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Steel Designers' Manual - 6th Edition (2003)
The
Steel Construction
Institute
Silwood Park, Ascot, Berks SL5 7QN
v
t
=
È 1
Í 1
+
Ê l ˆ
Í 1
Î 40 Ë x ¯
˘
˙
˙
˚
= 087 .
The beam segment is stable.
Worked examples 649
Subject Chapter ref.
DESIGN OF CONTINUOUS
COMPOSITE BEAM
LLT = 48 . m fi l =
4800
39. 7
= 121
G. 241 . . l = nuv l n = 10 . , u = 088 .
Design code Made by RML Sheet no.
BS 5950: Parts 1 & 3 Checked by TCC
fi l = 1. 0 ¥ 0. 88 ¥ 0. 87 ¥ 121 = 93 fi p = 155 N/ mm
TB b
fi M = p S fi M = 186 kNm
b b x b
h s
y =
È 1
&
˘
= 087 . , since a =
Í
2
1
+
Ê l ˆ
˙
2
Í 1
Î Ë x ¯
˙
40 ˚
t t
So adequate restraint is provided by a single lateral restraint 1.5 m
from the internal support. This is done by introducing a web
stiffener at this point.
Check natural frequency of beam
2
12
& b = 0,
from G.4.2 fi m = 054 .
For uniform member from G.2.1 (Fc = 0)
mM
M
t x
b
TB t t t
0. 54 ¥ 325
=
= 094 . < 10 .
186
To calculate the natural frequency of a continuous beam it is necessary
to consider the mode shape of vibration. Because of the influence
of the asymmetric inertial forces, the natural frequency is taken as the
same as for a simply supported beam.
2
21
13
(Table 16)
(Table G.1)