Civil Engineering Reference
In-Depth Information
E
XAMPLE
16.19
Calculate the reactions at the supports in the beamABCD shown
in Fig. 16.41. The flexural rigidity of the beam is constant throughout.
40 kN
5 kN/m
5 kN/m
B
C
A
D
EI
6m
5m
5m
6m
F
IGURE
16.41
Symmetrical
beam of Ex. 16.19
The beam in Fig. 16.41 is symmetrically supported and loaded about its centre line;
we may therefore use this symmetry to reduce the amount of computation.
In the centre span, BC,
M
BC
=−
M
CB
and will remain so during the distribution. This
situation corresponds to Case 3, so that if we reduce the stiffness (
K
BC
)ofBCto2
EI
/
L
there will be no carry over of moment from B to C (or C to B) and we can consider
just half the beam. The outside pinned support at A is treated in exactly the same way
as the outside pinned supports in Exs 16.17 and 16.18.
The FEMs are
6
2
12
5
×
M
AB
=−
M
BA
=−
=−
15 kNm
40
×
5
M
BC
=−
M
CB
=−
=−
25 kNm
8
The DFs are
K
BA
K
BA
+
3
EI
/
6
3
EI
/
6
DF
AB
=
K
BC
=
2
EI
/
10
=
0
.
71
+
Hence
DF
BC
=
1
−
0
.
71
=
0
.
29
The solution is completed as follows:
A
B
DFs
1
0.71
0.29
FEMs
15.0
15.0
25.0
Balance A
15.0
Carry over
7.5
1.78
0.72
Balance B
Final moments
0
24.28
24.28
