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and
¼ 1
:
0 for f ck 50 Mpa
ð 2
:
27 Þ
¼ 1
:
0 f ck 50
ð
Þ=
200 for 50
<
f ck 90 Mpa
It should be noted that, according to EC2, if the width of the compres-
sion zone decreases in the direction of the extreme compression fiber, the
value
f cd should be reduced by 10%.
2.3.6 Flexural Tensile Strength
The mean flexural tensile strength of reinforced concrete members depends
on the mean axial tensile strength and the depth of the cross section. EC2
[ 2.27 , 2.28 ] recommends the following relationship to be used in determin-
ing mean flexural tensile strength of reinforced concrete members:
28 Þ
where h is the total member depth in mm and f ctm is the mean axial tensile
strength following Table 2.9 . The relation given earlier also applies for the
characteristic tensile strength values.
f ctm , fl ¼ max 1
f
ð
:
6 h
=
1000
Þf ctm ; f ctm
g
ð 2
:
2.3.7 Confined Concrete
In cases where concretes are surrounded by a stiffer material, such as
concrete-filled steel tubular columns, the compressive strength and ductility
of concrete are improved significantly. In this case, the concrete is called
confined concrete, and depending on the yield stress and geometries of
the surrounding stiffer material, the mechanical properties of this concrete
improve considerably compared with unconfined concrete. To accurately
model confined concrete, improved mechanical properties must be consid-
ered in nonlinear structural analysis and in nonlinear finite element model-
ing. Current codes of practice provide guidelines to account for concrete
confinement. As an example, by adopting EC2 [ 2.27 , 2.28 ] , confinement
of concrete results in a modification of the effective stress-strain relationship,
achieving higher strength and higher critical strains. The other basic material
characteristics may be considered as unaffected for design. In the absence of
more precise data, the stress-strain relation shown in Figure 2.9 (compressive
strain shown as positive) may be used, with increased characteristic strength
and strains according to
f ck , c ¼ f ck 1
ð
:
0+5
:
0 s 2 =
f ck
Þ
for s 2 0
:
05 f ck
ð 2
:
29 Þ
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