Shear, bond and torsion
5.2 Anchorage bond
The reinforcing bar subject to direct tension shown in figure 5.7 must be firmly anchored
if it is not to be pulled out of the concrete. Bars subject to forces induced by flexure must
be similarly anchored to develop their design stresses. The anchorage depends on the
bond between the bar and the concrete, the area of contact and whether or not the bar is
located in a region where good bond conditions can be expected. Let:
lb;rqd ¼ basic required anchorage length to prevent pull out
¼ bar size or nominal diameter
fbd ¼ ultimate anchorage bond stress
fs ¼ the direct tensile or compressive stress in the bar.
L
Figure 5.7
Anchorage bond
F
ϕ
Considering the forces on the bar:
Tensile pull-out force ¼ cross-sectional area of bar direct stress
2
fs
4
anchorage force ¼ contact area anchorage bond stress
¼ ðlb;rqd Þ fbd
¼
therefore
ðlb;rqd Þ fbd ¼
2
fs
4
hence
lb;rqd ¼
fs 4fbd
and when fs ¼ fyd , the design yield strength of the reinforcement (¼ fyk =1:15) the
anchorage length is given by
lb;rqd ¼ ð=4Þ ½fyk =1:15=fbd
lb;rqd ¼ ðfyk =4:6fbd Þ
ð5:19Þ*
Basic anchorage length
Equation 5.19 may be used to determine the basic anchorage length of bars which are
either in tension or compression. For the calculation of anchorage lengths, design values
of ultimate anchorage bond stresses are specified according to whether the bond
conditions are good or otherwise.
117
118
Reinforced Concrete Design
Figure 5.8
Definition of good and
poor bond conditions
250 mm
direction of concreting
α
45 < α < 90o
for all values of h
h > 250 mm
300 mm
o
h
h
h < 250 mm
h > 600 mm
Good bond conditions for all bars
Good bond conditions in unhatched zone
Poor bond conditions in hatched zone
Good bond conditions are considered to be when (a) bars are inclined at an angle of
between 458 and 908 to the horizontal or (b) zero to 458 provided that in this second case
additional requirements are met. These additional conditions are that bars are
1. either placed in members whose depth in the direction of concreting does not exceed
250 mm or
2. embedded in members with a depth greater than 250 mm and are either in the lower
250 mm of the member or at least 300 mm from the top surface when the depth
exceeds 600 mm.
These conditions are illustrated in figure 5.8. When bond conditions are poor then the
specified ultimate bond stresses should be reduced by a factor of 0.7.
The design value of the ultimate bond stress is also dependent on the bar size. For all
bar sizes () greater than 32 mm the bond stress should additionally be multiplied by a
factor ð132 Þ=100.
Table 5.1 gives the design values of ultimate bond stresses for ‘good’ conditions.
These depend on the class of concrete and are obtained from the equation fbd ¼ 1:50fctk
where fctk is the characteristic tensile strength of the concrete.
Design anchorage length
The basic anchorage length discussed above must be further modified to give the
minimum design anchorage length taking into account factors not directly covered by
table 5.1.
Table 5.1
Design values of bond stresses fbd (N/mm2)
fck N/mm2
Bars 32 mm diameter and
good bond conditions
Bars 32 mm diameter and
poor bond conditions
12
16
20
25
30
35
40
45
50
55
60
1.6
2.0
2.3
2.7
3.0
3.4
3.7
4.0
4.3
4.5
4.7
1.1
1.4
1.6
1.9
2.1
2.4
2.6
2.8
3.0
3.1
3.3
Shear, bond and torsion
Table 5.2
Coefficients Value
of allows for the effect of:
1
The shape of the bars
Concrete cover to
the reinforcement
2
Type of anchorage
Reinforcement in
Tension
Compression
Straight
1.0
1.0
Other than straight
0.7 if cd > 3:0
or 1.0 if not
1.0
Straight
1 0:15ðcd Þ=
but 0:7 and 1:0
1 0:15ðcd 3Þ=
but 0:7 and 1:0
1.0
Other than straight
1.0
3
Confinement of transverse
reinforcement not welded to
the main reinforcement
All types of
reinforcement
1 K
but 0:7 and 1:0
1.0
4
Confinement of transverse
reinforcement welded to the
main reinforcement
All types, position and
sizes of reinforcement
0.7
0.7
5
Confinement by transverse
pressure, p (N/mm2)
All types of
reinforcement
1 0:04p
but 0:7 and 1:0
–
Note: the product 2 3 5 should be greater than or equal to 0.7
The required minimum anchorage length (lbd ) is given by
lbd ¼ 1 , 2 , 3 , 4 , 5 lb; rqd As; req =As; prov
ð5:20Þ*
where As; req , As; prov ¼ area of reinforcement required and provided at that section
(1 to 5) ¼ set of coefficients as given in Table 5.2
In Table 5.2:
P
P
119
cd ¼ concrete cover coefficient as shown in figure 5.9
K ¼ values as shown in figure 5.10
P
P
¼
Ast Ast;min =As
Ast ¼ the cross-sectional area of the transverse reinforcement along the design
anchorage length
Ast;min ¼ the cross-sectional area of the minimum transverse reinforcement
( ¼ 0:25As for beams and zero for slabs)
As ¼ the area of a single anchored bar with maximum bar diameter
This minimum design length must not be less than:
for tension bars: 0:3lb; rqd
for compression bars: 0:6lb; rqd
In both cases the minimum value must also exceed both 10 bar diameters and 100 mm.
Anchorages may also be provided by hooks or bends in the reinforcement. Hooks and
bends are considered adequate forms of anchorage to the main reinforcement if they
Reinforced Concrete Design
Figure 5.9
Values of cd for beams and
slabs (see table 5.2)
c1
a
a
c1
c
c
Straight bars
cd = min (a/2, c1, c)
Figure 5.10
Values of K for beams and
slabs (see table 5.2)
As
Looped bars
cd = c
Bent or hooked bars
cd = min (a/2, c1)
As
ϕt Ast
K = 0.1
As
ϕt Ast
K = 0.05
ϕt Ast
K=0
Figure 5.11
Equivalent anchorage lengths
for bends and hooks
90o < α < 150o r
> 5ϕ
120
α
ϕ
ϕ
l b.eq = α1l b.rqd
l b.rqd
Straight bar
Bend
> 5ϕ
r
>150o
ϕ
l b.eq = α1l b.rqd
Hook
ϕ
l b.eq = α1l b.rqd
Loop
Minimum internal radius of a hook, bend or loop = 2ϕ or 3.5ϕ for ϕ >16 mm
satisfy the minimum dimensions shown in figure 5.11. Bends and hooks are not
recommended for use as compression anchorages. In the case of the hooks and bends
shown in figure 5.11 the anchorage length (shown as lb; eq ) which is equivalent to that
required by the straight bar can be simply calculated from the expression:
lb; eq ¼ 1 lb; rqd where 1 is taken as 0.7 or 1.0 depending on the cover conditions
(see table 5.2).
The internal diameter of any bent bar (referred to as the mandrel size) is limited to
avoid damage to the bar when bending. For bars less than or equal to 16 mm diameter
the internal diameter of any bend should be a minimum of 4 times the bar diameter. For
larger bar sizes the limit is 7 times the bar diameter.
To give a general idea of the full anchorage lengths required for fck ¼ 30 N/mm2 and
fyk ¼ 500 N/mm2 , with bar diameters, 32 mm, lb; req can vary between 25 bar
diameters (25) and 52 bar diameters (52), depending on good and poor bond
conditions, and the value of the coefficients from table 5.2.
Shear, bond and torsion
121
EXA M PLE 5. 2
Calculations of anchorage length
Determine the anchorage length required for the top reinforcement of 25mm bars in the
beam at its junction with the external column as shown in figure 5.12. The reinforcing
bars are in tension resisting a hogging moment. The characteristic material strengths are
fck ¼ 30 N/mm2 and fyk ¼ 500 N/mm2 .
H25 bars
100 = 4ϕ
600
Anchorage length I bd
Figure 5.12
Anchorage for a beam
framing into an end column
effective span
Assuming there is a construction joint in the column just above the beam and, as the
bars are in the top of the beam, from figure 5.8 the bond conditions are poor and from
table 5.1 the ultimate anchorage bond stress is 2.1 N/mm2.
As the bars are bent into the column and the concrete cover coefficient, cd (figure 5.9)
is equivalent to 4, which is greater than 3, from table 5.2 coefficient 1 is 0.7. Also
from table 5.2, coefficient 2 ¼ 1 0:15ðcd 3Þ= ¼ 1 0:15ð4 3Þ= ¼ 0:85.
Hence the required anchorage length is
fyk
lbd ¼ 1 2
4:6fbd
500
¼ 31
¼ 0:7 0:85
4:6 2:1
¼ 31 25 ¼ 775 mm.
See also table A.6 in the Appendix for tabulated values of anchorage lengths.
5.3 Laps in reinforcement
Lapping of reinforcement is often necessary to transfer the forces from one bar to
another. Laps between bars should be staggered and should not occur in regions of high
stress. The length of the lap should be based on the minimum anchorage length modified
to take into account factors such as cover, etc. The lap length lo required is given by
lo ¼ lb; rqd 1 2 3 5 6
ð5:21Þ*
0:5
where 1 , 2 , 3 , and 5 , are obtained from table 5.2. 6 ¼ ð1 =25Þ (with an upper
and lower limit of 1.5 and 1.0 respectively) and 1 is the percentage of reinforcement
lapped within 0:65l0 from the centre of the lap length being considered. Values of 6
can be conveniently taken from table 5.3.
122
Reinforced Concrete Design
Table 5.3
Values of the coefficient 6
Percentage of lapped bars relative to the total
cross-sectional area of bars at the section being
considered
6
<25%
33%
50%
>50%
1
1.15
1.4
1.5
Intermediate values may be interpolated from the table
Notwithstanding the above requirements, the absolute minimum lap length is given as
lo; min ¼ 0:36 lb; reqd
ð5:22Þ
15 diameters 200 mm
Transverse reinforcement must be provided around laps unless the lapped bars are
less than 20 mm diameter or there is less than 25 per cent lapped bars. In these cases
minimum transverse reinforcement provided for other purposes such as shear links will
be adequate. Otherwise transverse reinforcement must be provided, as shown in
figure 5.13, having a total area of not less than the area of one lapped bar.
The arrangement of lapped bars must also conform to figure 5.14. Lapped bars may
be allowed to touch within the lap length and the clear space between lapped bars
should not be greater than 4 or 50mm otherwise an additional lap length equal to the
clear space must be provided. In the case of adjacent laps the clear distance between
adjacent bars should not be greater than 2 or 20 mm. The longitudinal distance
between two adjacent laps should be greater than 0:3lo . If all these conditions are
complied with then 100% of all tension bars in one layer at any section may be lapped;
otherwise, where bars are in several layers, this figure should be reduced to 50%. In the
case of compression steel, up to 100% of the reinforcement at a section may be lapped.
lo/3
Figure 5.13
Transverse reinforcement
for lapped bars
lo/3
As/2
As/2
<150 mm
As
lo
(a) tension lap
4ϕ
4ϕ
lo/3
lo/3
As/2
<150 mm
As/2
ϕ
As
lo
(b) compression lap
Shear, bond and torsion
>0.3lo
lo
Figure 5.14
Lapping of adjacent bars
ϕ
<4ϕ or 50 mm
>0.2ϕ or 20 mm
For more information on the practical application of the rules for providing
anchorage and lap lengths the reader is recommended to consult the Standard Method of
Detailing Structural Concrete (ref 30).
5.4 Analysis of section subject to torsional moments
5.4.1 Development of torsional equations
Torsional moments produce shear stresses that result in principal tensile stresses
inclined at approximately 458 to the longitudinal axis of the member. Diagonal cracking
occurs when these tensile stresses exceed the tensile strength of the concrete. The cracks
will form a spiral around the member as in figure 5.15.
Figure 5.15
Torsional reinforcement
A
123
A
T
cL
Crack lines
Reinforcement in the form of closed links and longitudinal bars will carry forces from
increasing torsional moment after cracking, by a truss action with reinforcement acting
as tension members and concrete as compressive struts between links. Failure will
eventually occur by reinforcement yielding, coupled with crushing of the concrete along
line A–A as cracks on the other face open up. It is assumed that once the torsional shear
stress on a section exceeds the value to cause cracking, tension reinforcement in the
form of closed links must be provided to resist the full torsional moment.
The equations for torsional design are developed from a structural model where it is
assumed that the concrete beam in torsion behaves in a similar fashion to a thin walled
box section. The box is reinforced with longitudinal bars in each corner with closed loop
stirrups as transverse tension ties and the concrete providing diagonal compression
struts. It is assumed that the concrete cannot provide any tensile resistance.
EC2 gives the principles and some limited design equations for a generalised shape of
a hollow box section. In this section of the text we will develop the equations that can be
used for the design and analysis of a typical solid or hollow rectangle box section.
Consider figure 5.16a. The applied torque (TEd ) at the far end of the section produces
a shear flow (q) around the perimeter of the box section at the near end of the diagram.
The shear flow is the product of the shear stress () and the thickness of the hollow