Civil Engineering Reference
In-Depth Information
it may not be sufficiently ductile to ensure that it will not fracture before
the design ultimate load for the beam is reached. The design tensile force
in this reinforcement is
N
s
=
A
s
f
sk
/
γ
S
=
A
s
f
yd
(4.2)
where
f
sk
is its characteristic yield strength.
If there were no tensile reinforcement, the bending resistance would be
that of the steel section,
M
pl
,
a
,
Rd
=
W
a
f
yd
=
N
a
z
a
(4.3)
where
W
a
is the plastic section modulus and
f
yd
is the design yield strength.
For rolled sections it is not necessary to calculate the forces
N
a
in the
stress blocks of depth
h
a
/2, nor the lever arm
z
a
, because values of
W
a
are
tabulated; but for plate girders
N
a
and
z
a
have to be calculated.
The simplest way of allowing for the force in the reinforcement is
to assume that the stress in a depth
x
a
of web changes from tension to
compression, where
x
a
is given by
x
a
t
w
(2
f
yd
)
=
N
s
(4.4)
provided that (as is usual)
x
a
≤
h
a
/2
−
t
f
The depth of web in compression is given by
α
d
=
d
/2
+
x
a
(4.5)
Knowledge of
α
,
d
/
t
w
and
f
y
enables the web to be classified, as shown in
Fig. 3.14 for
f
y
355 N/mm
2
. If, by this method, a web is found to be in
Class 4, the calculation should be repeated using the elastic neutral axis,
as the curve that separates Class 3 from Class 4 is based on the elastic
behaviour of sections. This is why, in Fig. 3.14, the ratio
=
ψ
is used, rather
than
.
Concrete-encased webs in Class 3 are treated as if in Class 2
(Table 3.1), because the encasement helps to stabilise the web.
The lever arm
z
for the two forces
N
s
in Fig. 4.1( b) is given by
α
z
=
h
a
/2
+
h
s
−
x
a
/2
