Civil Engineering Reference
In-Depth Information
For 0.9-mm sheeting, and 19-mm studs welded through the sheeting, EN
1994-1-1 gives the following upper limits for
k
t
, which govern here:
for
n
r
=
1,
k
t
=
0.85, so
P
Rd
=
51.2 kN
for
n
r
=
2,
k
t
=
0.70, so
P
Rd
=
42.1 kN
From Equation 3.114 the number of single studs needed in each half span
is
n
=
N
c
/
P
Rd
=
1303/51.2
=
25.4, say 26
For the detailing shown in Fig. 3.30, the lateral spacing of the two studs is
118 mm, or 6.2
d
. This exceeds the minimum given in EN 1994-1-1, which
is 4
d
. If the separation is large, Equation 2.17 should obviously be used
with
n
r
1, assuming that there is no interaction between the local stresses
around the two studs, and giving
k
t
=
=
0.85. If it is small (e.g., 4
d
), then
n
r
is presumably taken as 2, so
k
t
0.7; and if there are two studs side-by-
side on each side of the steel web,
n
r
=
4.
No upper limit for
k
t
is given in EN 1994-1-1 for
k
t
=
4, nor is 'large'
separation defined. In a situation not covered by the code being used, as
here, one has to use judgement or refer to research. Here, the separation
exceeds 4
d
, and
k
t
=
2.
There is one trough every 300 mm, or 16 in a half span, so it is assumed
that 32 studs are provided, one in each rib on each side of the web. In
Section 3.4.3, each stud was used to provide an anchorage force of 4.4 kN
for the sheeting, perpendicular to the direction in which resistance is now
required. The corrected values for
P
Rd
are as follows:
=
0.7 will be used for
n
r
≥
for
n
r
=
1,
P
Rd
=
51.2
2
−
4.4
2
so
P
Rd
=
51.0 kN
for
n
r
=
2,
P
Rd
=
42.1
2
−
4.4
2
so
P
Rd
=
42.0 kN
(3.116)
For 32 studs, the resistance is
N
Rd
=
32
×
42
=
1344 kN
This is sufficient, so
use 1 stud in each trough on each side of the web
.
Transverse reinforcement
Rules for the use of profiled sheeting as transverse reinforcement are
explained in Section 3.6.3.2. The cross-section in Fig. 3.30 illustrates
compliance with the rule that the sheeting should extend at least 2
d
do
beyond the centre of a stud welded through it, where
d
do
is an estimate
of the diameter of the stud weld, taken as 1.1
d
, or 20.9 mm here.
