Civil Engineering Reference
In-Depth Information
cantilever (without the cable) subjected to a uniform load
q
=
25 kN
combined with
F
x
89.4 kN at node
A
; where
F
x
and
F
y
are the forces exerted by the cable on the concrete member at time
t
1
.
The modulus of elasticity in Computer run 1 is equal to
E
c
(
t
1
)
=
178.9 and
F
y
=
−
25 GPa.
The results of Computer run 1 include the member end forces of
AB
at
time
t
1
:
=
{
A
(
t
1
)}
AB
=
{178.9 kN,
−
89.4 kN, 0,
−
178.9 kN,
−
160.6 kN,
355.6 kN-m}
The changes in the forces at end of member AB from the
fi
xed-end
status are (Equation (6.3) ).
{
A
D
(
t
1
)}
AB
=
{178.9 kN, 35.6 kN, 208.3 kN-m,
−
178.9 kN,
−
35.6 kN,
147.3 kN-m}
In Computer run 2 (Table 6.4) the structure is composed of the two
members
AB
and
AC
and the modulus of elasticity used is (Equation
(6.5) ):
25 GPa
E
c
(
t
2
,
t
1
)
=
0.8(2.0)
=
9.615 GPa
1
+
A transformed cross-sectional area equal to
A
s
E
s
/
E
c
is used for the
cable; a negligible value is entered for I. The end forces that can arti-
fi
cially prevent the time-dependent changes in displacements due to
creep at the two nodes of member
AB
are (Equation (6.6) ):
9.615
{
A
r
(
t
2
,
t
1
)}
creep
AB
=
−
25.0
(2.0){
A
D
(
t
1
)}
AB
{
A
r
(
t
2
,
t
1
)}
creep
AB
=
{
−
137.6 kN,
−
27.4 kN,
−
160.2 kN-m, 137.6 kN,
27.4 kN,
−
113.3 kN-m}
cially prevent the change in length due
to shrinkage of
AB
(Equation (6.7) ):
The axial force that can arti
fi
A
r
(
t
2
,
t
1
)
axial, shrinkage
AB
=
±[9.615 GPa (
−
300 ×10
−6
) (1.0 m
2
)
=
2884.5 kN
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