Analysis of time-dependent internal forces with conventional computer programs - Concrete Structures: Stresses and Deformations - page 194

creep and shrinkage of concrete and relaxation of prestressed steel.
Ignore cracking and presence of reinforcement in AB . Given data: E c ( t 1 )
=

25 GPa; φ ( t 2 , t 1 ) = 2; χ = 0.8; ε cs ( t 2 , t 1 ) = − 300 × 10 −6 ; ∆ σ pr = − 50 MPa.

Cross-sectional area properties for AB : A c = 1.0 m 2 ; I = 0.1 m 4 . For the

cable, A s 200 GPa.

Table 6.3 gives the input and the results of Computer run 1 using the
program PLANEF. During the tensioning, the change in cable length
can occur independently from the deformation of concrete; thus the
translation at the tip of the cantilever is not compatible with the elonga-
tion of the cable. For this reason, the analysis in Table 6.3 is for a

= 250 mm 2 ; E s =

Table 6.3 Input data and results of Computer run 1 using program PLANEF. Example
6.3, Fig. 6.9

Input data
Number of joints = 2; Number of members = 1; Number of load cases = 1
Number of joints with prescribed displacements = 1; Elasticity modulus = 25.0E+09
Nodal coordinates
Node
1
2

x y

0.0 0.0

10.0 0.0

Element information
Element
1

1st node
1

2nd node
2

a
.10000E+01

I
.10000E+00

Support conditions
Node Restraint indicators

u

Prescribed displacements

v u v

2 0 0 0 .00000E+00 .00000E+00 .00000E+00

Forces applied at the nodes
Load case
1

Node
1

F x
.17890E+06

F y
− .89440E+06

M z
.00000E+00

Member end forces with nodal displacement restrained
Ld.
case
1

Member
1

A r1
.0000E+00

A r2
− .1250E+06

A r3
− .2083E+06

A r4
.0000E+00

A r5
− .1250E+06

A r6
.2083E+06

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

u
.71560E − 04
.71560E − 10

v
.57534E − 03
.11853E − 08

.12200E − 03
− .14730E − 09

Forces at the supported nodes
Node
2

F x
− .17890E+06

F y
− .16056E+06

M z
.35560E+06

Member end forces
Member
1

F 1 *
.17890E+06

F 2 *
− .89440E+05

F 3 *
− .29104E − 10

F 4 *
− .17890E+06

F 5 *
− .16056E+06

F 6 *
.35560E+06