.1NST. TE
JUN 27 1942
LBRA4R
OF
MONOCOQUE
FUSELAGE
"
by
Captain Ismael Ndnez
-
Graduate from Army Hihg Technical
School - Argentina.-
Submitted in Pattial Fulfillment of the Requirements for
the Degree of
SCIENCES "
OF
in
'AERONAUTICAL
ENGINEERING
FROM
MASSACHUSETTS
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
FUSELAGE
of the requirements for
fulfillment
of
MONOCOQUE
MASTER
OF
",
in partial
the degree
SCIENCE in Aeronautical Engi nee-
ring.Respectfully,
Ismael Ndife2
Captain of Argentine Army
Air Corp
252540
It
B
N
I
E
Thesis Proper
No
xIt"
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
landing
34
8
Second Hypothesis
38
9
Forces &-'oments due to
Condition I
44
Forces and moments due
three point landing
46
11
Design for hypothesis II
47
12
Conclusions
50
10
"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 characteristics:
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.
AND
STABILITY
300
critical altitude)
speed (at
J.P.H.
CONTROL:
1)
Satisfactory longitudinal stability and contorl
center of aravitv positions.
for all
2)
Satisfactory lateral
3)
Sufficient rudder to maintain straight
1.20 V min. with an engine out.-
and control.
stability
flight
at
FOR THE PRESENT PROBLEH I HAD OBTAINED THE FOLLOWINGS
CHARACTERISTICS AND VALUES REQUIRES FOR kONOCO UE P'U.
SELAGE DESIGN:
C H T S
W E I
..--.
Gross. ................
Empty .
..
Useful
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
..
.
..
.
.
..
..
.
. ..
.
.
.
.
..
.
.
.
30,200
.
16,611
.
13,589
"A
Wfing
0
R E A
( sq. feet
-
-
.0
.
.0
"I
f-
)
.
0 *
. 548.0
0
72.0
Stabilizer. .
Elevator
.
0
0
0
0
.
24.0
Ft n
.
0
0
0
0
0
65.3
Rudder
0
"LOADINGS-
ON
DIFFERENTS
Wing
C lb/sq.ft.).
Span
(lb/ft.
). . .
Power
(
)...............
lb/Hp.
.
30.7
LARTS "
............
......
2
.
.
.
..
.
55.00
.
.
.
408.00
8.16
C 0 N D I T I 0 N
"
N
=
p435
8.16
p
=
p
.
Power Loading
-
30200 lbs.
Log.
=
2
/
i
=
x
1850
-
0.9118
0.3970
n
SEGURITY
3700 Hp.
=
30200 lb
=
55 lb/sq.ft.
Wing Loading
8.16
=
=
548sq.,ft.
.435 Log8.16.
.435
=
p
n
lb/Hp
Gross Weight
5s
s
302oo lb.
3700 lHp.
W
HP
32000
W f 9200
Ratio of total lift to gross weight
N or n
WJ
+
77
3.25
+
1
"t
I
1
4 35
(8.16)
+
(
3.25
2.49
=
2.49
.77
32000
+
30200 + 9200/
3.062
FACTOR
=
Ianeuverability Fac. =
=
3
1.50
1.5
4.60
x
3.062
Gnst Load Fantar
:
m
KUV
=
575 s
=
K
s 114
1
-
1(55(1
2
Log. 55
iLog. 55
4
(ss)
1/4
K
=
1,97402
=
0.4350
=
2.725
=
2.725
1. 3625
.
2
From Air Commerce Yanual in figure lla.
K for my case is :
K
1.20
U
30 ft/sec.
the value of
( Gust velocity
)
Velocity at horizontal flight
Power Plant Efficiency
e
0.60
7
52.7 (
d
)
p
d
Sdw
CDmi n
SDw
5.48
CDf
0.07
S
fus.
=
x
32.8 sq.ft.
4
SW
=
0. 010
x
548
CDnacel
0.10
Snacel.
20. 95
SDfr
0. 07
x
2
0.1
.
SDna c.
=
SD
x
5.48
32. 8
+
x
20. 95
2. 30
+
=
2.30
=
4.19
=
2560
4.19
11.97
d
=
30200 / 11.97
VL
=
52.7 ( .60
VL
=
303
1/s
When I
2560)
8.16
. .H.
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
equal to 6
P=
10 (aspect ratio)
m6
4.38
m
4.38/
41
3m
1
m
.4.
_6
10
87
67
curve when aspect ratio
is
=
An
1.20 x 30 x 303 x 4.87
575
A n'
ni
n,
GUST
=
1.455
S
1
=
2.455
LOAD
55
- An
1
=
FACTOR:
it
11
x
1.50
I
Total Lift
L
=
30200
q
-
Dynamic pressure
q
=
230
x
367
L / q.
CL
111000 /230.
CL
0.88
a
548
23015 characteristics.
I1I
Angle o,
attacw
From same airfoil data
c=
2.455
x
-
0.31
CD c o scr -
6
.111.,000
CLs i ncz
lbs.
V2 mph/391
CL
CN
C
1.455
3.668
=
From Airfoil
-
=
Foro
110
coser = 0.981
si nor, - 0. 191
C
0.31
cc
0.136
x
375
Fpr
0. 88
x
0.191
HPa/Va
375 g
Fpr
-
0. 981
x
x
0.60
3700 /300
2775
.
n3
In my
-
nlh2 +
nix 2 4 nx4 (h4 -h 2 )]
1
case due to the airplane characteristics:
h4
h
f
1
X3 ~2
0
-0
2
H
S
-0.16 x 7.4,ft.
- 0.16 C
C
.
x2
-
2
x3
-
1.184 ft.
+
H
C
=
3. 8 ft.
0. 0251
a
39. 8 -
1
(-1. 184)
-0.106
7
C 0.
0251 + 3. 67 (-1. 184) +0. 092 (0))
"S
P
A
I NO
T
I
R
B
U
T
I
0
N"
LOSS "
TIP
CN
S
I
D
N
= Constant =0.
R
-l-5 in.
" WITH
LOSS
TIP
C88. 8i 77.
1326. 2 in.
.-- --
"
t
0. 88
CN,
CN=
I
_
+-------
__
_
_
_
_
_
_
_
_
0.'70
CN
U
71
361.5 in.
8
__I
=44
I ~ ~ ~ ~ ~
X
322
in.__
_
_
I__-
2776"
W11Y6 D/A 1Y fO//
_
--
-_
_
_
_
_
----------
A.C t/leQ
T1/f
/
2
3
278
1V /425
.
2776
75
/20
/70
96
82
21V 3615 68
70
53
9000
/63
o
Y Cd Y'X
d CIY
26 2"
/ 00
/00
157O
/0D
4660
1.00
6741
XGA CAy Zc
0
/,
0
4660 /A
/07
7/9
a
;
a
V
z
J1
0
/4
C(Ldy C__c
0 u4zO l-/2
-
35,7/ 6,)'?4 v:P/
357/0
o
i
/
?j 7gCAl
/% aam
a
.
/
/3
1/ / /2
8
7
5
YP,, C4Y
Y
I
4
K
/5
_
Oaeqa Qo c
-
5?90
5.8A5,200
35,710
=
163.50
=
92,20
-293,100
35,710
=
Z (12)
0
C0
3 294,000
35,710
2(5)
Ky
=
(15)
.
2 13,
Kb
=
19, 794
3,294, 000
0.006
35, 710
1. 0
35,9710
/
sj-r
m /i mumis mi2i1
3
4
5
6
7
ff
8
/2
///
/3
/
/0
c,,
Gct
~STiPE
A.
4 b,
C
Caf
7$
/20
96190
f
2'7
.2
/4.2-
/70
S%
12
2776
62
70
361.5
68
53
/63/0
kzCta
/A00
9CV0
/0
/3/0
5740
/00
574
4660
0.706
37/
357/0
34760
14y
250 20O
?,32,.
66,OO
4
x0,e
/',,0
e zq4C4
xecjgy
/2S4k0
0
0
usaf %o
W/I&/7/7/) ZLY2JS
/
0 4w
o
.f Oaoo A
3.$
0
c24y
.
-
0
/2o
a o0 g
c
-
990 0
/
0
1oz 061
.
0
.4%6'
,,
/.
y
(8)
=
5,503,200
34,760
=
: (10)
=
293,100
=
8.43
=
92.20
2
-
.
158.2
34, 760
2
.
0
12)
z(7)
=
Z
13)
=
3,294,000
35, 710
=
0.006
=
34,760
2:( 5)
2
CY
Kb
15)
2 (13)
-7
35F710
5+
/3
=
0.973
A L ANCI
"B
NG
T A -B L E
0
C
tip
For
T
I
No
W
(
N
Groos weight
T A TI
C0MPU
D
I
"
III
T
lloss
0
oN
I
and
0
no
N
I
tip
loss
E
VL = 300 MPH
)
30200
0.00256 V 2
230
2
a
=
3
s
=
4
q/
s
5
n
6
CV
7
CL
0.880
8
C
0.136
9
n1
=
(8) x (4)
0.569
10
nx 4
=
Fpr/(1)
0.092
11
Cm
12
m
(11)
13
n3
(
14
n2
=
15
n
=
16
T
(
/
55
(2) /(3)
4.18
)
appl. wing load factor
(5)
(
/
3.67
0.878
(4)
Design moment Coeff.)
0.006
x (4)
0.025
Tail load factor
-()
)
-0.106
(13)
-3.564
(9) - (10)
-0.661
(1) x (13) (tail load)
-3200
-
/4
AND
"FOR CES
DISTANCES
C2 e
J
ON
AIRFOIL
i
C
1244
CN q_
144
a
cMa
q
C
z4r4
--b
I-
-
-
r
C'
/5
-1
T
"
FOR
B
LOSS
TIIP
L
E
NO
AND
H
E
T
I
A
i
_If
I
TIP
LOSS
along
Stations
I
span
III
II
IV
27.8
142.5
277.6
361. 5
0.833
0.666
0.486
0.368
(fraction of chord)
0.15
0.15
0.15
0.15
(.
)
0.65
0.65
().
65
0.65
(3)
0.50
0.50
0.50
0.50
(fraction of chord)
0.238
0.238
0.238
0.238
"
)i
0.245
0.245
0.245
0.245
(unit wing weight
)
7.44
7.44
7.44
7.44
1
Distance from root inch.
2
C'
/
3
f
4
r
5
b
6
a
7
j
(
8
e
9
r
-
a
= (4)
10
a
-
f
=
11
r-j
=
12
j - f-=
13
C'1144b= (2)
144
"
"
r-f=
(4)-
t
-
(6)
0.412
0.412
0.412
0.412
(6)
-
(3)
0.088
0.088
0.088
0.088
(4)
-
(7)
0.405
0.405
0.405
0.405
(7(
- (3)
0.095
0.095
0.095
0. 095
1.666
1. 332
0.972
0. 736
(5)
/6
Yf
SI
Net running load on front spar
CN(r
. n 2 e (r-i)
a) + CiaJ
Net running load on rear spar
Tr-
Tc
yo
-
N(a -
-
Cy)
q + n2 e (j
qa
Net running chord load
c .q
+
/7
nx2 .e ) C' / 144
f)J
T
A
N
No
I
T
B
L
T
0
E
/
V"
P
L
0
S
III
SPAN
IV
ALONG
II
S
2708
142.5
277.6
361.5
0.88
0.88
0.88
0.88
14
CNb
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)
-
0.369
0. 369
0. 369
0. 369
18
(17) x q
85.0
85.0
85.0
85.0
19
n2
-10.75
-10.75
-10.75
-10.75
20
(1R)
-
74.25
74. 25
74.25
74.25
21
Y
=
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)
16. 68
16.68
16.68
16.68
27
Yr
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
nx 2
x
-4.92
-4.92
-4.92
-4. 92
31
(29)
-
-36.22
-36.22
-36.22
-36. 22
32
Y c
-
-30.2
-24.15
-17.60
-13. 34
C
Rb
I
STATIONS
I
E
Distance from root
x
E
Kb
(15)
x (8) x (11)
-
(19)
(20) x (13)
(25)
(26) x (13)
(8)
(30)
(31)
x (2)
/
"
T
"
N 0
A
B
L
E
TIP
STATIONS
No
I
H!
f
T
distance from root
- 14
'NI x
CfNb ~
Rb/Kb
I
V
"
LOSS
"
ALONG
II
SPAN
III
IV
27.8
142.5
277.6
361.5
0. 905
0.905
0. 905
0.905
15
CA(variation with span)
0. 006
0. 006
0. 006
0.006
16
(14) x (9)
0. 373
0. 373
0.373
0. 373
17
(26) + (15)
0. 379
0. 379
0.379
0. 379
18
(17) x a
87. P
87.2
87.2
87. 2
19
n2 x (8) x (11)
-10. 75,
-10. 75
-10. 75
-10. 75
20
ai8)
+ (19)
76.45
76.45
76.45
76.45
21
Y
= (20)
127.3
102. 0
74. 30
43.40
22
(14) x (10)
0. 079
0. 079
0. 079
0.079
23
(22) -
0. 074
0. 074
0. 074
0.074
24
(23) x q
16. 94
16. 94
16. 94
16. 94
25
n2 x (8) x (12)
-2.52
-2.52-
-2.52
-2.52
26
(24) + (25)
14.42
14.42
14.42
14.42
27
Yr
24. 06
19. 20
14. 00
7.860
x (13)
(15)
(26) x (13)
28
c. (variation with span)
-0.136
-0.136
-0.136
-0.136
29
(28) x q
-31. 30
-31. 30
-31. 30
-31. 30
30
nx2
-4. 92
-4. 92
-4. 92
-4.92
31
(29) + (30)
-36.
2
-36. 22
-36. 22 .-36.22
32
Y
-30. 20
-24. 15
-17. 64
x (8)
= (31) x (2)
/9
-13. 36
D&1/#/6
ZOADS
/7
fDOQ6/T
JD/ Q
37a 2 "
286"
-
375'
"4
/60"
i
'if
I
II T
-~______________
__________________
/851
I1
I
0 --
70"
7.5"
246"
c%"
3262"
- --
/Y6
4
C8.8"
4/5"
I
7//? Z&S
~~~1
DUM##Y6
-
/e
//
ZO6I5
8
J70. 2"
"I
"
_____________________
I.________________________
I
I
It
/92?/I>>
/
.1
A
I.
I 'r
5
I
75" ~
5g42
+
6 in
-7
--
f(//T
4,
70"
~~1I-
NI-
W/IT#
1
I
Sallg
245"
I
/17D
ZOSS
88.8"
" TOTAL
LOAD
" N
II
.II
IV
TIP
L 0 A D
SECTION
(1).
I
0
x 185.4 -
LOSS
FRONT
13910
170 x 148.5-= 25260
III
IV
(4)
.529.,000
160. 0
4, 040, 000
82
x 108.3 =
8880
286. 0
2, 540, 000
88
x 82
=
7220
370.2
2, 600,900
"
II
HOMENT
37.5
9,702,000
55,270
I
SPAR
"
ARM
(3)
(2)
75
ON
kIOMENT
AND
75
WITH
TIP
x 192.0 = 14400
LOSS I
37.5
540, 000
170 x 153.0 = 26000
160. 0
4,160, 000
82
xlll.5
-
9150
286.0
2,716, 000
88
x 65.1
=
5730
370.2
2,221,000
55,280
22
9,637, 000
"
".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 hypothesis 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 placearound of the Gravity Centerwhere equipmentpilot
or crewmen necessarily must be place.-
In this
hypotesis (spar go thru the
fuselage) the condition which give me the high load
over fuselage is
three point landing condition,for
that reason I'LL design due to these loads...
23
"THREE
"CALCULATION
OF
LOAD
CONDITION"
LANDING
POINT
FACTOR
n
In figure 24 of the "Civil Aeronautic Manual
I got the following expression:
=
2.80
+.
=
2.80
+
=
3.063
Gross Weight
=
30200.00
Langding Gear
=
141000
n1
Weight in landing =
28,790 lb.
Design Load for
3 point landing
28,790
t
11
"
-
x
9000
+ 4000
9000
30200 + 4000
3.063
x
1.50
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
WEIGHT
ON
FRONT
WREELS
=
132,400
x
500
560
118,200 lbs.
WEIGHT
ON
TAIL
WHEEL
=
132,400
x
60
560
14,200 lbs.
On each front of the landing gear 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.-
25
7Z6/SfL//16f
tX/f Q/7zl L 7C, 2/QCtS
ZZ4~62/2j
JIOZJQ
-
5~9/c4Q
I
I
p-
75"
I
-~
6~9/O67~ 7~5~"
"
LOAD
BREAKING
DUE
AT
RIffG
TIREE E
TAGETIA.L
TO
ACC'ORDING
LANDING
TO
OF
O/IfT
OA C
OiCOCO Uz"
C. (.
20*
4---
5'no
/SIAe
S5q/(* ,(ob -r:?c wheel)
d
27
4
FORCES
ON
MONOCOQUE
RING
LANDING
28
"
DUE
TO
THREE
1OIN
AND
"FORCES
FOEMNT
POINT
ON
-Q
"
From point 0....3600
From point F....180 0
SHEAR
FORCE
ON
0:
Due to tangential at 0 .
-. 23(-58,825)
=
Due to tangential i
-.08( 58,825)
=
F =
+
Total SHEAR force
NORMAL FORCE
ON
It
ON
Z,730
0:
Due tangential force at
YOMENT
13,600
8,870 lbs.
Due tangential force at 0= +.50(-58,825)
Total NORYAL
+
F=
0
= + 28,550 lbs.
force
0:
does not exists
29
+
28,550
FORCES
AND
From point 0 .
From point F .
SHEAR
FORCE
ON
ON
MOMENT
.
.
.
.
.
.
POINT
A:
=
-. 02 (-58,825)
Due to tangential at F
=
-.05
Total SHEAR force
=
- 1,765 lbs.
FORCE
ON
(
58,825)
=
+
1,177
=
-
2,941
=
- 18,820
=
+
7,065
159,000
A:
Due to tangential at 0
.32 (-58,825)
Due to tangential at F
.12
Total NORMAL force
MOL'ENT
"
.330O
.150
Due- to tangential at 0
NORMAL
A
-
(
58,825(
11,770 lbs.
ON A:
Due to tangential at 0
=.06(45)(-58,825)
=
-
Due to tangential at F
=.04(45)( 58,825)
=
+ 106, 000
Total MOMENT
-
jo
53,000 in.lb.
POINT
ON
MOMENT
AND
" FORCES
B
"
From point 0.......3000
From point F.......1200
SHEAR
FORCE
ON
B:
Due to tangential force 0 =
.09(-58,825)
Due to tangential force F .
.02(
Total SHEAR force
NORMAL
FORCE
ON
-
58,825) =
5,320
+
1,182
- 4,138 lbs.
B:
Due to tangential force 0 =
.09(-58,825)
U
-
5,320
Due to tangential force F =
.15(
=
+
8,860
Total NORMAL force
MOMENT
ON
=
58,825)
+ 3,540 lbs.
B:
Due to tangential force 0 .. 04(45)(-58,825)=
- 106,600
Due to tangential force F =.04(45)( 58,825)=
+ 106,000
Total YOIENNT
0
5/
"
FORCES
AND
MOMENT
ON
POINT
C
"
From point 0......2700
From point F......
SHEAR
FORCE
ON
900
C:
Due to tangential at 0
.09(-58,825)
5,320
=
Due to tangential at F
.09( 58,825)
Total SHEAR force
0
NORMAL FORCE
ON
0
-.
Due to tangential at F
MOMENT
AORKLL force
ON
5,320
C:
Due to tanaential at
total
+
08 (-58,825)
-
+
4725
.08( 58, 825)
=
+
4725
+ 9,450 lbs.
C:
Due to
tangential at 0 .-. 015(45)(-58,825)n
+ 39,900
Due to
tangential at F
+ 39,900
Toatal YOYENT
.015(45)(
=
32
58,825)=
+ 79,800 in.lb.
FORCES AND
0
TUWE
MOMENT
YING
GOES
TO
THREE
ON
THRU
HYOCTHESIS
LANDING
7.
+
I
+a74
c
8
=
B
7
± ,765
N
-
-,765
=//1,770
//,770
-
64000
300
3
300
3
30300
3000=
300
S0
0
30
/,75
0300
7
H
N
ATp
N. -% ,.40
X.
n
=-400
0
7 =0
33
-
4670
_,765
+ //, 770
Y + 5!Looo
30o 0.4300 /
G
N -. //,770
A" . + .53,00
N =
7 =
300
30*
.+
DUE
= + -1,640
.0
0
V
I
"
V- -4,13
3,,14*N
8,670
28660
.
Alf
WHERE
o
D
V =
N
1/ =
MONOC04UE
N = +9,450
V =+ 4/38
y
IN
t-OINTS
v
N .+
RING
GOES
SPAR
WING
DESIGN
V
=
0
N
.
9,450 lb.
y
I
LANDING
C
or
I
79,800 in.lb.
17 S-T- Aluminium Alloy
5,,
ye"
I
I
HYPOTHESIS
POINT
AT
RING
THE
OF
IN
POINT
THREE
TO
DUE
THRU
THE
WHERE
ZONOCOUE
RING
THE
OF
"DESIGN
=
2(.25 x 4 x 2.375
-
+ .25
x 4.5
12
2
)
I
I
L
4
,,
i,
3
1,n.
I
A
=
2x4
A
=
3.125 in
Q
=
.25 x 4 x 2.375
Q
-
3.00
=79,800
x.25
x
+
2. 50
13.16
54
.25 x 4.5
+
2.25 x .25 x 1.125
'p
MTI
-
I
A
.
15,160 lb./sq.in.
f
=
15d'16O
f
.
18,185 lb/sq.in.
JY,
fc
.fc
OR
F
28,550 lb.
=
S
fs
0
9,450
3.125
0
.
AT
V
FOINT
ON
DESIGN
-
-8,870 lb.
S8,870
.25
b .I
=
8,080 lb/sq.in.
-
N/A
x
x
3.00
12.16
28,550 / 3.00
9,520 lb/sq.in.
I might design the ring with variable 7roment. of Inertia,
butseems to me that would be a bad remedy because if
eco-obmize some material,for other hand will raise the
35
I
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 inertia 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
I
2(.25 x 4 x 2.275 )
I
12.02 in4
Q
=
(.25 x 4 x 2.375).
.10 x 4.5
12
-
(2.25 x 0.10 x 1.125)
-
2.63
b
0.10
=
fs
b.Q
19,400 lb/sq.in.
36
0.10"
8,870
x
2.63
0.10
x
12.02
-7-"
T h0 /'
SHA PE
f
cozakmD
ADB
.4
/"
450
PA)r
J7
"
H
Y
P
0
T
H
E
S
I
I I
"Assuming
wings
spars
38
clamped
into
fuselage"
S"
"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
9 702,000 in.lb.
30
30
30
03
55.,280
-55,2 80
K
H
39
"FORCES
SHEAR
ON
ON
.MOVENT
AND
POINT
From point 0'.......
00
From point F .......
1800
0:
Due to tangentialO
a.-(-9, 702, 000)
45
)
"
.16(
45
)
-. 23(-55,280
'sue to tangential-F
-. 08(
Due to moment at 0
Due to moment at F
Total Shpar fotee
NORMAL
ON
=
103,500
34,500
=
*1
13,590
4,725
- 146,V865 lbs.
0
Due to moment to F
0
Due tangential
0
Due tangential
F
U
Total Normal Force
=
ON
+
)
"
Due to moment to 0
MOMfNT
-
5.50 (-55,280
)
U
0
=
.
.
I
.
0
j
0
.
0
.
27,640
0
.
.
0
0
27,640 lbs.
0:
Si.50
Due to moment to 0
4,851,000
(-9,702,000)
000
Due to moment to F
Due tangential
0
.
Due tangential
F
=
Total loment
0
:
.................
.....
*
. .
.*.
..
.........-----
0000..0
4,851,000 in.lb.
40
0*..*.......0000
............
" FORCES
SHEAR
.kOMEVT
_AND
From
point
0
From
point
F
"
"I
"I
"t F
tangential" 0
if
it
?,
F
Total Shear force
NORMAL
ON
=
+
949oo
*
)
-
+
25880
)
=
+
1105
)
.
-
2765
1 r(-9,702,000)
=
-
34500
-
-
34500
*
-
17710
-6C40
45
(
"'
=
45
-. 02(-55,280
=
-. 05(
= +119,120
'"
lbs.
A
45
F
=
Tangential"
0
.
45
.32(-55,280
)
)
F
=
.12 (
)
=
+
*
+ 2425000
.
-
776000
=
-
149400
-
+
"
i7
It"
MOMENT
ON
it
Tan gential
"
at 0
*
-. 25 (-9,702,000)
"F
.
-. 08(
=
. 06(45)
(-55,280)
=
.04(45)(
"
*
-1,599, 000. in.lb.
"
0
F
Total Moment
"
A
Due to moment
?I
-. 6(
-890,070 lbs.
Total normal force
"
1500
"
7? "
it
3300
-. 44 (-9,702,000)
due to m6ment at 0
"
.....
.....
A 11
A
ON
Due to moment a t 0
"t
I-OINT
ON
)
"
)
99600
"l FOR CES
From point 0
ON
SHEAR
/||.'.
3000
From point F .....
1200
B:
Due to moment at 0
=
-. 32 (-9; 702, 000)
*
45
"P
_0 (
45
.09 (-55,280
.
.02
(
.
-
72,874 lbs.
S
"
Tangential
"
"t
"t
TOTAL
SHEAR
NORMAL
0
"F
AT
B
?
)
)
7
"1
-
+
-
.
-
.
"
"
F
Tangential
" 0
"t
" F
KORMAL
AT B
TOTAL
IONENT
ON
"I
t
" Tangential
"I
TOTAL
.
.
4,980
+
1,106
.
..27(
45
-9,702,On)
-.27(
"'
S
.
)
.
-
58,200
-
58,201-
I
MOMENT
.
.09(
- '55,9280
"t
.15(
-
-
113,080
)
-
4,980
y
+
8,300
+
582,0'n
lbs.
B:
.
-. 06 (-9,702,000)
I"F
2
-.11
" 0
.
.04(45) (-s5.5,280 )
-
99,600
" F
=
.04(45)(
+
99,600
Due to moment at 0
t
.
-
45
"
69,000
B:
ON
Due to moment at 0
S
B "
IOINT
ON
MONkENT
_AND
AT
B=
(
"'
"I
- 486,000 in.lb.
42
)
y
=
-1,068, 000
AND
" FORCES
ON
SHEAR
C
I1OINT
ON
MOMENT
from point 0 .......
2700
frompoint F .......
900
C:
Due to moment at 0
.
-. 15(-9, 702, 000)
= + 32t950
45
F
"
- 22250
)(
PP
0
.
F
=
.09(
.
0
C
TOTAL SHEAR ON
S-._(
45
.09 ( -55,280
Tangential "
NORMAL
"
"
)
. - 4980
)
.
4980
C:
ON
Due to moment at 0
F
.32
(-9,702,000)
45
)
"
-.32(
=
=
-
69001
. -
69000
45
"
0
-. 08(
-55,280
)
F
-. 08(
"?
)
Tangen-dal "
TOTAL NORkAL AT
MOMENT
C
"
-
+ 4420
129,160 lbs.
=
.07(-9, 702, 090)
F
=-
-. 07(
Tangential "
0
=
-. 15 (5-55,
"
F
.
"P
-
C:
ON
Due to moment at 0
PP
U
+. 4420
P"
PP
TOTAL MOMENT AT C
.15(")4
"P)
. .
67900
- 679000
7,300
Ppqo)
- +
"n)b
. t 37,300
-1,28,",400 in.lb.
"
H
Y
P
0
T
H
E
S
I
S
7 = 0
N = -129,260 lb.
V-72,874
N=-113 080
.-
y8,0=
>.983,9400
B
D
V=-119, 120
N=-80, 070
Y=-1, 5 a 9, 0
.
V =72, 874
N =-113, 080
If =-486, 000
-9
A
V= -. 91, 2.
N=B0, 070
N-7
599 'P00
=
E
30---30
30
-146
- 27, 640
-
30
V= 46,875
N=~,7, (40
30
1
-9875
8Rs.., , 000
~98.51,000
,
30
30
0
30
30
V=-119 ,12
N=
80,Y 07(
Y= 1, 5 99,f ( 00
='1199120
N= BO, 070
V=-72, 874
H= 113, 080
H
M=/
1
f
" SHEAR
FUSELAGE
0
129,160
= 1,
V=7,
Y=486,94Oo
AND
NORI AL
FORCE
AND
L'OEERT
AROUND
WHERE
FRONT
WING
SPAR
1ASS-THRU
DUE - TO
CONDITION I "
-4i
87,4
8 00
FORCES
OVER
FUSELAGE
WHERE
FRONT
SIAR
I-A SS-THRU
DUE
TO
THE
THREE
1 OINT
LANDING
"
C
D
B
E
A
3O-*-3O
4,4 P
0
00
30
1
F
=.
4,
000
0
30
30
30
30
30
30
59,100 lb.
-59, 100 lb.
C
K
HI
"
H
Y
P
0
T
H
E
S
I
S
II
"t
V
N =-49660
1 =-22/,500
V =-27,362
N =-49,660
V=0
A =- 63,fso
If =-544aoo
27,362
N .- 22,1oo
C
D
B
V =- 5 4e48
N -- *3,370
V =-
= + 53,346
N
= - 43,370
S
E
-7 ,06-0
50
30
F
7443400
30'
3,0*
75oo
N =2a,6so
N -+ 2,215000
V
A =+ a .550o
0
I
t
,2/,000
'50*
300
+--300
-53,348
--
N
=+44-37o
N
= +700,400
K
S=+
4
G
J
H
V :- 27,3e2
N =+ 4,660
If
V =+ 27,362
v
N
+ 22,500
=
N =+ 4,660
N =++2215
y
"
SHEAR
FUSE"LAGE
TO
DUE
48
IV .+ 43 70
F + 700 oroo
NORMAL
WHERE
THREE
AROUND
MOLEJ T
AND
FORCES
1-ASS- THR U
SPAR
WING
FRONT
CONDITION "
LANDING
I-OINT
46
"
THE
OF
DESIGN
TO
DUE
"
DESIGN
17 S.T.
OF
THE
0
N
= '-27,
640 lb.
I
=
--
/4"1
7 in.
7
-
F "
42
-4,800,000 "in.Ib.
b
OR
146, 875 lb.
=
=
I I
Aluminium Allo.Y:
V
y
THE
I
CONDITION
I-OINTS
WRERE
HYEOTHESIS
IN
THRU
GOES
S.PAR
WING
L ONOCOQUE
RING
.-
I
0 i -n
.
(1.5 x 18 x 6.25')
+
I x I
12
I
-
2235 in 4
A
.
2(1.5 x 18)
A
2
65 in.
Q
1.5 x 18 x
Q
183.94 in
+
6.25
47
1.0 x 11.0
+
1.0 x 5.5 x 2.75
+-
.0
fb
f
7
4,800,000
S
=
15,000 lb/sq.in.
=
1Z6,875
1.0
r
x
183.94
2,235
s
=
12,040 lb/ sq.in.
" DESIGN
V
N
OF
=
0
=
-129,160
.
C
POINTS
I "
AND
lb.
in.1b.
-1,283,000
/IV
I
.
446.7 in 4
A
=
28 in?
n
-
530.
b
-
.7.0 in.
y
-
5.0 in.
=
1,283,000
=
14,350 lb.!
/0"
in
x
5
sq.in.
46
/
416.7
o
=
fb
+
14,250
=
N/A
+
16,780 lb/
129,160 / 53
sq.in.
49
"
C
0
N
C
L
U
S
I
0
I
S
N
Analizing both hypothesis it
enough that the hypothesis I
is
true
(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 hypothesis,but,why I got no satisfactory structure (from
stand view-point of the aircraft
hypothesis II
design ),in
The answer is neat.Due to, the fact
very long range,I designed for
created for
very high aspect ratio,getting
cuently I
that this
airlnane was
maximun L/D and
a very long span and conse-
have a big moment in the root chord and very high
clamped moment too.
Then for
se ) ,seems
its
to me is
will bei'tght
for
long range airplane ( this
ca
not commendable to have clamped wings
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 longitudinal 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
50
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 matter.-
5/