Civil Engineering Reference
In-Depth Information
Table 6.1
Input and results of Computer run 1 with program PLANEF. Example 6.1
Analysis results; load case No. 1
Nodal displacements
Node
1
2
u
.00000E+00
.00000E+00
v
.41668E
−
07
.12500E+00
.10417E
−
06
.16667E+00
Forces at the supported nodes
Node
1
F
x
.00000E+00
F
y
−
.10000E+01
M
z
−
.50000E+00
Member end forces
Member
1
F
1
*
.00000E+00
F
2
*
−
.10000E+01
F
3
*
−
.50000E+00
F
4
*
.00000E+00
F
5
*
−
.11102E
−
15
F
6
*
.55511E
−
15
The result of this computer run includes the member end forces
immediately after load application:
0.5
ql
2
, 0, 0, 0}
{
A
(
t
0
)}
=
{0,
−
ql
,
−
As expected, these are the forces at the ends of a cantilever. Apply
Equation (6.3) to obtain:
{
A
D
(
t
0
)}
=
{0,
−
0.5
ql
,
−
0.4167
ql
2
, 0, 0.5
ql
,
−
0.0833
ql
2
}
These are the changes in end forces produced by varying the nodal
displacements form null, when the nodal displacements are prevented,
to the values {
D
*} included in the results of Computer run 1. Creep
freely increases these displacements in the period
t
0
to
t
1
. The hypo-
thetical end forces that can prevent further increase in the period
t
1
to
t
2
are (Equation (6.11) ):
E
c
(
t
2
,
t
1
)
E
c
(
t
0
)
{
A
r
(
t
2
,
t
1
)}
=
−
[
φ
(
t
2
,
t
0
)
−
φ
(
t
1
,
t
0
)] {
A
D
(
t
0
)}
The age-adjusted elasticity modulus is (Equation (6.5) ):
E
c
(
t
1
)
E
c
(
t
1
)
E
c
(
t
2
,
t
1
)
=
(
t
2
,
t
1
)
=
0.8(2.45)
=
0.3378
E
c
(
t
0
)
1
+
χφ
1
+
Substitution in Equation (6.11) gives a set of self-equilibrating end
forces to be used as load input data in Computer run 2:
Search WWH ::
Custom Search