Civil Engineering Reference
In-Depth Information
Solving for the displacement
x
gives
x
=5
F
o
L
3
=
12
EI
ð2
:
116Þ
Then substituting Eq. (2.116) into the second and third equations of Eq. (2.114), 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
:
117Þ
Comparing these moment demands in Eq. (2.117) with the yield moments in Eq. (2.113) 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. (2.117), it is seen that the moment demand at PHL #2 has exceeded its maximum moment
capacity, and therefore PHL #2 is assumed to reach its softening stage (i.e. the second nonlinear
slope). According to Eq. (2.113), this gives
Þ
θ
0
1
,
m
2
=9
F
o
L
=
4−
EI
=
2
L
Þ
θ
0
2
m
1
=2
F
o
L
+2
EI
=
L
ð
ð
ð2
:
118Þ
Substituting Eq. (2.118) into Eq. (2.114) gives
2
3
<
=
<
=
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
9
F
o
L
=
4
4
5
6
EI
=
L
2
=
ð2
:
119Þ
4
EI
=
L
+2
EI
=
L
2
EI
=
L
:
;
:
;
6
EI
=
L
2
2
EI
=
L
4
EI
=
L
−
EI
=
2
L
Solving for Eq. (2.119) gives
<
=
<
=
<
=
2
4
3
5
−
1
12
EI
=
L
3
6
EI
=
L
2
6
EI
=
L
2
13
F
o
L
3
x
−
θ
0
1
−
θ
0
2
5
F
o
2
F
o
L
9
F
o
L
=
4
=
6
EI
−
F
o
L
2
6
EI
=
L
2
=
6
EI
=
L
2
EI
=
L
=
=
EI
ð2
:
120Þ
:
;
:
;
:
;
6
EI
=
L
2
−5
F
o
L
2
2
EI
=
L
7
EI
=
2
L
=
2
EI
Then substituting Eq. (2.120) into the second and third equations of Eq. (2.114), the moments
are calculated as
<
=
=
13
F
o
L
3
=
6
EI
6
EI
=
L
2
m
1
m
2
4
EI
=
L
2
EI
=
L
4
F
o
L
F
o
L
−
F
o
L
2
=
EI
=
ð2
:
121Þ
6
EI
=
L
2
2
EI
=
L
4
EI
=
L
:
;
−5
F
o
L
2
=
2
EI
These moments can similarly be obtained using Eq. (2.118), i.e.
F
o
L
2
EI
=4
F
o
L
2
EI
L
m
1
=2
F
o
L
+
ð2
:
122aÞ
5
F
o
L
2
2
EI
=
F
o
L
m
2
=
9
F
o
L
EI
2
L
4
−
ð2
:
122bÞ
