Civil Engineering Reference
In-Depth Information
W
X
EI
W
W
2
2
L
(a)
y
F
IGURE
13.6
Deflection of a
simply supported beam
carrying a concentrated load at
mid-span (Ex. 13.5)
x
G
y
(b)
Integrating we obtain
x
2
2
+
EI
d
v
W
2
d
x
=
C
1
From symmetry the slope of the beam is zero at mid-span where
x
=
L
/2. Thus
WL
2
/
16 and
C
1
=−
EI
d
v
W
16
(4
x
2
L
2
)
d
x
=
−
(iii)
Integrating Eq. (iii) we have
4
x
3
3
L
2
x
W
16
EI
v
=
−
+
C
2
and when
x
=
0,
v
=
0 so that
C
2
=
0. The equation of the deflection curve is, therefore
W
48
EI
(4
x
3
3
L
2
x
)
=
−
v
(iv)
The maximum deflection occurs at mid-span and is
WL
3
48
EI
v
mid-span
=−
(v)
=
0at
x
=
L
Note that in this problem we could not use the boundary condition that
v
to determine
C
2
since Eq. (i) applies only for 0
≤
x
≤
L
/
2; it follows that Eqs (iii) and
(iv) for slope and deflection apply only for 0
≤
x
≤
L
/
2 although the deflection curve
is clearly symmetrical about mid-span.
E
XAMPLE
13.6
The simply supported beam shown in Fig. 13.7(a) carries a con-
centrated load
W
at a distance
a
from the left-hand support. Determine the deflected
shape of the beam, the deflection under the load and the maximum deflection.
