Aircraft structures for engineering students - part 3 docx
Aircraft structures for engineering students - part 3 docx
... expressions for the greatest stresses due to bending at the centre and at the
ends
of
the minor axis.
3PU
-
3)
2~t3
-+ - +-
32
Am.
wO=
(d
a2b2 b4
f3pa2b2(b2
+
v2)
9(3b4
+
2a2b2
+
3d) ...
q(x,y)sin-sin- dxdy
.
mrx m'rx nry n'ry
sin
-
sin
-
dx
dy
=
m=l
2
n=l
T/:jIamsin-sin- a
a
b
b
-
ab
4
-
since
a
2
_-
-
w...
Aircraft structures for engineering students - part 4 docx
... supported
0
1
2
3
4
I5
I3
II
9-
7-
5,
k
-
-
-
a/b
(b)
k
4 0-
36
-
Clamped
edges
Simply supported
1 234 5
a/b
(C)
Fig.
6.16
(a) Buckling coefficients for flat plates in ...
u
=
AI
sin
-
,
8
=
A3
sin
-
L
L
L
(6.85)
1
(P-~)A~-PX~A~=O 2EIXX
(P-9)A1+PysA3=O
6.1
2
Flexural-torsional buckling
of
thin-walled columns...
Aircraft structures for engineering students - part 1 pps
...
e
-
E~
sin
9
cos
8
7=
Substitution
of
E,
and
E~+~~~
from Eqs (1 .31 ) and (1 .30 ) yields
-_
7
-
(E~-E~)sin2e-~,-os2e
2 2 2
(1 .34 )
vi
Contents
Reference
Problems
48
48
3 ... the point and their directions for each combination.
a,
N/m2
ay
N/m2
7.u
N/m2
(i) +54
+30
+5
(ii) +30
+54
-5
(iii)
-6 0 -3 6
+5
(iv)
+30
-5 0
+30
Am....
Aircraft structures for engineering students - part 2 pot
... 4000J2
-6 OOOOJ2
-2 J2pB,f /3
-2 J2 13
0
0
32 0J2
0
EF 4000
-6 0
000
-2 pB,f /3
-2 13
0
0
160
0
FD 4000J2
-8 0000J2
-d2PB.f /3
-
~2 13
0
0
640J2 13
0
CB 4000
80
000
PB,f /3 1 13
EB ...
6'FIaR (FL/A)aF/aR
AC
800J3
30
50
-
J3R/2 -J3 /3/ 2 -2 000
+
20J3R 48.2
CB
800
20
86.6
+
R/2
112 1 732
+
10R
87.6
-1 .8...
Aircraft structures for engineering students - part 5 pps
... AA.
Thus
4.5
g
mla = -3 g= 13. 5kN
Resolving forces parallel to the axis
of
the fuselage
N
-
T
+
mlacos 10"
-
4.5
sin 10"
=
0
N-
137 .1
+
13. 5~0~10 ~-4 .5sin1O0=O
1.e.
4.5 ...
h
from ESDU data sheets for aircraft having cloud warning
radar and integrating gives
9000
dh
6000
dh
-
=
3. 4
16000
110
-
=
14,
j3000
111)
-...
Aircraft structures for engineering students - part 6 potx
... of
4 .3 N/mm. In the wall 23
q 23
=
-6 9.0
x
1 0-~ 2
x
75ds2
-
34 .5
r
i.e.
q 23
=
-1 .04,~2
-
34 .5
Hence q 23 varies linearly from a value of -3 4.5 N/mm at 2 to -1 38 .5
N/mm
at 3 in ...
-2 8.9N/mm
qb,4j
=
-2 8.9
-
7.22
x
io-4(ioo
x
50)
=
-3 2.5N/m.m
qb,56
=
qb :34
=
-2 8.9
N/mm
(by symmetry)
qb.67
=
qb; 23
=...
Aircraft structures for engineering students - part 7 ppt
...
I1
[-2 50qI
+
qII(250
+
725
+
233
+
725)
-
233 q11~]
(ii)
de
1
dz
-
2
x
35 5
OOOGREF
_-
For
cell
111
(iii)
1
[-2 33 q11+ q111( 736
i-
233
+
736
+
36 8)]
d6’
-=
dz ...
and
-2 10
x
lo6
=
-0 .37 6~~
5.58
x
lo8
Yr
0z>r
=
Hence
Pi
=
P3
=
-P4
=
-P6
=
-0 .37 6
x
30 5
x
900
=
-1 032 12N
and
P2
=
-P...
Aircraft structures for engineering students - part 8 doc
... substitute in
Eq.
(1
1.28)
for
q
from
Eq.
(1
1.25)
and for
PB
from
Eq.
(1
1.26). Hence
1
#PA
-
Gt
(
PA
S,Z
PA)
-
-
-
-
-
-
-
-
-
-
-
2
dz2
dE
2B 2Bh A
Rearranging, ...
258 000
34 : 1015 0.559 3, 4 1 935
38 30 4 2. 030
25 30 4 1.625
Ans.
241.4mm.
P.10. 13
A
portion of a tapered, three-cell wing has singl...
Aircraft structures for engineering students - part 9 pps
...
dz
& ;-
T,(Z)
=
-+ -
-
dB
dz
TJ
=
GJ-
(Eq.
(9.60))
and
d30
Tr
=
-Er
-
dz3
(Eq.
(1
1.58))
so
that
Eq.
(1
1. 63)
becomes
d46' d2B
Er
-
-
GJ
-
-
T~(~)
d# ...
Stiffness matrix for a uniform beam
515
The beam may be idealized into two beam-elements,
1-2
and
2 -3 .
From Fig.
12.11
we see that
v1
=
v3
=
0,
F...
Aircraft structures for engineering students - part 10 pot
... virtual forces 7 1 -3
principle of virtual work 7 1 -3
reciprocal theorem 68, 10 3- 7
strain energy 68, 69, 142, 1 43
total complementary energy 68, 7 6-1 00,
total potential energy 68,71,7 3- 6 , 14 4-9 , ... plate 174
2 8 -3 2
Fabrication of structural components
22 5 -3 2
fuselages 232
integral construction 230
sandwich panels 230 , 23 1
sub-assembly 228, 229
Wings...
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