AAE 221 – Aerospace Structures II – Spring 2004
Chapter 3.2 – Analytical Solutions of Static Problems Using Energy Methods – Solution
1a.
Starting with the GDE,
( 29


=
3
2
'
'
'
'
L
x
P
x
EIw
δ
Boundary conditions:
( 29
( 29
( 29
( 29
( 29
0
'
'
'
0
'
'
0
'
0
'
'
0
0
=
=
=
=
L
EIw
L
EIw
Kw
EIw
w
Integrating,
( 29
( 29
( 29
( 29
4
3
2
2
3
1
3
3
2
2
1
2
2
1
1
2
6
3
2
6
3
2
2
3
2
2
3
2
'
3
2
3
2
'
'
3
2
'
'
'
C
x
C
x
C
x
C
L
x
stp
P
L
x
x
EIw
C
x
C
x
C
L
x
stp
P
L
x
x
EIw
C
x
C
L
x
Pstp
L
x
x
EIw
C
L
x
Pstp
x
EIw
+
+
+
+



=
+
+
+



=
+
+



=
+


=
Applying the boundary conditions, we get
The first BC above gives
0
4
=
C
.
The forth BC above gives
P
C
C
P
=
+

=
1
1
0
.
The third BC above gives
3
2
3
2
0
2
PL
C
PL
P
L
L

+
+


=
.
The second BC above gives
( 29
3
2
0
'
PL
Kw

=
, but since
( 29
( 29
EI
C
w
C
EIw
3
3
0
'
0
'
=
=
,
α
3
2
3
2
3
2
2
3
3
PL
K
PLEI
C
PL
EI
C
K

=

=

=
.
( 29
x
PL
PLx
Px
L
x
stp
L
x
P
x
EIw
α
3
2
3
1
6
1
3
2
3
2
6
2
2
3
3


+



=
In nondimensional form,
( 29


+



=
L
x
L
x
L
x
L
x
stp
L
x
PL
x
EIw
α
3
2
3
1
6
1
3
2
3
2
6
1
2
3
3
3
When
x = L
, the righthandside of the nondimensionalized equation is a constant,
i.e.,
( 29


=
α
3
2
81
14
3
EI
PL
L
w
Therefore, doubling the length increases the deflection by a factor of 8.
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1b.
Applying Castigliano’s theorems,
(
( 29
K
M
dx
EI
x
M
U
y
L
y
0
2
1
2
1
2
0
2
+
=
where
( 29
(
0
'
0
Kw
M
y
=
.
We can apply a dummy load
D
P
at the end of the beam and then inspect the free
body diagram at various sections of the beam.
For
L
x
L
≤
<
3
2
,
(
(
29
x
L
P
x
M
D
y

=
For
3
2
0
L
x
≤
≤
,
( 29
(
29



=
x
L
P
x
L
P
x
M
D
y
3
2
Substituting,
(
29
(
29
2
3
/
2
2
2
3
/
2
0
2
3
2
2
1
2
1
3
2
2
1

+

+



=
L
P
L
P
K
dx
x
L
P
EI
dx
x
L
P
x
L
P
EI
U
D
L
L
D
L
D
Integrating,
(
29
2
3
2
3
3
3
3
3
2
2
1
162
1
2
3
81
1
81
1
2
1

+
+
+

+

+
+

=
PL
L
P
K
EI
L
P
P
P
P
P
L
P
P
L
P
EI
U
D
D
D
D
D
D
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 Spring '04
 Lambros
 Trigraph, GDE, eiw, 2L P 2L, 2L L 2L

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