Chapter 8

Development,
Anchorage, and
Splicing of
Reinforcement

1
Tension Development Length
Basic tension development equation:
fy y ty ey s
 db
1.1 f  cb  ktr 
d '

 d 
c

 b 
d  300 mm

Basic y values are 1.

If the equation is written in terms of ℓd /db,
the results are in terms of bar diameters.
2
Reinforcement Location Factor
yt is a bar location factor given in ACI Table
25.4.2.4.

“Top bars” with more than 300 mm of fresh

concrete cast below do not bond well.
In this case yt = 1.3.

For other situations, yt = 1.0.

3
Bar Coating Factor
ye is an epoxy coating factor given in ACI
Table 25.4.2.4.
ye = 1.5 for epoxy coated bars with cover less
than 3db or clear spacing less than 6db.
ye = 1.2 for other epoxy coated bars.
ye = 1.0 for uncoated bars.
Epoxy coating used to prevent corrosion.
Product ytye need not to be greater than 1.7
4
Reinforcement Size Factor
ys is a reinforcement size factor given in ACI
Table 25.4.2.4.

Small bars require relatively less development

length.

ys = 0.8 for No 19 bars and smaller.

ys = 1.0 for other bars.

5
Lightweight Aggregate
 is the lightweight concrete modification
factor.
Normal weight concrete:  = 1
Lightweight concrete:  = 0.75
Sand-lightweight concrete:  = 0.85
When average splitting tensile strength fct
is specified,  shall be permitted to be
taken as: 
f ct
 1.0
'
0.56 f c
6
Effects of Spacing and Cover
Should the concrete cover or clear spacing
between the bars be too small, the
concrete may split.

This situation is accounted for

with the confinement term:
 cb  Ktr 
db

The confinement term shall not be taken

greater than 2.5.
7
Effects of Spacing and Cover
cb is the smaller of:
(a) the distance from center of a bar to
nearest concrete surface, and
(b) one-half the center-to-center spacing of
bars being developed.

Ktr is a transverse reinforcement index, used

to account for the contribution of stirrups
and ties across possible splitting planes.

8
Effects of Spacing and Cover
40 Atr
ACI Code Section 25.4.2.3b: Ktr 
sn
Atr is the total cross-sectional
area of all transverse
reinforcement within the
spacing s . Fig. Definition of Atr.

s is the maximum center to center spacing of

transverse reinforcement within ℓd.
n number of bars along plane of splitting.
9
Effects of Spacing and Cover

ACI Code Section 25.4.2.3. allows Ktr to be

taken equal to 0 to simplify calculations
even if there’s transverse reinforcement.

10
 Tension Development Length Restrictions

ACI Section 25.4.1.4: f c'  8.3 MPa

ACI Section 25.4.10.1 – Reduction of permitted

if area of steel provided exceeds area of
steel required: ℓd (As,required / As,provided)

Reduction cannot be used for the development at

supports of +ve reinf., or for development of
shrinkage and temperature reinf.

11
Tension Development Length –
Simplified Equations

Equations for Development lengths – SI Units.

12
Tension Development Length –
Simplified Equations
Explanation of
Cases 1 & 2

13
Compression Development Length
Compression-development lengths are
considerably shorter than tension-
development lengths.

ACI Code Section 12.3.2:

 0.24 f  
=max   y
 db ;  0.043 f y  db 
  f c' 
dc
 
 

14
Development Length for Bundled Bars
ld for individual bars within a bundle, in T or
C, shall be that of the individual bar, and
increased for 3-bar and 4-bar bundles:
- by 20% for three-bar bundle.
- by 33% for four-bar bundle.

When finding ye and cb replace the bundle

with a fictitious single bar having same
area as bundle of bars.

15
Example
Determine the
development length
required for the
epoxy-coated
bottom bars shown
assuming (a) Ktr = 0
and (b) Ktr is
computed.
Use fy = 420 MPa and
f’c = 21 MPa.
16
Solution)
Development length is given by:
fy y ty ey s
d
 f c'  4.6 MPa  8.3 MPa
db 1.1 f c'  cb  ktr 
 
 d b 
yt = 1.0 for bottom bars.
ye = 1.5 for epoxy-coated bars (with clear spacing bars
= 54.6 mm < 6 db = 6(25.4) = 152.4 mm)
Check yt .ye = (1.0)(1.5) = 1.5 < 1.7 O.K.
ys = 1.0 for No 25 bars.
17
 = 1.0 for normal weight concrete.

cb = smaller of
. side cover of bars = 80 mm,
. ½ of center to center spacing
of bars = 40 mm.

18
a) Ktr = 0
cb  ktr 40  0
  1.575  2.5
db 25.4

420 11.5 1

d

db 1.1 21 1.575 
d
 79.3 diameters
db
d  2015 mm

19
b) Using computed Ktr .
40 Atr 40  2  71
ktr    7.1
sn  200  4 
cb  ktr 40  7.1
  1.854  2.5
db 25.4
420 11.5 1
d

db 1.1 21 1.854 
d
 67.4 diameters
db
d  1715 mm

20
Development of Hooks in Tension
Hooks can be used to reduce tensile
development.
Hooks are ineffective for compression
development length.

ACI Code Section 25.4.3.1, for deformed bars:

0.24y e f y
dh  db  max  8db , 150 mm 
 f c'

21
Development of Hooks in Tension

ACI Code Section 25.4.3.2: ye = 1.2 for epoxy

coated bars,  = 0.75 for lightweight
concrete.
• For other cases, ye and  shall be taken as
1.0.

ldh is measured from the critical section to

the outside edge of the hook

22
Development of Hooks in Tension
Either the 90 or
the 180 Hook
may be used at
the free end.

Fig. Hooked bar details for development of standard hooks.

23
Development of Hooks in Tension
ACI 25.3.2 :Inside diameter of bend for
stirrups and ties shall not be less than 4db
for No. 16 bar and smaller.
For bars larger than No. 16, diameter of bend
shall be in accordance with ACI Table
25.3.1.
TABLE 25.3.1

24
 Development of Hooks in Tension
ACI Code Section 25.4.3.2, ldh can be multiplied
by modification factors:
Cover: No 36 bars or smaller, normal side cover 
65 mm and cover on bar extension beyond 90
hook  50mm …………………………… multiply by 0.7.
Ties or stirrups: No 36 bars or smaller, enclosed
in ties or stirrups, and ties or stirrups spaced
 3db ……………………………………………multiply by 0.8.
As ,required
Area of steel: Aprovided > Arequired, multiply by .
As , provided
25
Example
Determine the development
length for the epoxy-coated
bars in a cantilever beam for
(a) straight bars and Ktr = 0;
(b) bars end in a 180 hook;
(c) bars end in a 90 hook.
All the bars are top bars.
Use a concrete strength of
28 MPa and a steel yield
stress of 420 MPa.

26
(Solution)
(a) Development length for straight bars is given by:
fy y ty ey s
d
 f c'  5.29 MPa  8.3 MPa
db 1.1 f c'  cb  ktr 
 
 d b 
yt = 1.3 for top bars.
ye = 1.5 for epoxy-coated bars (with clear spacing bars
115 – 28.7 = 86.3 mm < 6 db = 6(28.7) = 172.2 mm)
yt .ye = (1.3)(1.5) = 1.95 > 1.7 Use yt .ye = 1.7
ys = 1.0 for No 29 bars.
27
 = 1.0 for normal weight concrete.
cb = smaller of
. cover of bars = 65 mm,
. ½ of center to center spacing
of bars = 57.5 mm.
cb  ktr 57.5  0
ktr  0    2.0  2.5
db 28.7
420 1.7 1.0 
d
  d  61.3 diameters
db 1.1 28  2.0  db
 d  1760 mm  300 mm

28
(b) Development length for 180 hook is given by:
0.24y e f y 0.24 1.2  420 
 db   28.7   656 mm
1.0   
dh
 f c
'
28

dh  660 mm  max 8  28.7  mm , 150 mm   230 mm

29
(c) Development length for 90 hook:

dh  660 mm

In (b) & (c): If stirrups were provided, multiply ℓdh by 0.8.

30
Bar Cutoffs in Flexural Members
Reinforcement is placed near the tensile face
of beams to provide the tension component
of the internal resisting couple.

Reinforcement is sized for maximum moment

For economy some of the bars can be

terminated or cut-off where they are no
longer needed.
31
Bar Cutoffs in Flexural Members

Fig. 9-1 Moments and reinforcement in a continuous beam.

At midspan, moments are +ve and reinforcement

is required at the bottom face.
The opposite is true at the supports.
32
Bar Cutoffs in Flexural Members
Location of Flexural
Cut-Off points

Flexural design
capacity varies
roughly with
reinforcement
area.
 a
 M n   As f y  d  
 2 Fig. Cut off points in a uniformly
loaded simple beam.

33
Bar Cutoffs in Flexural Members
If we take the abscissa x’ measured from the left
support, and the abscissa x measured from the
midspan, we have: l
x   x
2
The moment equation for any point on the beam can
be written:
wx w l  l  wl 2  x2 
M  l  x    x   x   1 2
2 22  2  8  (l / 2) 
 x2  M  x2 
M  M max 1  2
  1  2
 (l / 2)  M max  (l / 2) 
34
Bar Cutoffs in Flexural Members

Parabolic moment
diagram:
x12 2 x22 4
2
 ; 2

l 6 l 6
   
2 2

Bars must extend beyond

the theoretical cut-off Fig. Cut off points in a

point.
uniformly loaded simple beam.

35
Example
Determine the theoretical flexural points at each
end of the beam where 2 bars can be cut-off
and then determine where 2 additional bars
can be cut off.
Use f’c = 21 MPa and fy = 420 MPa.

36
(Solution)
For 6 No 29:

As  6  645 mm2   3870 mm2

a
As f y

 3870 mm 2
  420 MPa   200 mm
'
0.85 f b
c 0.85  21 MPa  455 mm 

 a
 M n   As f y  d  
 2 
 200  6
 M n  0.9  3870  420   685   10   855.8 kN.m
 2 

37
For 4 No 29:
As  4  645 mm2   2580 mm2

a
 2580 mm   420 MPa 
2

 133.4 mm
0.85  21 MPa  455 mm 

 133.4  6
 M n  0.9  2580  420   685   10 
 2 
 M n  603 kN.m

38
For 2 No 29:
As  2  645 mm2   1290 mm2

a
1290 mm   420 MPa 
2

 66.7 mm
0.85  21 MPa  455 mm 

 66.7  6
 M n  0.9 1290  420   685   10 
 2 
 M n  317.8 kN.m

39
 ΦMn Diagram.

317.8 kN.m

603 kN.m

855.8 kN.m

4 bars
6 bars

603 kN.m

317.8 kN.m

40
The location of the cut off points.
728 kN
80x
364x – 40x2

364 – 80x
364 kN 364 kN

364 kN
M  364 x  40 x 2  kN.m 

For M  603 kN.m  x  2.1 m  2 x1  4.9 m  x1  2.45 m

For M  317 kN.m  x  0.98 m  2 x2  7.14 m  x2  3.57 m

41
 Cut-off of Bars.

317.8 kN.m

603 kN.m

855.8 kN.m

4.9 m
4 bars
6 bars

7.14 m
603 kN.m

317.8 kN.m

42
Using the approximate procedure.

855.8 kN.m
603 kN.m

603 kN.m
x12 2
  x1  2.63 m

317.8 kN.m

317.8 kN.m
2
 9.1  6
 
 2 
x22 4
2
  x2  3.71 m
 9.1  6 6 bars
 
 2 
5.26 m
4 bars
7.42 m

The approximate solution yields fairly reasonable results .

43
Bar Cutoffs in Flexural Members

Actual Cut-Off Points

The flexural cut-off points must be modified

to account for shear, development, and
constructional requirements to get the
actual cut-off points used in construction

44
Bar Cutoffs in Flexural Members
From the envelope of the +ve and –ve
moments of a beam we can determine the
theoretical cut-off points.

ACI Code 9.7.3.3 requires that every bar be

continued at least a distance equal to
Max(d, 12db) beyond the point at which it
is theoretically no longer required to
resist stress, except at supports of simple
spans and at the free end of cantilevers.
45
Bar Splices in Flexural Members
Four types of splices: lapped, mechanical, welded,
and end-bearing splices.
ACI Code Section 25.5.2.1 distinguishes between 2
types of tension lap splices: Class A and Class B.
Table Tension lap splices.

46
Bar Splices in Flexural Members
The minimum splice length for each class is:
- Class A splice: 1.0 ℓd
- Class B splice: 1.3 ℓd

The splice length for both classes may not be less

than 300 mm.

47
Bar Splices in Flexural Members
In a compression lap splice, a portion of the force
is transferred through the bearing at the end
of the bar on the concrete. No cracks, so
shorter laps.

Splice may be by welding or by various mechanical

devices.

A common type of mechanical

splices is a steel sleeve.
Fig. 9-8 Steel sleeve.

48