Civil Engineering Reference
In-Depth Information
with
f
ck
in N/mm
2
units. The width of strut per unit length of beam is
1/
2
, and the force in the strut is
2
v
L,Ed
, so this condition is
v
L,Ed
<
0.6(1
−
f
ck
/250)
f
cd
h
f
/
2
(3.77)
This rule is for normal-density concrete. For lightweight concrete, with
oven-dry density
ρ
kg/m
3
, the expression is replaced by
v
L,Ed
<
0.5(0.4
+
0.6
ρ
/2200)(1
−
f
lck
/250)
f
lcd
h
f
/
2
(3.78)
This allows for the reduction in the ratio
E
cm
/
f
lck
as the density of the
concrete reduces.
These results are assumed to be valid whatever the length of the notional
struts in the slab (e.g., FG), and rely to some extent on the shear flow
v
L
being fairly uniform within the shear span, because the reinforcement
associated with the force 2
v
L
at A is in practice provided at cross-section
A, not at some point between A and mid-span. The type of cracking
observed in tests where shear failure occurs, shown in Fig. 3.22, is con-
sistent with the model.
Haunched slabs
Further design rules are required for the transverse reinforcement in
haunches of the type shown in Fig. 2.1(b). These are not discussed here.
Haunches encased in thin steel sheeting are considered below.
3.6.3.2
Transverse reinforcement in composite slabs
Where profiled sheeting spans in the direction transverse to the span of
the beam, as shown in Fig. 3.15(a), it can be assumed to be effective as
bottom transverse reinforcement where the sheets are continuous over the
beam. Where they are not, as in the figure, the effective area of sheeting
depends on how the ends of the sheets are attached to the steel top flange.
Where studs are welded to the flange through the sheeting, resistance to
transverse tension is governed by local yielding of the thin sheeting around
the stud. The design bearing resistance of a stud with a weld collar of
diameter
d
do
in sheeting of thickness
t
is given in Eurocode 4 as
P
pb,Rd
=
k
ϕ
d
do
tf
yp,d
(3.79)
where
k
ϕ
=
1
+
a
/
d
do
≤
6.0
(3.80)
