Basic HTML Version
112
IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 1
A variable limit for the instability parameter of wall-frame or core-frame bracing structures
Thus, inequality (46) denotes a value of 0,773 for
a
1
. In its turn, the
ABNT [8] code allows the coefficient
a
1
to be increased until 0,7 if the
bracing system is composed exclusively by shear walls or shear cores.
4.2 Substructures of the rigid plane frame type
In the case of bracing systems formed exclusively by rigid frames,
the equivalent bar of figure 4-a will deform predominantly by shear.
Also in this case, the efforts expression must take the deflected
shape into account. It can be proved that the infinitesimals
ds
and
dx
shown in figure 4-b are related by:
(47)
2
2 2
2
2
2
1
)
1(
f+ =
+
= + =
dx
dx dy
dx
dy dx
ds
The shear effort can be obtained from the derivation of equation (31) in
relation to the bar deflected axis. Introducing
ds
given by (47), results:
(48)
)(
1
)()
( )
(
)(
1
)(
2
2
x
x x q x w
x
dx dM
ds
dM xQ
f
f
f
+
- + -
=
+
-
= -=
l
l
It is an inclined shear effort, as shown in figure 4-b. The shear deforma-
tion caused by it has the same slope, given at the infinitesimal level by:
(49)
2
2
1
11
cos
f
f
f
+ =
+
=
=
dy
dy
ds dx
dy
dy
On establishing the differential equation of motion for this case,
two changes must be performed with relation to equation (16): to
introduce
dy
/cos
f
given by (49), in place of
dy
, and
Q(x)
given by
(48). On doing so, results:
(50)
(x)
S
(x) x)
q(
x)
w(
S
Q(x)
x
dx
dy
2
2
1
)(
1
f
f
f
+
- + -
=
=
+
l
l
Thus:
(51)
)] (
1[
)(
)(
2
x
S
x x)
q(
x)
w(
dx
dy
x
f
f
f
+
- + -
= =
l
l
In cases of “bland” geometric nonlinearity, as the ones treated by this work,
the rotations
f
(
x
) present valuesmuch lesser thanunit; therefore,
f
2
(
x
)may
be neglected in face of 1 and equation (51) may be put in the form:
(52)
)()
(
)
(
)(
x x q x w x S
f
f
- + - =
l
l
Isolating
f
(
x
):
(53)
)
(
)
(
)(
x q S
x w x
- -
-
=
l
l
f
Integrating equation (53) in relation to
x
and applying the condition
of zero displacement at support, leads to the displacements func-
tion. Integrating again, gives:
(54)
[
] [
]
{
}
C )x (
q
w )q (S x
q
Sw
x)
q( S x)
q( S
q
Sw Y(x)
+ +
- -
-- - - ×
- -
=
2
2
2
3
2
ln
1
ln
l
l
l
l
where
C
is the integration constant. Applying equation (54) for
x
=
0 and
x
=
l
, leads to the difference below:
(55)
÷÷
ø
ö
çç
è
æ
+
-
-
= -
q
S
q
w
q S
S
q
wS ) Y( )Y(
2
ln
0
3
2
l l
l
l
Thus, the bending moment at support can be expressed, substitut-
ing equation (55) into (32):
(56)
q
Sw
S q
ln
q
wS
q
S
w
q S
S ln
q
wS w )M(
l
l
l
l
l
l
-
-
=÷÷
ø
ö
çç
è
æ
+
-
-
+ =
/
1
1
2
2
0
2
2
2
2
2
Applying the condition expressed by inequality (1) to this bending
moment, results:
(57)
2
4111
41
41
411
1
ln
41
41
2
2 2
2
l
l
l
w ,
,
q,
Sw ,
Sq,
q ,
w S,
´ £
-
-
Performing the required algebraic transformations, inequality (57)
changes into:
(58)
770 1
411
1
ln
41
1
2
,
/Sq /Sq,
/S)
(q,
£ -
-
l
l
l
Taking the factor q
l
/Sas an unknown, inequality (58) can be solved
by means of trials, obtaining:
(59)
0962 ,0 /
£
S q
l