Aircraft Structures 1 2011 Part 7 potx
Aircraft Structures 1 2011 Part 7 potx
...
8000
i
x
1. 223
x
602
x
14 .5
N
=
1. 113
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)+0.31sin(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. 113
x
0. 075
=
1. 088
gi...
Aircraft Structures 1 2011 Part 5 potx
... conditions
for this particular case are
v
=
0
at
z
=
0
and
1.
Thus
A
=
0
and
Bsinpl=
0
For a non-trivial solution @e.
v
#
0)
then
sinpl=O or
pl=n.rr
wheren= 11 213 1
giving
or ...
03
4
2
$PCRTnZA:
U
+
V
=
-En
A,
-
-
n =1
41
n =1
413
(6. 51)
Assigning a stationary value to the total potential energy of Eq. (6. 51) with respect to
each coeffic...
Aircraft Structures 1 2011 Part 3 docx
...
d2pB.f
13 ~ 213
DC 4000 80
000
PB,f13
11 3
pDsf
1
32 013 320
PD,f
1 32 013 320
1
48 013 240
BA 4000 60
000
2pB,f/3
213
pD,f
0
0
0
0
FC 4000
10 0
000
0
0
C
=
12 68
C
= ...
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
20 R
1
40R 2 .1
AD
800
20 RI2 11 2 IOR
1
.o
C
=
-268
+
12 9.2R...
Aircraft Structures 1 2011 Part 6 ppt
...
7. 70 [1
+
0 .75 (b/d)']
The relationship between the diagonal tension factor and buckling stress ratio is
b
and
d
having their usual significance.
T/TCR
5
7
9
11
13 15
k
0. 37 ... Example 6 .1 and are listed below
A
=
600mm2
Zxx
=
1. 17
x
10 6mm4
J
=
800mm4
=
0. 67
x
10 6mm4
I?
=
2488
x
10 6mm6
Zo
=
5.32
x
10 6mm4
Problems
203
Fig.
P.6...
Aircraft Structures 1 2011 Part 9 doc
...
qb;23
=
(by
qb. 21
=
-7. 22
X
10 -4(250
X
10 0)
=
-18 .1N/mm
qb .18
=
-18 .1
-
7. 22
x
10 -4(200
x
30)
=
-22.4N/mm
qb: 87
=
qb: 21
=
-18 a1
N/mm
(by
Taking moments about the ... (9 .76 ) and (9 .77 ) we
have, from Eq. (9.38)
0
=
[qb, 81
X
60
x
480
+
2qbT12(240
x
10 0
+
70
x
240)
+
2qb,23
x
240
x
10 0
-2qb.43
x
12 0
x
10 0
-...
Aircraft Structures 1 2011 Part 10 ppt
...
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 0.3
-3
3
5
13 3
0
0.05
0
6 .7 17 7. 3
0
0.3 ... Table 10 .3
49
8
=
49
IO
=
ql6
1
=
q2
1
q32
=
qS7
‘?loll
=
q1 516
=
30.3
+
q 21
q43
=
q76
=
411
12
=
q1 415
=
53.5
+
9 21
q54
=...
Aircraft Structures 1 2011 Part 16 pps
...
214 -16
composite materials
2 18 -20
glass
218
history
211 -14
plastics
2 17 , 218
steel
216 , 2 17
titanium
2 17
analysis of pin-jointed frameworks
500 -7
analysis of space frames
5 07- 9 ... buckling
17 3
instability of stiffened panels
17 5 -7
interrivet buckling
17 7
Kirchhoff,
G.R.
13 3
maximum values of stress
13 7
neutral plane
12 2
plastic...
New Developments in Biomedical Engineering 2011 Part 7 potx
... Silhouette-basedHumanActivityRecognitionUsingIndependent
ComponentAnalysis,LinearDiscriminantAnalysisandHiddenMarkovModel 259
1 1
1 1
1 1
( ) ( ) ( )
( , )
( ) ( ) ( )
t ij j t t
t
q q
t ij j t t
i j
i a b O j
i j
i a b O j
(13 )
1 1
1 1
1 1
( ) ( ) ( )
( , )
( ) ...
NewDevelopmentsinBiomedicalEngineering258
1 1
1 1
1 1
(...
Mobile Robots book 2011 Part 7 potx
... every particle i do
9: compute weight using p(Z(k +1) |X
v
(k +1) )
10 : end for
11 : re-sample the particles
12 : if current observed feature exists in the map then
13 : for every particle i do
14 : ... p(Z(k +1) |X
v
(k +1)
8: end for
9: calculate robot pose & landmark position from particles & associated weight
10 : re-sample the particles
11 : until robot stop navigati...
Intro to Marine Engineering 2 2011 Part 7 potx
... and
ventilation
17 7
Figure 9 .10
Single-duct
system
Warmer
air
ducts
Fresh
air
_£E
Cooler
air
ducts
1
2
3
4
"i
'
,fc
12
Recircuiated
air
U13
=13 ;
" ;13
1
Mixing
box
2 Fan
3
... valve
6
Humidifier
7
Pre-heater
8
Zone heaters
9 Re
heater
10
Plenums
11
Warmer
air
plenums
12
Cooler
air
plenums
13
Pre
insulated spiro
ducting
14
Air
t...
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