Civil Engineering Reference
In-Depth Information
Using equation 6.26,
M
b
,
Rd
=
503.5
/
{
0.835
+
√
(
0.835
2
−
0.706
2
)
}
=
393.0kNm
>
384kNm
=
M
Ed
and the design moment resistance is adequate.
6.15.4 Example 4 - designing a braced beam
Problem
. Determine a suitable UB of S275 steel for the simply supported beam
of Section 6.15.1 if twist rotations are effectively prevented at the ends and if a
brace is added which effectively prevents lateral deflection
v
and twist rotation
φ
at mid-span.
Design bending moment.
Using Section 6.15.1,
M
Ed
=
433kNm.
Selecting a trial section.
The central brace divides the beam into two identical segments, each of length
L
=
3750 mm, and eliminates the effect of the load height.
Guess
f
y
=
275 N/mm
2
and
M
b
,
Rd
/
W
y
f
y
=
0.9.
Using equation 6.32,
W
pl
,
y
≥
(
433
×
10
6
/
275
)/
0.9 mm
3
=
1750 cm
3
.
Try a 457
×
191 UB 82 with
W
pl
,
y
=
1831 cm
3
>
1750 cm
3
.
Section resistance.
As in Section 6.15.3,
f
y
=
275 N/mm
2
,
M
c
,
Rd
=
503.5kNm
>
433kNm
=
M
Ed
and the section resistance is adequate.
Elastic buckling moment.
Using Figure 6.7,
α
m
=
1.75.
Using equation 6.3,
π
2
×
210000
×
1871
×
10
4
3750
2
Nmm
M
zx
=
81000
×
69.2
×
10
4
+
π
2
×
210000
×
0.922
×
10
12
3750
2
×
=
727.5 kNm
Using equation 6.4,
M
cr
=
1.75
×
727.5
=
1273kNm
(Using the computer program PRFELB [18] leads to
M
cr
=
1345 kNm.)
Member resistance.
Using equation 6.25,
λ
LT
=
√
(
503.5
/
1273
)
=
0.629
h
/
b
=
460.0
/
191.3
=
2.40
>
2
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