Civil Engineering Reference
In-Depth Information
The beam in Fig. 16.32 is not subject to lateral loads. Clearly, in practical cases,
unless we are interested solely in the effect of a sinking support, lateral loads will be
present. These will cause additional moments and shear forces at the ends of the beam.
Equations (16.23)-(16.26) may then be written as
2
θ
A
+
θ
B
+
L
(
v
A
−
v
B
)
2
EI
L
3
M
AB
M
AB
=−
+
(16.28)
θ
A
+
v
B
)
6
EI
L
2
2
L
(
v
A
−
S
AB
S
AB
=
θ
B
+
+
(16.29)
2
θ
B
+
v
B
)
2
EI
L
3
L
(
v
A
−
M
BA
M
BA
=−
θ
A
+
+
(16.30)
θ
A
+
v
B
)
6
EI
L
2
2
L
(
v
A
−
S
BA
S
BA
=−
θ
B
+
+
(16.31)
in which
M
AB
and
M
BA
are the moments at the ends of the beam caused by the applied
loads and correspond to
θ
A
=
0, i.e. they are
fixed-end moments
(FEMs). Similarly the shear forces
S
AB
and
S
BA
correspond to the fixed-end case.
θ
B
=
0 and
v
A
=
v
B
=
E
XAMPLE
16.15
Find the support reactions in the three-span continuous beam
shown in Fig. 16.33.
6kN
10kN
12 kN/m
A
B
C
D
EI
0.5 m
0.5 m
0.5 m
0.5 m
1.0 m
F
IGURE
16.33
Continuous
beam of Ex. 16.15
The beam in Fig. 16.33 is the beam that was solved using the flexibility method in
Ex. 16.7, so that this example provides a comparison between the two methods.
Initially we consider the beam as comprising three separate fixed beams AB, BC and
CD and calculate the values of the FEMs,
M
AB
,
M
BA
,
M
BC
, etc. Thus, using the
results of Exs 13.20 and 13.22 and remembering that clockwise moments are positive
and anticlockwise moments negative
6
×
1
.
0
M
AB
=−
M
BA
=−
=−
0
.
75 kNm
8
10
×
1
.
0
M
BC
=−
M
CB
=−
=−
1
.
25 kNm
8
1
.
0
2
12
12
×
M
CD
=−
M
DC
=−
=−
1
.
0kNm
