Civil Engineering Reference
In-Depth Information
Factor
C
4
is now found, using Fig. 4.6. From Fig. 4.11, for span BC,
M
0
=
23.7
×
9.3
2
/8
=
256 kN m
so
ψ
=
457/256
=
1.78
From Fig. 4.6,
C
4
17.4.
The effect of including in Equation 4.15 the St Venant torsion constant
for the steel section is negligible, so the simpler Equation 4.22 can be
used. It gives:
=
M
cr
=
(1.12
×
17.4/
π
)(0.463
×
210 000
×
6.01)
1/2
=
4742 kN m
The characteristic plastic bending resistance at support B is required for
use in Equation 4.23. The design value was found in Section 4.2.1.2 to
be 510 kN m. Replacing the
γ
S
factor for reinforcement (1.15) by 1.0
increases it to
M
pl,Rk
579 kN m.
The slenderness l
LT
is given by Equation 4.23, which is
=
l
LT
=
(
M
pl,Rk
/
M
cr
)
1/2
=
(579/4742)
1/2
=
0.35
This is less than 0.4, so
M
pl
,
Rd
need not be reduced to allow for lateral
buckling.
4.6.4
Shear connection and transverse reinforcement
For sagging bending, the resistance required, 449 kN m, is well below the
resistance with full shear connection, 829 kN m, so the minimum degree
of shear connection may be sufficient. For spans of 9.3 m, and an effect-
ive span of 9.3
×
0.85
=
7.9 m, this is given by Fig. 3.19 as
n
/
n
f
≥
0.49
In Equation 3.67,
N
c
/
N
c
,
f
may be replaced by
n
/
n
f
. This equation for the
interpolation method (Fig. 3.16) then gives
M
Rd
=
M
pl
,
a
,
Rd
+
(
n
/
n
f
)(
M
pl
,
Rd
−
M
pl
,
a
,
Rd
)
=
424
+
0.49(847
−
424)
=
631 kN m
which is sufficient.
From Equation 3.116, the resistances of the stud connectors are 51.0 kN
and 42.0 kN, for one and two studs per rib, (
n
r
=
1 and
n
r
=
2, respect-
ively). From Section 3.11.1,
N
c,f
=
2555 kN, so
N
c
=
0.49
×
2555
=
1250 kN
