# File - Swedish College Of Engineering & Technology

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Reinforced Concrete Design-I
Lec-07
Bond and Development Length
By
Dr. Attaullah Shah
Swedish College of Engineering and Technology
Wah Cantt.
− The basic assumption of the
RCC design is that the strain
in concrete and reinforcing
steel is the same. If the
reinforcing steel slips at its
ends, this is not valid. Hence
it must be ensured that
sufficient bond strength is
developed at the interface of
steel and concrete to avoid
slippage of the steel.
−
Bond Strength and Development length
− Two types of bond failure can
be expected in reinforcing bars:
− Direct pull out of the steel bars,
when ample concrete confinement
is provided in the form of large
spacing of bars or large concrete
cover
− Splitting of concrete along the bar
when cover confinement or bar
spacing is insufficient
b. Actual Distribution of Flexural Bond Stress
− Pure bending case
− Concrete fails to resist tensile
stresses only where the actual
crack is located. Steel T is
maximum and
T max = M / jd .
− Between cracks , concrete
does resist moderate amount of
tension introduced by bond.
− u is proportional to the rate of
change of bar force, and
highest where the slope of the
steel force curve is greatest.
− Very high local bond stress
6
− According to simple crack
sectional theory, T is proportional
to the moment diagram and u is
proportional to shear force
diagram.
− In actual, T is less than the
simple analysis prediction
everywhere except at the actual
cracks.
− Similarly, u is equal with simple
analysis prediction only at the
location where slopes of the steel
force diagrams are equals .If the
slope is greater than assumed,
bond stress is greater; if the slope
is less bond stress is less.
7
ULTIMATE BOND STRENGTH AND
DEVELOPMENT LENGTH
− Types of bond failure
− Direct pullout of bars
(small diameter bars are
used with sufficiently
large concrete cover
distances and bar
spacing)
− Splitting of the concrete
along the bar (cover or
bar spacing is
insufficient to resist the
lateral concrete tension
resulting from the
wedging effect of bar
deformations)
8
Depar
tment
a. Ultimate Bond Strength
− Direct pull out
− For sufficiently confined bar, adhesive bond and friction are overcome as the
tensile force on the bar is increased. Concrete eventually crushes locally ahead
of the bar deformation and bar pullout results.
− When pull out resistance is overcome or when splitting has spread all the
way to the end of an unanchored bar, complete bond failure occurs.
− Splitting
− Splitting comes from wedging action when the ribs of the deformed bars bear
against the concrete.
− Splitting in vertical plane
− Splitting in horizontal plane: frequently begins at a diagonal crack in
connection with dowel action. Shear and bond failures are often interrelated.
− Local bond failure
− Large local variation of bond stress caused by flexural and diagonal cracks
beam.
− Results small slip and some widening of cracks and increase of deflections.
− Harmless as long as the failure does not propagate all along the bar.
− Providing end anchorage, hooks or extended length of straight bar (development
length concept)
9
Consider a bar embedded in a
mass of concrete
P = s * [p*db2/4]
P = t*[Lb*p*db]
db
Lb
t = P / [Lb*p*db] < tmax
s = P/ [p*db2/4] < smax
P < tmax * [Lb*p*db]
P < smax * [p*db2/4]
To force the bar to be the weak link: tmax * [Lb*p*db] > smax * [p*db2/4]
Lb > (smax / tmax)* [db/4]
Development Length
− Ld = development length
− the shortest distance over which a bar can achieve it’s
full capacity
− The length that it takes a bar to develop its full
contribution to the moment capacity, Mn
Ld
Mn
0
Cc
Mn = (C or T)*(dist)
Ts
Steel Limit, smax
− Using the bilinear assumption of ACI 318:
smax = + fy
Lb > (fy / tmax)* [db/4]
Lb > fy * db / (4*tmax)
Concrete Bond Limit, tmax
− There are lots of things that affect tmax
−
−
−
−
−
−
The strength of the concrete, f’c
Type of concrete (normal weight or light weight)
The amount of concrete below the bar
The surface condition of the rebar
The concrete cover on the bar
The proximity of other bars transferring stress to
the concrete
− The presence of transverse steel
Concrete Strength, f’c
− Bond strength, tmax, tends to increase with concrete
strength.
− Experiments have shown this relationship to be
proportional to the square root of f’c.
Type of Concrete
− Light weight concrete tends to have less bond
strength than does normal weight concrete.
− ACI 318-08 introduces a lightweight concrete
reduction factor, l, on sqrt(f’c) in some equations.
− See ACI 318-08, 8.6.1 for details
Amount of Concrete Below Bars
− The code refers to “top bars”
as being any bar which has
12 inches or more of fresh
concrete below the bar
when the member is poured.
− If concrete > 12” then
consolidation settlement
results in lower bond
strength on the bottom side
of the bar
− See ACI 318-08, 12.2.4(a)
Surface Condition of Rebar
− All rebar must meet ASTM requirements for deformations
that increase pullout strength.
− Bars are often surface coated is inhibit corrosion.
− Epoxy Coating  The major concern!
− Galvanizing
− Epoxy coating significantly reduces bond strength
− See ACI 318-08, 12.2.4(b)
Proximity to Surface or Other Bars
− The size of the concrete “cylinder” tributary to
each bar is used to account for proximity of
surfaces or other bars.
Presence of Transverse Steel
− The bond transfer tends to cause a splitting plane
− Transverse steel will increase the strength of the splitting
plane.
b. Development Length
− Development length is the length of embedment necessary to develop
the full tensile strength of bar, controlled by either pullout or splitting.
− In Fig., let
− maximum M at a and zero at support
− fs at a T = Ab fs _
− Development length concept total tension force must be
transferred from the bar to the concrete in the distance ‘l ‘ by
bond stress on the surface.
− To fully develop the strength  T = Ab fy
 ld , development length
− Safety against bond failure: the length of the bar from any point of
given steel stress to its nearby end must be at least equal to its
development length. If the length is inadequate, special anchorage
can be provided.
ACI CODE PROVISION FOR DEVELOPMENT
OF TENSION REINFORCEMENT

Limit


(c + ktr) / db = 2.5 for
pullout case
√f’c are not to be
greater than 100 psi.
21
Depar
tment
For two cases of practical importance, using (c + ktr) / db = 1.5,
22
Example:
23
Continue:
24
Depar
tment
Continue:
25
ANCHORAGE OF TENSION BARS BY
HOOKS
In the event that the desired tensile stress in a bar can not
be developed by bond alone, it is necessary to provide
special anchorage at the end of the bar.
26
27
b. Development Length and Modification
Factors for Hooked Bars
28
29
Department of
Example
30
Depar
tment
ANCHORAGE REQUIREMENTS FOR
WEB REINFORCEMENT
31
DEVELOPMENT OF BARS IN
COMPRESSION
− Reinforcement may be
required to develop its
compressive strength by
embedment under various
circumstances.
− ACI basic development
length in compression
ldb = 0.02db fy/√f’c
32
Depar
tment
Determining Locations of Flexural
Cutoffs
Given a simply
supported beam with a
Determining Locations of Flexural
Cutoffs
Note:
Total bar length =
Fully effective length
+ Development length
Determining Locations of Flexural
Cutoffs
ACI 12.10.3
All longitudinal tension bars
must extend a min. distance
= d (effective depth of the
member) or 12 db (usually
larger) past the theoretical
cutoff for flexure (Handles
approximations,etc..)
Determining Locations of Flexural
Cutoffs
Development of flexural
reinforcement in a typical
continuous beam.
ACI 318R-02 - 12.10 for
flexural reinforcement
Bar Cutoffs - General Procedure
1. Determine theoretical flexural cutoff points for
envelope of bending moment diagram.
2. Extract the bars to satisfy detailing rules (from
ACI Section 7.13, 12.1, 12.10, 12.11 and 12.12)
3. Design extra stirrups for points where bars are
cutoff in zone of flexural tension (ACI 12.10.5)
Bar Cutoffs - General Rules
All Bars
Rule 1.
Bars must extend the longer of d or 12db past
the flexural cutoff points except at supports
or the ends of cantilevers (ACI 12.11.1)
Rule 2.
Bars must extend at least ld from the point
of maximum bar stress or from the flexural
cutoff points of adjacent bars (ACI 12.10.2
12.10.4 and 12.12.2)
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3. Structural Integrity
− Simple Supports At least one-third of the positive
moment reinforcement must be extend 6 in. into
the supports (ACI 12.11.1).
− Continuous interior beams with closed stirrups.
At least one-fourth of the positive moment
reinforcement must extend 6 in. into the support
(ACI 12.11.1 and 7.13.2.3)
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3. Structural Integrity
− Continuous interior beams without closed
stirrups. At least one-fourth of the positive
moment reinforcement must be continuous or
shall be spliced near the support with a class A
tension splice and at non-continuous supports be
terminated with a standard hook. (ACI 7.13.2.3).
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3.
Structural Integrity
− Continuous perimeter beams. At least onefourth of the positive moment reinforcement
required at midspan shall be made continuous
around the perimeter of the building and must be
enclosed within closed stirrups or stirrups with
135 degree hooks around top bars. The required
continuity of reinforcement may be provided by
splicing the bottom reinforcement at or near the
support with class A tension splices (ACI
7.13.2.2).
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 3. Structural Integrity
− Beams forming part of a frame that is the
primary lateral load resisting system for the
building. This reinforcement must be anchored
to develop the specified yield strength, fy, at the
face of the support (ACI 12.11.2)
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 4.
Stirrups
− At the positive moment point of inflection and
at simple supports, the positive moment
reinforcement must be satisfy the following
equation for ACI 12.11.3. An increase of 30 %
in value of Mn / Vu shall be permitted when the
ends of reinforcement are confined by
compressive reaction (generally true for simply
supports).
Bar Cutoffs - General Rules
Positive Moment Bars
Rule 4.
Mn
ld 
 la
Vu
Bar Cutoffs - General Rules
Negative Moment Bars
Rule 5.
− Negative moment reinforcement must be
anchored into or through supporting columns or
members (ACI Sec. 12.12.1).
Bar Cutoffs - General Rules
Negative Moment Bars
Rule 6. Structural Integrity
− Interior beams. At least one-third of the negative
moment reinforcement must be extended by the
greatest of d, 12 db or ( ln / 16 ) past the negative
moment point of inflection (ACI Sec. 12.12.3).
Bar Cutoffs - General Rules
Negative Moment Bars
Rule 6. Structural Integrity
− Perimeter beams. In addition to satisfying rule 6a,
one-sixth of the negative reinforcement required at
the support must be made continuous at mid-span.
This can be achieved by means of a class A tension
splice at mid-span (ACI 7.13.2.2).
Moment Resistance Diagrams
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.
Moment Resistance Diagrams
The nominal moment capacity of an under-reinforced
concrete beam is
a

M n  As f y d  
2

where, a 
As f y
0.85 fcb
To determine the position of the cutoff or bent point
Moment Resistance Diagrams
The ultimate moment resistance of one bar, Mnb is
a

M nb  Abs f y d  
2

where, Abs  area of bar
The intersection of the moment resistance lines with
the external bending moment diagram indicates the
theoretical points where each bar can be terminated.
Moment Resistance Diagrams
Given a beam with the 4 #8 bars and
fc=3 ksi and fy=50 ksi and d = 20 in.
Moment Resistance Diagrams
The moment diagram is
Moment Diagram
3000
2500
k-in
2000
1500
1000
500
0
0
2
4
6
8
10
ft
12
14
16
18
20
Moment Resistance Diagrams
The moment resistance of one bar is
a

M nb  Asb f y  d  
2

3.16 in 2   50 ksi 
As f y

a

 5.2 in.
0.85 f cb 0.85  3 ksi 12 in.
M nb
M ub
5.2 in. 

  0.79 in   50 ksi   20 in. 
 688 k-in.

2 

  M nb  0.9  688 k-in.  620 k-in.
2
Moment Resistance Diagrams
The moment diagram and crossings
Moment Diagram
3000
2480 k-in
2500
1860 k-in
k-in
2000
1240 k-in
1500
1000
620 k-in
500
0
0
2
4
6
8
10
ft
12
14
16
18
20
Moment Resistance Diagrams
The ultimate moment resistance is 2480 k-in. 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 termination of the bars.
Moment Diagram
3000
2480 k-in
2500
1860 k-in
k-in
2000
a
1500
1240 k-in
1000
620 k-in
c
500
b
0
0
2
4
6
8
10
ft
12
14
16
18
20
Moment Resistance Diagrams
Compute the bar development length is
la  12d b or d
 12 1.0 in. or 20 in.  20 in.
ld 
l f y d b
20 f c
50000 1.0 in.


20 3000
 45.6 in.  46 in.
Moment Resistance Diagrams
The ultimate moment
resistance is 2480 k-in.
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 termination
of the bars.
Moment Resistance Diagrams
It is necessary to develop
part of the strength of the bar
by bond. The ACI 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 12.11.1 specify that
1/3 of positive moment
reinforcement must be
continuous.
Moment Resistance Diagrams
Two bars must extend
into the support and
moment resistance
diagram Mub must
enclose the external
bending moment
diagram.
Example – Cutoff
For the simply
supported beam with
b=10 in. d =17.5 in.,
fy=40 ksi and fc=3 ksi
with 4 #8 bars. Show
where the reinforcing
bars can be terminated.
Example – Cutoff
Determine the moment capacity of the bars.
a
M nb
As f y
0.85 f cb
3.16 in   40 ksi 


 4.93 in.
2
0.85  3 ksi 10 in.
4.93 in. 

  0.79 in   40 ksi  17.5 in. 

2


 427.7 k-in.  35.64 k-ft.
2
Example – Cutoff
Determine the location of the bar intersections of
moments.
1 bar  35.64 k-ft.
3 bar  107 k-ft.
2 bar  71.3 k-ft.
4 bar  142.6 k-ft.
M  x   M0  mx
Example – Cutoff
Determine the location of the bar intersections of
moments.
1 bar  35.64 k-ft.
3 bar  107 k-ft.
2 bar  71.3 k-ft.
4 bar  142.6 k-ft.
 132.5 k-ft.  87.5 k-ft. 
107 k-ft.  132.5 k-ft.  
x
6 ft.


x  3.4 ft.  40.8 in. or 41 in.
Example – Cutoff
Determine the location of the bar intersections of
moments.
1 bar  35.64 k-ft.
3 bar  107 k-ft.
2 bar  71.3 k-ft.
4 bar  142.6 k-ft.
 87.5 k-ft.  0.0 k-ft. 
71.3 k-ft.  87.5 k-ft.  
x
5 ft.


x  0.93 ft.  11.1 in. or 11 in.
or 11 in. + 72 in. = 83 in. from center
Example – Cutoff
The minimum distance is
la  12d b or d
 12 1.0 in. or 17.5 in.  18 in.
ld 
l f y d b
20 f c
40000 1.0 in.


20 3000
 36.6 in.  37 in.
Example – Cutoff
The minimum amount of bars are As/3 or two bars
Example – Cutoff
The cutoff for the first bar is 41 in. or 3 ft 5 in. and 18 in
or 1 ft 6 in. total distance is 41 in.+18 in. = 59 in. or 4 ft
11 in.
Note error it is 4’-11” not 5’-11”
Example – Cutoff
The cutoff for the second bar is 83 in. + 18 in. 101 in. or
8 ft 5 in. (37-in+5-in+18-in+41-in= 101-in.)
Note error it is 4’-11” not 5’-11”
Example – Cutoff
The moment diagram is the blue line and the red line is
the envelope which encloses the moment diagram.
Bar Splices
Why do we need bar splices? -- for long spans
Types of Splices
1.
Butted &Welded
2.
Mechanical Connectors
3.
Lay Splices
Must develop 125%
of yield strength ACI
12.14.3.2 and ACI
12.14.3.4
Tension Lap Splices
Why do we need bar splices? -- for long spans
Types of Splices
1.
Contact Splice
2.
Non-Contact Splice (distance between the
bars  6” and  1/5 of the splice length
ACI 12.14.2.3)
Splice length (development length) is the distance
the two bars are overlapped.
Types of Splices
Class A Splice
(ACI
12.15.2)
Asprovided
When
 2 over entire splice
Asreq'd 
length.
and 1/2 or less of total reinforcement is
spliced win the req’d lay length.
Types of Splices
Class B Splice
(ACI 12.15.2)
All tension lay splices not meeting
requirements of Class A Splices
Tension Lap Splice (ACI 12.15)
where As (req’d)
ld
= determined for bending
= development length for bars (not
allowed to use excess reinforcement
modification factor)
ld must be greater than or equal to 12 in.
Tension Lap Splice (ACI 12.15)
Lap Splices shall not be used for bars larger than No. 11.
(ACI 12.14.2)
Lap Splices should be placed in away from regions of
high tensile stresses -locate near points of inflection
(ACI 12.15.1)
Compression Lap Splice (ACI 12.16)
Lap, req’d = 0.0005fy db
Lap, req’d = (0.0009fy -24) db
Lap, req’d  12 in
for fy  60000 psi
for fy > 60000 psi
For fc  3000 psi, required lap splice shall be multiply
by (4/3) (ACI 12.16.1)
Compression Lap Splice (ACI 12.17.2)
In tied column splices with effective tie area throughout
splice length  0.0015 hs factor = 0.83
In spiral column splices, factor = 0.75
The final splice length must be
 12 in.
Example – Splice Tension
Calculate the lap-splice length for 6 #8 tension bottom
bars in two rows with clear spacing 2.5 in. and a clear
cover, 1.5 in., for the following cases
a. When 3 bars are spliced and As(provided) /As(required) >2
b. When 4 bars are spliced and As(provided) /As(required) < 2
c. When all bars are spliced at the same location.
fc= 5 ksi and fy = 60 ksi
Example – Splice Tension
For #8 bars, db =1.0 in and  =  = g = l =1.0
3 fy
gl
ld

d b 40 f c  c  K tr 


 db 
3  60000 
1.0
 42.4  43 in.

40 5000  1.5 in.  0 


 1.0 in. 
Example – Splice Tension
The As(provided) /As(required) > 2, class A splice applies;
therefore lst = 1.0 ld >12 in., so lst = 43 in. > 12 in.
The bars spliced are less than half the number
The As(provided) /As(required) < 2, class B splice applies;
therefore lst = 1.3 ld >12 in., so lst = 1.3(42.4 in.) = 55.2 in.
use 56 in. > 12 in..
Class B splice applies and lst = 56 in. > 12 in.
Example – Splice Compression
Calculate the lap splice length for a # 10 compression
bar in tied column when fc= 5 ksi and
a) fy = 60 ksi
b) fy = 80 ksi
Example – Splice Compression
For #10 bars, db =1.27 in.
ld 0.02 f y

 0.003 f y
db
fc

0.02  60000 
 16.97 or 18
5000
ld  18 1.27 in.  22.86 in.  ld  23 in.
Check ls > 0.005 db fy = 38.1 in. So ls = 39 in.
Example – Splice Compression
For #10 bars, db =1.27 in. The ld = 23 in.
Check ls > (0.0009 fy –24) db
=(0.0009(80000)-24)(1.27in.) = 61 in.
So use ls = 61 in.
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