Analysis of time-dependent internal forces with conventional computer programs - Concrete Structures: Stresses and Deformations - page 199

Table 6.5 Abbreviated results of Computer run 1, Example 6.4. Space truss;
immediate displacements and forces at time t 1

Nodal displacements
Node u v w

1
2
3
4
5
6
7
8
9
10
11
12

.83225E − 03
.83225E − 03
− .92308E − 02
.71699E − 03
.71699E − 03
− .65934E − 02
.40567E − 03
.40567E − 03
− .23736E − 02
.41362E − 09
.41362E − 09
− .23736E − 08

.60029E − 03
− .60029E − 03
.10672E − 24
− .27823E − 03
.27823E − 03
.32702E − 23
− .26566E − 03
.26566E − 03
.99262E − 23
− .27428E − 03
.27428E − 03
.00000E+00

.44022E − 08
.44022E − 08
.14765E − 01
.29254E − 01
.29254E − 01
.39067E − 01
.48840E − 01
.48840E − 01
.52367E − 01
.55758E − 01
.55758E − 01
.00000E+00

Forces at the supported nodes
Node F x F y M z

1
2
9
10
11
12

.00000E+00
.00000E+00
.00000E+00
− .72000E+06
− .72000E+06
.14400E+07

.00000E+00
.00000E+00
− .45097E − 09
.00000E+00
.00000E+00
.00000E+00

− .24000E+06
− .24000E+06
.00000E+00
.00000E+00
.00000E+00
.00000E+00

Member end forces
Member F 1 F 2

1
2
3
4
5
6
7
...
31

.19209E+06
.51887E+06
.67612E+06
.19209E+06
.51887E+06
.67612E+06
.96046E+05

− .19209E+06
− .51887E+06
− .67612E+06
− .19209E+06
− .51887E+06
− .67612E+06
− .96046E+05

.43425E+04 − .43425E+04

E c ( t 2 , t 1 )
E c ( t 1 )

{ A r ( t 2 , t 1 )} creep = − φ ( t 2 , t 1 ) { A D ( t 1 )}

The values of { A D ( t 1 )} are calculated by Equation (6.3) using the results
of Computer run 1 and noting that { A r ( t 1 )}

= {0} for all members. The

arti cial restraining forces are calculated below for member 1 as

example:

fi

{ A D ( t 1 )} member 1 = {192.09, − 192.09} kN