Civil Engineering Reference
In-Depth Information
Then substituting Eq. (2.81) into the second and third equations of Eq. (2.79), the moments are
calculated as
=
F
o
L
3
12
EI
=
6
EI
=
L
2
6
EI
=
L
2
F
o
L
=
2
F
o
L
=
2
m
1
m
2
ð2
:
82Þ
Comparing these moment demands in Eq. (2.82) with the yield moments in Eq. (2.77) shows
that the plastic hinges have not yielded, which means the linear structure assumption is correct.
Therefore, the final displacement and moments are presented in Eqs. (2.81) and (2.82),
respectively.
Example 2.8 Nonlinear Column with Yielding in One Plastic Hinge
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
3
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
3
F
o
2
ð2
:
83Þ
6
EI
L
4
EI
L
2
EI
L
−θ
1
−θ
2
=
m
1
2
6
EI
L
2
EI
L
4
EI
L
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. (2.83) gives
12
EI
L
3
x
=3
F
o
ð2
:
84Þ
Solving for the displacement
x
in Eq. (2.84) gives
x
=
F
o
L
3
=
4
EI
ð2
:
85Þ
Then substituting Eq. (2.85) into the second and third equations of Eq. (2.83), the moments are
calculated as
=
F
o
L
3
4
EI
=
m
1
m
2
6
EI
=
L
2
6
EI
=
L
2
3
F
o
L
=
2
3
F
o
L
=
2
ð2
:
86Þ
Comparing these moment demands in Eq. (2.86) with the yield moments in Eq. (2.77) shows
that PHL #2 has yielded, but PHL #1 remains elastic. This means the original assumption that
the structure is linear is incorrect.
Now assume that PHL #2 has yielded while PHL #1 remains elastic. This gives
θ
0
1
=0,
m
2
=
F
o
L
+2
EI
=
L
Þ
θ
0
2
ð
ð2
:
87Þ
