Aircraft structures for engineering students - part 5 pps
Aircraft structures for engineering students - part 5 pps
...
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
kN ... substituting for
r(ue)
from Eq. (8 .51 )
3.23
1 05~ ;5. ’6
E(ue)
=
1000/~0
Equation (8 .54 ) then becomes
)
dtle
D,
=
-
1;
(
g)2
(
Su,e
7
SA :-)...
Aircraft structures for engineering students - part 1 pps
... 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.
(i)
aI
=
+55 N/mm2,
arI
=
+29N/mm2,
a1
at 11 .5& quot; ... load for a flat plate
51
51
59
61
63
65
65
68
68
70
71
73
76
77
85
100
103
103
107
109
110
110
122
122
1 25
129
137
141
14...
Aircraft structures for engineering students - part 9 pps
... matrix for a uniform beam
51
1
It is possible to write Eq. (12.44) in an alternative form such that the elements of
[KJ
are pure numbers. Thus
12 -6 -1 2 -6
-6 4 62
-1 2 6 12 6
-6 2 64
This form ...
Stiffness matrix for a uniform beam
51 5
The beam may be idealized into two beam-elements,
1-2
and
2-3 .
From Fig.
12.11
we see that
v1
=
v3
=
0,...
Aircraft structures for engineering students - part 2 pot
...
BC
3000
R
1
DA
3000 R
1
AC
50 00
-5 R/3
-5 13
1 25
000R/9 8 75
DB
50 00
-5 R/3
-5 13
1 25
000R/9 8 75
C
=
48
OOOR
Problems
67
Fig.
P.3 .5
P.3 .5
Determine the maximum shear stress ...
M
=
-
(L
-
z)
-
-Rz,
hence
-
=
-
-2
and between
B
and
C
P
fi
dM
d3
4
2
dR
2
M =-( L-z)
R(L-z),
hence
-=
(L-z)
Thus...
Aircraft structures for engineering students - part 3 docx
... zero then
d2W
-
a2W
d2W
-
d2W
ax2
-
8Y2
ay2
-
dX2
-
-v-
or
-
-v-
and the plate has curvatures of opposite signs. The case of
My
=
0
is illustrated in
Fig.
5. 3.
A
surface ...
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
_-
-
wh...
Aircraft structures for engineering students - part 4 docx
...
5
one simply supported
0
1
2
3
4
I5
I3
II
9-
7-
5,
k
-
-
-
a/b
(b)
k
4 0-
36
-
Clamped
edges
Simply supported
123 45
a/b
(C)
Fig.
6.16
(a) Buckling coefficients for ...
Flexural-torsional buckling
of
thin-walled columns
1 85
0
P
-
~EIJL~
-Pxs
P
-
~EI,,JL~
0
PYS
PYS
-
Pxs IOPIA
-
.rr2ET/L2
-
GJ
=O...
Aircraft structures for engineering students - part 6 potx
...
=
25mm of
4.3 N/mm. In the wall 23
q23
=
-6 9.0
x
1 0-~ 2
x
75ds2
-
34 .5
r
i.e.
q23
=
-1 .04,~2
-
34 .5
Hence q23 varies linearly from a value of -3 4 .5 N/mm at 2 to -1 38 .5
N/mm ... right-hand side
of
Eq. (ii)
we have
r=l
qb:23
=
qb,34
=
-7 .22
X
1 0-4 (400
X
100)
=
-2 8.9N/mm
qb,4j
=
-2 8.9
-
7.22
x
io-4(ioo
x...
Aircraft structures for engineering students - part 7 ppt
...
qb:27
=
-1 .07
x
qb,J6
=
-1 .07
x
x
3880
x
230
=
-9 5. 5N/mm
x
258 0
x
1 65
=
-4 5. 5N/mm
qb $5
-4 5. 5
-
1.07
x
io-4
x
258 0
x
(-1 65)
=
0
qb .57
=
1.07
X
qb;38
=
-1 .07 ...
For cell
I
For cell
I1
[-2 50 qI
+
qII( 250
+
7 25
+
233
+
7 25)
-
233q11~]
(ii)
de
1
dz
-
2
x
355
OOOGREF
_-
F...
Aircraft structures for engineering students - part 8 doc
...
12, 56 51 0 0 .55 9 1,6 6 45
I
93
000
23, 45 1 65 0.9 15 2,
5
1290
I1
258 000
34: 10 15 0 .55 9 3,4 19 35
38 304 2.030
25 304 1.6 25
Ans.
241.4mm.
P.10.13
A
portion of a tapered, three-cell ... 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)
-
-
-
-
-...
Aircraft structures for engineering students - part 10 pot
... aeroelasticity
For a non-trivial solution
(1
-
16d) -1 5Xw’
-(
1
-
6XJ)
Expanding this determinant we have
or
-( 1
-
16Xw2)(1
-
6XJ)
+
75( Xw2)’
=
0
21(AJ)’
-
22xw2
+
1
= ... the form
shown
in Fig. 13.10.
Substituting the second natural frequency
in
Eq.
(vii)
we have
-
-
~2
1
-
16Xw2
-
1
-
16
x
(1/21)
15
-...
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