Aircraft Structures 1 2011 Part 3 docx
Aircraft Structures 1 2011 Part 3 docx
...
16 0J2 /3
0
FB 4000J2 -20000J2
d2pB.f
13 ~ 2 13
DC 4000 80
000
PB,f 13
11 3
pDsf
1
32 0 13 32 0
PD,f
1 32 0 13 32 0
1
48 0 13 240
BA 4000 60
000
2pB,f /3
2 13
pD,f
0
0
0
0
FC 4000
10 0 ...
800J3
30
50
-
J3R/2 -J3 /3/ 2 -2000
+
20J3R 48.2
CB
800
20
86.6
+
R/2
11 2 17 32
+
10 R
87.6
-1. 8
BD
800J3 30 -J3R/2 -J3 /3/ 2 20J3R
CD
800...
Aircraft Structures 1 2011 Part 5 potx
...
thus integration of the above expression
yields
(5.47)
4ab
3c,
03
U+V=TC
D""
EA;
nz =1. 3, 5 n =1: 3, 5
ni =1, 3, 5
n =1, 3: 5
-
From the principle of the stationary value of the total ... principal stresses.
Ans.
13 .1 Nm/mm, 1. 9Nm/mm,
a
=
- 31 . 7",
a
=
+58 .3& quot;, *786N/mm2,
fl 14 N/mm2.
A plate 10 mm thick is subjected to bending momen...
Aircraft Structures 1 2011 Part 6 ppt
...
-
FB)
COS
/3
-
gttd
COS'
=
0
(6 .11 8)
Taking moments about
B
wz
-
FTd
COS
@
+
$gttd2
Cos2
a
=
0
(6 .1 19)
Solving
Eqs
(6 .11 7), (6 .11 8)
and
(6 .11 9)
for
q,
FT ...
0.48 0. 51
0. 53
Note that
a
is the angle of diagonal tension measured from the spanwise axis of the
Am.
1. 2mm, 13 0As/(l +0. 011 3As), 238 910 Nmm.
beam, as in the usual nota...
Aircraft Structures 1 2011 Part 7 potx
...
8000
i
x
1. 2 23
x
602
x
14 .5
N
=
1. 1 13
L
ipV2S
cL=
From Fig. 8 .10 (a),
a
=
13 .75" and
CM,cG
=
0.075. The tail arm
I,
from Fig. 8 .10 (b),
is
1
=4 .18 cos(a-2)+0 .31 sin(a-2) ... into Eq. (8 .12 ) we have
=nW
or dividing through by
4pV2S
We now obtain a more accurate value for
CL
from Eq. (iv)
1. 35
4 .12 3
CL
=
1. 1 13
x
0.075
=
1. 0...
Aircraft Structures 1 2011 Part 9 doc
...
idealization
33 3
Table
9 .1
0
Boom
0
B
[mm2)
0
uz
(N/mm2)
1
+660 640 278
x
lo6
35 .6
2 +600 600 216
x
IO6
32 .3
3
+420 600 10 6
x
IO6
22.6
4 +228 600 31
x
lo6 12 .3
5
+
25 ... AI2,
A 23
and
A34
from Eqs (i), (ii) and (iii) in Eq. (9.69) we have
2Ak
=
[
61
25
x
1. 15 ~1 dsl
+
2( 31 2 .5
-
4.02s2)2.5ds2
200
1
+
jy
2(
1 1
1. 5...
Aircraft Structures 1 2011 Part 10 ppt
... (kNm)
1
-10 0
0 .1
-0.05
-10
5
-10 1 .3
0.6
0 .3
3
-3
2
- 13 3
0
-0.05
0
6.7
-17 7 .3
0
0 .3
0 0
3
-10 0
-0 .1
-0.05
10
5
-10 1 .3 0.6
0 .3 -3 3
4
10 0
-0 .1
0.05
-10
5
10 1 .3 0.6 ... Table 10 .3
49
8
=
49
IO
=
ql6
1
=
q2
1
q32
=
qS7
‘?loll
=
q1 516
=
30 .3
+
q 21
q 43
=
q76
=
411
12
=
q1 415...
Aircraft Structures 1 2011 Part 16 pps
... planes 10 , 11 , 12 -14
tensile stress 3
Stress analysis of aircraft components
36 2- 432
cut-outs in wings and fuselages 415 -25
fuselages 37 4-80
bending 37 5, 37 6
shear 37 6-9
torsion 37 9, 38 0 ... section beams 30 7 -16
Bredt-Batho theory 30 7-9
condition for zero warping 31 5 , 31 6
displacements 30 9 -15
Neuber beams 31 6
shear flow distribution 30 7
war...
Supply Chain Management 2011 Part 3 docx
... Science, 35 , pp .3 21- 33 9.
Sterman J. D., (19 89b). Deterministic chaos in an experimental economic system, Journal of
Economic Behavior and Organization, 12 , pp .1- 28.
Sterman J. D., (19 89c). ... Operational Research, 17 7, pp .10 44 -10 61.
Wikner J., Towill D. R., Naim M., (19 91) . Smoothing supply chain dynamics, International
Journal of Production Economics, 22, pp.2 31...
Wireless Sensor Networks Application Centric Design 2011 Part 3 docx
... Science, Vol. 6, pp.
34 3 -35 5, 2006.
[2] Caine, N. (19 80). The rainfall intensity-duration control of shallow landslides and debris
flows, Geografiska Annaler, Vol. 62A, pp. 23- 27, 19 80.
[3] Christanto, ... parameters) 0. 03, 0.5
k
AP
(no. of encoded packets per AP) 36 00
R (bit-rate) 6, 11 , 12 , 24 Mbit/s
T
SF
(superframe duration) 10 0 ms
τ
HO
(handover time) 0.5 s
P
PL
(packet-l...