AAiT, School of Civil and Environmental Engineering
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CHAPTER 3. LIMIT STATE DESIGN FOR SHEAR
3.1. THEORETICAL BACKGROUND
3.1.1. DIAGONAL TENSION IN HOMOGENEOUS ELASTIC BEAMS
The stresses acting in homogenous beams can be derived from mechanics of elastic materials.
Shear stresses
 
VQ
Ib
(1)
act at any section in addition to the bending stresses
 
My
I
(2)
except for those locations at which the shear force V happens to be zero.
The role of shear stresses is easily visualized by the performance under load of the laminated
beam of Figure 3-1; it consists of two rectangular pieces bonded together along the contact
surface. If the adhesive is strong enough, the member will deform as one single beam, as shown
in Figure 3-1a . On the other hand, if the adhesive is weak, the two pieces will separate and slide
relative to each other, as shown in Figure 3-1b .
Figure 3-1 – Shear in homogeneous rectangular beams
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Evidently, then, when the adhesive is effective, there are forces or stresses acting in it that
prevent this sliding or shearing. These horizontal shear stresses are shown in Figure 3-1c as they
act, separately, on the top and bottom pieces. The same stresses occur in horizontal planes in
single-piece beams; they are different in intensity at different distances from the neutral axis.
Figure 3-1d shows a differential length of a single-piece rectangular beam acted upon by a shear
force of magnitude V. Upward translation is prevented; i.e., vertical equilibrium is provided by
the vertical shear stresses  . Their average value is equal to the shear force divided by the crosssectional area  av  V ab , but their intensity varies over the depth of the section. The shear
stress is zero at the outer fibers and has a maximum of 1.5 av at the neutral axis, the variation
being parabolic. If a small square element located at the neutral axis of such a beam is isolated as
shown in Figure 3-2b, the vertical shear stresses on it, equal and opposite on the two faces for
reasons of equilibrium, act as shown. However, if these were the only stresses present, the
element would not be in equilibrium; it would spin. Therefore, on the two horizontal faces there
exist equilibrating horizontal shear stresses of the same magnitude. That is, at any point within
the beam, the horizontal shear stresses of Figure 3-2b are equal in magnitude to the vertical shear
stresses of Figure 3-1d.
Figure 3-2 – Stress trajectories in homogeneous rectangular beam
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It is proved in any strength-of-materials text that on an element cut at 450 these shear stresses
combine in such a manner that their effect is as shown in Figure 3-2c. That is, the action of the
two pairs of shear stresses on the vertical and horizontal faces is the same as that of two pairs of
normal stresses, one tensile and one compressive, acting on the 450 faces and of numerical value
equal to that of the shear stresses. If an element of the beam is considered that is located neither
at the neutral axis nor at the outer edges, its vertical faces are subject not only to the shear
stresses but also to the familiar bending stresses. The six stresses that now act on the element can
again be combined into a pair of inclined compressive stresses and a pair of inclined tensile
stresses that act at right angles to each other. They are known as principal stresses (Figure 3-2e).
Since the magnitudes of the shear stresses  and the bending stresses f change both along the
beam and vertically with distance from the neutral axis, the inclinations as well as the
magnitudes of the resulting principal stresses also vary from one place to another. Figure 3-2f
shows the inclinations of these principal stresses for a uniformly loaded rectangular beam. That
is, these stress trajectories are lines which, at any point, are drawn in that direction in which the
particular principal stress, tension or compression, acts at that point. It is seen that at the neutral
axis the principal stresses in a beam are always inclined at 450 to the axis. In the vicinity of the
outer fibers they are horizontal near midspan.
An important point follows from this discussion. Tensile stresses, which are of particular concern
in view of the low tensile strength of the concrete, are not confined to the horizontal bending
stresses f that are caused by bending alone. Tensile stresses of various inclinations and
magnitudes, resulting from shear alone (at the neutral axis) or from the combined action of shear
and bending, exist in all parts of a beam and can impair its integrity if not adequately provided
for. It is for this reason that the inclined tensile stresses, known as diagonal tension, must be
carefully considered in reinforced concrete design.
3.1.2. BEHAVIOR OF BEAMS FAILING IN SHEAR
3.1.2.1. Behavior of beams without web reinforcement
The forces transferring shear across an inclined crack in a beam without web reinforcements are
illustrated in Figure 3-3.
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Figure 3-3 – Internal forces in a cracked beam without web reinforcements
Shear is transferred across line A-B-C by v cy , the shear in the compression zone, by v ay , the
vertical component of the shear transferred across the crack by interlock of the aggregate
particles on the two faces of the crack, and by v d , the dowel action of the longitudinal
reinforcement. Immediately after inclined cracking, as much as 40 to 60 percent of the total shear
is carried by v d and v ay together.
Considering D-E-F portion of the beam below the crack and summing moments about the
reinforcement at point E shows that v d and v a cause a moment about E that must be
equilibrated by a compression force C1 . Horizontal force equilibrium on section A-B-D-E
shows that T1  C1  C1 , and finally, T1 and C1  C1 must equilibrate the external moment at
this section.
As the crack widens, v a decreases, increasing the fraction of the shear resisted by v cy and v d .
The dowel shear, v d , leads to a splitting crack in the concrete along the reinforcement. When
 and C1 ,
this crack occurs, v d drops, approaching zero. When v a and v d disappear, so do v cy
with the result that all the shear and compression are transmitted in the depth AB above the
crack. At this point in the life of the beam, the section A-B is too shallow to resist the
compression forces needed for equilibrium. As a result, this region crushes or buckles upward.
Note also that if C1  0 , then T2  T1 , and as a result, T2  C1 . In other words, the inclined
crack has made the tensile force at point C a function for the moment at section A-B-D-E. This
shift in the tensile force must be considered in detailing the bar cut off points and in anchoring
the bars.
3.1.2.2. Behavior of beams with web reinforcement
Inclined cracking causes the shear strength of beams to drop below the flexural capacity. The
purpose of web reinforcement is to ensure that the full flexural capacity can be developed.
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Prior to inclined cracking, the strain in the web reinforcement is equal to the corresponding strain
of the concrete. Because concrete cracks at a very small strain, the stress in the web
reinforcements prior to inclined cracking will not exceed 3 to 6 ksi. Thus, web reinforcements do
not prevent inclined cracks from forming; they come into play after the cracks have formed.
The forces in a beam with web reinforcements and an inclined crack are shown in Figure 3-4.
Figure 3-4 – Internal forces in a cracked beam with web reinforcements
The shear transferred by tension in the web reinforcements, v s , does not disappear when the
crack opens wider, so there will always be a compression force C1 and a shear force Vcy
acting
on the part of the beam below the crack. As a result, T2 will be less than T1 , the difference
depending on the amount of web reinforcement. The force T2 will however, be larger than the
flexural tension T  M jd based on the moment at C
The loading history of such a beam is shown qualitatively in Figure 3-5.
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Figure 3-5 – Distribution of Internal shears in a beam with web reinforcement
The components of the internal shear resistance must equal the applied shear, indicated by the
upper 450 line. Prior to flexural cracking, the entire shear is carried by the uncracked concrete.
Between flexural and inclined cracking, the external shear is resisted by v cy , v ay , and v d .
Eventually, the web reinforcements crossing the crack yield, and v s stays constant for higher
applied shears. Once the web reinforcements yield, the inclined crack opens more rapidly. As the
inclined crack widens, v ay decreases further, forcing v d and v cy to increase at an accelerated
rate, until either a splitting (dowel) failure occurs, the compression zone crushes due to combined
shear and compression, or the web crushes.
Each of the components of this process except v s has a brittle load-deflection response. As a
result, it is difficult to quantify the contributions of v cy , v d , and v ay . In design, these are lumped
together as v c , referred to somewhat incorrectly as “the shear carried by the concrete.” Thus, the
nominal shear strength, v n , is assumed to be
Vn  Vc  Vs
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3.1.3. FACTORS AFFECTING THE SHEAR STRENGTH OF BEAMS WITHOUT WEB
REINFORCEMENT
Beams without web reinforcement will fail when inclined cracking occurs or shortly afterwards.
For this reason, the shear capacity of such members is taken equal to the inclined cracking shear.
The inclined cracking load of a beam is affected by five principal variables, some included in
design equations and others not.
1. Tensile strength of concrete
The inclined cracking load is a function of the tensile strength of the concrete, fct . The stress
state in the web of the beam involves biaxial principal tension and compression stresses. A
similar biaxial state of stress exists in a split-cylinder tension test, and the inclined cracking load
is frequently related to the strength from such a test. As discussed earlier, the flexural cracking
that precedes the inclined cracking disrupts the elastic-stress field to such an extent that inclined
cracking occurs at a principal tensile stress roughly half of fct for the uncracked section.
2. Longitudinal reinforcement ratio, w
When the steel ratio, w , is small, flexural cracks extend higher into the beam and open wider
than would be the case for large values of w . And increase in crack width causes a decrease in
the maximum values of the components of shear, v d and v ay , that are transferred across the
inclined cracks by dowel action or by shear stresses on the crack surfaces. Eventually, the
resistance along the crack drops below that required to resist the loads, and the beam fails
suddenly in shear.
3. Shear span to depth ratio, a/d
The shear span to depth ratio ,a/d or MV/d, affects the inclined cracking shears and ultimate
shears of portions of members with a/d less than 2.
4. Lightweight aggregate concrete
Lightweight aggregate concrete has a lower tensile strength than normal weight concrete for a
given concrete compressive strength. Because the shear strength of a concrete member without
shear reinforcement is directly related to the tensile strength of a concrete, equations for shear
capacity must be modified for members constructed with light weight concrete.
5. Size of beam
An increase in the overall depth of a beam with very little (or no) web reinforcement results in a
decrease in the shear at failure for a given fc , w , and a/d. The width of an inclined crack
depends on the product of the strain in the reinforcement crossing the crack and the spacing of
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the cracks. With increasing beam depth, the crack spacings and the crack widths tend to increase.
This leads to a reduction in the maximum shear stress that can be transferred across the crack by
aggregate interlock. An unstable situation develops when the shear stresses transferred across the
crack exceeds the shear strength. When this occurs, the faces of the crack slip, one relative to the
other.
6. Axial forces
Axial tensile forces tend to decrease the inclined cracking load, while axial compressive forces
tend to increase it. As the axial compressive force is increased, the onset of flexural cracking is
delayed, and the flexural cracks do not penetrate as far into the beam. Axial tension forces
directly increase the tension stress, and hence the strain, in the longitudinal reinforcement. This
causes an increase in the inclined crack width, which, in turn, results in a decrease in the
maximum shear tension stress that can be transmitted across the crack. This reduces the shear
failure load.
A similar increase is observed in prestressed concrete beams. The compression due to
prestressing reduces the longitudinal strain, leading to a higher failure load.
7. Coarse aggregate size
As the size (diameter) of the coarse aggregate increases, the roughness of the crack surfaces
increases, allowing higher shear stresses to be transferred across the cracks. In high strength
concrete beams and some light weight concrete beams, the cracks penetrate pieces of the
aggregate rather than going around them, resulting in a smoother crack surface. This decrease in
the shear transferred by aggregate interlock along the cracks reduces v c .
3.2. DESIGN OF BEAMS FOR VERTICAL SHEAR ACCORDING TO EN 1992-1-1-2004
For the verification of the shear resistance the following symbols are defined:
VRd ,c
is the design shear resistance of the member without shear reinforcement
is the design value of the shear force which can be sustained by the yielding shear
reinforcement
is the design value of the maximum shear force which can be sustained by the
VRd ,max
member, limited by crushing of the compression struts.
The shear resistance of a member with shear reinforcement is equal to:
VRd ,s
VRd  VRd ,s
(4)
In regions of the member where VEd  VRd ,c , no calculated shear reinforcement is necessary. VEd
is the design shear force in the section considered resulting from the external loading.
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In regions where VEd  VRd ,c , sufficient shear reinforcement should be provided in order that
VEd  VRd
The design shear force should not exceed the permitted maximum value VRd ,max , anywhere in the
member.
For members subject to predominantly uniformly distributed loading, the design shear force need
not be checked at a distance less that d from the face of the support. Any shear reinforcement
required should continue to the support. In addition it should be verified that the shear at the
support does not exceed VRd ,max
3.2.1. MEMBERS NOT REQUIRING DESIGN SHEAR REINFORCEMENT
The design value for the shear resistance VRd ,c is given by:
VRd ,c  CRd ,c k 1001fck 

with a minimum of
13
 k1 cp  bw d

VRd ,c  v min  k1 cp  bw d
(5)
(6)
where:
fck is in MPa
k  1
1 
Asl
bw
 cp
NEd
Ac
VRd ,c
200
 2.0 with d in mm
d
Asl
 0.02
bw d
(7)
(8)
is the area of the tensile reinforcement, which extends   lbd  d  beyond the
section considered
is the smallest width of the cross-section in the tensile area (mm)
 NEd Ac  0.2fcd [MPa]
is the axial force in the cross-section due to loading or prestressing in newtons (
NEd  0 for compression). The influence of imposed deformations on NE may be
ignored.
Is the area of concrete cross section [mm2]
is in newtons
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The values of CRd ,c ,  min and k1 for use in a country may be found in its National Annex. The
recommended value for CRd ,c is 0.18  c , that for v min is 0.035k 3 2  fck1 2 and that for k1 is 0.15.
The design of members with shear reinforcement is based on a truss model. The angle θ should
be limited. The limiting values of cot  for use in a country may be found in its National Annex.
The recommended limits are 1  cot   2.5
3.2.2. MEMBERS REQUIRING DESIGN SHEAR REINFORCEMENT
For members with vertical shear reinforcement, the shear resistance, v Rd is the smaller value of :
VRd ,s 
(9)
Asw
zfywd cot 
s
and
VRd ,max  c bw z fcd  cot   tan 
(10)
Where
Asw
s
fywd
is the cross-sectional area of the shear reinforcement
is the spacing of the stirrups
is the design yield strength of the shear reinforcement

follows from the expression below
For reinforced and prestressed members, if the design stress of the shear reinforcement is below
80% of the characteristic yield stress fyk ,  may be taken as:
(11)
for fck  60 MPa
  0.6
(12)
  0.9  fck 200  0.5
for fck  60 MPa
The value of tan c for use in a Country may be found in its National Annex. The recommended
value is 1 for non-prestressed structures.
The maximum effective cross-sectional area of the shear reinforcement Asw ,max is given by:
(13)
1
 c fcd
bw s
2
For members with inclined shear reinforcement, the shear resistance is the smaller value of
Asw ,max fywd
VRd ,s 

(14)
Asw
zfywd  cot   cot   sin
s
And

VRd ,max   c bw z fcd  cot   tan   1  cot 2 
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The maximum effective shear reinforcement Asw ,max follows from:
Asw ,max fywd
bw s
(16)
1
 c fcd sin 
 2
1  cos 
3.2.3. ADDITIONAL TENSILE FORCE IN LONGITUDINAL REINFORCEMENT
The longitudinal tension reinforcement should be able to resist the additional tensile force caused
by shear.
The additional tensile force, Ftd , in the longitudinal reinforcement due to shear VEd may be
calculated from:
Ftd  0.5VEd  cot   cot  
MEd z   Ftd
(17)
should be taken not greater than MEd ,max z
3.2.4. MINIMUM AREA AND MAXIMUM SPACING OF SHEAR REINFORCEMENT
The ratio of shear reinforcement is given by
w  Asw /  s  bw  sin 
(18)
where:
w
is the shear reinforcement ratio
w should not be less than w ,min
is the area of shear reinforcement within length s
Asw
s
is the spacing of the shear reinforcement measured along the longitudinal axis of
the member
is the breadth of the web of the member
bw

is the angle between shear reinforcement and the longitudinal axis
When, on the basis of the design shear calculation, no shear reinforcement is required, minimum
shear reinforcement should nevertheless be provided. The minimum shear reinforcement may be
omitted in members such as slabs (solid, ribbed or hollow core slabs) where transverse
redistribution of loads is possible. Minimum reinforcement may also be omitted in members of
minor importance which do not contribute significantly to the overall resistance and stability of
the structure.
The value of w ,min for beams for use in a Country may be found in its National Annex. The

recommended value is w ,min  0.08 fck
f
yk
The maximum longitudinal spacing between shear assemblies should not exceed sl ,max .
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The value of sl ,max for use in a country may be found in its National Annex. The recommended
value is smax  0.75d 1 cot  
where  is the inclination of the shear reinforcement to the longitudinal axis of the beam.
3.2.5. PROCEDURE FOR DESIGN

Step 1:

Step 2:

Step 3:

Step 4:
  0.5sin1  2VEd c bw zfcd  

Step 5:
If cot  is less than 1, re-size element, otherwise

Step 6:
Calculate amount of shear reinforcement required

Step 7:
Check min shear reinforcement and maximum spacing
Determine maximum applied shear force at support, VEd
V
Determine Rd ,max with cot   2.5
If VRd ,max  VEd cot   2.5 , go to step 6 and calculate required shear
reinforcement
If VRd ,max  VEd calculate required strut angle:
3.3. DEVELOPMENT, ANCHORAGE, AND SPLICING OF REINFORCEMENT
3.3.1. INTRODUCTION
‘Bond’ in reinforced concrete refers to the adhesion between the reinforcing steel and the
surrounding concrete. It is this bond which is responsible for the transfer of axial force from a
reinforcing bar to the surrounding concrete, thereby providing strain compatibility and
‘composite action’ of concrete and steel. If this bond is inadequate, ‘slipping’ of the reinforcing
bar will occur, destroying full ‘composite action’. Hence, the fundamental assumption of the
theory of flexure, viz. plane sections remain plane even after bending, becomes valid in
reinforced concrete only if the mechanism of bond is fully effective.
It is through the action of bond resistance that the axial stress (tensile or compressive) in a
reinforcing bar can undergo variation from point to point along its length. This is required to
accommodate the variation in bending moment along the length of the flexural member. Had the
bond been absent, the stress at all points on a straight bar would be constant, as in a string or a
straight cable.
Mechanisms of Bond Resistance
Bond resistance in reinforced concrete is achieved through the following mechanisms:
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1. Chemical adhesion — due to a gum-like property in the products of hydration (formed during
the making of concrete).
2. Frictional resistance — due to the surface roughness of the reinforcement and the grip
exerted by the concrete shrinkage.
3. Mechanical interlock — due to the surface protrusions or ‘ribs’ (oriented transversely to the
bar axis) provided in deformed bars.
Evidently, the resistance due to ‘mechanical interlock’ (which is considerable) is not available
when plain bars are used. For this reason, many codes prohibit the use of plain bars in reinforced
concrete — except for lateral spirals, and for stirrups and ties smaller than 10 mm in diameter.
Bond Stress
Bond resistance is achieved by the development of tangential (shear) stress components along
the interface (contact surface) between the reinforcing bar and the surrounding concrete. The
stress so developed at the interface is called bond stress, and is expressed in terms of the
tangential force per unit nominal surface area of the reinforcing bar.
Two Types of Bond
There are two types of loading situations which induce bond stresses, and accordingly ‘bond’ is
characterized as:
1. Flexural bond;
2. Anchorage bond or development bond.
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Figure 3-6 – Bond stress in a beam
‘Flexural bond’ is that which arises in flexural members on account of shear or a variation in
bending moment, which in turn causes a variation in axial tension along the length of a
reinforcing bar [Figure 3-6(d)]. Evidently, flexural bond is critical at points where the shear
V  dM dx  is significant.
‘Anchorage bond’ (or ‘development bond’) is that which arises over the length of anchorage
provided for a bar or near the end (or cut-off point) of a reinforcing bar; this bond resists the
‘pulling out’ of the bar if it is in tension [Figure 3-6(e)], or conversely, the ‘pushing in’ of the bar
if it is in compression.
These two types of bond are discussed in detail in the sections to follow.
3.3.2. FLEXURAL BOND
As mentioned earlier, variation in tension along the length of a reinforcing bar, owing to varying
bending moment, is made possible through flexural bond. The flexural stresses at two adjacent
sections of a beam, dx apart, subjected to a differential moment dM, is depicted in Figure 3-6(b).
With the usual assumptions made in flexural design, the differential tension dT in the tension
steel over the length dx is given by
dT 
dM
z
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where z is the lever arm.
This unbalanced bar force is transferred to the surrounding concrete by means of ‘flexural bond’
developed along the interface. Assuming the flexural (local) bond stress uf to be uniformly
distributed over the interface in the elemental length dx, equilibrium of forces gives:
uf   o  dx  dT
( 3-2)
where Σo is the total perimeter of the bars at the beam section under consideration [Figure
3-6(c)].
From the above equation, it is evident that the bond stress is directly proportional to the change
in the bar force. Combining the above two equations, the following expression for the local bond
stress uf is obtained:
uf 
dM dx
o z
( 3-3)
Alternatively, in terms of the transverse shear force at the section V  dM dx ,
uf 
V
o z
( 3-4)
It follows that flexural bond stress is high at locations of high shear, and that this bond stress can
be effectively reduced by providing an increased number of bars of smaller diameter bars (to
give the same equivalent Ast).
It may be noted that the actual bond stress will be influenced by flexural cracking, local slip,
splitting and other secondary effects — which are not accounted for in the above equation.
In particular, flexural cracking has a major influence in governing the magnitude and distribution
of local bond stresses.
Effect of Flexural Cracking on Flexural Bond Stress
From the above equation, it appears that the flexural (local) bond stress uf has a variation that is
similar to and governed by the variation of the transverse shear force V. In fact, it would appear
that in regions of constant moment, where shear is zero, there would be no bond stress developed
at all. However, this is not true. The tensile force T in the reinforcement varies between flexural
crack locations, even in regions of constant moment, as indicated in Figure 3-7. At the flexural
crack location, the tension is carried by the reinforcement alone, whereas in between the cracks,
concrete carries some tension and thereby partially relieves the tension in the steel bars. As local
bond stress is proportional to the rate of change of bar force, local bond stresses do develop in
such situations.
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Figure 3-7 – Effect of flexural cracks on flexural bond stress in constant moment region
The bond stresses follow a distribution somewhat like that shown in Figure 3-7(c), with the
direction of the bond stress reversing between the cracks. The net bond force between the cracks
will, of course, be zero in a region of constant moment. When the moment varies between the
flexural cracks, the bond stress distribution will differ from that shown in Figure 3-7(c), such that
the net bond force is equal to the unbalanced tension in the bars between the cracks.
Beam tests show that longitudinal splitting cracks tend to get initiated near the flexural crack
locations where the local peak bond stresses can be high. The use of large diameter bars
particularly renders the beam vulnerable to splitting and/or local slip.
Finally, it may be noted that flexural cracks are generally not present in the compression zone.
For this reason, flexural bond is less critical in a compression bar, compared to a tension bar with
an identical axial force.
3.3.3. ANCHORAGE (DEVELOPMENT) BOND
As mentioned earlier, anchorage bond or development bond is the bond developed near the
extreme end (or cut-off point) of a bar subjected to tension (or compression). This situation is
depicted in the cantilever beam of Figure 3-8, where it is seen that the tensile stress in the bar
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segment varies from a maximum
discontinuous end C.
 fs 
Reinforced Concrete I
at the continuous end D to practically zero at the
Figure 3-8 – Anchorage bond stress
The bending moment, and hence the tensile stress fs , are maximum at the section at D.
Evidently, if a stress fs is to be developed in the bar at D, the bar should not be terminated at D,
but has to be extended (‘anchored’) into the column by a certain length CD. At the discontinuous
end C of the bar, the stress is zero. The difference in force between C and D is transferred to the
surrounding concrete through anchorage bond. The probable variation of the anchorage bond
stress ua is as shown in Figure 3-8(b) — with a maximum value at D and zero at C.
An expression for an average bond stress uav can be derived by assuming a uniform bond stress
distribution over the length L of the bar of diameter  [Figure 3-8(c)], and considering
equilibrium of forces as given below:
L  uav


  2 4 fs  uav 
fs
( 3-5)
4L
This bond stress may be viewed as the average bond stress generated over a length L in order to
develop a maximum tensile (or compressive) stress fs at a critical section; hence, this type of
bond is referred to as ‘development bond’. Alternatively this bond may be viewed as that
required to provide anchorage for a critically stressed bar; hence, it is also referred to as
‘anchorage bond’.
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3.3.4. BOND FAILURE AND BOND STRENGTH
Bond Failure Mechanisms
The mechanisms that initiate bond failure may be any one or combination of the following:




break-up of adhesion between the bar and the concrete;
Longitudinal splitting of the concrete around the bar;
Crushing of the concrete in front of the bar ribs (in deformed bars); and
Shearing of the concrete keyed between the ribs along a cylindrical surface surrounding
the ribs (in deformed bars).
The most common type of bond failure mechanism is the pulling loose of the reinforcement bar,
following the longitudinal splitting of the concrete along the bar embedment [Figure 3-9].
Occasionally, failure occurs with the bar pulling out of the concrete, leaving a circular hole
without causing extensive splitting of the concrete. Such a failure may occur with plain smooth
bars placed with large cover, and with very small diameter deformed bars (wires) having large
concrete cover. However, with deformed bars and with the normal cover provided in ordinary
beams, bond failure is usually a result of longitudinal splitting. In the case of ribbed bars, the
bearing pressure between the rib and the concrete is inclined to the bar axis [Figure 3-9(b)]. This
introduces radial forces in the concrete (‘wedging action’), causing circumferential tensile
stresses in the concrete surrounding the bar (similar to the stresses in a pipe subjected to internal
pressure) and tending to split the concrete along the weakest plane. Splitting occurs along the
thinnest surrounding concrete section, and the direction of the splitting crack (‘bottom splitting’
or ‘side splitting’) depends on the relative values of the bottom cover, side cover and bar spacing
as shown in Figure 3-9(b).
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Figure 3-9 – Typical bond splitting crack patterns
Splitting cracks usually appear on the surface as extensions of flexural or diagonal tension cracks
in flexural members, beginning in regions of high local bond stress . With increased loads, these
cracks propagate gradually along the length of embedment (‘longitudinal splitting’) with local
splitting at regions of high local bond stress and associated redistribution of bond stresses.]. The
presence of stirrups offers resistance to the propagation of continuous longitudinal splitting
cracks [Figure 3-10]. However, in beams without stirrups, the failure due to bond can occur early
and suddenly, as the longitudinal split runs through to the end of the bar without the resistance
offered by the stirrups.
Figure 3-10 – Stirrups resisting tensile forces due to bond
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Bond Tests
Bond strength is usually ascertained by means of pull out tests or some sort of beam tests. The
typical ‘pull out’ test is shown schematically in Figure 3-11(a). A bar embedded in a concrete
cylinder or prism is pulled until failure occurs by splitting, excessive slip or pull-out. The
nominal bond strength is computed as P L  , where P is the pull at failure,  the bar
diameter and L the length of embedment. It may be noted, however, that factors such as cracking
(flexural or diagonal tension) and dowel forces, which lower the bond resistance of a flexural
member, are not present in a concentric pull out test. Moreover, the concrete in the test specimen
is subjected to a state of compression (and not tension), and the friction at the bearing on the
concrete offers some restraint against splitting. Hence, the bond conditions in a pull out test do
not ideally represent those in a flexural member.
Figure 3-11 – Bond tests
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Of the several types of beam tests developed to simulate the actual bond conditions, one test setup is shown in Figure 3-11(b) . The bond strength measured from such a test, using the same
expression P L  as for a pull-out test, is bound to give a lesser (and more accurate) measure
of the bond strength than the pull out strength. However, the pull out test is easier to perform,
and for this reason, more commonly performed.
From the results of such bond tests, the ‘design bond stress’ (permissible average anchorage
bond stress) is arrived at - for various grades of concrete. Tests indicate that bond strength varies
proportionately with fck  for small diameter bars and fck for large diameter deformed bars
Factors Influencing Bond Strength
Bond strength is influenced by several factors, some of which have already been mentioned. In
general, bond strength is enhanced when the following measures are adopted:









deformed (ribbed) bars are used instead of plain bars;
smaller bar diameters are used;
higher grade of concrete (improved tensile strength) is used;
increased cover is provided around each bar;
increased length of embedment, bends and /or hooks are provided;
mechanical anchorages are employed;
stirrups with increased area, reduced spacing and/or higher grade of steel are used;
termination of longitudinal reinforcement in tension zones is avoided;
any measure that will increase the confinement of the concrete around the bar is
employed.
Another factor which influences bond strength in a beam is the depth of fresh concrete below the
bar during casting. Water and air inevitably rise towards the top of the concrete mass and tend to
get trapped beneath the horizontal reinforcement, thereby weakening the bond at the underside of
these bars. For this reason, codes specify a lower bond resistance for the top reinforcement in a
beam.
3.3.5. SPLICING OF REINFORCEMENT
Splices are required when bars placed short of their required length (due to non-availability of
longer bars) need to be extended. Splices are also required when the bar diameter has to be
changed along the length (as is sometimes done in columns). The purpose of ‘splicing’ is to
transfer effectively the axial force from the terminating bar to the connecting (continuing) bar
with the same line of action at the junction. This invariably introduces stress concentrations in
the surrounding concrete. These effects should be minimized by:



using proper splicing techniques;
keeping the splice locations away from sections with high flexural/shear stresses; and
Staggering the locations of splicing in the individual bars of a group (as, typically in a
column).
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Splicing is generally done in one of the following three ways:
1. Lapping of bars (lap splice)
2. Welding of bars (welded splice)
3. Mechanical connections.
3.4. PROVISIONS OF EN -1992-1-1-2004 ON DEVELOPMENT, ANCHORAGE AND
SPLICING OF REINFORCEMENT
3.4.1. ANCHORAGE OF LONGITUDINAL REINFORCEMENT
Ultimate bond stress
The ultimate bond resistance shall be sufficient to prevent bond failure and the design value of
the ultimate bond stress, fbd , for ribbed bars may be taken as:
fbd  2.2512fctd
( 3-6)
Where:
fctd
is the design value of concrete tensile strength
1
is a coefficient related to the quality of the bond condition and the position of the bar
during concreting
1  1.0 when ‘good’ conditions are obtained and
1  0.7 for all other cases and for bars in structural elements built with slip-forms,
unless it can be shown that ‘good’ bond conditions exist
is related to the bar diameter:
2  1.0 for   32 mm
2
2  132    / 100 for   32 mm
Figure 3-12 – Description of bond conditions
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Basic anchorage length
The calculation of the required anchorage length shall take into consideration the type of steel
and bond properties of the bars.
The basic required anchorage length, l b,req , for anchoring the force As fyd in a bar assuming
constant bond stress equal to fbd follows from:
lb,rqd   4 sd fbd 
( 3-7)
where  sd is the design stress of the bar at the position from where the anchorage is measured
from at the ultimate limit state.
Design anchorage length
The design anchorage length, l bd , :
l bd  1 2 3 4 5 l b,rqd  l b,min
Where α1, α2, α3, α4 and α5 are coefficients given in Table 1:
1
2
3
4
is for the effect of the form of the bars assuming adequate cover
is for the effect of concrete minimum cover (see Figure 3-13)
is for the effect of confinement by transverse reinforcement
is for the influence of one or more welded transverse bars t  0.6  along the design
anchorage length l bd
is for the effect of the pressure transverse to the plane of splitting along the design
5
anchorage length
The product 235   0.7
l b,min
is the minimum anchorage length if no other limitation is applied:
- For anchorages in tension: l b,min  max 0.3l b,rqd ;10;100 mm
-
For anchorages in compression: l b,min  max 0.6l b,rqd ;10;100 mm
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Figure 3-13 – Values of Cd for beams and slabs
Table 1 – Values of α coefficients
Figure 3-14 – Values of K for beams and slabs
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3.4.2. LAPS
The detailing of laps between bars shall be such that:
 The transmission of the forces from one bar to the next is assured
 Spalling of the concrete in the neighborhood of the joints does not occur;
 Large cracks which affect the performance of the structure do not occur
Laps:
 Between bars should normally be staggered and not located in areas of high stress.
 At any one section should normally be arranged symmetrically
The arrangement of lapped bars should comply with Figure 3-15
 The clear transverse distance between two lapped bars should not be greater than 4 or
50 mm, otherwise the lap length should be increased by a length equal to the clear space
where it exceeds 4 or 50 mm;
 The longitudinal distance between two adjacent laps should not be less than 0.3 times the
lap length, l 0 ;
 In case of adjacent laps, the clear distance between adjacent bars should not be less than
2 or 20 mm.
Figure 3-15 – Adjacent laps
Lap length
The design lap length is:
l 0  1 2 3 5 6 l b,rqd  l0,min
( 3-8)
Where:
l 0,min  max 0.3 6 l b,rqd ;15;200 mm
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Values of 1, 2 ,3 , and 5 , may be taken from Table 1; however, for the calculation of  3 ,
 Ast ,min
should be taken as 1.0 As  sd fyd  , with As  area of one lapped bar.
 6   1 25 
0.5
but not exceeding 1.5, where 1 is the percentage of reinforcement lapped
within 0.65l0 from the center of the lap length considered (see Figure ). Values of  6 are given
in the table below.
Table 2 – Values of the coefficient  6
Figure 3-16 – Percentage of lapped bars in one section
Transverse reinforcement in the lap zone
a) Transverse reinforcement for bars in tension
Transverse reinforcement is required in the lap zone to resist transverse tension forces.
Where the diameter,  , of the lapped bars is less than 20 mm, or the percentage of lapped bars in
any one section is less than 25%, then any transverse reinforcement or links necessary for other
reasons may be assumed sufficient for the transverse tensile forces without further justification.
Where the diameter,  , of the lapped bars is greater than or equal to 20 mm, the transverse
reinforcement should have a total area, Ast (sum of all legs parallel to the layer of the spliced
reinforcement) of not less than the area As of one lapped bar
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 A
st
 1.0As  , assuming that
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the lapped bar is fully stressed. The transverse bar should be placed perpendicular to the
direction of the lapped reinforcement and between that and the surface of the concrete.
If more than 50% of the reinforcement is lapped at one point and the distance, a, between
adjacent laps at a section is  10 (see Figure 3-15) transverse bars should be formed by links or
U bars anchored into the body of the section.
The transverse reinforcement should be positioned at the outer sections of the lap as shown in
Figure 3-17.
b) Transverse reinforcement for bars permanently in compression
In addition to the rules for bars in tension one bar of the transverse reinforcement should be
placed outside each end of the lap length and within 4 of the ends of the lap length (Figure
3-17)
Figure 3-17 – Transverse reinforcement for lapped splices
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