Civil Engineering Reference
In-Depth Information
which is the equation of a straight line of gradient
w
as demonstrated for the can-
tilever beam of Fig. 3.12 in the previous paragraph. Furthermore, for this particular
example,
S
+
=
0at
x
=
L
so that
C
1
=−
wL
and
S
=−
w
(
L
−
x
) as before.
In the case of a beam carrying only concentrated loads then, in the bays between the
loads,
w
(
x
)
=
0 and Eq. (3.2) reduces to
S
=
C
1
so that the shear force is constant over the unloaded length of beam (see Figs 3.11
and 3.13).
Suppose now that Eq. (3.1) is integrated over the length of beam between the sections
X
1
and X
2
. Then
x
2
x
2
d
S
d
x
d
x
=+
w
(
x
)d
x
x
1
x
1
which gives
x
2
S
2
−
S
1
=
w
(
x
)d
x
(3.3)
x
1
where
S
1
and
S
2
are the shear forces at the sections X
1
and X
2
respectively. Equa-
tion (3.3) shows that the
change
in shear force between two sections of a beam is equal
to the area under the load distribution curve over that length of beam.
The argument may be applied to the case of a concentrated load
W
which may be
regarded as a uniformly distributed load acting over an extremely small elemental
length of beam, say
δ
x
. The area under the load distribution curve would then be
w
δ
x
(
=
W
) and the change in shear force from the section
x
to the section
x
+ δ
x
would
W
. In other words, the change in shear force from a section immediately to the
left of a concentrated load to a section immediately to the right is equal to the value
of the load, as noted in Ex. 3.6.
+
be
Now consider the rotational equilibrium of the element
δ
x
in Fig. 3.18(b) about
B. Thus
x
δ
x
2
M
−
S
δ
x
−
w
(
x
)
δ
−
(
M
+ δ
M
)
=
0
The term involving the square of
δ
x
is a second-order term and may be neglected.
Hence
−
S
δ
x
− δ
M
=
0
or, in the limit as
δ
x
→
0
d
M
d
x
=−
S
(3.4)
Equation (3.4) establishes for the general case what may be observed in particular in
the shear force and bending moment diagrams of Exs 3.4-3.9, i.e. the gradient of the
