Civil Engineering Reference
In-Depth Information
10.1, is used, and
I
is
increased by 10% (Section 3.9.1). Values corresponding to those from
Equations 3.120 and 3.121 are, from Table 4.5:
For vibration, the short-term modular ratio,
n
0
=
x
=
129 mm
10
− 6
I
b
=
1.1
×
751
=
826 mm
4
For the slab, the 'uncracked' value found in Section 3.4.5 is too low,
because
n
=
20.2 was used. Similar calculations for
n
0
=
10.1, with a 10%
increase, give
10
−6
I
s
=
20.5 mm
4
/m
From Equation 3.99 with
s
=
4 m,
L
=
8.6 m,
12
/
2
210 000
.
×
××
826
⎛
⎝
⎞
⎠
f
0b
=
.
=
62
Hz
512
4
8 6
4
From Equation 3.100,
12
/
210 000
.
×
20 5
⎛
⎝
⎞
⎠
f
0
s
=
.
356
.
=
20 4
Hz
512
×
4
4
From Equation 3.98,
f
0
=
5.9 Hz
This is below 7 Hz, so no check need be made for impulsive loads.
Response of composite floor
Following Section 3.9.2,
C
f
=
0.2 from Fig. 3.26. From Equation 3.103,
the vibrating width of slab is
14
/
210 000
.
.
×
20 5
⎛
⎝
⎞
⎠
S
=
.
45
.
=
17 7
m
512
×
5 9
2
or the actual dimension of the floor normal to the span of the beams, if
less. For a small value of
S
, the natural frequency would be higher than
f
0
as calculated here, so it is assumed that the building considered is more
than 17.7 m long, and this value is used.
With the critical damping ratio
ζ
=
0.03, Equation 3.106 gives the
response factor:
R =
(68 000
×
0.2)/(512
×
17.7
×
8.6
×
0.03)
=
5.8
