Civil Engineering Reference
In-Depth Information
be empirically modifi ed to agree with the results obtained from
experiments.
Where there is no shear reinforcement, the shear strength
relies on aggregate interlock and dowel action of the longi-
tudinal reinforcement. For this reason it is recommended that
shear reinforcement is always provided, except in members of
minor importance. Shear reinforcement may also be omitted
in slabs, provided the applied shear force is less than the shear
strength of the section without reinforcement.
Punching shear is often critical in fl at slabs. Unfortunately,
the truss theory described above does not fi t with the experi-
mental results for punching shear, and therefore the resistance
is based on an empirical approach in Eurocode 2. As a result
there are effectively three different approaches to shear design
when using Eurocode 2: for beams use the truss methodology,
for slabs where shear reinforcement is not required and for
punching shear use an empirical approach.
A further note on the shear resistance of concrete is that the
depth of the section is an important part of the calculation. This
is known as 'size effect' - the deeper the section the lower the
shear strength in terms of stress.
17.5.5.1 Beams
In some codes of practice the concrete strut is fi xed at 45°.
In Eurocode 2, the strut angle can be varied between 22° and
45°; a shallow angle means the concrete strut will cross more
shear links and therefore the area of shear reinforcement is
reduced - leading to a more effi cient design. When designing
to Eurocode 2, it will generally be found that a strut angle of
22° will give suffi cient shear capacity. It is therefore pragmatic
to determine the capacity of the section with a strut at 22° and
only if this is insuffi cient to determine the minimum required
strut angle.
Table 17.6
gives the shear stress resistance for
various concrete strengths. The design process is provided in
Figure 17.5
.
v
Rd
When cot
θ
= 2.5
(
θ
= 22°)
When cot
θ
= 1.0
(
θ
= 45°)
f
ck
25
3.10
4.50
Concrete strut in
compression
28
3.43
4.97
30
3.64
5.28
32
3.84
5.58
Shear reinforcement
in tension
35
4.15
6.02
40
4.63
6.72
45
5.08
7.38
Longitudinal reinforcement in tension
50
5.51
8.00
Figure 17.4 Shear resistance using truss theory
Table 17.6 Values for
v
Rd
Shear design
= 0.2 (1-
f
ck
f
ck
/250)
f
ck
f
ck
sin 2
v
Rd
v
Rd
ck
/250)
Determine the design shear stress:
v
Ed
v
Ed
= 0.138 (1-
f
ck
f
ck
/250)
f
ck
f
ck
when
ck
when
θ
= 22 deg
v
Rd
v
Rd
=
V
Ed
V
Ed
/(0.9
bbb ddd
= 0.200 (1-
f
ck
f
ck
/250)
f
ck
f
ck
when
ck
when
θ
= 45 deg
v
Rd
v
Rd
v
Ed
v
Ed
θ
= 0
θ
5
sin
-1
θ
= 0.
Determine concrete strut capacity,
f
ck
f
ck
ut capacity
v
Rd
v
Rd
,
0.20
f
ck
f
ck
1 -
at 22° (see Ta ble 17.6 or panel)
f
250
A
sw
bbb
v
Ed
v
Ed
=
f
ywd
f
ywd
cot
θ
s
Ye s
N
No
?
Is
Is
v
v
≤
≤
v
v
?
Ed
v
Ed
Rd
v
Rd
Rd
?
Cost
θ
= 2.5
Calculate
θ
Calculate area of shear
reinforcement,
A
sw
Carry out reinforcement detailing
Figure 17.5 Design process for shear reinforcement
