Rcc Lec 07 Bond and Dev Length

8/20/2019 Rcc Lec 07 Bond and Dev Length

1/83

1


2/83

By

Dr. Attaullah Shah

Swedish College of Engineering and TechnologyWah Cantt.

Lec-07

Bond and Development Length

Reinforced Concrete Design-I


3/83

− The basic assumption of theRCC design is that the strain

in concrete and reinforcingsteel is the same. f thereinforcing steel slips at itsends! this is not "alid. #ence

it must be ensured thatsufficient bond strength isde"eloped at the interface ofsteel and concrete to a"oidslippage of the steel.

−


4/83

Bond Strength and $e"elopment length− Two types of bond failure can

be e%pected in reinforcing bars&

− $irect pull out of the steel bars!when ample concrete confinementis pro"ided in the form of largespacing of bars or large concreteco"er

− Splitting of concrete along the barwhen co"er confinement or barspacing is insufficient


5/83


6/83

'

b. (ctual $istribution of )le%ural Bond Stress

− ure !ending case

− Concrete fails to resist tensilestresses only where the actualcrac* is located. Steel T isma%imum and

T max = M / jd .

− Between crac*s ! concretedoes resist moderate amount oftension introduced by bond.

− u is proportional to the rate ofchange of bar force! andhighest where the slope of the

steel force cur"e is greatest.

− +ery high local bond stressad,acent to the crac*.


7/83

-

− Beam under transverse loads"− (ccording to simple crac*

sectional theory! T is proportionalto the moment diagram and u isproportional to shear forcediagram.

− n actual! T is less than the simple

analysis prediction e"erywheree%cept at the actual crac*s.

− Similarly! u is eual with simpleanalysis prediction only at thelocation where slopes of the steel

force diagrams are euals .f theslope is greater than assumed!bond stress is greater/ if the slopeis less bond stress is less.


8/83

$epar tment

2

3T4(TE B56$ STRE67T# (6$$E+E3584E6T 3E67T#

− #$pes of !ond failure

− Direct pullout of bars9small diameter bars areused with sufficientlylarge concrete co"erdistances and barspacing:

− Splitting of the concrete along the bar 9co"er orbar spacing is

insufficient to resist thelateral concrete tensionresulting from thewedging effect of bardeformations:


9/83

;

a. ltimate Bond Strength− Direct pull out

− )or sufficiently confined bar! adhesi"e bond and friction are o"ercome as thetensile force on the bar is increased. Concrete e"entually crushes locally ahead

of the bar deformation and bar pullout results.− When pull out resistance is o"ercome or when splitting has spread all theway to the end of an unanchored bar! complete bond failure occurs.

− Splitting− Splitting comes from wedging action when the ribs of the deformed bars bear

against the concrete.

− Splitting in "ertical plane− Splitting in hori


10/83

Consider a bar embedded in a

mass of concrete

8 =τ>?3

b>π>d

b@

8 = σ > ?π>dbA@

τ = 8 ?3b>π>db@ D τma%

8 D τma% > ?3b>π>db@σ = 8 ?π>dbA@ D σma%

8 D σma% > ?π>dbA@

To force the bar to be the wea* lin*& τma% > ?3b>π>db@ σma% > ?π>dbA@

3b 9σma% τma%:> ?db@

3b

db


11/83

$e"elopment 3ength

− 3d = de"elopment length− the shortest distance o"er which a bar can achie"e itFs full

capacity− The length that it ta*es a bar to de"elop its full contribution

to the moment capacity! 4n

Cc

Ts

4n = 9C or T:>9dist:

4n

G

3d


12/83

Steel 3imit! σma%

− sing the bilinear assumption of (C H12&

σma% = I f y

3b 9f y τma%:> ?db@

3b f y > db 9>τma%:


13/83

Concrete Bond 3imit! τma%

− There are lots of things that affect τma%− The strength of the concrete! fFc− Type of concrete 9normal weight or light weight:

− The amount of concrete below the bar − The surface condition of the rebar − The concrete co"er on the bar

− The pro%imity of other bars transferring stress tothe concrete− The presence of trans"erse steel


14/83

Concrete Strength! fFc

− Bond strength! τma%! tends to increase with concretestrength.

− E%periments ha"e shown this relationship to be

proportional to the suare root of fFc.


15/83

Type of Concrete

− 3ight weight concrete tends to ha"e less bondstrength than does normal weight concrete.

− (C H12JG2 introduces a lightweight concrete

reduction factor! λ! on srt9fFc: in some euations.− See (C H12JG2! 2.'.1 for details


16/83

(mount of Concrete Below Bars

− The code refers to Ktop barsLas being any bar which has1A inches or more of freshconcrete below the barwhen the member is poured.

− f concrete 1AL thenconsolidation settlement

results in lower bondstrength on the bottom sideof the bar

− See (C H12JG2! 1A.A.9a:


17/83

Surface Condition of Rebar

− (ll rebar must meet (ST4 reuirements for deformationsthat increase pullout strength.

− Bars are often surface coated is inhibit corrosion.− Epo%y Coating The ma,or concernM− 7al"ani


18/83

8ro%imity to Surface or 5ther Bars

− The si


19/83

8resence of Trans"erse Steel

− The bond transfer tends to cause a splitting plane

− Trans"erse steel will increase the strength of the splittingplane.


20/83

b. $e"elopment 3ength

− $e"elopment length is the length of embedment necessary to de"elop the full tensile strength of bar! controlled byeither pullout or splitting.

− n )ig.! let− ma%imum 4 at a and


21/83

$epar tment

A1

(C C5$E 8R5+S56 )5R $E+E3584E6T5) TE6S56 RE6)5RCE4E6T

3imit3imit

9c I *9c I *tr tr : d: dbb = A.P for= A.P forpullout casepullout case QQfFc are not to befFc are not to be

greater than 1GG psi.greater than 1GG psi.


22/83

AA

%or t&o cases of practical importance" using 9c I *tr : db = 1.P!


23/83

AH

E%ample&


24/83

$epar tment

A

Continue&


25/83

AP

Continue&


26/83

A'

(6C#5R(7E 5) TE6S56 B(RS B0

#55Sn the e"ent that the desired tensile stress in a bar can notn the e"ent that the desired tensile stress in a bar can notbe de"eloped by bond alone! it is necessary to pro"idebe de"eloped by bond alone! it is necessary to pro"idespecial anchorage at the end of the bar.special anchorage at the end of the bar.


27/83

A-

b $ l t 3 th d 4 difi ti


28/83

A2

b. $e"elopment 3ength and 4odification)actors for #oo*ed Bars


29/83$e artment of

A;


30/83

$epar tment

HG

'(ample

(6C#5R(7E RERE4E6TS )5R


31/83

H1

(6C#5R(7E RERE4E6TS )5RWEB RE6)5RCE4E6T


32/83

$epar tment

HA

$E+E3584E6T 5) B(RS 6

C548RESS56− Reinforcement may be

reuired to de"elop itscompressi"e strength by

embedment under "ariouscircumstances.

− (C basic de"elopment

length in compression

l db = 0.02d b f y /√f c


33/83

Determining Locations of FlexuralDetermining Locations of FlexuralCutoffsCutoffs

Given a simply

supported beam with a

distributed load.


34/83


Note:

Total bar length =

Fully effective length+ Development length


35/83


ACI 12.10.3

ll longitudinal tension bars

must e!tend a min. distance

= d "effective depth of the

member# or $% d b "usually

larger# past the theoretical

cutoff for fle!ure "&andlesuncertainties in loads' design

appro!imations'etc..#


36/83


Development of fle!ural

reinforcement in a typical

continuous beam.

() *$,-% - $%.$ forfle!ural reinforcement


37/83

Bar Cutoffs - General ProcedureBar Cutoffs - General Procedure

Determine theoretical fle!ural cutoff points for

envelope of bending moment diagram.

/!tract the bars to satisfy detailing rules "from() 0ection 1.$*' $%.$' $%.$' $%.$$ and $%.$%#

Design e!tra stirrups for points where bars are

cutoff in 2one of fle!ural tension "() $%.$.3#

$.

%.

*.


38/83

Bar Cutoffs - General RulesBar Cutoffs - General Rules

4ars must e!tend the longer of d or $%d b past

the fle!ural cutoff points e!cept at supports

or the ends of cantilevers "() $%.$$.$#

All Bars

,ule $.

,ule %. 4ars must e!tend at least ld from the point of

ma!imum bar stress or from the fle!ural

cutoff points of ad5acent bars "() $%.$.%

$%.$.6 and $%.$%.%#


39/83


Structural Integrity

− Simple Supports t least one-third of the positivemoment reinforcement must be e!tend 7 in. intothe supports "() $%.$$.$#.

− Continuous interior beams with closed stirrups.t least one-fourth of the positive moment

reinforcement must e!tend 7 in. into the support

"() $%.$$.$ and 1.$*.%.*#

Positive Moment Bars

,ule *.


40/83



− Continuous interior beams without closedstirrups. t least one-fourth of the positivemoment reinforcement must be continuous or

shall be spliced near the support with a class

tension splice and at non-continuous supports beterminated with a standard hoo8. "() 1.$*.%.*#.


,ule *.

Bar Cutoffs General RulesBar Cutoffs General Rules


41/83



− Continuous perimeter beams. t least one-

fourth of the positive moment reinforcementre9uired at midspan shall be made continuousaround the perimeter of the building and must beenclosed within closed stirrups or stirrups with

$*3 degree hoo8s around top bars. The re9uiredcontinuity of reinforcement may be provided bysplicing the bottom reinforcement at or near thesupport with class tension splices "()1.$*.%.%#.


,ule *.


42/83



− Beams forming part of a frame that is the primary lateral load resisting system for thebuilding. This reinforcement must be anchored

to develop the specified yield strength' f y' at the

face of the support "() $%.$$.%#


,ule *.


43/83


Stirrups

− t the positive moment point of inflection andat simple supports' the positive moment

reinforcement must be satisfy the following

e9uation for () $%.$$.*. n increase of *

in value of ;n < u shall be permitted when theends of reinforcement are confined by

compressive reaction "generally true for simply

supports#.


,ule 6.


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,ule 6.

a

u

n

d

l M l +≤


45/83


− Negative moment reinforcement must beanchored into or through supporting columns ormembers "() 0ec. $%.$%.$#.

Negative Moment Bars

,ule 3.


46/83



− Interior beams. t least one-third of the negativemoment reinforcement must be e!tended by thegreatest of d' $% d b or " ln < $7 # past the negative

moment point of inflection "() 0ec. $%.$%.*#.


,ule 7.


47/83



− erimeter beams. )n addition to satisfying rule 7a'one-si!th of the negative reinforcement re9uired atthe support must be made continuous at mid-span.

This can be achieved by means of a class tension

splice at mid-span "() 1.$*.%.%#.


,ule 7.


48/83

Moment Resistance DiagramsMoment Resistance Diagrams

;oment capacity of a beam is a function of its depth'

d' width' b' and area of steel' s. )t is common

practice to cut off the steel bars where they are no

longer needed to resist the fle!ural stresses. s incontinuous beams positive moment steel bars may be

bent up usually at 63o' to provide tensile

reinforcement for the negative moments over the

support.


49/83


The nominal moment capacity of an under-reinforced

concrete beam is

To determine the position of the cutoff or bent point

the moment diagram due to e!ternal loading is drawn.

s y

n s y

c

where'% .3

A f a M A f d a f b

= − =


50/83


The ultimate moment resistance of one bar' ;nb is

The intersection of the moment resistance lines with

the e!ternal bending moment diagram indicates the

theoretical points where each bar can be terminated.

nb bs y bs where' area of bar %

a

M A f d A

= − −


51/83


Given a beam with the 6 > bars and

f c=* 8si and f y=3 8si and d = % in.


52/83


The moment diagram is

)oment Diagram

G

PGG

1GGG

1PGG

AGGG

APGG

HGGG

G A C ' 2 1G 1A 1C 1' 12 AG

ft

* -

i n


53/83


The moment resistance of one bar is

( ) ( )( ) ( )

( ) ( )( )

nb sb y

%

s y

c

%

nb

ub nb

%

*.$7 in 3 8si3.% in.

.3 .3 * 8si $% in.

3.% in.

.1? in 3 8si % in. 7 8-in.%

.? 7 8-in. 7% 8-in.

a M A f d

A f a

f b

M

M M φ

= −

= = =

= − = = = =


54/83


The moment diagram and crossings

)oment Diagram

G

PGG

1GGG

1PGG

AGGG

APGG

HGGG

G A ' 2 1G 1A 1 1' 12 AG

ft

* -

i n

'AG *Jin

1AG *Jin

12'G *Jin

A2G *Jin


55/83


The ultimate moment resistance is %6 8-in. The

moment diagram is drawn to scale on the basis bar

can be terminated at a' two bars at b and three bars at c.

These are the theoretical termination of the bars.)oment Diagram

G

PGG

1GGG1PGG

AGGG

APGG

HGGG

G A ' 2 1G 1A 1 1' 12 AG

ft

* - i n

'AG *Jin

1AG *Jin

12'G *Jin

A2G *Jin

a!

c


56/83


(ompute the bar development length is

( )

( ) ( )

a b

y b

d

c

$% or d

$% $. in. or % in. % in.

3 $. in.

% % *

63.7 in. 67 in.

l d

f d l

f

αβλ

=

= ⇒

= =

= ⇒


57/83


The ultimate momentresistance is %6 8-in.

The moment diagram

is drawn to scale on

the basis bar can be

terminated at a' two

bars at b and three bars

at c. These are thetheoretical termination

of the bars.


58/83


)t is necessary to develop

part of the strength of the bar

by bond. The () (ode

specifies that every barshould be continued at least

a distance d' or $%d b ' which

ever is greater' beyond the

theoretical points a' b' and c.0ection $%.$$.$ specify that

$


59/83


Two bars must e!tend

into the support and

moment resistance

diagram ;ub mustenclose the e!ternal

bending moment

diagram.


60/83

Example – Cutoff Example – Cutoff

For the simplysupported beam with

b=$ in. d =$1.3 in.'

f y

=6 8si and f c

=* 8si

with 6 > bars. 0how

where the reinforcing

bars can be terminated.


61/83


Determine the moment capacity of the bars.

( ) ( )( ) ( )

( ) ( )

%

s y

c

%

nb

*.$7 in 6 8si 6.?* in..3 .3 * 8si $ in.

6.?* in.

.1? in 6 8si $1.3 in. %

6%1.1 8-in. *3.76 8-ft.

A f a f b

M

= = =

= − = ⇒


62/83

ExampleExample – Cutoff – Cutoff

Determine the location of the bar intersections ofmoments.

$ bar *3.76 8-ft.

% bar 1$.* 8-ft.

=

=

( ) M x M mx= −

* bar $1 8-ft.

6 bar $6%.7 8-ft.

=

=


63/83



$ bar *3.76 8-ft.

% bar 1$.* 8-ft.

=

=

$*%.3 8-ft. 1.3 8-ft.$1 8-ft. $*%.3 8-ft.

7 ft.

*.6 ft. 6. in. or 6$ in.

x

x

− = − ÷

= ⇒

* bar $1 8-ft.

6 bar $6%.7 8-ft.

=

=


64/83



$ bar *3.76 8-ft.

% bar 1$.* 8-ft.

=

=

* bar $1 8-ft.

6 bar $6%.7 8-ft.

=

=

1.3 8-ft. . 8-ft.1$.* 8-ft. 1.3 8-ft.

3 ft.

.?* ft. $$.$ in. or $$ in.

or $$ in. + 1% in. = * in. from center

x

x

− = − ÷

= ⇒


65/83


The minimum distance is

( )

( ) ( )

a b

y b

d

c

$% or d

$% $. in. or $1.3 in. $ in.

6 $. in.

% % *

*7.7 in. *1 in.

l d

f d l

f

αβλ

=

= ⇒

= =

= ⇒


66/83


The minimum amount of bars are s


67/83


The cutoff for the first bar is 6$ in. or * ft 3 in. and $ in

or $ ft 7 in. total distance is 6$ in.+$ in. = 3? in. or 6 ft

$$ in.

Note error it is 4’-11” not 5’-11”

E lE l C ffC t ff


68/83


The cutoff for the second bar is * in. + $ in. $$ in. or

ft 3 in. "*1-in+3-in+$-in+6$-in= $$-in.#

Note error it is 4’-11” not 5’-11”

E lE l C t ffC t ff


69/83


The moment diagram is the blue line and the red line is

the envelope which encloses the moment diagram.

Bar plicesBar plices


70/83

pp

@hy do we need bar splicesA -- for long spans

Types of 0plices

$. 4utted B@elded

%. ;echanical (onnectors

*. Cay 0plices

;ust develop $%3of yield strength ()

$%.$6.*.% and ()

$%.$6.*.6


71/83

!ension Lap plices!ension Lap plices

@hy do we need bar splicesA -- for long spans

Types of 0plices

$. (ontact 0plice

%. Non-(ontact 0plice "distance between the

bars 7 and $


72/83

!"pes of plices!"pes of plices

Class A Splice "()

$%.$3.%#@hen over entire splice

length.

and $


73/83

!"pes of plices!"pes of plices

Class B Splice "() $%.$3.%#

ll tension lay splices not meetingre9uirements of (lass 0plices

! i L li #$C% &' &)*! i L li #$C% &' &)*


74/83

!ension Lap plice #$C% &'(&)*!ension Lap plice #$C% &'(&)*

wheres "re9Ed# = determined for bending

ld = development length for bars "not

allowed to use e!cess reinforcementmodification factor#

ld must be greater than or e9ual to $% in.

!ension Lap plice #$C% &' &)*!ension Lap plice #$C% &' &)*


75/83

!ension Lap plice #$C% &'(&)*!ension Lap plice #$C% &'(&)*

Cap 0plices shall not be used for bars larger than No. $$."() $%.$6.%#

Cap 0plices should be placed in away from regions of

high tensile stresses -locate near points of inflection"() $%.$3.$#

C i L li #$C% &' &+*


76/83

Compression Lap plice #$C% &'(&+*Compression Lap plice #$C% &'(&+*

Cap' re9Ed = .3f y d b for f y 7 psi

Cap' re9Ed = ".?f y -%6# d b for f y 7

psi Cap' re9Ed $% in

For f c * psi' re9uired lap splice shall be multiply

by "6


77/83

Compression Lap plice #$C% &'(&,('*Compression Lap plice #$C% &'(&,('*

)n tied column splices with effective tie area throughout

splice length .$3 hs factor = .*

)n spiral column splices' factor = .13

The final splice length must be $% in.

≥

≥


78/83

Example – plice !ensionExample – plice !ension

(alculate the lap-splice length for 7 > tension bottom

bars in two rows with clear spacing %.3 in. and a clear

cover' $.3 in.' for the following cases

@hen * bars are spliced and s"provided#


79/83


For > bars' d b =$. in and α = β = γ = λ =$.

( )

yd

b c tr

b

*

6

* 7 $.

6%.6 6* in.$.3 in. 6 3

$. in.

f l

d f c K

d

αβγλ =

+ ÷

= = ⇒+ ÷


80/83


The s"provided#


81/83

Example – plice CompressionExample – plice Compression

a# f y = 7 8si

b# f y = 8si

(alculate the lap splice length for a > $ compression

bar in tied column when f c= 3 8si and


82/83


For >$ bars' d b =$.%1 in.

( )

( )

ydy

b c

d d

.%.*

.% 7$7.?1 or $

3

$ $.%1 in. %%.7 in. %* in.

f l f

d f

l l

= ≥

= =

= = ⇒ =

(hec8 ls .3 d b f y = *.$ in. 0o ls = *? in.


83/83


For >$ bars' d b =$.%1 in. The ld = %* in.

(hec8 ls ".? f y J%6# d b

=".?"#-%6#"$.%1in.# = 7$ in.

0o use ls = 7$ in.

Documents

Rcc Lec 07 Bond and Dev Length