Civil Engineering Reference
In-Depth Information
Substituting Eq. (2.87) into Eq. (2.83) and extracting the first and third equations of the matrix
equation gives
=
12
EI
=
L
3
6
EI
=
L
2
x
−
θ
0
2
3
F
o
F
o
L
ð2
:
88Þ
6
EI
=
L
2
4
EI
=
L
+2
EI
=
L
Solving for Eq. (2.88) gives
=
−1
=
12
EI
=
L
3
6
EI
=
L
2
F
o
L
3
x
−
θ
0
2
3
F
o
F
o
L
=
3
EI
−
F
o
L
2
ð2
:
89Þ
6
EI
=
L
2
6
EI
=
L
=
6
EI
Then substituting Eq. (2.89) into the second and third equations of Eq. (2.83), the moments are
calculated as
=
=
6
EI
=
L
2
F
o
L
3
5
F
o
L
=
3
4
F
o
L
=
3
m
1
m
2
2
EI
=
L
=
3
EI
−
F
o
L
2
ð2
:
90Þ
6
EI
=
L
2
4
EI
=
L
=
6
EI
Comparing the moment
m
1
with the corresponding yield moment in Eq. (2.77) shows that PHL
#1 remains elastic. This means the original assumption that PHL #2 has yielded and PHL #1
remains elastic is correct. Therefore, in summary,
=
,
=
x
=
F
o
L
3
3
EI
θ
0
1
θ
0
2
0
F
o
L
2
m
1
m
2
5
F
o
L
=
3
4
F
o
L
=
3
,
ð2
:
91Þ
=
6
EI
Example 2.9 Nonlinear Column with Yielding in All Plastic Hinges
Consider again the SDOF column as shown in Figure 2.9(a) with the moment versus plastic
rotation relationships shown in Figure 2.9(b). Now let the lateral applied force be equal to
5
F
o
. The matrix equation used for solving the problem is similar to Eq. (2.79) except for
the applied load, i.e.
3
2
2
12
EI
L
6
EI
L
6
EI
L
x
5
F
o
2
−θ
1
ð2
:
92Þ
6
EI
L
4
EI
L
2
EI
L
=
m
1
2
6
EI
L
2
EI
L
4
EI
L
−θ
2
m
2
Again, the structure is first assumed to be linear, this gives
θ
0
1
=
θ
0
2
= 0. Then using the first
equation of Eq. (2.92) gives
12
EI
L
3
x
=5
F
o
ð2
:
93Þ
Solving for the displacement
x
in Eq. (2.93) gives
x
=5
F
o
L
3
=
12
EI
ð2
:
94Þ
