PART TWO

Simplified Win^ Stress AnalysisOr A Strut-Braced Monoplane

5—Distribution of Load Between Spars in a two-spar de-sign, such as this, is in inverse proportion to spar dis-tance from the center of pressure, at which point thebeam load is assumed to act:

Load on front spar:rear spar location—C.P.________

(in %) rear spar location—front spar location or:for PHAA: 71—24/71—14.8=84% Front spar.

Load on rear spar:C.P.—front spar location_______

(in %) rear spar location—front spar locationPITA A - 9d__IdR/T!__14 Q_1ft«- Root' ci

'v f i ̂ ai opai luuaiivju——nuiil opal iu\.aLluu Ifor PHAA: 24—14.8/71—14.8=16% Rear spar.

For PLAA: Front spar: 71—51/71—14.8 = 35%Rear spar: 51—14.8/71—14.8=65%

For NLAA: Front spar: 71—24/71—14.8 = 84%Rear spar: 24—14.8/71—14.8=16%

or:

BEAM LOADS

FlightCondition

PHAAPHAA

PLAAPLAA

NLAANLAA

Spor

FrontRear

FrontRear

FrontRear

TABLE 2PER INCH RUN ON SPARS

Net loodper in.

on Wing

2.572.57

2.572.57

2.572.57

LoodFactor

4.54.5

4.54.5

2.02.0

% carriedby Spor

84%16%

35%65%

84%16%

Net loadper in. run

9.711.85

4.057.52

—4.32—0.82

6—Moments, Shears and Reactions are computed easiestby placing a unit load on the spar and then increasingthe results obtained in direct proportion to the actualnet loads per inch on the spar for each condition asin Fig. 2:

1 Ib. per inch

59. 5" 95"

Fig. 2

M1 = (59.52 x .5/2)+ (59.52 x .5/2 x 3) = 885+ 295= 1180 in. Ib.

M2=.0 (Pin jointed).8—1=59.5 x (l + .5)/2= 44.6S+l= —1180/95—95 x »/2 = -59.9

104.5=R1

S—2=95 x %—1180/95= 35.1=R2Check: R+R=104.5 + 35.1 = 139.6 Ib.

Loading: 44.6+95 =139.6 Ib. (OK)

Actual moments and reactions are computed in Table3 by taking "Net load per inch run" from Table 2, foreach flight condition and multiplying by the abovevalues at 1 Ib, per inch loading for Mt and R;

38 DECEMBER 1963

TABLE 3Calculation of Actual Moments and Reactions

FlightCondition

PHAAPHAA

PLAAPLAA

NLAANLAA

Spar

FrontRear

FrontRear

FrontRear

Net loadper in.

9.711.85

4.057.52

—4.32—0.82

M under1 Ib.

per in.

11801180

11801180

11801180

ActualMoment

(Mt)

11,4582,183

4,7798,875

—5,098— 968

R under1 Ib.

per in.

104.5104.5

104.5104.5

104.5104.5

ActualReaction

<v1,015

193

423786

—451— 86

Note that NLAA flight loads are negative as they areacting down, in inverted flight. These loads put com-pression in the wing struts and as they are fairly long,this is usually the critical design condition which willbe solved for next.

7—Lift Strut LoadsAt the point where the lift strut attaches to the wing,the reactions, (R,) computed in Table 3, pull up inthe PHAA & PLAA conditions and push down in theNLAA condition. This puts the struts in tension whenpulling up and in compression when pushing down.Strut loads are obtained by multiplying the reaction,(Rj) by the length of the strut, (L) divided by the (V)component as listed in Table 4. The axial load on thespar can be obtained by multiplying the strut loadby its horizontal component, (H) divided by its length,(L). If the struts slant either away or toward eachother so that they are not parallel to the spars theywill impose drag and anti-drag loads on the wing. Inthis design, the struts are out of parallel with thespars by a very small amount but the loads are figuredin the last column of Table 5 in order to demonstratehow it should be done.

StrutFrontRear

V H42.25 84.0342.25 84.62

TABLED V2.92 1785

1.70 1785

4H*

70617161

D2 L2.84 8846.84

2.89 8948.89

L94.0694.60

FlightCondition Spar

PHAAPHAAPLAAPLAANLAANLAADIVEDIVENOTE:

FrontRear

FrontRear

FrontRear

FrontRear

TABLE

ReactionR!

1,015193423786

—451— 86—451

568

5

Loadin strut

(RjXL/V)

2.260T432T942T

1.760T1.004C

193C1.004C1.271T

Axial Dragspar load load-.

(Strut D/Lx(StrutloodxH/L) load)

2.019C386C842C

1.576C897T173T897T

1.137CNegative sign, ( — ) in last column indicatesacting forward, or anti-drag loads.

228

1032

—10— 4—10

23loads

Nose-Dive Condition is illustrated in Fig. 3, where loadson the rear spar are seen to be 1.26 times the loads onthe front spar and acting upwards:

r ,; —— B

- 30.375*T~~*i^>^•

Tail Poat

116 .83"———1

Fig. 3

Load on Rear Spar:F.S. x 116.83 + 30.375 = 1.26 x F.S.

116.83

For equilibrium it is evident that the rear spar loadsmust equal front spar loads plus the load on the tail.Front spar loads for Dive are taken as final designloads on that spar for the NLAA condition.

Net Loadper inch run

M,R,

NLAA ft NDFront Spar

—4.32—5,098— 451

ND Rear Spar(F.S. x 1.26)

5.446,424

568

8—Drag Truss Loads consist of two parts, first the designchord loads computed in paragraph 4 of the precedingtext, and distributed uniformly along the entire actualwing span, and second the drag and anti-drag loadsdue to the drag component of the lift struts, computedin Table 5 of the preceding text. The latter is a con-centrated load at the point of strut attachment to thespar, whereas, the distributed chord loads are con-sidered concentrated at the panel points, (compressionstrut location). The load on half a panel at either sideof a panel point is considered as applied at the panelpoint. Loads from wing tip to nearest panel point areconsidered as applied to that panel point. The solutionof the drag truss for PHAA, PLAA and Dive is thesame, except that the distributed loads act aft forPLAA and Dive instead of forward, as in PHAA. Thedrag truss would need to be solved for the NLAA con-dition in a case where the lift struts are out of parallelwith the spars enough to impose a heavy drag load.NLAA will not be solved for in this design due to thelow maximum concentrated load of 10 Ibs. and no dis-tributed drag load is involved in the NLAA condition.

Drag Truss SolutionThe PHAA condition only will be explained in detail

as the other conditions are solved by applying the samereasoning. First, sketch the drag truss outline, as in Fig.4, showing physical dimensions and then obtain lengths

—J 9.7T—P .s ' [

154.5"(H)46" (H)48" ( K ) 4 5 "

(V) 30.375'

R.3.1/V 1.82H/V 1.52

(L) 56.8"

L/V 1.87H/V 1.58

(I.)

Fig. 4

of the drag and anti-drag wires either by scaling off thelength from the sketch after being careful to draw it ac-curately to scale or by using the formula of: Wire lengthequals the square root of the sum of the vertical di-mension squared plus the horizontal dimension squared.Note that as each drag bay is a different length, this com-putation, (or scaling of the length) will have to be madefor each bay. Also compute the values of L/V and H/Vto obtain the factors used in computing loads in wires andloads in spar sections. Load in wire equals the verticallyapplied load times the L/V ratio, and load in spar equalsthe vertically applied load times the H/V ratio. These re-lationships will be worked out a little later on in the detaildrag truss analysis. Next, show the known loads in thecorrect location and direction and assign letters at panelpoints and spar junctions to identify members as in Figs.5, 6 and 7. The concentrated drag loads imposed by thelift struts are obtained from the last column of Table 5;the PHAA condition applying 22 Ibs. in an aft direction atthe front spar strut connection and 8 Ibs. aft at the rearspar strut fitting, as shown in Fig. 4.

The distributed drag loads are computed from the di-mensions shown in Fig. 4 and the "design chord load" perinch, listed in the last column of table in paragraph 4 onpage 13 of November, '63 SA. As an example, at panelpoint A m Fig. 5, the distributed load of 113.6 Ibs. is ob-tained by taking the tip overhang of 9.75 in. plus half the46 in. dimension of the adjacent panel, or 23 in., equal-ing 32.75 in., times the load of —3.47 Ibs. per inch runobtained from paragraph 4, page 13 of November SA,for the PHAA condition.

Now, let's begin the solution of drag truss loads forindividual members comprising the truss. Begin at panelpoint B in Fig. 5 and note that the 113.6 Ib. load has topass directly into member B-A, since member B-C is awire and cannot take compression and spar section B-Dis at right angles to the load and therefore cannot take iteither. At panel point A we have 113.6 Ibs. pushing up.The wire A-D must pull down to maintain equilibriumin the vertical plane. The load is, therefore, 113.6 x L/Vor 113.6 x 1.82 giving 207 Ibs. tension in wire A-D. Thisexerts a horizontal force at panel point A which is bal-anced by the spar section A-C pushing outward. The forcenecessary is 113.6 x H/V or 113.6 x 1.52 giving —173 Ibs.The minus sign, (—) is used to indicate compression.

Considering panel point D we have the vertical forceof 113.6 Ibs. exerted by wire A-D in addition to the ver-tical panel load of 155.1 Ibs., equaling a total of 268.7 Ibs.This vertical load is direct compression in strut memberD-C. The horizontal component of the wire A-D, 173 Ibs.is taken in tension by spar section D-F. At panel pointC we have the vertical load of 268.7 Ibs. pushing up, par-tially offset by the 22 Ibs. concentrated strut drag loadpushing down, so 268.7 — 22 gives 246.7 Ibs. This is thevertical load which must be taken by wire member C-F.

Load in wire is 246.7 x 1.87 giving462 Ibs. The horizontal component ofthe wire is 246.7 x 1.58 giving 390Ibs., which must be resisted in com-

" 5.75" pression by spar section C-E. But thismember must also resist the compres-sion force of 173 Ibs. transmitted byspar section A-C, so that total com-pression on spar section C-E is 173plus 390 giving —563 Ibs.

Considering panel point F, the hori-• 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

(Continued on page 40}

SPORT AVIATION 39

SIMPLIFIED WING STRESS . . .(Continued from page 39)

be added the tension transmitted by spar member D-F of173 Ibs., or a total of 563 Ibs. The vertical force of 246.7Ibs. exerted by wire C-F must be added to the panel pointload of 161.4 Ibs. to give a total of —408.1 Ibs. taken bystrut member F-E in compression. At panel point E thevertical force 408.1 Ibs. is taken in tension by wire E-H,equal to 408.1 x 1.79 or 731 Ibs. The horizontal componentdue to the wire, is 408.1 x 1.48 giving 604 Ibs., to whichmust be added the 563 Ibs. compression transmitted fromspar section C-E to give a total of 1167 Ibs. compressionin spar section E-G. This is held in equilibrium by the re-action of —1167 Ibs. provided by the front spar root ofthe opposite wing panel.

Considering panel point H, the horizontal componentof the tension in wire E-H exerts a tension of 604 Ibs.which must be added to the 563 Ibs. tension transmittedby spar section F-H to give 1167 Ibs. tension at panelpoint H, balanced by an equal tension in the rear spar rootof the opposite wing panel. The vertical component of408.1 Ibs., of wire E-H added to the panel point load of98 Ibs. gives 506.1 Ibs. compression in strut member H-Gwhich is held in equilibrium by an opposite reaction of506.1 Ibs. provided by the cabane attachment structureof the fuselage.

The PLAA and Dive conditions are solved in the samemanner and the values have been shown on Figs. 5, 6 and7, respectively.9—Summary of Total Drag Loads should now be shown in

table form such as Table 6, and the maximum loadsselected to be the design loads as listed in the lastcolumn.

TABLE 6Summary of Drag Truss Load

Member

Drag Wires

Anti-drag Wires

Compression Struts

Front Spar

Rear Spar . . . . . . . . .

NOTE: (— ) indicates

B-CD-EF-GA-DC-FE-HA-BC-DE-FG-HA-CC-EE-GB-DD-FF-H

PHAA

000

207462731

—114—269—408—506—173—563

—11670

173563

PLAA

105342474

000

—58—151—265—315

088

377— 88

—377—769

Dive

136365539

000

—75—172—301—366

0114422

—114—422—868

DesignLoad

136365539207462731

—114—269—408—506—173—563

—1167—114—422—868

compression.

22<f PHAA R= 506.1

E -1167 0 f R= 1167

-113.

R= 11C7

113.6*

57.63 92.7* PLAA

Cf 88 E lF.S.

-57

R.S.

Fig. 6

74.7* 64.4*

F.S.

-74

R.S.

Fig. 7

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40 DECEMBER 1963