Civil Engineering Reference
In-Depth Information
N
c
=
0.51
×
2555
=
1303 kN
(3.114)
Since, for
N
c,f
,
x
=
80 mm
x
c
=
0.51
×
80
=
40.8 mm
With reference to the stress blocks in Fig. 3.15(c), with
N
a,pl
=
2698 kN,
(Equation 3.111),
N
ac
=
2698
−
1303
=
1395 kN
Assuming that there is a neutral axis within the steel top flange, the
depth of flange in compression is
1395/(0.178
×
2
×
355)
=
11.0 mm
This is less than
t
f
(12.8 mm) so the assumption is correct, and the stress
blocks are as shown in Fig. 3.29(b). Taking moments about the top sur-
face of the slab,
M
pl,Rd
=
2698
×
0.353
−
1303
×
0.020
−
1395
×
0.156
=
709 kN m
(3.115)
which exceeds
M
Ed
(563 kN m).
The interpolation method gives
M
pl,Rd
0.51, so
the equilibrium method is significantly less conservative. For this example,
n
/
n
f
=
630 kN m with
n
/
n
f
=
=
0.51 will be used.
Number and spacing of shear connectors
It is assumed that 19-mm stud connectors will be used, 125 mm long. The
length after welding is about 5 mm less, so the height of the studs is taken
as 120 mm. The design shear resistance
P
Rd
is given by Equation 3.1 as
60.2 kN per stud.
For the sheeting used here, the width
b
0
(Fig. 2.14) is 162 mm, from
Fig. 3.9, and the other dimensions that influence the reduction factor
k
t
for
the resistance of studs in ribs are:
h
p
=
70 mm
h
=
120 mm
So from Equation 2.17,
k
t
=
(0.7/
n
r
)(162/70)[(120/70)
−
1]
=
1.16
(
n
r
=
1)
=
0.82
(
n
r
≥
2)
