Basic HTML Version
113
IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 1
R. J. ELLWANGER
Substituting
S
by expression (18), results:
(60)
3848 0
2
,
/EJ q
£ ×
l
l
According to the ABNT [8] code, the physical nonlinearity might be
considered, substituting
EJ
by (
EI
)
sec
given by (12). However, the
(
EI
)
sec
/
E
CS
I
C
ratio of the frame bars assemblage, as a function of
the individual bar (
EI
)
sec
/
E
CS
I
C
ratios cannot be considered a con-
stant value; it can vary due to many factors, such as the numbers
of floors and spans, story heights, span lengths, relation between
the cross section dimensions of beams and columns etc. Pinto and
Ramalho [11] show that the influence of physical nonlinearity in the
frame lateral stiffness depends mainly on the reinforcement ratios
and the loading magnitude; they obtained (
EI
)
sec
/
E
CS
I
C
ratios vary-
ing from 0,51 until 0,75 for the ultimate limit state.
On the other hand, Schueler [12] states that the contribution of beams
flexibility for the lateral deflections of a rigid frame can reach 65%,
remaining 35% due to columns flexibility. Furthermore, in a slender
frame the beam reinforcements
A
s
and
A
s
’ tend to be the same, due
to the predominance of wind effects. Thus, in this case equations (10)
and (11) may be employed to relate the components of
y
NL
(frame
horizontal displacements including physical nonlinearity), due to the
beams (
y
NL
BEAMS
) and columns (
y
NL
COLUMNS
), with the corresponding
components (
y
L
BEAMS
) and (
y
L
COLUMNS
) of the horizontal displacements
resulting from linear analysis (
y
L
). Simultaneously, the above-men-
tioned share factors (35% and 65%) of these components in the total
displacements may be used, leading to the following expressions:
(61)
941 ,0
35,0
941 ,0
L
L
NL
COLUMNS
COLUMNS
y
y
y
=
=
(62)
588 ,0
65,0
588 ,0
L
L
NL
BEAMS
BEAMS
y
y
y
=
=
Performing the sum of the components expressed by (61) and
(62), leads to the following relation between the total horizontal
displacements
y
NL
and
y
L
:
(63)
677 ,0
588 ,0
65,0
941 ,0
35,0
L
L
L
NL
y
y
y
y
=
+
=
As the frame lateral stiffness is inversely proportional to these dis-
placements, it may be written:
(64)
C CS
IE
EI
677 ,0 ) (
sec
=
Considering this expression of (
EI
)
sec
and following the same de-
ductive sequence that led to the inequalities (45) and (46), results:
(65)
510,
/
C CS
k
tot
£
´
IEN H
It can be noticed that this inequality is coherent with ABNT [8] code,
which appoints the value of 0,5 for the coefficient
a
1
, if the bracing
system is constituted exclusively by rigid frames. In fact, in order to
obtain
a
1
= 0,5, (
EI
)
sec
must satisfy the following:
(66)
CCi
C CS
IE
IE
EI
552 ,0
650 ,0 ) (
sec
=
=
4.3 Associations of rigid frames with shear walls
and/or shear cores
The same model of figure 3 is adopted and the same definitions
of section 2.2 are considered. In order to deduce the differential
equation of motion for the frames assemblage (figure 3-c), equa-
tion (50) is applied, adding the terms due to the wall-frame interac-
tion to the shear effort, as was done in equation (19):
(67)
)(
1)(
)(
1
)()
(
)(
)
(
2
2
1
1
x
x
x
S
x x q d u Q x w
S
Q(x)
x
T
f
f
f
f
x x
+
=
+
- +
+ + -
=
ò
l
l
l
Considering that
f
2
(
x
) may be neglected in face of 1 and isolating
the terms due to the interaction forces:
(68)
)()
(
)
(
)(
)(
1
1
x x q x w x S d u Q
x
T
f
f x x
- - - -
=
+
ò
l
l
l
In order to deduce the differential equation of motion for the walls
assemblage (figure 3-b), the bending moment given by (31) is in-
troduced, added to the terms due to the interaction forces, as was
done in equation (21):
(69)
[
]
2
2
2
2
2
2
)()
( )( )(
)
()(
)
(
2
)
(
dx
ydEJ
xyx
xY Yq dx
u x
Q x w
x
T
=
- - -
+ -
- - -
-
ò
l
l
l
l
l
x xx
Deriving equation (69) in relation to
x
, gives:
(70)
ò
- -
+ + - -=
=
l
l
l
x
T
x x
q d u
Q x w
dx
d EJ
dx
yd EJ
)()
(
)(
)
(
2
2
2
2
2
3
3
2
f
x x
f