Minimum shear reinforcement for thick plates and two-way slabs

Full text

Minimum shear reinforcement for thick plates and two-way slabs
E. Rizk
a,1
, H. Marzouk
b,
⇑
a
Faculty of Engineering and Applied Science, Memorial University of Newfoundland, St. John’s, Newfoundland, Canada A1B 3X5
b
Department of Civil Engineering, Faculty of Engineering, Architecture and Science, Ryerson University, Toronto, Ontario, Canada M5B 2K3
article info
Article history:
Received 29 July 2010
Revised 4 July 2012
Accepted 6 July 2012
Available online 3 September 2012
Keywords:
Thick plates
Punching resistance
Shear reinforcement
Sandwich model
Compression field
Diagonal cracking and size effect
abstract
Thick reinforced concrete plates of over 250 mm thickness usually exhibit brittle shear failure under con-
centrated loads. Potential punching failures in the vicinity of concentrated loads have already been taken
into consideration by conventional design methods. The current research is a continuation to previous
work published by the authors on minimum flexural reinforcement for thick plates. The main focus of
this research is to design adequate shear reinforcement for thick concrete plates. In this paper, new mod-
els are developed to calculate minimum shear reinforcements for thick plates, concrete plates and two-
way slabs. Design guidelines as well as a design example for minimum shear reinforcement that the
design engineers can follow are given. A comparison between the proposed formulae with experimental
tests carried by different researchers is conducted. A comprehensive review of different design codes
formulae is discussed and compared with the proposed formulae. The proposed formulae are simple
and account for member size effect through fracture mechanics concepts. For all equations, (MPa) is used
for stresses and (mm) for lengths unless otherwise noted.
Crown Copyright Ó2012 Published by Elsevier Ltd. All rights reserved.
1. Introduction
Potential shear failures of a slab that include a wide beam and
punching failures in the vicinity of concentrated loads have been
already taken into consideration by conventional design methods.
Most design codes try to avoid minimum shear reinforcement
requirements for slabs by limiting nominal shear stresses at
well-defined critical sections to guard against such failure modes.
Nominal punching shear stress is defined by design codes as the
punching shear force divided by a control surface around the
loaded area. With the extensive use of thick plates that are more
than 250 mm in thickness and made of high strength concrete
(HSC) for offshore structures, different guidelines must be used
for minimum shear reinforcement requirements. The main objec-
tive of the current research is to develop a formula to calculate
minimum shear reinforcement for thick plates, concrete walls
and slab-column connections that is required to prevent brittle
failure in the vicinity of concentrated loads.
2. Previous research
A shear stud reinforcement assembly is composed of vertical
bars. The top ends of the vertical bars consist of forged anchor
heads. The bottom ends are welded to a steel strip. The shear rein-
forcement assembly was originally developed at the University
of Calgary. Extensive tests that were conducted in Canada and
Germany [1–3] on full-size slab-column connections verified that
stud type reinforcements can substantially increase the strength
of slabs and prevent brittle failure. Marzouk and Jiang [4] conducted
an experimental investigation on six HSC plates that are reinforced
with different types of shear reinforcement to prevent brittle
punching shear behavior. The types of shear reinforcement tested
were single-bend, U-stirrups, double-bend, shear-stud and
T-headed shear reinforcement. It was concluded that double bend,
shear-stud and T-headed shear reinforcements are the most effi-
cient shear enhancement for concrete plates. The punching shear
failure of slabs provided with shear reinforcement was eliminated
and the punching failure mode was transformed into flexural fail-
ure for the HSC plates. In the meantime, both ductility and energy
absorption of the two-way slabs was significantly increased by
using shear reinforcement.
In 1993, thick concrete slabs were investigated by the U.S. Army
Engineer Waterways Experiment Station [5]. Thirteen one-way
reinforced concrete slabs were statically loaded. The study empha-
sized primary parameters that affect the large-deflection behavior
of a one-way slab, such as support conditions, quantity and spacing
of principal reinforcements, quantity and spacing of shear rein-
forcements, and span to effective depth (L/d) ratios. The tests ver-
ified that shear reinforcement has a significant contribution to the
ultimate resistance, and lacing and single-leg stirrups are about as
equally as effective.
0141-0296/$ - see front matter Crown Copyright Ó2012 Published by Elsevier Ltd. All rights reserved.
http://dx.doi.org/10.1016/j.engstruct.2012.07.006
⇑
Corresponding author. Tel.: +1 416 979 5000x6451; fax: +1 416 979 5122.
E-mail addresses: emad.rizk@mun.ca (E. Rizk), hmarzouk@ryerson.ca
(H. Marzouk).
1
Tel.: +1 709 800 1468.
Engineering Structures 46 (2013) 1–13
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Kordina and Meichsner [6] tested slabs with depths of 150, 250
and 450 mm. The tests included slabs without any shear reinforce-
ment. Slabs provided with minimum shear reinforcement were
calculated in accordance to Eurocode 2 (EC 2) [7] provisions. Con-
ventional U-stirrups were used. The researchers concluded that
the shear failure loads remained sudden and unaffected with extra
stirrups. This is probably due to the lack of anchorage of stirrups.
Headed studs have two advantages over conventional stirrups as
shear reinforcement for slabs that are: anchorage with heads can
develop the full yield strength of the stud immediately adjacent
to the head, resulting in more efficient confinement of the concrete;
elimination of the hooks and the overlaps of stirrups combined with
the possibility of using a smaller number of studs that reduces the
congestion of the reinforcement and the cost of installation.
Twenty-eight large-scale tests were conducted by Jaeger and
Marti [8] to investigate the shear strength and deformation capac-
ity of orthogonally reinforced concrete slabs. The test parameters
included the slab thickness, in-plane and transverse reinforcement
ratios and deviation of the principal shear (and moment) direction
from the direction of the in-plane reinforcement. It was concluded
that all tests without transverse reinforcement exhibit brittle shear
failures. The addition of transverse reinforcements with reinforce-
ment ratios of approximately 0.3% and 0.6% changed the failure
modes to ductile flexural failures; the tests without transverse
reinforcement showed a significant influence of slab thickness on
shear strength.
Vaz et al. [9] performed a study that aimed to define the mini-
mum shear reinforcement of flat slabs that lead to a punching
shear failure surface that crosses the shear reinforcement, in order
to avoid a sudden failure. In the attempt to define the minimum
punching reinforcement of the slabs, a parameter k, which is equal
to the total force in the transverse reinforcement inside a truncated
cone bounded by the shear crack divided by the punching strength
of a similar slab without shear reinforcement, was used. The k
values of the slabs varied between 0.27 and 1.03 and the punching
failure surface crossed the shear reinforcement when kwas smaller
than approximately 0.70. The results of the analyzed tests pointed
out that the value of kthat corresponds to the minimum reinforce-
ment should be around 0.5 to 0.7.
3. Failure modes of slabs with shear reinforcement
Design of slabs with punching shear reinforcement typically
considers several potential failure modes [10]:
1. Crushing of compression struts near the column region, this
failure mode is governing for high amounts of flexural and shear
reinforcement ratio, where large compressive stresses can
develop in the concrete near the column region.
2. Punching outside the shear-reinforced zone, this failure mode
may be governing when the shear-reinforced zone extends over
a small region.
Nomenclature
A
c
area of concrete cross section
A
s
area of reinforcement within the effective embedment
thickness
A
sl
area of tensile reinforcement which extends beyond the
section that is considered taking account of the ‘‘shift
rule’’
A
sw,min
area of a link leg (or equivalent) required for minimum
shear reinforcement
A
v,min
minimum web reinforcement area
A
z,min
area of minimum shear reinforcement
bwidth of the section (also plate span)
b
o
perimeter of critical section for shear in slabs and foot-
ings
b
w
minimum effective web width
cthickness of cover element
deffective depth to the centric of the tensile reinforce-
ment
d
v
effective shear depth of the core, is given by d
v
=hc
E
c
modulus of elasticity of concrete
F
cw
diagonal compressive force in concrete
F
sz
tensile force in the web reinforcement (stirrups)
f0
cuniaxial compressive strength of concrete (cylinder
strength)
f
ct
direct tensile strength of concrete
f
ck
characteristic compressive strength of the concrete in
(MPa)
f
r
modulus of rupture for concrete
f
sk
characteristic reinforcement strength defined as yield
stress or 0.2% proof stress in (MPa)
f
tk
expected lower characteristic tensile strength of the
concrete in (MPa)
f
y
yield stress of steel
G
f
fracture energy
hsection height
ksize effect factor or shear reinforcement index which is
equal to the total force in the transverse reinforcement
inside a truncated cone bounded by the shear crack di-
vided by the punching strength of a similar slab without
shear reinf.
l
ch
characteristic length
M
x
bending moment per unit length
M
y
bending moment per unit length
M
xy
torsional moment per unit length
N
x
in plane axial applied force per unit length
N
y
in plane axial applied force per unit length
N
xy
in plane shearing force per unit length
sspacing of shear reinforcement
s
max
maximum spacing of shear reinforcement
S
r
spacing of shear links in the radial direction
S
t
spacing of shear links in the tangential direction
v
c
nominal shear strength provided by concrete
V
cr
cracking shear force
v
f
factored shear stress
v
u
shear strength of slab as defined by the Canadian code
CSA-A23.3-04
V
o
principal shearing force per unit length
V
xz
shearing force per unit length
V
yz
shearing force per unit length
v
o
nominal shear stress
a
angle between stirrups and longitudinal axis of the
beam
bmember size effect factor
hangle between the diagonal compression field and the
XY plan
/
v
shear strength reduction factor
kmodification factor of lightweight concrete
q
z,min
minimum shear reinforcement ratio
r
cp
average stress in concrete section due to normal force
2E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

3. Punching within the shear-reinforced zone, such failure devel-
ops for low to medium amounts of shear reinforcement ratio.
Shear strength is thus governed by the contribution of concrete
and of the transverse reinforcement.
4. Minimum shear reinforcement for beams
4.1. North American codes
The loads observed in tests vary widely from the values given
by the design equations. Therefore, minimum shear reinforcement
requirements are given in the Canadian standard code CSA-A23.3-
04 [11] to guard against shear failure of a beam without web rein-
forcement in a sudden and brittle manner. CSA-A23.3-04 [11]
requires a minimum amount of web reinforcement, A
v,min
,tobe
provided if the factored shear stress (
v
f
), exceeds 0.5
v
c
+/
p
v
p
, equal
to:
A
v
;min
¼0:06 ffiffiffiffi
f
0
c
pb
w
s
f
y
ð1Þ
where b
w
is the minimum effective web width, sis the spacing of
shear reinforcement that shall not exceed 0.7 d
v
or 600 mm, d
v
is
the effective shear depth, taken as the greater of 0.9dor 0.72h,d
is the effective depth to the centroid of the longitudinal tension
reinforcement, his the member overall depth (thickness),
v
p
is a
component in the direction of the applied shear of the effective pre-
stressing stress, prestressing refers to main reinforcement or to
members reinforced with a combination of prestressed reinforce-
ment and nonprestressed deformed bars, /
p
is a resistance factor
for prestressing tendons, f
0
c
is the specified compressive strength
of concrete (MPa), and f
y
is the specified yield strength, (MPa).
The same requirements were adopted later by ACI 318-08 [12].
v
c
is given by:
v
c
¼0:17 ffiffiffiffi
f
0
c
qð2Þ
The ACI 318-08 [12] code does not include the influence of either
the flexural reinforcement or the size effect on the limiting shear
stress. The CSA-A23.3-04 [11] accounts for the beam effective
depth. If the effective depth, d, used exceeds 300 mm, the value of
v
c
shall be multiplied by b= 1300/(1000 + d), this size effect factor
is not effective for beams having an effective depth less than
300 mm. Fracture mechanics concepts suggest that the size effect
factor is not related to the member depth only but must be related
to the mechanical properties of concrete as well.
4.2. European codes
The Norwegian code NS 3473 E [13] requires a minimum area of
web reinforcement if the capacity is not sufficient enough without
shear reinforcement, and the area of shear reinforcement shall be
provided for the direction that has the greatest requirement.
Stirrups shall be provided along the entire length of a beam irre-
spective of the magnitude of the acting shear forces. NS 3473 E
[13], requires in section 18.1.6 that reinforced-concrete slabs be
provided with a minimum shear reinforcement that is the same
ratio necessary for beams. This stirrup reinforcement shall have a
cross-sectional area corresponding to:
A
s
0:2A
c
f
tk
f
sk
sin
a
ð3Þ
where A
c
is the concrete area of a longitudinal section of the beam
web,
a
is the angle between the stirrups and longitudinal axis of the
beam, f
tk
is the lower 5% fractile characteristic tensile strength of
the concrete and f
sk
is the characteristic reinforcement strength de-
fined as yield stress or the 0.2% proof stress in (MPa). The angle shall
not be taken to be less than 45°. The f
tk
shall not have a lower value
than 2.55 (MPa). The maximum distance between the stirrups shall
neither exceed s
max
¼0:6h
0
ð1þcot aÞh
0
nor exceed 500 mm,
where h
0
is the distance between the centroids of the tension and
compression reinforcement. The stirrups shall enclose all tensile
reinforcement bars, and if necessary, by means of spliced stirrups.
EC 2 [14] requirements, as summarized below, illustrate that
the minimum percentage of shear reinforcement depends on the
concrete and steel grades, while the maximum spacing of shear
reinforcement depends on the level of shear force. EC 2 [14]
requires that shear reinforcement should form an angle
a
of be-
tween 45°and 90°to the longitudinal axis of the structural ele-
ment. The shear reinforcement may consist of a combination of:
links enclosing the longitudinal tension reinforcement and the
compression zone; bent-up bars; cages, ladders, etc., which is cast
in without enclosing the longitudinal reinforcement but are prop-
erly anchored in the compression and tension zones. The ratio of
shear reinforcement is given by the following expression:
q
w
¼A
sw
sb
w
sin
a
ð4Þ
where
q
w
is the shear reinforcement ratio,
q
w
should not be less
than
q
w,min
,A
sw
is the area of shear reinforcement within length s,
sis the spacing of the shear reinforcement measured along the lon-
gitudinal axis of the member, b
w
is the width of the web of the
member and
a
is the angle between shear reinforcement and the
longitudinal axis of the beam. The value of required
q
w,min
for
beams is given by:
q
w;min
¼0:08 ffiffiffiffiffi
f
ck
p
f
yk
ð5Þ
where f
ck
is the characteristic compressive cylinder strength of con-
crete at 28 day and f
yk
is the characteristic yield strength of
reinforcement.
EC 2 [14] indicates that this minimum area of shear reinforce-
ment may be subject to change in a National Annex. The maximum
longitudinal spacing between shear assemblies should not exceed
s
max
. The value of s
max
is given by:
s
max
¼0:75dð1þcot
a
Þð6Þ
where dis the effective depth of the beam cross-section.
5. Development of minimum shear reinforcement ratios for
Plates
5.1. North American code requirements
Canadian standard code CSA-A23.3-04 [11] allows the use of
shear reinforcements, consisting of headed shear reinforcement,
stirrups, or shear heads to increase the shear capacity of slabs
and footings. Shear reinforcement should be extended to the sec-
tion where
v
f
is not greater than 0:19kffiffiffiffi
f
0
c
p, but at least a distance
of 2dfrom the column face, kis a factor to account for low-density
concrete. In the zone reinforced by headed shear reinforcement,
the shear stress resistance of the concrete
v
c
, shall be 0:28kffiffiffiffi
f
0
c
p.
Stirrups could be used as shear reinforcement provided that the
overall thickness of the slab is not less than 300 mm. For slabs
without shear reinforcement, the value of ffiffiffiffi
f
0
c
pthat calculates the
factored shear stress resistance (
v
r
=
v
c
+v
s
), shall not exceed 8
(MPa), where;
v
r
refers to total resistance shear stress as defined
by CSA-A23.3-04 and
v
s
is shear stress resistance provided by shear
reinforcement.
ACI 318-08 [12] and 421.1R-08 [15] provide out the same limits
for shear reinforcements design of slabs. ACI gives recommenda-
tions for the design of shear reinforcements by using shear studs
E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13 3

in slabs. The code [12] adopts the recommendations of ACI 421.1R-
08 [15] for the headed shear reinforcement in slabs. Shear studs
have proven to be effective in increasing the strength and ductility
of slab-column connections. ACI 421.1R-08 [15] suggests treating a
shear stud as the equivalent of a vertical branch of a stirrup. ACI
421.1R-08 [15] recommends the following allowable values for
v
n
,
v
c
, and s:
v
n
¼
v
c
þ
v
s
60:66 ffiffiffiffi
f
0
c
qð7Þ
v
c
¼0:25 ffiffiffiffi
f
0
c
qð8Þ
The strength of the shear studs
v
s
is calculated as follows:
v
s
¼A
v
f
y
v
b
o
sð9Þ
where f
yv
is the specified yield strength of shear reinforcement, b
o
is
perimeter of critical section for shear in slabs and footings and
v
c
is
the nominal shear strength of concrete in absence of shear rein-
forcement as defined by ACI 318-08 [12]. The distance between
the column face and the first peripheral line of studs, s
o
, could be
taken as (0.35d<s
o
<0.4d), while the spacing between successive
peripheral lines of studs, s, is given by the following limits [15]:
s0:75dwhen
v
u
=/0:5ffiffiffiffi
f
0
c
p
0:50dwhen
v
u
=/>0:5ffiffiffiffi
f
0
c
p
(ð10Þ
The shear-reinforced zone has to extend such that the maximum
shear stress
v
u
at d/2 from the outermost peripheral line of studs
is less than 0:17 ffiffiffiffi
f
0
c
p. The justification for these higher values is
mainly due to the almost slip-free anchorage of the studs and the
fact that the mechanical anchorage at the top and bottom of the
studs is capable of developing forces in excess of the specified yield
strength at all sections of the stud stem.
ACI 421.2R-10 [16] recommends a minimum amount of shear
reinforcement in earthquake-resistant flat plates at their connec-
tions. However, the design guide [16] does not consider the size ef-
fect factor. The design procedure recommended in this guide [16]
was developed based on numerical studies (finite element method)
and experimental research on reinforced concrete slabs subjected
to cyclic drift reversals that simulate seismic effects. The design
guide gives several design examples for different flat plate-column
connections provided with the minimum amount of shear
reinforcement.
5.2. European codes requirements
EC 2 [14] requires in Section 9.2.2 that reinforced-concrete slabs
subjected to punching loads should be provided with a minimum
shear reinforcement that is the same ratio necessary for beams,
and this is based on a simplified truss model (Fig. 1). Where shear
reinforcement is required, the area of a link leg (or equivalent),
A
sw,min
, is given by:
A
sw;min
1:5 sin
a
þcos
a
S
r
S
t
0:08 ffiffiffiffiffi
f
ck
p
f
yk
ð11Þ
where S
r
is the spacing of shear links in the radial direction, S
t
is the
spacing of shear links in the tangential direction, f
ck
is the character-
istic compressive cylinder strength of concrete at 28 days ((MPa)),
f
yk
is the characteristic yield strength of reinforcement ((MPa)). A
slab in which shear reinforcement is provided should have a depth
of at least 200 mm.
6. Shear sandwich model
Generally, slab elements are subjected to eight stress resultants,
i.e., three membrane force components (N
x
,N
y
, and N
xy
=N
yx
); two
transverse shear force components (V
xz
, and V
yz
); two flexural mo-
ments (M
x
and M
y
); and twisting moment (M
xy
=M
yx
); see Fig. 2a.
Marti [17] introduced a sandwich model whose covers are assumed
to carry moments and membrane forces, while the transverse shear
forces are assigned to the core; see Fig. 2b. As a simple approxima-
tion, Marti [17] assumed that the middle planes of the cover ele-
ments will coincide with the middle planes of the reinforcing
meshes close to the slab surfaces. Assuming equal thicknesses of
the cover elements at the top and bottom c, the lever arm of the
in-plane forces in the cover elements (d
v
) that is equal to the effec-
tive shear depth of the core, is given by d
v
=hc, where his the slab
thickness. Rizk and Marzouk [18] recently recommended an equa-
tion to calculate minimum flexural reinforcement for plates and
two-way slabs that is based on the modified sandwich shear model
by Marti [17]. The proposed model accounts for the torsional
Fig. 1. Truss model and notation for shear reinforced members [14].
4E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

moment effect, which also utilizes fracture mechanics concepts to
account for the size effect.
6.1. Uncracked core
Provided that the nominal shear stress,
v
o
, which is due to the
principal shear force (V
o
/d
v
) does not exceed the limit of 0:17 ffiffiffiffi
f
0
c
p,
one may assume that there are no diagonal cracks in the core. In
this case, a state of pure shear develops within the core as shown
in Fig. 2c, and the transverse shear force at a section has no effect
on the in-plane forces in the sandwich covers. Thus, no transverse
reinforcement has to be provided, and the in-plane reinforcement
need not be strengthened to account for transverse shear.
6.2. Cracked core
If V
o
/d
v
exceeds 0:17 ffiffiffiffi
f
0
c
p, diagonal cracking of the core must be
taken into consideration. Transverse reinforcement is necessary
and the amount of in-plane reinforcement must be increased to
account for the transverse shear. The horizontal component of
the diagonal compression in the core (V
o
cot h), must be compen-
sated by membrane forces in the sandwich covers that can be
determined from the free-body diagram as shown in Fig. 2d, his
the angle of inclination of the diagonal compressive stress field
with the XY plane.
Using transverse reinforcements that are normal to the plane of
the slab, the necessary transverse reinforcement ratio amounts to:
q
z;min
¼V
o
tan h
d
v
f
y
ð12Þ
7. Size effect
For design engineers, the size effect is a useful concept that is
based on fracture mechanics. The size of the fracture process zone
is represented by a material property called the characteristic
length, l
ch
. It relates the fracture properties of the concrete, such
as the modulus of elasticity; E
c
, the fracture energy; G
f
and the
tensile strength; f
t
and is calculated as:
l
ch
¼E
c
G
f
f
2
t
ð13Þ
where G
f
is defined as the amount of energy required to cause one
unit area of a crack, hence it can be obtained as the area under the
‘‘load-crack width opening curve’’. It should be noted that, the char-
acteristic length has no physical correspondence, i.e., it is not a real
length that can be measured. A higher value of l
ch
reflects that the
material is less brittle and a smaller value means that the material
is more brittle. In an earlier investigation by Marzouk and Chen
[19], the fracture energy, G
f
was determined experimentally for
high-strength concrete to be 160 N/m co(MPa)red to 110 N/m for
normal strength concrete. The characteristic length, l
ch
, was esti-
mated to have an average value of 500 and 250 mm for normal
and high-strength concrete, respectively. Consideration of l
ch
neces-
sitates a method for experimental determination of G
f
similar to
Nyx V
xz
N
x
Mx
dv-c
1
1
d
1
d
d
o
v
o
vcot
o
o
v
cot
(a) (c)
(b) (d)
o
o
v/sin
o
v
o
-
-
45°
1
Top Cover
c
Nx
2
Z
X
Y
Z
X
Y
c
1
2v cot
c
1
2v cot
V
xz
V
yz
c
Bottom Cover
-
Mx
dv
Nyx
2-Myx
dv
Ny
2-My
dv
Nx
2+Mx
dv
Nyx
2+Myx
dv
Ny
2+My
dv
Z
X
Y
Myx
Nxy
V
yz
Ny
Mxy
My
Fig. 2. Statics of slab elements: (a) stress resultants; (b) sandwich model; (c) pure shear in uncracked core; (d) diagonal compression field in cracked core.
E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13 5

that recommended by the CEB-FIP [20]. The values of G
f
and l
ch
for a
wide range of concrete strength have been reported earlier by
Hillerborg [21].
7.1. Empirical equations to predict the characteristic length
Several researchers have proposed empirical code-like equa-
tions to predict the characteristic length by knowing only the com-
pressive strength of the concrete. Two of such equations are those
of Hilsdorf and Brameshuber [22]; and Zhou et al. [23]. Hilsdorf
and Brameshuber [22] suggested the following formula:
l
ch
¼600ðf
0
c
Þ
0:3
ðmmÞð14Þ
The work of Zhou et al. [23], based on curve fitting, proposed the
following simple equation:
l
ch
¼3:84f
0
c
þ580 ðmmÞð15Þ
The results of these investigations suggest that the characteristic
length is based solely on concrete compressive strength. In the cur-
rent research, it is found that the use of Eq. (15) gives more consis-
tent results.
8. Transverse shear force in a thick concrete plate
8.1. Shear transfer in the interior of a slab
Using Cartesian coordinates with axes Xand Yin the plane of
the slab, and Zperpendicular to this plane, Fig. 3 illustrates the
positive directions for slab element internal forces (V
xz
and V
yz
).
Note that these slab element internal forces are forces per unit
length that act on the mid-surface of the slab element. The inner
slab core must transmit transverse shear forces as assumed by
the sandwich model. In the following, we shall analyze the behav-
ior of the core with unit length along the Xand Yaxes, where thick-
ness of the core is d
v
, subjected to shear forces orthogonal to V
xz
and V
yz
, as shown in Fig. 2b. In the following, we will distinguish
two possible mechanisms for the shear force transfer.
8.2. Slabs without shear reinforcement
When dimensioning slabs, we usually limit the nominal shear
stress,
v
o
, that act on critical defined sections so as to omit shear
reinforcement steel. Most design codes present formulae, where
the design punching load is produced by the nominal shear
strength of the design and the area of a chosen control surface.
Depending on the method used, the critical shear section for check-
ing punching shear in slabs is usually situated between 0.5 and 2
times the effective depth from the edge of the load or the reaction.
For example, ACI 318-08 [12] requires that the nominal shear
stress for slabs without shear reinforcement to be not more than:
v
c
¼0:33 ffiffiffiffi
f
0
c
qð16Þ
where f
0
c
is the specified compressive strength of concrete,
(MPa).
8.3. Slabs with shear reinforcement
When Eq. (16) cannot be applied, the resisting mechanism
should be analogous to that of a beam (Fig. 2d), which is locally ori-
ented in accordance to the principal shear direction. The diagonal
compression field (
v
o
/sin h) makes an angle hwith the XY plan,
and is the resultant of the sum of two component forces:
v
o
/sin h
parallel to the XY plane, and
v
o
, parallel to the Zaxis. Assuming that
vertical shear studs are used, the following equations must be
verified:
f
cw
¼
v
o
sin hf
2
d
v
cos hð17Þ
where f
cw
is the diagonal compressive stress in concrete, and f
2
is
the crushing strength of diagonal cracked concrete given by the
following equation:
f
2
¼f
0
c
0:8þ170
e
1
0:85f
0
c
ð18Þ
where
e
1
is the principal tensile strain in cracked concrete due to
the factored loads.
Tensile stresses in the web reinforcement (headed studs):
f
sz
¼
v
o
A
sz
f
yz
sd
v
cot hð19Þ
Additional truss axial stress in the tension and compression covers
(outer layers of the model) is given by:
D
f¼
v
o
cot hð20Þ
Angle his subjected to the same limitations that apply to line ele-
ments subjected to shear forces. According to part 1 of EC 2 [14],
angle hcan be chosen freely within the limits of 22°6h645°
(1 6cot h62.5). The selection of hmust be based primarily on
practical considerations in detailing. A low value of hallows for
large shear reinforcement spacing and facilitates the casting of con-
crete, but requires more longitudinal reinforcement.
9. Minimum shear reinforcement ratio-analysis based on
compression field theory
Fig. 4 illustrates the stress fields in the core of the two-way slab
before and after cracking. Prior to cracking, the shear is carried
equally by diagonal tensile and diagonal compressive stresses at
45°[24].
V
cr
¼f
2
ðbd
v
cos hÞsin hþf
1
ðbd
v
sin hÞcos hð21Þ
Just prior to cracking:
f
1
¼f
2
¼f
ct
ð22Þ
where f
1
and f
2
are the principal tensile and compressive stresses
and f
ct
is the direct tensile strength of concrete as determined by
the direct tension test [19] or any other fracture mechanics test,
f
ct
¼0:33 ffiffiffiffi
f
0
c
p.
Cracking is assumed to occur when the principal tensile
strength reaches the tensile strength of concrete in biaxial ten-
sion–tension.
vxz
X
Y
Z
vxz
v
y
z
vyz
Fig. 3. Slab element internal forces V
xz
and V
yz
.
6E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

V
cr
¼2f
ct
bd
v
sin hcos hð23Þ
V
cr
¼A
z;min
f
y
¼
q
z;min
bh cot hf
y
ð24Þ
q
z;min
bh cot hf
y
¼2f
ct
bd
v
sin hcos hð25Þ
q
z;min
bh cot hf
y
¼1:44f
ct
bh sin hcos hð26Þ
q
z;min
¼1:44f
ct
f
y
sin
2
hð27Þ
where d
v
is the effective shear depth taken as the greater of 0.9dor
0.72h.
Introducing the term (l
ch
/h)
0.33
to account for the size effect,
Eq. (27) can be written as follows:
q
z;min
¼1:44f
ct
f
y
sin
2
hðl
ch
=hÞ
0:33
ð28Þ
For
h¼30

q
z;min
¼0:12 ffiffiffiffi
f
0
c
p
f
y
ðl
ch
=hÞ
0:33
ð29Þ
The angle of the failure plane (h) normally varies between 22°and
45°based on experimental findings [25,26], tending to the smaller
value for the minimum shear reinforcement ratio. Hallgren et al.
[27] found that the shear crack propagated from the plane of the
flexural reinforcement to the slab-column root was at an angle of
about 50°to 60°, measured between the shear crack and the hori-
zontal plane, this is based on tests on column footings. This is a con-
siderably steeper angle than the shear crack angles observed in
punching shear tests of more slender slabs. Based on these experi-
mental findings the angle of failure plane hcould be assumed to have
a value of 22°for normal slab thicknesses of less than 250 mm, 30°
for medium slab thicknesses of 250–500 mm and 45°for thick slabs
that are greater than 500 mm. In the current investigation, his taken
as equal to 30°.
10. Minimum shear reinforcement ratio-analysis based on the
diagonal cracking load
Consider a concrete slab with a rectangular cross section sub-
jected to shear. From the elastic shear stress distribution (second
degree parabola, Fig. 5), the shear crack appears when
s
max
= 1.25
f
ct
[25], where f
ct
is the tensile strength of concrete, f
ct
¼0:33 ffiffiffiffi
f
0
c
p;
and f
0
c
is the cylinder compressive strength of concrete in (MPa).
According to the compression field theory approach, the shear
crack appears when
s
max
=2 f
ct
[24]. Hence, the associated shear
force is calculated as:
V
cr
¼2
3bh
s
max
¼0:83bhf
ct
ð30Þ
where bis the unit slab width, his the height of the slab and f
ct
is
the direct tensile strength of concrete.
Considering the failure plane to be inclined at an angle hto the
slab axis and the shear studs at the right angle to the slab axis, from
the equilibrium of forces in the vertical direction along a projected
length of the failure plane (hcot h), the minimum shear reinforce-
ment is calculated as follows:
V
cr
¼A
z;min
f
y
ð31Þ
A
z;min
¼
q
z;min
bh cot hð32Þ
q
z;min
¼A
z;min
bh cot h¼V
cr
bh cot hf
y
ð33Þ
q
z;min
¼0:83 f
ct
f
y
tan hð34Þ
where A
z,min
is the area of minimum shear reinforcement and
q
z,min
is
the minimum shear reinforcement ratio. ACI 318-08 [12] design code
does not account for the fact that the shear stress that can cause fail-
ure of members without shear reinforcement decreases as the depth
of the member increases, which is known as the size effect factor.
Modern European codes of practice such as EC 2 [14] provide a factor
k¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þð200=dÞ
pto account for size effect and as the depth of the
member increases the value of kas well as the predicted punching
shear resistance decreases. This means that by increasing the mem-
ber size, the behavior of the member becomes more brittle, and hence
Fig. 4. Stress fields in the core of reinforced concrete slab.
E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13 7

more shear reinforcement is required to enhance the behavior of
thick members. The previous equation, Eq. (34), could be modified
to account for the size effect factor by the inclusion of the brittleness
ratio (h/l
ch
) as proposed earlier by Marzouk et al. [28]. However, it
should be noted that any size effect factor that is derived using Linear
Elastic Fracture Mechanics (LEFM) could be applied.
The size of the fracture process zone, characteristic length, l
ch
could be determined using tension test similar to that recom-
mended by the CEB-FIP [20] or could be assumed to have an aver-
age value of 500 and 250 mm for normal and high-strength
concrete, respectively as determined by Marzouk and Chen [19].
However, for simplicity the characteristic length, l
ch
, could be
determined using the empirical equation (Eq. (15)) suggested by
Zhou et al. [23]. Introducing the term (l
ch
/h)
0.33
to account for the
size effect, Eq. (34) can be written as follows:
q
z;min
¼0:83 f
ct
f
y
tan hðl
ch
=hÞ
0:33
ð35Þ
For
h¼30

q
z;min
¼0:16 ffiffiffiffi
f
0
c
p
f
y
ðl
ch
=hÞ
0:33
ð36Þ
11. Comparison of proposed formula with different design
codes formulae
Current Code practices for minimum shear reinforcement (in
US, Canada and Europe) are based on one-way/beam shear. For
example, EC 2 [14] requires that reinforced-concrete slabs sub-
jected to punching loads be provided with a minimum shear rein-
forcement that is the same ratio necessary for beams. This amount
is relevant for punching failure mode that develops outside the
shear-reinforced zone. Hence, this practice is not convenient for
punching within the shear-reinforced zone that is desirable, since
it allows flexural reinforcement to develop full radius of yield. In
Fig. 5. Shear stress distribution in a slab cross section just before the formation of shear cracks.
0.00
0.05
0.10
0.15
0.20
0.25
0.30
020 40 60 80 100 120 140
Concrete strength (MPa)
ρz,min%
AASHTO LRFD (2004)
ACI 318 (2008)
CSA-A23.3 (2004)
CSA-S6 (2006)
EC 2 (2004)
Fig. 6. Comparison of minimum shear reinforcement requirements for beams by various design codes as a function of concrete compressive strength f
0
c
.
8E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

that situation, punching shear failure is eliminated and is trans-
formed into ductile flexural failure.
To verify the validity of the proposed formula, a comparison
between the proposed formula (Eq. (29)) with different design codes
formulae for calculating minimum shear reinforcement for beams
and two-way slabs is presented in Figs. 6 and 7 for three different
heights (250, 500 and 1000 mm). A summary of the minimum shear
reinforcement requirements provided by most popular design con-
crete codes is presented in Table 1. In this comparison, yield strength
of shear reinforcement, f
yv
, is assumed to be equal to 400 (MPa).
Analyzing the results of the comparison, it is obvious that
q
z,min
increases as the concrete strength increases. The amount of shear
(a)h=250 mm
(b)h=500 mm
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0 20406080100120140
Concrete strength (MPa)
ρz,min%
EC 2 (2004)
Proposed Formula
0.00
0.05
0.10
0.15
0.20
0.25
0 20 40 60 80 100 120 140
Concrete strength (MPa)
ρz,min%
EC 2 (2004)
Proposed Formula
(c) h=1000 mm
0.00
0.05
0.10
0.15
0.20
0.25
0 20 40 60 80 100 120 140
Concrete strength (MPa)
ρz,min%
EC 2 (2004)
Proposed Formula
Fig. 7. Comparison of minimum shear reinforcement requirements for two-way slabs by EC 2 and proposed formula as a function of concrete compressive strength f
0
c
: (a)
h= 250 mm; (b) h= 500 mm; (c) h= 1000 mm.
E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13 9

reinforcement calculated by the proposed formula (Eq. (29))is
slightly higher than the amount of shear reinforcement calculated
by EC 2 [14] design code requirement. It could be noted also that
as the beam height increases the value of minimum shear reinforce-
ment required by the proposed formula decreases. The ACI 318-08
[12] and CSA-A23.3-04 [11] expressions give similar amounts of
shear reinforcement, while the EC 2 [14] expression gives a larger
amount of minimum shear reinforcement for all concrete strengths.
Both ACI 318-08 [12] and CSA-A23.3-04 [11] require that a mini-
mum amount of shear reinforcement be provided in all reinforced
concrete flexural members where the factored shear force exceeds
one-half the factored shear resistance of the concrete, except for
slabs and footings. In contrast, EC 2 [14] requires that all beams that
contribute significantly to the overall resistance and stability of the
structure must contain at least a minimum amount of shear rein-
forcement. The following section presents an experimental valida-
tion required to verify the proposed theoretical analysis.
12. Experimental validation of the proposed formula
Shear stresses in vicinity of slab-column connections may cause
failure by punching with load below that of flexural failure. Shear
reinforcement is generally used in this region to increase the
strength and the ductility of the slabs and one of the types of this
reinforcement is the one consisting of studs. The following analysis
is done using experimental work conducted by different research-
ers [4,9,29]. Test results used were for either shear stud rails or
T-headed shear studs since these are considered to be the only
practical shear reinforcement types to be used in concrete slab
construction. Details of tested slabs are shown in Table 2.Table 3
presents a comparison of: the actual shear reinforcement ratio
and the ratio of minimum shear reinforcement by volume; k=
P
z,actual
/P
Reference
and k
min
=P
z,min
/P
Reference
value of failure loads of
test slabs compared to failure load of companion test slabs without
shear reinforcement, krefers to shear reinforcement index.
Analyzing test results presented in Table 3, it could be concluded
that using shear reinforcement results in increased shear capacity
compared to the slabs without shear reinforcement. It could also
be noted that the efficiency of shear reinforcement increases with
increasing the slab effective depth. Comparison of actual shear rein-
forcement index, k, with the minimum shear reinforcement index,
k
min
, is shown in Table 3. The minimum shear reinforcement index,
k
min
, is the value that defines the amount of minimum shear rein-
forcement of flat slabs that leads to a punching shear failure surface
that crosses the shear reinforcement, in order to avoid a sudden
Table 1
Minimum shear reinforcement requirements in different design codes.
Code Minimum area of shear reinforcement Maximum spacing of shear reinforcement Shear strength
reduction factor
Shear type
ACI 318-08 [12] A
v
min
¼0:062 ffiffiffiffi
f
0
c
p
bs
f
y

0:35b
w
s
f
y
s
d
2
600 mm /
v
= 0.75 One-way shear
If /
v
V
s
>0:33/
v
ffiffiffiffi
f
0
c
pb
w
d
Then s
d
4
300 mm
CSA-A23.3-04 [11] A
v
min
¼0:06 ffiffiffiffi
f
0
c
p
b
w
s
f
y
s0:7d
v
600 mm /
c
= 0.65 One-way shear
If V
f
>0:125k/
c
f
0
c
b
w
d
v
Then s0:35d
v
300 mm
CAN/CSA-S6 [30] A
v
min
¼0:15 f
rbs
f
y
s0:75 d
v
600 mm /
c
= 0.75 One-way shear
f
r
¼0:6kffiffiffiffi
f
0
c
pIf V
f
>0:10/
c
f
0
c
bd
v
Then s0:33 d
v
300 mm
NS 3473 E [13] A
s
0:2A
cf
tk
ðMPaÞ
f
sk
sin as
max
¼0:6h
0
ð1þcot aÞh
0
One-way shear
s
max
¼500 mm
EC 2 [14] A
sw;min
1:5 sin aþcos a
S
r
S
t
0:08 ffiffiffiffiffiffi
f
ck
p
f
yk
s
t;max
0:75d600 mm One-way shear and two-way shear
s
max
¼0:75dð1þcot aÞ
Table 2
Details of test specimens by different researchers.
Test
no.
Slab
thickness h
(mm)
Slab depth
d(mm)
Column
side C(mm)
Concrete comp.
strength f
0
c
(MPa)
Flexure
reinforcement ratio
q
(%)
Stud yield
strength f
yv
(MPa)
Stud
no.
Stud Area
(mm
2
)
Stud
spacing s
(mm)
2 160 124 250 27.7 1.54 465 8 71 62 Birkle and
Dilger [29]4 160 124 250 36.1 1.54 393 8 71 93
8 230 190 300 35.0 1.30 460 8 71 95
9 230 190 300 36.1 1.30 460 8 71 143
11 300 260 350 30.0 1.10 409 8 127 130
12 300 260 350 33.8 1.10 409 8 127 195
HS22 150 120 250 60.0 1.09 450 12 200 90 Marzouk and
Jiang [4]HS23 150 120 250 60.0 1.09 450 12 100 90
L2 130 89 150
a
39.0 1.38 624 8 20 42 Vaz et al. [9]
L3 130 87 150
a
39.0 1.38 624 8 20 42
L4 130 89 150
a
39.0 1.38 624 8 20 42
L6 130 91 150
a
38.9 1.38 708 4 14 67
L7 130 89 150
a
39.1 1.38 708 5 14 42
L8 130 90 150
*
39.2 1.38 708 5 14 67
L9 130 91 150
a
39.4 1.38 708 6 14 67
a
Circular column.
10 E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

failure. All slabs having a shear index, k, less than 0.55 failed in a
ductile failure with punching shear failure surface crosses shear
reinforcement. In the meantime all slabs having a shear index, k,
greater than 0.55 failed in a brittle failure with punching shear
failure surface outside the shear reinforcement region.
13. Discussion
Most design codes usually provide simple or general methods
for beams along with simple rules on the required minimum shear
reinforcement, and such requirements are usually sufficient en-
ough for beam designs. The situation is different for slab designs,
as most design codes generally try to avoid shear reinforcement
requirements. However, such simple and conservative design
methods provided by different design codes are not sufficient for
thick plates where there is a good possibility of brittle failure due
to punching shear cracks. A consistent procedure for design along
with minimum shear reinforcement requirements for such thick
plates is practically sufficient. Minimum shear reinforcement is
provided for plates to avoid shear cracks, help maintain the aggre-
gate interlock after shear cracking, reduce shear crack openings and
increase the strength by 10–30% [9]. One more reason to provide
shear reinforcement requirements is to allow using a slab with a
large amount of flexural reinforcement ratio. Tests performed by
Jaeger and Marti [8] revealed that the addition of transverse rein-
forcements with geometrical reinforcement ratios of approxi-
mately 0.3% and 0.6% (minimum shear reinforcement ratios)
changed the failure modes to ductile flexural failures; the tests
without transverse reinforcements showed a significant influence
of slab thickness on shear strength. No such size effect was ob-
served for the tests with transverse reinforcement. Codes thus spec-
ify upper limits to the shear capacity, which is a function of the
concrete compressive strength. It should also be noted that due to
the difficulties of anchoring shear reinforcement in thin slabs, in
EC 2 [14] a minimum slab thickness of 200 mm is stipulated. Below
this minimum slab thickness, one is not permitted to increase the
shear capacity by providing shear reinforcement. In order that the
presence of shear reinforcement may enhance the shear strength
of a section, it is necessary that it should raise the shear capacity
above the shear cracking load. Codes of practice give the minimum
amount of shear reinforcement that is necessary to satisfy the
requirements for beams, but in general, this does not apply to slabs.
This is because, very often, in past practice, shear reinforcement has
not been provided in slabs that have performed satisfactorily in ser-
vice. In addition, it is thought that a slab has the ability to redistrib-
ute shear forces from weak to adjoining strong areas.
14. Design guidelines
Minimum shear reinforcement is recommended for thick con-
crete plates, slab-column connections and concrete walls to pre-
vent brittle shear failure in vicinity of concentrated loads. The
existence of minimum transverse reinforcement is needed to force
punching shear cracks to develop within the shear-reinforced zone.
Shear strength is thus governed by the contribution of concrete
and of the transverse reinforcement. This failure type is desirable,
since it allows flexural reinforcement to develop full radius of
yield. In that situation, punching shear failure is eliminated and
is transformed into ductile flexural failure. To achieve this, the
shear reinforcement needs to be well anchored and to have enough
ductility (i.e. headed studs).
The minimum amount of shear reinforcement is recommended
for slabs and walls thicker than 250 mm. This value is a reflection
of the measured characteristic length for high strength concrete of
70 (MPa). This recommendation is also based on and supported by
previous research done at Memorial University. For slabs, the shear
reinforcement is recommended in the vicinity of connections with
columns. In walls, the shear reinforcement is recommended for
the area that can be subjected to a significant concentrated trans-
verse load (transverse ice pressure). Two models are presented in
Table 3
Comparison between the calculated values using proposed formula with the measured experimental values for test specimens by different researchers.
Test no.
q
z,actual
(%)
q
z,min
(%) A
z,min
f
yv
, (kN) P
Test
(kN) k¼
P
z;actual
P
Reference
k
min
¼
P
z;min
P
Reference
P
z;min
P
Test
P
Test
P
Reference
Failure surface
2 0.92 0.18 103 634 0.55 0.21 0.21 1.31 Outside
4 0.61 0.24 173 574 0.46 0.36 0.27 1.19 Inside
8 0.50 0.18 186 1050 0.32 0.23 0.23 1.27 Inside
9 0.33 0.18 283 1091 0.32 0.34 0.27 1.32 Inside
11 0.56 0.17 255 1620 0.40 0.24 0.24 1.55 Inside
12 0.37 0.18 403 1520 0.40 0.38 0.25 1.45 Inside
HS22 0.95 0.26 208 605 2.11 0.41 0.14 1.18 Outside
HS23 0.67 0.26 208 590 1.41 0.41 0.34 1.15 Outside
L2 0.51 0.17 52 321 0.69 0.26 0.26 1.12 Outside
L3 0.51 0.17 52 325 1.03 0.26 0.16 1.14 Outside
L4 0.51 0.17 52 357 1.03 0.26 0.16 1.25 Outside
L6 0.11 0.15 83 300 0.27 0.29 0.29 1.05 Inside
L7 0.22 0.15 52 303 0.51 0.18 0.17 1.06 Inside
L8 0.14 0.15 83 309 0.34 0.29 0.27 1.08 Inside
L9 0.17 0.15 83 315 0.41 0.29 0.27 1.10 Inside
Fig. 8. Arrangement of T-headed minimum shear reinforcement.
E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13 11

the current research that can be used to calculate minimum shear
reinforcement required to prevent brittle shear failure for thick con-
crete plates and walls in the vicinity of concentrated loads. Both
models account for the slab size effect by using principles of
fracture mechanics.
A proposed arrangement of shear studs in a cross pattern is rec-
ommended for slab-column connections, however, a radial distri-
bution of shear studs such as that recommended by European
standards (i.e. EC 2 [14]) could be used. For walls, headed bars or
legs of stirrups, are recommended to be arranged in an orthogonal
directions such as tests conducted by Jaeger and Marti [8].
For slabs, the distance between the column faces and the inner-
most peripheral line of studs should not exceed 0.5d; the spacing
between peripheral lines could be taken equal to 0.5dor 0.75d;
as recommended by ACI 421.1R-08 [15]. For walls or thick plates,
the spacing between orthogonal transverse reinforcement could
be taken equal to 0.75d. The higher limit for sis more desirable,
especially for offshore concrete structures where steel congestion
represents a significant challenge. Nonetheless, test results for
higher shear reinforcement spacing, s=dor more, are needed
and further research investigation for thick plates is required.
It is recommended to extend the distance between the column
faces and the outer-most peripheral line of shear stud reinforce-
ment to a distance of not less than 3.5din order to force shear
cracks to develop within the shear reinforced zone. According to
Birkle and Dilger [29], all specimens with shear reinforced zone
that extended to approximately 3.5dfrom the column faces failed
in punching inside the shear reinforced zone.
15. Summary of design procedure
For thick slabs more than 250 mm thickness, the designer can
chose one of proposed two models (Eqs. (29) or (36)) to calculate
the required minimum shear reinforcement area. The punching
failure load using design equation proposed by CEB-FIP [20] model
code or EC 2 [14] cab be used. These equations are recommended,
since these two design codes account for the size effect shear reduc-
tion factor. For plates and slabs that do not meet the requirements
for punching shear, slab thickness could be increased or shear rein-
forcement can be recommended.
For reinforced concrete thick plates over 250 mm thickness, the
spacing between orthogonal transverse reinforcement could be
taken equal to 0.75d. The higher limit for shear reinforcement spac-
ing of dis more desirable for the case of heavy slab reinforcement. It
is recommended to extend the distance between the column faces
and the outer-most peripheral line of shear stud reinforcement to a
distance of not less than 3.5din order to provide adequate shear
reinforcement. Design example is given in Appendix A.
16. Conclusions
The size effect and the minimum shear reinforcement for thick
concrete slabs are ignored by the ACI 318-08 [12] design code.
Europeans codes like EC 2 [14] code recommend the minimum
shear reinforcement for the size effect. Hence ACI 318-08 [12]
design code must be revised to include such critical require-
ments.
Two models are presented that can be used to calculate mini-
mum shear reinforcement required to prevent brittle shear fail-
ure for thick concrete plates and walls in the vicinity of
concentrated loads. The first model is based on the modified
compression field theory while the second model is based on
the diagonal shear cracking load. Both models account for the
slab size effect by using principles of fracture mechanics.
It is recommended that a minimum shear reinforcement is pro-
vided for a slab thickness that is 250 mm and greater. This value
is a reflection of the measured characteristic length for high
strength concrete of 70 (MPa). For slabs, the shear reinforce-
ment is recommended in the vicinity of connections with col-
umns. In walls, the shear reinforcement is recommended for
the area that can be subjected to a significant concentrated
transverse load.
A proposed arrangement of shear studs in a cross pattern is
recommended for slabs. For slabs, the distance between the col-
umn faces and the inner-most peripheral line of studs should
not exceed 0.5d; the spacing between peripheral lines could
be taken equal to 0.5dor 0.75d; as recommended by ACI
421.1R-08 [15]. For walls or thick plates, the spacing between
orthogonal transverse reinforcement could be taken equal to
0.75d. The higher limit for sis more desirable, especially for off-
shore concrete structures where steel congestion represents a
significant challenge.
It is recommended to extend the distance between the column
faces and the outer-most peripheral line of shear stud reinforce-
ment to a distance of not less than 3–4d, in order to force shear
cracks to develop within the shear reinforced zone.
Acknowledgments
The authors are grateful to the Natural Sciences and Engineer-
ing Research Council of Canada (NSERC) for providing the funds
for the project.
Appendix A. Design of shear reinforcement for a thick-plate
It is required to design to design the minimum shear reinforce-
ment for a thick concrete slab as follows: slab thickness, h=
350 mm; square column dimension, C= 400 mm; structural depth,
d= 262.5 mm; f
0
c
= 65.4 MPa; and f
yv
= 400 MPa. For further slab
details and full scale test results of the strengthened slab [31].
A.1. Minimum shear reinforcement ratio
A.1.1. Characteristic length, l
ch
The characteristic length, l
ch
could be determined using tension
test similar to as determined by Marzouk and Chen [19]. However,
for simplicity the characteristic length, l
ch
, could be determined
using the empirical equation (Eq. (15)) suggested by Zhou et al.
[23] as follows:
l
ch
¼3:84f
0
c
þ580 ¼329 mm
The required minimum shear reinforcement ratio could be calcu-
lated according to the compression field theory model (Eq. (29))
as follows:
q
z;min
¼0:12 ffiffiffiffi
f
0
c
p
f
y
ðl
ch
=hÞ
0:33
q
z;min
¼0:12 ffiffiffiffiffiffiffiffiffiffi
65:4
p
400 ð329=350Þ
0:33
¼0:24%
Required minimum shear reinforcement ratio according EC 2 [14]:
q
w;min
ðEC2Þ¼0:08 ffiffiffiffiffi
f
ck
p
f
yk
¼0:08 ffiffiffiffiffiffiffiffiffiffi
65:4
p
400 ¼0:16%
A.2. Details of shear studs reinforcement
The shear reinforcement ratio in the cross-pattern arrangement
(Fig. 8) is equal to the cross-sectional area of the studs on a periph-
12 E. Rizk, H. Marzouk / Engineering Structures 46 (2013) 1–13

eral line divided by dtimes the periphery of the column as calcu-
lated by the following equation:
q
z
¼A
v
4Cd
where dis the slab effective depth, Cis the column side length, the
perimeter of the critical section is 4C and A
v
is the area of the studs
on a peripheral line. The same equation is used for flat plate or a
wall plate and the critical perimeter is determined based on the
engineering judgment.
A
v
;min
¼0:24
100 4400 262:5¼1008 mm
2
=peripheral line
Twelve shear studs each of cross-sectional are 130 mm
2
are used
per each peripheral line (Fig. 8), the shear reinforcement should
be extended to a distance of 3.5dfrom the column face. The spacing
between the column face and the first peripheral line could be taken
equal to s
o
= 0.5d= 130 mm, and the spacing between peripheral
lines could be taken equal to s= 0.75d= 190 mm.
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