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Mechanism of bond transfer
Bond stresses provide mechanism of force transfer betweenconcrete and reinforcement.
Equilibrium Condition for Rebar
Note: Bond stress is zero at cracks
avgis the value at bond failure in a beam test
2
by b b
y b
d
F 0 Bond Force 0
0
4
4
avg
avg
T
df d l
f dl
= =
=
=
= bond stress
(adhesion,
friction, bearing,
etc.)
( )c
bar
k f
k f
=
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Simplifiedld (A23.3 Table 12.1)
Cases
Case 1: Member containing minimum
stirrups or ties within ld(A23.3 Cl.11.2.8.4
or A23.3 Cl.7.6.5)
Case 2: Slabs, walls, shells, or folded plates
having clear spacing between bars being
developed not less than 2db
Other cases
Minimum developmentlength, ld(mm)
1 2 3 4 y
b'
0.45 d
c
k k k k f
f
1 2 3 4 y
b'0.6 d
c
k k k k f
f
k1 is a bar location factor.
k2 is a coating factor.
k3 is a concrete density factor.
k4 is a bar size factor.
ldis the development length, mm
db is the diameter of the bar being developed , mm
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Factors used in expressions for
Development Length (A23.3 Table 12.1)
1- k1 = Bar location factor
1.3 Horizontal reinforcementmore than 300 mm of
fresh concrete is cast in the member below the
development length or splice (considered top rft)
1.0 Other reinforcement2- k2 = Coating factor
1.5 Epoxy-coated bars or wires with cover less than3dbor clear spacing less than 6db
1.2 All other epoxy-coated bars or wires
1.0 Uncoated reinforcement
3- k3 = Low-density aggregate concrete factor
1.3 Low-density concrete1.2 Semi-low-density concrete1.0 Normal-density concrete4- k4 = Bar size factor
0.8
No.20 and smaller bars and deformed wires1.0 No.25 and larger bars
The product k13k2 need not be taken greater than 1.7
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Excess Flexural Reinforcement
Reduction (A23.3 Cl.12.2.5)
Reduction = (As reqd ) / (As provided )
- Except as required for seismic design (see A23.3
Cl.12.2.5)
- Good practice to ignore this provision, since use
of structure may change over time.
- final ld 300 mm.
Development Length for Bars inCompression (A23.3 CL.12.3)
Compression development length,
ldc= ldb3 applicable modification factors 200 mm.
A23.3 Cl.12.3.3 gives the modification factor for excess
reinforcement and enclosure by spiral or ties
Basic development length for compression, ldbc
b y
db b'
0.240.044 d y
c
d fl f
f=
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Compression development length:
The basic development length ldb, may be multipliedby applicable factors, with a cumulative value of
not less than 0.6, for
(a)Reinforcement in excess of that required byanalysis (As reqd ) / (As provided )
(b)Reinforcement enclosed within spiralreinforcement of not less than 6 mm diameter and
not more than 100 mm pitch or within No.10 ties in
conformance with Cl.7.6.5 and spaced at not morethan 100 mm on centre 0.75
Example
For the shown cross section of a
simply supported beam
reinforced with 4 No.15 bars that
are confined with No.10 stirrup
spaced at 150 mm, determine the
development length of the bars if
the beam is made of normalweight concrete, uncoated
reinforcement bars, given that fc
= 30 MPa and fy= 400 MPa
300
4No.15
60 60
560
500
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1 2 3 4 yd b
'
1 3
2 4
0.45 d
1.0 1.0
1.0 0.8
c
k k k k f l
f
k k
k k
=
= =
= =
The development length is:
1 2 3 4 y
d b'
0.45 d
1 1 1 0.8 4000.45 16 420.6530
425 mm 300mm
c
d
d
k k k k f l
f
l
l
=
= =
=
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Standard Hooks for Tension Anchorage
Use of Standard Hooks for Tension Anchorage
Hooks provide additional anchorage when
there is insufficient length available to
develop a bar.
Note: Hooks are not allowed to developed
compression reinforcement.
Standard Hooks
A hook is used at the end of a
bar when its straight
embedment length is less
than the necessary length, ld.
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Standard hooks are
defined in A23.1 Cl.
12.2.2
Hooks resists tension by
bond stresses on bar
surface and bearing on on
concrete inside the hook.
Design of Standard Hooks for Tension
Anchorage (A23.3 Cl.12.5)
Development length for hooked bar, ldh.
multipliers
where, 8 or 150 mm whichever is smaller
dh hb
dh b
l l
l d
=
Basic development length for hooked bar = lhbwhen
fy = 400 MPa
'
100
hb b
c
l df
=
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Conditions
For No.35 or smaller bars, where the side cover
(normal to plane of hook) is not less than 60 mm
and for 90o hooks where the cover on the bar
extension beyond the hook is not less than 50 mm
Multiplier
fy /400
0.7
For bars with fy other than 400 MPa
Conditions
For No.35 or smaller bars, where the hook is
enclosed vertically or horizontally within at
least three ties or stirrup ties spaced along a
length at least equal to the inside diameter of
the hook, at a spacing not greater than 3db ,
where db
is the nominal diameter of the
hooked bar
Multiplier
0.8
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Conditions
Anchorage or development for fy is not
specially required, for reinforcement in
excess of that required by analysis
Structures low density concrete
Multiplier
As(reqd) /
As(provided)
1.3
Conditions
Epoxy-coated Reinforcement
Multiplier
1.2
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ExampleCompute the development length required for the top 3
No.20 bar of the cantilever beam that extend into the
column support if:
The bars are confined by #10 stirrups spaced at
200mm, and clear cover = 70 mm, and clear
spacing = 50 mm and fc =35 MPa and fy =400 MPa
(a)Anchored by hooksinto the column
(b)Developed in thebeam, b=300 mm
90o Hook
(a)3 No. 20 bars anchored in the column by standard90ohook
From the size of the column, the space available for
a hook is 450-50-0.5320=390 mm
The development length is the product of the basic
hook length lhb and the factors in A23.3 Cls.
12.5.3 and 12.5.4
( )
hb
b c
hb
100 10016.9
35
16.9 20 mm 338 mm 340 mm
l
d f
l
= = =
= =
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Cl.12.5.3 (a) does not apply...31.0
Cl.12.5.3(b) the bars are No.35 or smaller, the side
cover to the hooks is 70 mm, and the tail cover is 50
mm. Therefore, 12.5.3 (b) applies...30.7
Cl.12.5.3(c) does not apply.31.0
Cl.12.5.3(d) will be ignored 31.0
Cl.12.5.3(e) does not apply .31.0
Cl.12.5.3(f) does not apply ..31.0
180o Hook
Check weather Cl.12.5.4 requires joist ties. Side
cover=70mm. Top cover extends into the column
over the hook. Therefore extra ties are not needed in
the joint.
ldh= 238 mm > 8db = 160 mm or 150 mm.
ldh 390 mm , the hook will anchor the bar
hb
dh
340 mm
340 0.7 238 mm
l
l
=
= =
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In-beam development
(b) 3 No. 20 bars anchored in the beam
Development length available in the beam
=1200-50=1150 mm
v
2
V
0.06 35 300 200Min. A for s=200 mm=
400
A 53.2 mm
=
The spacing is less than the maximum spacing fromCl.11.2.11. therefore , the beam has more than
minimum stirrups and this is Case 1 in Table 12-1 of
A23.3
K1=1.3 ( more than 300 mm of fresh concrete below
the bars when the concrete is placed
K2=1.0 bars are not coated
K3=1.0 concrete is normal- density
K4=0.8 bars are No.20
1 2 3 4 y
d b'
d
0.45 d
1.3 1.0 1.0 0.8 400l 0.45 20 633 mm
35
c
k k k k f
lf
=
= =
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Bar Cutoff PointsWhy do you want to provide cut off points?
Cost!
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Factors Affecting Bar Cut-off Points
Bars no longer needed to resist tensile forces or
remaining bars are adequate (Use moment andshear envelopes)
Bars must be extended on each side of section to
develop bar force at that section.
1.)
2.)
Major stress concentrations occur when tension
bars are cutoff in regions of moderate to high
shear forces. Leads to cracking.
Code specified construction requirements (good
practice)
3.)
4.)
Load uncertainties (Seismic Considerations)5.)
Determining Locations of Flexural
Cutoffs
A23.3 Cl.12.10.2
All longitudinal tension
bars must extend a min.
distance = d(effective depth
of the member) or12 db
(usually larger) past the
theoretical cutoff for flexure
(Handles uncertainties inloads, design approximations,
etc.)
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Bar Cutoffs - General Rules
! Simple Supports At least one-third of the positive momentreinforcement must be extend 150 mm into the supports(A23.3 Cl.12.11.1).
! Continuous interior beams with closed stirrups. At leastone-fourth of the positive moment reinforcement mustextend 150 mm into the support (A23.3 Cl.12.11.1)
Positive Moment Bars
Rule 1-P
! Continuous interior beams without closed stirrups. Atleast one-fourth of the positive moment reinforcement
must be continuous or shall be spliced near the supportwith a class A tension splice and at non-continuous
supports be terminated with a standard hook. (A23.3 Cl.
12.11.2).
Rule 2-P
Bars must extend the longer of d or 12db past the flexural
cutoff points except at supports or the ends of cantilevers
(A23.3 Cl.11.3.8.1)
Bars must extend at least ld from the point of maximum
bar stress or from the flexural cutoff points of adjacent
bars (A23.3 Cls.12.10.2 and 12.10.4
Rule 3-P
d
ra
f
Ml lV
+ A23.3 Eq.12-6
lais the longer of the effective depth , d, or 12 db
Rule 4-P
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Bars must extend the longer ofdor12db past the flexural
cutoff points except at supports or the ends of cantilevers
(A23.3 Cl.11.3.8.1)
Bars must extend at least ld from the point of maximum
bar stress or from the flexural cutoff points of adjacent
bars (A23.3 Cls.12.10.2 and 12.10.4)
Rule 2-N
Rule 3-N
Negative Moment Bars
Rule 1-N
At least one-third of the negative moment reinforcement must
be extended by the greatest ofd, 12 db or ( ln/ 16 ) past the
negative moment point of inflection (A23.3 Cl.12.12.2).
Negative moment reinforcement must be anchored into or
through supporting columns or members (A23.3 Cl.12.12.1).
Rule 4-N
Critical Sections in FlexuralMembers
The critical sections for development of reinforcement in
flexural members are:
At points of maximum stress;
At points where tension bars within span are
terminated or bent;
At the face of the support;
At points of inflection at which moment
changes sign.
1.
2.
3.
4.
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Moment Capacity Diagram
Moment capacity of a beam is a function of its depth,
d, width, b, and area of steel, As. It is common
practice to cut off the steel bars where they are no
longer needed to resist the flexural stresses. As in
continuous beams positive moment steel bars may be
bent up usually at 45o, to provide tensile
reinforcement for the negative moments over the
support.
The resisting moment capacity of an under-reinforced
concrete beam is
To determine the position of the cutoff or bent point
the moment diagram due to external loading is drawn.
s y
r s y '
1
where,2
s
s
c c
A faM A f d a
f b
= =
The ultimate moment resistance of one bar, Mrb is
The intersection of the moment resistance lines with
the external bending moment diagram indicates the
theoretical points where each bar can be terminated.
rb bs y bswhere, is the area of one bar2
s aM A f d A =
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Example
Given a beam with the 4 #25 bars and
fc=30 MPa and fy=400 MPa and
d = 500 mm
6.0 m
60 kN/m
300
500
4 No.2560
Factored Moment Diagram
The moment diagram is
Moment Diagram
0
50
100
150
200
250
0 1 2 3 4 5 6
m
kN.m
300
500
4 No.25
60
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The factored moment of the beam is
The moment resistance of the beam is
f
60 36M 270 kN.m
8
= =
( )( )( )( )
( )( )
rb sb y
2
s y
1 c
2
r
2
0.85 2000 mm 400 MPa156.43 mm
0.6 0.805 30 MPa 300 mm
156.43 mmM 0.85 2000 mm 400 MPa 500 mm 286.8 kN.m
2
s
s
c
aM A f d
A fa
f b
=
= = =
= =
The moment resistance of one bar is
( )( )( )( )
( )( )
rb sb y
2
s y
1 c
2
rb
2
0.85 500 mm 400 MPa39.11 mm
0.6 0.805 30 MPa 300 mm
39.11 mm0.85 500 mm 400 MPa 500 mm 81.68 kN.m
2
s
s
c
aM A f d
A fa
f b
M
=
= = =
= =
The moment resistance of two bars is
( )( )( )( )
( )( )
rb sb y
2
s y
1 c
2
r
2
0.85 1000 mm 400 MPa78.215 mm
0.6 0.805 30 MPa 300 mm
78.215 mmM 0.85 1000 mm 400 MPa 500 mm 156.7 kN.m
2
s
s
c
aM A f d
A fa
f b
=
= = =
= =
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The moment resistance of three bars is
( )( )( )( )
( )( )
rb sb y
2
s y
1 c
2
rb
2
0.85 1500 mm 400 MPa117.32 mm
0.6 0.805 30 MPa 300 mm
117.32 mm0.85 1500 mm 400 MPa 500 mm 225.08 kN.m
2
s
s
c
aM A f d
A fa
f b
M
=
= = =
= =
Compute the bar development length is
( )a b
d
12 or d
12 25 mm or 500 mm 500 mm
4000.45 1 1 1 1 25 821.58 mm
30
825 mm
l d
l
=
=
= =
=
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Moment Capacity Diagram
The ultimate moment
resistance is 270 kN.m
The moment diagram
is drawn to scale on
the basis A bar can be
terminated at a, two
bars at b and three bars
at c. These are the
theoretical terminationof the bars.
60 kN/m
6.0 m
81.68
156.7
225.08
270
Moment Resistance Diagrams
It is necessary to develop part
of the strength of the bar by
bond. The A23.3 Code
specifies that every bar
should be continued at least a
distance d, or 12db , which
ever is greater, beyond the
theoretical points a, b, and c.Section 12.11.3.8.1 specify
that 1/3 of positive moment
reinforcement must be
continuous at least 150 mm
into the support.
Two bars must extend
into the support and
moment resistance
diagram Mub must
enclose the external
bending moment
diagram.
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Bar Splices
Why do we need bar splices? -- for long spans
Types of Splices
1. Butted &Welded
2. Mechanical Connectors
3. Lap Splices
Must develop 120%
of yield strength of
the bar but not less
than 110% of theactual yield strength
of the bar used.
A23.3 12.14.3.1
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Types of Splices
1-Splices for deformed bars and wires in Tension
The minimum length of lap for tension lap splices
shall be as required for a Class A or B splice, but
not less than 300 mm, where
(a) Class A splice 1.0ld
(b) Class B splice ....1.3 ld
Where ld is the tensile development length for the
specified yield strength without the modification
factor of Cl. 12.2.5
Tension Lap Splice (A23.3 Cl.
12.15.2 Table 12-2)
As (provided)
As (required)
Equal to or
greater than 2
Less than 2
Maximum percent of As spliced
within required lap length
50 100
Class A
Class B
Class B
Class B
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Compression Lap Splice (A23.3 12.16)
The minimum length of the lap for compression lap
splices shall be not less than 0.073fydb, nor
(0.133fy-24)db for fy greater than 400 MPa, nor
300mm
*Welded splices or mechanical connections used in
compression shall meet the requirements of
Cl. 12.14.3.3 or 12.14.3.4
Example
Calculate the lap-splice length for 6 No.25 tension
bottom bars in two rows with clear spacing 60 mm and a
clear cover, 40 mm, for the following cases
When 3 bars are spliced and As(provided) /As(required) >2
When 4 bars are spliced and As(provided) /As(required) < 2
When all bars are spliced at the same location.fc= 40 MPa and fy = 400 MPa
a.
b.
c.
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The As(provided) /As(required) > 2, class A splice applies;
therefore lst = 1.0 ld >300 mm, so lst = 657.27 mm >
300 mm. The bars spliced are less than half the
numberThe As(provided) /As(required) < 2, class B splice applies;
therefore lst = 1.3 ld >300 mm, so lst = 1.3(657.27 mm) =
854.45 mm use 860 mm > 300 mm
For No.25 bars, db =25.2 mm
d
4000.45 1 1 1 0.8 25 657.27 mm
30l = =
Example
a) fy = 400 MPa , and b) fy = 500 MPa
Calculate the lap splice length for a No. 30 compression
bar in tied column when fc= 30 MPa and
For No.30 bars, db =29.9 mm
'
0.240.044
0.24 29.9 400524.06 0.044 29.9 400 526.24
30
0.24 29.9 500655.077 0.044 29.9 500 657.8
30
b y
db b y
c
db
db
d fl d f
f
l
l
=
= = =
= = =
Use ldb=530 mm for (a) and ldb=660 mm for (b)