Civil Engineering Reference
In-Depth Information
Integration of Eq. (ii) yields
L
2
x
x
3
3
EI
d
v
d
x
w
2
Lx
2
=−
−
+
+
C
1
When
x
=
0 at the built-in end,
v
=
0 so that
C
1
=
0 and
L
2
x
x
3
3
EI
d
v
w
2
Lx
2
d
x
=−
−
+
(iii)
Integrating Eq. (iii) we have
L
2
x
2
Lx
3
3
+
x
4
12
w
2
EI
v
=−
2
−
+
C
2
and since
v
=
0 when
x
=
0,
C
2
=
0. The deflection curve of the beam therefore has the
equation
w
24
EI
(6
L
2
x
2
4
Lx
3
x
4
)
v
=−
−
+
(iv)
and the deflection at the free end where
x
=
L
is
wL
4
8
EI
v
tip
=−
(v)
which is again negative and downwards. The applied loading in this case may be easily
expressed in mathematical form so that a solution can be obtained using Eq. (13.5),
i.e.
d
4
v
d
x
4
w
EI
=−
(vi)
in which
w
=constant. Integrating Eq. (vi) we obtain
EI
d
3
v
d
x
3
=−
wx
+
C
1
We note from Eq. (13.4) that
d
3
v
d
x
3
S
EI
=−
(i.e.
−
S
=−
wx
+
C
1
)
When
x
=
0,
S
=−
wL
so that
C
1
=
wL
Alternatively we could have determined
C
1
from the boundary condition that when
x
=
L
,
S
=
0.
Hence
EI
d
3
v
d
x
3
=−
w
(
x
−
L
)
(vii)
