Civil Engineering Reference
In-Depth Information
Rearranging Eq. (iii) we have
L
x
d
x
L
L
2
L
WL
EI
0
v
B
=−
(
x
−
L
)d
x
+
+
0
0
Hence
x
2
2
−
Lx
x
)
L
0
WL
EI
0
L
2
log
e
(
L
v
B
=−
+
−
so that
log
e
2
WL
3
EI
0
1
2
+
v
B
=−
−
i.e.
0
.
19
WL
3
EI
0
v
B
=−
13.4 D
EFLECTIONS DUE TO
U
NSYMMETRICAL
B
ENDING
We noted in Chapter 9 that a beam bends about its neutral axis whose inclination to
arbitrary centroidal axes is determined from Eq. (9.33). Beam deflections, therefore,
are always perpendicular in direction to the neutral axis.
Suppose that at some section of a beam, the deflection normal to the neutral axis (and
therefore an absolute deflection) is
ζ
. Then, as shown in Fig. 13.17, the centroid G is
displaced to G
. The components of
ζ
,
u
and
v
, are given by
u
=
ζ
sin
αv
=
ζ
cos
α
(13.11)
The centre of curvature of the beam lies in a longitudinal plane perpendicular to the
neutral axis of the beam and passing through the centroid of any section. Hence for a
y
Loaded
Unloaded
u
G
Neutral axis
ν
a
a
z
G
F
IGURE
13.17
Deflection of a beam
of unsymmetrical
cross section
