3RM3 Module 6 Development Length

8/2/2019 3RM3 Module 6 Development Length

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CE 3RM3 Inspection, Repair & Maintenance of

Reinforced Concrete Structures, W.W. El-

Dakhakhni, 2012

Mechanism of bond transfer

Bond stresses provide mechanism of force transfer betweenconcrete and reinforcement.

Equilibrium Condition for Rebar

Note: Bond stress is zero at cracks

avgis the value at bond failure in a beam test

2

by b b

y b

d

F 0 Bond Force 0

0

4

4

avg

avg

T

df d l

f dl

= =

=

=

= bond stress

(adhesion,

friction, bearing,

etc.)

( )c

bar

k f

k f

=


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Simplifiedld (A23.3 Table 12.1)

Cases

Case 1: Member containing minimum

stirrups or ties within ld(A23.3 Cl.11.2.8.4

or A23.3 Cl.7.6.5)

Case 2: Slabs, walls, shells, or folded plates

having clear spacing between bars being

developed not less than 2db

Other cases

Minimum developmentlength, ld(mm)

1 2 3 4 y

b'

0.45 d

c

k k k k f

f

1 2 3 4 y

b'0.6 d

c

k k k k f

f

k1 is a bar location factor.

k2 is a coating factor.

k3 is a concrete density factor.

k4 is a bar size factor.

ldis the development length, mm

db is the diameter of the bar being developed , mm


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Factors used in expressions for

Development Length (A23.3 Table 12.1)

1- k1 = Bar location factor

1.3 Horizontal reinforcementmore than 300 mm of

fresh concrete is cast in the member below the

development length or splice (considered top rft)

1.0 Other reinforcement2- k2 = Coating factor

1.5 Epoxy-coated bars or wires with cover less than3dbor clear spacing less than 6db

1.2 All other epoxy-coated bars or wires

1.0 Uncoated reinforcement

3- k3 = Low-density aggregate concrete factor

1.3 Low-density concrete1.2 Semi-low-density concrete1.0 Normal-density concrete4- k4 = Bar size factor

0.8

No.20 and smaller bars and deformed wires1.0 No.25 and larger bars

The product k13k2 need not be taken greater than 1.7


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Excess Flexural Reinforcement

Reduction (A23.3 Cl.12.2.5)

Reduction = (As reqd ) / (As provided )

- Except as required for seismic design (see A23.3

Cl.12.2.5)

- Good practice to ignore this provision, since use

of structure may change over time.

- final ld 300 mm.

Development Length for Bars inCompression (A23.3 CL.12.3)

Compression development length,

ldc= ldb3 applicable modification factors 200 mm.

A23.3 Cl.12.3.3 gives the modification factor for excess

reinforcement and enclosure by spiral or ties

Basic development length for compression, ldbc

b y

db b'

0.240.044 d y

c

d fl f

f=


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Compression development length:

The basic development length ldb, may be multipliedby applicable factors, with a cumulative value of

not less than 0.6, for

(a)Reinforcement in excess of that required byanalysis (As reqd ) / (As provided )

(b)Reinforcement enclosed within spiralreinforcement of not less than 6 mm diameter and

not more than 100 mm pitch or within No.10 ties in

conformance with Cl.7.6.5 and spaced at not morethan 100 mm on centre 0.75

Example

For the shown cross section of a

simply supported beam

reinforced with 4 No.15 bars that

are confined with No.10 stirrup

spaced at 150 mm, determine the

development length of the bars if

the beam is made of normalweight concrete, uncoated

reinforcement bars, given that fc

= 30 MPa and fy= 400 MPa

300

4No.15

60 60

560

500


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1 2 3 4 yd b

'

1 3

2 4

0.45 d

1.0 1.0

1.0 0.8

c

k k k k f l

f

k k

k k

=

= =

= =

The development length is:

1 2 3 4 y

d b'

0.45 d

1 1 1 0.8 4000.45 16 420.6530

425 mm 300mm

c

d

d

k k k k f l

f

l

l

=

= =

=


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Standard Hooks for Tension Anchorage

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.

Standard Hooks

A hook is used at the end of a

bar when its straight

embedment length is less

than the necessary length, ld.


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Standard hooks are

defined in A23.1 Cl.

12.2.2

Hooks resists tension by

bond stresses on bar

surface and bearing on on

concrete inside the hook.

Design of Standard Hooks for Tension

Anchorage (A23.3 Cl.12.5)

Development length for hooked bar, ldh.

multipliers

where, 8 or 150 mm whichever is smaller

dh hb

dh b

l l

l d

=

Basic development length for hooked bar = lhbwhen

fy = 400 MPa

'

100

hb b

c

l df

=


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Conditions

For No.35 or smaller bars, where the side cover

(normal to plane of hook) is not less than 60 mm

and for 90o hooks where the cover on the bar

extension beyond the hook is not less than 50 mm

Multiplier

fy /400

0.7

For bars with fy other than 400 MPa

Conditions

For No.35 or smaller bars, where the hook is

enclosed vertically or horizontally within at

least three ties or stirrup ties spaced along a

length at least equal to the inside diameter of

the hook, at a spacing not greater than 3db ,

where db

is the nominal diameter of the

hooked bar

Multiplier

0.8


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Conditions

Anchorage or development for fy is not

specially required, for reinforcement in

excess of that required by analysis

Structures low density concrete

Multiplier

As(reqd) /

As(provided)

1.3

Conditions

Epoxy-coated Reinforcement

Multiplier

1.2


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ExampleCompute the development length required for the top 3

No.20 bar of the cantilever beam that extend into the

column support if:

The bars are confined by #10 stirrups spaced at

200mm, and clear cover = 70 mm, and clear

spacing = 50 mm and fc =35 MPa and fy =400 MPa

(a)Anchored by hooksinto the column

(b)Developed in thebeam, b=300 mm

90o Hook

(a)3 No. 20 bars anchored in the column by standard90ohook

From the size of the column, the space available for

a hook is 450-50-0.5320=390 mm

The development length is the product of the basic

hook length lhb and the factors in A23.3 Cls.

12.5.3 and 12.5.4

( )

hb

b c

hb

100 10016.9

35

16.9 20 mm 338 mm 340 mm

l

d f

l

= = =

= =


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Cl.12.5.3 (a) does not apply...31.0

Cl.12.5.3(b) the bars are No.35 or smaller, the side

cover to the hooks is 70 mm, and the tail cover is 50

mm. Therefore, 12.5.3 (b) applies...30.7

Cl.12.5.3(c) does not apply.31.0

Cl.12.5.3(d) will be ignored 31.0

Cl.12.5.3(e) does not apply .31.0

Cl.12.5.3(f) does not apply ..31.0

180o Hook

Check weather Cl.12.5.4 requires joist ties. Side

cover=70mm. Top cover extends into the column

over the hook. Therefore extra ties are not needed in

the joint.

ldh= 238 mm > 8db = 160 mm or 150 mm.

ldh 390 mm , the hook will anchor the bar

hb

dh

340 mm

340 0.7 238 mm

l

l

=

= =


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In-beam development

(b) 3 No. 20 bars anchored in the beam

Development length available in the beam

=1200-50=1150 mm

v

2

V

0.06 35 300 200Min. A for s=200 mm=

400

A 53.2 mm

=

The spacing is less than the maximum spacing fromCl.11.2.11. therefore , the beam has more than

minimum stirrups and this is Case 1 in Table 12-1 of

A23.3

K1=1.3 ( more than 300 mm of fresh concrete below

the bars when the concrete is placed

K2=1.0 bars are not coated

K3=1.0 concrete is normal- density

K4=0.8 bars are No.20

1 2 3 4 y

d b'

d

0.45 d

1.3 1.0 1.0 0.8 400l 0.45 20 633 mm

35

c

k k k k f

lf

=

= =


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Bar Cutoff PointsWhy do you want to provide cut off points?

Cost!


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Factors Affecting Bar Cut-off Points

Bars no longer needed to resist tensile forces or

remaining bars are adequate (Use moment andshear envelopes)

Bars must be extended on each side of section to

develop bar force at that section.

1.)

2.)

Major stress concentrations occur when tension

bars are cutoff in regions of moderate to high

shear forces. Leads to cracking.

Code specified construction requirements (good

practice)

3.)

4.)

Load uncertainties (Seismic Considerations)5.)

Determining Locations of Flexural

Cutoffs

A23.3 Cl.12.10.2

All longitudinal tension

bars must extend a min.

distance = d(effective depth

of the member) or12 db

(usually larger) past the

theoretical cutoff for flexure

(Handles uncertainties inloads, design approximations,

etc.)


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Bar Cutoffs - General Rules

! Simple Supports At least one-third of the positive momentreinforcement must be extend 150 mm into the supports(A23.3 Cl.12.11.1).

! Continuous interior beams with closed stirrups. At leastone-fourth of the positive moment reinforcement mustextend 150 mm into the support (A23.3 Cl.12.11.1)

Positive Moment Bars

Rule 1-P

! Continuous interior beams without closed stirrups. Atleast one-fourth of the positive moment reinforcement

must be continuous or shall be spliced near the supportwith a class A tension splice and at non-continuous

supports be terminated with a standard hook. (A23.3 Cl.

12.11.2).

Rule 2-P

Bars must extend the longer of d or 12db past the flexural

cutoff points except at supports or the ends of cantilevers

(A23.3 Cl.11.3.8.1)

Bars must extend at least ld from the point of maximum

bar stress or from the flexural cutoff points of adjacent

bars (A23.3 Cls.12.10.2 and 12.10.4

Rule 3-P

d

ra

f

Ml lV

+ A23.3 Eq.12-6

lais the longer of the effective depth , d, or 12 db

Rule 4-P


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Bars must extend the longer ofdor12db past the flexural

cutoff points except at supports or the ends of cantilevers

(A23.3 Cl.11.3.8.1)

Bars must extend at least ld from the point of maximum

bar stress or from the flexural cutoff points of adjacent

bars (A23.3 Cls.12.10.2 and 12.10.4)

Rule 2-N

Rule 3-N

Negative Moment Bars

Rule 1-N

At least one-third of the negative moment reinforcement must

be extended by the greatest ofd, 12 db or ( ln/ 16 ) past the

negative moment point of inflection (A23.3 Cl.12.12.2).

Negative moment reinforcement must be anchored into or

through supporting columns or members (A23.3 Cl.12.12.1).

Rule 4-N

Critical Sections in FlexuralMembers

The critical sections for development of reinforcement in

flexural members are:

At points of maximum stress;

At points where tension bars within span are

terminated or bent;

At the face of the support;

At points of inflection at which moment

changes sign.

1.

2.

3.

4.


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Moment Capacity Diagram

Moment capacity of a beam is a function of its depth,

d, width, b, and area of steel, As. It is common

practice to cut off the steel bars where they are no

longer needed to resist the flexural stresses. As in

continuous beams positive moment steel bars may be

bent up usually at 45o, to provide tensile

reinforcement for the negative moments over the

support.

The resisting moment capacity of an under-reinforced

concrete beam is

To determine the position of the cutoff or bent point

the moment diagram due to external loading is drawn.

s y

r s y '

1

where,2

s

s

c c

A faM A f d a

f b

= =

The ultimate moment resistance of one bar, Mrb is

The intersection of the moment resistance lines with

the external bending moment diagram indicates the

theoretical points where each bar can be terminated.

rb bs y bswhere, is the area of one bar2

s aM A f d A =


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Example

Given a beam with the 4 #25 bars and

fc=30 MPa and fy=400 MPa and

d = 500 mm

6.0 m

60 kN/m

300

500

4 No.2560

Factored Moment Diagram

The moment diagram is

Moment Diagram

0

50

100

150

200

250

0 1 2 3 4 5 6

m

kN.m

300

500

4 No.25

60


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The factored moment of the beam is

The moment resistance of the beam is

f

60 36M 270 kN.m

8

= =

( )( )( )( )

( )( )

rb sb y

2

s y

1 c

2

r

2

0.85 2000 mm 400 MPa156.43 mm

0.6 0.805 30 MPa 300 mm

156.43 mmM 0.85 2000 mm 400 MPa 500 mm 286.8 kN.m

2

s

s

c

aM A f d

A fa

f b

=

= = =

= =

The moment resistance of one bar is

( )( )( )( )

( )( )

rb sb y

2

s y

1 c

2

rb

2

0.85 500 mm 400 MPa39.11 mm

0.6 0.805 30 MPa 300 mm

39.11 mm0.85 500 mm 400 MPa 500 mm 81.68 kN.m

2

s

s

c

aM A f d

A fa

f b

M

=

= = =

= =

The moment resistance of two bars is

( )( )( )( )

( )( )

rb sb y

2

s y

1 c

2

r

2

0.85 1000 mm 400 MPa78.215 mm

0.6 0.805 30 MPa 300 mm

78.215 mmM 0.85 1000 mm 400 MPa 500 mm 156.7 kN.m

2

s

s

c

aM A f d

A fa

f b

=

= = =

= =


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The moment resistance of three bars is

( )( )( )( )

( )( )

rb sb y

2

s y

1 c

2

rb

2

0.85 1500 mm 400 MPa117.32 mm

0.6 0.805 30 MPa 300 mm

117.32 mm0.85 1500 mm 400 MPa 500 mm 225.08 kN.m

2

s

s

c

aM A f d

A fa

f b

M

=

= = =

= =

Compute the bar development length is

( )a b

d

12 or d

12 25 mm or 500 mm 500 mm

4000.45 1 1 1 1 25 821.58 mm

30

825 mm

l d

l

=

=

= =

=


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Moment Capacity Diagram

The ultimate moment

resistance is 270 kN.m

The moment diagram

is drawn to scale on

the basis A bar can be

terminated at a, two

bars at b and three bars

at c. These are the

theoretical terminationof the bars.

60 kN/m

6.0 m

81.68

156.7

225.08

270

Moment Resistance Diagrams

It is necessary to develop part

of the strength of the bar by

bond. The A23.3 Code

specifies that every bar

should be continued at least a

distance d, or 12db , which

ever is greater, beyond the

theoretical points a, b, and c.Section 12.11.3.8.1 specify

that 1/3 of positive moment

reinforcement must be

continuous at least 150 mm

into the support.

Two bars must extend

into the support and

moment resistance

diagram Mub must

enclose the external

bending moment

diagram.


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Bar Splices

Why do we need bar splices? -- for long spans

Types of Splices

1. Butted &Welded

2. Mechanical Connectors

3. Lap Splices

Must develop 120%

of yield strength of

the bar but not less

than 110% of theactual yield strength

of the bar used.

A23.3 12.14.3.1


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Types of Splices

1-Splices for deformed bars and wires in Tension

The minimum length of lap for tension lap splices

shall be as required for a Class A or B splice, but

not less than 300 mm, where

(a) Class A splice 1.0ld

(b) Class B splice ....1.3 ld

Where ld is the tensile development length for the

specified yield strength without the modification

factor of Cl. 12.2.5

Tension Lap Splice (A23.3 Cl.

12.15.2 Table 12-2)

As (provided)

As (required)

Equal to or

greater than 2

Less than 2

Maximum percent of As spliced

within required lap length

50 100

Class A

Class B

Class B

Class B


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Compression Lap Splice (A23.3 12.16)

The minimum length of the lap for compression lap

splices shall be not less than 0.073fydb, nor

(0.133fy-24)db for fy greater than 400 MPa, nor

300mm

*Welded splices or mechanical connections used in

compression shall meet the requirements of

Cl. 12.14.3.3 or 12.14.3.4

Example

Calculate the lap-splice length for 6 No.25 tension

bottom bars in two rows with clear spacing 60 mm and a

clear cover, 40 mm, for the following cases

When 3 bars are spliced and As(provided) /As(required) >2

When 4 bars are spliced and As(provided) /As(required) < 2

When all bars are spliced at the same location.fc= 40 MPa and fy = 400 MPa

a.

b.

c.


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The As(provided) /As(required) > 2, class A splice applies;

therefore lst = 1.0 ld >300 mm, so lst = 657.27 mm >

300 mm. The bars spliced are less than half the

numberThe As(provided) /As(required) < 2, class B splice applies;

therefore lst = 1.3 ld >300 mm, so lst = 1.3(657.27 mm) =

854.45 mm use 860 mm > 300 mm

For No.25 bars, db =25.2 mm

d

4000.45 1 1 1 0.8 25 657.27 mm

30l = =

Example

a) fy = 400 MPa , and b) fy = 500 MPa

Calculate the lap splice length for a No. 30 compression

bar in tied column when fc= 30 MPa and

For No.30 bars, db =29.9 mm

'

0.240.044

0.24 29.9 400524.06 0.044 29.9 400 526.24

30

0.24 29.9 500655.077 0.044 29.9 500 657.8

30

b y

db b y

c

db

db

d fl d f

f

l

l

=

= = =

= = =

Use ldb=530 mm for (a) and ldb=660 mm for (b)

3RM3 Module 6 Development Length

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