Civil Engineering Reference
In-Depth Information
to the parameters illustrated in Tables 4.1 and 4.2. The moments
m
1
and
m
2
and shear force
V
can be expressed in the terms of
θ
1
00
,
θ
2
00
and
τ
00
as:
<
m
1
= −91
:
95 + 8
:
6×10
4
θ
0
1
m
2
= −91
:
95 + 8
:
6×10
4
θ
0
2
ð4
:
37Þ
:
V
=61
:
3+4
:
0×10
4
τ
00
where the minus sign in Eq. (4.37) denotes that the negative cracking capacities has been
exceeded. Therefore, Eq. (4.35) becomes:
−
0.9586
−
0.9586
0.6391
0.6391
0.002
F
′′
−
0.9586
1.9587
+
0.86
0.9171
−
0.9586
−θ
−
91.95
1
5
×
10
=
′′
−
0.9586
0.9171
1.9587
+
0.86
−
0.9586
−θ
−
91.95
2
′′
0.6391
−
0.9586
−
0.9586
0.6391
+
0.40
−τ
61.3
ð4
:
38Þ
Solving for the plastic deformations
θ
1
00
,
θ
2
00
and
τ
00
in Eq. (4.38) gives:
<
=
<
=
θ
0
1
θ
0
2
τ
00
−0
:
1952 rad
−0
:
1952 rad
0
:
2804m
×10
−3
=
ð4
:
39Þ
:
;
:
;
Substituting Eq. (4.39) back into Eq. (4.35) and solving for the restoring force
F
, the moments
m
1
and
m
2
at the RHs and the shear force
V
gives:
<
=
<
=
72
:
4767 kN
−108
:
7150 kN-m
−108
:
7150 kN-m
72
:
4767 kN
F
m
1
m
2
V
=
ð4
:
40Þ
:
;
:
;
Comparing these demands in Eq. (4.40) with the corresponding capacities in Tables 4.1 and 4.2,
the structure is in the plastic domain and the RHs and SH behave on the hardening branches,
which means that the latest assumption is made correctly and the solution in Eq. (4.40) are
the final results.
Example 4.4 Static Analysis of SDOF Column with Input Displacement (2)
Consider again the SDOF column as shown in Figure 4.9 with the same parameters in
Tables 4.1 and 4.2. Now let the input displacement be equal to 20 mm.
According to results obtained in Example 4.3, first assume that the SDOF column is in the
plastic domain and the two RHs and one SH behave in the hardening branches. Then Eq. (4.38)
can be rewritten here by specifying the total displacement
x
as 0.02:
