Civil Engineering Reference
In-Depth Information
l
w
A
K
B
l
a
L
(a)
ve
s
k
a
b
ve
S
K
IL
(b)
k
1
a
1
b
1
m
F
IGURE
20.7
Shear
force and bending
moment due to a
moving uniformly
distributed load
ve
M
K
IL
(c)
If we consider an elemental length
δ
l
of the load, we may regard this as a concentrated
load of magnitude
w
δ
l
. The shear force,
δ
S
K
, at K produced by this elemental length
of load is then from Fig. 20.7(b)
δ
S
K
=
w
δ
ls
The total shear force,
S
K
, at K due to the complete length of load is then
l
S
K
=
ws
d
l
0
or, since the load is uniformly distributed
w
l
0
S
K
=
s
d
l
(20.15)
Hence
S
K
=
w
×
area under the projection of the load in the S
K
influence line.
Similarly
w
l
0
M
K
=
m
d
l
(20.16)
so that
M
K
=
w
×
area under the projection of the load in the M
K
influence line.
Maximum shear force at K
It is clear from Fig. 20.7(b) that the maximum positive shear force at K occurs with
the head of the load at K while the maximum negative shear force at K occurs with the
