Chapter-7
Bond Development Length
& Splices
Lecture Goals
• Slab design reinforcement
• Bar Development
• Hook development
Flexural Reinforcement in Slabs
For a 1 ft strip of slab is designed like a beam
As(req’d) is in units of (in2/ft)


12 in
As / ft  Ab 

 bar spacing in inches 
The table is A-9 from
MacGregor’s book.
Flexural Reinforcement in Slabs
The minimum spacing of the bars is given as:
S max
3t slab thickness
 smaller of 
18 in.


ACI Sec. 7.6.5
Also, check crack control - important for exterior
exposure (large cover dimensions) - ACI Sec. 10.6.4
Flexural Reinforcement in Slabs
Maximum & Minimum reinforcement requirements
• Thin slabs shrink more rapidly than deeper beams.
• Temperature & shrinkage (T&S) steel is provided
perpendicular to restrain cracks parallel to span.
(Flexural steel restrains cracks perpendicular to
span)
Flexural Reinforcement in Slabs
Maximum & Minimum reinforcement requirements
T&S Reinforcement (perpendicular to span) ACI Sec 7.12
As min   0.0020 * 12"* t
 0.0018 * 12"* t
 60 
 0.0018 *   * 12"* t
f 
 y
 0.0014 * 12 "* t
f y  40 or 50 ksi
f y  60 ksi
f y  60 ksi
Flexural Reinforcement in Slabs
T&S Reinforcement (perpendicular to span) ACI Sec 7.12
S max
 5t
 smaller of 
18"
Flexural Reinforcement (parallel to span) ACI Sec 10.54
As min   As min T &S
As max   0.75 As bal 
Smax from reinforced spacing
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
A. Concept of Bond Stress and Rebar Anchorage
Internal Forces in a beam
Forces developed in the beam
by loading.
Forces in Rebar
Bond stresses provide mechanism
of force transfer between concrete
and reinforcement.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Equilibrium Condition for Rebar
 F  0.  T  Bond Force  0
.
 d b2
4
.  ld 
f y   d b lb m  0
fydb
4m
Note: Bond stress is zero at cracks
m = bond stress
(coefficient of
friction)  k
fc
k  f  bar 
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Sources of Bond Transfer
(1) Adhesion between concrete & reinforcement.
(2) Friction
Note: These properties are quickly lost for tension.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Sources of Bond Transfer
(3)Mechanical Interlock.
The edge stress concentration
causes cracking to occur.
Force interaction between the
steel and concrete.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Splitting cracks result in loss of bond transfer.
Reinforcement can be used to restrain these cracks.
Splitting Load is Affected by:
1. Minimum edge distance and spacing of bars
(smaller distance= smaller load)
2. Tensile strength of concrete.
3. Average bond stress along bar.(Increase in bond
stress
larger wedging forces)
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Typical Splitting Failure
Surfaces.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
General splitting of
concrete along the
bars,either in vertical
planes as in figure (a) or
in horizontal plane as in
figure (b). Such splitting
comes largely from
wedging action when the
ribs of the deformed bar
bear against the concrete.
The horizontal type of splitting frequently begins at a diagonal crack.
The dowel action increases the tendency toward splitting. This
indicates that shear and bond failure are often intricately interrelated.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
B. ACI Code expression for development length for
bars in tension/in compression.
Development Length, ld
Shortest length of bar in which the
bar stress can increase from zero to
the yield strength, fy.
( ld used since bond stresses, m,
vary along a bar in a tension zone)
Development Length for Bars in Tension
Development length, ld  12” ACI 12.2.1
fc  10000 psi for Ch. 12 provisions for development length in ACI Codes.
Development length, ld (simplified expression from ACI
No. 6 and smaller No. 7 and larger
12.2.2)
bars and deformed bars
wires
Clear spacing of bars being developed or
spliced not less than db, clear cover not less
than db, and stirrups or ties throughout ld not
less than the code minimum
or
Clear spacing of bars being
developed or spliced not less than 2db and
clear cover not less than db.
Other cases
f y
ld

d b 25 f c
f y
ld

d b 20 f c
ld 3 f y

d b 50 f c
ld 3 f y

d b 40 f c
Development Length for Bars in Tension
Development length, ld
ACI 12.2.3
ld
3 fy


in which
d b 40 f c  c  K ct 


 db 
2.5
 c  K ct 

  2.5
 db 
limit to safeguard against pullout type failure.
Factors used in expressions for
Development Length (ACI 12.2.4)
  reinforcement location factor
where  < 1.7
Horizontal reinforcement so placed that more than 12 in of fresh concrete
is cast in the member below the development length or splice
1.3
Other reinforcement
1.0
  coating factor (epoxy prevents adhesion &
friction between bar and concrete.)
Epoxy-coated bars or wires with cover less than 3db or clear spacing less
than 6db
1.5
All other epoxy-coated bars or wires
1.2
Uncoated reinforcement
1.0
Factors used in expressions for
Development Length (ACI 12.2.4)
g  reinforcement size factor (Reflects more favorable
performance of smaller  bars)
No.6 and smaller bars and deformed wire
0.8
No. 7 and larger bars
1.0
  lightweight aggregate concrete factor (Reflects lower
tensile strength of lightweight concrete, & resulting
reduction in splitting resistance.
When lightweight aggregate concrete is used.
1.3
However, when fct is specified, shall be permitted to be taken as 6.7 f c f ct
but not less than
1.0
When normal weight concrete is used
1.0
Factors used in expressions for
Development Length (ACI 12.2.4)
c = spacing or cover dimension, in.
Use the smaller of either
(a) the distance from the center of the bar or wire to
the nearest concrete surface.
or
(b) one-half the center-to-center spacing of the bar or
wires being developed.
Factors used in expressions for
Development Length (ACI 12.2.4)
Kct = transverse reinforcement index (Represents the contribution
of confining reinforcement across potential splitting planes.)
K tr 
Atr f yt
1500 * s * n
Atr =
Total cross-section area of all transverse reinforcement within the spacing s,
which crosses the potential plane of splitting along the reinforcement being
developed with in the development length, in2.
fyt =
Specified yield strength of transverse reinforcement, psi.
s =
maximum center-to-center spacing of transverse reinforcement within ld in.
n =
number of bars or wires being developed along the plane of splitting.
Note: It is permitted to use Kct =0 as a design simplification
even if transverse reinforcement is present.
Excess Flexural Reinforcement
Reduction (ACI 12.2.5)
Reduction = (As req’d ) / (As provided )
- Except as required for seismic design (see ACI 21.2.14)
- Good practice to ignore this provision, since use of
structure may change over time.
- final ld
12
in.
Reduction 
Mu
M n req'd 
M n provided


M n provided
Development Length for Bars in
Compression (ACI 12.3)
Compression development length ldc = ldbc * applicable
reduction factors 8 in.

Basic Development Length for Compression, ldbc
 0.02 d b f y

ldbc  larger of 
fc
0.0003 d b f y

Development Length for Bars in
Compression (ACI 12.3)
Reduction Factors (ACI 12.3.3)
- Excessive Reinforcement Factor = (As req’d)/(As provided)
- Spiral and Ties
If reinforcement is enclosed with spiral
reinforcement  0.25 in. diameter and  4 in. pitch or
within No. 4 ties according to 7.10.5 and spaced  4 in.
on center. Factor = 0.75
Note ldc < ld (typically) because
- Beneficial of end bearing is considered
- weakening effect of flexural tension cracks is not
present for bars in compression.
Hooked Bar at Discontinuous
Ends (ACI 12.5.4)
If side cover and top (or bottom cover)  2.5 in.
Enclose hooked bar w/ ties or stirrup-ties:
Spacing  3db
db = of hooked bar
Note: Multiplier for ties or
stirrups (ACI 12.5.3.3)
is not applicable for
this case.
Hooked Bar at Discontinuous
Ends (ACI 12.5.4)
Table A-11, A-12, A-13 (Back of textbook) - Basic
Development lengths
Others
Mechanical Anchorage
ACI (12.6)
Welded Wire Fabric
ACI (12.7)
Bundled Bars
ACI (12.4)
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
C. 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.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
C. Use of Standard Hooks for Tension Anchorage
Standard Hooks are
defined in ACI 7.1.
Hooks resists tension by
bond stresses on bar
surface and bearing on on
concrete inside the hook.
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Development Length for Hooked Bar, ldb.
ldh  lhd * multiplier s
where ldb  8 d b and ldb  6 in .
Basic Development Length for Hooked Bar = lhb
when fy = 60,000 psi
1200d b
lhd 
fc
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Bar Yield Strength
Bars with fy other than 60,000 psi
Concrete Cover for 180 Degree Hooks
For No. 11 bars and smaller.
Side cover (normal to plane of hook)  2.5 in.
Concrete Cover for 90 Degree Hooks
For No. 11 bars and smaller.
Side cover (normal to plane of hook)  2.5 in.
Cover on bar extension beyond hook tail  2 in.
Multiplier
fy /60,000
0.7
0.7
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Excessive Reinforcement
Where anchorage or development for fy is not
specified required.
Lightweight Aggregate Concrete
Ties or Stirrups
For No. 11 bar and smaller.
Hook enclosed vertically or horizontally within ties
or stirrup-ties spaced along full ldh no farther apart
than 3db, where db is diameter of hooked bar.
Multiplier
As(req’d) /
As(provided)
1.3
0.8
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Epoxy-coated Reinforcement
Hooked bars with epoxy coating
Multiplier
1.2
Example
Example 4
GIVEN: A #5 Grade 40 bar is in tension as shown below. Use LIGHTWEIGHT
concrete with f’c = 4000 PSI.
REQUIRED: Determine the min. required hook dimensions “X”, “Y” and “Z”