Civil Engineering Reference
In-Depth Information
Therefore
EI
d
v
d
x
w
24
(6
Lx
2
4
x
3
L
3
)
=
−
−
(iii)
Integrating again gives
w
24
(2
Lx
3
x
4
L
3
x
)
EI
v
=
−
−
+
C
2
Since
v
=
0 when
x
=
0 (or since
v
=
0 when
x
=
L
) it follows that
C
2
=
0 and the
deflected shape of the beam has the equation
w
24
EI
(2
Lx
3
x
4
L
3
x
)
v
=
−
−
(iv)
The maximum deflection occurs at mid-span where
x
=
L
/2 and is
5
wL
4
384
EI
v
mid-span
=−
(v)
So far the constants of integration were determined immediately they arose. However,
in some cases a relevant boundary condition, say a value of gradient, is not obtainable.
The method is then to carry the unknown constant through the succeeding integration
and use known values of deflection at two sections of the beam. Thus in the previous
example Eq. (ii) is integrated twice to obtain
Lx
3
6
−
x
4
12
w
2
EI
v
=
+
C
1
x
+
C
2
The relevant boundary conditions are
v
=
0at
x
=
0 and
x
=
L
. The first of these
wL
3
/
24. Thus the equation of
gives
C
2
=
0 while from the second we have
C
1
=−
the deflected shape of the beam is
w
24
EI
(2
Lx
3
x
4
L
3
x
)
v
=
−
−
as before.
E
XAMPLE
13.5
Figure 13.6(a) shows a simply supported beam carrying a con-
centrated load
W
at mid-span. Determine the deflection curve of the beam and the
maximum deflection.
The support reactions are each
W
/2 and the bending moment
M
at a section X a
distance
x
(
≤
L
/
2) from the left-hand support is
W
2
x
M
=
(i)
From Eq. (13.3) we have
EI
d
2
v
d
x
2
W
2
x
=
(ii)
