Civil Engineering Reference
In-Depth Information
Considering the vertical equilibrium of the length AX of the beam gives
w
L
S
AB
−
wx
+
2
=
0
i.e.
w
x
L
2
S
AB
=+
−
(i)
S
AB
therefore varies linearly along the length of the beam from
−
wL
/2 at A (
x
=
0) to
+
wL
/2 at B (
x
=
L
). Note that
S
AB
=
0 at mid-span (
x
=
L
/2).
Now taking moments about X for the length AX of the beam in Fig. 3.15(b) we have
wx
2
2
wL
2
M
AB
+
−
x
=
0
from which
wx
2
(
L
M
AB
=+
−
x
)
(ii)
Thus
M
AB
varies parabolically along the length of the beam and is positive (sagging)
at all sections of the beam except at the supports (
x
=
0 and
x
=
L
) where it is zero.
Also, differentiating Eq. (ii) with respect to
x
gives
w
L
x
d
M
AB
d
x
=
2
−
(iii)
From Eq. (iii) we see that d
M
AB
/d
x
L
/2, so that the bending
moment diagram has a turning value or mathematical maximum at this section. In this
case this mathematical maximum is the maximum value of the bending moment in the
beam and is, from Eq. (ii),
=
0 at mid-span where
x
=
wL
2
/8.
+
The bending moment diagram for the beam is shown in Fig. 3.15(d) where it is again
drawn on the tension side of the beam; the deflected shape of the beamwill be identical
in form to the bending moment diagram.
Examples 3.4-3.7 may be regarded as 'standard' cases and it is useful to memorize the
form that the shear force and bending moment diagrams take including the principal
values.
E
XAMPLE
3.8
Simply supported beam with cantilever overhang (Fig. 3.16(a)).
The support reactions are calculated using the methods described in Section 2.5. Thus,
taking moments about B in Fig. 3.16(a) we have
R
A
×
2
−
2
×
3
×
0
.
5
+
1
×
1
=
0
