Civil Engineering Reference
In-Depth Information
B is a simple support and allows rotation of the beam, there is a moment at B; this is
produced by the loads on the cantilever overhang BC. Rewriting Eq. (iii) in the form
M
AB
=
x
(1
−
x
)
(iv)
we see immediately that
M
AB
=
0at
x
=
0 (as demonstrated above) and that
M
AB
=
0at
x
1m, the point D in Fig. 3.16(e). We shall see later in Chapter 9 that at the point in
the beam where the bending moment changes sign the curvature of the beam is zero;
this point is known as a
point of contraflexure
or
point of inflection
. Now differentiating
Eq. (iii) with respect to
x
we obtain
=
d
M
AB
d
x
=
1
−
2
x
(v)
and we see that d
M
AB
/d
x
=
0at
x
=
0
.
5m. In other words
M
AB
has a turning value or
mathematical maximum at
x
0
.
25 kNm. Note that this is
not the greatest value of bending moment in the span AB. Also it can be seen that for
0
<
x
<
0
.
5m,d
M
AB
/d
x
decreases with
x
while for 0
.
5m
<
x
<
2m,d
M
AB
/d
x
increases
negatively with
x
.
=
0
.
5m at which point
M
AB
=
Nowwe consider themoment equilibriumof the lengthX
2
C of the beam in Fig. 3.16(c)
about X
2
2
2
(3
x
)
2
M
BC
+
−
+
1(3
−
x
)
=
0
so that
x
2
M
BC
=−
12
+
7
x
−
(vi)
from which we see that d
M
BC
/d
x
is not zero at any point in BC and that as
x
increases
d
M
BC
/d
x
decreases.
The complete bending moment diagram is therefore as shown in Fig. 3.16(e). Note
that the value of zero shear force in AB coincides with the turning value of the bending
moment.
In this particular example it is not possible to deduce the displaced shape of the beam
from the bending moment diagram. Only three facts relating to the displaced shape
can be stated with certainty; these are, the deflections at A and B are zero and there is a
point of contraflexure at D, 1m from A. However, using the method described in Sec-
tion 13.2 gives the displaced shape shown in Fig. 3.16(f). Note that, although the beam
is subjected to a sagging bending moment over the length AD, the actual deflection is
upwards; clearly this could not have been deduced from the bending moment diagram.
E
XAMPLE
3.9
Simply supported beam carrying a point moment.
From a consideration of the overall equilibrium of the beam (Fig. 3.17(a)) the
support reactions are
R
A
=
M
0
/
L
acting vertically upward and
R
C
=
M
0
/
L
acting
