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Table of Contents
                            Title
Table of Contents
Preface and Acknowledgments
How to Use This Book
Codes and References Used to Prepare This Book
List of Tables
List of Figures
Nomenclature
1 Materials
2 Design Specifications
3 Flexural Design of Reinforced Concrete Beams
4 Serviceability of Reinforced Concrete Beams
5 Shear Design of Reinforced Concrete
6 Columns and Compression Members
7 Continuous One-Way Systems
8 Two-Way Slab Systems
9 Development of Reinforcement
10 Prestressed Concrete
11 Seismic Design of Reinforced Concrete Members
12 Practice Problems
Index
                        
Document Text Contents
Page 2

Professional Publications, Inc. • Belmont, California

Concrete Design
for the Civil PE and Structural SE Exams
Second Edition

C. Dale Buckner, PhD, PE, SECB

Page 92

9
Development of Reinforcement

In calculating the strength of reinforced concrete mem-
bers, an implicit assumption is made that a complete
bond exists between steel and concrete at the limit
state. This requires that reinforcement must develop
the design strength of the reinforcement.

1. Development of Reinforcement in Tension

ACI 318 permits three ways to develop bars in tension.

• straight embedment of the bar beyond the point
of maximum stress

• extending the bar a sufficient distance beyond the
point of maximum stress and providing a properly
detailed 90◦ or 180◦ hook

• providing mechanical anchorage in the form of a
properly welded cross bar or plate

Figure 9.1 illustrates each of the methods.

ld

ldh

T

straight
embedment

ACI 90� hook

mechanical
anchorage

T

T

Figure 9.1 Representative Methods to
Develop Bars in Tension

A. Straight Embedment

ACI Sec. 12.2 gives a general equation for the straight
embedment length in tension.

ld =
3
40

(
fy

λ


f ′c

)






ψtψeψs(
cb + Ktr

db

)





⎠ db 9.1

ψt equals 1.0 for bottom bars and 1.3 for horizontal bars
with 12 in or more of fresh concrete cast beneath. ψe
is 1.0 for uncoated bars and 1.2 for epoxy-coated bars
unless the cover is less than 3db or clear spacing is less
than 6db, in which case ψe is 1.5. ψs is 0.8 for no. 6
or smaller bars and 1.0 for bars larger than no. 6. λ is
1.0 for normal weight concrete and 0.75 for lightweight
concrete. The parameters cb and Ktr are dependent on
the cover and transverse reinforcement surrounding the
longitudinal bars.

For practical designs, minimum transverse reinforce-
ment in the form of ties or stirrups is present, and the
spacing and cover provided is sufficient to permit a sim-
plification of the general equation to

ld =

(
ψtψeψsfy

20λ


f ′c

)

db 9.2

The code sets an upper bound on the value of


f ′c of
100 psi and a lower bound on ld of 12 in. ACI defines
the development length of single bars in bundles as

• for two-bar bundles, the length of the individual
bar

• for three-bar bundles, the length of the individual
bar increased by 20%

• for four-bar bundles, the length of the individual
bar increased by 33%

--- 67 ---

Page 93

68 Concrete Design for the Civil PE and Structural SE Exams

Example 9.1
Development Lengths for Grade 60 Bars

Generate a table giving the development lengths of no. 3
through no. 11 grade 60 rebars (yield strength of
60,000 psi), assuming normal weight concrete (λ = 1)
with a compressive strength of 3000 psi and uncoated
bottom bars (ψe = ψt = 1).

Solution:
For bars no. 6 and smaller, ψs equals 1.0 and the devel-
opment length equation (Eq. 9.2) is

ld =

(
ψtψeψsfy

25λ


f ′c

)

db

=
(1)(1)(1)

(
60,000

lbf
in2

)
db

(25)(1)


3000
lbf
in2

= 43.8db

For no. 7 and larger bars, ψs equals 1.25, and the same
equation is

ld =

(
ψtψeψsfy

25λ


f ′c

)

db

=
(1)(1)(1.25)

(
60,000

lbf
in2

)
db

(25)(1)


3000
lbf
in2

= 54.8db

The development lengths for these conditions are tabu-
lated as shown.

bar no. db ld
(in) (in)

3 0.375 16.0
4 0.500 21.9
5 0.625 27.4
6 0.750 32.9
7 0.875 48.0
8 1.000 54.8
9 1.128 61.8

10 1.270 70.6
11 1.410 77.3

Similar tables of development lengths are available (for
example, Design Aid 11.2.9 in the PCI Design Hand-
book) and are more commonly used than the code equa-
tions.

ACI Sec. 12.2.5 permits a reduction in the develop-
ment length when the area of steel furnished, As,provided,

exceeds the area theoretically required at a section,
As,required. The development length from the table or
formula may be multiplied by the ratio As,required/
As,provided to give the allowed development length.

Example 9.2
Selecting Reinforcement to Ensure Development

A continuous wall footing is loaded to produce a uni-
formly distributed upward pressure of 4 kip/ft2 under
design factored loading. The width of footing is 6.0 ft
and the concrete wall above is 12 in wide. The con-
crete is normal weight with a compressive strength of
3000 psi; reinforcement steel is grade 60. Determine the
required area of flexural steel per foot of wall length and
select appropriate reinforcement.

h � 12 in

3 in cover

d � 8.5 in

wu � 4 kip/ft
2

2.5 ft

Pu

Solution:
Maximum bending moment occurs at the face of sup-
port.

Mu =
wua

2

2
=

(
4

kip
ft2

)
(2.5 ft)2

2

= 12.5 ft-kip/ft
Calculate the flexural steel required using Eq. 3.14.

φMn = Mu

= φρbd2fy

(
1 − 0.59 ρ

(
fy
f ′c

))

(
12.5

ft-kip
ft

) (
12

in
ft

) (
1000

lbf
kip

)

= 0.9 ρ
(

12
in
ft

)
(8.5 in)2

(
60,000

lbf
in2

)

×




⎝1 − 0.59 ρ






60,000
lbf
in2

3,000
lbf
in2











ρ = 0.00337

As = ρbd = (0.00337)
(

12
in
ft

)
(8.5 in)

= 0.34 in2/ft

Professional Publications, Inc.

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