Civil Engineering Reference
In-Depth Information
Equations (v)-(vii) then become respectively
EI
d
v
W
(
a
−
L
)
[3
x
2
d
x
=−
+
a
(
a
−
2
L
)]
(
x
≤
a
)
(xii)
6
L
EI
d
v
d
x
Wa
6
L
(3
x
2
a
2
2
L
2
)
x
=−
−
6
Lx
+
+
≥
a
)
(xiii)
W
(
a
−
L
)
[
x
3
EI
v
=−
+
a
(
a
−
2
L
)
x
]
x
≤
a
)
(xiv)
6
L
Wa
6
L
[
x
3
3
Lx
2
(
a
2
2
L
2
)
x
a
2
L
]
x
EI
v
=−
−
+
+
−
≥
a
)
(xv)
The deflection of the beam under the load is obtained by putting
x
=
a
into either of
Eq. (xiv) or (xv). Thus
Wa
2
(
a
L
)
2
−
v
C
=−
(xvi)
3
EIL
This is not, however, the maximum deflection of the beam. This will occur, if
a
<
L
/
2,
at some section between C and B. Its position may be found by equating d
v
/d
x
in Eq.
(xiii) to zero. Hence
3
x
2
a
2
2
L
2
0
=
−
6
Lx
+
+
(xvii)
The solution of Eq. (xvii) is then substituted in Eq. (v) and the maximum deflection
follows.
For a central concentrated load
a
=
L
/2 and
WL
3
48
EI
v
C
=−
as before.
E
XAMPLE
13.7
Determine the deflection curve of the beam AB shown in Fig.
13.8 when it carries a distributed load that varies linearly in intensity from zero at the
left-hand support to
w
0
at the right-hand support.
To find the support reactions we first take moments about B. Thus
2
w
0
L
L
1
R
A
L
=
3
which gives
w
0
L
6
R
A
=
