Civil Engineering Reference
In-Depth Information
of the height of the structure in metres,
h
, and the number of columns in
the plane frame considered,
m
, as follows:
φ
=
α
h
α
m
/200
(5.4)
where
α
h
=
2/
h
with
2/3
≤
α
h
≤
1
(5.5)
and
α
m
=
(0.5
+
1/
m
)
1/2
(5.6)
Thus, in the frame in Fig. 5.1(b),
h
=
36 m,
α
h
=
2/3,
m
=
3,
α
m
=
0.816, and
φ
=
0.67
×
0.816/200
=
1/366
This initial sway applies in all horizontal directions, and is uniform over
the height of the frame. In this example, the overall out-of-plumb of each
column is assumed to be 36 000/366
98 mm.
Let the total design ultimate gravity load on the frame, for a particular
combination of actions, be
G
=
Q
per storey. The imperfections can then
be represented by a notional horizontal force
+
Q
) at each floor level
- but there may or may not be an equal and opposite reaction at founda-
tion level.
To illustrate this, we consider the single-bay single-storey unbraced
frame ABCD shown in Fig. 5.6(b), with pin joints at A and D, and
assume sin
φ
(
G
+
at B and C
is associated with the assumption that the loads
N
still act along the
columns, as shown. There are obviously horizontal reactions
N
φ
=
φ
, cos
φ
=
1. The use of additional forces
N
φ
φ
at A and
D; but the vertical reactions
N
are replaced by reactions
N
(1
±
2
φ
h
/
b
) at
angle
φ
to the vertical. The total horizontal reaction at A is therefore
N
φ
−
N
φ
(1
−
2
φ
h
/
b
)
=
2
N
φ
2
h
/
b
≈
0
The maximum first-order bending moment in the perfect frame is zero.
The imperfection
φ
increases it to
N
φ
h
, at corners B and C, which may not
be negligible.
If there are pin joints at these corners, the frame has to be braced
against side-sway by connection to the top of a stiff vertical cantilever EF
(Fig. 5.6(c)). The external reactions now do include horizontal forces
N
φ
at A and D, with an opposite reaction 2
N
φ
at F; and the vertical reactions
at A and D are independent of
φ
.
