This document discusses bond and development length in reinforced concrete. It defines bond as the adhesion between concrete and steel reinforcement, which is necessary to develop their composite action. Bond is achieved through chemical adhesion, friction from deformed bar ribs, and bearing. Development length refers to the minimum embedment length of a reinforcement bar needed to develop its yield strength by bonding to the surrounding concrete. The development length depends on factors like bar size, concrete strength, bar location, and transverse reinforcement. It also provides equations from design codes to calculate the development length for tension bars, compression bars, bundled bars, and welded wire fabric. Hooked bars can be used when full development length is not available, and the document discusses requirements for standard hook geome

1. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 1
CE 370
REINFORCED CONCRETE-I
Prof. A. Charif
Bond and Development Length
Bond
• RC design assumes perfect bond between concrete and steel
• There must be no slippage between steel bars and surrounding
concrete
• Bond stress results from variation of axial force in bars, caused
by moment variation
• Bond stress is affected by development of tensile cracks.
• At a point where a tensile crack crosses a reinforcing bar, all
the tensile force is carried by the reinforcement.
• At a short distance from the crack, tension is resisted by both
the reinforcement and uncracked concrete.
• Bond stress is zero at the crack, but may be large a short
distance away.
2

2. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 2
Bond
• Note: If there is no bonding between the two
materials and if the bars are not anchored at
their ends, they will pull out of the concrete.
As a result, the concrete beam will act as an
unreinforced member and will be subject to
sudden collapse as soon as the concrete
cracks.
3
Mechanism of bond transfer
• Bond between concrete and steel bars is due to chemical
adhesion, friction and bearing of reinforcement ribs on the
concrete
• Adhesion and friction are lost when the bar is loaded in tension as
a result of diameter reduction due to Poisson’s ratio effect
• Adhesion and friction are neglected
• Bond is transferred mainly by bearing of the deformed bars
• Plain bars have a poor bond transfer and are no longer used
• Steel bars subjected to axial compression develop a better bond
because of Poisson’s ratio effect
4

3. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 3
Mechanism of bond transfer
Bar Bearing
5
6
Mechanism of bond transfer

4. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 4
Types of Bond Failures
7
 Forces in concrete have longitudinal and radial
components
 Wedging action of ribs against concrete produces
tension in a cylindrical section around the bars
(similar to a concrete pipe filled with water)
 Wedging action causes cracks (splits) to occur
around the reinforcement
 Bond failure occurs by splitting of concrete parallel
to the bar
 Splitting failure surface depends on cover and bar
spacing values
8
Typical splitting failure surfaces

5. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 5
9
Bond stress
Bond stress results from
variation of axial force in
bars, caused by moment
variation
jdd
V
jdd
dx
M
V
jd
dxd
jd
M
TTT
b
b
b













armLever
12
Development length
 Development length is the minimum embedment length of a bar
in concrete necessary to develop the yield stress in the bar, plus
some additional distance to insure toughness
 If the distance from a point where the bar stress is equal to yield
stress fy to the end of the bar (zero stress) is less than the
development length, then the bar will pull out of the concrete
10
u
by
d
dbu
y
b
yb
df
l
ld
f
d
fATT



4
4
0
2
2




6. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 6
Tension development length
 As bond in tension bars is less than bond in compressed
bars, the development length expressions are different.
 Many factors affect the development length:
 Bar size
 Bar location
 Type of concrete (normal or lightweight)
 Bar coating (if any)
 Concrete cover and bar spacing
 Transverse reinforcement effect (confinement)
 ACI , SBC and all codes give simplified expressions
based on experimental data
11
12
Tension development length
• Bar diameter effect: Bond strength is better with larger bars
• Bar location effect: Bars with more than 300 mm of fresh
concrete cast below them will experience some gravity
migration of mortar resulting in reduction of bond strength
• Bar coating effect: Epoxy-coated bars (to protect them against
corrosion) have a reduced bond
• Concrete type effect: Lightweight concrete has a lower bond
strength than normal concrete
• Transverse steel effect: Stirrups and ties improve the bond
strength
• Cover and bar spacing effect: Bond strength is reduced with
smaller values of cover and bar spacing.

7. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 7
13
SBC / ACI Tension development length ld
General Equation
5.2with
10
9
'


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

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
 

b
tr
b
b
trc
y
d
d
Kc
d
d
Kcf
f
l (1)

mml
K
c
d
l
d
tr
b
d
300
factorsteelTransverse
effectspacingbarandCover
factordiameterBar
factorconcretetLightweigh
factorcoatingEpoxy
factorlocationlocationBar
mmindiameterBar
mminlengthtDevelopmen


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

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



MPaff
l
f
l
cc
d
c
d
4.69
3
25
:ofequationsallIn
ofin termsexpressed
lengthtDevelopmen
''
'

14
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0.1:22
8.0:20
:factorsizeBar
1.3:concretetLightweigh
1.0:concreteNormal
:factorconcretetLightweigh
1.2:coatingepoxyofcasesOther
1.5:6thanlessspacingclearor3thanlesscoverwithcoatingEpoxy
1.0:entreinforcemUncoated
:factorcoatingEpoxy
1.0:casesOther
bond)weakeningmigrationmortarwithsteeltopof(case
1.3:(splice)bardevelopedbelowcastconcretefreshofmm300thanMore
:factorlocationlocationBar




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







mmd
mmd
dd
b
b
bb
SBC / ACI Tension development length

8. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 8
15
SBC / ACI Tension development length
presentissteeletransversifevenzerotoequaltakenbemay
planesplittingalongdevelopedbeing(wires)barsofNumber
withinsteeletransversofspacingcenter-to-centerMaximum
barsdevelopedalongplanesplittingpotentialthecrosseshichw
spacingwithinsteeletransversallofareasection-crossTotal
10
factoreffectentreinforcemTransverse
tr
d
tr
yttr
trtr
K
n
ls
sA
sn
fA
KK




c = Minimum of :
(a) Smallest distance measured from concrete surface to
center of bar
(b) One half of center-to-center bar spacing
16
SBC / ACI Tension development length
In addition to the previous general equation (1), SBC and
ACI propose other conservative simplified equations for
tensile development length.

9. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 9
17
Bar diam 20 mm or less
and deformed bars (a)
Bar diam 22 mm
or more (b)
Case 1: Clear spacing of bar
not less than db, cover not less
than db, stirrups or ties
throughout ld not less than
minimum or
Case 2: Clear spacing of bar
not less than 2db and cover
not less than db
Other cases
(S1a)b
c
y
d
f
f








'
25
12 
(S2a)b
c
y
d
f
f








'
25
18 
(S2b)b
c
y
d
f
f








'
10
9 
(S1b)b
c
y
d
f
f








'
5
3 
SBC / ACI Tension development length ld given
by simplified equations, not less than 300 mm
18
SBC / ACI Tension development length








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

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
1.3:concretetLightweigh
1.0:concreteNormal
:factorconcretetLightweigh
1.2:coatingepoxyofcasesOther
1.5:6thanlessspacingclearor3thanlesscoverandcoatingEpoxy
1.0:entreinforcemUncoated
:factorcoatingEpoxy
1.0:casesOther
1.3:bardevelopedbelowcastconcretefreshofmm300thanMore
:factorlocationlocationBar
mmindiameterBar
mminbarsdeformedoflengthtDevelopmen










bb
b
d
dd
d
l

10. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 10
19
SBC / ACI Compression development length
(2)








 by
c
by
dc df
f
df
l 043.0,
24.0
Max:unitsmmandNewtonIn
'
Development length for bundled bars
• For bundled bars, the development length shall be that of an
individual bar increased by :
 20 % for three-bar bundle
 33 % for four-bar bundle
• The value of bar diameter d0 to consider in equations shall be
that of a hypothetical bar having the same area as the bundle.
)4or3(
44
0
22
0
 nndd
d
n
d
b
b
20
Development length for wire fabric
Deformed welded wire fabric
The development length is that
of deformed bars, times a wire
fabric factor.
The minimum value is 200 mm spacingWire
0.1,
5
,
240
Max
,.,





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w
w
b
y
y
bardwireweldd
s
s
d
f
f
ll

 (3)
Plain welded wire fabric
For plain welded wire fabric,
the development length
shall be determined by :
factorconcretetLightweigh
areasectioncrossWire
spacingWire
3.3
',






w
w
c
y
w
w
wireplaind
A
s
f
f
s
A
l (4)

11. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 11
Hooked Anchorages
 When sufficient space is not available to provide straight
development lengths, hooks are used.
 Hooks must satisfy geometric conditions
 Hooks are not effective in compression
 Standard Hooks: As defined in SBC 304:
21
Development Length for Standard Hooks
22
The development length ldh is measured from the critical
section of the bar to the outside end or edge of the hooks.

12. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 12
Development length for a hook
23
The development length needed for a hook is given by:
casesspecialsomeinexcept1.0
1.0toequalareandcasesotherallFor
concreteaggregatetlightweighfor1.3toequaltaken
barscoated-eopxyfor1.2astaken
.8ormm150thanlessbenotshallequationabovefromobtained
24.0
'


Factor
dl
Factord
f
f
l
bdh
b
c
y
dh




(5)
Development length for a hook
24
0.1:taketoveconservatiisIt
0.1:casesotherallIn
8.0
3spacingwith(stirrups)lar tiesperpendicu
withenclosedbars,36andHook180
8.0
3spacingwith(stirrups)tiesparallelwithin
enclosedor(stirrups)lar tiesperpendicu
withenclosedbars,36andHook90
7.0
hookbeyondcoverHook with90
mm60coverside,36
:
24.0
0
0
0
'






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
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
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



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
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



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


Factor
Factor
d
mmd
d
mmd
mmd
Factor
Factord
f
f
l
b
b
b
b
b
b
c
y
dh (5)


13. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 13
25
Development length for a hook
26
Development length for a hook
Correction factor values
Hook
Type
Hooked
bar db
Side
cover
Tail
cover
Stirrups
or ties
Factor
value
1800 Any Not required 0.7
900 Not required 0.7
900 Any Any Perpendicular 0.8
900 Any Any Parallel 0.8
900 or
1800
Any Any Perpendicular 0.8
60
50
36
bds 3
6036
36
36
36
bds 3
bds 3
For all other cases : Factor = 1.0

14. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 14
27
Problem-1: Anchorage of a Straight Bar
A 400 x 450 mm cantilever beam frames into a 400 mm thick wall.
The three 25-mm top bars are assumed to be yielding at 420 MPa.
point A (face of the wall). Compute the minimum embedment of the
bars inside the wall. 20 MPa lightweight concrete is used.
28
Problem-1: Anchorage of a Straight Bar
Construction joints are located as shown. The beam has closed 10-mm
stirrups with 300 MPa yield stress and 180 mm spacing.
Cover is 40 mm. The three 25-mm bars (in tension) are inside 14-mm
bars of grade 420 at 300 mm in each face of the wall

15. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 15
29
Solution-1a: General Equation
mm22ismm)(25diameterbarbecause1.0
usedisconcretehlightweigtbecause1.3
coatedepoxynotarebarsbecause1.0
barstopthebelowcastconcretefreshof
mm300thanmorebewilltherebecause3.1
10
9
'









 






(1)b
b
trc
y
d d
d
Kcf
f
l
c = Minimum of :
(a) Smallest distance measured from concrete surface to
center of bar
(b) One half of center-to-center bar spacing
No stirrups in wall but 14-mm bars play role of transverse steel
30
c = Minimum of :
(a) Smallest distance measured from concrete surface to
center of bar in wall = 40 + 14 + 25/2 = 66.5 mm
(b) One half of center-to-center bar spacing
Solution-1a: General Equation
mmc
mm
5.66
75.66
2
5.662400
5.0






 

36.14
330010
4208.307
3:anchoredbeingbarsofNumber
300:withinsteeletransversofspacingMaximum
face)eachatmm300atbarmm14vertical(onesplitting
ofplanecrossingsteeletransversallofareasection-crossTotal
8.307
4
14
2420
10
2
2







tr
d
tr
tryt
yttr
tr
Knn
mmsls
-
A
mmAMPaf
sn
fA
K


16. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 16
31
resultthechangenotwill,0Taking:Note
walltheintom1.5barsExtend
4.142825
5.2
3.10.10.13.1
20
420
10
9
5.25.22.3
25
36.145.66
10
9
'

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
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
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
tr
d
b
tr
b
tr
b
b
trc
y
d
K
mml
d
Kc
d
Kc
d
d
Kcf
f
l (1)

Solution-1a: General Equation
32
Solution-1b: Simplified Equations
(S1b)equationsimplifiedusethuswe
,25withand(2)caseisThis
barscover toclear2spacingbarClear
4.40.110
2
253-52.52-400
spacingbarClear
mm52.55.2140barsthecover tosideClear







mmd
dd
dmm
b
bb
b
equationgeneralrepect towith
veconservatieryequation vSimplified:Note
walltheintom2.5barsExtend
7.238025
205
3.10.13.14203
5
3
'






 









 mmld
f
f
l db
c
y
d (S1b)


17. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 17
33
Problem-2: Development of a Bar in a Cantilever
Previous example with same data. The beam has closed 10-mm
stirrups with 300 MPa yield stress and 180 mm spacing.
Check whether there is sufficient distance in the span for the
development of the three 25-mm bars.
If not what is the maximum bar diameter that can be used ?
34
Solution 2
The maximum yield stress in the bars is at point A (face of the wall).
The bars must extend a distance ld into the wall support and a
distance ld into the span. The span is 1.5 m long.
The development length for 25-mm bars has been determined in
example 1. The only difference is in the transverse reinforcement.
We use the simplified equations.
Clear bar spacing determined in example 1 is 91.5 mm = 3.66 db
Stirrup spacing (180 mm) is less than maximum spacing value
d/2 = 385/2 = 192.5 mm
For 180 mm spacing, the minimum stirrup area is:
min,
2
2
2
'
min,
0.157
4
10
2:areastirrupActual
0.80
300
180400
3
1
,
16
20
Max
3
1
,
16
Max
vv
y
wc
v
AmmA
mm
f
sbf
A























18. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 18
35
Solution 2
Since the clear bar spacing is at least equal to db
and stirrup area exceeds minimum value, this in fact corresponds to
case 1, and 25-mm bars, equation (S1b) is to be used. We find the
same solution as in solution 2 of example1:
mmld
f
f
l db
c
y
d 7.238025
205
3.10.13.14203
5
3
'





 









 (S1b)

The maximum distance for bar extension is equal to the span length
minus the cover : 1500 – 40 = 1460 mm
This distance is insufficient for the straight development length of
2381 mm. Thus 25-mm bars cannot be used.
We must use smaller diameter bars.
Three 25-mm bars (1472 mm2) can be replaced by six 18-mm bars
(1526 mm2).
36
Solution 2
Three 25-mm bars replaced by six 18-mm bars (1526 mm2).
(S1a)equationsimplifiedusethuswe
)20(,18withand(1)caseisThis
value,minimumexceedsareastirrupand,spacingbarClear
16.28.38
5
816-942-400
spacingbarClear
mm94940wallin thebarsthecover tosideClear







mmmmd
d
dmm
b
b
b
mml
ld
f
f
l
d
db
c
y
d
1.1371
18
2025
3.10.13.142012
25
12
'






 









 (S1a)

The available distance (1460 mm) is sufficient for this development
length. 18-mm bars can thus be used for this cantilever beam.

19. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 19
37
Problem-3: Hooked Bar Anchorage into a Column
The end of 400 x 600 mm beam frames into a 650 x 650 mm column.
The column has four 36-mm bars. The top reinforcement of the beam
consists of four 25-mm bars. 20 MPa normal concrete and 420 MPa
steel are used. Design the anchorage of the four 25-mm bars.
38
Problem-3: Hooked Bar Anchorage into a Column

20. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 20
39
Solution 3
As the column reinforcement provides considerable confinement,
we will use general equation (1) for a straight development length :
mm22ismm)(25diameterbarbecause1.0
usedisconcretenormalbecause1.0
coatedepoxynotarebarsbecause1.0
barstopthebelowcastconcretefreshof
mm300thanmorebewilltherebecause3.1
10
9
'









 






(1)b
b
trc
y
d d
d
Kcf
f
l
40
c = Minimum of :
(a) Smallest distance measured from concrete surface to
center of bar = 125 + 40 + 10 + 25/2 = 187.5 mm
(b) One half of center-to-center bar spacing
mmc
mm
83.45
83.45
3
)5.121040(2400
5.0






 

6.41
451410
4207.2035
4:anchoredbeingbarsofNumber
514)2/361040(2650:steeletransversofSpacing
bars)mm36vertical(twosplitting
ofplanecrossingsteeletransversallofareasection-crossTotal
7.2035
4
36
2420
10
2
2







tr
tr
tryt
yttr
tr
Knn
mmss
-
A
mmAMPaf
sn
fA
K

Solution 3

21. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 21
41
Solution 3
mml
d
Kc
d
Kc
d
d
Kcf
f
l
d
b
tr
b
tr
b
b
trc
y
d
8.109825
5.2
0.10.10.13.1
20
420
10
9
5.25.25.3
25
6.4183.45
10
9
'














 
 (1)

This development length exceeds the column width.
It is thus necessary to use hooks to anchor the bars.
42
mm563.5oflengthtdevelopmenhookfor thesufficientisdistanceThis
mm60050650coverTail-650hookforavailableDistance
5.563OK
144)8,150(Max5.56325
20
4200.10.124.0
0.1factorcorrectionunitconsiderWe
bars)forcoating-epoxynoandconcrete(Normal1.0
.8ormm150thanlessbenotshallequationabovefromobtained
24.0
'








mml
mmdmml
Factor
dl
Factord
f
f
l
dh
bdh
bdh
b
c
y
dh


(5)
Solution 3
The development length for a hook is given by equation (5):

22. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 22
43
Solution 3
heightmm600joint withcolumn-beamin thefitwilldistanceThis
40016
312
2
12hookofheightVertical
5.563
mmd
dddd
D
d
mml
b
bbbbb
dh



Bar Cutoff
• RC flexural design is usually performed at the
maximum moment point.
• At other points, the required reinforcement is smaller
and the number of bars may be reduced by stopping
some of them (bar cutoff)
• A bar can be stopped when it is no longer required. It
must however be extended at a sufficient distance.
• The bar must be extended by a distance equal to the
development length plus an extra distance (due to shear
effect) equal to Max(d , 12db) where d is the steel depth
and db is the bar diameter
44

23. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 23
Capacity and demand diagrams (offer > demand) with bar cutoff for
simply supported beam– BMD on tension side in RC
For real structures, bar cutoff must be performed using the
envelope of moment diagram using all load combinations
Bar Cutoff
46
Bar Cutoff

24. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 24
Bar Splices
• Reinforcing bars must frequently be spliced because of:
– Limited bar lengths available (bar length usually limited to
12 m and exceptionally to 18 m)
– Requirement at construction joints
– Changes from larger bars to smaller bars
• There are three splice types:
 Lapped splices
 Mechanical splices
 Welded splices
47
Lap Splices
• Lap splices are achieved by overlapping the bars over a certain
length, thereby enabling the transfer of axial force from the
terminating bar to the connecting bar through the mechanism of
anchorage (development) bond with the surrounding concrete.
• Lap splices are usually not permitted for very large diameter bars
(Ф > 36 mm), for which welded splices are recommended (except
at footing-column joints).
48

25. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 25
49
Tension Lap Splice
Tension Lap Splice
• Center-to-center distance between lapped bars not greater than
minimum of (one fifth of lap splice , 150 mm)
• There are two classes of tension lap splices:
 Class A splice : 1.0 ld
 Class B splice : 1.3 ld
• The required lap class is selected as shown :
50
Amount of steel spliced
Stress ratio
fs / fy
50 % or less
spliced
More than 50 %
spliced
0.5 or less Class A Class B
More than 0.5 Class B Class B

26. 15-Mar-13
CE 370: Prof. Abdelhamid Charifi 26
Welded Splices and
Mechanical Connections
• Welded splices and mechanical connections are
particularly suitable for large diameter bars.
• This reduces consumption of reinforcing steel.
51
52
Compression Lap Splice
Compression lap splices will be described
in the Column Chapter