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|lItilllu£lijg I|'.Ir.:‘

STRESS ANALYSIS OF THE WING STRUCTURE OF A
PRIMARY GLIDER

A Thesis Submitted to

The Faculty of
MICHIGAN STATE COLLEGE
of
AGRICULTURE AND APPLIED SCIENCE

Sven Gustaf Johnson

 

Candidate for the Degree of
Bachelor of Science
in

Civil Engineering

June 1941

 

"fHEiS'E

ACKNOWLEDGMENTS

This thesis is based on the theories presented in
"Aviation Handbook“ by Warner and Johnston (1931), and “Air-
plane Stress Analysis" by Alexander Klemin (1929) without
which this project could not have been accomplished. The
author wishes to extend grateful appreciation to The Few-
cett Publication who furnished the blueprints and to Bro-
fessors C. L. Allen and Jo E° Meyers for the assistance
given in the organization of this thesis.

SOG.J.
East Lansing, Michigan
June 1, 1941

TABLE OF CONTENTS
Introduction ...................................
A - Calculation of Weights
(a) Wings ..................................
(b) Tail Surfaces ..........................
(c) Fuselage 1 ..............................
B - General Data on Glider
(a) Gross Weight of Glider -; ..............
(b) Weight of wings ......................
(0) Net Weight of Glider ...................
(d) Total Span .............................
(0) Length of Sections ---------------------
(f) airfoil Section ........................
(g) Chord Length ---------------------------
(h) Angle of Incidence ----¢ ----------------
(i) Dihedral ...............................
(J) Area ot-Wings ...........................

(k) Location of Wing Spars in Per Cent From
Leading Edge of Wing --------- ---

(1) Center of Pressure in Per Cent of Chord
From Leading Edge ...............

(m) Load Factors Required ...................

(n) Ratio of Chord to Beam Component ........

(0) Discussion on Wing Loading .......... --—-
C - Wing Loads

(a) Effective Semi-span ---------------------

Pages
1 - 3

4 - 5
5 - 8

mmoommmmmmm

12

(b) Dead weight of Wings Per Lin. In. ------
(c) Normal Gross Beam Load

(d) Normal Net Load of Beams

(e) Check of Values

(f) Calculated Chord Leads

(3) Calculated Distribution of Loads on

the Spars ------------- 13
(h) Beam Loads Per Lineal Inch .............
(i) Moments, Shear and Reactions ----------- 15
(j) Nose Dive Condition (Moments, Shear
and Reactions) -------------
(k) Stresses in Tail Guide wires ------- ':--- 20
(1) Stresses in Wing Struts ............... 24
D‘- Investigation of Internal wing Structure
(a) Truss Solution By Method of Joints
(Case 2) ----------------- 27
(b) Truss Solution For Case 1 ............. 32
(0) Check or Truss wires - .................
E - Check the Design or the wing Spars
(a) Spar Constants (Front spar) ------------ 33
(b) n " (Rear spar) .............
(C) A1181 Loads on Spars (In bay) .......... 34

(d) Determining the Outer Point of Inflection _
(In Bay) --------------- 55

(e) Determining the Maximum Moment (In bay)- 36
and Computation of Stresses

Conclusion ....................................

Bibliography -a .................................

13
13
13
13
13

'14

15
18

19
23

32
33
33

34
34
35

36
42
43

44

-1-

’ INTRODUCTION

The glider under consideration in this thesis is one
that was presented in the "Flying Manual“ Vol. 2 - No. 6,
1941 publication. Having given some thought to building
a primary glider, I became interested in finding out Just
how well the wing structure reacted under loading. Since
this glider is intended to be of such nature that it could
be built by the average laymen, I was curious to know how
severe the stresses in the wing structure were as well as
become somewhat acquainted with the principles of airplane
stress anaylsis.

In stress analyzing the wings of a heavier than air
machine there are four main flight condition that must be
investigated. These are high incidence, low incidence, in-
verted flight, and nose dive condition. These conditions
are encountered when a plane is put into a steep dive and
then pulled out. This can best be illustrated by sketch
(8) p-e.

In the high incidence flight condition sketch (b)
there is a gross 11ft component as shown. From this is

subtracted the weight of wings giving the net lift comp-
onent. There is also induced drag making up the drag
component. These two forces form the resultant(R). If

we now break this resultant up into two components we have

f//'j/7z‘ COfld/f/O/AS /

fl? {‘7} // Inverted
/‘ F/l’y/It

.D/‘VC
Low
Ino/o’cnca /

#1‘9/2
Inc/a'encze

/

    

15"”

  
     
   

1"
Net A/ft /’ /

l
/

 

C7/055 .D ray

 
   
 

 

(Word
7 Componenzf ‘ H I
/r — F/ y. (D)
9r055 A/‘Fé
N’ct Lift // Qt
/__ C7‘r035 Dray

 

——f

the beam lift component and a chord component.The latter
acts‘T forward and will be considered negative.

In the low incidence sketch (0), the net lift component
and drag component are so nearly equal to the beam lift
component and chord load respectively that they are consi-
dered the same since the wing is almost horizontal in this
flight condition. The chord component acts backwards and
is considered positive.

In the inverted flight condition the loading is the
same as for the L. I. condition except the lift component
acts in the Opposite direction.

In the dive there is no lift component merely a ehord
component considered as the weight of ship distributed along
the beam.

With these considerations in mind we will proceed with

the wing structure anaylsis.

O

A - CALCULATING WEIGHTS
(a) WINGS
Length of one wing element=1849"= 225 in.
Width = 63 in.
Area =63x225= 1q175 sq. in.

Tip grass to be subtracted

 

 

 

 

 

 

 

 

”/I%\ 1/3bh.=13x6x9 = ............... 18.00 sq. in.
R /3 53 16 75 - '295 50 I a
\ii.‘ [3 l X x 0 " """"""""" e
/6 7.5” r
Space between ailerons lx98.75= -------- 98.70 “ "
" " ' end and wing preper

9/32X193 ------- 5. 30 " "

Total --------- 313 55 W“ n

Net wing area (13175- 416. 6)/144= 95. 58 sq. ft.

 

Total lifting area: 2195.58==191.16 sq. ft.

 

 

*From data on existing wings of this type weight
equals 1.10 lb. per sq. ft.

fling weight::l.10x191.16: --------------- 210.00 1b.

 

(b) TAIL SURFACES

 

 

(1) Rudder Area 1080.0 sq. in.
54x20 = ............
6x24 19/32 ........... 147.1 I n
Total ...... m n w
6. Subtract
I \\ e
S a Mama/exam: ----- 6.0 sq. in.
30’:
$\ A (3x20)/2 = ------- 30.0 ' "

 

 

 

*Alexander Klemin Airplane Stress Analysis"p.ll6

 

Rudder Area (Cont.)

 

 

 

 

\\§\ 8x20/2 --------- 80.0 sq. in.
‘éa§a>% Total ----- II870“" "
0" .
Net rudder area 1???.l-116.0 ---------- 1111.1 sq. in.

(2) Pin Area
1/2(h+Q)h 1/2(27+50)24 ............ 684,0 u u
(3) Stabilizer Area
32 1/2x102 ...................... 5315.0 u u
(4) Elevator area

1/2(equh 1/2(40+54)16 1/2 -------- 775.5 a a
Total tail surface as ................... 3880.6 n u

(5) Tail surface construction is usually taken as
weighing 1.0 lb. per sq. ft.

Tail surface weight:=5880.6/144xl = 40.4 lb.

 

(C) FUSELAGE WEIGHT

(a) Spruce -- average wt.= 27 lb. per cu. ft.

No. 10 gage steel 11:5.63 " " 33. ft.
N 12 N II II = 4 . 38 N N N
[I 14 n N H :3 1:3 H II I. II
N 20 3 N I! n -_- 1:50 I. N N I.

Bracin wire #30 Music wire =1.59 lb. per 100 ft.
** 1 3/16 O.D. steel tubing =1 lb. per ft.
3 7 strand flexible cable =2.45 lb. per 100 ft.
(1) Longeron Material li"x1}"x194.9"

1&x1ix194.9x27/1728 ------------- 4.73 15.

*Op. cit. p.4.

aanviation Mechanics Vol. 1 No.1, U- 80

e
‘-

 

-5-

FUSELAGE WEIGHT (Cont.)

- (2) truts Rea fusela e)
46) 27 1728 ---------------- 2.82 lb.

(3) Strut?k(Front) fuselage)

1 1/ 2%n209) 27/ 1728 ------------------ 15. 59 N
(4) Skid truss
(1&x11x228)27/1728 ................... 5.56 s

(5) Plywoo for Skid 5/52"
Area: (14x112)-102 2 = 29:52 sq. in.
O

Plywo -Gusset Pla as for
Fuselage Joints
Area 2(8x3x4) ----- 192 “ "
3l§4 " Total

* 5/32“ Plywood: 3. 2 oz. /sq. ft.
Glue (Casein)'=8. 0 oz. / "
Totat wt. --- fif§"" " "

Wt. of The Above Plywood
(11. 2x3124)/l6xl44 ................... 15.20 '

(6) Ash Wood

Ash Clamp

252x§x26x41/1728 .................... 1.99 "
Ash Clamp
2Q 3x%x9x41/1728 -------------------- 0.96 "
Ash Skid
. 2@3x%x44x4l/1728 .................... 4.91 “
Ash Blocks
aeSxeé x2x41/1728 -------------------- 4.80 "
Ash Bloch
l@6xleéx%x41/1728 -------------------- 0.53 "

nsh Rudder Bar
1519$x1§x7/8x41/1728 ------------------ 0.71 “

Ash Seat Board
lallxllx§x4l/ 1728 ------------------ 2.15 "

* Niles and Nowell, Airplane Structures p. 379

Fuselage Weight (Cont.)
(7) Control Unit
1 3/16" O.D. Steel Tubing

(21 11/16):1/12 ....................... 1.75
Torque ube 1 3/16" Steel Tubing
(25 5/ 1/12 ....................... 2.09
Co trol Stick Yoke 10 a. C.R.S.
[inéxzsyuflm/BD §14 5.63 -------- 1.17
Tor ue Tube Bea ng Strap loge. C.R.S.
(assay/14 5.63 ------------------ 0.75
Hoc r arm 10 ga. C .S.
(10x1 3/8x4)/l{g:4.38 -------------- 2.15
Eleva or Sheave Yo 9 Pattern 12 ga. C.h.S.
(1 5/8x4%)/14 4.38 ——————————————— 2.25
Six Sheaves 2 in. Die.
6x6/l6 -. ........................ 2.25
Wing Mount Plates 12 ga. 0.3.
45(2flet-im/14gx4cs ------------- 0.55
Longeron Joint Plate 0 ga C.S.
sags-an 1/8)/144 x5.63 ------------ 0.54

Front trut Clamps 10 ga. C.S. ’
w (2 5/813 3/16)/144]x5.65 -------- 0.65

hear Strut Clam 10 ga. C.S.
233[(3x5)/l44 3.5.63 ----------------- 1.17

Launching Hook 10 ga. 0.8.
as [1x8/144J x5.63 ................... 0. 63

wing Strut 1%" 20 ga. Steel Tube
4%!) 102%1'13.X0.056 ---------------- 2035

Brace hire Links 14 ga C.S.
205137/8x2 7/8)/144jx5.15 .......... 1.10

wheels Plus Tubing and axle ---------- 4'00

ID.

H

H

FUSELAGE WEIGHT (Cont.)
(7) Control Unit (Cont.)

Co trol Cables (7 strand flexible cable)

(72+68)/100]x2.45 --------------------- 3.45 15.
Braci hire #30 Nusic lire
[(96) 1ocflx169 ........................ 1.62 "
(8) Ian weight --------------------------- - 165.00 "
(9) Add for Incidentals ------------------- 57.00 “
Total Wt.of Structure ------------ .
Call It """"""""""""""""" 555000 "

 

 

B - GENERAL DATA ON GLIDERV
(a) Gross Weight of Glider (Man included)---- 555.00 1b.
(b) Wt. of Wings (1.10 lb. per sq.ft.) ------ 210.00 '
(0) Net Wt. of Glider(Gr0ss Wt.-Wing Wt.)---- 345.00 "

(d) Total Span ------------ ~ ................ 37'- 9"

(e) Lenght of Sections

(1) Lenght of Center Section ------------- 3 in.
(9) " " Bay ."“--9 ------ ~ --------- 116.0 in.
(3) " " Overhang ................... 109.0 u

(f) AirfofLSection Type .-------.---- Modified ”.3. 27

(g) 035d Length --------------------------- -- 63 in.

(h) angle of Incidence ................... --- o

(1) Dihedrel ........................... --- 1

(1) Area of Wings (Ailerons included) ------- 191.2 sq.ft.

(k) Location of Wing Spars in Percent From Leading
Edge of Wing:

Front spar 9/65xaoo -------------------- 12.8
Rear ' 44/63x100 .................... 70.0

(1) e Center of Pressure in Per Cent of Chord From
Leading Edge of wing.

By actual testing of different airfoils in
wind tunnel it has been found that the center of
pressure for an airfoil of the v.3. 27 type is

as follows:

He Ie -------- C--- -------------- 30
Io Fe --------------------------- 30
Le Ie --------------------------- 50

(m)_Loed Factors Required.
J In an airplane structure the load factor at
llow incidence should not be less than 3 nor should
N the load factor for inverted flight be less than 2.
The low incidence load factor is usually taken as
65 per cent of th:”€:Zidence value and that of in-
'?1§Zfit as 40 per cent of high incidence. Load

factor values then become:

H. I. 3/. 65 ---------- 4. 5 Use 5.00
ThOn Le Ie 5Xe 66 ....... --" 3e 25
lo 1". 5Xe40 .......... ZeUU

(n)*natio of Chord to Beam Lift Component for U.S.
27 Airfoil at:

5.1. - ------------------------ 0.175
LeIo ------------------- ---- 0.15
Io Fe ............. - ....... 0.0

(0) Discussion on Wing Loading.
Since there is no experimental data on the wing
the.

tip of wing under consideration the usual procedure is

to analyze the forces as follows:

 

e Alexander Klemin, Airplane Stress Analysis p. 118.

iffcc t/

V6
77/9

/Load C’urve

-10-

 

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flpp/fcof/OH "S! by [‘0
I .Fb/flt 9! 'Q‘o [u
b T ' ,/ ‘Q‘ 9 3310 $0
13 '3 I / 2'3 9.4 gm
3) ' 2‘ (a :3
bx ,, ‘ J 10
g £/// //// “/114/=o’eoo’ u/é of mes/J/Z’r/é’
M42 J lffc cE’ffE)§E’-.€’Z:T_§_€0_£W-_ 1.----.)
L77}, enzzflzf :43] 1
(fr 6(3/77/9 5pc”? __‘_‘__;
“Yr/51 /
fiffcot/Ve
/ Wing 77,0 // 400d Curve.
——————— c
l 1 Rf ’03
I RI 9| b
Outer 5trut Q! QJI [‘4
| _ flpP/Icot/b»? ! QID
' Point 0 0 [(3 W
' ’ IQIO “0 0 °)
5 3 0|
Q I ‘1): / 31“ E": Q
“I I Q! §l 0| lg
§// ”/7:// // //'7// 441.: dead Wt. 07‘ W/fiya/7}:7/ ‘K
GALA"), iffcctLge \ScQz/— Spa/7 ______________j
flake/75224:“ i
49ers”;- 3/0027 +4

 

 

 

* .
F/y. {3

e Warner and Johnston, Aviation Handbook, p. 620

-11-

If the wing is of a monOplane type varying neither in
plan form or thickness and is externally braced so that
outer strut point is approximately one chord length from
the wing tip use the loading shown in Fig. 1.

Since the glider wing under question has the outer
strut point located 109 inches from wing tip and 63s chord
length equalto 63 inches, it is commonly agreed that more
'severe stress will be set up due to this fact than the
previous loading diagram would give. when the structure
is of such nature that strut point is at a distance greater
than one chord length from wing tip use the loading shown
in Fig. 2 to design the spar sections outboard of the outer
inflection point infhe bay.

These loading conditions apply to a rectangular.wing
structure but since in our case we have a negative rake

I
on the wing tip of 17.5. But according to Warner, Airplane

 

 

7'0” 9 = /6'7f=.j/6

 

-— —— J3
) E2 .F75A.j
6‘ 1 a =/ZJ'°
.. _ T

 

Design unless the changing of tip form doesn't extend in-
ward more than one chord length from tip the lift increase

and center of pressure change is so small it can hardly be

-12-

given any consideration. Therefore for stress analysis the
slight negative rake will be neglected.

In summarizing the above loading conditions it can be
said that Fig. 1 loading is used in stress analyzing the
members inboard of the outer point of inflection. .For
stresses outboard of this inflection point use Fig. 2 load-
ing. For future reference call:

Case No./ ----- Fig. 1 loading
Cage NO e2 """ 2

In clarifying these loading condition it might be said,
what is actually done is to calculate an effective wing
which will naturally be smaller than the actual size due
to the fact that some air precaure is lost by air slipping
off the wing tip. To transform this effective wing to the
actual wing we have the slepe of load curve from w to 0.5w
and 0.8W in Fig. 1 and Fig. 2 respectively. Also the ef-
fective wing is somewhat less in length than the actual
since it is decreased 0.25Lt and 0.10L, in Fig. l and 2
irespectively.

Since the ovarhang from strut point to wing tip is
greater than one chord length (wing width) both cases 1
and 2 will have to be used.

C - WING LOADS
(a) Calculate Effective Semi-span S.

For case 1 --- 8.: 225- 0. 25x109 --------- 197. 62 in.
“ 2 -—- Se: 225- 0.10x63 --------- 218. 90 °

-13-

(b). Dead Weight of wings per lineal Inch
swam/22522 ---------- -----o 0.47 15. per lineal in.
(0) Normal Gross Beam Load

Case 1 : 5'9..555/(197. 62x2)d .4 40 1b. per.linea1 in.
Case 2 : w,- 555/(218. 90x2)d .m ' ' ' '

(d) Nonmal Net Load On Beam

Case 1 : is: 1. 40-0. 47 ----- 0.93 lb. per lineal in.
Case 2 : W,::1. 27- 0. 47 ----- 0.80 “

(9) Check These Values

Case 1 x 2(0.93x197.62)+210 -—--a ----- 577.56 1b.
Case 2 : 2(0.80x218.90)¢210 ---------- 560.00 '

These check values fire somewhat more than the actual~
gross load of 555 lb. but they produce more severe stresses
and are on the safe side. Therefore they will be used.

_ (r) Calculated Chord Loads

 

 

 

 

1 Case - l . ' .
'Beam Loa ad Chord Chord”. Ritual Cfiofd
Flight Per Lin. Beam. Load Per .Load Load Per Lin.
Condition in. Ratio Lin. in. Factor In.
::fH.I._' , 0.95 -0.175 -0.175*7* 5.00 3”fi'-0.865
L.I. 0.93 0.15 0.140 3.25 .0.455
I.F. 0.93 0.00 0.00 2.00 0.00
D13. ’ ' 1000 : :0e77

v —.‘—w w v -

 

In a dive the net weight of plane would be distributed
along the chord and the load factor would be 1.

In the above table the dive chord load '
(555-210)/'225x2 ----- ----- 0.77 15. per 115. in.

(g) Calculated Distribution of Loads 0n the Spars

\ECe/réer of Pressure

 

 

 

 

K E h g;
:9 E X' I x E '3
“ w—m ”1* -----*+ Q F.
8, %*_fl “ if- 1*-..4 a) g co. ‘9"
8 E : \ Q
Q K I Q S
c k ! 0 K
\1 i “r k
Front spar takes x/yxlOO =per cent
" " ' gear Spar location-center of pressure

 

Hear spar location-front spar location

Rear spar takes xVyxlOO a per cent

" " " Center pressure-front spar location
Hear spar locationnfront spar location

Then for Case 1 and 2 at H. I. and I. F.

Front spar takes 70-501100 _______ 69 93 0/0

Rear Spar takes 785%;100 ---- 50.07 o/<>

For L. I. Condition

near spar takes 50-12.8 ____ .
$6:l§T§'X1OO 05.03 o/o

The above values represents the percentages of the
load carried by each spar under the varies conditions.
The variation in the load on the spars is due to the
travel of the center of pressure toward the leading edge

of wing as the angle of incidence increases.

(h) Beam Loads Per Lineal Inch on Spars

-15-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

” Net Load o/o‘Cari‘Net Loaa
F11 ht Per in. Load ried by per Lin.
Cond tion Case Spar 0n Wing Factor Spar In.
3.1. 1 Front 0.93 5.00 69.93 3.25
Rear 0.93 5.00 30.07 1.40
2 Front 0.80 5.00 69.93 2.80
Rear 0.80 5.00 30.07 1.20
L.I. 1 Front 0.93 3.25 52.62 1.58
Rear 0.93 3.25 65.03 1.96
2 Front 0.80 3.25 52.62 1.37
Rear 0.80 3.25 65.03 1.69
I.F. 1 -Front 0.93 2.00 69.93 -l.29
REST 0.93 2.00 30.07 -0056
2 Front OeBO 2eOO 69993 'leO].
Rear # 0080 2.32 30e07 'Oe48
. Net load per in. = Column (41:51:65)
(1) Compute Moments, Shears and Reactions
Case No. 1 (Unit Loading)
/-/b For LII). [/7. A_J U
l T— ‘ 5
one M ’ J ’ J’. 3 ’ ’ B
. . 3 1.5;:- ,g- WV 6 '
AfiL——_ .—.-—--JL-~ —!——-——.{ :/ $9,723 Of! A 3 3 a.)
1 ,’ HflP/IZ' 25/0»? ! g 1 0x,
4' l ' A ' f 0
’ ’ - . I ! 1 \
; |/' I q i 3’,
Ii) 1 L i 1' 1 ’ (£25K
b--- ‘ .../52.211- __ __-_.- _ _- , ._ Jé: _ __ .2
-----__._--_~ 5 . 3.4:.-- _ - __ ,
\SGmPWS/Dafl =4°EJW We:

f5}? sf

-16-

By placing gunit loading of one pound per lineal inch
on spar

M,a(109x0.5x109/2) + (109x0.5/2x109/3)=3,960.33 in. 15.

 

u,.o

Shear right‘and left of reactions

7". 109 x(0.5+l)/2 ..................... 81.75 lb.

v,,u-5960.55/116 -' 116ml ................ -92.14 lb.
H.2Total --------------------------- .. 173.89 "

v“: 116 x g - 3960.33/116 ................ 23.86 "

7:0

(Chews) K,+ R,=1'I3.89 + 23.86 . 81.75 + 116211 4 197.75

c... No. 2 ( Unit Loading)

 

 

 

/- /b. Per Lu) /'/7- . a
1 T 1 { 1 f 1 i 1 b
, I ! ' ' i
f i } 5 ’ ' ‘ ’ I f N
, 1 1 ' 54‘1’701‘ ;6- Mfre 1 E g
g ’ ’ (what 01" 7 9
I l ¥,’ xl,°P//005Ibl7 , 1 I\
; 5 I ' ' w
1 I r ‘ ’ l ’
J 9 1 L I 1 j i "

 

 

 

 

 

1
Is
0\
t
5‘1;
‘R
I
ve—L—O
31

 

_—'—-—-———-V--A-o~——.——._— --—..-—-.»--—- -..-Mm .-. -— — , “<0 .._- ~——— A

.J3czn/—.5F%2n:.ez&f“

/:49. 6

-17-

'11,=63x.8(48+63/2) . 0.2163/2(46-63/3) + 46xlx46/2

 

-------------- 5,452.25 in. lb°
143:0
Shear right and left of reactions.
79:63 x(0.8+1)/2 + 4611 ............... 102.70 1b. "
Vg,=-5,452.25/116 - 116 x g ------------ 405.00 " "
R,:Total ....................... 207.70 " "
VL3=115 x 5 - 5452.25/116 R ------------ 11.00 ‘

Vgr-O

é

Check 11,. a£=207.70 . 11.00 =102.7 . 118 £218.70 lb.

To obtain the actual bending moments, cheer and re-
actions multiply the unit values just found by the loading

per lineal inch as shown in table under article (h).

-lc-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

’I‘j
I’ln'H’

 

munoson

HHH .oz flqmde

 

 

 

 

 

 

 

0.0 00.0m . me.0u 00.m 50.5mm 00.5 05.n5 em.0ew 00.5 00.55 0a.aom aeom
m5.55 05.000: 50.5: 50.05 00.50m 50.5 00.00 00.500 om.m 00.55 05.500 among m.
0.55 mm.a0 u 0m.0n 58.00 mm.0em 00.5 00.00 00.neu 00.5 00.nm 00.055 nemm
05.00 mm. mm: 00.5- 0a.an 05.55m 00.5 00.aa 05.000 0m.n 00.00 00.055 50055 5
w 5 .55 .055 m 5 .05 .055 m 5 .05 .055 m 5
m m .nmq 0005 m m Hen 0005 m. m non 0005 m m. 50mm memo
.m.5 .5.5 fl .5.: .05 .055\.05 5
mnoflpowox
55.00 00.0: 00.ma5 00.5 05.055 0w.5 0a.005 5:05
0a.005 50.5- 50.045 an.5 00.005 om.“ 05.005: 50055 N
00.50 00.0: 00.005 00.5 00.055 05.5 55.50 seem
00.055 00.5- 00.055 00.5 05.0. mm.n 05.“ .1 50055 5
5m .55 .855 c .05 .555 .05 .855
.> Mon 0005 HAP Hog 0005 HfiP Ann 0005 .nfi .n55\.mH a gnaw 0000
.5.5 .5.5 .5.m 5.xes. 5r> .
Hmonm
00.a50.mn 00.0: 00.55m.0 00.5 oa.mem.0 00.5 mm.mme.m seem
85.000.01 50.5: mm.moe.a a0.5 00.00m.05 00.m mm.mne.m 50055 m
ma.a5w.mu 00.0: mm.moa.a 00.5 0e.wem.m 05.5 00.0mm.n Heme
00.005.01 0m.5a mn.amm.0 00.5 a0.5am.m5 00.0 m.000.n among 5
.5 .05 .055 .5 .85 .855 .5 .05 .255
pom 0004 Hon 0005 com 0005 .na .q55\.na H Roam ammo
.a.5 . .5.5 .H.: .2
g L

 

 

 

~19-

(J) Hose Dive Condition

787/.Fbrcez

 

;
I
I
_.—_—

\jj” /j7”

 

 

 

+1 . l
F?o»t.5par fiflxn'lSpar' 7b// /313t’4yfl

xp/gh )7

In the nose dive condition the three forces shown in
Fig.‘1 keeps the structure in equilibrium. If we assume
the loading on the front spar being the same as for the

I.F. condition then:

ISM = 0

137(R.S.)==172 (F.S.)
8.3. = 1.255 (F.S.)
By using the worst loading on the front spar in I.F.
condition (Table II) the following results are obtained.

TABLE N0. IV

r #—
TL—

 

 

 

 

I.F. and Rose Dive
Nose Dive Rear Spar
(Front Spar) -1.255 F.S.
Case No. 1 Load /Lin. in. *1.29 1.58
m, ~5108.83 6,257.32
RI "224e32 274e75
a, 30.72 37.69
0888 No. 2 Lead/Lin. In. -l.01 1.24
M, -2,217.78 8,780.79
R: 209.78 257.55

R2 11.11 013.64

(k) Stresses In Tail Guide Wires.

(1) T0p wires

/¢77 .‘ .55 . " ~ 3’ '

 

   
 
 
    

Pear ~5por
of W/ny

  
  
 
 

21/
#J0 (Sroce

Wires

#JO larvae M/fres
Upper Lonyeron -

 

 

 

 

/7g7 :5
fl -; -

U: 2237.523 , 31,2/1781/ Post
Calculate the force P on fin and rudder

From -------- Sect. A(Art.b,1 and2)

R'Udder area ................. - ------ 111101 sawine

Fin n --------------------- ;684og_ .

Total vertical tail area ----- --- 1795.1 ' "

laForce on vertical surfaces 22.5 1b./sq. rt.
9 =1795.1x22.5/144 .....—....—-----4- 280.34 1b.

Area above upper longeron
3x26/2 39 sq. in.

21x26 546 "
2x20 40 n '
Total ----- 555 ' "
44X 3 232.55

X = 5.28 inches down from upper longeron is a
point where there is as much fin
and rudder area above as below.
This is the point or application
for the force P on tail post.

aAlexander Klemin, Airplane Stress Analysis p.202

 

U Tn” UPPG r Longeron

(u
5*, _._,____ 2490...»? u;

 

 

kl! W7oi/ POJZ'
\
N
A Var/Lower Lonyeroa
27U = 280.34x21.72
U = 280.34x21.72/27 225.52 lb.

 

”

l"
a

‘ 54.82 1b.

‘11“

Since the longeron is a flexible beam the slight
bending resistance offered by the butt Joint at trailing
edge of wing is neglected. The structure in Fig..8 is,
however, indeterminate; therefore, it will be investigat-
ed in the following manner.

Drop out tires No. l and 3 (Fig. 8)

....__..._....._ --.... I : [‘30 :l--...--*_.__-._._ gu;-_ -
X = 564’” 44 73/2‘” j l/éfl5‘7'

(7 iii5 _=_q *‘v" 2%;
f ~«w«~.2:-* A I *
U: 38.5.52 /5,

4 ‘
I = d/l2 (1%}2

30.2034

 

 

 

 

 

 

 

 

 

Fig. 9

a a --“- ““3
A x 25 (l - x): 225.5218x56fi1150 - 5611). .... 7.366 1 .
" 6E1]. 6x1. 2x10‘ x0. 2034x150 n

Then solve for the horizontal component at hflby

substituting into

 

a A
A : mik b
x
B” =, 3EI1 = 7.366x311.2110 x0.2034x130 4 3 0

(See Fig. 9’)

-22..

 

 

 

 

 

 

 

 

 

*___ /~ ARO” _ 1--..._-._._--
..., 1
C {I l H Fig. 9'
-jjjézfémmlil- b = 73/3 j,
( (3 f
2
f “w 3 3
i‘.\@ 92” s
b \ /721”
X eka—w A
£5»=4A3/Aa /
\ ® /
\ ‘ @ i.
\
\E‘g / / 1W Fig. 10
.\ / )\
____._J
\\ /flZf'
flyi‘r‘b \ H
U =ZBJTJZ’ /b. y r
Za=0
150A. - 41. 51x56$ - 225. 52x138= o
A ................................ 257. 57 lb.
Aqus. 5/117. 50)257. 57 -------------- 161.11 "
2:M,= o
41.51x73.5 - 1500,- 8x225.52 = o
c, .......... ----. .................. 9.46 "
(Check) 41.31 4 225.52 - --------- 268763" "

Force in wire No. 1 and a; #30 music wire

r.=(97.5/79. 5)41. 51 a 50. 59 1b.
53(154. 9/117. 5)257. 57 a 295.10 1b.

Stresses
S, = 50.59AOO505 ------ 1,005.7 lb./sq. in.
s, : 295.1/.00503----_ 5s,705.o ' .' '

Maximum allowable stress 165,000 " ' "

 

 

 

 

 

 

-23...
(2) Lower wires
fi WVhy'
{L . :QN J/ 71
Lower 1.0/7 eron
A I Ii . 9'
\ /]\\.
\\ / $9
‘§ \\ / ¢£¢u
\9 Wk 3 /
\
\ 4i \\ / F180 11
‘ H017 P/a/Ic
~ L.=¢f¢w6£’hé
Assume no bending moment at B.
ZM5=O
92.4Au- 54.821116.4=0
A” ---a ........................... 69.15 lb.
ZN,:O
54.82124 - 92.4BH=0
14.33 a
. R
55.74 "

5 race W/re

I1”: 69/.5' /b,

 

Fig. 12
P'= (145.93x69.15)1117.5 ----. ............ 85.88 1b,
Q : (28/145.95)x(145.95/117.5)x69.15 ----- 16.48 '
P : (85.88 + 16.48)5 ..................... 87,45 '

Stress = P/A . 87.45/.00503 ----17,400 15./66.111.
Allowable ----------- 155,000 lb./sq.in.

 

 

 

 

 

 

 

-24-
(l) Stresses in wing struts
If
Ab 112” H .5
\ a) I
a. .7 :
‘4 I [92%;]
.F/ . A] L
7 <22? (1 .5 :0
TABLE NO. V
Length and Components of fling_Struts
I I I I I I IL v - HI
Member IV ' B t D v v I D 1 - D c - L

 

I I

Front-~—(AC):45¢116.0: 0.012025113595.561

I 1 v o
Rear----(BD)c45.116.0:22.75:" 9

c
115620.561124.98

I I v
1517.56116138.121127.04

 

Front L/V 124.98/45 2.777
Strut H/L 116.0 /124.98 0.955

D/L

Loads in Wing

0.0 /124.98 0.0

Rear L/V 127.04/45 2.822
StrutH/L 116.0

127.04 0.917

D/L 22.75 127.04 0.179

Struts - Spar and Drag Loads

 

m

Load-
ing Con- tion In Strut Spar Loads
H L -le.5_w ”91.1

-_..w-‘—-. -w--—_~—,~-._r—n-...—...

Case 0
1 Front H.'.
L.I.
I.F.
N.b.
hear H.I.
L010
10?.
N.D.

2 Front Hole
L.I.

I.F.

N.D.

Rear H.I.
L.I.

I.F.

NoD.

 

~“u—‘p—‘m c...- ahr‘fi 'v—_- -.

Reac-

o
9
J
.

274.75
-224.32

243.45
540.82
~97.58
274.75

581.56
284.55
-209.78

249.24
351.01
-99.69
257.55

LC’LHE

762.98T

-623.61C

687.02T
961.79T
274.810
775.34?

1614.99T

790.2OT
581.960

705.36T
990.94T

726.80T

Axial

'6'.
711.770
581.85T

630.000
881.960
251.99T
710.990

1506.890

7572260
542.96T

644.980
908.690
257.97T
665.29C

 

7=h

Drag Loads

_ .199 .5)
0.00 ""
0.00

0.000

122.98
172.16
-49.19
138.79

0.00

125.90
177.27
‘46018
150.10

 

-25-

In the above table the front strut is in the same vert-

ical plane as the front Spar therefore it causes no drag

load due to lift.
(1) From TABLE No.7 (Front Strut)

Greatest strut tension force ------------- 1614.99 1b.
8 = 1614.99/0.1356 ....... - 12,052.10 lb./sq. in.
Allowable stress .......... 55,000 " “ "

Factor of Safety 55,000/12,081.1----4.57

Since this is a long slender tube Euler's Formula
will apply in the case of compression strength.

Greatest compression1orce ---------------- 625.61 1b.

P.=c 11281/1‘
C:=l

E = 28x10‘
I = 0.02467
1.3124.98

P = 1x x28x10‘x0.02467/124.98
4,564.6 1b. 7

Factor of Safety 4,564.6/625.61---- 7.02
(2) Check 3" strut pin for shear
v x(l6l4.99)/(O.521 -- 8,210 lb./sq. in.
Allowable = 35,000 " " "
Factor of Safety 55,000/8,210 = 4.26

(5) Check pin End or Tube for Shear and Bearing

‘ t .2 0.0.3.5- l7).
F/ottenea’ [Ind
L_

4’ fl PM) #0 le

 

.._._.—-—-q

 

 

.Area of a 20 gage It" O.D. tube ------------ 0.1536 sq. in.
Subtract 2" pin hole

Shear area = 0.1556 - (fix0.055x2) --- 0.0986 " "

v = 1614.9970.0986 = 16,590 lb./sq. in.

Allowable shear for mild carbon steel
equals ----------- 45,000 a I a

Factor of safety ==45,000/16,590 -- ------ 2.75
Bearing force =1614.99/gx2x0.055-—46,200 1b./sq. in.
Allowable bearing stress = 90,000 " “ "
Factor of safety'290,000/46,200 -------- 1.95

(4) Rear Strut

In discussing the rear strut let us consider the left
wing.: Refering to Fig.16 there might be a time when left
rudder is used which would put wires No. 2 and the lower
tail brace wires in tension as calculated. However, to
find the worst tension force on the rear strut these forces
are considered zero because there might be a condition
where right rudder was used putting the right hand wires
in'tenSion.

From TABLE NO V the greatest tensile force (rear strut)
equals --------- - 990.94 lb.

 

Since this force is less than the force on front
strut and from the fact that both struts are of the same
material and crossection the latter has a greater safety

factor in tension than the front strut.

 

1
\

-27-

In testing for compression strength, however, the ten-

sion in the lower tail brace wire produces a dOanard reac-

tion.
Q ::16.48 1h. (see Sect.-C- Art. k(2i]
P =»16.48xL/V= 16. 4812. 822 ---------- 46. 51 1b.
Load in strut ----------------- 281. 32 "
Total load in strut ------------- .

This force is also less than the compressive force
solved for in the case of the front strut. Therefore, no
quher investigation will be needed. A strut of less
crossection could have been used here. In calculating the
forces in the struts no consideration was made for the
stresses taken by the two wires (a) and (b) Fig. 13. These
wires no doubt take some stress but in the event that they
would become slack or broken the struts should be so de-
signed to carry the load. The wing structure is consider-
ed a rigid structure;therefore, it is difficult to predict
what deflection if any takes place in the wing thus putting

stress in these wires.

D -I!yESTIGATION 0F INTERHAL WING STRUCTURES
(a) Truss Solution By The Method of Joints (Case No. 2)
Refering to Fig. 14 and starting at joint No. (11)
the following solution applies:

 

0) .
@ 0/ / ZV = 0
/ 0 42.81-(11-12) = O
IQD 60 (11-12)---- 42.81 lb. 0.
«98/ EH 2 o '
(11-9)-59/55xv6rt.(11-10).o
(11-9) ----- 0.00

 

 

.131

 

A: 8.2. u m\ mesonasmeflzefinnxn.+8.51% ...-
.fi 8.3.6 ...m ...m
o . zw
2: 6.56 . SQWES-8.602186666336»;Ede-Sag .2

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unecoosoo phone seen up oedanaaass npcaw Hecma.uucsofl ucaoa Hosea

3on . ...-6k

-28-

 

 

 

 

 

 

 

 

 

«A .w: mQXQ. :43
be an. .0 0k 0.4.4} Eta +550
65 Q“ .8. omnuimsxmm. “3.4th
at .o\ .. ark he“ 1m fix.“ .Ikhbu 4K5?" \me New.”
hens-.53 ltd-beam. b 520.46% hen. amp .0 60.3.5 .560.th 6..
encfiNuunw.: a m
1.9 8: ©. 6 @1 @- x
\s $.05. Ms 7 $3.95- 7 , 09.3.0. \ , ,, 06 fire? 9.6.x... . 0 . a
A . a». a. . . .
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no to. W.- W- .V f. / , .f /
9, .7. r .- .. .9 9.1,. -.. . . .....L .. --..
. a\b\ , h<w<h. ,2 .aNMDV , 3 $1300. 2 NWNV / r
3.4....th @520 .15. New 0.. 6.5.3 @ .00 5.x- 6 50.5.4- 8 that- ® . .
hf .a5 .60 :ab .mh :mm. KOQKW/
WQNIN u Q WQQKK

QHHAnEd mwomOb 5:3 and.“ EH:
ZOHHHQZOO Brod: HOZHGHUZH waz .OZHmDn mmsmu. 62H B 20 waded

bn.¢n

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Hm.nmansm

.DH H©.anu.m

 

 

 

 

 

 

 

 

 

 

 

0H .wah no coausfioe can on headadm
ma 1.3.— 0\ MR
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0-0 a \ 0e 0.0? W840- 00-400
m 5.50 n .0.
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if. -MV . \x ,, a. o s. a
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ft /, l1 . .... / \ I I _9 /
z / 9 ,2 ,/A. awe ,/ .6” x./ .
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\ 0 ...: .. as a \0 .0: - © 00.53- .60 50 6.3..- © 50.007 ® 0.85.9.0 (-
® .60 ...-0 ( ...-5 .50 ...-5 /\0o\h7
ans-..- .. ms mast-K

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ea .ama no so-psmmm one 6» sea-s-m
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fi:.\e\ 0\0\Q--

   

 

 

 

 

 

 

 

 

ma «h
Sis-s 3?: E08
\\DQrmJ \Qbflux 00%: . merw 0.0-30%

ha. .... as... w: \. a
9 30.00.39me 00.000? @ 009-0-0- ® .00 00- @ 9056-.- ® -2
*Q \ H g I / \\ . \. /. .
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/ 1’ \\... IV / ‘0: . .. 1’ . / I. .H. 17 ./- a. 1\.-.\ . . / -\|.. r.- _
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GEM-En: mBomOh Hm; 4H4."- EH3
205:”ng SSH]: HUZHQHOZH BOA GZHmDo mmbmn. 62H; 20 named

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.
50.560.0- m. 00.0.: 000..- .5005. no 0.0. 00 .05. mi 05.
an «.06.. 5.26.3 max-65 a... ...6...) am. «65 mas-006-

 

QHHdmmd. muomOk HmHE mafia FHHR H>HQ wZHmDQ mwamn. 62:.- 20 mead-H

 

 

 

 

Proceding to joint No.(12)

@_

 

- 42-8!

 

®

69

CD

jZV'= o

42. 81 - 0

V,,_ _9 ----- V42. 81 lb. (tom)
ZH =

B...- 0(10-12) = 0
$2,39/55M¢,59/35x42.81:
47

(10-12)--47.56,lo.(comp.)

The remainder of Joints were solved in a similar

manner.

(b) Truss Solution For Case No. 1

If we solve tne ling truss for case No. 1 looding in

a similrr manner to the solution or truss for case No.2

loading (Fig. 14, 15, 16, 17) we will obtain the necessary

information to solve for the maximum bending momentin»

board of the inflection point.

The solution of truss for

Case No. 1 yields the following values. Only those values

used are listed.

 

 

3.1.
Front (2 - 4) 555.85 lb.
Spar (4 - 6) _ 119.87 '
a (6 - 8) -153.65 "
H.I.
Front (2 -4 ) ~205.57 1b. Rear (1 - 3) 142.24 lb.
Spar (4 - 6) -142.24 " spar (3 - 51 118.83 '
L.I.
Front (2 - 4) 1081.26 lb. Rear (1 - 5)-1616.27 lb.
Spar (4 - 6) 565.71 " Spar (3 - 5)-1081.26 “
(6 - 81_ 69.95 ' (5 -

7) :565.71 "

 

 

-53-

The truss for case No.1 loading was not solved as com-
pletely as for case No.2 because in checking the stresses
inboard of the inflection of the spare, we are interested
in the condition thetproduces the greatest bending moment
under the case No. 1 loading as stated in Sect.-B Art.(o).
In case of the rear spar if we inspect Tasgggnp. Villthe con-
dition that will produce the greatest bending moment occurs
under H.I. . Therefore to be sure we will investigate as
shown inTABLE NO. V11.

(c) Check Stresses In Truss Wires

(1) Drag wires (#30 music wire)

From Fig.(14,15,16,l7) the greatest load occurs in
wire 2 - 3 (Fig. 17).
P = 753.4 lb.

Area(#30 wire) = (670274)xfr -------- 0.00503 sq. in.
s xp/a . 755.4/o.oosos ....... 149,780 1b./eq. in.
Allowable -- ----------------- 165,000 " ‘ "
Factor of safety :165,000/149,780 ----- 1.01

(2) Antiédreg wires (#26 music wire)
F 8 114.8 lb.

Area (#26 wire) (6.632/4)TT ---------- 0.0031 sq. in.
s = P/A = 114.8/o.oo:51 ---------- 37,032 lb./sq. in.
Factor of safety =165,000/37,032 ----..- 4.45

E - CHECK THE DESIGN 0F4IHE hING SPAhS
(a) Spar Constants(Frcnt Spar)

(1) Position of neutral axis:

 

 

 

 

 

 

 

 

 

 

 

 

 

3
,¢%‘4
/
z
'5
t ...JZR .
5‘
JL_. x
. -%¢*:%J

__ Mcgtra/ Axis

'Since the t0p and bottom;n{ flanges are identical the

neutral axis occurs at center as shoin.4

(2) Homent of inertia:‘

1(2. 625;}--7. 961 in.

I(flanges) = 2[%/122 if. (
I (webs) = /12x§21(6 ........ -- 5. 575 "
I(tota1‘)t -- ------------- II?33E“‘
y/I = 5/11. 556 --------- --------- ..... o. 265
Area = 21% + 613/32 ----- - --------- 3-1. 688 sq.in.
at strut point and fuselage (spar solid) 7

' I = 1/12x x(e)’ ----- .--.--4- ----------- 15.5 in.
= 3/1 .5 ------ - ------------------ ' 0.222

Area = $16 ---------------------- 4.5 sq. in.

(b) Spar Constants (rear spar)

 

 

 

 

 

 

=1/12x%x(4. 5) = 5.65 (
y/Ia(2 1/8 /5. 65 = 0.577 4”
Area 2 £1 : 5.185 sq. in. gig
(0) Axial Load On Spars (In bay)
6359 N00 1 '
g. TABLE N0.'v11
average Due to ._1
Flight Spar Due to Due to Tail - Total
Condition . Dra Lift Wires
‘H.I. Front 107.g3 ~IZE4.242 "-1624.79
Beer ~523.52 -630.00 -122.98 -‘626.88
H.I. Front -159.55 ~1464.24 * ~1624.79
Rear 126063 ”630000 ‘122098 '626e55
L.I. Front 571.64 -711.77 . ~140.15
Rear ~1087.72 -881186 ~172.16 ~2141.84

 

 

*

-55-

 

 

 

 

Case No. 2
TABLE N9, V111
“ Average Due to -
Flight Spar Due to Due to Tail Total
.Conditicn Drag Lift Wires
(Fig. 14) Rear -330.34 -644.98 -326.52 -1301.84
H.I. Front _;154.17 -1506.89 -1661.06
(Fig. 15) Rear 122.39 -644.98 —41.31 -563.90
L.I. Front 561.72 -757.26 -165.54

Rear '1066067 “908069 -326e52 -2301e68

N.D. Front 623.40 542.96 1166.56
Rear -1150.61 -665.29 -526.52 -2142.42

Average axial spar load due to drag in TABLE NO.V11
(See Sect.-D, Art. (b)

Average axial spar load due to drag in TABLE NO.V111
(See Fig. 14,15,16,and 17)

Axial load due to lift (See TABLE N0. V)

Axial load due to tail wires (See Sect.-C, Art. k)

a In this particular flight condition the tail brace wire
No. (1) Fig. 10 produces enaxial spar load as shown at the

second panel point.

(d) Determining the Outer Point of Inflecticn

By referring to Bect.-C, Art.(i) it becomes obvious
that the outer point of inflection is determined by case
No. 1 loading because u, is less than in case No. 2. Also
the loads are larger per inch in case H0. 1 thus reducing
the moment to zero sooner in the bay.

Point of inflection equals:

I,:M,+V,,x+vx‘/2=0

-36-

From 3ect.-C, Art. (1)
---------------------- 3,960.33 1ne lbs

11.
v,, ---------------------- -92.14 1b.
" ---------------------- 1.00 "

3,960e33 " (-92e14X) "' ”2:0

184.28 - . - .
72

68.28 Or 116 . 00
-===F~

N N
u n

This formula neglects the effect of axial loading on
the spar but for the purpose of determining the inflection

point the above values are accurate enough.

(e) Determining the thimum Moments in Bay

Bed/27 iced/77y

 

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By inspection we can readily see that the wing beams
are not simple supported, but have a bending moment at
the strut point and also a uniform plus axial loading;:
therefore, to determine the maximum bending moment in
the bay of beam the precise moment equation will be made
use of. Calling strut point N0. 1 and the point where
wing is fastened to fuselage No. 2 then the moment between
these two points can be found from equations:

(1)*N.321782;é;;3L/1 sin x/j + D,sos x/j + wj‘

v ' _

 

D = M «11.1“,2
D = H - VJ
J = (El/Pyé

if

(2) Tan x/J = D‘- Dgcos #Z]
D,s n

e
(3) Maximum moment between points 1 and 2

MM, .2 D, sec x/j + wj‘
; I

:With these fiomulae in mind we will proceedtofie solve

for the maximum moments in the following tables.

a Niles and Nowell, Airplane Structures p. 201.

 

 

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-42..

In the TABLE§fx and x1 when the moments were plus
this indecates that there is no inflection point. If we
solve for (x),the distance from strut point to point of
maximum moment,and this is found to be a value greater
than (L), the length from strut point to fuselage,then
this indicates that M. at the strut point is the worst
moment. Thus:

In TABLE NO. 11 under rear Spar H010

 

Tan x/j: -19.733
x/j :- 92.9 = 92.9/57.3=l.621(radians)
x = l.6211’72.03 = 116.6 in.
which id greater than L (116.0 in.)
If we were to choose some point outboard of the strut
point it can be very easily seen that the axial force as
well as bending moment on spar would become less and snub“:
one to design the spar with a smaller crossection as the
section becomes nearer to the wing tip. However, since
the Spars are of the same crossection throughout in this
case there is no need to check the forces outboard of the

Strut pOInt o

 

 

-45..

CONCLUSION

In working through this problem in stress anaylsis the
forces in the ribs at panel points were solved for in the
solution of truss but no mention was made of as to what the
allowable stresses are. The strength of the ribs in a wing
structure are not computed mathematically but are tested in
a mechanical Jig. However, in our case the forces in ribs
are not severe as can be seen in Fig. 17. The worst com-
pressive stress being 503.19 pounds.

An inspection of the results found in this thesis re-
vsggz none of the members of the wing structure are over-
stressed. 0n the contrary most of the members have a high
factor of safety. Some of the safety factors are as high
as 4 or 5. The design could perhaps be made more conserv-
ative but this would not be recommended since this glider
is intended to be built by the average airminded person
who in most instances would have no means of testing the
structure or material, an.0peration which would be neces-
sary if one Wto adhere to the 1.5 safety factor which
is used in some aircraft design. Furthermore in a glider
of this nature we are not as much interested in attaining
high Speeds as we are in reliability under rough usage.

This thesis does not go into any great detail in the
application of aerodynamic principles.but I have merely

applied some of the fundamentals to one of the simpler

type of aircraft design.

BIBLIOGRAPHY ” , I g'.,}\ i

Niles and Nowell, Airplane structures, New York, John Wiley ~f
and Sons, Inc., 1929. ‘ «3

Warner and Johnston, Aviation Handbook, New York and London,
McCraw - Hill Book Co., Inc., 1931.

 

Klemin, Alexander, airplane Stress Analysis, New York, The
Renald Press 00., 1929.

Warner, Edward P., Airplane Design , New York, McGraw-
Hill Book Co., Inc., 1927.

Klemins, Alexander, Simplified Aerod namics, Chicago, The
Goodheart Wilcox Co., Inc., 19 O.

 

Trayer, George h., Wood In Aircraft Construction,
Washington .C.,‘The National‘fumber Manufacturers
Ass., 1950.

 

Rathbun, John B., AerOplane Construction and Operations,
Chicago, Stanton and'VanVIiet Co., 1918.

 

 

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BENO OvER '
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.— C‘. .R STEEL ——

 

 

 

 

 

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*.._ fi5555555.5555'555955 5.5     BLUEPRINT No A 5|2   7 "7m

SHEET NO 7

 

 

MICHIGAN STATE UNIVERSITY LIBRAR’FS

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3.030'6 007