal broken line in the right hand chart. For n/p values less than .287, the web will be critical in tension. The intersection of the sheet thickness lines with the broken horizontal line at .287’D/p value represents the maximum shear per inch that each particular web can carry; At the web would be on the verge of failure in both tension and hearing. In the chart the rivets are assumed to be in single shear. If they are in double shear, only half of the design shear per inch is applied to a given shear area. Therefore only half of the design shear is used on the left side of the chart. Since assuming one bearing area between the rivet and the web in drawing the right hand side of the curve, it follows that any variation from this must be compensated for. That is, in a double row of rivets there are two bearing areas available for a given shear per inch. Therefore allow for this by using only one-half the design shear per inch on the right hand side of the chart. Table I outlines the dif- ferent possibilities and indicates how to use the chart with each pat- tern. TABLE I Proper shear per inch to use with design chart for various rivet patterns. Rivet pattern Divid shear per inch by 9,3. Sjgg fiR.H. 81W Single row, single shear - - Single row, double shear 2. -— Double row, single shear ;2 2 Double row, double shear 4 2 In order to know whether a web is sheer resistant or acting as a tension field it is necessary to calculate the buckling stress. This is 5a. Q bXBmxhx \‘U\<\ ”Wk ”‘WXAMJ \‘V\<\ ~h§s Q‘M\\% i _ _ 4. A eh?) .i I i ‘0 MVLVXQNQ a _ w \\ m. fikhbhw \0 \f‘kfi: ~ . p . o 0\ . 03%“ WIIIi e... L. \ a :1 m. d 4 \\ .‘3 Q.“ 1\\ n. .6 3/ //// . . 4V . an generally done by using the dimensions of the web and Figure 9 which is taken from.ANC—5, Strength of Aircraft Elements. The critical stress which must not be exceeded if the web is not to buckle, is also given in AND-5 as, For a K E (t/b)2 Eq. 12 where Fcr is the critical shear stress for the web (f/Sq. in.) K is a constant from figure 9 E is 107 for alclad 2431* t is thickness of the web in inches b is the short side of the panel in inches If the shear stress is greater than Fcr tension field conditions result. Illustrative problem: Assume a wagner type beam, that is, the rails take the tension and com- pression. The web is then designed to take this shear. Let shear, V = 6,000 lbs. effective depth, d.=»7.94 inches assume a web thickness, t = .064 _Y... = Jim... = 15,720 lb. per sq. in. d t 7.94 x. .064 Far = 7.2 x 107 (.064/5)2 = 11,800 lb. per inth The shear stress is above F cr' therefore the web will buckle and tension field conditions exist. 6000 2 _t1 «—- = 25 700 lb. or in. From EQ. 6, Ft = 2 .064 x 7094 ’ p £1292 - l = 1.56 Margin in tension. 25,700 Design of web flange rivets. Shear per inch, 5.3.1.992 = 756 lb. per inch. 94 m Wksomk.‘ o\v moi i .u-o2< xii .25 m .y .. 3...... IMF V n-‘O—a §<°a Then from figure 8, (a single row of rivets in double shear) 756/2 = 578 lb. per inch on the left side of the figure and 756 lb. on the right side. Using 5/52" rivets, 7/8" pitch, the allowable bearing is 680 lb. per inch. This will be insufficient. By choosing a 1/8" rivet at a pitch of 1/2', the allowable bearing goes up to 880 lb. per inch, which is satisfactory. It is also observed that with 1/8" rivets at % inch spacing, a web of .051 could be used at a lower margin and a better weight efficiency. PART III CONSTRUCTION AND TEST OF A BEADED SPAR PART III Construction and Test of a Production Beaded Spar In the manufacture of airplane spars of recent design production men often ask the question, 'Could the spare be designed out of a sheet with formed bead stiffeners and thus eliminate the tedious assembly of adding the stiffener angles which are riveted to the web?" Having no figures for calculating exactly what would be the outcome, it was decided to make a test specimen to determine the feasibility of a re-design of the spars for the BZ4-E which is the Ford Motor Company version of the Consolidated Liberator Bomber. For a basis of consideration for a new spar, there could be no increase in the total weight allowed. With that in mind, a design with increased web thickness was used which would be of equal weight as the present web plus the weight of the stiffeners and rivet heads. W The static test was conducted on a cantilever spar to determine the strength characteristics. The spar was fabricated from aluminum.allqy material. The important characteristic of the spar was the .081 thick web which differed from the conventional stiffener web construction in that it had vertically beaded impressions spaced three inches apart and 3/6 inch bead depth. 30 added stiffeners were used. The web of 24S alumi— num alloy material was pressed out, using a forming die, in the "80' con- dition, then heat treated to the ST condition. The web was straightened after heat treat by attempting to stretch it, though little straightness was obtained by this method. The web was bolted to the upper and lower flanges. The wrinkles due to heat treat were eliminated after the web was bolted to the flange. The web appeared uniform in every respect. W The tested beaded web-Spar is considered unsatisfactory as compared to a stiffener web spar of equal weight. The beaded web spar has approximately 980% greater buckling strength over the flat panel shear resistant type when subjected to shear stresses. The beaded web spar failed under a load of 15,600 lbs., by the buckling of the entire web. A stiffener web Spar of the same weight failed at 27,500 lbs. However, it is noted that the critical shear resistant load for the stiffener web spar was approximately 8,000 lbs. A picture of the web tested is shown in figure 1. In order to investi- gate further this type of web construction, it is recommended that a web with greater bead depth and extension of the beads to the flanges be built. W The spar was loaded at one end by'a 20—ton hydraulic jack. The other end was bolted to an upright steel column. Deflection gages were mounted on the upper flange spaced at equal one-foot interVals along the spar. Elec- tric strain gages were placed at predetermined regions on the web and flanges totaling 48 gages.’ Gage locations are shown in Figure 2. The gages were connected to a«48 point automatic scanning recording apparatus which determined graphically the stresses in the various regions of the spar. The spar was loaded in increments of 2000 pounds until definite failure occurred. Reference: Article by William.T. Thomson in Aero Digest, May 1945, Airframe Stress Analysis by the Electrical Strain Gage. Figural Tag; results The spar failed due to buckling of the web under a load of 15,600 lbs. Just prior to this ultimate load, the web appeared perfectly'normal and showed no buckling characteristics. Diagonal web waves were pro- duced by the failing load.. The waves stretched from end to end on the web. W A complete stress analysis was successfully accomplished on the Spar by the use of electric strain gages. The direction of the principal stresses was approximately 56° measured from the horizontal base line. Being a single test, it is not possible to predict the possibilities other than they appear to be rather limited as indicated by the conclusions of this report. 1-5. 19.0 179ch = 0-110 1: .= I“? 0 a .=9== 91:13: 11b = d as s 2 F = K E (t/b) a = 70 inches or a/b = 12-7-2- : 2.92 b = 24 inches t = 0081 belles F = 5.5 x 10.5 . 106 (103.192 K = 5.5 (Ref. 1110.5) er 24 = 54.6 x 10° 1: .0000114 = 620 psi Shear stress = i- = $392 = 9182 psi P = 15000 1b. load just prior to 1.95 failure Per cent increase in strength due to the beads leoo=mxloo=980% 620 620 0 ate ese .8 o The load in the upper and lower chord flanges is given by the equation = A! M is spar moment h h is spar depth The stress then becomes Ms; c where A = area of chord flange The per cant difference between the calculated flange stress and the stress obtained from the test data is computed as follows: The slope of the (42 - 45) curve on page (0 is 1:502:43 8000 7 The slape of the calculated curve is £92: .475 8000 This is a 15.0% difference between the calculated stress and that obtained from the test data. Load 1: _x_ F = V35 1 = Location lbs o’V Span-1n h lbs . h APB,» gage 2000 86 5.16 6520 1580 #25 a #29 6000 86 5.18 18960 4500 10000 86 5.16 51600 7900 2000 54 1.97 5940 985 #42 a #45 6000 54 1.97 11820 2950 10000 54 1.97 19700 4920 2000 17.5 .656 1512 525 #18 a #22 4000 17.5 .656 2624 655 6000 17.5 .656 5956 984 A=(.625x5 2x2x2x.09 2x.081)=4.017in¥ h = 26.7 in. area of chord including flanges and web porticwl 9 a“: am ,. Calculations of the princi- v9 pal strains for the 60° M Rosette electrical strain t' gage. . ex T \ g. >\ ‘x \7 .I e. 40"" 6 )\ Y gu‘ 50‘ X Ref. 441/: ao’ Principal strains: 7 A’s e =e+A 0:2-0' P 5 fr. _ A ><\ 7‘”: °q - e - /\ ‘e es s 31- ‘1 :, u’ 6p = strain corresponding to 81) eq = strain corresponding to Sq lhere Sp and Sq are the principal stresses e = gp____eq s W , A a 22.2.0. = 31.1.? 2 5 2 cos 20' Tan 20' = 5 l - 2[§.L;_921 °:2 " °5 Knowing the principal strains from gages solve for the principal stresses 0” a‘qacl—f-zz— (¢q+ nap) u = Poisson's Ratio P = :51? (GP-i 116g), 3 = Hodulus of Elasticity laximlnshear=0’2_ (r9 2 See mmerical computation following tables of average strains. .33 no: 888 .m g .33 no: no 33 e333 .3983 D .33 not .«o 036 028:3 «3333 0 QB w .v ask-‘91 k0 taken IIIIIIIQ\|IIIIIIII' 3/ ufis rd Ham E d! 3 @i ~74 : n in m 3 San \\\ \\ \\\\\\\\lr\\\3\\\\\\ \L I %\ .L i ® é o new as 833: oasfinm I flOfluflOH . nmvrlllllnnrn noduaonnuso .3388 3&on «a nodpoonan one snow on commonpm .n on=Mfih ‘\ a \ I I m _ 4 .34/\ m .9 is .3 _ . “ o‘N/\ 4/“ .l q oVN/ I 1“ 001/ m- u- .. a \/ is a. e \ a m 53 be m .0.» \ m1< \ \ meanzm Emnmmz not 20, f. O". .‘t ..A' L C D O I .5 v I t . r V r r. . .r . . .. _ a. . . a i L. . l . _ .‘. o \ a . a1. \- . x... \. .0 . . . Ix |.,m n.. a x u . . - .\ .IJ lrh $.4 .C- .. . . \o .u .. . . . . . c . . . (K ‘\ .1: _o. a I x (2‘ pl... I. I . .0; r.. (K {1‘ a . \ x .v z . .: . -. \ . \. e . . ‘ p-IL .71\ (L II» .at\ .4!» .‘I‘M C r'\ (\ ’.. ~I.IV flp r . a! i‘! -11 3'9! I II. i. IIIuluIY' III {Ill-Iv u... 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O. . .II , . . d _ .IIIIIIIIIII; III. I . LII .II II ILLIIIII .....-II..£- III-III! III! I, IIIIII . I II... .C, O ..— - .—-.I.4- .... .L _ :J .I., .. I. .P @311...me "... ...1.I._;, 1. 74.; .13. V_. 3" ...,v' ‘/ ..Hr._“.—.. r—ooo-mo ‘*r ~m*:4—.— vv'-- --'.~ ' n. ‘s \. V\. \. .\ .‘ , - . - I . x ‘- ‘. I \» ‘\ \x \. . - ‘\. -\‘ \ "" ._ ‘ “ - ._ -__““ ssw-r». on- .—. ..- “4-....- .... ..- an 9. ‘wr‘w .-V‘..‘.~. v..- I I am i I l v— .- -..! ~'.—_-~—-___ . . ‘ .V-‘-.'~ I ~-. A. _ ~I‘ ...-um“- Q'- Tables and Computations 1 SPAR.DEFLECT10N TABLE Load Gage Gage Gage Gage Gage Gage Gage }Gage Lbs. I47 #48 #49 #50 #51 #52 #53 #54 In. In. .In.n_ In. In. In. .In. .In._i 0 0 0 0 0 0 0 0 0 2,000 .086 .075 .060 .049 .056 .026 .016 .009 4,000 .177 .155 .125 .100 .078 .054 .054 .018 6,000 .284 .245 .202 .162 .124 .089 .057 .051 8,000 .400 .386 .278 .250 .126 .127 .082 .048 10,000 .515 .445 .569 .299 .228 .166 .107 .059 12,000 .642 .560 .467 .578 .290 .211 .154 .074 15,600 Failure Tables and Computations 2 STRAIN GAGE TEST DAIA Loud Gage Gage Gage 633617 Lbs . I1 #2 #5 #4 1 2,000 90 - 80 15 40 4,000 175 -160 50 85 6,000 250 «250 70 145 8,000 560 -540 120 210 10,000 470 -440 180 280 12,000 520 -520 520 1100 Gage Gage Gage Gage #5 I6 #7 #8 2,000 - 10 155 - 20 - 140 4,000 - 10 285 - 60 - 280 6,000 ., 0 440 - 95 - 440 8,000 50 600 - 165 - 620 10,000 55 790 - 210 - 785 Gage Gage Gage Gaga #9 #10 #11 (12 2,000 110 20 - 80 100 4,000 210 40 - 180 180 6,000 ‘ 520 55 - 260 260 8,000 420 85 - 560 550 10,000 i 520 545 - 410 580 tension - compression all gage readings moasured in micro-inches per inch. STRAIN GAGE TEST DATA Tables and Computations Gage Load flange Gage Gage Lbs. #13 #14. #15 #16 2,000 - 65 + 60 0 0 4,000 - 140 + 100 0 + 5 6,000 - 200 + 150 o + 20 3,000 - 270 + 220 - 10 + 1.0 10,000 - 320 + 300 - 40 + 60 Gage Gage Gage Gage ##17 jfig: #QS’ ##20 2,000 0 - 10 + 90 - 35 4.000 O — 30 +190 -190 6,000 + 20 - 50 +280 -280 8,000 + 20 - 75 +390 —370 10,000 + 38 - 100 +1.89 4.65 Gage Gage Gage Gage #21 #22 #23 #2]. 2,000 + 20 + 30 - 100 0 4,000 + 30 + 60 - 200 0 6,000 + 45 + 80 - 335 - 5 8,000 + 76 + 1.10 - 460 - 20 10,000 + 115 + 140 - 600 - 1.5 *111 gage readings measured in nicro—inches per inch. 3 STRAIN GAGE TEST DATA 1‘ Tables and Computations :3? ”$5; “352 “$53 “$32 2,000 - 130 - 90 + 86 - 5 4,000 - 25o - 200 + 11.0 - 20 6,000 - 1.00 - 320 + 250 - 45 8,000 - 520 1.440 + 330 — 85 10,000 - 680 - 590 + 430 - 145 Gage Gage Gage 668° #22 #30 #31 #32 2,000 + 140 - 45 - 30 0 4,000 + 250 - 90 - 60 0 6,000 + 400 - 140 - 85 - 20 8,000 + 530 - 170 - 120 - 1.0 10,000 + 670 - 180 - 130 - 80 Gas. Gage Gage 688° #33 #31. #35 #36 2,000 g - 60 + 100 + 20 '- 85 4,000 - 120 + 180 + 1.0 - 180 6,000 - 165 + 280 + 60 - 280 8,000 - 235 + 360 + 90 - 350 10,000 - 240 + 470 + 125 - 480 *All gage readings measured in micro-inches per inch. 4. STRAIN GAGE TEST DATA Tables and Computations Load *Gaeo Gage Gage Gaze Lbs. #37 #38 #39 #40 2,000 + 80 0 - 110 + 60 4,000 + 160 0 - 220 + 120 6,000 + 250 0 - 350 + 180 8,000 + 340 - 5 - 1.80 + 235 10,000 + 1.30 - 20 - 680 + 270 Gage Gaze Gaze Cage #41 #42 #43 #44 2,000 - 30 - 50 + 80 - 20 4.000 - 50 - 130 + 165 - 1.5 6,000 - 80 - 200 + 250 — 60 8,000 - 100 - 300 + 31.0 - 105 10,000 - 11.0 - 1.15 + £20 - 110 688° 688° #45 #46 2,000 + 60 + 100 4,000 + 115 + 190 6,000 + 160 + 290 8,000 + 210 + 390 10,000 + 240 + 1.85 5 Tables and Computations 6 505 you «0:05:05? 5 35680.» swam a... _ am «No a + Ila”: gww am 3M + _ a ... g6.” 8mm 8“ 8.... +. 8m 1. 88 02 can + or." .. 68.» 83 8.4 8.4 + 8.4 .. 83 “8.: 3H + o3 .. 8.6 88 an 08 + 08... «R. 2.8 mm + 8 .. 88.4 83 and 8." + oma .. n5 «.3 8 + 3. . 8.8.8 n ma... mug. ‘ m 3.. om‘. 3 cud... 03 g .3 03 Tlldwqfll wfijlfiqll .Indr. nfianIIINdmIllar r on u magi 8mm 8... 8m + 8n .. an» 3. 8H u 8 .. 80.18 88 “8 08 + 88 .. ms. “.3 no .. o 8.8 33 5." 3H + on.“ .. can. an 8 .. S n 80.4 82. 8 8 + on .. m8 8 8 n S - 80.8 m 8%..de raw fifillllfll and“ .3 .3 .54.”... .3 88 . amnd 03 a ... com .. 9.9.. an...“ mm + n ma _ + _ media 88 “.8 ad + a. .. 8nd 8..” o8 + 3 + 806 .88 no 8 + on .. 8a 8 8H .+ 8 + 8..» 8n 3 8 + on u 5... 8 3H + m + 68:4 88 8 8 + S u 48 on 8 + o 8o.~ m. Mr 3 3 mm. mm. as“ ...“. Wm. lids fillemJIIdfi .. .8 113.118.1884 88 «.8 8 .. 3 .. $8 8 SH .. 8 + 80.» .5 S 8 .. o 3. 3 8 .. 8 + .8.» o o o o 08 “.8 3. u o 80.4 o o o o 0.3 ca om .. o 8.8 7.1.3.4 1.3 888 a an... a. 2.88 m 34 E. .23 55m a 3:. + o 56 .58 555 u 3. + o 58 .58.. 88 88 u «.2 82 u «.3 gnu finaONHmom no 95349500 TABLES 0F AVERAGE STRAINS Load *Gage Gage 2 + 936 = e2 81 - 82 02 - 83 ‘ e1 - e2 Lbs. #9 #36 2 82 - 83 2,000 + 110 - 85 97.5 - 17.5 + 87.5 - .20 4,000 + 210 - 180 195 - 25 + 175 - .1428 6,000 + 320 - 280 300 - 45 + 272.5 .- .165 8,000 + 420 - 360 390 - 40 + 345 - .1159 10,000 + 520 - 430 500 - 80 + 317.5 - .253 Gage Gage 810 + 938 = 83 #10 #38 2 2,000 + 20 0 10 4,000 + 1.0 0 20 6,000 + 55 0 27.5 8,000 + 85 — 5 1.5 10,000 + 345 - 20 182.5 Gage Gage °11 + 92; = 91 #11 #37 2 2,000 - 60 + 80 80 4,000 - 180 + 160 170 6,000 - 260 + 250 255 8,000 - 380 + 340 350 10.000 - g0 _+__4;0 520 *All gage readings in micro-inches per inch TABLES 0F AVERAGE STRAINS Load *Gage Gage a6 + 932 = 81 Gage Gage ' m = 81 Lbs. #6 #33 2 #8 #34 2 2,000 + 135 - 60 97.5 - 11.0 + 100 120 4,000 + 285 - 120 202.5 - 280 + 180 230 6,000 + 1.1.0 — 165 302.5 - 1.1.0 + 230 380 8,000 + 600 - 235 1.17.5 - 620 + 360 1.90 10,000 + 720 - 21.0 515 - 785 + 470 627 Gage Gage 87 + 925 = 82 Gage Gage 96 + 023 = 82 #7 #35 2 #6 #33 2 2,000 - 20 + 20 20 + 135 - 60 97 4,000 - 60 + 1.0 50 + 285 — 120 202 6,000 - 95 + 60 77.5 + 41.0 - 165 302 8,000 - 165 + 90 127.5 + 600 - 235 417 10,000 — 210 + 125 167.5 + 790 - 21.0 515 Gage Gage °8 + 931, = 03 Gage Gage a7 + 935 = .3 #8 #34 2 #7 #35 2 2,000 - 11.0 + 100 120 — 20 + 20 20 4,000 - 280 + 180 230 - 60 + 1.0 50 6,000 - 1.1.0 + 280 360 - 95 + 60 '77 8,000 — 620 + 360 1.90 - 165 + 90 127 10,000 - 785 + 470 627 - 210 + 125 167 *All gage readings in micro-inches per inch. TABLES 0F AVERAGE STRAINS Tables and Computations Load *Gage Gage 12 + 022 = e2 Gage Gage 91 "’ 026 = e2 Lbs. #12 #39 2 #1 #26 2 2,000 ’ + 100 - 100 105 + 90 - 90 90 4.000 + 180 - 220 200 + 175 - 200 187.5 6,000 + 260 - 350 305 + 280 - 320 300 8,000 + 330 - 1.80 1.05 + 360 - 1.1.0 1.00 10,000 + 380 - 680 530 + 1.70 - 590 530 Gage Gage °12 + 950 = e1 Gage Gage e2 "' 0g: = e1 #13 #40 2 #2 #27 2 2,000 - 65 + 60 62.5 - 80 + 86 83 4,000 .- 140 + 120 _ 130 .4 160 + 11.0 150 6,000 - 200 + 180 160 - 250 + 250 250 8,000 - 270 + 235 252.5 - 340 + 330 335 10,000 - 320 + 270 295 - 440 + 430 435 Gage Gage °1_l. 4‘ ‘5; = 83 Gage Gage °3 + .28 = 03 #11. #41 2 #3 #28 2 2,000 + 60 - 30 15 + 15 - 5 10 4,000 + 100 - 50 75 + 30 - 20 25 6,000 + 150 - 80 125 + 70 - 45 57.5 8,000 + 220 - 100 160 + 120 - 85 102.5 10,000 + 300 - 140 220 + 180 - 145 162.5 *All gage readings in micro-inches per inch. 9 TABLES OF AVERAGE STRAINS Tables and Computations Load *Gage Gage 912 + 923 = 02 Gage Gage °21 4’ 0g. = e3 Lbs. #19 #33 2 #21 #21. 2 2,000 + 90 - 100 95 + 20 0 10 4,000 + 190 - 200 195 + 30 o 15 6,000 + 280 - 335 307.5 + 45 0 22.5 8,000 + 390 - 1.60 1.25 + 76 - 5 40 10,000 + 1.80 - 600' 51.0 ' + 115 - 1.5 so Gage Gage e20 + ’45 = e1 #20 #46 . 2 2,000 - 85 + 100 92.5 4,000 - 190 + 190 190 6,000 — 280 + 290 285 8,000 - 370 + 31.0 355 10,000 - 1.65 + 1.85 1.75 *All gage readings measured in micro-inches per inch. 40.398 .833 swam on... 5 .380 05 you 33.89.00 8. a... 3.3 .593... «.33 sad 3 84 an 6285?... 8582. 88 38.38 a .363. dtoflvo noncommfiuam‘umo 58 a 08 u m d + o u a. mm. 688 8... 8.. a. u «6 news by» 98 «x n .no 83 38 88 8.8m 82. «.38 «..an «.88 68.8 3.: 88 88 m8 RR 848 «.8 6.8... 68.» 83 88 68« 2h 88 2.88 «.8 98m 86... 8.. 88 8.3 «.8 38 «.m«« 4.8 «.88 86.5 88 5. «R. 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