Civil Engineering Reference
In-Depth Information
rst column in the 3 × 3 matrix in
Equation (a) is
l
/6; this is equal to the total elastic load on the beam
which is the integral
The sum of the elements of the
fi
0. The sum of
the elements in the second and third column of the matrix is equal to
similar integrals.
The displacement at coordinate 1 in Fig. C1(a) is equal to the change
in length of the beam; thus
∫
ψ
d
l
when
ψ
1
=
1, while
ψ
2
=
ψ
3
=
D
1
=
ε
O
d
l
.
(e)
This integral is to be evaluated over the length of the beam for the
variable
0, and this procedure has to
be repeated two more times, each time setting one of the
ε
O
when (
ε
O
)
1
=
1 while (
ε
O
)
2
=
(
ε
O
)
3
=
ε
o
values equal
to unity and the others zero. Thus, summing the elements in each col-
umn of the matrix in Equation (a) and changing the variable
ε
O
, we
can express the displacement
D
1
in terms of the axial strain at the three
nodes:
ψ
to
l
6
2
l
3
l
6
D
1
=
{
ε
O
}.
Appendix C lists a series of expressions derived by the same pro-
cedure as Example 3.4. The variation of
is assumed either
linear or parabolic and the number of nodes either 3 or 5.
ε
O
and
ψ
Example 3.5 Simplified calculation of displacements
Use the values of the axial strain and curvature at mid-span and at the
ends of the post-tensioned simple beam in Fig. 2.7 to calculate the
vertical de
ection at point C, the centre of the span and the horizontal
movement of the roller at B at time
t
, after occurrence of creep, shrink-
age and relaxation. Assume parabolic variation of the axial strain and
curvature between the three sections.
We prepare the vectors {
fl
ε
O
} and {
ψ
} to be used in the equations of
Appendix C (see table p. 94):
The de
fl
ection at the centre (Equation (C.8) ) is given by
64
The minus sign indicates upward de
(18.6)
2
96
[1 10 1]
−
298
64
10
−6
=−
0.0103 m
=−
10.3 mm.
fl
ection.
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