Simplified Wing Stress Analysis of a Strut-Braced Monoplane

Simplified Wing Stress Analysis Or A Strut-Braced Monoplane

The following references have been used in TABLE 1 summarizing the data applied in this article: THE CORBEN C-l "BABY ACE" (Modified version by Paul H. Poberezny) No. 1—Stress Analysis of Commercial Aircraft Airfoil section Clark "Y" (1928), by A. Klemin; Gross weight, (Wg) ...... 828 pounds No. 2—Elementary Airplane Structural Analysis by Weight of wings, (Ww) ..... 123 pounds, (See Note 1) Graphic Methods, (1938) by J. P. Eames; Net weight, (Wg-Ww) ...... 705 pounds No. 3—Air Service Information Circular No. 520, Total wing span, (S) ...... 25 ft., 9 in. (309 in.) Stress Analysis of Lieut. Phillip's "Alouette" Length of lift strut bay ...... 95 in. (See Note 2) Biplane; Length of overhang ...... 59.5 in. (See Note 2) No. 4—Procedure Handbook for Aircraft Stress An- Chord of wing ...... 54 in. alysis (1940), by Nye-Hamilton-Eames; Incidence ...... 1V6 deg. No. 5—Aircraft Structures (1950), by D. J. Peery; Area, (incl. ailerons) ...... 112.3 sq. ft. No. 6—Aircraft Design (1954), by K. D. Wood; Location of spars from the leading edge: No. 7—ANC-18, Design of Wood Aircraft Structures; Front spar, in inches—8.COO in.— No. 8—Civil Aeronautics Manuals, Volume II, (incl. in % of chord—14.8% Rear spar, in inches—38.375 in.— supplements to May 1, 1958). in % of chord—71.0% The easiest way to assimilate something new is by Center of pressure in % of chord, (See Note 3): reference to an example. Hence, the methods of analysis PHAA, (Positive high angle of attack condition) 24% used in this report will be exemplified by a wing an- PLAA, (Positive low angle of attack condition) 51% alysis of the Corben C-l "Baby Ace." Most EAA members NLAA, (Negative low angle of attack condition) 24% are familiar with this design. Another reason for select- ing this airplane is its general similarity to the design Load factors, (See Note 4): of a majority of homebuilts, namely, a strut-braced, single PHAA—4.5 bay, rectangular wing monoplane. The analysis of this PLAA—4.5 design can be used by substituting values applying to NLAA—2.0 one's own design and thus solve for stresses and design Ratio of chord to beam components, (See Note 5): components which will be adequate in strength. PHAA— —.30 This article should make it possible, when applied to PLAA— +.15 a similar design, to determine the maximum stresses to be NLAA— 0 expected under various conditions of loading and to compute sizes of members to withstand such loads. Of more Note 1—The required weights must be computed as close- importance, possibly, is the consideration of what it will ly as possible from comparison with similar air- NOT enable you to do: planes or from average figures. An average weight schedule for wings which have been con- A—It will NOT provide a method of analysis which will structed in wooden designs are: guarantee FAA approval from the analysis alone, Lightly loaded biplanes--.8 to 1.0 Ib. per sq. ft. without static or sand-bag proof loading, - - and Heavily-loaded biplanes—1.0 to 1.2 Ib. per sq. ft. B—It does NOT attempt refinements of procedure en- Semi-cantilever monoplanes—1.1 to 1.3 Ib. per abling construction of the highest efficiency to stem sq. ft. from its application. Full-cantilever monoplanes—1.2 to 1.5 Ib. per A simplified stress analysis considers only that por- sq. ft. tion which is absolutely essential in calculating the re- Metal wings for each type would average .1 to .3 quirements for a safe structure. Nothing else is possible Ib. per sq. ft. increase over the above figures. in a short article of this type. In this design, 1.1 Ibs. per sq. ft. will be used to compute wing weight. DESIGN DATA Note 2—95 inches is approximately the mean of the ac- It is usual in discussing methods of stress analysis tual lift strut bay dimensions for the front spar, to review basic mechanics and compare the analytical (actual—94.66 in.) and the rear spar, (actual— and graphical methods of solution. However, it was felt 95.25 in.), while 59.5 in. is close to the mean of that this report would be more concise and of more im- the actual overhang for the front spar, (actual— mediate use to the amateur constructor by commencing 59.84 in.) and the rear spar, (actual—59.25 in.). the actual analysis and referring to the type of solution Use of the approximate mean values will not by example. materially affect results of the final computa- Actual wing analysis should begin by listing pertinent tions, but will simplify the work. data applying to the design in question. This information is listed in TABLE 1 for the "Baby Ace." The source of Note 3—The current method most commonly used is to some of this information will be covered in appropriate compute the loading based on the elastic axis. notes following the table: • However, this is a refinement such as referred

12 NOVEMBER 1963 to in sub-paragraph B on the first page. The 4—Compute the normal chord loads. method used here, (center of pressure location) 5—Compute distribution of loads between spars. uses values obtained from airfoil wind tunnel 6—Compute moments, shears and reactions on each data from sources referred to in our report No. spar. 1, such as NACA Technical Report No. 824, etc. For the PHAA and NLAA conditions the C.P. is 7—Compute loads in lift struts. considered to b2 at the most forward location 8—Solve for drag truss loads. given in the wind tunnel data charts. For the 9—Summarize final total loads in all members. PLAA condition the C.P. is obtained by observ- ing its location at the point where the lift co- 1—Effective Span is the distance from wing tip to wing tip efficient, (C,) is 25% of its maximum value. For less any portion covered by the fuselage and an allow- some of the most commonly used sections, the ance for tip loss. The full span of the wing is not ef- values following are expressed in % of chord fective because the lift forces are greatly reduced at length from the leading edge: the tips due to air slipping off the tip portion. The decrease in effective semi-span, to allow for tip loss, C.P. Location C.P. Location is shown in Fig. 1 for the externally braced wing Airfoil for PHAA & NLAA for PLAA where the distance from the outer strut point to the Clark Y 51% tip is substantially the same as the wing chord. A Gottingen 398 30% 50% length equal to ¥4 of the overhang, (.25 L,) is sub- USA 27 29% 50% tracted from the actual semi-span to obtain the effec- USA 35B 30% 49% tive semi-span, or: USA 45 27% 38% Effective semi-span, (Se) = (95 + 59.5)—(.25x59.5)= 139.6 in. NACA M6 25% 40% 2—Normal Gross Beam Load equals gross weight of the NACA M12 25% 40% airplane divided by the effective span, or: Note 4—The load factor to be used is selected by the de- Gross beam load, (Wgb) = Wg/Se x 2 = 828/139.6 x 2 = signer. FAA specifies a minimum positive man- 2.97 Ib. per in. euvering load factor of 4.4 for utility aircraft or 3—Normal Net Beam Load equals the gross beam load 6.0 for acrobatic. The minimum negative man- minus the dead weight of the wings, dead weight of euvering load factor is .4 times the positive fac- wings, (Wwd) per inch run is computed by dividing tor for utility and .5 times the positive factor weight of wings, (Ww) by the actual, (not effective) for the aerobatic category. A positive load fac- span, (S), or: tor of 4.5 is selected for this design, as it is in- Wwd = Ww/S = 123/309 = .4 Ib. per in. run. tended to be non-aerobatic, thus giving a nega- Net beam load, (Wn) = Wgb—Wwd = 2.97 — .4 = tive load factor of 1.8 (4.5 x .4). However, to sim- 2.57 Ib. per in. run. plify computations and err on the safe side, we Check on work: 2(Wn x Se) + Ww = Wg shall use a value of 2.0. 2(2.57 x 139.6) + 123 = 840 Ibs. Note 5—Ratio of chord to beam components are posi- 840 Ibs. is 12 Ibs. more than 828 Ibs., but is satisfactory, tive ( + ), when chord loads are applied toward as it loads the wing more severely and is, therefore, the rear of the drag truss and negative (—), on the safe side. when applied in a forward direction. For the PLAA condition, CAM No. 8 in Appendix B, 4—Normal Chord Loads are computed by multiplying the paragraph .2121 states, "Although no aft acting net beam load by the chord/beam ratio: chordwise loading is specified, the structure Flight Net beam Chord/beam Chord load Load Design should be capable of sustaining aft chordwise Condition lood Rotio. per inch Factor Chord load loads." It has been the practice for many years PHAA 2.57 —.30 —0.77 4.5 —3.47 to apply the value given, ( + .15) for this condi- PLAA 2.57 + .15 0.39 4.5 1.76 tion in the case of most all wing sections. NLAA 2.57 0 — — __ Dive — — 2.28 1 2.28 SEQUENCE OF WING ANALYSIS For the Dive condition, net weight of the airplane, 1—Compute the effective span of the wing. (705 Ibs.) is distributed uniformly along the actual span. No load factor is required, hence, 705/3C9 = 2—Compute the normal gross beam loads. 2.28 Ib. per in. run. 3—Compute the normal net beam loads. (CONTINUED NEXT MONTH)

Effective wing tip Wn beaFLad Plane of Actual side of win<; tip Ww(Wing weight) k\ fuselage (8e)Effective Semi-span

Lift Strut

SPORT AVIATION 13 PART 3 Simplified Wing Strut Analysi I Strut-Brace f O s d Monoplane

Brj Noel J. Becar, EAA 725 l Rosy De Wa a6 31 n MateoSa , Calif.

HIS REPOR a continuatio s Ti f o n is belo e "criticalwth "a value s i t i , fo a fronr t strut e wilw , l investigate T Part 2 in the December, 1959 short column. If above, it is consid -7 in.—SA8 e . th 413ENo 0 steel tube issu f SPORo e T AVIATION, n thai t ere da lon g columne lengtth s f i h o L . originally called for d havin,an a g it investigates design procedures for the column in inches and p, (the let- 2.69 . majoin 7 r axis a 1.14, 3 minor the structural members analyzed for ter "rho" in the Greek alphabet) sym- axis, and a wall thickness of .065 in. stresses in that report. The refer- bolizes the radius of gyration. e w 3 Fro . mNo r referenco 2 . No e ences consulte e preparatioth n i d f o n e radiuTh f gyratio o se com b n -ca n f majofino p d r axip L/ = .406s d 2an this paper are as follows: puted as p = I/A, where, I = mo- 4 = 21 87/.406 f o p = 2214L/ n .A No. 1—Stress Analysis of Commer- men f inertie columo t th f o a n section s i greatl 0 whicn i exces15 y f ho s cial Aircraft (1928) by A. Klemin; and A = cross-sectional area of the should not be exceeded. However, No. 2—Procedure Handbook for colum n squari n e inches. Practically this plan s shorha e t jury struts con- Aircraft Stress Analysis (1940) by all of these values can be obtained necting the long lift struts to the Nye-Hamilton-Eames; from tables in various handbooks or spar at a point 35 in. inboard from No. 3—ANC-5, Strengt f Metao h l references, No. 2, No. 3 or No. 5 and e outeth r strut poin thio s t s restraint Aircraft Elements (1955 GPOy b ) . therefore do not have to be computed makes it possible to compute the value No. 4—ANC-18, Design of Wood in practice. The ratioe w L/ f pI . shoulin 7 placn i 8 d . f in o eno 2 t5 s a oL f Aircraft Structures (1951) by GPO. exceed 150 and should be kept as substitute this value in our L/p equa- No. 5—Airplane Structures, Vol. 1, small as possible. tion we get: 52/.4062 = 128, which d editio3r n (1943 y Nilesb ) . "Local instability s i failur" y b e s obviousli y satisfactory. Desig f Lifno t Struts crushing of the walls of the column. Step No. 2—As the critical L/p val- e previouth In s e analysith f o s This occurs when the wall thicknes . 413No 0sr steefo e lu from reference wing e stresseth , s which were first of the column or tube is too small No. 1, No. 2 or No. 3 is 91, this is obtained in complete form were those a value for its diameter. Tubes do not considered a long column. imposed on the external lift struts. need to be investigated for this condi- = . 2/.063—ASte t No p= D/ 5s e willW , therefore, consider firse th t e diameteth tio f i n r (D), dividey b d 30.7, this tub- ein wile t neeb no lo t d design procedure for these struts. In e walth l thickness (t) s les,i s tha. 50 n vestigated for local instability as the the following table, "computed stress- e sequencTh f operationo e s neces- t valuD/ s beloi e . (Thw50 e2 =D e mosth te severar " f thoses o e e list- sary to select a proper size tube is e equivalenvaluth s i e t round tube ed on Table 5 of the preceding ar- as follows: diameter from which the streamline ticle. 1—Arbitrarily choose a standard size tube is formed, also listed in refer- The Civil Aeronautics Manuals, Vol. tube. ences No. 1, No. 2 or No. 3). , pag 3 , specifie. 27 e No M s CA n i , 11 2—Determine whethe a lon s r i o g t i r . Ste4—ThNo p e allowable stress that a safety factor of 1.5 be imposed, short column. (Fc) for long columns is equal to: in addition to the load factor, to take 3—Determine if column should be in- Fc = 286x 1Q6 care of variation in individual lots vestigated for local instability, (L/p)2 of material from the test values given (D/t over 50). To err on the safe side, we shall for strength. This beeha s n dono t e 4—Solv r allowablfo e e stress. conside p =-L/ _r 214 e origina,th l obtai e valueth n s indicatee th n i d 5—Solv r actuafo e l stress. value obtaine d ignor an de actio th e n last column of Table A, and will be 6—Determin e margith e f safetyo n . e jurth ofy struts, = hence c F : applied to other stress values ob- If margin of safety, (M.S) is nega- 286,000,000/(214)2 = 286,000,000/45,796 tained from the previous article. tiv n valuei e , choos a strongee r 613= 5 psi. A member carrying compression is tube and repeat the calculations. Step No. 5—The actual stress (fc) is customarily a "colreferre s a -o t d n excessivela s ha t i Ify positive e desig th wherA equas i P/ n o P et l umn" d sinc e an method ,th een f o d margin, choose a smaller tub e etube th e arei nth f . o as i A stres d an s attachment affect e probleth s m con- order to save weight and recalcu- Hence, 1506/.3951 = 3812 psi. siderably, the first step is to deter- late. Step No. 6—Margin of safety is min e coefficienth e f o fixityt- re , Application of the foregoing to the Fc/fc—1 or, 6135/3812—1 = .609 ferred to as "c." "c" equals 1 for pin- "Baby Ace" design follows: and as this is a positive value, it is ended columns, and "c" equals 2 for Ste . 1—TpNo opropee checth r krfo satisfactory. restrained columns such as welded size of streamline tube to be used (Continued on next page) section f tubino s a fuselag n gi e structure. The length of the column is ex- TABLEA tremely important and the ratio L/p Typ« of Computed DESIGN STRESS is used to designate a column as Strut Stress Condition Stress in K (Comp. stres 1sx . S) "short r "long"o " , base n whao d s i t calle e criticath dp valueL/ l . This Front Compression NLAA 1,004 1,506 critical value is listed in references Front Tension PHAA 2,260 3,390 r differenfo , 3 . tNo d an 2 . No No , 1 . Rear Compression NLAA 193 289 kinds of materials. If the L/p value Rear Tension PLAA 1,760 2,640

SPORT AVIATION 13 e requirementth f thio s s stress-analy- TABLEB d giv an a epositiv s si e M.S. Note, DESIGN OF DRAG TRUSS MEMBERS however, thae arbitrarilw t y selected Computed Design r loaou ds a facto 5 r thi4. fo rs exam- Member Stress Stress Material Fc fc M.S. ple although 4.4 is all that the FAA 3 Compression Strut A-B — 114 — 171 /4"xy4" Spruce 750 psi 1 52 psi 3.93 requires. Hence, if the 4.4 factor had Compression Strut C-D —269 — 404 3/4"x3/4 " Spruce 750 psi 359 psi 1.09 been use r PHAAfo d e computeth , d Compression Strut E-F — 408 — 612 3/4"x3/4 " Spruce 750 psi 544 psi 0.38 Compression Strut G-H — 506 —759 3/i"x%" Spruce 750 psi 675 psi 0.11 stresse senoug w woullo e proo ht b d - vide a positive M.S. Therefore, this Anti-drog Wire A-D 207 311 #6-40 Tie rod 1000# 311# 2.21 wire is no doubt strong enough as Anti-drag Wire C-F 462 693 #6-40 Tie rod 1000# 693# 0.44 Anti-drag Wire E-H 731 1097 #6-40 Tie rod 1000# 1097# —0.09 originally computed by the designer. Design of Spars Drag Wire B-C 136 204 #6-40 Tie rod 1000# 204* 3 90 When a spar is subjected to com- Drag Wire D-E 365 548 #6-40 Tie rod 1000# 548# 0.82 Drag Wire F-G 539 809 #6-40 Tie rod 1000# 809* 0.24 bined bending and compression, the stress at any section is the sum of the stresses due to bending and com- Simplified Wing Stress ... No. 4 contain Figs. 41 and 2-6, respec- pression. Whe a nspa r deflects under (Continued from preceding page) tively, which give allowable column bending e compressivth , d loaen ed Performing the same steps for the r solistressefo di woops n si d columns. increases the bending stress. As it rear lift strut, which was originally Referrin o ANC-18t g , (Referenc. No e is stressed mor t i deflecte s more, in.—SA9 designe8 n a . 413 s da ENo 0 4), Fig. 2-6, we find that a spruce and the end load times the additional steel streamline tube havin a gmajo r column with an L/p value of 137m deflection increase- strese de th sd san axis of 1.686 in., a minor axis of .71 4n allowabla s ha e column stress value flection still more, until a point of in. and a wall thickness of .049 in., of 750 psi. equilibrium is reached. To determine we compute: Step No. 3—Solve for actual stress, the bending stress at this point of equilibrium we have to use the so- Step No. 2—p of major axis = .2509 . squar,in % e (fc)e a arestru f Th o :a t called "Precise Formula." 89/.250= p L/ 9= 355 . This, agains i , is .5625 sq. in., hence, fc = P/A or, in e cas th f compressioo e n strut A-B: Befor e "Precisth e e Equationn ca " greatly in excess of 150, so we apply be solved, it will be necessary to know e sam th jure eth yreasoninf o e us o gt 171/.5624 psi30 . However= 5 s a , e momenth e th f o f inertiao t ) (I , struts, and recompute for a length ther e doublar e e strut t eaca s h location e actuath , l stres. in n eac o s % h spar cross-section. Therefore, a spar in.2 r o52/.2505 o f, = 9207 . This i s section must be designed which will stil lp ratiot excessivL/ bu n , a s a e stru s hale computei t th f d amountr o , 152 psi. The other three strut loca- e winfith t g profile propeth t a er lo- e factoe on th e rs i th allowin e us s it g catio d havnan a elarg e enough value extremel w desiglo y n compression tions are computed in the same man- e sixtlisted th an hn di r columne f no of I. This is best estimated from pre- load of 289 lbs. vious experience, otherwise a guess Step No. 3—D/t = 1.25/.049 = Table B. Step No. 4—Compute Margin of must be made and the precise equa- 25.5, well under 50. tion solved after which another at- Step No. 4—Again we shall err on Safety: In the case of strut A-B, M.S. = 750/152—1 = 3.93. The remaining tempt mus e e stresmadb tth - f i eob s the safe side by using the original tained is too great. L/p value of 346, thereby giving: struts have their M.S. computen i d same manner. We will now examine the spar Fc = 286,000,000/(355)2 = 286,OCO,000/ propertie e sectionth f o sn thi - i s de s 126,00 = 0227 0 psi. Anti-Drag and Drag Wires sign. The front and rear spars have Ste5—f. pNo = 289/.184c 9= 156 2 As these members are in pure ten- the dimensions shown in Fig. 1. psi. sion, it is only necessary to look up In applyin precise gth e equationo t s Step No. 6—M.S. = 2270/1562—1 their rated strength, which can be determine maximum momene th n i t = .453. done in Reference No. 1, Table 10 span, M'--, the average axial drag load e th musw f i no t e e checse W o t k , r Referenco 50 , Tabled 2 . an 9 No es4 in the bay is used. Later, in de- struts will withstand maximum ten- giving an allowable tensile load, (F,) signin e cross-sectioth g o takt ne th e sion loads e allowablTh . e tensile yield of 1,000 lbs. The margin of safety is axial load s listea s e n Tabli dth , F e 413. No 0 E steestres SA 75,00s i lr sfo 0 computed as for the compression true axial load at each cross-section psi , which A hence r x fo ,P , F= t struts, or: F,/f, — 1 = M.S. These will be used. the front strut = 75,000 x .3951 = value e listear s n columi d . Anti7 n - Computations from here on will be 29.63 2r desig ou lbs s n.A stress i s drag wire, E-H, has a negative mar- mad . slidin e 0 ewit1 rul a ho wil s e l only 3,390 lbs., this tube is well over gin of safety, which means the next only have that degree of accuracy strengt tensionn i h . (M.S = 29,632. / larger size should be specified to meet possible to attain on such a rule. 3,390— = 17.74) . Rear strut com- putes as: 75,000 x .1849 = 13,867 lbs., compared with a design stress of 2,640 lbs., again showin n ovea g r strength condition in tension. (M.S. = 13,867/2,640—1 = 4.25). Compression Struts Step No. 1—Compute the L/p value: a squar r r rectangulaFo o e r section, p = .288d where d is the shortest side, hence, .288 x .75 = .216 and L = 29.625 in. Therefore, L/p = 29.62S/ .216 = 137. . Ste2—SolvNo p r allowablfo e e stress, (Fc): References No. 1 and

4 1 SEPTEMBER 1965 The safest and easiest method of TABLE C analyzing beams with combined ten- AXIAL LOADS ON SPARS sion and bending is to neglect the Fiight Average du e Due to decrease e tenth n bendini -o t e du g Condition Spar to Drag Lift Total sile d loainvestigatan d e onle th y PHAA Front —865 —2,019 —2,884 bending stress. This would apply Rear 368 — 386 — 18 e fron th e abovecas th f t o en spai , r PLAA Front 233 — 842 — 609 in dive condition. However, the load Rear —573 —1,576 —2,149 per inch run s foun a ,a precedin n i d g DIVE Front 268 897 1,165 article, is only —4.32 lbs., which is Rear —645 —1,137 —1,782 less than half that under the PHAA condition e investigatedb t wilo s , no l . If these I beaspar r o m sx werbo f o e Normally ,n erroi thi e b rs t wilno l the front spar is: .75 x (5.140)3/12 = construction, wit a hheavie r compres- more than 1/10 of 1 percent, which .7 x 5135.7/1 e 2th =r Fo 8.48 . in 0 sion flange on top, it would be advis- s deemei e accuratb o t d e enougr fo h rear x (3.312)3/1spar 5 .7 : x 25 =.7 abl o chec- t e in e strengt th ke th n i h these types of engineering calcula- 36.329/1 = 22.27 . Are in 0f fron o a t verted flight or dive condition as the tions. spar = b x d = .75 x 5.140 = 3.855 loa s reversei d n directioi e d th d an n The moment of inertia, (I) of a. Are in f rea . 3.31ao x sq r5 spa.7 2 = r bottom tension flange might prove rectangl : bd3/12is e . Thereforer fo I , = 2.484 sq. in. e weab compressionn o i kt . However, as these spars are of solid, rectangu- TABLE E lar construction, this reversed loading COMPUTATION OF MAXIMUM MOMENT IN SPAN chec e necessarykb t wilno l . From her, onle fronon eth y e t th spa n i r Computation Front Spar—PHAA PHAA condition will be investigated Design load) (w , 9.71 as an example, to reduce the work of J2 = EI/P 3,800 computing. wj2 36,898 M- 11,458 TABLE D 2 wj D= M— ' —25,440 FRONT SPAR LOADD SAN M« 0 PROPERTIES______D- = M- — wj2 —36,898 L/j (in radians) 1.542 Flight condition PHAA L/j (in degrees) 88.36C Total axial load, (P) . . . —2,884 Cos L/j .029 Momen f inertiao t ) (I , ..... 8.480 j L/ s DCo I —738 El (E - 1,300,OCO for j L/ s DCo — -' —36,160 spruce) 11,024,000 Sin L/j .999 j= 2EI/ P ...... 3,800 j L/ n DSi ' —25,414 j ...... 61.6 j L/ / Dn j >Si L/ = s DD=— >j Co X/ n Ta 1.422 Spa ) (L n ...... 5 9 . Tan2 X/j — 1 3.022 L/j (in radians) 1.542 Sec X/j = vTan2 X/j — 1 1.738 L/j in degrees (57.3 L/j Dl Sec X/j —44,214 n radiansi ) ...... 88.36° M' — * = DI Sec X/j — wj2 —7,316 Loa r incn pe d...... hru . 9.71 LOCATION OF MAXIMUM MOMENT IN SPAN = 1.42 j j Ta=X/ X/ n2 54.9 ° Converte o radianst d : 54.9/57. = 3.95 8 radians = 61. j 6, (fro m, .95 or Tabl 8 j . 61. x x Thereforee D) j 6 = X/ = X , 59.01 inches inboard from outer strut point.

TABLE F COMPUTATION OF STRESSES IN FRONT SPAR

Front Spar — PHAA Item Mi out Mi in M,--.. Moment (M) 11,458 11,458 7,316 y/I (y = 2.648, I = 8.480) .3122 .3122 .3122 fb = My/I (bending stress) 3,577 3,577 2,284 P (Axial load) —173 —2,582 —3,186 A (Are f sparo a ) 3.855 3.855 3.855 (CompressioA P/ f= t. n stress) —45 —670 —827 ft = fh + fc (Required stress) 3,622 4,247 3,111 Modulus of Rupture 9,400 9,400 9,400 p (46/2.18L/ 95/2.186& 6 ) 21.0 43.5 43.5 V't .99 .84 .73 F, (total allowable stress) 9,30i 0ps 8,60i ps 0 8,000 psi ft (total required stress) 3,62i 2ps 4,24i ps 7 3,111 psi Ft/ ft — 1 (M.S.) 1.56 1.02 1.57 e lowesth s A t margi f e strusafetno th t ta ypoin s 1.02i t e fron,th t spas i r definitely satisfactory. Now about that stupid homebuilt project you've been planning . . . SPORT AVIATION 15 PART TWO Simplified Win^ Stress Analysis Or A Strut-Braced Monoplane

5—Distribution of Load Between Spars in a two-spar design, such as this, is in inverse proportion to spar dis- TABLE 3 tance from the center of pressure, at which point the Calculation of Actual Moments and Reactions beam load is assumed to act: M under Actual R under Actual Flight Net load 1 Ib. Moment 1 Ib. Reaction Load on front spar: Condition Spar per in. per in. (Mt) per in. rear spar location—C.P.______

1 Ib. per inch TABLE 4 Strut V H D V2 H* D2 L2 L 59. 5" 95" Front 42.25 84.03 .92 17857061 .84 8846.84 94.06 Rear 42.25 84.62 1.70 17857161 2.89 8948.89 94.60

Fig. 2 TABLE 5 M (59.52 x .5/2)+ (59.52 .5/2 3) = 885+ 295 Axial Drag 1= x x Load spar load load-. = 1180 in. Ib. Flight Reaction in strut (Strut D/Lx(Strut M =.0 (Pin jointed). Condition Spar (RjXL/V) loodxH/L) load) 2 R! 8—1=59.5 x (l + .5)/2= 44.6 PHAA Front 1,015 2.260T 2.019C 22 S+l= —1180/95—95 x »/2 = -59.9 PHAA Rear 193 432T 386C 8 104.5=R1 PLAA Front 423 942T 842C 10 S—2=95 x %—1180/95= 35.1=R2 PLAA Rear 786 1.760T 1.576C 32 Check: R+R=104.5 + 35.1 = 139.6 Ib. NLAA Front —451 1.004C 897T —10 Loading: 44.6+95 =139.6 Ib. (OK) NLAA Rear — 86 193C 173T — 4 DIVE Front —451 1.004C 897T —10 Actual moments and reactions are computed in Table DIVE Rear 568 1.271T 1.137C 23 3 by taking "Net load per inch run" from Table 2, for Negative sign, ( — ) in last column indicates loads each flight condition and multiplying by the above NOTE: acting forward, or anti-drag loads. values at 1 Ib, per inch loading for Mt and R;

38 DECEMBER 1963 Nose-Dive Condition is illustrated in Fig. 3, where loads of the drag and anti-drag wires either by scaling off the on the rear spar are seen to be 1.26 times the loads on length from the sketch after being careful to draw it ac- the front spar and acting upwards: curately to scale or by using the formula of: Wire length equals the square root of the sum of the vertical dimension squared plus the horizontal dimension squared. i > Tail Poat r ; —— B T~~* ^ ^ Note that as each drag bay is a different length, this computation, (or scaling of the length) will have to be made - 30.375* 116 .83"———1 , • for each bay. Also compute the values of L/V and H/V to obtain the factors used in computing loads in wires and loads in spar sections. Load in wire equals the vertically Fig. 3 applied load times the L/V ratio, and load in spar equals Load on Rear Spar: the vertically applied load times the H/V ratio. These re- F.S. x 116.83 + 30.375 = 1.26 x F.S. lationships will be worked out a little later on in the detail 116.83 drag truss analysis. Next, show the known loads in the correct location and direction and assign letters at panel For equilibrium it is evident that the rear spar loads points and spar junctions to identify members as in Figs. must equal front spar loads plus the load on the tail. 5, 6 and 7. The concentrated drag loads imposed by the Front spar loads for Dive are taken as final design lift struts are obtained from the last column of Table 5; loads on that spar for the NLAA condition. the PHAA condition applying 22 Ibs. in an aft direction at the front spar strut connection and 8 Ibs. aft at the rear NLAA ft ND ND Rear Spar spar strut fitting, as shown in Fig. 4. Front Spar (F.S. x 1.26) The distributed drag loads are computed from the di- Net Load mensions shown in Fig. 4 and the "design chord load" per per inch run —4.32 5.44 inch, listed in the last column of table in paragraph 4 on M, —5,098 6,424 page 13 of November, '63 SA. As an example, at panel R, — 451 568 point A m Fig. 5, the distributed load of 113.6 Ibs. is obtained by taking the tip overhang of 9.75 in. plus half the 46 in. dimension of the adjacent panel, or 23 in., equal- 8—Drag Truss Loads consist of two parts, first the design ing 32.75 in., times the load of —3.47 Ibs. per inch run chord loads computed in paragraph 4 of the preceding obtained from paragraph 4, page 13 of November SA, text, and distributed uniformly along the entire actual for the PHAA condition. wing span, and second the drag and anti-drag loads due to the drag component of the lift struts, computed Now, let's begin the solution of drag truss loads for in Table 5 of the preceding text. The latter is a con- individual members comprising the truss. Begin at panel centrated load at the point of strut attachment to the point B in Fig. 5 and note that the 113.6 Ib. load has to spar, whereas, the distributed chord loads are con- pass directly into member B-A, since member B-C is a sidered concentrated at the panel points, (compression wire and cannot take compression and spar section B-D strut location). The load on half a panel at either side is at right angles to the load and therefore cannot take it of a panel point is considered as applied at the panel either. At panel point A we have 113.6 Ibs. pushing up. point. Loads from wing tip to nearest panel point are The wire A-D must pull down to maintain equilibrium considered as applied to that panel point. The solution in the vertical plane. The load is, therefore, 113.6 x L/V of the drag truss for PHAA, PLAA and Dive is the or 113.6 x 1.82 giving 207 Ibs. tension in wire A-D. This same, except that the distributed loads act aft for exerts a horizontal force at panel point A which is bal- PLAA and Dive instead of forward, as in PHAA. The anced by the spar section A-C pushing outward. The force drag truss would need to be solved for the NLAA con- necessary is 113.6 x H/V or 113.6 x 1.52 giving —173 Ibs. dition in a case where the lift struts are out of parallel The minus sign, (—) is used to indicate compression. with the spars enough to impose a heavy drag load. Considering panel point D we have the vertical force NLAA will not be solved for in this design due to the of 113.6 Ibs. exerted by wire A-D in addition to the ver- low maximum concentrated load of 10 Ibs. and no dis- tical panel load of 155.1 Ibs., equaling a total of 268.7 Ibs. tributed drag load is involved in the NLAA condition. This vertical load is direct compression in strut member D-C. The horizontal component of the wire A-D, 173 Ibs. Drag Truss Solution is taken in tension by spar section D-F. At panel point The PHAA condition only will be explained in detail C we have the vertical load of 268.7 Ibs. pushing up, par- as the other conditions are solved by applying the same tially offset by the 22 Ibs. concentrated strut drag load reasoning. First, sketch the drag truss outline, as in Fig. pushing down, so 268.7 — 22 gives 246.7 Ibs. This is the 4, showing physical dimensions and then obtain lengths vertical load which must be taken by wire member C-F. Load in wire is 246.7 x 1.87 giving 462 Ibs. The horizontal component of the wire is 246.7 x 1.58 giving 390 154.5" Ibs., which must be resisted in com- —J 9.7T— (H)46" (H)48" (K)45" " 5.75" pression by spar section C-E. But this P .s ' [ member must also resist the compression force of 173 Ibs. transmitted by spar section A-C, so that total com- (V) 30.375' (L) 56.8" (I.) pression on spar section C-E is 173 plus 390 giving —563 Ibs. 1/V 1.82 L/V 1.87 Considering panel point F, the hori- R.3. H/V 1.52 H/V 1.58 • zontal component of the tension in 1 wire C-F exerts a tension of 390 Ibs. •p- in spar member F-H, to which musl Fig. 4 (Continued on page 40}

SPORT AVIATION 39 SIMPLIFIED WING STRESS . . . The PLAA and Dive conditions are solved in the same (Continued from page 39) manner and the values have been shown on Figs. 5, 6 and 7, respectively. be added the tension transmitted by spar member D-F of 9—Summary of Total Drag Loads should now be shown in 173 Ibs., or a total of 563 Ibs. The vertical force of 246.7 table form such as Table 6, and the maximum loads Ibs. exerted by wire C-F must be added to the panel point selected to be the design loads as listed in the last load of 161.4 Ibs. to give a total of —408.1 Ibs. taken by column. strut member F-E in compression. At panel point E the vertical force 408.1 Ibs. is taken in tension by wire E-H, equal to 408.1 x 1.79 or 731 Ibs. The horizontal component TABLE 6 due to the wire, is 408.1 x 1.48 giving 604 Ibs., to which Summary of Drag Truss Load must be added the 563 Ibs. compression transmitted from Design spar section C-E to give a total of 1167 Ibs. compression Member PHAA PLAA Dive Load in spar section E-G. This is held in equilibrium by the re- Drag Wires B-C 0 105 136 136 action of —1167 Ibs. provided by the front spar root of D-E 0 342 365 365 F-G 0 474 539 539 the opposite wing panel. Anti-drag Wires A-D 207 0 0 207 Considering panel point H, the horizontal component C-F 462 0 0 462 of the tension in wire E-H exerts a tension of 604 Ibs. E-H 731 0 0 731 Compression Struts A-B —114 —58 —75 —114 which must be added to the 563 Ibs. tension transmitted C-D —269 —151 —172 —269 by spar section F-H to give 1167 Ibs. tension at panel E-F —408 —265 —301 —408 point H, balanced by an equal tension in the rear spar root G-H —506 —315 —366 —506 of the opposite wing panel. The vertical component of Front Spar A-C —173 0 0 —173 C-E —563 88 114 —563 408.1 Ibs., of wire E-H added to the panel point load of E-G —1167 377 422 —1167 98 Ibs. gives 506.1 Ibs. compression in strut member H-G Rear Spar ...... B-D 0 — 88 —114 —114 which is held in equilibrium by an opposite reaction of D-F 173 —377 —422 —422 506.1 Ibs. provided by the cabane attachment structure F-H 563 —769 —868 —868 of the fuselage. NOTE: (— ) indicates compression.

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40 DECEMBER 1963