Analysis of time-dependent internal forces with conventional computer programs - Concrete Structures: Stresses and Deformations - page 196

The restraining forces for creep and shrinkage are entered on separate
lines as load data (for member 1) in Computer run 2 (Table 6.4).
The relaxation in cable AC is a loss of tension presented in Computer
run 2 by an axial compressive force in the member; thus the member
end forces to be used in the input (Equation (6.8) ):

Table 6.4 Input data and results of Computer run 2 using program PLANEF.
Example 6.3, Fig. 6.9

Input data
Number of joints = 3; Number of members = 2; Number of load cases = 1
Number of joints with prescribed displacements = 2; Elasticity modulus = 9.615E+09
Nodal coordinates
Node
1
2
3

x y

0.0 0.0

10.0 0.0

10.0

− 5.0

Element information
Element
1
2

1st node
1
1

2nd node
2
3

a
.10000E+01
.52000E − 02

I
.10000E+00
.10000E − 06

Support conditions
Node

Restraint indicators
u

Prescribed displacements

v u v

2
3

0 0 0 .00000E+00 .00000E+00 .00000E+00

0 0 0 .00000E+00 .00000E+00 .00000E+00

Forces applied at the nodes
Load case
1

Node
1

F x
.00000E+00

F y
.00000E+00

M z
.00000E+00

Member end forces with nodal displacement restrained
Ld.
case
1
1
1

Member
1
1
1

A r1
− .1376E+06
− .2885E+07
.1250E+05

A r2
− .2740E+05
.0000E+00
.0000E+00

A r3
− .1602E+06
.0000E+00
.0000E+00

A r4
.1376E+06
.2885E+07
− .1250E+05

A r5
.2740E+05
.0000E+00
.0000E+00

A r6
− .1133E+05
.0000E+00
.0000E+00

Analysis results; load case No. 1
Nodal displacements
Node
1
2
3

u
.31270E − 02
.31270E − 08
.12014E − 08

v
.38513E − 02
.30455E − 08
− .24022E − 08

− .16116E − 03
− .49711E − 09
− .56920E − 09

Forces at the supported nodes
Node
2
3

F x
− .15478E+05
− .15478E+05

F y
− .77388E+04
.77388E+04

M z
.77890E+05
.19580E+01

Member end forces
Member
1
2

F 1 *
− .15478E+05
.17305E+05

F 2 *
.77388E+04
.32547E+00

F 3 *
− .16808E+01
.16808E+01

F 4 *
.15478E+05
− .17305E+05

F 5 *
− .77388E+04
− .32547E+00

F 6 *
.77890E+05
.19580E+01