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FIGURE 3.14
Cantilevered beam.
This is a direct application of Betti's reciprocal theorem. Thus, when applied to the beam
of Fig. 3.14b,
P
·
w
|
x
=
0
,
due to
P
1
=
P
1
·
w
|
x
=
a,
due to
P
(1)
As noted in Fig. 3.14a,
w
|
a
due to
P
is
x
=
P
6
EI
(
a
3
3
L
2
a
2
L
3
w
|
=
−
+
).
x
=
a
From (1),
x
=
a,
due to
P
=
P
1
P
w
P
1
6
EI
(
a
3
3
L
2
a
2
L
3
w
|
x
=
0
,
due to
P
1
=
−
+
)
(2)
This is the requested result.
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