Civil Engineering Reference
In-Depth Information
The matrix equation used for solving the problem can be obtained, just like the one in
Eq. (2.92), as:
3
2
2
12
EI
L
6
EI
L
6
EI
L
x
5
F
o
2
6
EI
L
4
EI
L
2
EI
L
−θ
1
=
m
ð2
:
103Þ
1
2
6
EI
L
2
EI
L
4
EI
L
−θ
2
m
2
The structure is first assumed to be linear, this gives
θ
0
1
=
θ
0
2
= 0. Then using the first equation of
Eq. (2.103) gives
12
EI
L
3
x
=5
F
o
ð2
:
104Þ
Solving for the displacement
x
in Eq. (2.104) gives
x
=5
F
o
L
3
=
12
EI
ð2
:
105Þ
Then substituting Eq. (2.105) into the second and third equations of Eq. (2.103), the moments
are calculated as
=
5
F
o
L
3
12
EI
=
6
EI
=
L
2
6
EI
=
L
2
m
1
m
2
5
F
o
L
=
2
5
F
o
L
=
2
ð2
:
106Þ
Comparing these moment demands in Eq. (2.106) with the yield moments in Eq. (2.102) shows
that both PHLs #1 and #2 have yielded. This means the original assumption that the structure is
linear is incorrect.
Now assume that both PHLs #1 and #2 have yielded. Based on the moment demands in
Eq. (2.106), it is also assumed that PHL #1 is at the first nonlinear slope and PHL #2 is at
the second nonlinear slope. According to Eq. (2.102), this gives
m
1
=2
F
o
L
+
EI
=
ð Þ
θ
0
1
,
m
2
=3
F
o
L
=
2+
EI
=
ð Þ
θ
0
2
ð2
:
107Þ
Substituting Eq. (2.107) into Eq. (2.103) gives
<
=
<
=
2
4
3
5
12
EI
=
L
3
6
EI
=
L
2
6
EI
=
L
2
x
−
θ
0
1
−
θ
0
2
5
F
o
2
F
o
L
3
F
o
L
=
2
6
EI
=
L
2
=
ð2
:
108Þ
4
EI
=
L
+
EI
=
L
2
EI
=
L
:
;
:
;
6
EI
=
L
2
2
EI
=
L
4
EI
=
L
+
EI
=
L
Solving for Eq. (2.108) gives
<
=
<
=
<
=
2
4
3
5
−
1
12
EI
=
L
3
6
EI
=
L
2
6
EI
=
L
2
7
F
o
L
3
x
−
θ
0
1
−
θ
0
2
5
F
o
2
F
o
L
3
F
o
L
=
2
=
6
EI
−2
F
o
L
2
6
EI
=
L
2
=
5
EI
=
L
2
EI
=
L
=
=
3
EI
ð2
:
109Þ
:
;
:
;
:
;
6
EI
=
L
2
−5
F
o
L
2
2
EI
=
L
5
EI
=
L
=
6
EI
Then substituting Eq. (2.109) into the second and third equations of Eq. (2.103), the moments
are calculated as
