Civil Engineering Reference
In-Depth Information
Solving for the displacements at the DOFs gives
<
=
<
=
x
1
x
2
x
3
0
:
0782
−0
:
0117
−0
:
0117
=
ð7
:
58Þ
:
;
:
;
Then, substituting Eq. (7.58) back into the last six equations of Eq. (7.53) gives the moments
<
=
2
3
<
=
m
1
m
2
m
3
m
4
m
5
m
6
1479
:
9 2027
:
30
1479
:
9 3892
:
20
1479
:
9
92
:
00
70
:
19
92
:
00
70
:
19
−70
:
19
−70
:
19
4
5
<
=
0
:
0782
−0
:
0117
−0
:
0117
0
2027
:
3
=
=
ð7
:
59Þ
1479
:
9
0
3892
:
2
:
;
:
;
:
;
0
4000
2000
0
2000
4000
Based on the setup of the frame as shown in Figure 7.6, the column axial force can be updated
by computing the shear force at the two ends of the beam using equilibrium of the beam, i.e.
P
1
=
P
+
m
5
+
m
6
L
b
= 200 +
−
70
:
19
−
70
:
19
4
= 164
:
9kN
ð7
:
60aÞ
m
5
+
m
6
L
b
= 200−
−
70
:
19
−
70
:
19
4
P
2
=
P
−
= 235
:
1kN
ð7
:
60bÞ
Now, to check whether any plastic hinge has yielded, it appears that PHL #3 has the largest
combination of axial force and moment. Using Eq. (7.54a) for PHL #3 gives
2
2
235
:
1
600
92
:
00
100
PHL#3
:
+
=1
:
00
ð7
:
61Þ
which indicates that PHL #3 has reached yielding at a lateral force of
F
o
= 73.28 kN that causes
a lateral displacement of
x
1
= 7.82 cm.
The analysis continues with PHL #3 yielded. The lateral force is now applied up to
F
o
= 75.42 kN. Since updated information on the column axial forces has been obtained in
Eq. (7.60), the stiffness matrices are updated accordingly and Rows 1, 2, 3, and 6 are now
extracted from Eq. (7.53) while making use of Eq. (7.55a) with an addition of stiffness
K
t
,
c
= 800 kN/m to the appropriate diagonal term:
2
3
<
=
<
=
1379
:
9 1483
:
4 1476
:
3 1476
:
3
1483
:
4 7911
:
3
x
1
x
2
x
3
−
θ
0
3
75
:
42
0
0
92
:
00
4
5
2000
0
=
ð7
:
62Þ
1476
:
3
2000
7873
:
0 2032
:
3
:
;
:
;
1476
:
3
0
2032
:
3 4673
:
0
Note that in Eq. (7.62),
m
yc
,3
is computed by using Eq. (7.56) as
q
1− 235
:
1
=
600
2
m
yc
,
3
= 100 ×
ð
Þ
=92
:
00 kNm
ð7
:
63Þ
