Civil Engineering Reference
In-Depth Information
Now instead, let the lateral displacement of the mass be monotonically applied and be equal
to
m
p
L
2
/2
EI
at time step
k
. It follows from Eqs. (3.43) and (3.63a) that
m
p
L
2
2
EI
m
k
+
3
EI
L
θ
0
k
=
3
EI
ð3
:
66Þ
L
2
First, assume that
θ
0
k
= 0, i.e. the column remains linear. It follows from Eq. (3.66) that
m
k
=3
m
p
=
2
ð3
:
67Þ
Since the calculated
m
k
is greater than
m
p
, the assumption that the column remains linear is
incorrect and the column has yielded. Now, let
m
k
=
m
p
according to Eq. (3.63b). It follows
from Eq. (3.43) that
m
p
+
3
EI
L
θ
0
k
=
3
EI
m
p
L
2
2
EI
ð3
:
68Þ
L
2
Solving for
θ
0
k
in Eq. (3.68) gives
θ
0
k
=
m
p
L
ð3
:
69Þ
6
EI
which is the final answer for
θ
0
k
.
Note that a positive moment
m
k
=
m
p
induces a positive plastic rotation
θ
0
k
=
m
p
L
=
6
EI
. This is
an important requirement in dynamic analysis for keeping track of the plastic hinge develop-
ment, which will be seen in the next example.
Example 3.3 Force Analogy Method with Input Displacement after Yielding
Consider the SDOF system as shown in Figure 3.5 with
L
p
= 0 has reached a lateral
displacement of
m
p
L
2
/2
EI
at time step
k
,where
m
p
is the yield moment of the plastic
hinge. According to Example 3.2, the moment versus plastic rotation equation is
θ
00
ðÞ=0
m
ðÞ=
m
p
m
ðÞ≤
m
p
m
ðÞ>
m
p
,
ð3
:
70Þ
if
then
and the plastic hinge response was calculated as
θ
0
k
=
m
p
L
6
EI
,
m
k
=
m
p
ð3
:
71Þ
Now let the lateral displacement of the mass at time step
k
+1 be equal to
x
k
+1
=2
m
p
L
2
/3
EI
.
Since the loading is an increase in displacement from the previous time step, incremental
analysis presented in Eq. (3.59) must be used. Based on the new displacement at time
step
t
k
+1
, it follows from Eqs. (3.59) with stiffnesses given in Eq. (3.63a) that
