Civil Engineering Reference
In-Depth Information
length over a simply-supported span
L
. The materials are assumed to be
concrete with a characteristic cube strength of 30 N/mm
2
and mild steel,
with a characteristic yield strength of 250 N/mm
2
. Creep is neglected (
k
c
=
1) and we assume
n
=
10, so for the concrete
E
c
=
E
c
=
20 kN/mm
2
, from
Equation 2.19.
The dimensions of the beam (Fig. 2.16) are so chosen that the trans-
formed cross-section is square:
L
=
10 m,
b
=
0.6 m,
h
c
=
h
s
=
0.3 m. The
steel member is thus a rectangle of breadth 0.06 m, so that
A
a
=
0.018 m
2
,
I
a
10
−4
m
4
.
The design of such a beam on an ultimate-strength basis is likely to
lead to a working or 'service' load of about 35 kN/m. If stud connectors
19 mm in diameter and 100 mm long are used in a single row, an appro-
priate spacing would be 0.18 m. Push-out tests give the ultimate shear
strength of such a connector as about 100 kN, and the slip at half this load
is usually between 0.2 and 0.4 mm. Connectors are found to be stiffer in
beams than in push-out tests, so a connector modulus
k
=
1.35
×
150 kN/mm will
be assumed here, corresponding to a slip of 0.33 mm at a load of 50 kN
per connector.
The distribution of slip along the beam and the stresses and curvature at
mid-span are now found by partial-interaction theory, using the results
obtained in Section A.1, and also by full-interaction theory. The results
are discussed in Section 2.7.
=
First
α
and
β
are calculated. From Equation 2.22 with
I
c
=
nI
a
(from the
shape of the transformed section) and
k
c
=
1,
I
0
=
2.7
×
10
−4
m
4
.
From Equation 2.20 with
A
c
=
nA
a
and
k
c
=
1,
A
0
=
0.009 m
2
.
From Equation 2.21, 1/
A
′
=
0.3
2
+
(2.7
×
10
−4
)/0.009
=
0.12 m
2
.
From Equation 2.23, with
k
=
150 kN/mm and
p
= 0.18 m,
150
×
××
.
0 12
α
=
.
.
=
185
m
2
−
2
018
.
200
027
whence
α
=
1.36 m
−1
. Now
L
=
10 m, so
α
L
/2
=
6.8 and sech(
α
L
/2)
=
0.002 23. From Equation 2.24,
018
×
××
. .
03
β
=
.
=×
−
30
10
6
m/kN
0 12
.
150
1000
×
10
−4
m. An expression for the slip in terms of
x
is now given by Equation
2.27 with
We assumed
w
=
35 kN/m, so
β
w
=
1.05
×
10
−4
and
β
w
/
α
=
0.772
ε
c
=
0:
10
4
s
=
1.05
x
−
0.0017 sinh(1.36
x
)
(2.28)
