Civil Engineering Reference
In-Depth Information
A
B
B
C
C
D
F
IGURE
16.34
Moments and
reactions at the ends
of the spans of the
continuous beam of
Ex. 16.15
1.15
1.15
1.4
1.4
R
AB
R
BA
R
BC
R
CB
R
CD
R
DC
1
.
4
1
.
0
=
R
CD
=−
R
DC
=
1
.
40 kN
Therefore, due to the end moments
only
, the support reactions are
R
A,M
=−
1
.
15 kN
R
B,M
=
1
.
15
−
0
.
25
=
0
.
9kN,
R
C,M
=
0
.
25
+
1
.
4
=
1
.
65 kN
R
D,M
=−
1
.
4kN
In addition to these reactions there are the reactions due to the actual loading, which
may be obtained by analysing each span as a simply supported beam (the effects of
the end moments have been calculated above). In this example these reactions may
be obtained by inspection. Thus
R
A,S
=
3
.
0kN
R
B,S
=
3
.
0
+
5
.
0
=
8
.
0kN
R
C,S
=
5
.
0
+
6
.
0
=
11
.
0kN
R
D,S
=
6
.
0kN
The final reactions at the supports are then
R
A
=
R
A,M
+
R
A,S
=−
1
.
15
+
3
.
0
=
1
.
85 kN
R
B
=
R
B,M
+
R
B,S
=
0
.
9
+
8
.
0
=
8
.
9kN
R
C
=
R
C,M
+
R
C,S
=
1
.
65
+
11
.
0
=
12
.
65 kN
R
D
=
R
D,M
+
R
D,S
=−
+
=
1
.
4
6
.
0
4
.
6kN
Alternatively, we could have obtained these reactions by the slightly lengthier
procedure of substituting for
θ
A
,
θ
B
, etc., in Eqs (16.29) and (16.31). Thus, e.g.
6
EI
L
2
S
AB
=
R
A
=
(
θ
A
+
θ
B
)
+
3
.
0(
v
A
=
v
B
=
0)
which gives
R
A
=
1
.
85 kN as before.
Comparing the above solution with that of Ex. 16.7 we see that there are small
discrepancies; these are caused by rounding-off errors.
Having obtained the support reactions, the bending moment distribution (reverting
to the sagging (positive) and hogging (negative) sign convention) is obtained in the
usual way and is shown in Fig. 16.35.
