##### 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.

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.