Civil Engineering Reference
In-Depth Information
Alternatively we have
L
L
Wx
2
EI
(
−
Wx
)
EI
v
B
=
x
d
x
=−
d
x
0
0
which gives
WL
3
3
EI
v
B
=−
as before.
Note that if the built-in end had been selected as the origin for
x
, we could not have
determined
v
B
directly since the term
x
B
(d
v/
d
x
)
B
in Eq. (ii) would not have vanished.
The solution for
v
B
would then have consisted of two parts, first the determination of
(d
v/
d
x
)
B
and then the calculation of
v
B
.
E
XAMPLE
13.12
Determine themaximumdeflection in the simply supported beam
shown in Fig. 13.14(a).
y
w
B
x
A
C
EI
wL
2
wL
2
R
A
R
B
L
/2
L
/2
(a)
Centroid of area of left-hand
half of
M
/
EI
diagram
w
L
2
8
EI
F
IGURE
13.14
Moment-area
method for a simply
supported beam
carrying a uniformly
distributed load
5
82
L
(b)
From symmetry we deduce that the beam reactions are each
wL
/2; the
M
/
EI
diagram
has the geometry shown in Fig. 13.14(b).
If we take the origin of axes to be at A and consider the half-span AC, Eq. (13.10)
becomes
x
C
d
v
d
x
x
A
d
v
d
x
C
M
EI
x
d
x
C
−
A
−
(
v
C
−
v
A
)
=
(i)
A
In this problem (d
v/
d
x
)
C
=
0,
x
A
=
0 and
v
A
=
0; hence Eq. (i) reduces to
L
/
2
M
EI
x
d
x
v
C
=−
(ii)
0
