Page 316 - Advanced Design Examples of Seismic Retrofit of Structures

Example of a Steel Frame Building Retrofitted Chapter 5 305

635 b f
(5.21)
L C ¼ p ffiffiffiffiffi
F y
14 10 5
(5.22)
L C ¼
d=A f F y
where:
b f ¼width of the compression flange;
A f ¼area of the compression flange; and
d ¼height of the flexural member;
As an example and for IPE 140 we have:
635 7:3
635:b f
ffiffiffiffiffi ¼ p ffiffiffiffiffiffiffiffiffiffi ¼ 95:43 cmð Þ
L C ¼ p
2360
F y
14 10 5 14 10 5
ð
¼ ¼ 347:45 cmÞ
L C ¼
d=A f F y ð 14=8:2Þ 2360
) L C ¼ min 95:43, 347:45ð Þ ¼ 95:43 cmÞ
ð
All the beams that carry gravity loads are braced, because the distance of the
joists in the diaphragms is smaller than L c . For the beams that do not carry grav-
ity loads, and where the conditions of braced section are not satisfied, the
expected flexural capacity of the section, Q CE , shall be computed in accordance
with Eq. (5.23):
Q CE ¼ M CE ¼ 1:1F ye S (5.23)
where S ¼the elastic section modulus of a member.
The expected flexural capacity of the beams is presented in Table 5.8.
If the beam strength is governed by the shear strength of the unstiffened web
418
and h ffiffiffiffip , then V CE shall be calculated in accordance with Eq. (5.24):
t w
F y
TABLE 5.8 Expected Flexural Capacity of Beams
M CE (tonm)
3
2
3
Profile Z (cm ) S (cm ) F ye (kg/cm ) Braced Unbraced
2INP240 829 708 2360 19.56 18.38
INP240 411 354 2360 9.70 9.19
2IPE140 172 154.6 2360 4.04 4.01
IPE140 88 77.3 2360 2.07 2.01
IPE180 166 146 2360 3.91 3.79
IPE240 366 324 2360 8.63 8.41