U N I V E R S I T Y OF C A L G A R Y
Reinforced Concrete Beam Design for Shear
by
Hongge (Gordon) Wang
A THESIS
S U B M I T T E D TO T H E F A C U L T Y OF G R A D U A T E STUDIES
IN P A R T I A L F U L F I L L M E N T OF T H E R E Q U I R E M E N T S F O R T H E
D E G R E E OF M A S T E R OF E N G I N E E R I N G
D E P A R T M E N T OF CIVIL E N G I N E E R I N G
CALGARY, ALBERTA
N O V E M B E R , 2002
© Hongge (Gordon) Wang 2002
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ABSTRACT
The two methods for design of shear adopted by the present C S A Standard A23.3
are either too simple or too complicated. That presents the need for ongoing research to
establish a new design guideline for shear design.
Recent studies by Dr. Loov and others have shown that shear design can be based
on the shear resistance along potential inclined crack and slip planes. Because the basic
equations for this shear design method are derived from "shear friction" theories, we call
it "the shear friction method".
In this thesis an entire review of shear design methods has been given and a
method of shear design based on shear friction theories has been introduced. From
comparison calculations with present code methods it is proved that "the shear friction
method" provides a simpler and more accurate approach for shear design.
iii
ACKNOWLEDGMENTS
I am extremely grateful to my supervisor, Dr. Robert E. Loov for his endless
patience and guidance throughout the course of this program.
I would also like to thank my current employer Kassian Dyck & Associates for
giving me the chance to finish this thesis.
Finally I wish to thank my wife Candy for her support and encouragement.
iv
T A B L E OF C O N T E N T S
Cover Page
i
Approval page
ii
Abstract
iii
Acknowledgements
iv
Table of Contents
v
List of Tables
viii
List of Figures
ix
Notation
xiii
CHAPTER ONE: INTRODUCTION
1
1.1 General
1
1.2 Code Review
2
1.3 Scope of Study
3
1.4 Thesis Organization
3
C H A P T E R T W O : B A S I C S H E A R THEORIES
5
2.1 Homogeneous Beam
5
2.2 Beam Cracking Modes
9
2.3 Shear Transfer Mechanisms
10
2.4 Shear Failure Modes
12
2.4.1 Beams without Shear Reinforcement
12
2.4.2 Beams with Shear Reinforcement
16
2.5 Factors Affecting the Shear Strength
16
2.5.1 Tensile Strength of Concrete
16
2.5.2 Longitudinal Reinforcement
17
2.5.3 Shear Span-to-depth Ratio, a/d
17
2.5.4 Size of Beams
19
v
2.5.5 Axial Forces
20
2.5.6 Web Reinforcement
20
C H A P T E R T H R E E : S H E A R DESIGN - C S A S T A N D A R D A23.3-94
22
3.1 General
22
3.2 Simplified Method
22
3.2.1 Shear Supported by Concrete, V
3.2.2 Shear Supported by Stirrups, V
23
c
25
s
3.3 General Method
25
3.3.1 Shear Supported by Concrete, V
3.3.2 Shear Supported by Stirrups, V
26
c g
26
s g
C H A P T E R F O U R : S H E A R DESIGN - S H E A R FRICTION M E T H O D
27
4.1 General
27
4.2 General Equations for Beams Based on Shear Friction
27
4.2.1 Shear Friction Strength
28
4.2.2 Basic Shear Design Equations Based on Work by Loov
31
4.2.3 Approximate Shear Capacity of Concrete
36
4.2.4 Approximate Shear Capacity of Stirrups
40
4.2.5 Approximate Shear Design Equations for Beams with
T>T
40
opl
4.2.6 Critical Shear Failure Angle
41
4.2.7 Beams with Longitudinal Reinforcement T<T ,
op
C H A P T E R F I V E : E X P E R I M E N T A L STUDIES A N D C O M P A R I S O N
44
45
5.1 Application of Shear Friction Method
45
5.2 Test Results in Literature
47
5.2.1 Yoon, Cook and Mitchell's Tests, 1996
48
5.2.2 Saram and Al-Musawi's Tests, 1992
58
vi
5.2.3 Summary of Tests from Literature
71
5.2.3.1 Beams with Shear Reinforcement
79
5.2.3.2 Beams without Shear Reinforcement
93
C H A P T E R SIX: P R O P O S E D C O D E C L A U S E S F O R S H E A R D E S I G N
106
6.1 Proposed Code Clauses for Shear Design
106
6.1.1 Required Shear Resistance
106
6.1.2 Factored Shear Resistance
106
6.1.3 Determination of V
c s f
106
6.1.4 Determination of V
s s f
107
6.1.5 Determination of 0
107
6.1.6 Determination of \|/
108
6.1.7 Limiting Shear Failure Angle
108
6.2 Design Examples
108
C H A P T E R S E V E N : DISCUSSION A N D C O N C L U S I O N
7.1 Conclusions and Recommendations
7.2 Future Research
BIBLIOGRAPHY
111
Ill
112
113
vii
LIST OF T A B L E S
TABLE
5.1 Details of Beam Specimens (Yoon)
5.2 Test Results and Comparison of Predictions (Yoon)
5.3 Details of Beam Specimens (Sarsam)
5.4 Details of Materials (Sarsam)
5.5 Test Results and Comparison of Predictions (Sarsam)
5.6 Details of Specimens with Stirrups
5.7 Details of Specimens without Stirrups
5.8 Comparison of Predictions for Beams with Stirrups
5.9 Comparison of Predictions for Beams without Stirrups
viii
LIST OF FIGURES
FIGURE
2.1
Internal Forces in Beam
5
2.2 Distribution of Flexural Shear Stresses
6
2.3 Principal Stresses
7
2.4a Stress Trajectories
8
2.4b Potential Crack Pattern
8
2.5 A Cracked Beam without Shear Reinforcement (MacGregor, 2000)
9
2.6 A Cracked Beam with Shear Reinforcement (Peng, 1999)
10
2.7 Internal Forces in a Cracked Beam
11
2.8 Effect of a/d on Shear for Beams Without Shear Reinforcement (MacGregor,
2000)
14
2.9 Shear Failure Modes (Pillai, 1983)
15
2.10 Shear Strength vs. Longitudinal Reinforcement (MacGregor, 2000)
17
2.11 Shear Strength vs. a/d (Kani, 1979)
18
2.12 Influence of Member Size on Shear Strength (CSA A23.3-94)
19
2.13 Effect of Axial Loads in Inclined Cracking Shear (MacGregor, 2000)
20
2.14 Distribution of Internal Shears of Beam with Shear
Reinforcement
(MacGregor, 2000)
21
3.1
Comparison of Shear Design Methods and Test Results ( C S A A23.3-94)
24
4.1
Shear Friction Concept (CSA A23.3-94)
28
4.2 Reinforcement Inclined to Potential Failure Cracks ( C S A A23.3-94)
29
4.3
Push-off Test Results (Loov, 1998)
30
4.4 Free Body Diagram of Beam (Loov, 1998)
31
4.5
34
Shear Strength vs. Crack Angle (Loov, 1998)
ix
4.6 Three-dimensional surface of shear strength along all possible failure planes
for beam 544 (Loov, 1998)
34
4.7 Possible Critical Shear Failure Planes (Loov, 1999)
35
4.8 A Tested Beam with Critical Shear Cracks (Peng, 1999)
35
4.9
37
Shear Strength vs. cotO
4.10 Shear Strength vs. cot9 by Eq. 4-15 and Eq. 4-19
38
4.11 Shear Strength vs. Longitudinal Reinforcement of Beam
39
4.12 The Shear Contributions of Concrete and Discrete Stirrups (Loov, 1998)
42
5.1
Details of Beam Specimens and instrumentation (Yoon, 1996)
48
5.2
Effect of Concrete Strength on the Shear Friction Method (Yoon)
51
5.3
Effect of Concrete Strength on the Simplified Method (Yoon)
52
5.4 Effect of Concrete Strength on the General Method (Yoon)
52
5.5
Effect of Stirrup Spacing on the Shear Friction Method (Yoon)
53
5.6 Effect of Shear Reinforcement on the Shear Friction Method (Yoon)
54
5.7 Effect of Stirrup Spacing on the Simplified Method (Yoon)
54
5.8
55
Effect of Shear Reinforcement on the Simplified Method (Yoon)
5.9 Effect of Stirrup Spacing on the General Method (Yoon)
55
5.10 Effect of Shear Reinforcement on the General Method (Yoon)
56
5.11 Effect of Concrete Strength on the Shear Friction Method (Yoon)
57
5.12 Effect of Concrete Strength on the Simplified Method (Yoon)
57
5.13 Effect of Concrete Strength on the General Method (Yoon)
58
5.14 Details of Beam Specimens and Instrumentation (Sarsam,1992)
59
5.15 Effect of the Ratio of Shear span on the Shear Friction Method (Sarsam)
62
5.16 Effect of the Ratio of Shear span on the Simplified Method (Sarsam)
63
5.17 Effect of the Ratio of Shear span on the General Method (Sarsam)
63
5.18 Effect of Concrete Strength on the Shear Friction Method (Sarsam)
64
5.19 Effect of Concrete Strength on the Simplified Method (Sarsam)
65
5.20 Effect of Concrete Strength on the General Method (Sarsam)
65
x
5.21 Effect of Stirrup Spacing on the Shear Friction Method (Sarsam)
66
5.22 Effect of Shear Reinforcement on the Shear Friction Method (Sarsam)
67
5.23 Effect of Stirrup Spacing on the Simplified Method (Sarsam)
67
5.24 Effect of Shear Reinforcement on the Simplified Method (Sarsam)
68
5.25 Effect of Stirrup Spacing on the General Method (Sarsam)
68
5.26 Effect of Shear Reinforcement on the General Method (Sarsam)
69
5.27 Effect of Longitudinal Reinforcement on the Shear Friction Method
(Sarsam)
70
5.28 Effect of Longitudinal Reinforcement on the Simplified Method (Sarsam)
70
5.29 Effect of Longitudinal Reinforcement on the General Method (Sarsam)
71
5.30 Predicted Results by the Shear Friction Method (with stirrups)
80
5.31 Predicted Results by the Simplified Method (with stirrups)
81
5.32 Predicted Results by the General Method (with stirrups)
82
5.33 Effect of the Ratio of Shear Span on the Shear Friction Method (with
stirrups)
84
5.34 Effect of the Ratio of Shear Span on the Simplified Method (with stirrups)
85
5.35 Effect of the Ratio of Shear Span on the General Method (with stirrups)
85
5.36 Effect of Concrete Strength on the Shear Friction Method (with stirrups)
86
5.37 Effect of Concrete Strength on the Simplified Method (with stirrups)
86
5.38 Effect of Concrete Strength on the General Method (with stirrups)
87
5.39 Effect of Stirrup Spacing on the Shear Friction Method (with stirrups)
87
5.40 Effect of Stirrup Spacing on the Simplified Method (with stirrups)
88
5.41 Effect of Stirrup Spacing on the General Method (with stirrups)
88
5.42 Effect of Shear Reinforcement on the Shear Friction Method (with stirrups)
89
5.43 Effect of Shear Reinforcement on the Simplified Method (with stirrups)
89
5.44 Effect of Shear Reinforcement on the General Method (with stirrups)
90
5.45 Effect of Longitudinal Reinforcement on the Shear Friction Method (with
stirrups)
90
xi
5.46 Effect of Longitudinal Reinforcement on the Simplified Method (with
stirrups)
91
5.47 Effect of Longitudinal Reinforcement on the General Method (with stirrups)
91
5.48 Effect of Beam Depth on the Shear Friction Method (with stirrups)
92
5.49 Effect of Beam Depth on the Simplified Method (with stirrups)
92
5.50 Effect of Beam Depth on the General Method (with stirrups)
93
5.51 Predicted Results by the Shear Friction Method (without stirrups)
95
5.52 Predicted Results by the Simplified Method (without stirrups)
96
5.53 Predicted Results by the General Method (without stirrups)
97
5.54 Effect of the Ratio of Shear Span on the Shear Friction Method (without
stirrups)
99
5.55 Effect of the Ratio of Shear Span on the Simplified Method (without stirrups)
99
5.56 Effect of the Ratio of Shear Span on the General Method (without stirrups)
100
5.57 Effect of Concrete Strength on the Shear Friction Method (without stirrups)
100
5.58 Effect of Concrete Strength on the Simplified Method (without stirrups)
101
5.59 Effect of Concrete Strength on the General Method (without stirrups)
101
5.60 Effect of Longitudinal Reinforcement Ratio on the Shear Friction Method
(without stirrups)
102
5.61 Effect of Longitudinal Reinforcement Ratio on the Simplified Method
(without stirrups)
102
5.62 Effect of Longitudinal Reinforcement Ratio on the General Method
(without stirrups)
103
5.63 Effect of Longitudinal Reinforcement Strength on the General Method
(without stirrups)
103
5.64 Effect of Beam Depth on the Shear Friction Method (without stirrups)
104
5.65 Effect of Beam Depth on the Simplified Method (without stirrups)
104
5.66 Effect of Beam Depth on the General Method (without stirrups)
105
xii
NOTATION
a
shear span, distance from centre of support to point load
a
clear shear span, distance between outer edge of plate for concentrated
c
load and inner edge of plate at support
A
area of potential shear failure plane
A
area of longitudinal reinforcement in tension zone
A
area of one stirrup
s
v
b
width of beam web
c
coefficient of the cohesion between a potential shear failure plane
Cb
concrete cover at top of beam
c,
concrete cover at bottom of beam
C
concrete strength of beam
w
C
factored concrete strength of beam
C. O. V.
coefficient of variation
r
d
distance from the extreme compression fibre to the centroid of the
longitudinal tension reinforcement
d
diameter of a reinforcing bar
d
distance measured perpendicular to the neutral axis between the resultants
b
v
of the tensile and compressive forces due to flexure
d
effective length of stirrup in the shear friction method
fry
specified yield strength of the stirrups
f
specified yield strength of the longitudinal reinforcement or stirrups
ev
y
f'
specified compressive concrete strength
h
overall height of member
k
factor for relating shear strength and normal stress determined from
c
experiments
n
number of stirrups crossed by a potential shear failure plane
R
normal force acting on potential shear failure plane
xiii
5
spacing of stirrups
S
shear force on potential shear failure plane
T
longitudinal reinforcement strength of beam
T
opi
force in longitudinal reinforcement for peak shear strength
T
r
factored longitudinal reinforcement strength of beam
T
v
tension force in a stirrup
T
vr
factored tension resistance in a stirrup
v
average shear stress on potential shear failure plane according to Loov's
equations
V
factored shear resistance attributed to the concrete
V
factored shear resistance attributed to the concrete for the C S A general
c
cg
method
V f
CS
factored shear resistance attributed to the concrete for the shear friction
method
V
dowel force in the longitudinal reinforcement
Vf
factored shear force at section
d
V
shear resistance of beam using C S A A23.3-94 general method
V
factored transverse component of prestress of beam
V
factored shear resistance of beam
V
rg
factored shear resistance of beam using C S A A23.3-94 general method
V
s
factored shear resistance provided by the shear reinforcement
Vf
factored shear resistance for the shear friction method
V
factored shear resistance provided by the stirrups for the C S A general
g
p
r
s
sg
method
V
sim
shear resistance of beam using C S A A23.3-94 simplified method
V
s¡
shear resistance provided by one stirrup
Vf
factored shear resistance attributed to the reinforcement for the shear
ss
friction method
xiv
V
ultimate shear resistance of beam measured from test
V
shear resistance of concrete on a 45° plane
a
angle between transverse reinforcement and the shear plane
a/
angle between shear friction reinforcement and longitudinal axis
(8
factor that depends on the average tensile strains in the cracked concrete
t
4S
using C S A general method
j3
calibration factor for shear friction method
6
angle between longitudinal axis and potential shear failure plane
v
9
minimum shear failure angle for the shear friction method
X
factor to account for low density concrete
\i
coefficient of friction
p
longitudinal tension reinforcement ratio
p
transverse reinforcement ratio
o
average normal stress on potential shear failure plane
min
v
<j)
resistance factor for concrete
¢¡.
resistance factor for reinforcement
y/
factor that depends on the ratio of longitudinal reinforcement strength to
c
optimum tension for the shear friction method
xv
1
CHAPTER 1
INTRODUCTION
1.1 General
Failure in shear of reinforced concrete takes place under combined stresses
resulting from an applied shear force, bending moments and, where applicable, axial
loads and torsion as well. Because of the non-homogeneity of material, non-uniformity
and non-linearity in material response, presence of cracks, presence of reinforcement,
combined load effects, etc., the behavior of reinforced concrete in shear is very
complicated, and the current understanding of and design procedures for shear effects are
based on analyses of results of extensive tests and simplifying assumptions rather than on
an exact universally acceptable theory.
The best-known model for the expression of the behavior of beams with web
reinforcement failing in shear is the truss model. The truss model is a helpful tool to
visualize the nature of stresses in the stirrups and in the concrete, and to base simplified
design concepts and methods on. It may also be used to derive equations for the design of
shear reinforcement. However, it does not recognize fully the actual action of web
reinforcement and its effect on the various types of shear transfer mechanisms.
A shear-friction model has been developed to predict the shear strength of beams
by Loov
( 1 7 ) ( 1 8 ) ( I 9 )
and many others O W i X W M W W W S ) . Because shear friction works well
for composite beams, it might also predict the shear strength of beams which also have
potential
major
cracks
along which
slip can occur. Stirrups and longitudinal
reinforcement provide a clamping force thereby increasing the friction force which can be
transferred across a crack along a potential failure plane. This model is based on the shear
strength after cracking so that no diagonal tension strength is included. In this thesis, the
shear friction model has been investigated and developed for the purpose of shear design
of beams.
2
1.2 Code Review
Prior to its 1984 revision, C S A Standard A23.3 recommended a method for shear
and torsion design based on the traditional method adopted by the A C I code.
The
procedure is called the "V + V" approach. The term V is referred to as the "shear carried
c
c
by the concrete", while the term V is referred as the "shear carried by the stirrups". A23.3
s
assumes that V is equal to the shear strength of a beam without stirrups and further
c
simplifies V to equal the shear at inclined cracking. V relies on the tensile strength of the
c
s
transverse reinforcement. The stirrups and the inclined compressive struts are assumed to
act as members of a 45-degree truss and the term V is calculated based on this model.
s
The 1984 revision of the Canadian Standard, C A N 3 A23.3-M84, recommended
two alternative methods for shear design.
The first of these, termed the "simplified
method" (CAN3 A23.3-M84 (11.3)) is a shortened version of the traditional method
followed by A C I and previous Canadian codes. In the simplified method, the transverse
reinforcement is designed for the combined effect of shear and axial load i f any, while the
longitudinal reinforcement is designed for the combined effect of flexural and axial load.
The second method is termed as "general method" for shear design (CAN3
A23.3-M84(l 1.4)). In this method, the truss analogy has been used in a more direct
manner to account for the influence of diagonal tension cracking on the diagonal
compressive strength of concrete, and the influence of shear on the design of longitudinal
reinforcement. The code requires that deep beams, parts of members with deep shear
span, brackets and corbels, and regions with abrupt changes in cross-section (such as
regions of web openings in beams) be designed by the general method only. But we will
find later in this thesis that the general method is not suited to the design of deep beams,
brackets and corbels.
C S A Standard A23.3-94 recommends three alternative methods for shear design.
Regions of members in which it is reasonable to assume that plane sections remain plane
shall be proportioned for shear and torsion using either the general method or the
simplified method (if member is not subjected to significant axial tension) or the strut-
3
and-tie model. Regions of members in which the plane section assumption of fiexural
theory is not applicable shall be proportioned for shear and torsion using the strut-and-tie
model.
The simplified method of shear design described in C S A Standard A23.3-94 is
not simple. The designer is required to check numerous equations and limits. On the other
hand, the general method is extremely complex so engineers rarely use it in engineering
practice.
1.3 Scope of Study
The main objective of this study is to introduce the shear-friction method for
engineering design. After reviewing shear design theories and shear design methods
which are used by recent C S A Standard A23.3-94, a method of shear design based on
shear friction theories has been applied to predict the shear capacity of reinforced
concrete beams.
A comparison of the shear-friction method and recent code methods with the test
results of beams from the literature has been presented in this thesis. Proposed code
clauses for shear design based on the shear-friction method have been developed and
design examples based on the shear friction method are also included in this thesis.
1.4 Thesis Organization
Chapter 2 contains the review of basic shear theories. The factors of shear
strength are listed and shear failure mechanisms and modes are discussed in this chapter.
In Chapter 3 C S A Standard A23.3-94 for shear design has been introduced and
the design methods have been discussed.
Chapter 4 introduced the shear friction methods by Loov and others. In Chapter 5
a modified equation of the shear friction method has been introduced and a comparison of
the shear-friction method and recent code methods with the test results of beams from the
literature has been presented in this chapter.
4
Proposed code clauses for shear design based on the shear-friction method with
design examples have been put in Chapter 6. Conclusions and recommendations are
given in Chapter 7.
5
CHAPTER 2
B A S I C S H E A R THEORIES
2.1 Homogeneous Beam
In order to gain an insight into the causes of shear failure in reinforced concrete,
the stress distribution in a homogeneous elastic beam of rectangular section will be
reviewed briefly. From the free-body diagram as shown in Fig.2-1, it can be seen that
Where
dM = the bending moment change from section to section
dx = the distance between sections
V = the shear force on the section
M+dM
Fig. 2-1, Internal Forces in Beam
By the traditional theory for homogeneous-elastic-uncracked beams, the shear
stresses, v, and the flexural stress, f , at a point in the section distant y from the neutral
x
axis are given by
(2-2)
6
(2-3)
Where
Q = the first moment about the neutral axis of the part of the cross-sectional
area above the depth y
I - the moment of inertia of the cross section
b = the width of the beam
The distribution of these stresses is as shown in Fig. 2-2. Considering an element
at depth y (Fig. 2-3), the fiexural and shear stresses can be combined using Mohr's circle
into equivalent principal stresses, f¡ and f , acting on orthogonal planes inclined at an
2
angle a, where
ÍZ7
i
f
u
~
2
f
x
±
Í{2
f
(2-4)
x
and
(2-5)
tan(2«) = —
Fig. 2-2, Distribution of Fiexural Shear Stresses
Fig. 2-3, Principal Stresses
The principal stress trajectories in the uncracked beam are plotted in Fig. 2-4a.
Stress trajectories are a set of orthogonal curves, whose tangent at any point is in the
direction of the principal stress at that point. The compressive stress trajectories are steep
near the bottom of the beam and flatter near the top.
In concrete, which is weak in
tension, tensile cracks would occur at right angles to the tensile stresses and hence the
compressive stress trajectories indicate potential crack patterns (see Fig.2-4b). (Note that
if in fact a crack is developed, the stress distributions assumed here are no longer valid in
that region and redistribution of the internal stresses takes place.) The location of the
absolute maximum principal tensile stress will depend on the variation off and v, which
x
in turn depends on the shape of the cross section and on the span and loading.
It is seen that the general influence of shear is to induce tensile stresses on an
inclined plane. Failure of concrete beams in shear is triggered by the development of
these inclined cracks under combined stresses. To avoid a failure of the concrete in
compression, it is also necessary to ensure that the principal compressive stress,/^, is less
than the compressive strength of concrete under the biaxial state of stress.
Fig. 2-4a, Stress Trajectories
Fig. 2-4b, Potential Crack Pattern
9
Although several theories of failure have been used for concrete shear design, for
the traditional method of shear and torsion design, the principal tensile stress theory has
been followed.
2.2 Beam Cracking Modes
The cracking pattern in a test beam is shown in Fig.2-5. Two types of cracks can
be seen. The vertical cracks occurred first, due to fiexural stresses. These start at the
bottom of the beam where the fiexural stresses are the largest. The inclined cracks at the
ends of the beam are due to combined shear and flexure. These are commonly referred to
as inclined cracks or shear cracks. Such cracks must exist before a beam can fail in shear.
Several of the inclined cracks have extended along the reinforcement toward the support,
weakening the anchorage of the reinforcement.
Fig. 2-5, A Cracked Beam without Shear Reinforcement (Ref. 27)
Although there is a similarity between the planes of maximum principal tensile
stresses and the cracking pattern, fiexural cracks generally occur before the principal
tensile stresses at midheight become critical. Once such a crack has occurred, the
10
principal tensile stresses across the crack drops to zero. To maintain equilibrium, a major
redistribution of stresses is necessary. As a result, the onset of inclined cracking in a
beam cannot be predicted from the principal stresses unless shear cracking precedes
flexural cracking. This very rarely happens in reinforced concrete but does occur in some
prestressed beams (such as I-section beam).
The cracking pattern in a test beam with shear reinforcement is shown in Fig.2-6.
It is obvious that inclined cracks are almost straight lines instead of curves that we have
seen in the test beam without shear reinforcement in Fig.2-5. Another evidence we can
see is that inclined cracks bypass as many stirrups as possible. These evidences are useful
to predict possible beam shear failure planes.
Fig. 2-6, A Cracked Beam with Shear Reinforcement (Ref. 35)
2.3 Shear Transfer Mechanisms
There are several mechanisms by which shear is transmitted between two planes
in a concrete member. Fig. 2.7 shows a free body of one of the segments of a reinforced
concrete beam separated by an inclined crack. The major components contributing to the
shear resistance are:
11
(1) The shear strength, V , of the uncracked concrete;
cz
(2) The vertical component, V^, of the aggregate interlock shear, V ;
a
(3) The dowel force, V , in the longitudinal reinforcement;
d
(4) The shear, V , carried by the shear reinforcement.
s
Fig. 2-7, Internal Forces in a Cracked Beam
The aggregate interlock, V , is a tangential force transmitted along the plane of the
a
crack, resulting from the resistance to relative movement (slip) between the two rough
interlocking surfaces of the crack, much like frictional resistance and transverse rebar
dowel effects. So long as the crack is not too wide, the force V may be very significant.
a
The dowel force in the longitudinal tension reinforcement is the transverse force
developed in these bars functioning as a dowel across the crack, resisting relative
transverse displacements between the two segments of the beam.
12
Each of the components of this process except V has a brittle load-deflection
s
response. So it is difficult to quantify the contributions of V , V¿¡, and V . In design,
cz
ay
these are lumped together as V , referred to as "the shear carried by the concrete". Thus
c
the nominal shear strength, V , is assumed to be
n
V =V +V
n
c
(2-6)
s
In North American design practice, V is traditionally taken equal to the failure
c
capacity of a beam without stirrups.
2.4 Shear Failure Modes
2.4.1 Beam without Shear Reinforcement
In beams without shear reinforcement, the breakdown of any of the shear transfer
mechanisms may lead to failure. In such beams there are no stirrups enclosing the
longitudinal bars and restraining them against splitting failure and the value of V is
d
usually small. The component V
ay
also decreases progressively due to the unrestrained
opening up of the crack. The spreading of the crack into the compression zone decreases
the area of uncracked concrete section contributing to V .
cz
However, in relatively deep
beams (a/d < 1), tied-arch action may develop following inclined cracking (see Fig. 2-9
(b)), which in turn will transfer part or all of the shear load at the section directly to the
supports thereby the shear capacity of the beam does not totally rely on V
ay
and V .
cz
Because of the uncertainties in all these effects, it is difficult to predict precisely the
behavior and strength beyond diagonal cracking of beams without shear reinforcement.
In beams without shear reinforcement, the shear failure load may equal or exceed
the load at which inclined cracks develop, depending on several variables such as the
ratio M/(Vd), thickness of web, influence of vertical normal stresses, concrete cover and
resistance to splitting (dowel) failure. Further, the margin of strength beyond diagonal
cracking fluctuates considerably.
Hence, for beams of normal proportions (M/(Vd) >
about 2.5), as a design criterion, the shear force, V , causing the formation of the first
cr
13
inclined crack is generally considered as the usable ultimate strength for beams without
shear reinforcement.
The moments and shears at inclined cracking and failure of rectangular beams
without web reinforcement are plotted as a function of the shear span, a, to the depth, d,
in Fig.2-8. The shaded areas in this figure show the reduction in strength due to shear, so
web reinforcement has to be provided to ensure that the full fiexural capacity can be
developed.
Typical shear failure modes of reinforced concrete beams, and the influence of the
a/d ratio, are illustrated in Fig. 2-9 with reference to a simply supported rectangular beam
subjected to symmetrical two-point loading.
In very deep beams (a/d < 1) without web reinforcement, inclined cracking
transforms the beam into a tied-arch (Fig. 2-9b).
The tied-arch can fail by either a
breakdown of its tension element, or by a breakdown of the concrete compression chord
by crushing.
In relatively short beams, with a/d in the range of 1 to 2.5 (Fig. 2-9c), the failure
is initiated by an inclined crack, usually a flexural-shear crack. The actual failure may
take place by crushing of the reduced concrete section above the head of the crack under
combined shear and compression, or cracking along the tension reinforcement resulting
in loss of bond and anchorage of the tension reinforcement. This type of failure usually
occurs before the fiexural strength of the section is attained.
Normal beams have a/d ratios in excess of about 2.5. Such beams may fail in
shear or in flexure. The limiting a/d ratio above which fiexural failure occurs depends on
the tension reinforcement ratio, yield strength of reinforcement and concrete strength.
V
v
a
a
(a) Beam.
Deep
*
H
Slender
Very.Shorty
short
c
1
re
^¾¾^
c
Failure
03
O
E
o
,
y / A / e r y slender
" ' V
1
/
/
/
/
^ ^ C ^
^
^ s ^ ^
^
^
Flexural capacity
Inclined cracking
and failure
^*».
2
•
1.0
Inclined
cracking
i
2.5
6.5
a/d
(b) Moments at cracking and failure.
<T3
C>t
Flexural capacity
JO
CO
nclined cracking and failure
1.0
2.5
6.5
a/d
(c) Shear at cracking and failure.
Fig. 2-8, Effect of a/d on Shear for Beams Without Shear Reinforcement (Ref. 27)
15
i
(a)
Shear-tension failure
( )
c
Diagonal tension failure
l«a/d<2
5
Shear-compression failure
(d) 2.5 < a / d < * * 6
T
( e ) Web-crushing failure
Fig. 2-9, Shear Failure Modes (Ref. 36)
16
For beams with a/d ratios in the range of 2.5 to 6, fiexural tension cracks develop
early on; however, before the ultimate fiexural strength is reached the beam may fail in
shear by the development of inclined flexure-shear cracks, which, in the absence of web
reinforcement, rapidly extend right through the beam as shown in Fig. 2-9d. This type of
failure is usually sudden and without warning and is termed diagonal-tension failure.
Addition of web reinforcement in such beams leads to a shear-compression failure or a
fiexural failure.
In addition to these different modes, thin webbed members such as I-beams with
web reinforcement may fail by crushing of the concrete in the web portion between
inclined cracks under the diagonal compression forces (Fig. 2-9e).
2.4.2 Beam with Shear Reinforcement
In members with shear reinforcement the shear resistance continues to increase
even after inclined cracking until the shear reinforcement yields and V can increase no
s
more. Any further increase in applied shear force leads to increases in V , V , and V^.
cz
d
With progressively widening crack width (which is no longer restrained because of
yielding of the shear reinforcement),
begins to decrease forcing V and V to increase
cz
d
at a faster rate until either a splitting (dowel) failure occurs, or the concrete in the
compression zone fails under the combined shear and compression forces.
Thus, in
general, the failure of shear-reinforced members is more gradual (ductile).
2.5 Factors Affecting the Shear Strength
2.5.1 Tensile Strength of Concrete
The inclined cracking load is a function of the tensile strength of the concrete.
The stress state in the web of the beam involves biaxial principal tension and
compression stresses as discussed before. 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 test.
17
2.5.2 Longitudinal Reinforcement
Fig. 2-10 shows the shear capacities of simply supported beams without stirrups
as a function of the steel ratio, p. When the steel ratio, p, is small, flexural cracks extend
higher into the beam and open wider, as a result inclined cracking occurs earlier and the
beam shear strength tends to be lower.
2.5.3 Shear Span-to-depth Ratio, a/d
As discussed earlier, the shear span-to-depth, a/d, has effects on the inclined
cracking shears and ultimate shears of "deep" beam, while for longer shear spans with a/d
greater than 3 it has little effect on the inclined cracking shear.
0.005
0.010
0.015
0.020
0.025
0.030
0.035
Fig. 2-10, Shear Strength vs. Longitudinal Reinforcement (Ref. 27)
0.040
Fig. 2-11, Shear Strength vs. a/d (Ref. 14)
19
2.5.4 Size of Beam
As the overall depth of a beam increases, the shear stress at inclined cracking
tends to decrease for a given f' , p, and a/d. As the depth of the beam increases, the
c
crack widths at points above the main reinforcement tend to increase. This leads to a
reduction in aggregate interlock across the crack, resulting in earlier inclined cracking. In
beams with web reinforcement the web reinforcement holds the crack faces together so
that the aggregate interlock is not lost as much as that of beams without web
reinforcement.
Fig. 2-12, Influence of Member Size on Shear Strength (Ref. 7)
20
2.5.5 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 fiexural cracking is delayed and the fiexural cracks do not penetrate as far into
the beam. So a larger shear is required to cause principal tensile stresses equal to the
tensile strength of the concrete.
•
•
_
••
Vu
•
^Fcbwd
-
Eq. 6 - 1 7 a
/
(ACI Eq. 11-4)
1500
1000
•
•
-/-—
^**"
Eq. 6 - 1 7 b
(ACI Eq. 11-8)
500
Compression
A x i a l stress, NJA
g
(psi)
Fig. 2-13, Effect of Axial Loads in Inclined Cracking Shear (Ref. 27)
2.5.6 Web Reinforcement
Prior to inclined cracking, the strain in the stirrups is equal to the corresponding
strain of the concrete. Since concrete cracks at a very small strain, the stress in the
stirrups prior to inclined cracking will be very small. Thus stirrups do not prevent
inclined cracks from forming. They come into play only after the cracks have formed.
Following the development of inclined cracking, stirrups intercepted by the cracks
resist a portion of the shear. The web reinforcement contributes significantly to the
21
overall shear strength by the direct contribution of V to the shear strength. Secondly, web
s
reinforcement crossing the inclined cracks restricts the widening of the crack and thereby
helps maintain the aggregate interlock resistance of shear. The web reinforcement also
can improve the longitudinal tension reinforcement dowel action and provide another
dowel action of itself crossing inclined cracks.
Flexural
cracking
Inclined
cracking
Yield of
stirrups
Failure
Applied shear
Fig. 2-14, Distribution of Internal Shears of Beam with Shear Reinforcement (Ref. 27)
22
CHAPTER 3
S H E A R DESIGN - C S A S T A N D A R D A23.3-94
3.1 General
The C S A Standard A23.3-94 recommends two alternative methods for shear
design. The "Simplified Method" is a short version of the traditional method followed by
A C I and previous Canadian Codes. In the Simplified Method, a 45-degree truss model
has been used and the transverse reinforcement is designed based on that.
The second method is the "general method" for shear design. In this method, the
truss analogy has been used in a more direct manner to account for the influence of
diagonal tension cracking on the diagonal compressive strength of concrete, and the
influence of shear on the design of longitudinal reinforcement.
Both simplified method and general method are sectional methods and can be
applied only to the flexural region of beams, in which it is reasonable to assume that
plane sections remain plane and that shear stresses are distributed in a reasonably uniform
manner over the depth of the beam. Because of this, both methods are not appropriate for
regions of members near static or geometric discontinuities, the code requires regions
with abrupt changes in cross-section (such as regions of web openings in beams) and
brackets and corbels, to be designed by the strut-and-tie method, which is capable of
more accurately modeling the actual flow of forces in these regions.
3.2 Simplified Method
For flexural members not subjected to significant axial tension, the Canadian code
allows shear design based on the simplified method.
Required shear resistance for beam is:
V >V,
r
(3-1)
23
Where Vf is the factored shear force at a section, and V is the sum of the
r
contribution attributed to the concrete and transverse reinforcement.
V = V + V,
r
(3-2)
c
But V is limited to:
r
V,iV +0.8ty JfJb d
e
e
(3-3)
w
This upper limit is intended to ensure that the stirrups will yield prior to crushing
of the web concrete and that diagonal cracking at specified loads is limited.
3.2.1 Shear Supported by Concrete, V
c
V =0.2ty Jf¡b d
c
e
(3-4)
w
This equation can be used only for beams with minimum transverse reinforcement
given by Clause 11.2.8.4 i f ^exceeds 0.5 V + <j) V :
c
p
p
A =0.06jf ^f
v
(3-5)
c
J V
The minimum transverse reinforcement restrains the growth of inclined cracking
and increases ductility to provide a warning of failure.
For beams without transverse reinforcement, Clause 11.3.5.2 shall be used to
account for the reduced strength of beams deeper than 300 mm.
24
V,
260
1000+ d
(3-6)
At Jfìb d>0.lA&4fÌb d
c
w
w
Studies have shown that the equations for V above are more appropriate for
c
beam with a/d ratios greater than three. It results in overly conservative design for beams
with a/d ratios less than 2.5 (see Fig. 3-1).
0.30
|«
a
0.25
»|
|——a
»»j
//.illi.l
f
0.20
Clause 11.5:
Strut-and-tie model
bdf,
-
610mm
•
f ' = 27.2 MPa
c
max. agg. = 19 mm
0.15
d = 538 mm
Experimental
result
b= 155 mm
0.10
A, = 2277 m m
Clause 11.4:
Sectional model
r = 3 7 2 MPa
y
A =0
v
0.05
3
4
a/d
Fig. 3-1, Comparison of Shear Design Methods and Test Results (Ref.7)
2
25
3.2.2 Shear Supported by Stirrups, V
s
<l> Avf d
s
y
(3-7)
s
Here the transverse reinforcement is assumed to be perpendicular to the
longitudinal axis of the member.
Additional
maximum spacing (Clause
11.2.11) and minimum transverse
reinforcement requirement (Clause 11.2.8) have been patched onto the basic equation in
order to obtain satisfactory behavior under various conditions.
3.3 General Method
Shear resistance for beam is:
(3-8)
Where V
cg
is the factored shear resistance contributed by concrete at a section,
and V is the factored shear resistance contributed by transverse reinforcement.
sg
But V„ shall not exceed
V = 0.25
c
Áfcf¿b d
w
v
(3-9)
Where d is the distance measured perpendicular to the neutral axis between the
v
resultants of the tensile and compressive forces due to flexure, but need not be taken less
than 0.9d.
This upper limit is intended to ensure that the stirrups will yield prior to crushing
of the web concrete and that diagonal cracking at specified loads is limited.
26
3.3.1 Shear Supported by Concrete, V
V =UA&ft
cg
4f!b d
c
w
(3-10)
v
Where p is determined in accordance with Clause 11.4.4.
3.3.2 Shear Supported by Stirrups, V
sg
S
Where 0is given in Clause 11.4.4. Obviously i f 0 = 45° both simplified method
and general method will have the same shear resistance contributed by transverse
reinforcement. Again assume that the transverse reinforcement is perpendicular to the
longitudinal axis of the member.
For members with transverse reinforcement inclined at an angle a to the
longitudinal axis, V shall be computed from
sg
_
faA f d (cot0
v y
v
+ cota)sina
27
CHAPTER 4
S H E A R D E S I G N - S H E A R FRICTION M E T H O D
4.1 General
The Clause 11.1.3 in C S A Standard A23.3-94 states that shear friction shall be
used to design "Interfaces between elements such as webs and flanges, between
dissimilar materials, and between concrete cast at different times or at existing or
potential major cracks along which slip can occur..." Because beam shear failure
normally comes with a major crack and slip between the crack, it would seem that shear
friction can also be applied to predict the shear strength of beams. In 1997, Loov
(19)
presented the rudiments of a procedure , which applied this concept to the shear design
of beam. In recent years, Loov, Peng, Tozser, Kriski, and others, have shown that it is
possible to use a simpler method for shear design that is based on the shear friction
theory.
(16)(17)(18)(21)(23)(24){25)(26)(35)
It is encouraging that some of the resulting equations
derived by Loov match those equations derived by a number of people, including
(5)
Braestrup , Nielsen
(33)
and Zhang
(45)
based on theories of plasticity.
4.2 General Equations for Beam Shear Based on Shear Friction:
The shear-friction concept for concrete-to-concrete interfaces is based on the
assumption that a crack will form and shear will be transferred across the interface
between the two parts that can slip relative to one another. If the crack faces are rough
and irregular, this slip is accompanied by separation of the crack faces. The separation
will stress the reinforcement crossing the crack until the reinforcement reaches its yield
point. Thus the reinforcement provides a clamping force across the crack interface.
28
Shear displacement
t î t î î t î 1111
Compression
in concrete = T
(i) Shear Tension Causing Crack Opening
i
Shear stress
Tension in
reinforcement = T
(ii) Free-Body-Diagram
Fig. 4-1, Shear Friction Concept (Ref.7)
4.2.1 Shear Friction Strength:
There are many equations that have been developed for predicting shear friction
strength.
Fig.4-1 illustrates the
shear
friction
concept for the case where the
reinforcement is perpendicular to the potential failure plane. Because the interface is
rough, shear displacement will cause a widening of the crack. This crack opening will
cause tension in the reinforcement balanced by compressive stresses, a, in the concrete
across the crack. The shear resistance of the face is often assumed to be equal to the
cohesion, c, plus the coefficient of friction, ju, times the compressive stress, a, across the
face. That is,
v =À&(c+Mff)
(4-1)
r
If inclined reinforcement is crossing the crack, part of the shear can be directly
resisted by the component, parallel to the shear plane, of the tension force in the
reinforcement. See Fig.4-2. Clause 11.6 of C S A Standard A23.3-M94 suggests that the
factored shear stress resistance of the shear plane shall be computed as:
v =A,fc(c+fia)+ûp fcosa/
r
v
(4-2)
29
Where a is the angle between the shear friction reinforcement and the shear
f
plane.
\ \ \
Fig. 4-2, Reinforcement Inclined to Potential Failure Cracks (Ref.7)
C S A Standard A23.3-M94 also gives an alternative equation for shear friction
strength, which is based on the work of Loov and P a t n a i k
(20)(22)
.
(4-3)
Where
& = 0.5 for concrete placed against hardened concrete
k = 0.6 for concrete placed monolithically.
In this method, the shear resistance is a function of both the concrete strength and
the amount of reinforcement crossing the failure crack. Fig. 4-3 shows how this equation
compares with the results from various push-off tests.
a Mattock (uncracked)
• Mattock (cracked)
A Walraven (cracked)
v/0~fHMPa)
Fig. 4-3, Push-off Test Results (Ref.22)
31
Fig. 4-4 shows a free body diagram of the end portion of a simple beam with
loads applied somewhere to the right of the section. Two equilibrium equations relate the
normal force, R, and the shear force, S, to T, the force in the main tension reinforcement,
nT , the total force in the stirrups crossing the plane and V, the end reaction. The forces
v
on a potential failure plane vary with the angle 0 between the axis of the beam and the
plane. When the loads between the reaction and the plane in question are negligible, then
V is equal to the vertical shear on the inclined plane.
R = Tsin0-(V-ZT )cosd
n
(4-4)
S = Tcos0-(V-ZT )sin0
(4-5)
v
v
32
Where T = Af
y
and T = AJ^. Here A is the area of longitudinal reinforcement
v
s
and f is its yield strength, while A is the total area of all legs of a stirrup and f
y
v
vy
is the
stirrup yield strength.
Using the relationship from Eq. 4-3, the shear friction stress is
v=
kjtf
(4-6)
While
R
and
a =
Where A is the area of the inclined failure plane,
A=
bh
w
(4-7)
sine?
Where b is the width of web, h is the total depth, and 6 is the angle between the
w
longitudinal reinforcement and the crack.
The shear force is therefore proportional to the square root of the normal force, R
S = k4Rf^4
(4-8)
The equations shown above (Eq. 4-4 to Eq. 4-8) can be combined to give a
general equation for the shear strength
2
V = 0.5k C
2
2
0.25 k C
+ cot
0-cotO (1 + cot 6)-Tcot0
2
+ Yjv
4
9
(")
Where
C -
f'Xh
(4-10)
33
This equation is similar to that derived by Braestrup
( 5 )
and by Nielsen
( 3 3 )
with
plasticity theory.
For design, the factored values should be used thus
2
V
r
2
=0.5k C,
0.25k'C.
• + cot
0-cotO (1 + cot 0)-T cot0
2
r
+ J]T
vr
(4-11)
n
Where
(4-12)
(4-13)
T
r=<t>Asfy
T
vr
(4-14)
A
= <t>s vf,y
A l l planes between the inside edge of the support and the edge of the load to a
maximum angle of 90° should be considered to be potential failure planes. The shear
strength on each plane is calculated and the lowest strength, when comparing all possible
failure planes, is the governing shear strength. Under some circumstances it may be
extremely unlikely that a crack will form along particular failure planes so that choosing
the absolute
lowest strength
without regarding to location may be excessively
conservative. This aspect has been investigated by Zhang
( 4 5 )
.
Fig. 4-5 shows the change
in predicted shear strength as the failure plane angle is varied. When a crack intercepts a
stirrup, the shear strength increases by T , the force that can be developed in the stirrup.
v
Fig. 4-6 shows a three-dimensional surface plot, which was obtained by analyzing
beam tests by Kani
( 1 4 )
. The test beams had only one stirrup but in different locations to
determine the effects of stirrup location. The test result shows that it is not necessary to
check every potential failure plane. The planes with the lowest strength have the flattest
possible angle while intersecting a minimum number of stirrups.
Fig. 4-6, Three-dimensional surface of shear strength along all possible failure
planes for beam 544 (Réf. 18)
35
Fig. 4-7 shows a beam with possible critical shear failure planes. Fig. 4-8 is a
photograph of a beam indicating that the actual cracks correspond to the expected failure
planes.
essala
V
T-
Fig. 4-8, A Tested Beam with Critical Shear Cracks (Ref. 35)
B - 7
36
4.2.3 Approximate Shear Capacity of Concrete
If the shear failure plane bypasses the stirrups, the strength along the weakest
plane depends on the longitudinal reinforcement and the angle of the failure plane, but is
unaffected by the stirrup strength. From Eq. 4-9 we can obtain
2
V = 0.5k C
2
2
0.25k C
2
+ cot 6 -cot6 (1 + cot
(4-15)
0)-Tcot6
Beams depend on longitudinal reinforcement and the anchorage of longitudinal
reinforcement to develop shear capacity. The optimum tension in the longitudinal
reinforcement, by which the maximum shear capacity will be developed, can be obtained
by differentiating Eq. 4-15
2
dV
(l + cot 6)
— - = . '
c
/
2
dT
=-cotO
(4-16)
2
4t /0.25k C +cot 0
2
T
opt
(4-17)
2
= 0.25k C(2 +tan 6)
Substitute Eq. 4-17 into Eq. 4-15, the shear strength of beams will be
V = 0.25k''Ctond
(4-18)
c
Eq.4-18 gives the shear capacity of beams with longitudinal reinforcement tension
capacity f A
y
s
>T
2
opt
2
- 0.25k C(2+tan
9). It is assumed that anchorage for longitudinal
reinforcement to develop such tension capacity is sufficient.
Fig. 4-9 shows VjC
vs.cot# for different ratios of longitudinal reinforcement. It
is clear that Eq. 4-8 represents the upper bound value of shear capacity of beams. For
beams with longitudinal reinforcement tension capacity less than T t> the ^
op
longitudinal reinforcement tension capacity, the less the shear capacity.
e s s
the
37
0.50
0.40
0.30
>
0.20
0.10
0.00
Fig. 4-9, Shear Strength vs. cotO
For beams with longitudinal reinforcement tension capacity f A
y
s
less than Jopt,
Eq. 4-15 can be substituted approximately by a simple equation as following:
2
V, = 0.25k sin
\
n T
2T
°t* J
Ctand
(4-19)
38
Fig. 4-10, Shear Strength vs. cotO by Eq. 4-15 and Eq. 4-19
The curves from Eq. 4-15 and Eq. 4-19 have been plotted on Fig. 4-10 for
comparison. The graph shows that Eq. 4-19 is a useful approximation for the shear
capacity of concrete.
For factored design, we should use:
2
V
cr
= 0.25 k C tane
r
2
V„ = 0.25k sin
n T
C tan6
r
2T
V
°p>
When
T>Topt
(4-20)
When
T<Tapt
(4-21)
J
Where
(4-22)
39
T = <t> AJ
s
T
opt
(4-23)
y
2
= 0.25k C (2
r
2
+ tan 9)
(4-24)
Fig. 4-11 plots the beam shear strength of concrete vs. the beam longitudinal
reinforcement for a particular plane in the beam based on shear-friction equations of Eq.
4-18 and Eq. 4-19. It shows that the variation of the beam shear strength of concrete
increases as the beam longitudinal reinforcement increases. When the beam longitudinal
2
reinforcement reaches f A
y
s
2
= 0.25k C(2 +tan 9), the beam shear strength of concrete
reaches its peak value and will not increase even though the beam longitudinal
reinforcement increases.
A f (kN)
s
y
Fig. 4-11, Shear Strength vs. Longitudinal Reinforcement of Beam
40
4.2.4 Approximate Shear Capacity of Stirrups
The usual equations for the shear strength of stirrups are overly optimistic. Fig.4-7
shows several possible failure planes with zero, one and two stirrups crossing them. Fig.
4-8 is a photograph of a beam indicating that the actual cracks correspond to the expected
failure planes. To ensure a conservative prediction, the number of stirrups that are
considered to cross the shear plane should be the number of stirrup spaces crossed by the
crack minus one. M a r t i
( 2 8 )
correctly accounted for this in his work. Therefore, because of
the nature of shear failure planes that tend to avoid stirrups the proper estimate of the
stirrup contribution may be
ÚL cote
(4-25)
For factored design, we shall use:
V
y
sr
- V
r
'd cote
}
ev
(4-26)
si
Where
(4-27)
vy
Equation 4-27 is one of the most significant discoveries by M a r t i
( 2 8 )
and Loov
( 1 8 )
in shear design, because this corrects a basic mistake that has been used for years in shear
design.
4.2.5 Approximate Shear Design Equations for Beams with T>T
opt
Using the " V + V " approach, the approximate shear strength along a plane at an
c
s
angle 0 to the beam axis is
41
V =V tanû
r
4S
'décote
+ V
sl
v
s
-1
(4-28)
j
Where
2
V = 0.25k C
4}
(4-29)
r
Further, Eq. 4-31 can be written as:
V =WJ 4TXh
45
(4-30)
v
Where
2
J3 = 0.25k 4/:
(4-31)
V
The coefficients k and fi are calibration factor that can be adjusted to match the
v
equation with test results.
The shear strength of beams without stirrups is governed by the first term in
Equation (4-28), where 0 is the angle of the failure plane with the lowest slope that can
be expected to occur.
V =V tanG
r
45
(4-32)
4.2.6 Critical Shear Failure Angle
Although theoretically we have only an integer number of possible shear failure
planes such as 1, 2 and 3 in Fig. 4-7 and Fig. 4-8, it is convenient to treat Eq. 4-28 as a
continuous function of 6 when deriving the critical shear failure angle. It is notable that
the effects of stirrup spacing will be ignored and Eq. 4-28 will form a lower bound of the
shear capacity, when Eq. 4-28 is considered to be a continuous function of 0 (see Fig. 412).
42
Fig. 4-12, The Shear Contributions of Concrete and Discrete Stirrups (Ref. 18)
The critical angle 0 corresponding to the minimum strength can be found by
differentiating Eq. 4-28.
ñ
—V
d0
r
V
=—½
cos 0
2
V
d
"
=0
sin 0 s
ev
IK,
tane =i^L^-
v
]
a
(4-33)
2
s
Substituted Eq. 4-34 into Eq. 4-28,
(4-34)
43
+v
V. = V45
\v
-i
sl
45
y45
(4-35)
s
So
d„
V =2AV V ^--V
r
45
sl
s¡
(4-36)
Eq. 4-36 is a direct solution for the shear strength of reinforced concrete beams. It
combines the contribution of the web stirrups and concrete corresponding to the
minimum strength of the combination.
From Eq. 4-36 we can solve directly to obtain the maximum stirrup spacing.
s<
(4-37)
(Vf+K,)
2
Eq. 4-37 can be used for design of stirrup spacing, while Eq. 4-36 is used to
calculate the shear capacity of a beam with known stirrup spacing.
Eq. 4-36 and Eq. 4-37 do not apply in cases where the shear failure angle is not
determined by Eq. 4-34. The shear failure crack can only be formed between the beam
support and load, so the beam shear span limits the minimum shear failure angle to:
tanO>
(4-38)
a..
The strength along this steeper plane can be obtained directly using Eq. 4-28.
However, Eq. 4-36 and Eq. 4-37 are conservative i f the shear failure angle becomes
steeper under the limitation of beam shear span.
44
4.2.7 Beams with Longitudinal Reinforcement T <T.opt
For beams without stirrups, the shear capacity can be derived from Equation (421) as following:
K
=yV tanO
When
45
T<Topt
(4-39)
Where
TV T
y/ - sin
(4-40)
2 T
Accordingly, Eq. 4-36 and Eq. 4-37 need to be modified as follows:
V; = 2
\ V V -^-V
s
¥
V
s<
45
sl
(v +v r
f
sl
(4-41)
(4-42)
sl
It is worthy to notice that Eq. 4-41 and Eq. 4-42 may generate conservative results
for beams with short shear span.
45
CHAPTER 5
E X P E R I M E N T A L STUDIES A N D C O M P A R I S O N S
By testing the proposed shear friction method against available experimental
results from different authors, the shear friction method for shear design of beams will be
evaluated in this chapter. A comparison study of the simplified method and the general
method is also conducted in this chapter to choose a more accurate method for shear
strength prediction.
5.1 Application of Shear Friction Method
Using the "V + V" approach as discussed in Chapter 4, the total shear capacity of
c
a beam is:
=V
V
Y
y
sf
csf
T
(5-1)
+V
ssf
y
The shear capacity of concrete, V ^ can be calculated from
cs
(5-2)
Vcs =¥V tane
f
45
Where
(5-3)
=i
w
if/ = sin
tanO =
n T
2T
o
When
T<Topt
(5-5)
(5-7)
y
45
(5-4)
s
*,AJ
T =V (2
T>T,opt
(5-6)
V
4S
T =
When
2
+
tan 0)
(5-8)
46
(5-9)
IT
The value of V / shall be computed from
ss
d„., cot 6
Y
ssf
y
si
-I
(5-10)
Eq. 5-2 is derived from the equation in Chapter 4 with some modifications. The
value of P from Eq.4-37 is:
v
P =0.25k 4fl
2
(5-11)
v
It has been found that k becomes smaller as the concrete strength increases
( 2 2 ) ( 3 7 )
.
The equation found from a least-squares fit of tests is:
k=2.0(/:)-
0
(5-12)
4
Substituted Eq. 5-12 into Eq. 5-11 and get:
/
A
=0.36
\0.30
'30^
\f'c
(5-13)
J
To consider the effects of beam depth as discussed in Chapter 2, Eq. 5-13 needs to
be modified. According to the researches by Tozser and Loov
( 2 5 ) ( 2 6 )
, the shear strength of
025
beams decreases when the depth of beams increases in proportion to h~ . Finally, the
equation for calculating fi is presented by Eq. 5-14.
v
47
0.30
/3 =0.36
V
h
(5-14)
There are two limitations for the cracking angle 0. First, for beams with short
shear spans the shear cracking angle may be limited by the ciç/h ratio as mentioned in
Chapter 4. Second, from pictures of crack patterns of specimens from literature
( 1 4 ) ( 4 5 )
, it
is observed that when the shear span is greater than 2.5, the shear cracking angle 0 stays
at a limiting angle even with increasing shear span. Based on the analysis of the test
results from literature, the minimum shear cracking angle 9 is about 2 i f - ^ / ^ J
degree.
So the two limitations for the failure angle are:
tanO>
(5-15)
6>21 'fit"
\30)
(5-16)
5.2 Test Results in Literature:
Experimental data from the literature were examined to verify whether the shear
friction method is a rational approach for estimating the shear capacity of beams. Tests
from two series of tests from the literature are presented and discussed in detail. The
results predicted by the shear friction method were then compared with the test results
from a total of 113 beams with stirrups and 105 beams without stirrups. A l l selected
beams were simply supported rectangular beams subjected to a symmetrical single or
two-point load. The effects of concrete strength, shear span ratios, amount of longitudinal
reinforcement and stirrup spacing are discussed. Notice that the limitation on maximum
stirrup spacing by the C S A A23.3-94 clauses 11.2.11 for the simplified method and the
general method was ignored during the analysis.
48
5.2.1 Yoon, Cook and Mitchell's Tests, 1996
( 4 4 )
:
Yoon, Cook and Mitchell investigated six full-scale beam specimens
( 4 4 )
. The six
beams having different amounts of shear reinforcement at each end were tested to
provide a total of 12 shear tests. Fig. 5-1 shows the details of the 375 mm wide x 750 mm
deep specimens that were tested with a clear shear span of 2000 mm and shear span ratio
of a/d = 3.28 and a^h = 2.67. The fiexural tension reinforcement for all of the specimens
consisted of 10-No.30 bars in two layers, giving a reinforcement ratio of p = 0.028. A
symmetrical single point load had been applied at midspan. Table 5-1 lists the details of
the beam specimens.
75
11
|350|
1
2000 mm dear
-
150
1—»
1/¾
2150
2150^
5000
y
1
|350i
ELEVATION VIEW
2-No.lO
Strain 9«9e«
on
Stirrup
rwrtforoefnent
vanea
fWNrafONIMflt
r"*""""4
J a/2
LVOT»
on
concreto
10- No 30
cover MOmm
I
SECTION A-À
650
J6JL
INSTRUMENTATION
Fig. 5-1, Details of Beam Specimens and instrumentation (Yoon, Cook and Mitchell)
49
Table. 5-1, Details of Beam Specimens
(,
Shear reinforcement
/ / . Stirrup size and spacing,
MPa
mm
Specimen
36
Comments
8.0 mm diameter at 325
9.5 mm diameter at 465
9.5 mm diameter at 325
0.00
0.35
0.35
0.50
No stirrups
Min /4,, s - d/2
Min A,, s = 0.7d
> Min A,, s = d/2
8.0 mm diameter at 325
0.00
0.35
M-Series:
Ml-S
Ml-N
67
M2-S
M2-N
9.5 mm diameter at 325
9.5 mm diameter at 230
No stirrups
AC1 83, ACI 89,*
CSA 84
Min A^ s = d/2
0.50 CSA 94 min /1,, s = d/2
0.70 ACI 89t mm A,, s < d/2
8.0 mm diameter at 325
0.00
0.35
H-Series:
Hl-S
Hl-N
H2-S
H2-N
87
w
MPa
N-Se ríes:
Nl-S
Nl-N
N2-S
N2-N
b s*
9.5 mm diameter at 270
9.5 mm diameter at 160
No stirrups
ACI 83, ACI 89,*
CSA 84
Min Ay, s = d/2
0.60 CSA 94 min A„ s < d/2
1.00 ACI 89t min A* s < d/2
•Lower amount of minimum *4 provided when *jf ií a/69 MPa in design.
V
tUpper amount of minimum A provided when Jf/
v
c
> a/69 MPa in design.
2
Noterj^. for all stirrups is 430 MPa; area of 8.0-mm-diamcter bar = 50 mm ; area of
2
9.5-mm-diameter bar = 7! mm .
The purpose of the paper was to evaluate the minimum shear reinforcement
requirements in normal, medium, and high-strength reinforced concrete beams. Therefore
the tested beams were reinforced with minimum shear reinforcement, except three
50
specimens without shear reinforcement ( N l - S , M l - S and H l - S ) . Here these test data are
used for evaluation of shear design methods under the effects of concrete strength, stirrup
spacing and shear reinforcement,
Table 5-2 gives the test results and a comparison of predicted and measured shear
capacities of specimens. The predictions using the shear friction method and the
simplified method agree well with the experimental results with value of C.O.V. 6.8%
and 12.5% respectively, while the prediction using the general method results in a higher
value of C.O.V. 23.7%.
Table. 5-2, Test Results and Comparison of Predictions
V,
Vf
Vsim
Vg
Specimen
(kN)
(kN)
(kN)
(kN)
Vt/V,,
v /v
Nl-S
249
208
232
204
1.199
Nl-N
N2-S
N2-N
Ml-S
Ml-N
M2-S
M2-N
Hl-S
Hl-N
H2-S
H2-N
457
363
483
296
405
552
689
376
354
436
270
411
468
567
302
432
534
712
318
318
418
316
403
525
576
360
209
208
404
278
269
455
504
1.216
1.025
1.108
1.096
0.985
1.180
1.214
1.075
1.436
316
1.082
447
606
708
300
526
625
1.117
1.119
1.012
m
a
327
483
598
721
S
C.O.V.
t
sim
1.143
1.156
0.937
1.006
1.051
1.196
0.908
1.082
0.986
Vt/V
g
1.223
2.188
1.741
1.194
1.066
1.508
1.213
1.367
1.034
1.018
1.612
1.136
1.154
1.11
1.08
1.37
0.075
6.8%
0.135
12.5%
0.325
23.7%
51
The analyses of the 9 beams with shear reinforcement are illustrated from Fig. 5-2
to 5-10. In Fig. 5-2, the ratios of test results to the results predicted by the shear friction
method against concrete strength, f' , are plotted to demonstrate the effect of concrete
c
strength on the shear friction method. It shows no obvious trend in the prediction of shear
capacity for beams with different concrete strength. Fig. 5-3 and Fig 5-4 present the
analysis results of the effects of concrete strength using the C S A simplified method and
general method respectively. A downward trend exists for both of methods.
2.5
2
V
t
Vrf
•••
1.5
•
•
•
1
•
•
-tr
0.5
30
40
50
60
70
80
f c (MPa)
Fig. 5-2, Effect of Concrete Strength on the Shear Friction Method
90
2.5r
1.5-
•
Vsim
•••
1
•
V
t
•
-B-
0.5
0
30
40
50
60
70
80
90
f c (MPa)
Fig. 5-3, Effect of Concrete Strength on the Simplified Method
2.5r
V
t
•••
-s•
•
1-5
1
0.5
0
30
40
50
60
f
c
70
80
(MPa)
Fig. 5-4, Effect of Concrete Strength on the General Method
90
53
The ratios of test results to the results predicted by the shear friction method
against the ratios of s/d and the web reinforcement index Pyfyy are plotted in Fig. 5-5 and
Fig. 5-6 respectively. The shear friction method demonstrates a consistent accuracy of the
prediction of shear capacity for beams with different stirrup spacing and different
amounts of shear reinforcement.
In Fig. 5-7 to Fig. 5-10, the measured/calculated ratios of shear capacity versus
the ratios of s/d and the web reinforcement index / V v y by
m
e
C S A simplified method
and general method are plotted. There is a larger scatter for these results than the scatter
when shear strength is predicted by shear friction. Notice that the scatter gets
significantly larger around / V v y
=
0.3 ~ 0.4. The reason is that some specimens are just
under the minimum shear reinforcement requirement by the code and the application of
different equations creates inconsistent conservative results. Fig.5-9 also shows that when
the ratio of s/d increases the general method tends to be more conservative.
2.5
2
•
Vrf
• ••
o
1
•
•
o
0.4
0.5
•
0.5
0
0.2
0.3
0.6
s
d
Fig. 5-5, Effect of Stirrup Spacing on the Shear Friction Method
0.7
1
2.5r
V
t
:
I
r
r
i
r
i
15
•
•
•
Vsf
• •D
I
•
1
0.5
0
0.2
J
I
I
I
I
I
I
L
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
f
p - vy(MPa)
v
Fig. 5-6, Effect of Shear Reinforcement on the Shear Friction Method
2.51
V
t
15
•
•
Vs i•m
•••
1
v
-a-
0.5
0
0.2
0.3
0.4
0.5
0.6
s
d
Fig. 5-7, Effect of Stirrup Spacing on the Simplified Method
0.7
55
2.5r
v
r
i
1.5-
t
Vs i•m
•
•
v
•••
1
0.5
0
0.2
J
L
0.3
0.4
0.5
0.6
J
L
0.7
0.8
J
0.9
I
1
f
P - v y (MPa)
v
Fig. 5-8, Effect of Shear Reinforcement on the Simplified Method
2.5r
V
t
•••
•
•
1-5
-B-
1
0.5
0.2
0.3
0.4
0.5
0.6
Fig. 5-9, Effect of Stirrup Spacing on the General Method
0.7
1.1
56
2.5r
•
•
•
—Q-
v, 14
S
ODD
1
•
O.5-
0
_!_
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
Pv-fyy (MPa)
Fig. 5-10, Effect of Shear Reinforcement on the General Method
Fig. 5-11, Fig. 5-12 and Fig 5-13 present the analysis results of the effects of
concrete strength for 3 beams without shear reinforcement using shear friction method,
the C S A simplified method and the general method respectively. The results by shear
friction method are slightly more conservative than the other methods. Both the
simplified method and the general method show a slightly larger downward trend when
f'
c
increases.
2.5r
V
1.5-
t
V¡f
•••
1
0.5
o'
1
30
1
40
1
50
1
60
f c (MPa)
1
70
80
90
Fig. 5-11, Effect of Concrete Strength on the Shear Friction Method
2.5
V
t
1-5
Vs i•m
•••
1
v
•
0.5
0
30
40
50
60
f
c
70
80
(MPa)
Fig. 5-12, Effect of Concrete Strength on the Simplified Method
90
58
2.5r
V
t
• •*•
V
1.5•
1
1
•
0.5"
30
40
50
60
f
c
70
80
90
(MPa)
Fig. 5-13, Effect of Concrete Strength on the General Method
5.2.2 Sarsam and Al-Musawi's Tests, 1 9 9 2
(40)
:
A total of 14 beams had been tested and all failed in shear. Fig. 5-14 shows the
details of the 180 mm wide x 270 mm deep specimens. A l l beams have the same 4 mm
diameter stirrups with different spacings. The shear span ratios of a/d = 2.5 and 4 had
been tested and different concrete strengths had been used. Different fiexural tension
reinforcement had been provided to test the effects on prediction of shear capacity. A
symmetrical two-point load had been applied at midspan. Table 5-3 and Table 5-4 list
the details of beam specimens.
The tests were designed to evaluate the effects of concrete strength, shear span
ratios, amount of longitudinal reinforcement and stirrup spacing.
59
25mm
wtlding
cross
bars
with
to alt
As bars
stirrups
/d s
positive
( typ.)
25*100*180 mm
plat*
(typ)
4 mm
spacing
through
out
î
j 22 $4 -2280
150mm
mm (a/d
4'fm//> J * / 5 « / - /575 mm
for As
( a/d
for
bars:
As
3-20
or
î
on
mm
2.25mm
2 5 mm
+/_ 16 mm
or
) i (min )
a:
2 - 10mm
stirrups
:2.5
25 mm cover
bars:
¿mm
, /SOmoi
* A)
on
3.25mm
Section
cover
As
A-A
Fig. 5-14, Details of Beam Specimens and Instrumentation (Sarsam and Al-Musawi)
60
Table. 5-3, Details of Beam Specimens:
Specimen
AL2-N
AL2-H
AS2-N
AS2-H
AS3-N
AS3-H
BL2-II
mm
235
235
235
232
235
235
233
BS2-H
BS3-H
BS4-H
CL2-H
CS2-H
CS3-H
CS4-H
ft*
4
4
2.5
2.5
2.5
2.5
4
N/mm*
40.4
75.3
39.0
75.5
40.2
71.8
75.7
mm
943
943
943
943
943
943
1181
p»
0.0223
0.0223
0.0223
0.0226
0.0223
0.0223
0.0282
233
233
233
233
2.5
2.5
2.5
4
73.9
73.4
80.1
70.1
233
233
233
2.5
2.5
2.5
70.2
74.2
75.7
1181
1181
1181
1470
1470
1470
1470
0.0282
0.0282
0.0282
0.0351
0.0351
0.0351
0.0351
aid
2
Spacing
of 4mm
stirrups,
mm
N/mm
150
0.76
150
0.76
150
0.76
150
0.76
100
1.14
100
1.14
150
0.76
150
100
75
150
0.76
1.14
1.53
0.76
kN
114.7
122.6
189.3
201.0
199.1
199.1
138.3
223.5
228.1
206.9
147.2
150
100
75
0.76
1.14
1.53
247.2
247.2
220.7
2
Table. 5-4, Details of Materials:
Reinforccmeni
diameter, mm
Description
Application
N/mm
4
High-yield, colddrawn smooth wire
Stirrups
820
10
Hot-rolled
deformed bar
Top
reinforcement
450
16
Hot-rol led
deformed bar
Tension
reinforcement
525
20
Hot-rolled
deformed bar
Tension
reinforcement
495
25
Hot-rolled
deformed bar
Tension
reinforcement
543
2
61
Table 5-5 shows that all methods are conservative with high values of C.O.V.
from 21.0% to 23.1%. The small scale of the specimens may cause the somewhat greater
variation. Comparing the two same size beams CS2-H and CS4-H, they have the same
ratio of a/d, the same longitudinal reinforcement, while CS4-H with higher concrete
strength and two times more shear reinforcement (half of the stirrup spacing of CS2-H),
but the test results show that CS2-H has higher shear capacity than CS4-H. The same
thing happened on beam BS2-H and BS4-H. Because of the accuracy of the tests this
group of test samples will not be included in the further analysis, even though it is a good
example for demonstrating variables of beam shear capacity.
Table. 5-5, Test Results and Comparison of Predictions:
v
t
Vf
s
Vsim
Vg
Specimen
(kN)
(kN)
(kN)
(kN)
Vt/V
AL2-N
114.7
97.1
86.1
75.5
1.182
AL2-H
AS2-N
AS2-H
AS3-N
AS3-H
BL2-H
BS2-H
122.6
189.3
201.0
104.6
97.3
106.2
123.5
132.1
104.7
106.4
105.8
85.2
104.5
102.2
120.2
104.9
104.0
92.7
70.5
90.1
90.3
106.0
91.8
86.1
1.172
1.945
1.894
BS3-H
BS4-H
228.1
206.9
147.2
247.2
247.2
220.7
132.4
157.6
119.8
139.0
102.1
102.2
120.1
103.7
124.1
CL2-H
CS2-H
CS3-H
CS4-H
199.1
199.1
138.3
223.5
103.7
105.7
132.6
156.6
136.9
88.2
84.8
104.0
122.3
m
a
C.O.V.
rf
1.613
1.508
1.321
2.100
v /v
t
sim
1.331
1.159
2.222
1.923
1.948
1.656
1.319
2.149
Vt/V
g
1.520
1.323
2.685
2.231
2.204
1.878
1.506
2.596
2.340
1.865
1.409
1.905
1.489
1.442
2.420
2.058
1.612
2.199
1.668
1.669
2.914
1.63
1.76
2.04
0.35
21.2%
0.37
0.47
21.0%
23.1%
1.723
1.313
1.419
2.377
1.804
62
In Fig. 5-15, the ratios of V/V f against the ratios of shear span a/d are plotted.
s
The figure shows the prediction of shear capacity tends to be more conservative and more
scattered when the ratios of shear span a/d = 2.5. For the specimens with a/d = 4.0 the
prediction by the shear friction method agrees well with the test results. The prediction
results by the C S A simplified method and general method are plotted in Fig. 5-16 and
Fig. 5-17 respectively for comparison. The figures also show that the predictions tend to
be more conservative and more scattered for both methods when the ratios of shear span
a/d = 2.5.
1
1
1
1
1
1
1
1
•
•
1
•
•
•
•
•
2
Vrf
2.2
1
•
•
•
1
1
1
1
1
1
1
1
1
2.4
2.6
2.8
3
3.2
a
3.4
3.6
3.8
4
d
Fig. 5-15, Effect of the Ratio of Shear span on the Shear Friction Method
4.2
i
1
i
r
r
ft
•
a
•
•
sim
2.2
J
L
2.4
2.6
2.8
J
I
L
3
3.2
a
3.4
3.6
J
L
3.8
4
4.2
d
Fig. 5-16, Effect of the Ratio of Shear span on the Simplified Method
1
•
1
1
1
1
1
1
1
1
•
•
•
il
ft
•
•
•
2.2
1
1
1
1
1
1
1
1
1
2.4
2.6
2.8
3
3.2
a
3.4
3.6
3.8
4
d
Fig. 5-17, Effect of the Ratio of Shear span on the General Method
4.2
64
In Fig. 5-18, the ratios of V/V f against concrete strength, f' , are plotted to
s
c
demonstrate the effect of concrete strength on the shear friction method. It shows no
obvious trend in the prediction of shear capacity for beams with different concrete
strength. The prediction results by the C S A simplified method and general method are
plotted in Fig. 5-19 and Fig. 5-20 respectively for comparison. There is also no obvious
trend.
35
40
45
50
55
60
65
70
75
80
f c (MPa)
Fig. 5-18, Effect of Concrete Strength on the Shear Friction Method
85
l
i
l
1
i
1
1
1
1
•
•
•
•
2
o
•
•
n
•
v
sim
DDO
35
•
•
•
•
•
1
1
1
1
1
1
1
1
1
40
45
50
55
60
65
70
75
80
f
c
85
(MPa)
Fig. 5-19, Effect of Concrete Strength on the Simplified Method
1
1
1
1
1
1
1
b
•
•
•
• •
•
2
•
o
•
35
1
•
•
•
•
1
1
1
1
1
1
1
1
1
40
45
50
55
60
65
70
75
80
f c (MPa)
Fig. 5-20, Effect of Concrete Strength on the General Method
85
66
The ratios of test results to the results predicted by the shear friction method
against the ratios of s/d and the web reinforcement index pjvy are plotted in Fig. 5-21
and Fig. 5-22 respectively. The shear friction method demonstrates no obvious trend in
the prediction of shear capacity for beams with different stirrup spacing and different
amounts of shear reinforcement. The comparison results by the C S A simplified method
and general method are plotted in Fig. 5-23 to Fig. 5-26. There is no obvious trend.
1
1
•
•
s
1
1
1
1
•
n
a
D
•
•
•
•
't
V f
1
a
o
•
• OD
0
0.3
1
1
1
1
1
1
1
0.3
0.4
0.4
0.5
s
0.5
0.6
0.6
d
Fig. 5-21, Effect of Stirrup Spacing on the Shear Friction Method
0.7
67
V.t
V f
s
1
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
,f
Pv vy(MPa)
Fig. 5-22, Effect o f Shear Reinforcement on the Shear Friction Method
3i
1
1
1
i
r
•
v
c
sim
0.3
0.3
0.4
0.4
0.5
0.5
0.6
s
d
Fig. 5-23, Effect of Stirrup Spacing on the Simplified Method
0.6
0.7
68
1
i
1
i
i
i
i
i
•
B
•
•
v
sim
a
•
•
•
•
•
0.7
i
0.8
i
0.9
i
1
i
1.1
i
1.2
i
1.3
i
1.4
i
1.5
1.6
f
P - v y (MPa)
v
Fig. 5-24, Effect of Shear Reinforcement on the Simplified Method
1
1
1
1
1
d
1
•
•
2
•
•
•
•
•
0.3
1
1
1
1
1
1
1
0.3
0.4
0.4
0.5
0.5
0.6
0.6
s
d
Fig. 5-25, Effect of Stirrup Spacing on the General Method
0.7
69
0.7
I
I
I
I
0.8
0.9
1
1.1
I
I
I
1
I
1.2
1.3
1.4
1.5
1.6
f
P v ' v y (MPa)
Fig. 5-26, Effect of Shear Reinforcement on the General Method
In Fig. 5-27, the measured/calculated ratios of shear capacity versus the ratios of
beam longitudinal reinforcement p by the shear friction method are plotted. There is a
slight trend up when beams are reinforced with more bottom reinforcement. The
prediction results by the C S A simplified method and general method are plotted in Fig. 528 and Fig. 5-29 respectively for comparison. There is also an up-trend shown in both
Fig. 5-28 and Fig. 5-29 when increasing beam longitudinal reinforcement.
70
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
P (%)
Fig. 5-27, Effect of Longitudinal Reinforcement on the Shear Friction Method
1
1
1
1
1
1
1
•
•
•
•
•
Vt
V
•
•
•
c
sim
0
•
•
•
•
1
1
1
1
1
1
I
2.2
2.4
2.6
2.8
3
3.2
3.4
P (%)
Fig. 5-28, Effect of Longitudinal Reinforcement on the Simplified Method
3.6
71
( )
.
I
I
1
I
I
I
I
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
P (%)
Fig. 5-29, Effect of Longitudinal Reinforcement on the General Method
5.2.3 Summary of Tests from Literature:
Test data from a total of 113 beams with stirrups and 105 beams without stirrups
are selected from literature. A l l selected beams were simply supported rectangular beams
subjected to a symmetrical single or two-point load. The selected beams had different
concrete strengths from about 20 M P a up to more than 100 M P a . The selected beams also
had different shear span ratios, different amounts of longitudinal reinforcement, different
shear reinforcement and different stirrup spacing. Table 5-6 gives the details of
specimens with stirrups and Table 5-7 gives the details of specimens without stirrups.
Table 5-6, Details of Specimens with Stirrups
h
NAME
SPECIMEN
Rodriguez
et al
E2A1
E2A2
E2A3
C2A1
C2A2
E3H1
E3H2
C3H1
C3H2
Al
Bl
CI
Dl
D2
F5
F6
B50-3-3
B50-7-3
B50-11-3
B50-15-3
B100-3-3
B100-7-3
B100-11-3
B100-15-3
Debaiky &
Eliniema
Mphonde
B150-3-3
B150-7-3
B150-11-3
B150-15-3
(mm)
368
368
368
368
368
368
368
368
368
300
300
300
300
300
300
300
337
337
337
337
337
337
337
337
337
337
337
337
d
(mm)
152
152
156
154
157
152
152
152
152
120
120
120
120
120
120
120
150
150
150
150
150
150
150
150
150
150
150
150
(mm)
318
321
321
318
321
318
311
318
323
260
260
260
260
260
260
260
298
298
298
298
298
298
298
298
298
298
298
298
a
(mm)
274
274
274
274
274
267
267
267
267
234
234
234
234
224
232
232
282
282
282
282
276
276
276
276
276
276
276
276
(mm)
1295
1295
1295
1295
1295
864
864
864
864
975
775
975
975
975
975
975
1067
1067
1067
1067
1067
1067
1067
1067
1067
1067
1067
1067
(mm)
1193
1193
1193
1193
1193
762
762
762
762
900
700
900
900
900
900
900
965
965
965
965
965
965
965
965
965
965
965
965
fc
P
fy
(MPa)
25.5
19.3
20.1
22.6
22.1
24.8
27.5
22.6
22.8
24.5
24.5
28.0
29.8
30.6
20.2
20.5
22.1
39.8
59.7
83.0
27.9
47.1
68.6
81.9
28.7
46.6
69.5
82.7
(%)
1.57
1.57
1.57
1.57
1.57
1.57
1.57
1.57
1.57
2.89
2.89
2.89
2.89
2.89
2.41
1.92
3.4
3.4
3.4
3.4
3.4
3.4
3.4
3.4
3.4
3.4
3.4
3.4
(MPa)
320
325
331
318
324
380
377
385
410
408
408
408
408
408
408
408
448
448
448
448
448
448
448
448
448
448
448
448
2
(mm )
142
142
142
142
142
253
253
253
253
47
47
47
47
100
57
57
16
16
16
16
36
36
36
36
51
51
51
51
fvy
S
(MPa)
346
347
351
354
349
331
316
318
318
318
318
318
318
318
314
314
303
303
303
303
266
266
266
266
284
284
284
284
(mm)
254
254
254
254
254
152
191
152
191
200
200
200
100
200
200
200
89
89
89
89
89
89
89
89
89
89
89
89
(kN)
130.5
119.7
128.9
99.8
122.8
213.7
189.3
189.3
173.9
72.1
67.5
71.1
81.9
73.5
66.2
61.3
76.1
93.9
97.9
111.2
95.2
120.5
151.7
115.7
139.0
133.4
161.5
149.9
Table 5-6, Details of Specimens with Stirrups (cont.)
NAME
SPECIMEN
Elzanaty
et al
G4
G5
G6
No.l
No.2
No.4
No.5
No.7
No.8
1
2
3
4
5
6
7
9
10
A-l
A-2
B-l
B-2
C-l
C-2
NNW-3
NHW-3
NHW-3a
NHW-3b
NHW-4
Johnson
& Ramirez
Roller
& Russell
Bresler &
Scordelis
Xie
et al
h
b
d
d
a
a
f'c
(mm)
305
305
305
610
610
610
610
610
610
635
679
718
718
743
870
870
870
870
561
559
556
561
559
559
254 .
254
254
254
254
(mm)
178
178
178
305
305
305
305
305
305
356
356
356
356
356
457
457
457
457
307
305
231
229
155
152
127
127
127
127
127
(mm)
267
267
267
539
539
539
539
539
539
559
559
559
559
559
762
762
762
762
466
464
(mm)
242
242
242
535
535
535
535
535
535
502
520
547
547
572
724
724
724
724
436
434
461
466
464
464
203
198
198
198
198
431
436
434
434
185
185
185
185
185
(mm)
1067
1067
1067
1670
1670
1670
1670
1670
1670
1397
1397
1397
1397
1397
2286
2286
2286
2286
1827
2288
1821
2286
1831
2289
610
594
594
594
792
(mm)
967
967
967
1505
1505
1505
1505
1505
1505
1194
1194
1194
1194
1194
2083
2083
2083
2083
1611
2072
1605
2070
1615
2073
508
492
492
492
690
(MPa)
62.7
40.0
20.7
36.4
36.4
72.3
55.8
51.3
51.3
120.1
120.1
120.1
120.1
120.1
72.4
72.4
125.3
125.3
24.1
24.3
24.8
23.2
29.6
23.8
42.9
103.4
94.7
108.7
104.0
w
ev
c
P
(%)
3.3
2.5
2.5
2.5
2.5
2.5
2.5
2.5
2.5
1.65
3.04
4.56
6.08
6.97
1.73
1.88
2.35
2.89
1.8
2.28
2.43
2.43
1.8
3.66
3.2
4.54
4.54
4.54
4.54
fy
(MPa)
434
434
434
525
525
525
525
525
525
472
431
431
431
462
464
483
483
464
555
555
555
555
555
555
421
421
421
421
421
2
(mm )
63
63
63
63
63
63
63
63
63
63
253
396
396
396
143
143
143
143
63
63
63
63
63
63
63
63
63
63
63
fvy
S
(MPa)
379
379
379
479
479
479
479
479
479
407
448
458
458
458
445
445
445
445
325
325
325
325
325
325
324
324
324
324
324
(mm)
191
191
191
133
267
267
133
267
267
216
165
127
89
64
381
197
197
133
210
210
191
191
210
210
102
99
76
64
99
(kN)
147.2
113.2
77.6
338.5
221.9
315.9
382.7
280.8
258.1
297.3
1097.4
1655.0
1940.0
2234.5
665.4
787.9
749.4
1172.2
233.5
244.7
222.4
200.2
155.7
162.4
87.1
102.4
108.2
122.5
93.7
Table 5-6, Details of Specimens with Stirrups (cont.)
NAME
SPECIMEN
Anderson
& Ramirez
Wl
W2
W3
NIN
N2S
N2N
MIN
M2S
M2N
H1N
H2S
H2N
1
3
5
7
8
9
10
B-l
B-2
B-3
B-4
B-5
B-6
B-7
BM100
BM100D
Yoon
et al
Kriski
Peng
Podgorniak
& Stanik
h
b
w
d
dey
(mm)
406
406
406
750
750
750
750
750
750
750
750
750
400
400
400
400
400
400
400
320
320
320
320
320
320
320
1000
1000
(mm)
406
406
406
375
375
375
375
375
375
375
375
375
360
360
360
360
360
360
360
280
280
280
280
280
280
280
300
300
(mm)
343
343
343
655
655
655
655
655
655
655
655
539
345
345
345
345
345
345
345
274
274
274
274
274
274
274
925
925
(mm)
292
292
292
638
632
632
638
632
632
638
632
632
327
327
327
327
327
327
327
247
247
247
247
225
225
225
870
870
a
(mm)
914
914
914
2150
2150
2150
2150
2150
2150
2150
2150
2150
1050
1050
900
1050
900
1050
900
950
950
950
950
950
950
950
2700
2700
a
c
(mm)
812
812
812
2000
2000
2000
2000
2000
2000
2000
2000
2000
900
900
750
900
750
900
750
848
848
848
848
848
848
848
2548
2548
f'c
(MPa)
29.2
32.2
32.3
36.0
36.0
36.0
67.0
67.0
67.0
87.0
87.0
87.0
28.9
28.9
30.1
74.3
77.8
77.0
76.3
31.3
31.8
32.7
33.0
32.4
29.3
32.2
47.0
47.0
P
(%)
2.275
2.275
2.275
2.85
2.85
2.85
2.85
2.85
2.85
2.85
2.85
2.85
2.01
2.01
2.01
2.01
2.01
2.01
2.01
2.7
2.7
2.7
2.7
2.7
2.7
2.7
0.76
0.76
fy
(MPa)
434
434
434
400
400
400
400
400
400
400
400
400
433
433
433
433
433
433
433
478
478
478
478
478
478
478
550
550
2
(mm )
285
285
143
100
142
142
100
142
142
100
142
142
51
51
51
51
51
51
51
51
51
51
51
200
200
200
142
142
fvy
S
v,
(MPa)
544
544
544
430
430
430
430
430
430
430
430
430
600
600
600
600
600
600
600
587
587
587
587
456
456
456
508
508
(mm)
178
178
89
325
465
325
325
325
230
325
270
160
150
150
150
150
150
150
150
355
300
250
195
355
300
250
(kN)
458.2
548.9
504.4
457.0
363.0
483.0
405.0
552.0
689.0
483.0
598.0
721.0
249.0
224.5
293.0
304.5
391.0
242.0
390.5
114.0
119.0
121.0
143.0
181.0
191.0
187.0
600
600
342.0
461.0
Table 5-6, Details of Specimens with Stirrups (cont.)
NAME
SPECIMEN
Clark
Al-1
Al-2
Al-3
Al-4
Bl-1
Bl-2
Bl-3
Bl-4
Bl-5
Cl-1
Cl-2
Cl-3
Cl-4
Dl-1
Dl-2
Dl-3
YB2000/9
YB2000/6
YB2000/4
DB120M
DB140M
DB165M
DB180M
DB0.530M
SE100A-M-69
SE100B-M-69
SE50A-M-69
SE50B-M-69
Yoshida
Angelakos
Collins
& Kuchma
h
b
w
d
dev
a
a
f'c
(mm)
457
457
457
457
457
457
457
457
457
457
457
(mm)
203
203
203
203
203
203
203
203
203
203
203
203
203
203
203
203
300
300
300
300
300
300
300
300
295
295
169
169
(mm)
390
390
390
390
390
390
390
390
390
390
390
390
390
390
390
390
1890
1890
1890
925
925
925
925
925
920
920
459
459
(mm)
335
335
335
335
335
335
335
335
335
335
335
335
335
335
335
335
1814
1814
1814
860
860
860
860
860
896
896
447
447
(mm)
914
914
914
914
762
762
762
762
762
610
610
610
610
457
457
457
5400
5400
5400
2700
2700
2700
2700
2700
4600
4600
2500
2500
(mm)
825
825
825
825
673
673
673
673
673
521
521
521
521
368
368
368
5100
5100
5100
2548
2548
2548
2548
2548
4448
4448
2424
2424
(MPa)
24.6
23.6
23.4
24.8
23.4
25.4
23.7
23.3
24.6
25.6
26.3
24.0
29.0
26.2
26.1
24.5
36.0
36.0
36.0
21.0
38.0
65.0
80.0
32.0
71.0
75.0
74.0
74.0
457
457
457
457
457
2000
2000
2000
1000
1000
1000
1000
1000
1000
1000
500
500
c
P
(%)
3.1
3.1
3.1
3.1
3.1
3.1
3.1
3.1
3.1
2.07
2.07
2.07
2.07
1.63
1.63
1.63
0.74
0.74
0.74
1.01
1.01
1.01
1.01
0.50
1.03
1.03
1.03
1.03
fy
(MPa)
321
321
321
321
321
321
321
321
321
321
321
321
321
335
335
335
455
455
455
550
550
550
550
550
483
483
475
475
A
v
2
(mm )
142
142
142
142
142
142
142
142
142
142
142
142
142
142
142
142
645
284
129
142
71
71
71
71
200
200
51
51
f vy
S
(MPa)
331
331
331
331
331
331
331
331
331
331
331
331
331
331
331
331
467
467
467
508
508
508
508
508
522
522
593
593
(mm)
183
183
183
183
191
191
191
191
191
203
203
203
203
152
152
152
2700
1350
590
600
300
300
300
300
440
440
276
276
(kN)
222.5
209.2
222.5
244.7
278.9
256.6
284.8
268.1
241.4
277.7
311.1
245.9
285.9
301.1
356.7
256.6
474.0
551.0
659.0
282.0
277.0
452.0
395.0
263.0
516.0
583.0
139.0
152.0
76
Table 5-7, Details of Specimens without Stirrups
h
NAME
SPECIMEN
Kani et al
24
25
26
85
87
94
100
27
28
b
w
d
a
a
fc
v
P
fy
(MPa)
27.9
24.5
27.1
25.5
27.2
25.3
27.2
29.8
29.2
(%)
1.88
1.88
1.88
2.80
2.80
2.80
2.80
1.88
1.88
24.5
25.2
26.1
26.1
27.5
27.4
27.4
27.4
30.3
25.3
25.3
27.2
26.2
1.88
1.88
1.88
1.88
2.80
2.80
2.80
2.80
2.80
2.80
2.80
2.80
2.80
(MPa)
396
396
396
381
366
352
366
396
396
350
350
491
491
343
343
342
364
372
338
335
366
366
155
285
425
555
695
985
1235
1505
29.6
29.6
29.6
29.6
29.6
29.6
31.0
31.0
2.07
2.07
2.07
2.07
2.07
2.07
2.01
2.01
465
465
465
465
465
465
465
465
388.3
259.9
147.1
81.6
60.3
60.8
62.3
65.7
699
699
699
699
699
699
762
762
30.3
31.0
31.0
31.5
21.2
21.6
2.17
2.15
2.22
2.37
1.62
1.63
313
313
313
313
313
313
60.1
66.7
75.6
71.2
56.3
60.1
36.7
25.8
1.89
313
1.89
313
57.8
52.3
30.7
1.89
313
52.0
30.9
1.89
313
51.2
c
29
30
35
36
81
83
84
91
93
95
96
97
99
(mm)
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
305
Leonhardt
& Walther
1
2
3
4
5
6
7
8
320
320
320
320
320
320
320
320
Moody et al
Al-A
A2-A
A3-A
A4-A
Bl-A
B2-A
305
305
305
305
305
305
Bl-B
B3-B
305
305
(mm) (mm) (mm)
152
271
407
152
271
543
152
271
543
154
274
272
154
269
272
153
273
543
153
270
544
152
271
678
152
271
678
152
271
1221
152
271
1221
155
269
953
153
273
953
153
274
1628
156
271
814
151
271
1085
154
269
1628
155
273
1763
153
275
678
153
275
1085
152
276
815
152
272
679
190
270
270
190
270
400
190
270
540
190
270
670
190
270
810
190
270
1100
190
278
1350
190
278
1620
178
262
800
178
267
800
178
268
800
178
270
800
178
800
267
178
268
800
152
268
914
152
268
914
B5-B
305
152
268
914
762
B7-B
305
152
268
914
762
(mm)
305
441
441
170
170
441
442
577
577
1119
1119
851
851
1526
712
983
1526
1662
577
983
714
578
t
(kN)
181.9
104.1
78.1
233.5
239.5
110.5
111.9
51.4
54.3
42.9
46.3
44.9
51.6
51.2
64.9
55.4
51.0
53.8
72.7
56.3
62.5
77.2
77
Table 5-7, Details of Specimens without Stirrups (cont.)
h
NAME
SPECIMEN
Van Den Berg
A4-1
A4-2
A4-3
A4-4
A4-5
A4-6
A4-7
A4-8 .
A4-9
A4-10
A4-12
Dl
D2
D3
D4
D5
D6
D7
D8
DIO
DU
D12
D13
D14
D15
D16
D17
D18
D19
D20
Mphonde
& Frantz
AO-3-3b
AO-3-3c
AO-7-3a
AO-7-3b
AO-11-3a
AO-11-3b
AO-15-3a
AO-15-3b
AO-15-3c
b
w
d
a
a
c
fc
(mm) (mm) (mm) (mm) (mm) ( M P a )
419
229
359
991
889
43.6
419
229
359
1372 1270
38.9
419
229
359
1448 1346
41.8
419
229
359
1524 1422
38.9
419
229
359
1257 1156
39.6
419
229
359
1524 1422
44.9
419
229
359
1257 1156
50.3
419
229
359
1600 1499
42.8
419
229
359
1676 1575
47.6
419
229
359
1753 1651
35.4
419
229
359
991
889
44.0
419
229
359
1257 1156
49.8
419
229
359
1257 1156
43.0
419
229
359
1257 1156
36.1
419
229
359
1257 1156
35.5
419
229
359
1257 1156
43.0
419
229
359
1257 1156
41.3
419
229
359
1257 1156
32.2
419
229
359
1257 1156
25.5
419
229
359
1257 1156
26.7
419
229
359
1257 1156
19.1
419
229
359
1257 1156
23.3
419
229
359
1257 1156
20.8
419
229
359
1257 1156
23.9
419
229
359
1257 1156
22.3
419
229
359
1257 1156
25.9
419
229
359
1257 1156
22.2
419
229
359
1257 1156
24.4
419
229
359
1257 1156
27.4
419
229
359
24.2
1257 1156
337
152
298
1074 973
20.8
337
152
298
1074 973
27.1
337
152
298
1074 973
37.7
337
152
298
1074 973
41.6
337
152
298
1074 973
74.9
337
152
298
1074 973
74.6
337
152
298
1074 973
81.3
337
152
298
1074 973
93.7
337
152
298
1074 973
91.8
P
fy
v
(%)
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
(MPa)
(kN)
310
310
310
310
310
310
310
310
310
310
310
310
310
310
310
310
310
310
310
177.9
133.4
134.3
135.0
133.4
142.3
142.3
124.5
131.2
122.3
177.9
151.2
131.2
129.0
144.6
131.2
140.1
140.1
117.9
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
4.53
310
310
310
310
310
310
310
310
310
310
310
126.8
109.0
106.8
99.2
106.8
102.3
111.2
104.5
104.5
115.7
106.8
3.36
2.32
3.36
3.36
3.36
414
414
414
414
414
3.36
3.36
3.36
414
414
414
64.6
66.8
82.2
82.8
89.7
89.4
93.5
100.0
3.36
414
97.9
t
Table 5-7, Details of Specimens without Stirrups (cont.)
h
NAME
SPECIMEN
Ahmad et al
Al
A2
A3
A7
A8
A9
Bl
B2
B3
B7
B8
B9
Cl
C2
C3
C7
C8
C9
Bresler
& Scordelis
OA-1
OA-2
OA-3
Xie et al
Yoon et al
K
d
a
a
v
fc
P
fy
(mm) (mm)
813
711
610
508
549
447
832
731
624
522
562
460
807
705
605
503
545
443
832
731
624
522
562
460
737
635
552
451
497
396
826
724
620
518
558
456
(MPa)
(MPa)
414
414
414
414
414
414
414
414
414
414
414
414
414
414
414
414
414
414
(kN)
60.8
60.8
60.8
60.8
60.8
60.8
67.0
67.0
67.0
67.0
67.0
67.0
64.3
64.3
64.3
64.3
64.3
64.3
(%)
3.93
3.93
3.93
1.77
1.77
1.77
5.04
5.04
5.04
2.25
2.25
2.25
6.64
6.64
6.64
3.26
3.26
3.26
1829
2286
3200
1613
2070
2985
22.5
23.7
37.6
1.81
2.27
2.74
555
555
555
166.8
177.9
189.0
648
648
546
546
39.7
104.2
2.07
2.07
421
421
36.7
45.7
655
655
2150
2150
2000
2000
36.0
67.0
2.80
2.80
400
400
249.0
296.0
655
2150
2000
87.0
2.80
400
327.0
NNN-3
NHN-3
(mm) (mm) (mm)
254
127
203
254
127
203
254
127
203
254
127
208
254
127
208
254
127
208
254
127
202
254
127
202
254
127
202
254
127
208
254
127
208
254
127
208
254
127
184
254
127
184
254
127
184
254
127
207
254
127
207
254
127
207
556
310
461
561
305
466
556
307
462
254
127
216
254
127
216
N1S
MIS
750
750
375
375
HIS
750
375
c
t
57.8
68.9
68.9
46.7
48.9
80.1
51.2
68.9
100.1
44.6
46.7
80.1
54.3
75.6
68.9
45.4
44.5
45.4
79
5.2.3.1 Beams with Shear reinforcement:
Table 5-6 gives the details of specimens with stirrups. Table 5-8 gives the
summary of the comparison of predictions by three different methods for beams with
stirrups. The predictions by the shear friction method are slightly conservative. The
average value of V/V fis
s
1.09 and the coefficient of variation (C.O.V.) is 16.6%. The
predictions by the simplified method have a larger scatter. The average value of V/V
sim
is
1.16 and the coefficient of variation (C.O.V.) is 20.4%. The most conservative
predictions with the largest scatter are obtained by the general method. The average value
of V/V is 1.35 and the coefficient of variation (C.O.V.) is 24.6%.
g
Table. 5-8, Comparison of Predictions for Beams with Stirrups
v /v
t
vw
sf
t
NAME
No.
m
a
C.O.V
m
Rodriguez
9
1.18
0.098
8.3%
Debaiky
7
1.07
0.113
Mphonde
12
1.09
Elzanaty
3
1.11
Johnson
v /v
t
sim
e
C.O.V
m
a
C.O.V
1.04 0.112
10.7%
1.05
0.160
15.2%
10.6%
1.24 0.143
11.6%
1.39
0.195
14.0%
0.106
9.7%
1.23 0.116
9.4%
1.47
0.184
12.5%
0.223
20.0%
1.19 0.140
11.8%
1.37
0.172
12.6%
6
0.94 0.081
8.6%
1.02 0.076
7.5%
1.16
0.130
11.2%
Roller
9
1.13
0.208
18.3%
1.07 0.228
21.4%
1.26
0.314
25.0%
Bresler
6
1.12
0.047
4.2%
1.33 0.070
5.2%
1.61
0.079
4.9%
Xie
5
0.96 0.035
3.7%
1.08 0.053
4.9%
0.049
4.1%
Anderson
3
1.09
0.115
10.5%
1.11
0.074
6.7%
1.20
1.12
0.078
6.9%
Yoon
9
1.11
0.081
7.3%
1.12 0.130
11.6%
1.46
0.326
22.4%
Kriski
7
1.10
0.189
17.1%
1.20 0.190
15.9%
1.39
0.230
16.5%
Peng
7
1.16
0.170
14.7%
1.07 0.058
5.4%
1.17
0.980
8.4%
Podgorniak
2
0.99 0.147
14.8%
1.09 0.162
14.8%
1.75
0.260
14.8%
Clark
16
1.29
0.152
11.7%
1.49 0.184
12.4%
1.57
0.252
16.0%
Yoshida
3
0.86
0.021
2.4%
0.94 0.298
31.7%
1.74
0.814
46.7%
Angelakos
5
0.78
0.128
16.3%
0.80 0.180
22.7%
1.09
0.410
37.7%
Collins
4
0.90 0.049
5.5%
0.80 0.039
4.9%
0.91
0.045
4.9%
Total
113 1.09
16.6%
1.16 0.236
20.4%
1.35
0.331
24.6%
0.182
a
80
Fig. 5-30 to Fig. 5-32 illustrate the results of predictions by the shear friction
method, the C S A simplified method and the C S A general method respectively.
Comparing the figures it is clear that the shear friction method gives the best prediction
of the three methods.
2500
(kN)
V r f 1250
•oo
a
0
1250
2500
V
(kN)
t
Fig. 5-30, Predicted Results by the Shear Friction Method
81
2500
(kN)
•
•
simi250
•
•
»
D
D
• •
o
>
1250
2500
V
(kN)
t
Fig. 5-31, Predicted Results by the Simplified Method
82
2500
(kN)
V
g 1250
o
• •
•
•
1250
v.
Fig. 5-32, Predicted Results by the General Method
2500
(kN)
83
In Fig. 5-33, the ratios of V/V f against the ratios of shear span a/d are plotted.
s
Generally speaking, the prediction by the shear friction method agrees well with the test
results.
In Fig. 5-34 and Fig. 5-35, the measured/calculated ratios of shear capacity versus
the ratios of shear span a/d by the C S A simplified method and general method are plotted
respectively. There is a larger scatter of the results predicted by the general method
compared to that predicted by shear friction.
In Fig. 5-36, the ratios of V/V j- against concrete strength, f' , are plotted to
s
c
demonstrate the effect of concrete strength on the shear friction method. It shows no
obvious trend in the prediction of shear capacity for beams with different concrete
strength. The predicted results by the simplified method and the general method against
concrete strength, f' , are plotted in Fig. 5-37 and Fig. 5-38 respectively.
c
The ratios of test results to the results predicted by the shear friction method
against the ratios of s/d and the web reinforcement index f\f y
V
are plotted in Fig. 5-39
and Fig. 5-42 respectively. The shear friction method demonstrates no obvious trend in
the prediction of shear capacity for beams with different stirrup spacing and different
amounts of shear reinforcement. Fig. 5-40 and Fig. 5-43 give the results by the simplified
method against the ratios of s/d and the web reinforcement index / V y y - The results by the
general method against the ratios of s/d and the web reinforcement index Pyf y are shown
V
in Fig. 5-41 and Fig. 5-44. It is notable that the lowest ratios occur with small / V v y
t 0
a
^
three methods.
In Fig. 5-45, Fig. 5-46 and Fig. 5-47, the measured/calculated ratios of shear
capacity versus the ratios of beam longitudinal reinforcement p by the shear friction
method, the simplified method and the general method are plotted respectively. When p >
1.2%, all three methods show that there is no obvious trend in the predictions of shear
capacity for beams with different amounts of longitudinal reinforcement. But, when p <
1.2%, the lowest ratios occur to all three methods while the general method has a larger
scatter in this range.
84
Fig. 5-48 to Fig. 5-50 illustrate the effects of beam depth to predictions of shear
capacity by the three methods. There is no obvious trend in the predictions of shear
capacity for beams with different beam depth for all three methods.
Fig. 5-33, Effect of the Ratio of Shear Span on the Shear Friction Method
85
Fig. 5-34, Effect of the Ratio of Shear Span on the Simplified Method
•
•
li
•
a
•
•
•
B
S
•
•
•
UP
m
• • a
B° S
Q
1.5
2.5
•
•
•
•
•
•
•
•
•
•
o
•
3
3.5
•
4.5
a
d
Fig. 5-35, Effect of the Ratio of Shear Span on the General Method
•
•
5.5
86
60
80
100
f c (MPa)
Fig. 5-36, Effect of Concrete Strength on the Shear Friction Method
1
V
1
1
1
1
1
B
t
•
_ O g1 0 0^ §
• • •
n
Bü
Q•
% m
° g3
sim
D
•
•
^
•
• ft •
° •
°
•
0
1
i
i
i
20
40
60
80
f
c
i
100
i
120
(MPa)
Fig. 5-37, Effect of Concrete Strength on the Simplified Method
•
140
87
1
1
1
1
1
1
•
B
•
n
Dn
u
rfio
V
g
• •
k
i
o
• •
D
•
a
a
D
•
a „
an
Û On
o
°
• *
•
rj 11
OO
D
œ
•
0
D
D
D
g
1=1
•
a •
u
°
•
n
•
•
1
1
1
1
1
1
20
40
60
80
100
120
f
c
(MPa)
Fig. 5-38, Effect of Concrete Strength on the General Method
Fig. 5-39, Effect of Stirrup Spacing on the Shear Friction Method
140
88
1
1
1
1
1
1
1
2
•
•
sim
g
f
Bn
1
í
• LP
•
•
U
•
a
n
1
1
!
U
n
•
•
•
•
D
Bo
•
0
1
1
1
1
1
1
1
0.2
0.4
0.6
0.8
s
1
1.2
1.4
d
Fig. 5-40, Effect of Stirrup Spacing on the Simplified Method
Fig. 5-41, Effect of Stirrup Spacing on the General Method
1.6
89
1
31
v
1
1
1
1
1
1
rn
2
3
4
5
6
7
8
t
"
0
1
Pv-fvy(MPa)
Fig. 5-42, Effect of Shear Reinforcement on the Shear Friction Method
31
I
0
1
I
2
I
3
I
1
4
5
1
6
1
7
8
f
p - v y (MPa)
v
Fig. 5-43, Effect of Shear Reinforcement on the Simplified Method
•
a
•
•M
ŒJ
1
• OD
•
•
-O
•
B-
•
•
0
0
1
2
3
4
5
6
7
f
P - v y (MPa)
v
Fig. 5-44, Effect of Shear Reinforcement on the General Method
Fig. 5-45, Effect of Longitudinal Reinforcement on the Shear Friction Method
91
1
sim
1
•
1•
D
D
13
Í
1
•
a
s
s Ha¡3l
u
D
3
1
1
•
•
PS
•
i
•
•
a
1
0
1
•
j
•
•
1
1
2
i
i
i
3
4
5
1
i
7
i
6
p (%)
Fig. 5-46, Effect of Longitudinal Reinforcement on the Simplified Method
•
-Qv
u
•
B
D
t
•
•
Beg
•
B-O
•
•
•
a
2
3
4
P (%)
Fig. 5-47, Effect of Longitudinal Reinforcement on the General Method
92
r
i
v
t
°B
Vsf
1
a
B
-cr
l
•
•
0
200
J
L
400
600
J
800
1000
L
1200 1400
h (mm)
J
1600
L
1800
2000
Fig. 5-48, Effect of Beam Depth on the Shear Friction Method
i
E B
v sim
c
•
Ë
• D
-B-
r
•
•
•
a
•
0
200
400
J
L
J
600
800
1000
I
1200 1400
h (mm)
1600
1800
Fig. 5-49, Effect of Beam Depth on the Simplified Method
2000
93
3
1
1
1
1
1
1
1
1
1
•
•
•on
HfitWmnmnrrrn
2
B
a
•
•
•
200
U
• n
• •
• n
B
"
D
D
°
o •
°b
•
n
•
•
Ü
•
1
1
1
1
400
600
800
1000
•
1
1
1200 1400
h (mm)
1
1
1
1600
1800
2000
Fig. 5-50, Effect of Beam Depth on the General Method
5.2.3.2 Beams without Shear reinforcement:
Table 5-9 gives the summary of the comparison of predictions by three different
methods for beams without stirrups. The predictions by the shear friction method are
conservative. The average value of the measured/calculated ratios of shear capacity,
V/V A is 1.41 and the coefficient of variation (C.O.V.) is 15.9%. The predictions by the
S
simplified method and by the general method are more conservative than the shear
friction
method
and
have
much
larger
scatters.
The
average
value
of
the
measured/calculated ratios of shear capacity by the simplified method, V/V , is 1.58
sim
and the coefficient of variation (C.O.V.) is 59.2%. The average value of the
measured/calculated ratios of shear capacity by the general method, V/V , is 1.80 and the
g
coefficient of variation (C.O.V.) is 62.1%.
94
Table. 5-9, Comparison of Predictions for Beams without Stirrups
NAME
No.
m
V,/V
a
Kani
22
1.35
0.193
Leonhardt
8
1.29
Moody
10
1.26
0.226 17.6%
0.082 6.5%
Van Den Berg
30
1.58
0.101
Mphonde
9
1.55
0.142
Ahmad
Bresler
18
1.35
3
Xie
sf
C.O.V
m
14.3%
2.01
2.51
V,/V
a
v,/v
sim
e
m
1.328 66.2%
2.26
1.500 66.4%
3.13
1.39
2.523 80.5%
1.26
2.049 81.7%
0.090 7.2%
0.096
6.9%
6.4%
1.40
0.124
8.8%
1.50
0.132
8.8%
9.2%
1.29
0.168
13.1%
1.69
0.185
10.9%
0.265
1.49
0.399 26.7%
1.62
0.449 27.8%
1.47
19.6%
0.080 5.4%
0.096
7.1%
1.90
0.133
7.0%
2
0.96
0.030
3.1%
1.35
0.94
1.09
3
1.13
0.052
4.6%
0.97
1.11
0.076
0.083
7.0%
Yoon
13.1%
0.073 7.5%
Total
105
1.41
0.225
15.9%
1.58
0.936 59.2%
1.80
1.119 62.1%
0.123
a
C.O.V
C.O.V
7.5%
Fig. 5-51 to Fig. 5-53 illustrate the results of predictions by the shear friction
method, the C S A simplified method and the C S A general method respectively.
Comparing the figures it is clear that the shear friction method gives the best prediction
of the three methods.
95
200
V
t
Fig. 5-51, Predicted Results by the Shear Friction Method
400
(kN)
96
400
(kNJ
•
V
• • •
s i m
200
• o
•
a
poo
en
•
200
400
v.
(kN)
Fig. 5-52, Predicted Results by the Simplified Method
97
400
(kN)
•
•
V 200
• na
g
200
400
(kN)
Fig. 5-53, Predicted Results by the General Method
98
In Fig. 5-54, the ratios of V/V s against the ratios of shear span a/d are plotted.
s
The figure shows that the predictions of shear capacity by the shear friction method are
within a very narrow range to the test results. Notice that the ratios of V/V f does not
s
change significantly even when the ratios of shear span, a/d, are less than 2.5.
The measured/calculated ratios of shear capacity versus the ratios of shear span,
a/d, by the C S A simplified method and the C S A general method are plotted in Fig. 5-55
and Fig. 5-56. There are larger scatters of the results than the scatter predicted by the
shear friction method. Notice that the ratios of V/V
sim
a
n
¿ V/V get significant larger
g
when the ratios of shear span, a/d, are less than 2.5.
In Fig. 5-57, the ratios of V/V f against concrete strength, f' , are plotted to
s
c
demonstrate the effect of concrete strength on the shear friction method. It shows no
obvious trend in the prediction o f shear capacity for beams with different concrete
strength. The predicted results by the simplified method and the general method against
concrete strength, f' , are plotted in Fig. 5-58 and Fig. 5-59 respectively. There is no
c
obvious trend in the prediction of shear capacity for both of the methods.
The ratios of test results to the results predicted by the shear friction method, the
simplified method and the general method against the ratios of longitudinal reinforcement
ratio, p, are plotted in Fig. 5-60, Fig. 5-61 and Fig. 5-62 respectively. N o obvious trend
can be found in any of the figures.
In Fig. 5-63, the ratios of V/V j- against the ratios of longitudinal reinforcement
s
strength, T/T , are plotted. The figure shows that the predictions of shear capacity by
opt
the shear friction method are within a very narrow range to the test results and there is no
obvious trend against T/T .
opt
Fig. 5-64 to Fig. 5-66 show the effects of beam depth to predictions o f shear
capacity by the three methods. There is no obvious trend in the predictions of shear
capacity for beams with different beam depth for all three methods.
9
8
7
6
v.i
5
Vrf
4
• • •
3
2
•co
D
A
1
3
O
4
a
Fig. 5-54, Effect of the Ratio of Shear span on the Shear Friction Method
9
8
7
-o-
6
h
5
^sim
• • •
4
•
•
3
•
2
1
00
1
2
3
4
5
6
a
d"
Fig. 5-55, Effect of the Ratio of Shear span on the Simplified Method
'
100
9
8
7
•
•
6
5
4
• • •
3
TO
2
-B
faff
i
1
0
BD t
•
n
jp fln°
a
0
a
7
Fig. 5-56, Effect of the Ratio of Shear span on the General Method
9
i
1
r
r
8
7
6
v
t
Vrf
• • a
5
4
3
•
2
0
10
20
30
40
o
-EL
•
1
50
J
L
60
70
f
(MPa)
c
80
-tr
J
I
90
100
Fig. 5-57, Effect of Concrete Strength on the Shear Friction Method
110
101
9
8
7
6
_ Ï L
^sim
• • •
5
4
dpoü
3
o
2
fea a l i
1
0 10
20
30
40
50
• •
_Q
•—
60
70
f
(MPa)
c
80
90
100
110
Fig. 5-58, Effect of Concrete Strength on the Simplified Method
9
8
7
-QCL
6
v
t
•
5
4
• • D
• •
-on—
3
2
•
o r
•
10
20
30
40
in
Ü-|J
1
0
•
50
60
70
f
(MPa)
c
•
80
_L
_L
90
100
Fig. 5-59, Effect of Concrete Strength on the General Method
110
102
8
7
6
v
t
Vrf
• on
5
4
3
2
1
1
-B•
•
_L
0
2
3
4
5
6
7
P (%)
Fig. 5-60, Effect of Longitudinal Reinforcement Ratio on the Shear Friction Method
9
8
-B-
7
6
5
Vs i•m
v
4
ODD
3
• •
-¾—cr
2
1
0
s
9 |
DCL
•
•
•
•
•
J
4
P (%)
Fig. 5-61, Effect of Longitudinal Reinforcement Ratio on the Simplified Method
103
9
8
7
6
•
5
4
• • •
3
•
2
-rJ3„ qj
n
-e-
•
"XT
-R-
1
•
"ET
•
•
_D_
0
Fig. 5-62, Effect of Longitudinal Reinforcement Ratio on the General Method
9
8
7
6
V.1
5
Vrf
• • •
4
3
2
•
1
Q
0.5
cP •1
•
1.5
an
IB
•
a
2
2.5
^
•
•
•
3.5
T
opt
Fig. 5-63, Effect of Longitudinal Reinforcement Strength on the Shear Friction Method
1
300
%0
400
500
h (mm)
600
800
700
Fig. 5-64, Effect of Beam Depth on the Shear Friction Method
-B-
§ a
• I
-BJL
>00
300
400
500
600
700
h (mm)
Fig. 5-65, Effect of Beam Depth on the Simplified Method
800
105
•
•
•
~o—
-B-fl-
%0
300
400
500
600
700
h (mm)
Fig. 5-66, Effect of Beam Depth on the General Method
800
106
CHAPTER 6
PROPOSED C O D E C L A U S E S FOR S H E A R D E S I G N
6.1 Proposed Code Clauses for shear design
6.1.1 Required Shear Resistance
Member subjected to shear shall be proportioned so that
V >V
sf
(6-1)
f
6.1.2 Factored Shear Resistance:
The factored shear resistance shall be determined by
(6-2)
r -V +V++V,
¥
t¥
But V shall not exceed
sf
Kr=^JXh
+V
(6-3)
p
6.1.3 Determination of V
csf
:
The value of V shall be computed from
csf
V^=y/V tane
45
(6-4)
Where #is given in Clause 6.1.5, i//is given in Clause 6.1.6 and
(6-5)
107
/
\ 0.30 ,
0.25
i 30)
(500^
(6-6)
P =0.36
v
\fc
J
6.1.4 Determination of V :
ssf
To determine V.
ss/
(a) For members
with
transverse
longitudinal axis, the value o f V
ssf
V
y
ssf
=V
r
(décote
reinforcement
perpendicular
to the
shall be computed from
(6-7)
-1
si
Where #is given in Clause 6.1.5 and
(6-8)
Ki = AAAvy
(b) For members with transverse reinforcement inclined at an angle a to the
longitudinal axis, the value of V
ssf
shall be computed from
d (cot 0 + cot a)
ev
sina
(6-9)
6.1.5 Determination o f &.
The value o f 0 for each design section of a member shall be computed from
tan 6
K d e v
V
4i
s
(6-10)
108
Where a is the distance from the face of the support to the applied concentrated
c
load location, or from the face of the support to the design section for uniform load.
6.1.6 Determination of yr.
The value of y/for each design section of a member shall be computed from
1// = 1
When T>T
(6-11)
When T<T
(6-12)
opi
f
-
\
y/ = sin n T
opt
2 T
Where
T = AA f
s
(6-13)
y
2
T =V (2 + tan 0)
opl
4S
(6-14)
6.1.7 Limiting Shear Failure Angle:
The failure angle 0 shall satisfy the following conditions:
tan6>
(6-15)
a,.
IL
6>2l
K
(6-16)
30j
6.2 Design Examples
Design a simply supported beam for shear to span 4 m. The beam carries one 300
k N concentrated live load at midspan. Use 25 M P a concrete and 400 M P a steel. The
beam will have 6-No.25 longitudinal reinforcing bars at bottom and will use No. 10
stirrups.
109
(1) Factored shear force:
Vf = 1.5x300/2 = 225 k N
(2) Beam size:
Choose beam size b x h = 400 mm x 600 mm
. n 25
\ OJO ,
/
30
¡3 =0.36
0.36 x
V
V n )
V
45
= ^c¡3v^TcbJi
^30^
(
y 25 y
\600 j
500^
= 0.36
= 1.0x0.6x0.36x425x400x600x
3
10~ =259 k N
(3) Check maximum shear capacity:
3
Ka* = -<Pcf'cKh = -x0.6x25x400x600x
4
4
>V =
10~ = 900 k N
225 k N , ok!
f
(4) Stirrup spacing:
Assume T > T
opt
d
ev
= h-(c,+c
+4d ) = 600-(40 + 40 + 4x11.3) = 475 mm
b
b
3
Ki =<¡> AJ =
s
4V
s 4sV d
s<—
sl
0.85 x 100 x2x 400 xlO'
vy
4x259x68x475
, = —;
-—
ev
= 390 mm
2
(Vf+Vsi)
(225 + 68)
Choose stirrup spacing at 350 mm.
(5) Calculate shear angle 6 :
x
tanO
V
45
6 = 31.0°
s
259
= 0.60
350
3 f ì n
=68
m
(6) Calculate yr.
T = fcAJ =0.85x6 x 500x400 xl0~
3
=1020 k N
y
2
2
T
=V (2 + tan 0) = 259x(2 + tan (31.0)) = 612 k N
r>r
o p
opl
45
„ S o i// = i
(7) Check shear failure angle:
V
68
ev
tan0 = ' "
^
5
\ 1x259
0 = 31.0° >21
v
475
•x
350
600
a,
7600
= 20.3°
30j
y
h
1 5
2f]
27 x
50
=o.60>
K
(8) Check shear capacity of beam:
v =v
sf
csf
+ v +v
ssf
p
y/V tan0 + Vsi
45
(d cot0
^
ev
i
= 1x259 x 0.60 + 68x1
I
4
7
5
x
1
6
7
- ?
350
= 155.4 + 86.1 = 241.5 k N > V, = 225 kN, ok!
Ill
CHAPTER 7
CONCLUSIONS A N D RECOMMENDATIONS
The following conclusions and recommendations were made based on the
analyses of the test data from a total of 113 beams with stirrups and 105 beams without
stirrups from the literature. A l l selected beams were simply supported rectangular beams
subjected to a symmetrical single or two point load. The selected beams had different
concrete strengths from about 20 MPa up to more than 120 MPa. The selected beams also
had different shear span ratios from about 1 up to 7, different amounts of longitudinal
reinforcement, different shear reinforcement and different stirrup spacing.
7.1 Conclusions and Recommendations:
1. The shear friction method changes the way of beam shear design by
simplifying design procedure and increasing accuracy. The results of the
comparison show that the predictions by the shear friction method had less
scatter with test results compared to the predictions by the C S A simplified
method and general method.
2. The simplified method and the general method are only suitable for beams
with shear span, a/d, greater than 2.5. But the shear friction method may be
used for short shear span situations, such as deep beams.
3. According to the analysis in Chapter 5, stirrup spacing of beams has no
obvious effects on the predictions of the shear capacity of beams by all three
methods. Notice that the limitation on maximum stirrup spacing by the C S A
A23.3-94 clauses 11.2.11 for the simplified method and the general method
was ignored during the analysis. Further investigation should be conducted to
determine whether the limitation on maximum stirrup spacing needs to be
revised.
112
4. The calibration factor, f3 , accurately predicts the effects of concrete strength
v
for the ranges of 20 M P a to 125 M P a and beam height for the ranges of 250
mm to 2000 mm.
5. The shear capacity of beams relies on not only the longitudinal reinforcement
of beams, but also the anchorage of such reinforcement. So any excessive bar
cut offs near beam supports shall be avoided.
6. The equation for calculating minimum shear failure angle, 0 ,
mm
is appropriate
in the prediction of the shear capacity of beams without stirrups. It is not
necessary for shear calculations for beams with the usual amounts of shear
reinforcement.
7.2 Future Research:
1. Different types of beams should be included in the analysis to determine the
range of applicability of the shear friction method, including such beams as
continuous beams, T-beams, prestressed beams and cantilevered beams.
2. The beams with longitudinal reinforcement ratio from 0.5% to 1.5% are very
popular in practice. But there is a lack of test data in that reinforcement range
and more beams in that reinforcement range should be included in the
analysis..
3. Because distributed loads are very common in beam design, it is important to
include tests with distributed loads in the analysis to determine
applicability of the shear design methods.
the
113
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