Civil Engineering Reference
In-Depth Information
tail of the load at K. Note that the shear force at K would be zero if the load straddled
K such that the negative area under the load in the
S
K
influence line was equal to the
positive area under the load.
Maximum bending moment at K
If we regard the distributed load as comprising an infinite number of concentrated
loads, we can apply the criterion of Eq. (20.14) to obtain the maximum value of
bending moment at K. Thus the load per unit length of the complete beam is equal to
the load per unit length of beam to the left of K and the load per unit length of beam
to the right of K. Therefore, in Fig. 20.8, we position the load such that
w
ck
1
a
1
k
1
=
w
dk
1
k
1
b
1
or
ck
1
a
1
k
1
=
dk
1
k
1
b
1
(20.17)
From Fig. 20.8
fc
hk
1
=
a
1
c
a
1
k
1
so that
a
1
k
1
−
hk
1
=
1
hk
1
a
1
c
a
1
k
1
ck
1
ck
1
a
1
k
1
fc
=
hk
1
=
−
a
1
k
1
Similarly
1
hk
1
dk
1
b
1
k
1
dg
=
−
Therefore, from Eq. (20.17) we see that
fc
=
dg
and the ordinates under the extremities of the load in the
M
K
influence line are equal.
It may also be shown that the area under the load in the
M
K
influence line is amaximum
when fc=dg. This is an alternative method of deducing the position of the load for
maximum bending moment at K. Note that, from Eq. (20.17), K divides the load in
the same ratio as it divides the span.
l
c
d
k
1
a
1
b
1
F
IGURE
20.8
Load
position for maximum
bending moment at K
f
g
M
K
IL
h
