Civil Engineering Reference
In-Depth Information
3.8.6.5 Design Equations for the Evaluation of Headed Stud Capacities
As mentioned previously, Section 2.6 of Chapter 2 , headed stud shear con-
nectors are widely used in steel-concrete composite constructions owing to
many advantages including rapid installation, equal strength, and stiffness in
shear in all directions normal to the axis of the stud and high ductility.
Therefore, the development of design equations used in evaluating headed
stud capacities is discussed in the next paragraphs. The capacities depend on
the type of the concrete slab used with the steel beam to form the composite
interaction.
(a) Composite beams with solid reinforced concrete slabs
The strength of shear studs in solid reinforced concrete slab was first
determined by Ollgaard et al. [2.48] and was presented in terms of an empir-
ical formula after carrying out 48 push-off tests. The ultimate shear force
resistance Q u (in N units) of the headed studs was given as follows:
5 A s
p
76 Þ
where A s is the cross-sectional area of the stud diameter d (mm 2 units), f c is
the concrete cylinder compressive strength (N/mm 2 ), and E c is the static
modulus of elasticity of the concrete (N/mm 2 ). This equation, which was
adopted in CP 117 [2.34,2.35], assumes concrete crushing failure rather than
a shear failure of the headed stud. Later, in BS 5950 [2.36], data presented by
Menzies [2.46] were used to develop the characteristic shear force resistance
Q K . There is no theoretical basis to these data and values given in BS 5950
[2.36] reflect only the size of the stud and strength of the concrete. Cur-
rently, the commonly used design equations for headed stud shear connec-
tors are given in EC4 [ 3.6 ] . The resistance of headed stud ( P Rd ) is defined
using two equations. The equations represent concrete and stud failures.
The lower of the following values should be used in design:
Q u ¼ 0
:
f c E c
ð 3
:
8 f u pd 2
0
:
=
4
P Rd ¼
ð 3
:
77 Þ
g v
29 ad 2
p
0
:
f ck E cm
P Rd ¼
ð 3
:
78 Þ
g v
whichever is smaller with
for 3 h sc =
h sc
d +1
0
:
2
d 4
ð 3
:
79 Þ
1 for h sc
=
d
>
ð 3
:
80 Þ
4
 
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