Aircraft structures for engineering students - part 1 pps
Aircraft structures for engineering students - part 1 pps
...
10 3
10 3
10 7
10 9
11 0
11 0
12 2
12 2
12 5
12 9
13 7
14 1
14 2
14 9
14 9
15 2
15 2
15 6
16 0
16 2
16 5
16 9
17 3
17 4
1. 1
6
Experimental measurement
of
surface strains
3
1
Also
i.e. ...
Since
Q,
=
0,
Eqs
(1. 1 1) and (1. 12) reduce to
Problems
33
Am.
or
=
10 0.2N/mm2,
0
=
24" 11 '
a 11
=
-2 0.2 N/IIUII~,
6'
=
11 4...
Aircraft structures for engineering students - part 5 pps
...
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-
13 7 .1
+
13 .5~0 ~10 ~-4 .5sin1O0=O
1. e.
4.5
kN ... variation of
0 .1,
K,,
=
1. 45
for six specimens,
K,,
=
1. 445
for
10
specimens,
K,,
=
1. 44
for
20
speci-
mens and for
10 0
specimens or more
K,,...
Aircraft structures for engineering students - part 9 pps
...
vj
o
12 1~~ -6 1~ ~
o
-1 21~ ~ -6 1~ ~
o
-1 21~ ~ 61~ ~
o
12 1~~ 61~ ~
00
000
0
0
-6 /L2 4/L
0
6/L2 2/L
00
000
0
0
-6 /L2
2/L
0
6/L2 4/L
,
(12 .45)
We may deduce the transformation ... Thus
12 -6 -1 2 -6
-6 4 62
-1 2 6 12 6
-6 2 64
This form of Eq. (12 .44) is particularly useful in numerical calculations for an assem-
blage o...
Aircraft structures for engineering students - part 2 pot
...
Area
@
F
AB -R/2
-1
12 ~14 ~~
-R/2
-1 12
~14 ~~
AB
=I2
BC
Ll2
CD
LI2
DE
LIZ
BD
LIZ
EB
LI2
A€
LI2
AB
A
R
1
RIA
A R
1
RIA
A -R
-1
RIA
A -R
-1
RIA
A R
1
RIA
The ... 4000 20
000
2PB,f 13 213
0
0
16 013
0
0
0
-
16 0J2/3
0
FB 4000J2 -2 0000J2
d2pB.f
13 ~ 213
DC 4000 80
000
PB,f13
11 3
pDsf
1
32 013 320
PD,f
1 32 0...
Aircraft structures for engineering students - part 3 docx
...
thin plates
12 3
Fig.
5 .1
Plate subjected
to
pure bending.
are given
by
1
Z
Ex =-,
Ey -
Px
PY
Referring to
Eqs (1. 47)
we
have
1
1
&, =-( cJx-vcry);
E
Ey =-( cry-Vcrx)
E
(5.2) ...
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
_-
-...
Aircraft structures for engineering students - part 4 docx
...
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
18 3
or
UB=U+(YS-YB)@
Similarly ... member
6 .1
2
Flexural-torsional buckling
of
thin-walled columns
18 5
0
P
-
~EIJL~
-Pxs
P
-
~EI,,JL~
0
PYS
PYS
-
Pxs IOPIA
-
.rr2ET/L2
-
GJ
=O
(...
Aircraft structures for engineering students - part 6 potx
...
qb.67
=
qb;23
=
(by
qb. 21
=
-7 .22
X
1 0-4 (250
X
10 0)
=
-1 8.1N/mm
qb .18
=
-1 8 .1
-
7.22
x
1 0-4 (200
x
30)
=
-2 2.4N/mm
qb:87
=
qb: 21
=
-1 8a1
N/mm
(by
Taking moments ...
(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
+
25 620 0.4
x
lo6...
Aircraft structures for engineering students - part 7 ppt
...
0.09
1. 78
x
10 '
345.6
7 .1
Cell
11
0 .14 9 0 .11 2
367
37 .15 18 .76
6.63 4.97
0.67 0.25
0 .12 0.07
435.6
3.09
x
10 '
5. 51
x
IO*
8.9
Cell
I 11
0 .12 1
15 5
41. 10
2 .10
0.56 ...
For cell
I
For cell
I1
[-2 50qI
+
qII(250
+
725
+
233
+
725)
-
233q 11~ ]
(ii)
de
1
dz
-
2
x
355
OOOGREF
_-
For
cell
11 1
(i...
Aircraft structures for engineering students - part 8 doc
... we 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 ... ~~
12 " 10 84 1. 220
1
2L 216 0 1. 625
14 ,23 12 7 0. 915
34u 797 0. 915
34L 797 0. 915
I
10 8400
I1
202
500
I 11
528
000
An...
Aircraft structures for engineering students - part 10 pot
... the form
shown
in Fig. 13 .10 .
Substituting the second natural frequency
in
Eq.
(vii)
we have
-
-
~2
1
-
16 Xw2
-
1
-
16
x
(1/ 21)
15
-
v1
15 Xw’
-=
~2
1
-
16 h2 1- 16 ... simplifies to
(13 .15 )
556
Elementary aeroelasticity
For a non-trivial solution
(1
-
16 d) -1 5Xw’
-(
1
-
6XJ)
Expanding this determi...
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