Civil Engineering Reference
In-Depth Information
C
W
5
W
4
W
3
W
2
W
1
B
A
K
R
A
R
B
0.105 m
0.105 m
20 m
(a)
9kN
15kN
15kN
8kN
8kN
F
IGURE
20.14
Determination of
absolute maximum
bending moment in
the beam of
Ex. 20.6
2 m
2.3 m
2.7 m
2.3 m
W
5
W
4
W
3
W
2
W
1
(b)
The first step is to find the position of the centre of gravity of the set of loads. Thus,
taking moments about the load
W
5
we have
+
+
+
+
¯
x
=
×
+
×
+
×
+
×
(9
15
15
8
8)
15
2
15
4
.
3
8
7
.
0
8
9
.
3
whence
x
¯
=
4
.
09m
Therefore the centre of gravity of the loads is 0
.
21m to the left of the load
W
3
.
By inspection of Fig. 20.14(b) we see that it is probable that the maximum bend-
ing moment will occur under the load
W
3
. We therefore position
W
3
and the centre
of gravity of the set of loads at equal distances either side of the mid-span of the
beam as shown in Fig. 20.14(a). We now check to determine whether this position
of the loads satisfies the load per unit length condition. The load per unit length on
AB
=
55/20
=
2
.
75 kN/m. Therefore the total load required on AK
=
2
.
75
×
10
.
105
=
27
.
79 kN. This is satisfied by
W
5
,
W
4
and part (3
.
79 kN) of
W
3
.
Having found the load position, the bending moment at K is most easily found by
direct calculation. Thus taking moments about B we have
R
A
×
20
−
55
×
10
.
105
=
0
which gives
R
A
=
27
.
8kN
