Auburn University Libraries
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' 3 1706025851275 0
AIR CORPS INFGRMA ION CIRCULAR
PUBLISHED BY THE CHIEF OF THE AIR CORPS, WASHINGTON, D. C.
Vol. VI November I, 1927 No. 599
DESIGN OF
TAPERED INTERNALLY BRACED WINGS
(AIRPLANE BRANCH REPORT)
Prepared by E. C. Friel
Materiel Division, Air Corps
Wright Field, Dayton, Ohio
April 15, 1927
UNITED STATES
GOVERNMENT PRINTING OFFICE
WASHINGTON
1928
Ralph Brown Draughon
UBRARY
MAY 2 8 2013
Non·Depoitorv
Auburn University
I N DE X
Article No. Title P age
1. Design characteristics ____________________________________________ _________________________ l
2. Computation of loading cur ves _____________________________________________________________ 2
3. Distribution of loads bet ,Yeen pars  
4. Computations of shears and moments_______________________________________________________ 9
5. Computations of t rial spar sizes _________________________________________________ ·___________ 13
6. Spar properties ______________________________________ _ __ ______________________ 14
7. Computations of deflections at the N strut__ ____________________________________ _____________ 16
8. Computation of Nstrut equali zing forces____________________________________________________ 20
9. Computation of stresses in N strut ____________________________ ______ ____ ____________________ 25
10. Design of N strut__ _______________ ______ _____________________________ __ ______________ __ ___ 26
11. Computation of drag t russ loads____ ________________________________________________________ ' 27
12. Design of drag wires ______________ ____ ___________________________  ___  31
13. Design of drag st ruts or compression ribs____ _____________________ ___________________________ 33
14. Computat ion of loads in cabane structure______________________________________________ ______ 34
15. Design of cabane members__________ ___ __ __________________________________________________ 38
16. Design of spars_ __ ____ ____ __ _________ __ ___ ________________________________________________ 39
17. Investigat ion of spars for horizontal shear ___________________________________________________ 42
18. Conclusion__ _____ _____ __ __ ______ __ __ __ ___________________________________________________ 42
CERTIFICATE: By direction of the Secretary of War, the matter contained herein is published as administrative
information and is required for the proper t ransaction of the public business.
(n )
DESIGN OF TAPERED INTERNALLY BRACED WIN
SCOPE OF REPORT
D uring the pa t few y ars, more or less data have
been p ublished on the design and anal \"Sis· of extern all v
braced biplane wi ngs. Very little, h~wever, ha bee~
published on the design of tapered internal ly braced
wings, especially such data as would include the latest
rules for computing loads, as well as recent method
of design. Many of the points of design and methods
of analysis used for external!_\' braced wings a re equallv
applicable to the design of internally braced wing ..
I n general, however, the methods of anah· is a re d ist
inctly different, and consequently it wa~ consiclcrecl
desirable to publish a separate report on the problems
involved in the design and analysis of tapered internally
braced wing . It is interesting to note that two
a irplanes of comparatively recent design, namely, the
Curtiss P 1 and Boeing PW 9 pursuit a irplane ,
though of the externally braced biplane type, have
structures of such a character that some of the methods
used in the de ign of internally braced wings can be
advantageously combined with those used for externally
braced wings in making an analysis of t he struct
ure. A number of the methods of analysis and design
given in this report are not exclusive featu res of the
internally braced wing design, but are applicable to
any design. A few of these a re: T ripod cabane
analysis by the " Method of determinants"; analytical
analysis of parallel chord tru ses; internal drag t rut
design, etc.
Many of the method of analysis used in this report
have already been publi heel in Air Service Information
Circula rs No. 213, " Deflection of Beam of onuniform
Section," by . S. Niles, and No. 440, " Design
of Internally Braced Bip lane Wings," by A. S. Niles.
T he latter circular gives an application of the methods
of design and analysis of internally braced wings to
the design of the PW l A a irp lane. Since the publication
of this circular several years ago, a number of
changes and expansions of the ru les for computing
loads have been made, thus requiring modifications
of the method of anal.vsis used there. It is the object
of this report to present as clear!.\· as pos ible an
example of design in which all such changes are incorporated
and in which a ll rules that are appl icable to
t he design are in accordance with those specified in
P ar t VI of t he append ix to Volume I of Airpla ne
Design. It shoulcl be noter! at this point that a ll
references to ru les for Loads will be made to Volume
I of Airplane Design rather than to t.he fl andhook for
Airplane De ·igncrs, a · it is bclic,·cd that the former
is more generally acccssih!c than the latter.
It was considered ctc. irablc to select as · n example
for design and ::i.nalyHi an a irplane wh ich had most
of the feature~ typical o.f the inte rnall_,. bra('e'l biplane
type as weU as some that are not usually incorpor.i terl
in such design. For example, most airplanes of th is
type have a variation of both wing chord and depth
of spar with span. The latter featu re i sometimes
obtained by varying the chord but keeping the same
section throughout the " ·ing, the ordinates at anv
section being the same in percentages of chord as thos~
at the root section. It oRen happens, however, t hat,
as in the U. S. A. 35 and U. . A. 45 a irfo ils, the ai rfo
il section is not kept con tant; that is, the t hi ckneos
ratio varies along the span. This introduces a compl
ication d ue to the fact that a variation in the t hi ckncs
· ratio is believed to involve a variation in the lift
coefficient. I n computing the loadiJ1g cu rve a long
the span, it is, therefore, 11eccssary to take account
of two variable instead of one, na~ely the variation
of lift due to variation in area per inch run and t hat.
due to variation in lift C'Oefficient. VVliat the true
variation of lift coefficient along the span actually is
in the case of a \\;ng with double taper is imknown, but
tests ham indicated that it i not in direct proportion to
the th ickness ratio as pecificd in the loading rule
used in this report. The most important data on
t his problem are t ho c given in N. A. C. A. Report
229, " Pressure D istribution over T hick Tapered Airfoils,
N. A. C. A. 1, . S. A. 27C Modified, and U.S. A.
35," by El li ot~ G. Reid. T hese data and thei r bearings
on the rules for computing loads are discussed in
paragraph 67 of erial Report 2'7 7, "Study of Pressure
Distribution Data,'' by A. S. Niles, to which tb.e reader
is referred. Although the rule that the lift coefficient
be assumed to var.1· in direct proport ion to the thicknc
s ratio "·ill probably be changed in t he near future,
it is probable that there wi ll always be certain types
of design for which it will be specified that the lift
coefficient shall be as urned to vary according to some
definite rule. In such cases the method of analysis
will be practicall.v the same as that shown in this
report, the chief difference being in the shapes of the
cun·c of variation of lift coefficient used. As it w a~
desired to illustrate the method of analvsis to be used
in such cases, the cell ulc chosen for. the numerical
example was of the double tapered type and the ex isting
loading rules gi,·en in Part VI of the appendix to
Airplane Design were fo llowed. This cell ule is that
of a hypothetical de ign simi lar in externa1 features
to the HuffDaland AT 1 training a irplane. It is
sufficie ntly different in a number of respects. however,
t hat many of the conclusions formulated from t he
result of the analysis would not be per t inent if appliect
to the AT1 clcs'.gn.
ARTICLE 1. DESIGN CHARACTERISTICS
In the analysis of an.1· airplane there i a large amount
of data such a . \Ying areas, weights, dimensions, etc.,
which arc necessary for a number of the preliminary
(1)
2
computations. The first step in making an analysis
should be the collection of these data, usually called
the design characteristics, in a table for ready reference.
Table 1 gives these characteristics for the design
selected. As stated before, the external dimensions are
similar to tho e of the A T 1 airplane.
'l'able 1. Design Characteristics
Name _____ ___ _____ ____ ___ • ___ _
Type ______ __________ ___ ____ _ _
Gross weight_ __________ __ _____ _
Disposable load _________ ______ _
Weight empty ____ ___ ___ ______ _
Weight per square foot _ _______ _
Engine ________________ _______ _
Type ___________ ____ __ ___ ____ _
Rated horsepower ___ _____ ____ _ _
Weight per rated horsepower_ __ _ _
Weight of engine p er hor epower__
Weight of power plant per horsepower
_______ __________ ____ _ Wing section ___ __ __ _______ ____ _
Vfo1g area (in cluding ailerons) :
Upper __ _____ ____________ _
Lower __ ___ __ ___ __________ _
Span:
Upper wing ____________ ___ _
Lower wing __________ ____ _
Chord:
" I carus."
Training.
2,338 pounds.
744 pounds.
1,589 pounds.
10.98 pounds.
Wright model E.
v, water cooled.
180 at 1,800 revolut
ion per minute.
12.99 pounds.
2.66 pounds.
4.27 pounds.
. S. A. 35.
138 square feet.
75 square feet.
29 feet, 4 inches.
23 feet, 8~ inches.
Upper wing __________ _____ 43.3 in ch e s60
inches.
Lower wing _____________ ___ 33.3 in ch e s  53
r
Stablizer area ______________ __ _
Elevator area _______ ______ ____ _
Fin area __ ______ __ __ _____ __ ___ _
Rudder area __________ ________ _
Mean gap ________ ____ ________ _
Stagger (distance from L. E.) __ _ _
Stabilizer setting with reference to
inch es.
23 square feet.
20 square feet.
2 square feet.
8 square feet.
59 inches.
12 inches.
propeller axis _ _ _ _ _ _ _ _ _ _ _ _ 0°.
Location of C. G. in per cent M.
A. C    ~     39.4.
Width of fuselage ___ __ ____ ____ _ 2 feet, 6 inches.
Distance from C. G. to elevator
hinge (fully loaded) _____ ______ 16 feet, 5 inches.
Distance from C. G. to rudder
post_ _ _________ _____ ________ 16 feet, 17 inches.
ARTICLE 2. COMPUTATION OF LOADING
CURVES
The loading curves of beam and chord forces a re
computed according to t he rules outlined in Part
VI of Volume I of Airplane Design. The necessary
steps in the computations are given below in the
same order as in the text mentioned above.
1. Compute or estimate gross weight of airplane.
2. Compute effective span of each wing.
3. Compute or estimate dead weight of wings.
4. Determin e " equivalent wings. "
5. Compute relative efficiency of wings.
6. Compute normal gro s beam load on each wing.
7. Compute normal net beam load on each wing.
8. Compute normal chord load on each wing.
Item No. 5, the relat ive effic iency of wings, has
been changed slight ly from the original rule for tapered
wings in order to simplify computations and more
closely approximate what are believed to be t he actual
conditions. The comp utation and determination of
the above items will be taken up in the following
paragraphs.
The estimated gross weight of the a irplane is taken
as being equal to the actual gross weight of the AT 1
a irplane, 2,33 pounds. This is a logical assumption
to make, because, as stated before, the external dimensions
were assumed to be very similar to those of the
A T 1. In actual practice the estimation of weight is
a eparate proce s involving a number of assumptions
and computations. The description of thi p roce s
is not given here because of lack of space. Chapter IV
of Airplane Design however, gives a fairly complete
discussion of the problem.
The method of computing the decrease in effective
span to allow for t ip loss is given in article A42 of the
appendix to Airplane Design, Figure A 7 being used
for externally braced wings and Figure A 8 for
internally braced wings. Figures 1 and 2 of this report,
which follow, give the principal dimensions of
the wings as well as the computations of the effective
spans and equivalent wings.
The estimated weights of wings are obtained by
assuming the weight per square foot and multiplying
by the areas of the wings. The weight per square foot
is some assumed value comparable to that of a design
similar to the one under consideration . For this design,
the unit weight was taken as 1.33 pounds per square
foot. Therefore the weight of the upper wing is
138.0X 1.33= 183 .5 pounds, and that of the lower
wing is 75.0X l.33 = 99.7 pounds, where 138.0 and
75.0 are the areas in square feet (obtained from Table 1)
of the upper and lower wings, respectively.
When a wing varies in thickness ratio along the span,
the lift coefficient and, consequently, the air loads are
assumed to vary in the same proportion. (See article
A 42 of Airplane Design, p. 446.) When computing
the loading curves an "equivalent wing" is used
instead of the actual wing. The equivalent wing is a
wing of constant thickness ratio, the span of which is
the same as the actual wing, but the chord at any
section is equal to the t rue chord multiplied by the
thickne s ratio at t he section and divided by the
thickness ratio at the root section, or section of maxim
um chord. The variations in thickness ratio for the
upper and lower wings are plotted in Figure 3.
The chords of the equivalent wing are computed for
four sections, AC, DE, GH, and IJ (the section of
maximum chord or root) . The computations of these
chords are given in Table 2.
3
tlj)jler Winq
1Yf'echve Span <$ L'tjviva/enl Milt{'
lip Len9lhLI
As
Fron/Spar
G 
~ Ef'f'ec/Jve SemiSp.m·/70.8"
~ ~:/76'~
Ef'f'eclive Sj)On
11,P LenqllrLI. :IJF=.lJE = AC cor : .JJ.81.511..500 = 0.314. COTIJ: 3..5/.8 = 00361
1COTa:COTIJ' I a: 176.0  3/.5 , 176
4,J..}  ~ "o. " l0./.34QOJ6/  o8Z79  .52..J W.52.
.£1'techveSemi5pa17 = j01L1 =/760./x5Z =176.5.Z ' =/ 708."
.Ef'f'ecliveSpan: 2x t7o.8"=J41.6"
Efvi vo/enl W!i79
Chords o/'l:tjvivo/m} W!(lq"Heasured Iiom L.E. of' Aclva/ M/Jq.
.l)osh Line Shows TE o/".Ewiva/enl Winq."
Flo. 1
1. 66.Z'"~+
6  ').1f_f::~0f,.  a/e 1r /ifl!J._.   •' •11:1 I
. tdq;J.SF' ?' f 1\l I
rraih'!J. s: ,, __ !:: I _J_ I
1T+=== ~_....__ !~::,/~.+   ;'+
~ I
• ~ I
I
I I
I A.2 I A.f
H I  _} ~·
l/?3.8Z5=J''/Y.:..._<~NJ"
iJ =14zj ·~·i
Ef'rechve S/Jo/7
7/n Le,n/7'>i;=l}'r=nr= AC corcr = ?~iz5a.5oo 01· 2 ~~ c:or·A 7. 7s 1.t1o a,,,..166
'.I' " 'f 'rll I ij,t;, /COTa:COTJJ, 14225Ji'.625 = . ..,..,..,, Jo/ /~ZZS/4 . .375 .vr•
J.3.J J3.3 =40.5"
1o.1:J.J1ao466 aBzoo
El'f'eclive Semi 5pon = ja/L; =/42.25'!.4. 05'::1.38. Z "
.E!Yec/ive 5po11 =2x/.38.2=276.4"
L:y>v1va/e/ll W!i'l~ "
C"7ort/J o/' E<!uiva/e/11 Mi"l7,'.' ore #et7svreo' from i..f. or Ac/val Mo0
.!Jo.5h hoe Shows TE o.rL:pu1ya/enJ Mi"?§'."
FIG. 2 Lower wing
4
TABl,E 2.Conipulations of tlze chord~ of lhe equivalent this assumption entails but a small error is shown at
wings section GII, where the trailing edge (straight line be
UPPER WING tween the equivalent chords at sections DE and IJ)
Section I
,\ ctual Thick:
Station chord ness ratio
(inches) 3 ~i~;cdeviates
but little from the actual equivalent chord
'rhick Cbord or point at that section, the chord measured to the trailnessratio
eq ui va ing edge being 56.3 inche , while the computed chord is
at root lent wing
section (inches) 55.5 inches.
AC' __ ____ _______ ___ DE __ ___ ______ ____ li6. 000 4:J. 3 o n ________ ____ ____ 124. 000 52. 0 70. 000 61. 5 ]J_ _________ _______ 31. 500 68.0
LOWER WTNG
.\C ___ __________ ___ 14.2. 250 I
DE ........ _____ _._ JOI. 750
OH .... ____________ 65.000
lJ _    • 32. 625
I 43.3 X l1.56   6. ·1
18
_
18
 21. we 1es.
I
33. 3
40. 5
47. 25
53.00
U.56
13. 95
16. 40
18.18
Jl. 56
13. 65
15. 55
l .18
!
IJ
18.18
18.18
18.18
18.1
18.1
18. 18
1 . 18
18. 18
I
I 27. 6
39. 9
55. 5
68.0
21. 2
30. 4
40. 4
53.00
!F/7 ~
The problem of determining the relative effi ciency of
wings is much more difficult of solution for tapered
wings than for con tantchord, constantsection wings.
sufficient data are available regarding the effect of wing
interference on the distribution of lift between wings
of equal and constant chords to permit the formulation
of a rule for effi ciency of such wings that gi\·es fairly
satisfactory results. The only question that might
be raised about t his rule i why the gapchord ratio and
stagger a rc measured at a section midway between the
side of the fuselage and the effective t ip .· The effecth·e
tip is selected as one point of reference because the lift
I
      ,_  > 
~ '~ ,., nt • I p, ~~,~ ,,JJ, "'  ,,.,, ~ I ~
I <;,  < ..,, LJ . ..J ,/
_ , . ,,.,, ;.,_ l/.R V.R/ ~"'. ~
7/1 : ,,
I I TJn iP/ I/,;., .., _ '/ ' A ~'i ~,q ,,,, J J.. T I_ ,c,, ~,;., 1// ift. ~~ Ir
I rL r I ~ .....
I I r..._ ~  : 11  u ..   ~ 
 ~ I  I     r L,... r~ rY I I
~ J.4 I I .;.. K. ~ 1c? r1 I I
~ ~ ~ I I rr ~ r:
'.~ "' ~I  ~II
. 11 r,....._ I  I 
~ ,~ I  11
~I t r9 >
,,, ,§ ~: ~ I I I
~"' r;:i , "; I I I
11 t
' ! 1
R ~I 11 1  t.:: :; I ~I i

 ~~ 1 ~ 1
,c. I v I  I .3 I
 5? I S I ~ I s:
~I ct I ~ I

I d.. .!: ~
I I ;. I r I : ,_ I
? I t.: I ~ 1 .
! I I I : I II
II
i) I ~
f I
tf1 'I? c~? (.,'{) JI? o? .s~ A IO /. 'O l.V J'. lO ;.ro /; 0 A O I V1 l76
I kJ,: ,{7, •ce in /np~t ,, tro1 ?J 'la; e Jr s~ m1 ?el ry
FIG. 3.
Figures 1 and 2 show t hese chords of the equivalent
wing plc?tted at the corresponding ections. It " ·ill
be noted that the trailing edge of the equivalent wing
a d etermined by t he above chords does not have a
traight line variation . This is due to the fact that
the true wing is tapered both in plan and thickness
ratio. (Note that t he product of two straight li ne
variations is a parabolic variation .) For ease in computing
the loading curv es, howeYer, straight line variations
a re a umed between the tip and section DE
and between DE and the maximum ection IJ. That
over all of the portion inboard of it may be considered
uniform. The gap is measured at the midway section,
thus obtaining the average gap, in order to take care
of any variation that might result from a difference in
dihedral a ngles of the wings. In general, the rule had
to be made defi nite and yet to cover as many cases as
po sible.
There a re, however, few or no data available on
the effects of interference on tapered wings. Because
of this lack of data, t hen, t he problem mu t be handled
by finding some condition of constant chord wing
which may reasonably be conside red equivalent to
that of the existing tapere<l wing design. The original
rule given in Part VI of the appendix to Airplane
Design is an approximation of that condition. The
rule states that the gapchord ratio and stagger shall
be measured at a section determined by the i11 tersection
of the a irfoil sections with a secant plane
parallel to the plane of symmetry and passing through
the centroid of the effective area of the smaller wing;
that is, the area inboard of the effective t ip, rath er
than by a secant plane passing through the midway
point. The rule is somewhat ambiguous in thait
there are three possible interpretations as follows:
1. The centroid may be considered as being that
of the effective area of the actual wing, and the effective
chords those determined by the intersections of the
secant plane t hrough the centroid with the actual wings.
2. The centroid may be considered as being that
of the effective area of t he equivalent wing, and the
effective chords those of the equival en t wing at t he
section through the centroid.
5
3. The chords mii,y be considered as t hose of the
actual wing at the ection made by the secan t plane
passing through the centroid of the effective area
of the equivalent wing.
The first of the above interpretations is based upon
the assumption that the interference effect is a function
of the taper and independent of the thickness ratio.
The last two interpretations assume that the interference
effect is a function of both taper and thickness
ratio. Inasmuch as the effect, if any, of thickness
ratio upon interference is unknown and must certainly
be small compared to that of taper on gapchord ratio ,
the first interpretation will be assumed correct (the
interference is independent of the thickness ratio) .
Consequently, the rule given on page 445 of Part VI
of the appendix was changed to read as given in the
footnote on page 4461 and the rule at present reads as
follows : " In computing the relative effic iency, the
gapchord ratio and stagger shall be determined by
the dimensions of the intersections of the actual wings,
not the equivalent wings, with a vertical plane parallel
to the plane of symmetry and passing through the
geometrical centroid of the effective area of the smaller
actual wing."
The relative efficiency is computed below as specified
by the modified rule. The computations for the other
cases mentioned are not shown, as they are of no particular
interest.
From Figure 2, the chord at effective t ip of actual
lower wing= 34.0"; the theoretical chord at side of
fuselage=56.3"; the ratio of chords=;~:~= l.65.
Knowing this ratio, the distance from the center
line (C. L.) to the center of gravity (C. G.) may be
found from Table A 4 of the appendix to Airplane
Design, which gives the location of the centroid of a
trapezoid.
For ratio 1.65, the distance from the effective tip to
the C. G. in percentage of span is given as 54.08. The
effective span is 123.825 inches. Therefore, the distance
from the effective tip to the C. G. is equal to
0.5408X 123.825=66.99 inches, and the distance from
the C. L. to the C. G. is (123.825 + 14.375)  66.99=
71.21; say, 72 inches.
Chord of actual lower wing at station 72 = 46
inches.
Chord of actual upper wing at station 72 = 61
inches.
Mean chord=0.5 (46 + 61) = 53 .5 inches.
Mean gap = 62.0 inches.
Tan 8 =lr'2"= 0.113, 8 = 6° 30'.
Gap/chord ratio= M Gaph d
62
·
0
1.16.
ean c or 53.5
From Figure A 5, page 440, Airplane Design
Ratio Load~ng upper "~ng l.lO.
Loadmg lower wmg
Figure 4 is a diagram of the intersections of t he wings
with a vertical plane passing through the geometrical
centroid of the effective area of the lower wing. The
FIG. 4
stagger as measured from the leading edge is 12 inches,
while that measured from the third points of the chords
as shown in the figure is only 7 inches. It is, therefore,
very important that the stagger be obtained according
to the method shown in the figllJ'e in order to obtain
the correct relative efficiency.
The formula for normal gross beam load for a \\·ing
tapered in plan is given in Part VI of the appendix,
page 446, as follows:
W1g=eA~+.Ai' and W 0.=e W1•
where Au and A1 are the effective areas in sq uare feet
of the actual upper and lower wings, respectively, and
w1• and W "" are expressed in pounds per square foot.
When the wing is tapered in thickness ratio also, the
above formulas are used, but A. and A1 are the effective
areas of the upper and lower equivale.nt wings respectively.
The effective areas of the equivalent wings are found
by computing the effective areas A1, A2, etc., of the
five segments into which the wings ar~ arbitrarily
divided and obtaining the total area by summation.
The dividing lines bet'Yeen the segments are shown in
Figur.es 1 and 2,. the lines being denoted by the letters '
DE, GH, IJ, and LM. The area of the t ip section
extends only to the effective t ip. The computations of
the effective areas are given below. The chords of
the eq uivalent wing are obtained by measurement
from Figures 1 and 2, but in a few places, as explained
before, t here is a difference due to t he assumption of a
straigh t line variation in chord length.
Effecti ve area of UJJper equivalent wing
Square
inches
Arca A1 = 0.5X l3.75 (42.5+42.0) = 581. 0
Arca A2=0.5X l 7.75 ('l2.0+68.0) = 976. O
Area Aa=0.5X38.50 (68.0+56.3) = 2, 393. O
Area A,=0.5XM.OO (56.3+ 39.9) =2, 597. 0
Area A; = (525.2) (39.9+ 28.8)X0.5= 1, 608. 0
Total a rea =8, 155. O
Therefore, Au=8,155+ 144= 56.63 square feet.
Effective area of lower equivalent wing
Square
inches
Area A1=0.5X9.25 (40.5+40.l) = 373. o
Area A2=0.5X9.00 (40.1+53.0) = 419. O
Area A1=0.5X32.375 (53.0+ 42.4) =I, 544. 0
Arca A,=0.5X36.750 (42.4+30.4) =I, 338. 0
Area A,=0.5X36.450 (30.4+ 2"2.2) = 959. O
Total area =4, 633. 0
Therefore, A1= 4,633 + 144 = 32.17 sq uare!eet.
w 2 338
y=2=1,169 pounds, e= l.10
Then, from the formula given above
6
1,169
(l.10X 56.63) +32.17 12.38 pounds per square
foot
and
W ug = eW1. = l.lOX 12.38 = 13.62 pounds per square foot
The ruling on t he "Effect of taper in t hickness
ratio," given in article A 42 of Airplane Design, page
446, has been modified by t he following addition to
paragraph [(12)]: " In certain cases, such as those of
metal or plywood covered wings where the weight of
the covering forms a large percentage of the total
weight of the wing , the Materiel Division may require
that the dead weight of the wings be assumed to be
distributed uniformly over the actual a rea of the wing
instead of being varied in proportion to the thickness
ratio." ·
From this modified rule it is evident t hat there are
two methods of computing the normal net beam loads.
Non;nally these loads are obtained by assuming that
the weight of wings per square foot varies as the thickness
ratio and subtracting the weight of the wing in
pounds per sq uare foot of equivalent wing area from
the normal gross beam loads in pounds per sq uare foot.
The alternative method is to consider the weight of
wing independent of the thickness ratio, the loading
curve being computed upon the basis of the actual wing
area. The normal net beam loads in this case however
can not be obtained by subtracting t he weight of wing~
per square foot from the normal gross beam loads,
since these loads are computed upon the basis of the
equivalent wing area and not the actual area. The
procedure used is to compute curves of normal gross
beam loads and weight of wings in units of pounds per
inch run and, by subtracting of ordinates, obtain the
normal net beam loads. It was decided to use t he latter
method for this design inasmuch as it gives lightly
more co nservative result. and as it illustrates the more
involved type of co mputations which must sometimes
be used. Ordinates for the curves of normal gro, s beam
loads and weight of wing. in pounds per inch run are
computed below for both upper and lower wings. The
normal gross beam load is obtained by multiplying the
chord of t he equivalent wing at the section where the
load ing is desired by the load per square foot and dividing
by 144. The weight of wing per inch run is obtained
by multiplying the chord of t he actual wing by t he
weight of the wing in pounds per square foot and dividing
by 144. . The computations of the normal gross
beam loads per inch run are given in T able 3, and the
weight of wing per inch run is computed in Table 4.
T ABLE 3.Computation of gross beam loads per inch run
. chordX w.. 13. 62
Upper wrng TV.. 144 chordX144= 0. 0946Xchord
. chordX lf'tc 12 38
Lower wmgWi,= 144 ch ord X~ 0 .0860 Xchord
Station
0
13. 75
31.50
70. 00
124. 00
170. 80
14. 375
23. 625
32. 625
65.000
IOI. 750
138. 200
PPER WING
Section
Center line._·        · 
Ll\L ..   ·      ·       · ·
IJ Gl· l· ·_·__ ·__ ____ _________·__·__·_ _·_______·_ _
DE   · ·   ·· ·    ·  
Effective tip.·     ·· ·  · ···
LOWER WING
Root. . ... _. __ _ ._. ____ . ______ ___ __ _ •..
1
1
LM .. ·   · ·    ·  ·  · · · IJ
   ·            · ·GH
. . .   ·      ·    ·DE
····  ···· ·     ·   
Effective tiP    · ·  1
Chord of :r~obe.~
equivalent load per
wing inch run
42.5
42. 0
68. 0
56. 3
39.9
28.8
4400.. 5I 1
53.0
42.4
30.4
22. 2 I
4.02
3. 97
6.43
5. 33
3. 77
2. 72
3.48
3. 45
4.56
3. 65
2. 61
I. 91
TABLE 4.Computation of weight of wing per inch run
Upper and lower wingschord
X Ww 1.33
lVuw= W1. 144 =chordX 144 0.00924Xchord
UPPER WING
Station Section
0 Center line .. . ..   ·      · ·
13. 75 LM . . ·       ··     · 
31. 50 IL   ·  .... · ··_. __
70. 00 GH . . . .  · · ··
124. 00 I DE···· · · ·    ·     · 
170. 80 Effective tiP·   ·   ···
LOWER WING
14. 375 RooL  ··   ·         · ··
23. 625 LM· ·       · ··· 
32. 625 IJ _ • . ·· ..     ·   ·
65. 000 GH .... ··    ·  · ·  
113081.. 725000 DEfEfe .c.t. ive tip. ______________·_ ______·__ _
Wu.,orw1w
Chord of weight of
actual wing wing per
inch run
42. 5
42. 0
68. 0
61. 5
52. 0
44.0
40.5
40.1
53. 0
47.25
40.5
34.0
o. 393
.388
. 628
. 568
. 480
. 407
0. 374
. 371
. 490
.437
. 374
. 314
The curves of Figure 5 are plotted from the ordinates
of the above tables, normal gross beam load and weight
of wing per inch run being plotted against span. In
plotting such curves the first step is to draw a base
line AB and plot, on ordinates below that line, the curve
DC (weight of wing per inch run). Then the curve
DC, in turn, is used as a base line from which values
of normal gro beam loads are plotted above the line
on ordinates perpendicular to AB to give the curve
EF. The values of normal net beam loads may now
be obtained by graphfoal subtraction; that is, the normal
net beam load at any station is the value of the
7
/
Area imder net beam load curve of lower wing
A 1=0.5 [40.5 (1.22 + 2.236)+ 69.125 (2.236 +4.07)+
9.00 (4.07+3.079).1 9.25 (3.079 +3.106)]
= 0.5 [139.68+ 435.90+ 64.34 + 57 .22]
= 348.57 pounds.
Actual net weight of airplane
Gross weight = 2,....33 . 0
Weight of wings= 2 3. 2
Net weight = 2, 054. pound
ordinate from base line AB to curve EF at that station . Check
It is desirable at this point to check the curves by 2 [A +A ] t 1 t · ht
computing the areas under the normal net beam " 1  ac ua ne weig = er ror.
load curves of both wings, adding them, multiplying 12 [684.65+348.57)  2,054. = 2,066.44  2,054.8 = 11.64,
the sum by 2, and ubtracting t his product from the or 12 pounds.
nq.s
Normal Gross G' #el ./3eam Loads and
Wei9hl ol Mi7q Per .1nch Run.
llorizonlo/ Sco/er :i'O"
Verlica/ Scale 1"=2.0#
Q3T4
"~°
"i
~
~ Ii:)
§
~
~
~ \)
~ ~
~ "l
")
~
Q437
~ ~ f ~ \) ); 'l ~ ~ ~
~ ~ \) ~
'f ") ")
~ ~
l>Jw '.l)
049 a371 a37I
Lower W!i79
F IG . 5
actual net weight of the airplane, which value is equal
to gross weight of ai rplane minus weight of wings. The
difference indicates the magnitude of the error; that
is, a difference of zero indicates an exact chec k. T he
following computations illustrate the cheek outlined
above:
Area under net beam load curve of uvper wing
Au= 0.5 [52 (l.76 +3.29)+ 92.5 (3 .29 + 5. 02)+ 17.75
(5.802+3.582) + 13.75 (3 .582 +3.627)]
= 0.5 [262.60+841.02+ 166.56 + 99.12]
=684.65 pounds.
81700282
The small error of 12 pounds indicate that no large
error has been made in computing the loading cu rve .
The normal chord loads per inch run are found by
multiplying the normal net beam loads by the ratios
of the chord to beam component (C/B ratios) for the
high and low incidence co nditions. The C/B ratios for
various airfoils a re given in Table A 3 , page 448, of
Part VI of the appendix.
For the . S. J\. 35 airfoil
C/B in II. I. =  0. 166
C/B in L. I. = 0.150
C/B in I. F. = 0.000
TABLE 5.Compittation of normal chord loads
UPPER WING
Normal chord loads
Station Section
Net
beam
loads H. I.= 0.166B L. I. =0.15B
0
13. 75
31. 50
70.00
124. 00
170. 80
Center line ______ _
LM ______ ________ _
JJ_ _______________ _
GR ______________ _
DE ______________ _
Effective tip _____ _
3. 63
3. 58
5.80
4. 76
3. 29
2.31
LOWER WING
14. 375 Root _____________ _
23. 625 LM ______ ________ _
32. 625 IL ____ ____ _______ _
65. 000 GH ______________ _
101. 750 DE ______________ _
138. 200 Effective tip_
Sia 116 110.0
3.11
3.08
4.07
3. 21
2. 34
1.60
0.603
. 594
. 963
. 790
. 546
. 383
o. 516
. 511
. 676
.533
.388
. 266
124.0
o. 545
. 537
. 870
. 714
.494
. 347
0.466
. 462
. 611
. 482
.351
.240
8
2. Divide the a rea into a number of segments, t he
most suitable width of each segment being 10 inches,
as the later computations of shears and bending
moments are simplified by the use of this interval.
3. Plot the curve of normal net beam loads below
the· planform of the wing, using the same horizontal
scale as is used for the wing. Any convenient vertical
scale may be used provided that it be of sufficient
magnitude to permit a fair degree of accuracy in
scaling off the ordinates.
4. Construct ordinates on the loading curve to
repre8ent the average loading on each segmen t of the
wing. For most cases these ordinates may be taken
halfway between the side of each segment, the error
involved through the use of this point rather than the
center of gravity of the area of the segment (which is
usually trapezoidal in shape) being negligible. These
ordinates a re shown in Figures 7 and 9 by dotted lines,
the full lines representing the segment divisions.
700 .315 13.75
U,o,Per Wi/79 1 Horizon/a) Sco/e1"=20"
Vedica) Sco/el''=O.o#
'
I
10/. 7.50 65000
LowerW!/77
FIG. 6.Normal chord loads per inch runhigh incidence
These values of normal chord loads are plotted in
Figure 6 to give curves of normal chord load per inch
run . Although the values of the chord loads for low
incidence as well a. high incidence were computed, it
was considered s ufficient to plot high incidence curves
only, since t he low incidence chord load at any station
may readily be obtained by multiplying the high
in cidence value obtained from Figure 6 by the ratio of
0.150
the C/B values or 0_166=0.904.
ARTICLE 3. DISTRIBUTION OF LOADS
BETWEEN SPARS
Distribution of load between spars of tapered wings
is perhaps best obtained by the following procedure:
1. Lay out the planform of the wing to scale, locating
the spars and the center of pressure curves for high
incidence and low incidence as shown in Figures 7 and 9.
5. Redraw the planform of the wing, or at least that
portion of the wing which will include both the center
of pressure curves and the spars, using a scale for the
chord direction that will give 100 units between the
spars. The advantage of such an arrangement is due
to the fact that it permits a direct reading, in percentage,
of the load carried by each spar. Any convenient
horizontal scale may be used. With reference to
Figures 8 and 10, for obtaining the high incidence
loads on the spars, scale No. 1 and the high incidencecenter
of pressure curve are used for the front spar,
and scale No. 2 and the same curve for the rear spar.
For obtaining the low incidence loads, the procedure is
the same except that the low incidencecenter of pressure
curve is used.
6. The remaining computations are best made in
tabular form. Referring to Table 6
Column 1 gives the stations at which the load distributions
are computed.
Column 2 gives the interval in inches between
stations.
Column 3 gives the load per inch run obtained from
the normal net beam load curve of Figure 7.
Column 4 gives the net load considered as concent
rated midway between sections, and each value is
the product of the corresponding values in column 2
and column 3.
Column 5 and column 6 give the percentages of load
taken by the front spar in high incidence and the rear
spar in low incidence, respecti,·ely, the values being
read directly from Figure 8.
Columns 7, 8, 9, and 10 give the loads on the front
and rear spars in high incidence and low incidence,
column 7 being the product of coiumns 4 and 5, and
column 10 the product of columns 4 and 6, while columns
8 and 9 are obLained by subtracting columns 10 and 7,
respectively, from column 4.
The computations for the distribution of load between
the spars of the lower wing are similar to those
of the upper wing and, hence, need no explanation.
T able 7 gives the computed results.
TABLE 6.Computation of loads on front and rear spars
UPPER WINO
.Q
0 gJ .s '"O' Per cent Load on front Load on rear
.g .... § spar load spar spar
.s """ "0 " .s "' .... .... 'O .,;
<i " μi ...i " :s .,
~ I> 0 .s .... "" . il .... .... .... .... .... ., 21 ~ =
li' " 0 00 .:i z !>..<.. p":l μi ...i μi ...i
     
1 2 3 4 5 6 7 8 9 10    
176
12 1.94 23.28 55.0 77.6 12. 80 5. 22 10. 48 18. 06
164
10 2. 25 22. 50 54.2 80.1 12. 20 4. 48 IO. 30 18. 02
154
10 2. 55 25.50 53.5 82. 5 13. 64 4. 46 11. 86 21.04
144
10 2. 84 28.40 52. 9 84. 7 15. 02 4.35 13.38 24. 05
134
10 3.13 31. 30 52.1
124
87.0 16. 31 4. 07 14.99, 27. 23
2 3.30 6. 60 51. 7 88.4 3. 41 0. 77 3.19 5.83
122
10 3.48 34. 80 51.3 89. 7 17. 85 3. 58 16. 95 31.22
112
10 3. 75 37. 50 50. 5 92. 0 18. 94 30. 0 18. 56 34. 50
102
10 4.02 40.20 49.9 94. 2 20. 06 2.33 20.14 37. 87
92
10 4. 30 43.00 49. 2 96.5 21.16 1.50 21. 84 41.50
82
10 4. 57 45. 70 48. 5 98.8 22. 16 0. 55 23. 54 45. 15
72
10 4.84 48.40 47. 8 101. 0 23.14 0. 48 25.26 48. 88
62
10 5.11 51.10 47.1 103. 3 24. 07 1.69 27.03 52. 79
52
10 5.39 53.90 46.5 105.5 25. 06 2.96 28.84 56.86
42
10. 5 5.66 59.43 45. 8 107. 9 27. 22 4.69 32. 21 64. 12
31. 5
8. 5 5. 26 44. 71 45. l 110. 0 20.16 4.47 24. 55 49.18
23
9. 25 4. 15 38. 39 44. 5 112. 0 17. 08 4.61 21. 31 43. 00
13. 75
13. 75 3. 60 49.50 43. 7 114. 7 21. 63 7.28 27. 87 56. 78
0  c c
684. 21 331.91 352. 30
9
TABLF. 7. Computation of loads on front and rear spars
LOWER WINO
I .c Per cent Load on Load on "' 0 ."g .: 'O"' spar load front spai; rear spar .: .... " " " 
.: "" 0
"' "" ....
'O .,,; ;ii
,..;
c Cl ":s = ...i
~ t 0 .s ~ " ... ,_; ,_; = 21 "" ~ 0 "' ,_; ...;
00 ..c.. i z" ~ p":l μi ...i ~ ...i
   ·1  
l 2 3 4 5
_ 6 _
1
_ 1_·_ '_8 _ __ 9_
10
 
127. 875 I
JO I. 36 13. 6 61. 3 71. 4 8.34 3.89 5. 26 9. ii
117. 875
10 J.60 16. 0 60. 8 73. 5 9. 73 4. 24 6. 27 11. 76
107. 875
9. 5 J.84 17. 4 60. 25 75. 7 JO. 53 4. 25 6. 95 13. 23
98. 375
11 2.10 23. I 59. 9 78. 0 13. 84 5. 08 9. 26 18. 02
87. 375
9. 375 2. 36 22. 12 59. 25 80. 3 13. LI 4. 36 9. 01 17. 76
78
JO 2. 61 26. 1 58. 9 82. 5 15. 37 4. 57 10. 73 21. 53
68
10 2.88 28. 8 58.4 84. 7 16. 82 4. 41 11. 98 24. 39
58 110 3. 14 31.4 57. 9 86. 9 18.18 4. 11 13. 22 ~ 7 . 29
48
110 3. 41 34.. 1 57. 4 89. l 19. 57 3. 72 14. 53 30. 38
38
10 3. 67 36. 7 57. 0 91. 4 20.92 3. 16 15. 78 33. 54
28
9. 75 3. 94 3 . 42 56. 5 93.5 21. 71 2. 50 16. 71 35. 92
18. 25
9 3. 56 32. 04 56.0 95. 5 17. 94 I. 44 14. 10 30. 60
9. 25
9. 25 3.09 28.58 55. 5 97.6 15.86 0. 69 12. 72 27. 89
0  ~ ~
348.44 201.92 146. 52
ARTICLE 4. COMPUTATIONS OF SHEARS AND
MOMENTS ON SPARS
The complete computations of shears and moments
on the front upper spar in high incidence are given in
Table 8. One set of computations was considered
sufficient to illustrate the methods involved and, therefore,
only the results of the computations for other
loading conditions and spars are given. Table 9 gives
the e results, the moments and M/I values, for the
upper wing in high incidence and low incidence and
for a 100pound load acting at the Nstrut point,
station 122. Table 10 gives the same for the lower
wing, the station of the Nstrut load, however, being
98.375 instead of 122. The method of determining
the moment is described on pages 12 to 18 of Airplane
De ign. The computations involved in this design
are, however, simpler than those used in the example
of the above reference due to the fact that th ere are
no concentrated loads acting on the spars.
Referring to Table 8
Column 1 gives the station or section at which the
shear and moment are computed.
Column 2 gives the distance between adjacent
sections.
Column 3 gives the load between adjacent sections
as given in Table 6.
Column 4 gives the shear to the left of the section;
that is, the sum of the values of column 3 for segments
t o the left of the section in question.
'
I h '; !
. I ~ I: /if' ,.., ,d''"'
fn. io1 l,,.V,Y, 7
10
I I I I\
I I I I I : ~
I I I I I I I I I I r..
,_,+tt0+0>it+.~;r;J,_.._ __ j 111++:+++l+' ++hl+++'I 'H+,l .Hc+.1++Hl+++H++rl +t,~~+++++l
,_,+o+t1+ifititil..,____,•+ I lrir11rrrr11rtrl rtri~l rrt'lrrtr•l_rtrr1 rti:trrrrlrrn1 ' ii+"i"'F't11
.,, __ 1,. , _ _I I I I I I I I I I : I II I
......  t I + 
~ .. ~ > •111+lr++H:+++':+IH ll+tH: H++;+++1~ H+~ltt1':t++.l+t+lr+++i
1 ~ rr. P/71 ~ ra1 +'l'll,1'4+'l'+l+41+ ++,f+++'l l++1...++++'+ +l'1'+1_,__.._1+ ++1"+++'l+*l ++++1~ 141
,~· I "'" I II I I I I I I I I I I :
FLl/.d~~ ~~+'14+++++!jf+I++  ..,..  I I I 1 I I l'lo.
5: .., J A L l '   ,.,,~,; ~ ..,.. I I I I I I I I 11 '
. ·v·, J: . T"" 1 : I +'1t+
1+l++•'+++'+_,_11t+.1+1+<1>+_.1,+<+:i1_,_+1
~ i.. I I I I I I I : I I I I I
l?t '"" ,__   I : I : ; I : I I I II I
~  I I i I I I I I I I I
1: I I I I I I I : I I : I
/, '{}
FIG. 7
I l'Ji ~ 17 I  ~   ~ _ J r:;, "'" ~; '>r  Im I ·  "' _,,Ac· I  l
lrfn 1. 11,~~ ' I ri I I I  MO
I I .,
~ · '..,_ l ... H'" I I I I I
I I I u.  rT'" I I I I I I I ·
I ~ l..J.. I I I I
 ~    ~ I I I I I I
/ ... 11 ~
~, I I I I I I I I I I 
I I  ;. I I I
I I I I I I I I I I r~
~  ._.  I I I I I I I
I I I I I I I 1.
.~  · _L I L I I I I I I I I I I I I · ~
~' f f ...._ I I I I I I I I I I I  /
~ I  I I I I I I I I I I I I I ~ /
~ I I I I I I I I I I I I I I  '"" ~
~  I _L L. ,.j ' .} ;p, I
I
I
I I I I I · c.i
~  I . I I I I I I I I I ~ ., I I ~  I I I I. I I I I I I I I Ar  ~ ~
~  I I I I I I I
I
I
I I I I  ~ I
' I I I
I I
 I I I I I I I ,., ,. ~
~ 'V I I I I I I I I I I I J I I  r\i
~ I I I I I I I I I I ,,,.. I I I I ,~,. .,
<>  I I l+ I I I I I I I
,.._
~ I I I I I
I I I I I
· I I I I I ')  I I I 11 I I I I I I I I I I I v
I I I
/f l(')
I I I I 11; v J l..t >.:: ,_I I I I I I I I I I "l/'l
/,'(} ;, 'J.'} ; :19 1W I~ ~ II/ /0. ~l ~l 7l ~l ~l w ~?. If ' "7..f. r ii w ~ ~
h't ~/i'.· '(')f ~lo. )/ l < "4'1 :? 1~ ?O ' liV/. cu J ( cc ve tj =~ 4"'
Fro. 8
11
I ;; •/?lQ / ·~ ·d ht? ,..;, ".L ev>1il ' "" r,')
..... u ..... w, .,,,,  J
I I ! ' ['\,., ~k 1fN •//,H I I
I I I t\. I I
r· r II"' I ! I
I I I I I I  I
I Vj I I I 1,.. ~I I I I I I
If f+ ~ I
.. ·       ,• J I I I ,, ~· I I I I I I I I /,
I ...,, ,
1~
1,1 .. ~~l; ..·y. h. !I I I I I I ,J ~
I I I I I I I I I II
I I I
I N.... I
I I )/~ re ii ~~r I I I I
I
/ .. ;,:, fi._L .... I I I I : I
I I II
'"22 ~""
I ., I
~. " ' r''"" 'll ~ ~ ¢8 V? ?.!? ,,_  ~ '

,J._ i I I'\
I I ,.,_  I I'\ I
>'N ,,, 11 /I /pf ,!/t 1L ~,v i.rr I I ..,. I I I >nn ~ I
"' ~ ..... rr I I I I ,,, I I ~ I I
..,.~ .~ h"" I u I
I
I I I I
 I I I
 ~ ,.. h"' I ,  ~·
I I
I
I I • I I 1 ' .: l.l I ~ .v :ii ,~; ( ~: "' <f)I \ ., ( ii "' '~· ' I " I Q)) ~ <: '• IS \:) ~ ,., . "•.
~: r;' 'f;; "' 'll ,., ..... ~ , J: ll "I i1 I I I
. ~ ..IZ i?.B, 5° I ~"" lf !?.1 '?J. ,Z.8, '5"' ~6; ro v.J 63 ~ ¢..} .u Z./?j ;. i/{5 !.~ ?,f :;1
FIG. 9
r   rID· 7 1 h  c
~. iP [:;; 1F 1 P.or ,o, ta """
11~ ,,,.,,
~L ~ ria ~ ?£... r.m ke ,, ,,:,..,,
11n b/p r .?1 IJ" ///'
II J.· ,"), ""/ I> dt>"
II? " "" fl}'
I I I I I I I I I I I I _
_,
l//1 I I I I I I I I I li f"" I Q/1
l I I I I I I I _lj... T : I I t:;
.~ 17/1
I I I
I
I ~ ,1 I I I I I R/J I ~
h_J I I I R_ cJ b.lH....... I I
I I I I I I !:...
~ [q,i Lrr I ~· 1 I I
I I I I •m ,,..,co
'  I I I I I I I I I I I ~
~ ld/l ~+ I I I I I I I I I 'L/ I°'\~
~ I I I I I I + I I I I I . ~
·!:;i l.<rr I I I I I I I l"!i': ~
Sjl I I I I I I I I I I I <:l I
~ IL/1 I I Ir I? ~ l< JV: I I I I '4/
...
~ r
I ~ I I I I I I I I I I I \'<
I
~ 17/J I I I I I I I I I I I ['l/'. ~
1rSi I I I I I
I I I I I e I
~~ I I I I I I I [?/'. J
I I I I
I I I I I I I I I I
1   ~ I
[Qr. I I I I I I I I I I II ~/: : I I I I I I I I II I I
VN I I I I I I 11nw Si J<!?i I I I I ()
.M?.~ Z5 v.c.: r7.f m '?f« IP.8. lf tfZ6. lf fZJ 63 w !t3 .iu ..:11/47 ~;:: .10.J
'
FIG. 10
12
Column 5 gives the change in moment between article 6, pages 38 43, in clusive. The values of M/I
sections due to the load acting between sections. are used in article 7 in computing the deflections.
Column 6 gives the change in moment bet\\·een In computing the values of column 5, x is the
sections due to shear at the section to the left. distance to the center of gravity of the loading curve
Column 7 gives the total moment at the section. between adjacent sections. This value is taken as
Columns 8 and 9 giv e t he I and M/I values of the "dL/2," as the error involved in so doing is negligible
spar at the section. The values of I are obtained from . and the computations are much simplified.
TABLE 8.Shears and moments front upper sparhigh incidence
I I
Station Interval Load Moment Moment Total I I M / I dL S=WdL Shears 8. x sdL moment
 
1 2 3 4 5 6 7 8 9
 
176 0 0 0 0   
12 12. 80
164 12. 80 76. 8 0 76. 80 1. 00 76. 80
10 12. 20
154 I . 25.00 61. 0 128. 00 265. 80 1. 67 159. 16
10 13. 64
I 144 38. 64 68.2 250. 00 584. 00 2. 55 229. 02
lO 15. 02
134 53. 66 75. 1 386.40 1, 045. 50 3. 73 280. 29
10 lG. 31
124 69. 97 81. 55 536. 60 1, 663. 65 6. 08 273. 63
2 3. 41
122 73. 38 3. 41 139. 94 1, 807. 00 6. 70 269. 70
10 17.85
112 91. 23 9.25 733. 80 2, 630. 05 10. 09 260. 66
I
10 18. 94
102 110. 17 94. 70 912. 30 3, 637. 05 H.72 247. 0
10 20. 06
92 130. 23 100. 30 1, 101. 70 4,839. 05 20. 56 235. 36
IO 21.16
82 151. 39 105. 80 I, 302. 30 6, 247. 15 28. 31 220. 67
10 22.16
I
72 173. 55 110. 80 I, 513. 90 7, 871. 85 37. 76 208. 47
10 23.14
62 196. 69 115. 70 I, 735. 50 9, 723. 05 49.28 197. 30
10 24.. 07
52 220. 76 120. 35 1, 966. 90 11, 810. 30 63. 40 186. 28
10 25.06
12 (245. 82)86. 09 12.fi. 30 2, 207. 60 14, H3. 20 79. 74 177. 37
10.5 27. 22
31. 5 (273. 04)58. 87 14.2. 91 903. 95 13, 382. 16 88. 67 150. 92
8. 5 20.16
23 (293. 20)3 . 71 85. 68 500. 40 12, 967. 44 93. 32 138. 96
9.25 17. 08
13. 75 (310. 28)21. 63 79. 00 358. 07 12, 688. 37 93. 32 135. 97
13. 75 21. 63
0 (331. 91)0 14.8. 71 297. 4.1 12, 539. 67 93.32 134.. 37
1, 684.. 56 10, 855. 11
TABLE 9.Bending momentsupper wing
I
Bending moments M /I
M. of I.
Station High incidence Low incidence 100 ponnds H igh incidence Low incidence 100 ponnds
Front Rear Front Rear F rorneta ra nd Front Rear Front Rear Front I Rear Front Rear
 
1 2 3 4 5 6 7 8 9 IO 11 12 13 14

li6 0 0 0 0             164 76. 80 62.88 31. 32 108. 36   1.00 0.84 76.80 74.86 31.32 129. 0   154 265.80 219.18 105. 92 379. 06  1. 67 I. 53 159.16 143. 25 63. 66 247. 75   14.4 584.. 00 486. 28 225. 22 845.06   2. 55 2. 61 229. 02 186. 31 88.48 323. 78   134 1,04.5. 50 879. 58 388. 57 I, 536. 51   3. 73 4.14 280. 29 212.46 104.. 28 371.14     124 I, 663. 65 I, 414. 73 594. 02 2,4.84.36   6. 08 6.30 273. 63 224.56 97. 77 394. 34.   122 1, 807. 00 1, 539. 94 639. 95 2, 706. 99 0 6. 70 6. 76 269. 70 227. 80 95. 57 400. 4.4 0 0
112 2, 630. 05 2, 266. 69 891. 35 4, 005. 39 1,000 10. 09 9.86 260. 66 229. 89 88.38 406.23 99.11 101.42
102 3, 637. 05 3, 170. 99 1, 175. 65 5, 632. 39 2, 000 14.. 72 13. 79 247. 08 229. 95 79. 9 408. 44 135. 87 145. 03
92 4, 839. 05 4, 268. 79 1, 486. 60 7, 621. 34. 3,000 20.56 18.88 235. 36 226.10 72. 32 403. 67 145. 91 158. 90
82 6, 247.15 5, 576. 49 1, 816. 70 10, 007. 24 4, 000 28. 31 25.18 220. 67 221.47 64.19 397. 43 14.1. 29 158. 86
72 7, 871. 85 7, UL 09 2, 157. 05 12, 826. 39 5,000 37. 76 32. 76 208. 47 217. 07 57.14 391.53 132. 42 152. 63
62 9, 723. 05 8, 889. 69 2, 4.97. 75 16, 115. 69 6,000 49.28 4.2. 10 197. 30 211.16 50.69 382.80 121. 75 142. 52
52 11, 810. 30 10, 929. 74 2, 27. 60 19, 913. 34 7,000 63.40 53. 71 186. 28 203. 50 4.4. 61 370. 76 110. 41 130. 33
42 14., 143. 20 13, 249.14 3, 134. 20 24, 259. 24 8,000 79. 74 66.93 177. 37 197. 96 39. 31 362.4.6 100. 33 119. 53
31. 5 13, 382. 16 16, 005. 02 a. 330. 61 29, 457. 58 ,000 9, 050 88.67 74. 00 150. 92 216. 28 37. 57 398.08 90.22 122. 30
23 12, 967. 4.4 18, 477. 21 3, 450. 67 34., 147. 29 ,000 9,900 93.32 77. 59 138. 96 238.14 36. 98 I «o. lo 85. 73 127. 59
13. 75 12, 688. 37 18, 120. 85 3, 539. 33 33,4.23.20 ,000 9,900 93. 32 77. 59 135. 97 233. 55 37. 93 430. 77 85. 73 127. 59
0 12, 539. 67 17, 929. 25 3, 589. 38 33, 032. 83 8,000 9,900 93. 32 77. 59 134. 37 231.08 38.47 425. 74. 85. 73 127. 59
I
13
TABLE 10.Bending momentslower wing
Bending moments M/I
100 M. of I.
Station High incidence Low incidence pounds High incidence Low incidence 100 pounds
Front Rear Front Rear Front
and rear Front I Rear Front Rear Front Rear Front Rear

1 2 3 4 5 6 7 8 9 10 11 12 13 14 1   
127. 875 0 0 0 0 
117. 875 41. 70 26. 30 19. 45 48. 55 ~: 8! 8: ~~ 77. 22 77. 35 36. 02 142. 79
107. 875 173. 75 110. 25 79. 55 204. 45 167. 07 157. 50 76. 49 292. 07 __ __ ii ____
98. 375 395. 44 252.80 176. 9 471.26
_____ ii ____
1. 75 1.18 225. 97 214.24 101.13 399. 37
87. 375 786.16 507. 01 341.10 952. 07 1, 100 2. 98 2.09 263. 81 242. 59 ll4. 46 455. 54 369. 13 ~26. 32
78 1, 245. 77 809.49 525. 32 1, 529. 95 2, 037. 5 4. 50 3.19 276. 84 253. 76 ll6. 74 479. 61 452. 78 638. 71
68 1, 878. 72 1, 231. 04 1 766. 57 2, 343. 20 3, 037. 5 6. 78 4.83 277.10 254. 87 ll3. 06 485. 13 448. 09 628.
58 I 2, 672. 62 1, 766. 14 1, 052. 72 3, 386. 05 4, 037. 5 9.68 6. 99 276. 10 252. 67 108. 75 484. 41 417. 10 577. 61
48 3, 641. 52 2, 427. 24 1, 381. 47 4, 687. 30 5, 037. 5 13. 33 9. 67 273. 18 251. 01 103. 64 484. 73 377. 91 520. 94
38 4, 799.17 3, 227. 09 1, 749. 37 6, 276. 90 6, 037. 5 18. 03 13.19 266. 18 244. 66 97. 03 475. 88 334. 86 457. 73
28 6, 159. 27 4, 178. 49 2, 151. 67 8, 186. 90 7, 037. 5 24. 19
17. 62 1
254. 62 237. 14 . 95 464. 59 290. 93 399.40
18. 25 I 7, 693. 19 5, 264. 49 2, 571. 51 10, 386. 19 8, 012. 5 30.87 22. 81 249. 21 230.80 83.30 455. 33 259. 56 351. 27
9. 25 9, 287. 54 6, 405. 60 2, 976. 78 12, 716. 38 8, 912. 5 38.81 28. 75 239. 31 222. 80 76. 70 442. 31 229. 64 310. 00
0 ll, 082. 50 7, 702. 45 3, 403.16 15, 381. 81 9, 837. 5 48.19 35. 82 ~:g~ I 215. 03 70. 62 429. 42 204. 14 274. 64
C. L. 11, 082. 50 7, 702. 45 3, 403. 16 15, 381. 81 9, 837. 5 48.19 35.82 215. 03 70. 62 429. 42 204. 14 274. 64
ARTICLE 5. COMPUTATIONS OF TRIAL SPAR
SIZES
After having determined the total moments on the
spars for the various conditions, it is necessary to
make some trial computations of spar sizes beforethe
properties of the spars can be determined and the
columns of I and M/I of Tables 8 to 10 of article 4
filled out. In making these computations, the spars
are considered as pure cantilevers subjected only to
the air load moment already computed. It has been
found through experience that such computations of
spar sizes, based upon the high incidence condition for
all the spars, usually give satisfactory trial sizes.
The moments in high incidence are, therefore, used for
both front and rear spars, upper and lower wings, in
determining their sizes. The sections of maximum
moments only are investigated. The spars are treated
as trusses with continuous webs and the average load,
P, in each chord or flange is found by dividing the
moment by the distance between the centroids of the
flanges. Since this distance is not known (the depth
of the spar at the section being the only definitely
known quantity), the thicknesses of the flanges must
be assumed. The allowable compressive stress for
spruce is 5,000 pounds per square inch. 'l'herefore,
Pf A is equal to 5,000, or A is equal to P/5,000. But,
since both P and the thickness t0 , of the compression
flange are known, the width, B, of the spar may be
solved for directly, or B is equal to P/5,000 times t0 •
Computations of the size of all four spars at the sections
of maximum moment are given below in Table 11.
It should be noted that another method of determining
the spar dimension would be to assume the width of
spar, B, and the distances between the centroids of the
flanges and solve for the thicknesses of the flanges t0
and t.. This method is not advocated, however, since
it involves the as umption of two variables instead of
one. In fact, it is quite possible that it would require
more than one trial because of the improbability of
the actual distance between centroids checking suffi ciently
close to the assmned value in the first trial.
In the computations of Table 11,
h= Depth of par.
t0= Thickness of compression flange.
t,= Thickness of tension flange.
H= Distance between the C. G. of the flanges.
=h0.5 (t0 +t,) .
P= Average load on flange= M/H.
p
B= Width of spar ·
5,000X t.
The computations of Table 11 below show how the
width of the spar may be determined, knowing the
depth and assuming the thicknesses of the compression
and tension flanges. In this de ign the dimensions
as determined in the table were not used, the
dimensions given in article 6 being obtained from a
design of similar external dimensions and gross load.
The width of the upper spars at station 0, for example,
is 2 inches for both the front and rear, it
being considered desirable at times to keep the width
of the spars the same. The dimensions given in
article 6 are then, in reality, second t rial sizes.
TABLE ILComputations of trial spar sizes
Spar Sta Basic Tota l moment h t. Hht.+t, M E=p " ti on moment P= M=8.0Xbasic M  2 H 5,000Xt,
2 4 6 7 8 10
Inchpounds Inchpounds Inches Inches Inches Inches Pounds Inches
Front upper ________________   0 12, 539. 67 100, 317 9.44 1. 75 1.00 8. 065 12, 439 1. 42
Do.    42 14, 143. 2 113, 146 8.84 1. 75 1. 00 7. 465 15, 157 1. 73
Rear upper ___                        0 17, 929. 3 143, 434 9.08 1. 50 .875 7. 8925
I
18, 173 2.42
FronDt olower _______________________ ________ 23 18,477.2 147, 818 9.08 1. 50 .875 7. 8925 18, 729 2.50 0 11, ()l!2. 5 ,660 . 125 1. 40 1. 00 6. 925 12,803 1.83
Rear lower_      0 7, 702. 5 I 61,620 7. 16 1.50 1.00 5. 910 10,426 1. 39
14
ARTICLE 6. SPAR PROPERTIES The complete computations of the moment of
The dimensions of the spar of the upper and lower inertia for the front upper par are given in Table 12,
~ings, as obtained from a design of similar external using the above given formula. The dimensions necdimensions
and gross weight, are plotted in Figures 11 essary for use in the formula are obtained from Figures
and 12, respectively. Referring to the spar section on
F~gure 11, A is the height or depth of the spar, B is the
\Vidth of spar exclush·e of webs, and C and D are the
thicknesses of the compre sion and tension flanges,
r~ ·pectively.
Mr. Fokker proposes the follo\\ing formula for the
precise determination of the moment of inertia of box
spars having unequal flanges:
B' A 3 Bl13
I= 12 12 ae Fi
"' here
B= width of flange.
B'= 1ddth of flange plus thickness of one \Yeb.
A= depth of spar.
fl= clear distance between flanges=A(C+ D).
F,= gross area= B' A.
F=area of hole= BH.
fa\= net area= F1  F .
C= thickress of upper flange.
D= thickness of lower flange.
CD
e=z.
F F
a= Fi  FXe= F. Xe.
IA _ LJ ..... r=::t...
~
In, _ _/
l/J
11 and 12. Complete computations of the moments
f inertia of the other spars are not given, but the
computed values of I a well as certain other essential
spar properties are given for all four spars in Table 13.
The values of I for the front and rear spars of both
wings are plotted against span in Figure 13.
t
6l? to fi 'O .f.'O
FIG. 11
/
15
I.Pi: ., 4.P
l.J) · · ~~ .,..~ ,,_ r. ii' ' l/i...: 'r.<; ,. ; , .. /. ,,,.. ,.,,,., ,,..,
~
OJ':
!{{, ...... le"" fAiro
f""....._ v I/" "Ai:i
f"".., ~ I/ ,.
7/J r...._ , lt?r ~;: 'n
......... ......... ht /  ..
......... /  r.n
~ ~r... l/ ......... K v 
11.l /./'. /  r Z"'
~ ...;;::::: r...... k:: ) <... v  .._ ~ ....._
c:::J ,.,;: I""'~ r;::: ~ ,/ ~ .v... ..... ...... K.
"1il ......... .;::::: .;:::;;: v ........ K ........
r.... [ ' ""' ~ 
~ If tJ £: .d/J ......... ~ r::::; ~ r......._ ~r...
~
·:::: r. ......... F::::: t=::::: r. r... ~  i:::.. t:... ~ ::::..
t=:::::
·.:::: ~ ~ ~ ~ ........ r..... I"...._ ~ ''I t'.  ~
~ ~ v  ~~ ~ t::::::: ~ ...._ I"...._ .............. ~
~ VJ.'i :?!' / r. r.,_ r.;::: tr:.:.;:: i:::::.. ~ ......... I"...._
L ~T, il"l' './','n  r. r.~
...;;;::: s::: ~ ~
........
.. .....:: r ""' ...;;::::: ..;;: l /J ~  ....,;:_
r 11
(~ ,2 ') 10 60 ~o /(~0 J.,. 'O PO
' i;Jc 7J7 Ji d es
F ro. 12
F in .,~
Ill>' .
" i<'J .,,,, ..1,; t> /'] r ln . ?r
• U F
.........
l.<>r " ~ ,I :;'n ~ \ I >  '!I, ~n >r ,;;..,,..,
'" \ I  .'i Y > v J
I ~ 17,.;
.s: I\.. v I
I 1.u ' '\I
~ 'I J'
~ I"/ '\ 'I
~ I\. '\. \
<::i 1,y ' ~. 1 ~n'I~ ' 1"
~ r... I/ ~ I'
h ~7 '·· r;n,
~ '!/" "~ N  .,,,..., Ir~ ~ :". ~
~ "I'... N ... ....... 0
?/. I'( .... , .... ~,..
............ .......... ' ~
~/J
........ "' ........ r... ["": ~
r...:::: :.  ~~ 
() 2t7 sit;/ 60 6 ? 1l 'O 1, v 1~0 hO
'JD or. ~ r;c '7e>
FIG. 13
81700283
16
TABLE 12.Computation of moment of inertiafront upper spar
Station A B I B' c D fc+1n) Fi F F, e a B'A' BH• F I I l2 l2 ae 1
  _ _ I_ 
023 9.44 2.00 2.125 1. 74 1. 00 6. 70 20. 06 13.40 6. 66 0.370 0. 744 148. 97 50.13 5. 52 93. 32
I 31. 5 9.24 2.00 2.125 1. 74 1.00 6. 50 19. 64 13. 00 6.64 .370 . 724 139. 70 45. 77 5.26 88. 67
42 8.84 2.00 2.125 1. 74 1. 00 6.10 I ~6:~~ 12. 20 6. 59 . 370 . 685 122. 33 37.83 4. 76 79. 74
52 8.38 1. 88 2.005 1. 60 .93 5. 85 11. 00 5.80 . 335 . 635 98.33 31. 36 3. 57 63.40
62 7. 91 1. 75 1.875 1.46 . 86 5. 59 14. 83 9. 78 5.05 .300 . 581 77. 33 25. 47 2.58 49. 28
72 7.45 1.63 1. 755 1. 32 • 78 5. 35 13. 07 8. 72 4. 35 . 270 . 541 60. 47 20.80 1. 91 37. 76
82 7. 00 1. 50 1. 625 1.18 • 70 5. 12 11. 38 7.68 3. 70 .240 .498 46. 45 16. 78 1. 36 28. 31
92 6. 51 1.37 1. 495 1. 04 .63 4.84 9. 73 6. 63 3.10 .205 .438 34. 37 12. 94 .87 20.56
102 6. 06 1. 25 1. 375 .90 • 55 4. 61 8.33 5. 76 2. 57 .175 • 392 25. 50 10. 21 • 57 14. 72
112 5. 60 1.12 1. 245 • 76 .47 4.37 6. 97 4.89 2.08 . 145 .341 18. 22 7. 79 .34 10.09
122 5.13 1. 00 1.125 . 62 .40 4.11 5. 77 4.11 1.66 . 110 . 272 12. 66 5. 79 .17 6. 70
124 5. 04 . 97 1. 095 . 59 .38 4. 07 5. 51 3.95 1. 56 . 105 . 266 11. 68 5. 45 .15 6.08
134 4.58 .84 .965 .45 . 30 3.83 4.42 3. 22 1. 20 . 075 • 201 7. 73 3. 93 . 07 3. 73
144 4.11 • 72 .845 .44 .30 3.37 3.47 2.43 1.04 .070 . 164 4.89 2. 30 .04 2. 55
154 3. 65 . 59 . 715 .44 . 30 2. 91 2. 61 1. 72 .89 .070 .135 2.90 1. 21 . 02 1. 67
164 3. 20 .46 .585 • 44 . 29 2. 47 1. 87 1.14 . 73 • 075 . 117 1.60 .58 . 016 1. 00
TABLE 13.Moment of inertia of spars
Front spar I Rear spar
Station I A I H I F, I a I I I A I H I F, I a I I
UPPER WING
023 9. 44 6. 70 6. 66 o. 744 93. 32 9. 08 6. 745 5. 81 0. 679 77. 59
31. 5 9. 24 6.50 6. 64 . 724 88.67 8.90 6. 565 5. 78 . 665 74.00
42 8.84 6.10 6. 59 .685 79. 74 8.53 6.195 5. 74 .630 66.93
52 8. 38 5. 85 5:80 .635 63.40 8.10 5. 92 5. 11 . 588 53. 71
62 7. 91 5. 59 5. 05 . 581 49.28 7. 66 5. 64 4. 49 .550 42.10
72 7.45 5. 35 4. 35 . 541 37. 76 7. 21 5. 33 3. 95 . 507 32. 76
82 7.00 5.12 3. 70 .498 28.31 6. 79 5.06 3.44 . 452 25.18
92 6. 51 4. 84 3.10 .438 20. 56 6.35 4. 76 2.97 .406 1 . 88
102 6.06 4. 61 2. 57 .392 14. 72 5.91 4.48 2.53 .365 13. 79
112 5. 60 4.37 2.08 . 341 IO. 09 5.48 4. 20 2.12 . 310 9.86
122 5. 13 4.11 I. 66 .272 6. 70 5. 01 3.88 1. 76 . 254 6. 76
124 5. 04 4. 07 1. 56 .266 6.08 4.94 3. 83 1. 69 . 253 6. 30
134 4.58 3.83 1. 20 . 201 3. 73 4. 50 3.54 1. 37 .195 4.14
144 4.11 3. 37 1. (}i .164 2. 55 4. 05 3.24 1. 09 . 139 2. 61
154 3. 65 2. 91 .89 . 135 1. 67 3.60 2.93 . 84 .093 1. 53
164 3.20 2.47 • 73 .117 1.00 3.17 2. 65 . 63 .039 .84
LOWER WING
0 8. 125 5. 725 4. 61 0.373 48. 19 7.16 4. 69 4.60 0. 360 35.82
9. 25 7. 69 5. 43 4.17 . 351 38.81 6. 77 4.45 4.14 . 336 28. 75
18. 25 7. 25 5.12 3. 76 . 338 30. 87 6.38 4. 20 3. 72 . 318 22. 81
28 6. 80 4. 80 3. 37 . 305 24.19 5. 95 3.90 3.33 . 287 17. 62
38 6. 31 4. 47 2. 94 .285 18.0"J 5. 53 3. 64 2.90 . 272 13. 19
48 5.84 4. 14 2. 57 . 261 13.33 5.10 3. 35 2. 53 .250 9.67
58 5.37 3. 82 2.22 . 232 9.68 4. 68 3.08 2. 19 .225 6. 99
68 4. 90 3. 50 1.88 . 204 6. 78 4. 25 2. 79 1. 86 . 205 4. 83
78 4. 41 3.16 1. 58 . 189 4. 50 3. 81 2. 51 1. 54 . 187 3. 19
87. 375 3. 96 2.84 1. 32 .157 2. 98 3.41
I
2.24 1. 28 .160 2. 09
98. 375 3. 44 2. 49 I. 05 .147 1. 75 2. 93 1.94 I. 01 .143 1.18
107. 875 3. 00 2.1 . 85 .117 I. 01 2.53 I. 67 .81 . 117 . 70
117. 875 2. 51 1.85 .63 .099 .54 2.09 I. 38 . 60 .105 . 34
ARTICLE 7. COMPUTATIONS OF DEFLECTIONS flections for all the spars under every loading
AT THE "N" STRUT condition arc given in Table 18 for both basic and deThe
deflections at the N strut are computed by the
method outlined in articles 46 and 47 of Airplane
De ign (Vol. I ) . Several tables of complete deflection
computations are included herein. Table 14 gives the
computations of the net deflection at the ~ strut for
the front upper spar in high incidence, and Table 15
gives the same for the front lower spar in high incidence.
Tables 16 and 17 give the computations of the net
deflections at the N strut due to a load of 100 pounds
acting at the strut for the rear upper spar and the front
lower spar, respectively. These four tables are sufficient
to show the method of computation. Tables of
computations of deflectious for the other spars and
other conditions are not included, but the final design
loads. Figure 14 shows how the net deflection
at the N strut is obtained. The deflection curve shown
in that figure is for the front upper spar and is a much
exaggerated representation of the actual deflection
curve. Since the direction of the slope of the curve is
known for only one point, the center lin e of the airplane,
the deflection in every case is measured from a
tangent to the curve at that point. Referring to Figure
14, AC represents the total deflection at the N
strut, BC the deflection at the inner support point,
and AB the net deflection at the N strut. The value
of AB as clearly shown by the figure, is the difference
between the total deflection at the N strut and the
deflection at the inner support point. The strut is
located at station 122 for the upper wing and at station
17
98.375 for the lower wing. The inner support of the
upper wing is at station 42 for the front spar and at
station 23 for the rear spar, while for the lower wing
the side of the fu elage, station 0, i the inner support
point.
The values of M/I used in the following computations
are obtained from the tables of article 4. As explained
in article 46, page 54, of Airplane Design, these values
may be treated as ordinates of a loading curve and
moments figured from them. Referring to any of the
following tables, 14 to 17 inclusive.
Column 1 gives the station points.
Tola/ .De/lee/ion al H Slral =AC
lJe/leclion al Sllj)jJOr/ Polnl=./3 C
Column 7 gives the product of the values of column
5 and the distance to the center of gravity of the
segment. This distance, x, is assumed equal to dL/2,
as the error involved in using this distance rather
than the true distance to the center of gravity is
negligible.
Column 8 gives the product of columns 2 and 6.
Column 9 gives values corresponding to the total
moment of article 4 and, hence denoted by " Moment."
They are actually values of Ey or !:iE, where y or !:i
is the deflection. They are equal to the cumulative
sums of columns 7 and 8.
Ne/ l)el/ecfion af N Slrvl == A.B I
orLJ~AC;/JC·A8 1 Su,.o)Jorls
.Derleclion Curvelironl V,.oper S,Par
# Slral Pol/JI I
FIG. 14
Column 2 gives the distance between adjacent
stations.
Column 3 gives the values of M/I at the station
points. These values correspond to the values of w
of a loading curve and, hence, are denoted by the
symbol "w,'' the quotation marks being used to differentiate
the notation from that for an actual load.
Column 4 gives the sum of the adjacent values of
column 3.
Column 5 gives the area of the M/I curve between
adjacent stations. The values are obtained by multiplying
the values of column 4 by onehalf the value in
column 2. This value corresponds to !:is of the tables of
article 4 and, hence, is denoted by !:i "s." the quotation
marks being used for the same reason as in column 3.
Column 6 gives the summation of the values of
column 5.
Column 10 gives the deflection from the tangent to
to the elastic curve at the center line. It is obtained
for any section by dividing the "moment" at that
section by the modulus of elasticity, E, which for
spruce is 1,300,000. The net deflection of the strut
points from a line through the supports is given near
the bottom of the column. As this is the only deflection
that will be used in subsequent computations, the only
values in column 10 are those needed for its computation,
though it would be a simple matter to compute
the deflections of the other points if they were desired
for any reason; for example, to make an accurate plot
of the elastic curve.
The totals of columns 7 and 8 are given so that their
sum can be checked against the last value in column 9.
This check should always be exact.
18
T ABLE 14.Defleclion of front u pper sparhigh incidence
Station I Interval M ___ iJ.L I~''w" Sum I A"s" '"s" .6."s" .x "s".dL "Moment" Deflection , ___ 
1 2 3 4 5 6 7 8 9 10

0 134. 37 0 0 0
13. 75 270. 34 1, 858. 59
13. 75 135. 97 1, 858. 59 12, 777. 80 0 12, 777. 80
9. 25 274. 93 1, 271. 55
23 138. 96 3, 130.14 5,880. 92 17, 191. 96 35,850. 68
8. 50 289.88 1, 231. 99
31. 5 150. 92 4, 362.13 5,235. 96 26,606. 19 67, 692. 83
I 10. 50 328. 29 1, 723. 52
42 177. 37 6,085.65 9,048.48 45,802. 37 122, 543. 68 o. 09426
10 363. 65 1, 818. 25
52 186.28 7, 903. 90 9, 091. 25 60,856. 50 192, 491. 43
10 383. 58 1, 917. 90
62 197. 30 9, 821. 80 9, 589. 50 79, 039. 00 281, 119. 93
10 405. 77 2, 028. 85
72 208. 47 11, 850. 65 10, 144. 25 I 98, 218. 00 3 9,482.18
JO 429.14 2, 145. 70
82 220. 67 13, 996. 35 1.0, 728. 50 118, 506. 50 518, 717. 18
10 456. 03 2, 280.15
92 235. 36 16, 276. 50 11, 400. 75 139, 963. 50 670, 081. 43
10 482. 44 2,412. 20 . 102 247. 08 18, 688. 70 12, 061. 00 162, 765. 00 844, 907.43
10 507. 74 2, 538. 70
112 260. 66 21, 227. 40 12, 693. 50 186,887. 00 1, 044, 487. 93
10 530. 43 2, 652.15
122 269. 77 23, 79. 55 13, 260. 75 212,274.00 1, 270, 022. 68 . 97694
I
121, 912. 66
I
1, 148, llO. 02
Net =0. 88268
TABLE 15.Deflection of front lower sparhigh incidence
Station Interval },[ " ,, iJ.L y w Sum fl."&" "s" .6.''s".x "s".dL ''.M oment" D eflection 
1 2 3 4 5 6 7 8 9 10

C.L. 229. 98 0 0 0 0
14. 375 459. 96 3,305. 96
0 229. 98 3, 305. 96 23, 761. 6 0 ZS, 761. 6 o. 01828
9. 25 469. 29 2, 170. 47
9. 25 239. 31 5, 476. 43 10, 038. 4 30, 580. 1 64, 380.1
9 488. 52 2, 198. 34
18. 25 249. 21 7, 674. 77 9, 892. 5 49, '187. 9 123, 560. 5
9. 75 503. 83 2,456.17
'18 254. 62 10, 130. 94 11, 973. 8 74,829. 0 210, 363. 3
10 520. 80 2,604. 00
38 266.18 12, 734. 94 13, 020. 0 101, 309. 4 324, 692. 7
10 539. 36 2,696. 80
48 273.18 15,431. 74 13, 484. 0 127, 349. 4 465, 526. 1
10 549. '18 2, 746.40
58 276. 10 18, 178. 14 13, 732. 0 154, 317. 4 633, 575. 5
10 553. 20 2, 766. 00
68 277. 10 20, 944. 14 13, 830. 0 181, 781. 4 829, 186. 9
10 553. 9.J 2, 769. 70
78 276. 84 23, 713. 84 13, 848. 5 209,441. 4 1, 052, 476. 8
9. 375 540. 65 2, 534. 30
87. 375 263. 81 26, 248, 14 11, 879. 5 222, 317. 3 1, 28G, 673. 6
11 489. 78 2, 693. 79
98. 375 225. 97 '18, 941. 93 14, 815. 8 288, 729. 5 1, 590, 218. 9 I. 22325
150, 276. 1 1, 439, 942. 8
Net= 1. 20497
19
T ABLE 16.Dejlection of rear itpper spar100 pounds at station 122
Station Interval dL .!1! .."w" Sum .6."s" "s" .6."s".x "s".dL "Moment" Deflection
 
1 2 3 4 5 6 7 8 9 JO
 
0 127. 59 0 0 0 0
13. 75 255. 18 1, 754. 36
13. 75 127. 59 I, 754. 36 12, 061. 2 0 12,061. 2
9. 25 255.1 1, 180. 21
23 127. 59 2, 934. 57 5, 4 . 5 16, 227. 8 33, 747. 5 0. 02596
8.50 249. 9 1, 062. 03
31.5 122. 30 3, 996.60 4, 513.6 24, 943. 8 63, 201. 9
10.50 241. 83 1, 269. 61
42 119. 53
10 249. 86 I, 249. 30
5, 266. 21 6, 665. 5 41, 964. 3 Ill, 834.. 7
52 130. 33 6, 515. 51 6, 246. 5 52, 662.1 170, 743. 3
10 272. 85 I, 364. 25
62 142. 52
10 295. 15 1, 4.75. 75
7, 879. 76 6,821. 2 65, 155. 1 242, 719. 6
72 152. 63 9, 355. 51 7, 378. 8 78, 797. 6 328, 896. 0
10 311. 49 I, 557. 45
82 158. 86 10, 912. 96 7, 787. 3 93, 555. 1
I
430, 238. 4
10 317. 76 I, 588. 80
92 158. 90 12, 501. 76 7, 94.4.. 0 109, 129. 6 547, 312. 0
10 303. 93 1, 519. 65
102 H5.03 14, 021. 41 7, 598. 2 125, 017. 6 679, 927. 8
10 246. 45 1, 232. 25
112 101.42 15, 253. 66 6, 161. 3 140, 214. I 826, 303. 2
10 101.42 507.10
122 0 15, 760. 76 253. 6 152, 536. 6 979, 093. 4 . 75315
I
!
78, 9. 7 900, 203. 7 I Net=. 72719
T ABLE 17.Dejlection of front lower spar100 poimds at station 98.375
Station IntdeLrv al ~1 "w" S um a"s" "s" fl."s".x "s".dL "1'1oment" Deflection
I

1 2 3 4 5 6 7 9 10
 I
C.L. 204. 14 0 0 0 0
14. 375 408. 28 2, 934. 51
0 204. 14 2, 934. 51 21,091.8 0 21,091.8 0. 01622
9. 25 433. 78 2, 006. 23
9. 25 229. 64 4, 940. 74 9, 278.8 27, 144. 2 57, 514. 8
9 489. 20 2, 201.40
18. 25 259. 56 7,1'!2.14. 9, 906. 3 44, 466. 7 11J,887. 8
9. 75 550. 49 2, 683. 64
28 290. 93 9, 825. 78 l!l, 082. 7 69, 635. 9 194, 606. 4
10 625. 79 3, 128. 95
38 334. 86 12, 954 . 73 15, 644. 7 98, 257. 8 308, 508. 9
10 712. 77 3, 563.85
48 377. 91 16, 51 . 58 17, 819. 3 129, 547. 3 455, 875. 5
10 795. 01 3, 975. 05
58 417. 10 20, 493. 63 19, 75. 2 165, 185. 8 640, 936. 5
10 865. 19 4, 325. 95
68 448. 09 24, 819. 58 21, 629. 8 204, 936. 3 867, 502. 6
lO 900. 87 4, 504. 35
78 452. 78
9. 375 821. 91 3,852. 70
29, 323. 93 22, 521. 7 248, 195. 8 1, 138, 220. 1
87. 375 369.13 33, 176. 63 18, 059. 5 274, 911. 8 l, 431, 191. 4
11 369.13 2, 030. 22
98. 375 0 35, 206. 85 11, 166. 2 364, 94.2. 9 1, 807, 300. 5 I. 390'23
180, 076. 0 1, 627, 224. 5
Net=!. 37401
81700 284
20
TABLE 18.Summary of deflections at N strut
Low incidence Nose dive
Deflection
1~~~1~1·~11,.~1 for
High incidence Inverted !light
Basic load L. F.=5.5 Basic load L. F.=3.5 Basic load L. F.=3.5 lOO pounds
Spar
Basic load L. F.=8
Upper fron t __________ __________ __ Upper rear _______________________ I 0.88293 7. 06344 0. 23608 1. 29844 1. 21939 9. 75512 2. 22532 12. 23926
Lower front 1. 20497 9. 67576 .41870 2.30285
Lower rear   1. 11688 8. 93504 2. 18947 12. 04209
0.88293  3. 09026
1. 21939 4. 26787
1. 20497 4. 21740
1. 11688  3. 90908
0.88293

1.20497
 
3. 09026
' + 4.61248
4. 21740
'+6.11398
0. 53010
. 72719
I. 37401
1. 88097
1 A slight error was found on checking the original com pn tations,ihe effect of which was to cb_jlnge the basic deflection on the upper front spar
in high incidence from 0.88293 inch to 0.88268 inch. As the difference is so small, it was not considered worth while to recompute the remaining
computations to correct for this small error.
'From c·omputations on p. 66.
ARTICLE 8. COMPUTATION OF "N " STRUT
EQUALIZING FORCES
The free deflection of a spar is defined as that
deflection which would occur if the N strut were not
present. The free deflections of the spars at the r_
strut point are given in Table 18. The upper front spar,
for example, under a load factor of 8.0 would deflect
7.063 inches and the lower front spar, 9.676 inches.
In order to equalize these deflections, the connecting
strut must exert equal and opposite forces on the
spars. The resulting force in the spar is called the
equalizing force. The computations of these forces for
the high incidence loading condition follow.
Equal,izing f orces for high incidence
The difference in free deflections between the lower
and upper front sp.ars is 9.675767.06344, or 2.61232
inches. Under a load of 100 pounds at the Nstrut
points, the upper spar deflects 0.53010 inches and the
lower spar 1.37401 inches. The total deflection of the
pair of spars, then, is equal to 0.53010+ 1.37401, or
1.90411 inches. The force required to equalize the
deflections is, therefore, equal to:
F 100X2.61232 137 1938 d C
• 1.90411 · poun s '
where C denotes compression and 7' tension . The
sign of the equalizing force can best be determined
by visualizing the direction of the force required to
eq ualize the deflections. Certain rules for determining
the sign of the force might be derived, but they would
be cumbersome and difficult to remember. The direction
and character of the force should not be confused.
In this example, the character of the force, compression
or tension, is denoted by C or T , respectively, while
the direction of the force is denoted by plus or minus
for upward or downward action, respectively. Thus,
as in inverted flight, it is possible to have a minus T
force.
Equalized deflection of front spars
Since a force of 100 pounds causes the upper front
spar to deflect 0.53010 inch, the increase in deflection
due to the equalizing force is equal to the product of
that force per 100 pounds by the deflection 0.53010
inch. This product, added to the free deflection, gives
the equalized deflection, or
dF.U
137i~g38 X0.53010+ 7.06344=
0,72726 +7.06344=7.79070 inches,
Similarly, the decrease in deflection on the lower
front spar is found by multiplying the deflection per
100 pounds by the equalizing force per 100 pounds.
This product subtracted from the free deflection gives
the equali zed deflection. Therefore,
aF L.=9.67576 137i~g38 x 1.37401=
9.67 5761.88506= 7. 79070 inches.
The rear spars are treated in the same manner as
the front spars, the computations being given below.
Force required to equalize deflections of rear spars
F 100 (9. 755128.93504)
0.72719+ 1.88097
82
·
0080
2.60816 =31 . 4429T
Equalized deflection of rear spars
da.u. =9.755120.314429X 0.72719=
9.755120.22865= 9.52647 inches.
da.L. =8.93504+0.314429X 1.88097 =
8.93504+0.59143=9.52647 inches.
Force quired to deflect front spars 1 inch
If 100 pounds deflects the upper front spar 0.53010
inch, the force required to deflect it 1 inch is equal to
100
0_53010= 188.6437 pounds. Similarly, the force required
to deflect the lower front spar 1 inch is equal to
100
1.37401=72.7797 pounds. The fo rce required to deflect
the pair of spars 1 inch is equal to the sum of
the two forces, or 188.6437 + 72. 7797 = 261.4234 pounds,
and the deflection per 100 pounds equals 21if.~~34, or
0.38252 inch.
Force required· to deflect rear spars 1 inch
100 100
F= o. 72719+ 1.88097=137 .5156 + 53.1641 = 190. 6797
pounds. The deflection in inches per 100 pounds
100
equals 190_6797 0.52444 inch.
Using the values of deflections ju8t obtained, the
force required to equalize the deflections of the two
pairs of spars is found by computations similar to
those used for the front spars.
21
Force required to equalize the deflections of the two pairs
of spars
F 100 (9.526477.79070)
0.38252+ 0.52444
173.5770
O. 90696 = 191.3833 pounds.
Equalized deflections of two pairs of spars
dF=7.79070+ 1.913833 X 0.38252=
7. 79070+ 0. 73208= 8.52278 inches.
da=9.52647 l.913833 X 0.52444=
9.526471.00369= 8.52278 inches.
The upper portion of Table 19 gives the computations
of the net equalizing forces and net deflections on
the spars for complete equalization. The net equalizing
fo rce is, for the front upper spar, for example,
the difference between its free deflection and the equalized
deflection of the two pairs of spars multiplied by
the force required to deflect it one inch, or F= (8.52278
 7.06344)X188.6437=+275.2953 pounds. If the
free deflection of the spar under consideration is less
than the equalized deflection of the two pairs of spars,
the net equalizing force must act upwards to equalize
the deflection, and, hence, the sign of the force should
be positive. Similarly, if the free deflection is greater
than the equalized deflection of the two pairs of ;;pars,
the sign of the net equalizing force should be negative.
The ne~ deflection is equal to the free deflection of the
spar plus the net equalizing force per 100 pounds multiplied
by the deflection of the spar for 100 pounds.
Thus, for the front upper. par, d=7. 06344+
275i6g53
X 0.53010, or d= 7.06344+ 1.45934= 8.52278 inches.
If this deflection is the same as the equalized deflection
of the two pairs of spars, already obtained, the computation
of the net equalizing force is checked.
The lower portion of Table 19 gives the computations
of the net equalizing forces and net deflections
under the assumption that the force in the N strut is
only 80 per cent of that required for complete equalization
of the two pairs of spars. Between the upper
and lower spar~ of both the front and rear pair, the
deflections are completely equalized with the exception
of the slight deformation of the connecting strut.
Between the front and rear pairs of spars, however.
the equalization of the deflections is not complete,
due to deformation of the drag trusses and the presence
of some torsion in the structure. From the results of
two static tests on internally braced biplanes, it was
indicated that the force in the center member was
approximately 90 per cent of that required for complete
equalization. The results from these two tests
are by no means conclusive, however, and until more
data are obtained on this subject an arbitrary design
value of 80 per cent will be used. This value is believed
to be very representative of actual conditions and,
moreover, is in q uite close accordance with the Handbook
ruling on least work; that is, " In computing the
net loads, the effect of redundant members shall be
increased or decreased 25 per cent, whichever is the
more conservatiYe, in order to allow for play in joints,
variations in crosssection, etc." The computations
of the force required for and the deflections of 80 per
cent equalization follow:
Forrefor 80 per cent equalization of the two pairs of spars
F=0.80X 191.3833pounds= 153.1066 pounds.
80 per cent equalization of deflections of two pairs of spars
dF= 7. 79070+ 0.80 X 1. 913833 X 0.38252= 7. 79070
+ 0.58566=8.37636 inches.
da=9.526470.80X 1.913 33X0.52444=9.52647
 0.80295= 8.72352 inches.
T ,rnLE 19.Net equalizing forces and deflections on sparshigh incidence
COl\1PLETg EQUALIZATION
Spar Net equalizing force (F)
Upper fronL (8. 5227 7. 06344) Xl88. 6437=+275. 2953 pounds.
Lower fronL (9. 675768. 5227 )X 72. 7797=  83. 9135 pounds.
Upper rear  (9. 755128. 52278) Xl37. 5156=  169. 4660 pounds.
Lower rear (8. 935048. 52278)X 53.1641=  21. 9174 pounds.
80 PER CENT E QUALIZATION
Upper fronL   (8. 376367. 06344)Xl88. 6437=+247. 6741 pounds.
Lower !ronL   (9. 675768. 37636)X 72. 7797=  94. 5699 pounds.
Upper rear   (9. 755128. 72352) Xl37. 5156= 141. 8611 pounds.
Lower rear  (8. 930548. 72352)X 53.1641= 11. 0060 pounds.
Net deflection (d)
1. 06344+
2i5i~53 xo. 53010=8. 52278
9. 67576 i~135 xi. 37401=8. 52278
9 75512169· 4 . 1006 GOXO . 72719=8 . 5227
8. 93504
2 \~~Xl. 88097=8. 52278
1. 06344+
247ig~41 xo. 53010=8. 37636
9. 67576 
94i~99 Xl. 37401=8. 37636
141. 8611
9. 75512woXO. 72719=8. 72352
11. 0060
8. 93054 lOOXl. 88097=8. 72352
Checkcomplete equalization: +275.295383.9135169.466021.9174 = +275.2953275.2969= 0.0016 pound error.
Check80 per cent equalization: +247.674194.5699141.861111.0060=+247.6741247.4370=+0.2371 pound error.
•
22
Equalizing forces for low incidence Equalized deflections of rear spars
The computations of the equalizing forces for low
incidence are made in the same manner as those for
high incidence, explanations of which were previously
given; hence no comments are required on the following
computations:
dR.U. = 12.23926 0.075597X 0.72719= 12.23926
0.05497= 12.1 429 inches.
du.L. = 12.04209 + 0.075597X 1. 8097= 12.04209
+ 0.14220= 12.l 429 inches.
Force lo equalize deflection of two pairs of spars
Force to equalize deflections of front s71ars
F= 100 (2 . 3028~ 1. 29844) 100.4410
0.53010+ 1.37401 1.90411
= 52.7496 pounds C.
Equalized deflections of front spars
dF.U. = l.29844 + 0.527496 X 0.53010 = 1.29844
+0.27963 = 1.57807 inches.
dF.L. =2.302 50.527496X 1.37401=2.30285
 0.7247 = 1.57 07 inches.
Force lo equalize deflections of rear spars
F= 100 (12.2392612.04209) = 19.7170
0.72719+ 1. 097 2.60816
=7.5597 pounds T.
F=lOO (12.18429J.57Jl07)
0.38252+ 0.52444
= 1169.4253 pounds.
1060.622
0.90696
Equalized deflections of two pairs of s71ars
dy= 1.57807 + 11.694253 x 0.38252 = 1.57807
+4.47329=6.05136 inches.
dn = 12.78429 11.694253 X 0.52444=12.18429
6.13293= 6.05136 i'nches.
Force f or 80 per cent equalization of two 71airs of spars
F = ll69.4253X 0.80 = 935.5402 pounds.
Equalized deflections of two pairs of spars
dF = 1.57807 + 9 .355402 X 0.38252 = 1.57 07
+3.57863 = 5.15670 inches.
du= 12.18429  9.355402X 0.52444= 12.18429
4. 90635 = 7 .27794 inches.
TABLE 20.Net equalizing forces and deflections on sparslow incidence
COMPLETE EQUALIZATION
Spar Net equalizing force (F) ' et dellection (d)
Upper front_ __                (6. 051361. 29844) Xl88. 6437=+896. 6084 pounds. 1. 29 44+
896j:84xo. 53010=6. 05136 inches.
Lower front_ _             ·     (6. 051362. 30285)X 72. 7797= + 272. 8154 pounds. 2. 30285+27~!154 x 1. 37401=6. 05136 in ches.
Upper rear __ _          (12. 239266. 05136) Xl37. 5156= 850. 932S pounds. 12. 2392685~i;28 xo. 72719=6. 05136 inches.
Lower rea<          (12. 042096. 05136)X 53.1641=  318. 491 pounds. 12. 04209
318ji:18x l. 88097=6. 05136 inches.
SOPER CR 'T E Q ALIZATION
Upper front_ ___________  (5.1 5670 1. 29844)XI . 6437= + 727. 8364 pounds. 1. 29844+
727i:64 xo. 53010=5.15670 in ches.
Lower fron t_ __ ·       (5.156702. 30285)X 72. 7797=+ 207. 7023 pounds. 2. 30285+
207j~0023 x l. 37401=5. 15670 inches.
ppcr rear __             (12. 239267. 27794)Xl37. 5156= 682. 2589 pow1ds. 12. 2392668~·0~89 xo. 72719=7. 27794 inches.
Lower rear____ _____ __ ___ ___ ___ __ _______ ______ __ (12. 042097. 27794)X 53.1641= 253. 2812 pounds. 12. 04209253i:12xL 88097=7. 27794 in ches.
•
CheckComplete equalization : +896. 6084+272. 8154850.9328318.49JS=+l 169.4238ll69.4246= 0.0008 pound error.
Check 80 per cent equalization: +727 .8364+207 .7023682.2589253.2812= +935.5387935.5401 = 0.0014 pound error .
23
Equalizing forces for inverted flight
The computations of the equalizing forces for inverted
flight are made in the same manner as for the
other conditions. The signs of the deflections are, of
course, negative, and consequently the signs of the
equalizing forces will be affected.
Force lo egualize deflections of front spars
F 100(4.21740(3.09026))
0.53010+ 1.37401
= 59.1951 pounds T.
112.714
1.90411
Equalized deflections of front spars
dF·U·= 3.09026+ (0.591951X0.53010) = 3.09026
0.31379= 3.40405 inches
dF·L·= 4.21740(0.591951X 1.37401) =  4.21740
+0.81335= 3.40405 inches.
Force to equalize deflection of rear spars
F=lOO [4.26787(3.90908)]= 35.879=l3 0.72719 + 1. 097 2.60 16 . 7564
pounds C.
Equalized deflections of rear spars
dfl.u.= 4.26787(0.137564X0.72719) = 4.26787
+0.10004= 4.16783 inches.
da·L·= 3.90908+ (0.137564X 1.88097) = 3.90908
0.25875= 4.167 3 inches.
Force to equalize deflections of two pairs of spars
F 100 [4.16783 (3.40405))
0.38252+0.52444
pounds.
76.378
0.90696 84.2132
Equalized deflections of two pairs of spars
dF= 3.40405+ (0.842132X 0.3 252) = 3.40405
0.32213=3.72618 inches.
dR= 4.16783(0.842132X 0.52444) = 4.16783
+0.44165= 3.72618 inches.
Force for 80 per cent equalization of two pairs of spars
F= 84.2132X 0.80= 67.37056 pounds.
Equalized deflections of two pairs of spars
dF= 3.40405+ (0.6737056X0.38252) = 3.40405
0.25771= 3.66176 inches.
dR = 4.16783 (0.6737056 X 0.52444) = 4.16783
+0.35332= 3.81451 inches.
TABLE 21.Net equalizing forces and deflections on sparsI. F.
COMPLETE EQUALIZATION
Spar Net equalizing force (F) Net deflection (d)
U pper front_____________________________ [ 3. 72618 ( 3. 09026 )] Xl88. 6437= 119. 9623 poun"~" 3. 09021619. 9~6X23O . 53 01 0=3. 72 61 8 m. e h es.
Lower front_ _____________________ _____ __ [4. 21740(3. 72618)]X 72. 7797=+ 35. 7508 pounds__ 4. 21740+ 3\::i08xi. 37401= 3. 72618 inches.
Upper rear.  [4. 26787(3. ~2618)]Xl37. 5156=+ 74.4908 pounds .. 4. 26787+
7
\:
08
xo. 72719=3. 72618 inches.
Lo [ 08 ( )] + d +
wer rear.       3. 909  3. 72618 X 53. 1641= 9. 7.237 poun 9. 7237 9 . h s__ 3. 90908 moXI. 8801!_ =3. 72618 mt es.
80 PER CENT EQUALIZA'l'ION
Upper front..     [3. 66176(3. 09026)]X188. 643.7=107. 8099 pounds .. 3. 09026107i~~XO. 53010= 3. 66176 inches.
Lower front     [4. 21740(3. 66176)]X 72. 7797=+ 40. 4393 pounds__ 4. 21740+ 4oi~~93 x1. 37401=3. 66176 inches.
Upper rear     [4. 26187(3. 8145l)]Xl37. 5156=+ 62. 3441 pounds__ 4. 26787+ 62i~~1 xo. 72719=3. 81451 inches.
Lower rear  [3.90908(3.8145l)]X 53.1641=+ 5.0277 pounds .. 3.90908+ 5i~~7XI.88097=3.81451 inches.
Checkcomplete equalization : 119.9623+35.7508+74.4908+9.7237= 119.9623+ 119.9653= 0. 0030 pound error.
Check80 per cent equalization : 107 .8099+40.4393+62.3441 +5.0277 = 107.8099+107 .8111 = +o. 0012 pound error.
24
Deflections and net equalizing forcesnose dive
The nose dive loading condition imposes a down
load on the front spar and an up load on the rear.
The magnitudes of these loads are obtained from the
following Handbook rule: "In the nose dive condition
the beam loads on the front spars shall be the same as
the de ign ultimate loads on those spars for the inverted
flight condition. The load on each front spar
shall be assumed to be held in equilibrium by a similarly
distributed up load on the corresponding rear
spar and a down load acting at the center of pressure
of the horizontal tail surfaces or at the tail post."
(See p. 447 of Airplane Design.)
Computation of beam loads on rear spars
FIG. 15
PFu=front upper beam force=331.91X3 .5=1161.685
pounds. (See Table 6, p. 9.)
PFL=front lower beam force=201.98X3 .5=706.930
pounds. (See Table 7, p. 9.)
Pau=rear upper beam force= i~~:~x PFu (taking moments
about center line of elevator hinge)=
210.5
183.5X 1161.685= 1332.614 pounds.
193.5 193.5
PaL=rear lower beam force= 170.5XPFL= 170.5X
706.930=802.293 pounds.
PT=force on tail required to balance beam forces
= (PFu  Pau) + (PFL  PaL) = (1161.685 
1332.614) + (706.930  802.293) = 170.929
95.363= 266.292 pounds.
OTE.Minus sign indicates downward force.
Computation of free deflections of rear spars
By the expression "similarly distributed" in the
preceding rule for nose dive, it is meant that the load
on the rear spar should be distributed in a manner
similar to the distribution on that spar in the high
incidence condition. Therefore, since both the load
and deflection on the rear spars are known for high
incidence, the free deflection on the rear spars may be
found by simple proportion, as the loads on the rear
spars have already been determined. Thus,
ec Table 6, column 9, p. 9, and p. 20 for load and
deflection, high inciden ce .)
daL=
8i°i6~5963 x + 1.1168 = +6.11398 inches.
(See Table 7, column 9, p. 9, and p. 20 for load and
deflection, high incidence.)
Force to equalize deflections of front spars
F=l00[4.21740(3.09026)] 112.714
0.53010+ 1.37401 1.90411
= 59.1951 pounds T.
Equalized deflections of f ront spars
dF. U. = 3.09026+ ( 0.591951X0.53010) = 3.09026
0.31379= 3.40405 inches.
dF.L. = 4.21740(0.591951X 1.37401) = 4.21740
+0.81335= 3.40405 inches.
Force to equalize deflections of rear spars
F= 100(6.113984.61248) =150.150
0.72719+ 1.88097 2.60816
= + 57 .5693 pounds C.
Equalized deflections of rear spars
da.u. =4.61248+0.575693 X 0.72719=4.61248
+0.41 64=5.03112 inches.
dR.L. =6.113980.575693X 1.8 097=6.11398
1.08286=5.03112 inches.
Force to equalize deflections of two pairs of spars
F=100[5.03112 (3.40405)]=843.517
0.38252+0.52444 0.90696
= 930.0487 pounds.
Equalized deflections of two pairs of spars
dF= 3.40405+9.300487 X0.3 252= 3.40405
+3.55762= +0.15357 inches.
da = + 5.031129.300487 X 0.52444 = + 5.03112
4.87755=+0.15357 inches.
Force for 80 per cent equ_alization of two pairs of spars
F = 930.0487X 0.80=744.0390 pounds .
Equalized deflections of two pairs of spars
dF= 3.40405+7.440390X0.3 252= 3.40405
+2.84610= 0.55795 inches.
da= + 5.031127.440390X 0.52444= +5.03112
3.90204= + 1.12908 inches
25
TABLE 22.Net equalizing forces and deflections on spars . D.
COMPLETE EQUALIZATIO
Spar Tet equalizing force (Fl Net deflection (d)
Upper front_ ____   (3.090260.15357) X188.6437=+611.9281 pounds. 3.0 9026 + 6111.092801 xo.530!0= + 0.15357.m ch es.
Lower front_ __  (4.217400.15357) X72.7797=+318.1l79 pounds.
Upper rear __  _        (+4.612480.15357) XI37 .5156= 613.1697 pounds.
4.21740+
31 ~~79 x 1.37401=+0.15357 inches.
Lower rear ___  (+6.113980.15357) X53.1641=316.8798 pounds.
+4.6142861 ;·~97
X0.72719=+0.15357 inches.
+6.11398316i~~98 xr. 097=+0.15357 inches.
80 PER CENT EQUALIZATION
Upper front_ ___  (3.09026 (0.55795)]X188.6437 =+477.7043 pounds. 09026+
477.7043 . b
3. WO X0.53010= 0.55795 mc es
Lower front_ __   [4.21740 (0.55795)] X72.7797=+266.3337 pounds. 4.21740+ ~·~37
Xl.37401= 0.55795incbes.
Upper rear ___  (4.612481.12908) Xl37.5156= 479.0218 pounds . + 4.6124 8 479w.0o218X 0.72719= + 1.12908 m. e b es.
Lower rear ___   ( +6.113981.12908) X53.1641 = 265.0177 pounds. +6.1139826~~77
XL 097=+l.12908incbes.
Check for complete equalization: +611.9281+318.1179613.1697316.8798=+930.046930.0495=0.0035 pound error .
Check for 80 per cent equalization: +477.7043+266.3337479.0218265.0177=+744.0380744.0395=0.0015 pound error.
ARTICLE 9. COMPUTATION OF STRESSES IN the Nstrut members, and Table 24 gives the stresses
N STRUT in the members anc;! their components.
TABLE 23.Dimensions of Nstrut members
Member c B s Le(nLg)t h C/B S/B L/B
The r strut shown diagrammatically in Figure 16 is
considered a truss whose external forces arc equilibrants
of the equalizing forces just computed. These
forces form a couple which is resisted by equal and
opposite forces in the drag trusses. The t russ is solved
by any of the ordinary methods, that of joints being,
perhaps, the most convenient. The vertical component
of the stress in the center member is equal to the force
required to equalize the deflections of the front and
rear pairs of spars. Table 23 gives the dimen ions of
 
Member
Inches Inches Inches
Front ______ 17. 0 59. 5 9. 25
Rear _______ 13. 0 59. 5 9. 25
Center_ __ __ 10. 0 59. 5 9. 25
C=component in chord direction.
B=component·in beam direction.
S=component in side direction.
TABt,E 24.Stresses and com71onents in N strut
Inches
62. 6
61. 6
61.0
[Ugh incidence Low incidence Inverted flight
o. 286 0.155 1.052
. 218 .155 1.035
.168 .155 1. 025
Nose dive
JOO per cent 0 per cent 100 per cent 80 per cent lOOper cent 80 per cent 100 per cent 80 per cent
F ront str ut:
C component_ _______      ________ ___ 78. 7 70. 8 256.4 208.2 34.3 30.8 175. 0 136. 6 B component_ ____ ______     ____________ 275. 3 247. 7 896.6 727. 8 120.0 107. 8 611.9 477. 7
S component_  __ 42. 7 38. 4 139. 0 112. 8 18. 6 16. 7 94. 74. 0
Stress ____                         289.6 200.6 943. 2 765.6 +126. 2 +113.4 +643. 7 502. 5
Rear strut:
C component _____               4.8 2.4 69.4 55. 2 2.1 1. 1 69.3 58.1
B component _________              21.9 11. 0 31 .5 253.3 9. 7 5.0 318. l 266.3
S component 3. 4 1. 7 49.4 39. 3 1. 5 .8 40. 3 41. 3
Stress_                     22. 7 11.4 329.6 262. 2 +10.0 +5.2 329. 2 275.6
CentCer c sotmrupto:n ent _____________ ' ______________ 32. 2 25. 7 196. 5 157. 2 14. 1 ll.3 156. 2 125. 0
B component_ ____  191. 4 153. l 1, 169. 4 935. 5 84. 2 67.4 930. 0 744.0
S component_  29. 7 23. 7 181. 3 145. 0 13.1 10.4 144. 2 115.3
Stress ______  ____             +100. 2 +156. 9 +1, 108. 6 +958.9 86.3 69. l +953. 3 +762. 6
NoTE.Plus <+l denotes tension; minus () denotes compression.
The magnitude of the equal and opposite drag forces 2:M=275.3X4083.9X23169.5X1359.5 Cu=O,
which resist the couple formed by the equilibrant forces or
is found by adding algebraically the C component of
the stresses in the Nstrut members. Thus, for high
incidence, complete equalization, Cu=CL=7 .7+4. +
32.2= 115.7 pounds, and for 0 per cent equalization,
Cu=CL=70.8+2.4+25.7= 9 .9 pounds. As a check
59.5Cu=11,0121929.72203.5=6878.8
C u=6s 97x8.8= 115 .~r poun d s
on these drag forces, moments of the vertical forces Similarly for 80 per cent equalization:
shown on the diagrams of Figure 16 may be taken l:M=247.7 X 4094.6X23141.9X1359.5 Du = O,
about one of the spars and the drag force obtained by I or
dividing the sum of these moments by the gap. For 59.5 D u=99082175.  1 44.7=58 7.5
example, taking moments about the lower rear spar of
the equilibrant forces in high incidence, for complete
equalization:
5887.5
Du= 59_
5
=98.9 pounds.
•
26
ARTICLE 10. DESIGN OF N STRUT Tubes, Figure A45, Volume I , Airplane Design. In
obtaining these loads, the moment of inertia and
The N strut is built up of Specification 10225 length scales of the chart are used. The values of
streamline tubes, the properties of which are given in the allowable tensile loads, based on an ultimate
(247.7)#
Z75..3 #
(/41.9)#
169..56
(98..9"9
, ..... it...,Cv=l/5.7#
Z.3"
~.J.Z#)
1~<Cv=.50..5#
~.J.Z#)
(L=50.S;;
(7Z7.8'J
8.96.6#
(68?..J#)
8.50.9'9'"
(;Z0.6"")
+ Cv=5ZZ.J"'
(¢7Z/#)
611..9 '*
t.____:.
Low
/1Jcide!7ce
{.J/9. 7#)  ·j\
Q400..5# l
..318./#
(266..Jrl)
.Nole:.8mclels, ( )J/Jtlicole Torces Corres,,oondin; lo
80% .& vo/izolion.
Fm. 16
Table A23, page 379, Volume I , Airplane Design. tensile strength of 55,000 pounds per square inch, are
The size 1.Ys inch 0.035, for example, indicates the found from Table A 13, page 365, Volume I , Airplane
size of the basic round tube from which the streamline Design. Table 25 (p . 27) gives the computations of the
tube is drawn. The allowable compressive loads are allowable loads and margins of safety in the Nstrut
found from the Euler omographic Chart for Steel members.
27
TABLE 25. Allowabl,e loads and margins of safety in strnt
Maxjmum load Allowable load
Member Size Length i\f.S.
rrension Compression
Tension Compress
ion

Front strut_ ___              
Pounds
+126.2
+10. 0 I +J, l98.6
Pounds
943.2
lnch'8 Inches
1%X0.049 62. 6 Pounds Pounds
0. 01532 +n, 230 1,090 +0.16 c.
Rear strut_ ___ ______        329.6 lYsXO. 035 61. 61 .00608 +6,590 4[>() +0. 37 c.
Center strut_ ________________              80.3 l Ys XO. 035 61.0 . 00608 +6,590 455 +4. 50 c.
In the coluillll of margin of safety in the above table,
only the M. S. of the critical load is given. The
letter "C" indicates that t he compression load is
critical. It will be noted that the center strut has a
much greater margin than the front or rear strut·
The size of strut, however , is the minimum drawn at
the time of writing of this article. It is probable
that, in practice, one size would be used for all three
members.
ARTICLE 11. COMPUTATION OF DRAG TRUSS
LOADS
The determination of the loads in the drag structure
may, perhaps, be best obtained by the following
procedure:
l. Computation of external loads on structure :
a. Panel loads.
b. Drag or antidrag reactions.
2. Determination of internal loads in structure by
analytical method.
1A. Computation of panel loads
The computations of the chord loads per inch run
are given on page 8, and the curves of chord load per
Chore! Loocl Cun/e
floriz. Scale :!"=.CO"
Veri Scale .f'=Q,5#
I
I I I
~ ;s ~I ~: ~I
'f:I C§:
tj I::) C:il \J I
.17611a8 I .f.?4.0 I
Nf.5
c,rr.5.7#
..f0(.7
/.17.44# (74..58#
inch run for high incidence are shown in Figure 6,
page 8. For t he nosedive condit ion, the Handbook
states that " The chord components of t he air load on
the wings shall be proportional to the chord loads in
low incidence and their total magnitude shall equal the
weight of the airplane."
Gross weight of airplane= W = 2,338 pounds.
w 2=1,169 pounds
If the summation of the panel loads for low incidence
on onehalf of . the upper and lower wings be denoted
by "IP, the ratio of the panel loads for nose dive to
those for low incidence equals 0.5W/2:P, or panel loads
for nose dive= pa11el loads for low incidence X 0.5W /"IP.
Figure 17 shows the method of dividing up the loading
curve into areas to obtain the loads acting at the
panel points. The rule given on page 447 of Airplane
Design states that " The chord loads shall be considered
as concentrated loads at the panel points of the
internal wing trusses. The load at each such joint
shall be the product of the normal chord load per inch
run at t he joint and the distance between midpoints
of the adjacent panels of the internal truss." In other
I
I I
I I
I I
<:'.) I ~ ~I ~I ~ ~ ~I
C:i ~I \JI ~ C:i
I I I
7ao 1 .31.S j ,J.7.f
6T"' .Jf.5
I I
I
1'/.J.60# I ltf1.06"'
I
F10. 17
28
words, the panel load, P, is equal to the area under
the chordload curve between the midpoints of adjacent
panels of the truss. The panel loads for the upper
wing high incidence condition are computed in detail
below. For the lower wing in high incidence and both
the upper and lower wings in the other loading conditions,
the computations for the panel loads are not
given. The loads, however, for all the conditions are
given in Tables 26 and 27, the panel loads on the upper
wing being given in Table 26 and those on the lower
wing in Table 27.
Computation of panel loads on iipper wingH. I. L. F.
=8.0
Note: Plus ( +) indicates drag direction of load.
Minus (  ) indicates antidrag direction of load .
Station 161:
P 8·0 (0.43~+ 0 ·365 )X(176141.5)=117.44
pounds.
Station 122 :
p 8.0 (0.48~ + 0.546) x (141.5124.0)
+ 8·0 co.54~+ 0 ·644)x (124 102.5) = 12.24
102.34 = 17 4.58 pounds.
Station 83:
P= 8.0 <0·64i+0·823) (102.562.5) = 234.72 pounds.
Station 42:
P 8·0 (0.323+o.957) (62.532.5) = 213.60 pounds.
2
Station 23:
P =8.0 (0.95710.963) (32.531.5)
+ (0.96310.594) (31.513.75)
+ <0·594! 0·603) (13.750) =8.0 (0.960
+ 13.818+8.229) = 184.06 pounds.
Check 2:P=area under curve.
2:P= 117.44174.58234.72213.60184.06
• = 924.40 pounds.
A 8(0.365+0.546) (176124)
2
_ 8.0(0.54~+0.963) (12431.5)
8.0(0.96;+o.594) (31.5 _ 13.75)
_ 8.0(0.59~+0.603) 13.75= 189.488
558.330110.54765.835= 924.20
pounds.
Error= 924.4 (924.20) = 0.20 pound.
In addition to the panel load acting at station 122,
there is an additional antidrag force acting from the
N strut. This force for the upper wing is shown as
Du in Figure 16 and is115.7 pounds for high incidence.
The direction of the force is opposite from that
shown in Figure 16, or in the antidrag direction,
because the force shown in Figure 16 is the external
force acting on the strut, which means that the
strut in turn must exert an equal and opposite force
on the drag truss.
TABLE 26. Panel loads on upper wing
High incidence Low incidence Nose dive•
p Du p Du p Du
Pounds Pounds +12. 96 +97. 91
Pounds
161 117. 44
122 174. 58
83 234. 72
42 213. 60
23 .184.06
115. 70 +108. 40 522. 30 +145. 47 400. 5
+145. 31 +195. 01
+132. 50 +111. 82
+114. 37 +153. 48
924.40  , +573. 54 1 +769. 69 
1 From p. 27, panel loads in nose dive are equal to: Panel load! in
1169 1169
L. I . x 0.5 W/EPL.(, Thus, PN.o.=573.54+297.20 PL.1.870.74 PL.I.~
1.342 PL.I.
TABLE 27.Panel loads on lower wing
High incidence Low incidence I Nose dive •
Stntion 1
p _DL_ , _ p_ _DL_ _p_ DL 
Pounds Pound• Pounds Pound• I Pounds Pound•
98.375 (112.75).  161. 92 ~~~~~~~ti~:~ \~~~~~~ tm: ~ +400.6
42.375 (56.75) ___ 222. 27
0 (14.375) _______ 96. 49 _________ +59. 95 _________ +so. 45

480. 68  +297. 20 1· ·1+398. 84
(C. L.)
1 Two station references are given: (I) Station 0 at side of fuselage
(this method or:referencing used when computjng defiectionsat strut).
(2) Station 0 at C. L. (reference used in chordload diagrams) .
• From p. 27, panel loads in nose dive arc equal to: Panel loads in
1169 1169
L. r.xo.5 lV/EPL.l Thus, PN.D· 573.54+297.20 PL . l.~870.74 PL.1.=
1.342 PL.I
1B. Determination of drag or antidrag reactions
In normal design it makes little difference in the
computation of the drag tru,s loads whether the
reactions are assumed to be taken by the front or by
the rear members of the cabane, as they intersect the
spars at the same station. In this design, however,
the tripod of the cabane strut intersects the front spar
at tation 42, while the rear cabane strut intersects
the rear spar at station 23. The only drag reaction
which can exist at station 23 is the drag component of
the load in the rear strut, which load is determined
by the lift forces on the wing. Therefore, in computing
the drag reaction, that drag component of the load in
the rear strut is considered as an additional panel
load at st11.tion 23, and the total drag reaction is then
taken by the front tripod at station 42. From Figure
23, page 34, the dimensions of the rear strut, D, are
obtained, and the computations of the components
of the length and their coefficients are given in
Table 28. The loads in the strut are given in Table 29.
The drag component of the load, which is considered
as an additional panel load, is called P' and is given
in column 10 of Table 29. The load in the strut is
determined from the total load acting at station 23
where the total load is equal to the air load on the
rear spar plus or minus the equalizing force acting at
the strut point.
29
TABLE 28.Components and length coe_fficients nf rear
cabane strut
Length components Length Length coefficients
D v s L ~I~ S/L
In ches I11clies Inches l n<hes 3. 5 26. 0 5. 5 26.80 0.131 I 0. 970 0. 205 I
It may be noted from Table 29 that the d rag component
is a maximum when an 80 per cent equalizing
force is used at the N strut rather than when a 100 per
cent force is used. The drag or antidrag forces from
the N strut acting at the N strut are, however, critical
for the 100 per cent equalization condition. Therefore,
in order to have consistency, the drag component for
the 100 per cent equalizing condit ion will be used.
The error involved is small and affects only one drag
wire.
TABLE 29.Load in rear cabane strut
Equalizing force Vertical component Load Drag comFonent
(V) (D = ')
Loading condition Air load
lOO percent 80 per cent lOOpercent 80 per cent lOOpercent 80 per cent
JOO percent 80 per cent equaliza equaliza equaliza equaliza equaliza equalizat
ion tion tion tion tion tlon
 
I 2 3 4 5 6 7 8 9 10

H.L ..      +2, 818. 40 169. 5 141.9 +2, 648. 9 +2, 676. 5 +2, 730. 82 +2, 759. 28 357. 74 361. 47
L . L  +3, 718. 55 850. 9 682. 3 +2,867. 65 +3, 036. 25 + 2, 956. 34 +3, 130.15 387.28 410. 05
I. F  1, 233. 05 +74.5 +62. 3 1, 158. 55  1, 170. 75 1, 194. 38 1, 206. 96 + 156.46 + 1ss.11
N. D   + 1, 332. 61 613. 2 479. 0 +719. 41 +853.61 + 741. 66 + 0. 01 97. 16 115. 28
NOTE. Plus<+) indicates tensile load; upward vertical force; backward or drag force. Minus() indicates compressive load; downward
vertical load; forward or antidrag force.
The resultant reaction acting at station 42 is equal
to the summation of the external forces acting on the
truss, or Ru = 2:P+Cu+ P' . Table 30 below gives the
reactions on the upper drag truss.
TABLE 30.Reactions at station 42upper drag truss
Loading condition D u P'
High incidence ____________ 924. 40  115. 70  357. 74
Low incidence _____________ +573. 54  522. 30 387. 28
Nose dive _________________ +769. 69  400. 50 97. 16
Ru
+1, 397. 84
+336. 04
+272. 03
The inverted flight condition is not critical for the
qrag truss, since 2:P = O and Du and P' are small.
The reaction for the lower drag truss, RL=2:P+DL.
This reaction is, however, taken directly by the fuselage
and, hence, is not computed.
2Determination of internal drag truss loads by analytical
method
The following procedure gives, perhaps, the simplest
method of obtaining the loads in a parallel chord drag
truss.
1. A scale or dimensioned diagram of the drag truss
is drawn and all external forces, including the reaction
and the Nstrut drag load are shown acting at the
panel points. Using for an example the upper drag
truss in high incidence, Figure 17 shows the truss
diagram with all of the external forces.
2. Two geometrically similar right triangles are now
drawn for each bay of the truss. Each of these triangles
has its hypotenuse coinciding with the drag
wire, and its two legs parallel to the chord and the
vertical compression strut, respectively, of the drag
bay. One of these triangles is called the "dimension
triangle" and the other the "force triangle." The
length of the drag wire and its drag and side components
are given upon the hypotenuse and the proper
legs of the "dimension triangle." The load in the
drag wire and its drag and side components are given
upon the proper sides of the force triangle. Figure
18A shows the diagram of Figure 17 with the addit
ions of the "dimension" and "force" triangles. The
"dimension" triangles are shown complete. The
magnitudes of the drag components only of the loads
in t he wires are shown on the "force" triangles. The
determination of these components is given below as
the third step in the procedure. It should be noted,
however, that although the use of two triangles ii!
suggested, the " dimension" triangle may be omitted
if the dimensions of all members are plainly shown.
3. The drag components of the loads in the wires are
obtained by the method of shear, explained in article 8,
page 6, of Airplane Design, Volume I. This method
consists of taking a section, through the bay under
consideration, which will cut the two chord members
and the diagonal or wire. Since the chord members of
a parallel chord truss have no drag components, the
drag component of the wire is, t hrough the relationship
2:D=O, equal to the shear to the left of the section.
For example, the drag component of the wire in the
second bay from the tip is equal to (117.44 pounds
+ 115.70 pounds+174.58 pounds), or 407.72 pounds.
Knowing the drag components of the loads in the wires,
the loads in the drag struts may be easily determined
by either the method of shear or the method of joints,
working from the t ip inward. Figure 18A shows the
drag components of the force triangles and the loads
in t he drag struts.
4. Since the force and "dimension" triangles are
similar , and the drag components of the loads in the
wires have been determined, the loads in the wires and
their side components may be determined by simple
30
proportion, a sliderule operation. For example, the
load in the wire of the second bay from the tip ancl the
side component of that load are found as follows·
Load = f.
Side component=S.
Then,
f 47.44
407 . 72=~· or ! = 716.36 pounds.
s 39
407
.
72
= 27 , or = 588.93 pounds.
It should be noted that, since both 407.72 and 27 are
common to both computations, the values of both f
and Scan be found by one sliderule operation. Figure
18 B hows the completely solved force triangles.
5. The loads in the chords are now readily determined
by the method of joints. ince the side components
of the loads in the wires are given on the force
~J.5Sia
. .161
3.9"
lJu=.11.5.70#
.171.58#
.39 ..
f2Z
moments. For example, the load in the tension chord
of the second bay from the center line is found by
taking moments about tation 23 of the front spar.
I:Msta, 23=0.
Therefore,
8.J
117.44 x (161 23) + 115.70 (122  23) + 174.58
(122  23) + 234.72 ( 323) + 213.60 (4223)
1,397. 4 (4223)SX 27 = 0.
117.44 x 13 + 115.70 x 99 + 174.5 x 60 + 234.72
X 60 + 213.60X 19  1,397.84X 19  27S = O.
16,206.72 + 11,454.30 + 17,283.40 + 14,083.20
+4,d58.40  26,558.96  27S= 0.
27S= 63,0 6.02  26,55 .96 = 36,527.06.
S 36'~~7·06 = + 1,352. 5 pounds.
,,?/.3.6011!
42 .2.J
..Lf.9:18A Pu=JJ97.84#
.Ou=.fff 10# ..3.57.71
~
..tJ'tl.58# 2.J.f.72# ..18~06"
0 7.58..J"T# T
./ZZ F.2 2.J
Puz.f.J9UP"
FIGS. 18A and 18B
t riangles, the process of determining the loads in the
chord members resolves itself into simple addition or
subtraction. The loads in all of the members are
shown in Figure 18 B.
6. A check is obtained on the loads in the vertical
web members if the load in the innermost drag strut
equals the external forces acting at that strut; that is,
load in drag strut at station 23 is 541.80 pounds C. and
sum of external forces at that station is (357.74+
184.06) or 541.80 pounds. A check on the loads in
the chord members is made by computating the load
in the chord of one of the inner bays by the method of
As the load obtained by method of joints is + 1,352.87
pounds, the error is only 0.02 pounds. which is negligible.
7. The direction of the drag wires of the diagram is
determined by the direction of the shear at the panel
points. Whenever an indicated compressive load is
obtained in a wire, the direction of that wire should be
changed in such a manner that the shear may be taken
by the tension of the wire.
The drag loads for the upper wing for low incidence
and no e dive and for the lower wing for high incidence,
low incidence, and nose dive are computed in a manner
similar to that outlined for the upper truss in high
31
incidence. Figure 19 gives the loads in the upper
truss in low incidence and nose dive, and Figure 20
gives the load in the lower trus for high incidence,
low incidence, and nose dive. Figure 21 gives r eference
diagrams of the upper and lower trusses, and
Table 31 gives a summary of the drag loads in both
wings.
TABLE 31.Summary of drag loads in wings
I Loads, upper wing Loads, lower wing
~:~ , __________ , _________ _
H. I. L. I . N. D. H . I. L. I. N. D.
!______ 0
2___ ___ +169. 64
3______ +758. 57
4 ______ +1, 352. 87
5 ______ +1, 352. 87
6______ 169. 64
7______ 758. 57
8 ______ 1, 734.13
9 ______ 1, 734.13
10 _____ 1, 352. 87
ll_____ 117. 44
12_____ 407. 72
13_____ 642. 44
14_ ____ 1,397.84
15_____ 541. 80
16_ ____ 0
17 _____ +206.35
18_____ 0
19__ ___ +116. 36
201 0
2L ____ +1, 168. 05
22_____ +662. 19
23_____ 0
24 _____ 1 0
25_____ 0
1 See Figure 25.
105.39
105. 39
+387.08
+492.10
+492.10
0
387.08
684.15
684.15
492.10
72. 96
522. 30
340. 94
132. 50
272. 91
+ 128.19
0
0
+599.03
0
+355.68
+333. 56
0
0
0
14.1. 43
141.43 +27. 98
+27. 98
+67.61
0
85. 52
85. 52
67. 61
67. 61
97. 91
400. 50
195. 01
272.03
56. 32
+ 172. 03
0
0
+276.06
+68.89
0
0
+68. 83
0
0
0
+112. 54
112. 54
607. 20
161. 92
268.49
0
+121. 66
0
+562. 78
1,515. 07
2, 914. 46
0
+1, 515.07
622. 26
759. 55
+1,637. 90
0
+1, 592. 08
0
1,301. 76
2,626. 24
0
+1,301. 76
534. 65
718. 89
+1,407. 29
0
+1, 506. 85
0
ARTICLE 12. DESIGN OF DRAG WIRES
Table 31 of article 11 gives the loads in all of the
members of the drag trusses. Table 32 below gives the
size of wires and the margins of safety in the wires for
the ma::dmum loads. It will be noted that, since the
loads in all of the wires are fairly small, the minimum
ize wire allowed by pecification is used throughout.
The member numbers refer to the diagrams of Figure 21.
TABLE 32.Design of drag truss wires
UPPER WING
Member Loading Maxi.mum Size or I Allow I
condition load wire able load M. S. ·· ____ , _______ ! __ _
1167______________________________ NH. .D . I.
1I 9__ _____________ H. I. 
20___ _______ _____ N. D.
2L ______________ H . I.
22_______________ H. I.
23_______________ N. D.
24 and 25 ___ _____   
Pounds
+172. 03
+206. 35
0
+116. 36
+68.89
+1, 168. 05
+662.19
+68.83
0
1032
1032
1032
1032
1032
l~~~ I
1032
1032 1
LOWER WING
81________________ HL.. II..
9________________ L. I.
10 ____________ ___ II. I.
+1. 637. 90 I +121.66
+1,592. 0
+562. 7
1032
1032
lo32
1032
Pounds
+2, 100
+2, 100
+2, 100
+2, 100
+2, 100
+2,100
+2,100
+2,100
+2, 100
+2,100
+2,100
+2, 100
+2,100
+11. 22
+9. 18
+1.93
+29. 50
+o.80
+2.17
+29. 50
+o.28 +16. 26
+0. 32
+2. 73
7296"' / 08.40# /¥.:5..J/# /.J.Z.70.,,,
..:l2i:g9.. iluss Looclst!p,.Q.er Wi179 low Incidence
40a5tJ# 2720.J#
/~S~7# /.9.:5.0/#
./Jra{J_ stressest.1,,())2.er M/JG?,Nose .f)ive
FIG. 19
/619?"'
0
32
h/i?a haclmce
Fm. 20.Drag truss loadslower wing
0 fronl Sμlr
{/jJ,Per Wia9
® .Fron! Sj)ar
Lower'Wi"rl!J
Fm. 21.Reference diagrams of drag trusses
@
33
ARTICLE 13. DESIGN OF DRAG STRUTS OR I Table 33 gives the computations of the values of
COMPRESSION RIBS PL. The member numbers refer to Figure 21. The
values of P are obtained from Table 32. The height,
From Part VI, Volume I, of Airplane Design,
the rule for the loading on internal drag
struts states, "The struts of the internal wing
trussing shall be designed to carry the design
axial load assumed to be applied to the strut
with an eccentricity equal to onesixth of the
spar depth in addition to any bending moments
due to the design of the fittings."
The struts for this design will be assumed of
the type shown in Figure 22A. This type,
which has proved quite satisfactory, is merely
an extra heavily trussed rib. Figure 22B,
an enlarged view of the rear spar with a
portion of the drag strut, shows the axial load,
P, acting with an eccentricity of onesixth
the spar depth, and the resulting axial loads,
P u and PL, in the chords of the strut.
Taking moments about point Q.
(
D H) P (3D+ H)
PLX D = P 2+5 • or PL= 6D ·
It is assumed that the chord members take
the axial load, P, and the web members carry
shear only. Then Pu = PPL. But, since the
eccentricity may be either above or below the
horizontal centerline of the spar section, the
design value of Pu must be taken as equal to
PL above, rather than the value of P  PL. '
In the above given equation,
P=axial load in strut.
Pu=axial load in upper chord of strut.
PL= axial load in lower chord of strut.
H =spar depth (including thicknesses of
trailing edge cap strips).
H t. ·t
6 =eccen r1c1 y.
h=depth of chord members.
D = Hh.
Ull14"'4~~~~~~~~~~~~~<1~l""'l"""x x
y
S~®AA
Fm. 22. Constructlon detail of drag strut
H , is equal to the actual height of spar H' plus twice
the thickness of the trailing edge cap strip. Assuming
this thickness as fr inch, H =H' + 0.375. The values
of H' are obtained from Figure 11, page 14. (H'=An
of figure .)
TABLE 33.Computation of drag strut chord loadPu
Member
Upper:
lL ___ ___      
12_            
13_              
14_            
15 _____              
Lower:
5 ___ ___ __             
o_      
H
Inches
3.675
5. 375
7.125
8.9().5
9.455
2.685
5.100
h
Inch
0. 75
. 75
. 75
. 75
. 75
. 75
. 75
D 3D
Inches
2.925 8. 775
4.625 13. 875
6.375 19.125
8. 155 24. 465
8. 7().5 26.115
]. 935 5.8().5
4.350 13. 050
3D+H
12. 450
19. 250
26. 250
33. 370
35. 570
8.490
18. 150
6D
17. 550
27. 750
38. 250
48. 930
52. 23
11. 610
26.100
3D+H
6D
o. 7094
. 6937
.6863
.6820
.6810
. 7313
.6954
p
Pounds Pounds
117. 44 83. 31
522. 30 362.32
642.44 440. 91
1,397.84  953.33
541. 80  368. 97
622.26 455. 06
759.55  528. 19
After the axial loads in the chord members of the putation, d, the distance between spars is taken as
strut are obtained, these members are t reated as col 27 inches for the upper wing and 23 inches for the lower,
umns for design. When considering failure in a lateral although these values are actually the distances bedirection,
the length L is the total distance between tween the center lines of the spars. Table ·34 gives the
spars, for a vertical failure L is considered as onehalf computations of the allowable loads and the margin of
the distance between spars. For simplicity in com safety in chord members of struts.
34
T ABLE 34. Design of chord members of drag stn tts
[Cri tical L/p= i2]
UPPER WING
I I l xx I L
Allowable lateral load
Member Size, inche.s PL
P .,/J y_y I
A
Area
1
I YY

]!_ _____ _ _ _ __ _ __ 12 __ ___ __ ____ ___ ~x~ 0. 3750 o. 0078 0. 0176
13 _____________ x~ . 5625 . 0264 . 0264 '4XU . 5625 . 0264 . 0264
1145 •_•__•_ ·___~_______ ulxXUu . 7500 .0625 . 0352 . 5625 . 0264 . 0264
5 ___ _______ ___ __
1
Ysx~ I o. 6563 I o. 0419 I o. 0308 I 6     · YsX"4. . 6563 . 0419 . 0308
For short columns, (L fp< 72) P= A[5,000 0. 5(L /p)•].
F or long columns, (L fp> 72) P
13
•
000
•
0001
 L'
27 13. 5 83. 31 0. 1442
27 13. 5 362. 32 . 2166
27 13. 5  440. 91 . 2166
27 13. 5 953.33 . 2887
2713. 5 368. 97 . 2166
LOWER 'YING
I 23 11. 5  455. 06 I
23 11. 5 j .528. 19 0.. 22552266 1
L /p Allow . I'

187. 2  139. 1
124. 7 470. 8
124. 7  470.8
93. 5 114. 5
124. 7 470.8
91. I 11, 029. 7 1
91. I  1, 029. 7
Allowable ver t ical load
p

o. 2165
. 2166
. 2166
. 2166
. 2166
0. 2166 1
. 2166
L /p Allow.P
  
62. 4  1, 144. 9
62. 3 1, 720. 9
62.3 1, 720. 9
62. 3  2, 294. 6
62. 3  1, 720. 9
53. 1 12. 356. 2 \
53. l  2, 356. 2
M. S.
(min .)
  
+v. 67
+.30
+. 07
+ . 17
+.28
+ l.26
+.95
ARTICLE 14. COMPUTATION OF LOADS IN
CABANE STRUCTURE
all members. These two views of the structure of t hi
example are shown in F igu re 23.
The cabane struct ure of this example consists of
four struts on each side of the fuselage, three of them
forming a tripod whose apex is at the point of a ttachment
of the front upper par. The fourth strut a t t
a ches to the r ear par and the fuselage, and acts
independently of the tripod. The stresses in the rear
strut have already been determined on page 29. The
stresses in the tripod structure are obtained by the fo llowing
procedure :
1. Draw front and sideview diagrams of the structure
showing the vertical, drag, and side dimensions of
fro/JI View
2. Compute the t rue lengths of t he members and the
dimension coeffi cients. These computations for the
tripod are given in Table 35.
3. Compute and tabulate the loads acting at the
vertex of the t ripod (p oin t of atta chment of tripod to
front upper spar). In order to simplify computa tion ,
the following assumptions a re made:
a. The vertical load, which acts perpendicular to the
propeller axis, is ~he same as the beam load.
b. The drag load, which acts parallel to the propeller
axis is t he same as the chord load.
27''~
Side ~ew
Fm . 23. Diagrams or cabane structure
T ABLE 35.Dimensions of tripod members
Member
A_                  B
                             .  . .  .       .  .   .        c 
D V S L engthL I
6. 0
4. 6
23. 5
34. 0
26. 0
26. 0
26. 625
37. 125
23. 125
43. 60
45.55
41. 92
D/L
0. 138
. 101
. 561
V/L I S/L
0: ;~~ 1,0: ~i~
. 620 . 552