Civil Engineering Reference
In-Depth Information
is the sum of the loads to the left of K and
n
W
j
,R
j
=
1
is the sum of the loads to the right of K, we have, from Eqs (20.5) and (20.6)
W
j
,L
L
n
n
W
j
,R
L
d
M
K
d
x
−
a
a
=
+
−
L
j
=
1
j
=
1
For a maximum value of
M
K
,d
M
K
/d
x
=
0 so that
W
j
,L
L
n
n
−
a
a
L
=
W
j
,R
L
j
=
1
j
=
1
or
n
n
1
a
1
W
j,L
=
W
j
,R
(20.12)
L
−
a
j
=
1
j
=
1
From Eq. (20.12) we see that the bending moment at K will be a maximum with one
of the loads at K (from the previous argument) and when the load per unit length of
beam to the left of K is equal to the load per unit length of beam to the right of K.
Part of the load at K may be allocated to AK and part to KB as required to fulfil this
condition.
Equation (20.12) may be extended as follows. Since
n
n
n
W
j
=
W
j
,L
+
W
j
,R
j
=
j
=
j
=
1
1
1
then
n
n
n
W
j
,R
=
W
j
−
W
j
,L
j
=
1
j
=
1
j
=
1
Substituting for
n
W
j
,R
j
=
1
in Eq. (20.12) we obtain
&
'
(
n
n
n
1
a
1
)
W
j
,L
=
W
j
−
W
j
,L
L
−
a
j
=
1
j
=
1
j
=
1
Rearranging we have
j
=
1
W
j
−
j
=
1
W
j
,L
j
=
1
W
j
,L
L
−
a
=
a
