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with
b
1
fb
1
F
=
(5.131a)
V
b
1
fb
0
(5.131b)
Once
X
is evaluated using Eq. (5.130), several responses of interest can be computed. For
example,
=−
=
=
+
p
b P
with
b
b
0
b
1
X
(5.132a)
v
=
fp
=
fbP
(5.132b)
b
T
v
b
0
v
b
0
fbP
V
=
=
=
(5.132c)
See Example 5.8 for a demonstration of the validity of
b
T
v
b
0
v
=
(5.133)
EXAMPLE 5.8 The Force Method: A Continuous Beam
Use the force method to analyze the beam resting on two supports of Fig. 5.31a. For this
continuous beam with two applied moments, find the internal bending moments and the
corresponding slopes at the supports.
As shown in Fig. 5.31b, the beam is considered to be formed of three elements (1, 2, and
3) with the ends designated by
a
,
b
,
c
, and
d
. For this beam the global and local coordinate
systems will coincide; consequently, no coordinate transformation will be introduced. The
variables that are sought are the deformations
v
and forces
p
.
a
M
a
M
b
M
b
M
c
M
c
M
d
θ
b
θ
b
θ
v
=
p
=
(1)
c
θ
c
θ
d
θ
FIGURE 5.31
A continuous beam.
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