Design of monocoque fuselage

.1NST. TE

JUN 27 1942

LBRA4R

MONOCOQUE

FUSELAGE "

by

Captain Ismael Ndnez

-

Graduate from Army Hihg Technical

School -

Argentina.-Submitted in

Pattial

_{Fulfillment of the Requirements for}

the Degree of

OF

SCIENCES

"

in

'AERONAUTICAL

ENGINEERING

MASSACHUSETTS

FROM

THE

INSTITUTE

OF

TECHNOLOGY

1 9 4 2

Signature of Author .. -..

Department of Aernautical Engineering !ay 16,194

Signature of Profesor

Signature of Chairman

of Departme

Committee on Graduate Students

..

..

Cambridge, Massachusetts.

May 16,1942

rofessor George W.Swett,

Secretary of the Faculty,

Massachusetts Institute of Technology

Cambridge,Yassachusetts.

Dear Sir:

I herewith submit a thesis entitled

" DESIGN OF MONOCOQUE FUSELAGE ", in partial

fulfillment of the requirements for the degree

of

MASTER

OF

SCIENCE in Aeronautical

ring.-Respectfully,

Ismael

Ndife2

Captain of Argentine Army

Air Corp

nee-N B E

No

Thesis Proper

Page

1

Theme

1

2

Condition I

3

3

Running Loads due

ConD.I

18

4

First Hypothesis

23

5

Three Points Landing

24

6

Forces & Yorments for

Three

Points landing

33

7

Design for three point

34

landing

8

Second Hypothesis

38

9

Forces &-'oments due to

Condition I

44

10

Forces and moments due

three point landing

46

11

Design for hypothesis

II

47

12

Conclusions

50

xIt"

"TAKING MY AIRPLANE DESIGN PROBLEM OF 16:14,

WHICH

THEIJE

HEREBY IS;I HUST DESIGN SO!E OF THE MOST LOADED RINGS

OF TEE M1ONOCOQUE FUSELAGE "

Twin-engines light bomber having the folloming charac-teristics:

PERFORMANCES:

1) Military load 300072b.

2) Crew four men @ 200 lbs. each (include parachute) 3) Range at economical speed 8000 mi lles

4) Take-off run (ground) 3000 ft.

5) Yax. speed (at critical altitude) 300 J. P.H.

STABILITY AND CONTROL:

1) Satisfactory longitudinal stability and contorl

for all center of aravitv positions.

2) Satisfactory lateral stability and control.

3) Sufficient rudder to maintain straight flight at

1.20 V min. with an engine

"A R E A ( sq. feet

Wfing

0 -Stabilizer. .

Elevator

Ft n

Rudder

- .0 . .0 . 0 * 0

.

548.0 72.0 . 0 0 0 0 . 24.0 . 0 0 0 0 0

65.3

0 . 30.7

"LOADINGS- ON DIFFERENTS LARTS "

Wing

C

lb/sq.ft.). ...- ... 55.00 Span

(lb/ft.

).

. .

. . . . .. . . . . 408.00 Power

(

lb/Hp.

)...

8.16 2 f- "I

)

"

C 0 N D I T

I 0 N

=

1 + 3.25 77 +

p435

32000

W

f 9200

Ratio of total lift to gross weight

= 302oo lb.

3700

lHp.

. Power Loading - 30200 lbs. Gross

Weight

= 1850 x 2

s

=

Wing Loading

Log. 8.16 - 0.9118

.435 Log8.16.

0.3970

=

(8.16)

4 35 n = 1 + 3.25 2.49

(

=

2.49

.77

N or n

p

p

W J

W

8.16 lb/Hp

=

3700 Hp.

HP

s i

/

5 = 30200 lb

548sq.,ft.

= 55 lb/sq.ft.

.435

p

+

32000

30200 _{+ 9200/} I "t

N

K

Log.

55

iLog.

55

4

1/4

(ss)

K

=

KUV

m

575 s

=

1 s 114

=

1,97402

=

0.4350

=

2.725 = 2.725

2

From Air Commerce Yanual in figure lla. the value of

K for

my case is :

1.20

30

ft/sec.

( Gust velocity

)

Velocity at horizontal

flight

Power Plant

Efficiency

0.60 52.7 ( d

)

p

CDmi

_{n x} 5.48 0.07

=

32.8 sq.ft.

SW

=

0. 010 x 548

4

- 1(55(1 2 .

1. 3625

K

U

e

7

d

Sdw

SDw CDf S

fus.

Gnst Load Fantar :

0.10

20. 95

0. 07

x

32. 8

=

2 x 0.1 x 20. 95

=

2.30 = 4.19 5.48 + 2. 30 + 4.19 11.97 = 30200 / 11.97 = 2560

1/s

VL = 52.7 ( .60 2560) 8.16

VL

= 303 . .H.

When I calculated this velocity in same problem of 16:14

using the method as is it in Technical Report 408,1 got

the same velocity.

M = An/4

R

m slope of lift curve when aspect ratio is

equal to 6 CDnacel .

Snacel.

SDfr

SDna c. SD

d

=

1.20 x 30 x 303 x 4.87

575 x 55

= 1.455

S

1

-

An

= 1

=

2.455

GUST LOAD FACTOR:

11

L

q

it I = Total Lift

=

30200 x 367 -

Dynamic

pressure

=

230

q

a

.111.,000

lbs. V2mph/391

CN

L /

q.

CL 111000

/230.

548 CL 0.88

From Airfoil 23015 characteristics.

I1I

Angle

o,

attacw

From same

airfoil

data

C - 0.31 c= _{CD c o scr - CLs i ncz}

6

An

A

n'

ni

n,

-

1.455

1.50

x

2.455

3.668

CL

Foro

C

cc

Fpr

Fpr

S

-

0.16 C

-2

+

H

C

0. 0251 =

3. 8 ft.

=

110 coser = 0.981 si nor, - 0. 191 0.31 x 0. 981 - 0. 88 x 0.191 0.136 375 g

HPa/Va

375 x 0.60 x 3700

/300

2775 . 1

f

- nlh₂ + nix₂ 4

nx4 (h

₄

-h

₂

)]

X3 ~2 1

case due to

the airplane characteristics:

0

-0

n

₃

In my

h

4 h 2 H

C

. - 1.184 ft. x 2

x 3

-0.16 x

7.4,ft.

"S P A N

I

NO

TIP D I S T R I B U T I 0 N" LOSS " CN

=

Constant

=0.

R

-l-5 in.

" WITH TIP LOSS "

.-- --

1326.

2

in.

t

CN, 0. 88

I

C88.

8i 77.

CN=

0.'70

_ _ _ _ _ _ _ _ _ _ _ _ _ 71 U __I +--- 361.5 in.

8

CN =44

I

~ ~ ~ ~ ~

X

322

in.__ _ _ _ _ _ _

I__-

--

-_ _ --- A.C t/leQ

T1/f

K

/

2

3

4

5

7

8

1/ / /2

/3

/4

/5

Y

YP,,

C4Y

d

CIY

Y

Cd

Y'X

XGA CAy

Zc

?j 7gCAl

C(Ldy

C__c

_

I

278

75

/20

9000

/

00

26 2"

/,

0

/%

aam

o

a

Oaeqa Q

o

c

1V

/425

/70

96

/63 o

/00

/

.

a

;

a

V

-

5?90

.

2776

82

70

157O

/0D

6741

/

i

0

u4zO l-/2

21V

3615

68

53

4660

1.00

4660 /A

0

-

0

357/0

35,7/ 6,)'?4

v:P/

/07

7/9

z

J1

5.8A5,200 35,710

=

163.50

=

Z

(12)

2(5)

= (15)

2 13,

-293,100

35,710

0

3 294,000

35,710

.

=

19, 794

3,294, 000

35, 710

35,9710

=

92,20

0.006

1. 0

C0

Ky

Kb

ff

m i mumis mi i1

/

2

3

4

5

6

7

8

///

/2

/3 / /0

~STiPE

A.

4

b,

C

Caf

kzCta

14y

x

₀

,e

xecjgy

e

zq4C4

c

2

4y

c,,

Gct

f

2'7

7$

/20

96190

/A00

9CV0

250 20O

/',,0 /2S4k0

0

0 / 0

4w

a

o0

g

c

.2

/4.2-

/70

S%

/63/0

/0

/3/0

?,32,.

.f

Oaoo A

o

- 990

0

12

2776

62

70

5740

/00

574

66,OO 3.$

.

0 1oz 061

.

/

361.5

68

53

4660

0.706

37/

4

0

-

0

0

.4%6'

,, /.

357/0

₃₄₇₆₀

_{usaf %o}

_/2o

W/I&/7/7/)

ZLY2JS

sj-r

-

2

(8) . : (10) = 5,503,200 34,760 = 293,100 34, 760

=

158.2

=

8.43

.

2

12)

z(7)

=

Z

13)

2:( 5)

=

3,294,000 35, 710 = 92.20

=

0.006

-7 5+ = 34,760 35F710 = 0.973

y

0

C

_Y

2

15)

2 (13) Kb

"B

A L

ANCI

NG C0MPU T A TI

oN

0 "

T A -B L E

III

C

0

N

D

I T

I

0

N

I

For

tip

lloss

and no

tip loss

No

I

T

E

VL

=

300 MPH

W

(

Groos weight

)

30200

2

a

= 0.00256 V2 230

3

s

=

/

55

4

q/

s

(2) /(3)

4.18

5

n

(

appl. wing load factor

)

3.67

6

CV

(5) / (4)

0.878

7

CL

0.880

8

C

0.136

9

n

₁ =

(8) x (4)

0.569

10 nx₄ = Fpr/(1) 0.092

11

Cm

(

Design moment

Coeff.)

0.006

12

m

(11) x (4)

0.025

13

n

₃

(

Tail load factor

)

-0.106

14

n

₂ =

-()

-

(13)

-3.

564

15

n

=

(9)

-

(10)

-0.661

16

T

(1) x (13) (tail load)

-3200

i AND DISTANCES J

a

C2 CN q_ 144

c

_Ma

r

e

C 1244

q

C z4r4

--

b

-C' I-

--1

"

T

A

B

L

E

FOR TIIP LOSS AND NO TIP LOSS

I T

E

H

Stations

Distance from root inch.

C'

/

144

f

(fraction of chord)

r

(.

"

"

)

b

r-f= (4)-

(3)

a

(fraction of chord)

j

(

t

"

)i

e (unit wing weight

)

r

-

a

=

(4)

-

(6)

a

-

f

=

(6)

-

(3)

r-j

=

(4)

-

(7)

j - f-=

(7(

- (3)

C'1144b= (2)

(5)

along

span

1

2

3

4

5

6

7

8

9

10

11

12

13

I

27.8

0.833

0.15

0.65

0.50

0.238

0.245

7.44

0.412

0.088

0.405

0.095

1.666

II 142.5 0.666

0.15

0.65 0.50 0.238 0.245 7.44 0.412 0.088 0.405 0.095 1. 332 III 277.6 0.486

0.15

().

65

0.50

0.238 0.245 7.44 0.412 0.088

0.405

0.095 0.972 IV

361.

5

0.

368

0.15

0.65

0.50

0.238

0.245

7.44

0.412

0.088

0.405

0. 095

0.

736

/6 _If i I

Yf

S

I

Net running load on front spar

CN(r

a) + Cia J

. n

2

e (r-i)

Net running load on rear spar

-

N(a

- -

C

y)

qa

q

+ n2 e

(j

f)J

Net running chord load

c

.q

+

nx2

.e

)

C'

/

144

Tr-Tc

T A B L E V"

N 0 T I P L 0 S S

STATIONS ALONG SPAN

No I T E E I II III IV

Distance from root 2708 142.5 277.6 361.5

14 CNb C x R b

/

Kb 0.88 0.88 0.88 0.88

15 Ca (variation with span) 0.006 0.006 0.006 0.006

16 (14) x (9) 0.363 0. 363 0.363 0. 363 17 (16) -

(15)

0. 369 0. 369 0. 369 0. 369 18 (17) x q 85.0 85.0 85.0 85.0 19 n₂ x (8) x (11) -10.75 -10.75 -10.75 -10.75 20

(1R)

-

(19)

74.25 74. 25 74.25 74.25 21 Y

=

(20) x (13) 123.6 99.00 72.20 54.60 22 (14) x (10) 0.077 0.077 O.077 0.077 23 (22) - (15) 0. 0835 0. 0835 0. 0835 0. 0835 24 (23) x q 19.2 19.2 19.2 19.2 25 n 2 x (8) x (12) -2.52 -2.52 -2.52 -2.52 26 (24) - (25) 16. 68 16.68 16.68 16.68 27 Yr (26) x (13) 27.8 22.2 16. 2 12.3

28 CC(uariation with span) -0.136 -0.136 -0.136 -0.136

29 (28) x q -. ?1. 30 -31. 30 -31.30 -31.30

30 nx₂ x

(8)

-4.92 -4.92 -4.92 -4. 92

31

(29)

-

(30)

-36.22 -36.22 -36.22 -36.

22

32

Y

c - (31) x (2) -30.2 -24.15 -17.60 -13. 34

" T A B L E "

N 0

TIP No I T f H!

-

14

15

16

17

18

19

20

21

22

23

24

25

26

27

STATIONS

I

27.8 0. 905 0. 006 0. 373 0. 379 87. P -10. 75, 76.45 127.3 0. 079 0. 074 16. 94 -2.52 14.42 24. 06

distance from root

CfNb ~

'NI

x Rb/Kb

CA(variation

with span)

(14) x (9)

(26)

+ (15)

(17) x

a

n2

x (8) x

(11)

ai8)

+

(19)

Y

=

(20) x (13)

(14) x

(10)

(22)

-

(15)

(23)

x

q

n2 x

(8) x (12)

(24) + (25)

Yr

(26) x (13)

c.

with span)

V " LOSS " -0.136

28

ALONG II

142.5

0.

905

0. 006 0. 373 0. 379 87.2 -10. 75 76.45 102. 0 0. 079 0. 074 16. 94 -2.52-14.42 19. 20 -0.136 SPAN III

277.6

0. 905 0. 006 0. 373 0. 379 87.2 -10. 75 76.45 74. 30 0. 079 0. 074 16. 94 -2.52 14.42 14. 00

-0.136

IV

361.5

0.

905

0.

006

0.

373

0. 379 87. 2

-10.

75

76.45

43.40

0.

079

0.

074

16.

94

-2.52

14.42

7.

860

-0.136

ZOADS

/7

fDOQ6/T

JD/

Q

37a 2 "

/60"

286"

-

375'

"4

/851

I

--0

246"

'if

I

70"

3262"

-~______________

I

__________________

I

T

c%"

4

- --

4/5"

I

/Y6

7//?

Z&S

I1 C8.8"

~~~1

i

7.5"

D&1/#/6

ZO6I5

//

f(//T

5g42

J70. 2"

- 86 in /e " /92?/I>> -- 75"

~

-7

I

/ 70" 245"

+

"

I

_____________________

I.___________________________

I It 5 I. I I 'r A .1 4, 1

~~1

I-NI-

88.8" Sallg

DUM##Y6

I

I

" TOTAL kIOMENT TIP L 0 A D

(2)

75 x 185.4 - 13910 170

x

148.5-= 25260 82 x 108.3 = 8880 88 x 82

=

7220 55,270 " WITH TIP I 75 x 192.0 = 14400 II 170 x 153.0 = 26000 III 82 xlll.5 - 9150 IV 88 x 65.1 = 5730 55,280 SECTION (1). I II .II I V HOMENT

(4)

.529.,000 4, 040, 000 2, 540, 000 2, 600,900 9,702,000 ARM (3) 37.5 160. 0 286. 0 370.2 LOSS 37.5 160. 0 286.0 370.2 540, 000 4,160, 000 2,716, 000 2,221,000 9,637, 000

22

AND LOAD " N 0 ON LOSS " FRONT SPAR " I

".FIRST' HYPO THESIS"

As usual I shall assume that the spars of the wing

(

two spars ),its going thru the fuselage.

Latter I'll assume another hypothe-sis considering the spars does not go thru the fuse -lage-and its will be clamped in the fuselagp.I never saw before structure like this in regular and standard airplane,but I know that from of stability view-point the best airplane is the middle wing , however most of the time the designers do not use that kind of aircraft due to the fact that the wings spar thru the fuselage deducts room right in the more important

pla-cearound of the Gravity Centerwhere equipmentpilot or crewmen necessarily must be

place.-CONDITION"

"CALCULATION OF LOAD FACTOR n

In figure 24 of the "Civil Aeronautic Manual I got the following expression:

=

2.80

=

2.80

+.

9000

+

4000

+

9000

30200 + 4000

=

3.063

Gross Weight

Langding Gear

Weight in landing

Design Load for

3 point landing

= 30200.00 = 141000

=

28,790 lb. - 28,790 x

3.063

x

1.50

t

11

"

_{132,400 lbs.}

Now the Design Load,Must he dinidedl betwppn

landing gear and tail wheel in inverse proportion to the distan ces,measured parallel to the ground line,from the C.G. of air -plane to the points above mencioned.

24

"THREE POINT LANDING

WEIGHT ON FRONT WREELS

=

WEIGHT

ON

TAIL

WHEEL

=

On each front of the landing gear

132,400 x 500

560

118,200 lbs.

132,400 x

60

560

14,200 lbs.

I'1l

have

118,200

=

59,100

lb.

2

These values of forces were obtained from my balance diagram and shortening the shock absorver and tires at middle

way.-tX/f

Q/7zl

L 7C, 2/QCtS

7Z6/SfL//16f

ZZ4~62/2j

JIOZJQ

I

5~9/c4Q

I

I

p-

75"

-~ -

6~9/O67~ 7~5~"

" BREAKING LOAD DUE TO TIREE E O/IfT

LANDING ACC'ORDING TO TAGETIA.L OA C

AT RIffG OF OiCOCO Uz"

C. (.

20*

/SIAe 5'no

4---FORCES ON MONOCOQUE RING DUE TO THREE 1OIN

LANDING "

28

"FORCES AND FOEMNT ON POINT From point 0....3600 From point F....1800 SHEAR FORCE ON 0:

Due to tangential at

Due to tangential

i

Total SHEAR force

NORMAL

FORCE

ON

0:

Due tangential force at

0 .

-. 23(-58,825) F = -.08( 58,825) + 8,870 lbs. =

+

13,600

=

0= +.50(-58,825)

Z,730

Due tangential force at F= Total NORYAL force

YOMENT ON 0:

It does not exists

= + 28,550 lbs.

-Q "

0

FORCES AND MOMENT ON POINT A "

From point 0 . . . . .330O

From point F . . . . .150

SHEAR

FORCE

ON

A:

Due- to tangential at 0

=

-. 02

(-58,825)

=

+

1,177

Due to tangential at F

=

-.05

(

58,825)

=

-

2,941

Total SHEAR force

=

- 1,765 lbs.

NORMAL

FORCE

ON

A:

Due to tangential at 0

Due to tangential at F

Total

NORMAL

force

.32

(-58,825)

=

-

18,820

.12

(

58,825(

=

+

7,065

-

11,770

lbs.

MOL'ENT

ON A:

Due to tangential at 0

=.06(45)(-58,825)

= - 159,000 Due to tangential at F =.04(45)( 58,825) = + 106, 000

Total

MOMENT

- 53,000 in.lb.

" FORCES AND MOMENT ON POINT

From point

0...3000

From point F...1200

SHEAR

FORCE ON

B:

Due to tangential force 0

=

Due to tangential force F

.

Total SHEAR force

.09(-58,825)

-.02(

58,825)

=

+

- 4,138 lbs.

NORMAL FORCE ON B:

Due to tangential force 0 =

Due to tangential force F

=

Total NORMAL force

=

.09(-58,825)

U

-.15( 58,825) = + + 3,540 lbs.

MOMENT

ON

B:

Due to tangential force 0 .. 04(45)(-58,825)=

-

106,600

B

"

5,320

1,182

5,320

" FORCES AND MOMENT

ON

POINT

C

"

From point 0...2700

From point F ... 900 SHEAR FORCE ON C:

Due to tangential at

Due to tangential at

Total SHEAR force

NORMAL FORCE ON C:

Due to tanaential at Due to tangential at

total AORKLL force

0

F

.09(-58,825)

.09( 58,825)

0

0

F

-. 08 (-58,

825)

.08( 58,

825)

+ 9,450 lbs. MOMENT ON C: tangential at 0 .-. 015(45)(-58,825)n tangential at F .015(45)( 58,825)= YOYENT

=

+

79,800 in.lb.

+ 39,900

+ 39,900

32

=

5,320

5,320

+

- + = +

4725

4725 Due to Due to Toatal

0 FORCES AND MOMENT ON RING MONOC04UE WHERE

TUWE YING GOES THRU IN HYOCTHESIS I

TO THREE t-OINTS LANDING "

DUE V = ± ,765 N - //,770 1/ = - 64000 7. + 8,670 . 28660 Alf 0 V .+ /,75 N -. //,770 A" . + .53,00

v

o

N = +9,450 V

=+

4/38

I

+a74 V- -4,13 8 N .+ 3,,14*N _{= + -1,640} y .0 c = D B 7 -,765 - N

=//1,770

3 300 3 300 30300 300 3000= S0 N

=

24,50 0 7 = - 4670 30 300 30* 0300 7 _,765 G 30o 0.4300 / + //, 770 Y + 5!Looo ATp H N N. -% ,.40 X. n 0

=-400

=0

"DESIGN OF THE RING ZONOCOUE WHERE THE

WING SPAR GOES THRU IN HYPOTHESIS I

DUE TO THREE POINT LANDING

DESIGN OF THE RING AT POINT C or I

V = 0 N . 9,450

lb.

y

79,800 in.lb.

17

S-T- Aluminium

Alloy

ye" I

2

=

2(.25

x 4 x 2.375

)

- 3 + .25 x 4.5 12

5,,

I

4

,, 1,n.

=

2 x4 x .25 +

.25

x 4.5

=

3.125 in

=

.25 x 4 x 2.375 + 2.25 x .25 x 1.125 - 3.00 =79,800 x 2. 50 13.16

54

I I A

A

Q

Q

I

L

i,

- MTI

I

A . 15,160 lb./sq.in. = 15d'16O _-

_9,450

3.125

f

.

18,185 lb/sq.in.

DESIGN ON

FOINT

0 OR F

.

0

28,550

lb. = -8,870 lb.

S

b .I

S8,870

x

3.00

.25

x 12.16

=

8,080

lb/sq.in. - N/A

28,550

/

3.00

'p

f

JY,

AT

V

fs

fc

price of manufacturing *and the different in weiGht (ve-ry important thing in aircraft structures ) in this case is too small;however I'll change the moment of the iner-tia reducing the thickness of web and keep same others dimensions

For instance on points 0 and F I'll choose the following shape:

Same others dimensions except web thickness 0.10"

2(.25 x 4 x 2.275

)

12.02 in4 (.25 x 4 x 2.375). 2.63

0.10

b.Q

- .10 x 4.5 12 - (2.25 x 0.10 x 1.125)

19,400 lb/sq.in.

36

I I

Q

=

b

fs = 8,870 x 2.63 0.10 x 12.02

-7-"

T h0 /' SHA PE

f _cozakmD

ADB

" H Y P 0 T H E

S

I S" I I

"Assuming wings spars

clamped into fuselage"

"FORC ES OVER FUSELAGE WHERE FRONT SPAR PASS TERU DUE TO CONDITION I

C D B E S 9,7 2,000 in.lb. 0 9 30 30 30 03 55.,280 -55,2 K H 702,000 in.lb. 80

"FORCES

From point

0'...

00

From point F ... 1800

SHEAR ON 0:

Due to moment at 0

Due to moment at F

Due to tangentialO

'sue

_{to tangential-F}

Total Shpar fotee

NORMAL ON 0

Due to moment to 0

Due to moment to F

Due tangential

0

Due tangential

F

Total Normal Force

MOMfNT

ON 0:

Due to moment to

Due to moment to

Due tangential Due tangential Total loment

0

F

0

F

= U =

a.-(-9,

702, 000)

45

.16(

"

)

45

-. 23(-55,280

)

-. 08(

"

)

- 146,V865 lbs.

0

5.50 (-55,280

0

j 27,640 lbs. - + =

*1

U =

)

.

Si.50

(-9,702,000)

. ...

...

0000..0

*

4,851,000 in.lb.

000 103,500

34,500

13,590

4,725

0 . 0 0 . 0

27,640

. 0 . 0 0

4,851,000

.*. ... ...---0*..*...0000

40

AND .MOVENT

ON

POINT

.

I

.

=

:

.

.

.

.

" FORCES

From point 0 ... 3300 From point F ... 1500

SHEAR

Due to

"t

"I

ON A

moment

"I

a

"t

" tangential"

if

?,

it

t 0

F

0

F

Total Shear force

NORMAL ON A

due to m6ment at 0 7? " " " F

" Tangential" 0 i7 _It"

_F

Total normal force

MOMENT

Due to

ON A

moment

at 0

-. 44 (-9,702,000)

45

*

(

"'

)

45

=

-. 02(-55,280

)

=

-. 05(

'"

)

=

+119,120

lbs.

1

r(-9,702,000)

45 =

-. 6(

"

)

45

.

.32(-55,280

)

=

.12

(

"

)

-890,070 lbs.

*

-. 25 (-9,702,000)

=

+ 949oo - + 25880 = + 1105 . - 2765 = - 34500 - - 34500 * - 17710 =

+

-6C40

* + 2425000

_AND

MONkENT

ON

IOINT B "

From point 0

/||.'. 3000 From point F ... 1200

SHEAR

ON

B:

Due to moment at 0

" Tangential "

0

=

-.

32

45

S

_0

45

*

.09

(-9; 702, 000)

(

"P

)

(-55,280

- + 69,000 - . . . .

)

-

4,980

. .02 ( "1

7

-

+

1,106

TOTAL

SHEAR

AT

B

_.

NORMAL ON B:

Due to moment at 0 S ? "

"

F "

Tangential

" 0

"t

_"

_F

S ..27( -9,702,On) 45 .

-.27(

"'

)

45 . .09( - '55,92 80

)

-

.15(

"t

.

y

- 58,200 - 58,201--

4,980

+

8,300

TOTAL

KORMAL

AT B

IONENT

ON

B:

Due to moment at 0

.

t

"I

_t

_I"

_F 2 " Tangential " 0 . "I I - 113,080 lbs. -. 06 (-9,702,000)

-.11

(

"'

)

=

.04(45) (-s5.5,280

)

"

F

= .04(45)( "I y

TOTAL MOMENT

AT

B=

- 486,000

in.lb.

42

"t

"t

"F

- 72,874 lbs.

+ 582,0'n

-1,068,

000

- 99,600

+

99,600

"l FOR CES

" FORCES

AND

from

from

MOMENT

ON

I1OINT

point

0 ...

2700

point F ... 900

SHEAR ON C:

Due to moment at

0

F

" Tangential "

0

F

TOTAL SHEAR ON

C

NORMAL

ON

C:

Due to moment at

" Tangen-dal "

TOTAL NORkAL AT

0

F

0

F

C

.

-.

15(-9,

702,

000)

45 S-._( PP

)(

45 . .09

(

-55,280

)

=

.09(

"

)

.

0

= U MOMENT ON C:

Due to moment at 0

=

PP "P P" PP F =-" Tangential "

0

= " F .

.32

(-9,702,000) 45 -.32( " ) 45 -. 08( -55,280

)

-. 08( "?

)

- 129,160 lbs. .07(-9, 702, 090) -. 07( "P) -. 15 (5-55, Ppqo)

.15(")4

"n)b

= + 32t950

-

22250

. - 4980 . 4980

=

- 69001 . - 69000 +. 4420 -

+

4420 . - 67900 . - 679000 - +

7,300

.

t

37,300

C

"

" H Y P 0 T H E S I S V

=72,

874 V-72, 874 7

=

0 N

=-113,

080 N=-113 080 N = -129,260 lb. If =-486, 000 .- y8,0 = >.983,9400 D B V= -. 91 2., V=-119, 120 _{. -9}

_N-7

N=B0, 070

599

'P00

N=-80, 070 A = Y=-1, 5 a 9, 0 E 30---30 30 ₃₀ V= 46,875 N=~,7, (40 30 ~98.51,000, -9875 1 30 30 3 0 00 V=-72, 874 _H H= 113, 080

M=/

AND NORI AL WHERE FRONT DUE - TO " SHEAR FUSELAGE

30

0

='119

9120

N= BO, 070

0 V=7, 87,4 1 129,160 8 00 f = 1, Y=486,94Oo

FORCE AND L'OEERT AROUND

WING SPAR 1ASS-THRU

CONDITION I " -4i -146 - 27, 640 - 8R s.., , 000 V=-119 ,12 N= 80,Y 07( Y= 1, 5 99,f (

FORCES OVER

I-A SS-THRU

FUSELAGE WHERE FRONT SIAR

DUE TO THE THREE 1 OINT

LANDING " C D _B E A 3O-*-3O

4,4 P

00

0

30

1

=.

4,

000

F

₀

30

30

30 30 30 30 C K

HI

-59, 100 lb. 59,100

lb.

"t

V =-27,362 N

=-49,660

N .- 22,1oo V

=-

5 _4e48 N -- *3,370 V

=-

75oo N =2a,6so N -+ 2,215000 -- -53,348 N

=+44-37o

N

= +700,400 V :- 27,3e2 N =+ 4,660 If + 22,500 D E

3,0*

30

F

I

N

V=0

V 27,362 A =- 63,fso N =-49660 If =-544aoo 1 =-22/,500 C B

V

= + 53,346 N = - 43,370 S 7443400

30'

50

-7

0

A =+

a

'50* +--300 K S=+ 4 IV .+ 43 F + 700 J V =+ 27,362 N =+ 4,660 = N =+ +2215 " SHEAR

NORMAL

FUSE"LAGE WHERE DUE TO THREE FORCES FRONT I-OINT AND MOLEJ WING SPAR LANDING T AROUND 1-ASS- THR U CONDITION " 46 ,2/,000 ,06-0 .550o

t

300 G H v 48 70 oroo

y

" DESIGN OF THE RING L ONOCOQUE WRERE THE

WING S.PAR GOES THRU IN HYEOTHESIS I I

DUE TO CONDITION I

" DESIGN OF THE I-OINTS 0 OR F "

17 S.T. Aluminium Allo.Y: = 146, 875 lb. = '-27, 640 lb. -4,800,000 "in.Ib. = 7 in. 42 /4"1 -- +- .0 7 0 i -n - .- .

I

=

(1.5 x 18 x 6.25')

+

I x I

12 - 2235 in 4 . 2(1.5 x 18) + 1.0 x 11.0 V N y b I I A

S 4,800,000 7 = 15,000 lb/sq.in. = 1Z6,875 r 183.94 1.0 x 2,235

=

12,040 lb/ sq.in. fb

f

s

" DESIGN OF POINTS C AND I "

= 0 = -129,160 lb. . -1,283,000 in.1b. . 446.7 in4

=

28 in? - 530. in - .7.0 in. /IV

/0"

o - 5.0 in. = 1,283,000 x 5

/

416.7

=

14,350 lb.! sq.in.

46

V N I A

n

b y

+ N/A = fb

=

14,250

+

129,160 / 53

16,780 lb/ sq.in.

I

" C 0 N C L U S I 0 N S

Analizing both hypothesis it is true

enough that the hypothesis I (wing span go thru fuselage)

is the best one and gives the most reasonables values,for that I have not chances to doIkeep the answer of this hy-pothesis,but,why I got no satisfactory structure (from

stand view-point of the aircraft design ),in hypothesis II

The answer is

neat.-Due to, the fact that this airlnane was

created for very long range,I designed for maximun L/D and

very high aspect ratio,getting a very long span and

conse-cuently I have a big moment in the root chord and very high clamped moment too.

Then for long range airplane ( this ca

se ) ,seems to me is not commendable to have clamped wings

its will bei'tght for pursuit airplanes and when the desia

ner has troubles for room near the C.O.

position.-In this thesis I did not paid attention to the running chord load or perpendicular ring -monocoque

component force,because,this load must be carry for

longi-tudinal monocoque fuselage structure and part o~f the skin.

I am very sorry about this,but I have not-time for further

calculati ons.

For calculate the other rings of the

monocoque structure I must proceiure in same way as I

did.

About openings,until now,there is not theory in order to calculate its,but the designers fo -llows the experiences and results from factories and la boratories that worked in this