Rebar Development Lengths_ppt

7/31/2019 Rebar Development Lengths_ppt

1/52

July 2, 2003

CVEN 444


2/52

Slab design reinforcement

Bar Development

Hook development


3/52

For a 1 ft strip of slab is designed like a beam As

(reqd) is in units of (in2/ft)

=

inchesinspacingbar

in12ft/ bs AA


4/52

The table will allow

to determine the

amount of steel per

each foot of slab.


5/52

The minimum spacing of the bars is given as:

Also, check crack control - important for exterior

exposure (large cover dimensions) - The spacing S of

reinforcement closest to the surface in tension ACI Sec.10.6.4

( )

( )max

ACI Sec. 7.6.5

3t slab thickness

S smaller of 18 in.

=

( )c

s s

12 365402.5s c

f f=


6/52

Thin slabs shrink more rapidly than deeper beams.

Temperature & shrinkage (T&S) steel is provided

perpendicular to restrain cracks parallel to span.

(Flexural steel restrains cracks perpendicular to

span)

Maximum & Minimum reinforcement requirements


7/52

Maximum & Minimum reinforcement requirements

T&S Reinforcement (perpendicular to span)

ACI Sec 7.12.2

( ) ( )

( )

( )

( ) t

ftf

ft

ftA

*"12*0.0014

ksi60*"12*60*0018.0

ksi60*"12*0018.0

ksi50or40*"12*0020.0

y

y

y

ymins

=

==

==


8/52

T&S Reinforcement (perpendicular to span)

ACI Sec 7.12.2.2

=18"5tofsmallermaxS

t thickness of the slab


9/52

Flexural Reinforcement (parallel to span)

ACI Sec 10.5.4

Smax from reinforced spacing

( ) ( )s min s min T&SA A=

max

3tsmaller of

18"S

=


10/52

A. Concept of Bond Stress and Rebar Anchorage

Internal Forces in a beam

Forces developed in the beam by loading.


11/52

A. Concept of Bond Stress and Rebar Anchorage

Forces in Rebar

Bond stresses provide mechanism of force transferbetween concrete and reinforcement.


12/52

Equilibrium Condition for Rebar

Note: Bond stress is zero at cracks

2

b y b b

y b

d

F 0 Bond Force 0

0

4

4

T

df d l

f dl

= =

=

=

= bond stress(coefficient offriction)

( )

c

bar

k f

k f

=


13/52

Sources of Bond Transfer

(1) Adhesion between concrete & reinforcement.

(2) Friction

Note: These properties are quickly lost for tension.


14/52

(3)Mechanical Interlock.

The edge stress concentration causes cracking

to occur.


15/52

(3) Mechanical Interlock (cont).

Force interaction between the steel and concrete.


16/52

Splitting cracks result in loss of bond transfer.

Reinforcement can be used to restrain thesecracks.


17/52

Splitting Load is Affected by:

Minimum edge distance and spacing of bars

( smaller distance = smaller load )

Tensile strength of concrete.

Average bond stress along bar. ( Increase in

bond stress larger wedging forces. )

1.

2.

3.


18/52

Typical Splitting Failure Surfaces.


19/52

Typical Splitting Failure Surfaces.


20/52

General splitting of

concrete along the

bars,either in vertical

planes as in figure (a) or in

horizontal plane as infigure (b). Such splitting

comes largely from

wedging action when the

ribs of the deformed barbear against the concrete.

The horizontal type of splitting frequently begins at a diagonal

crack. The dowel action increases the tendency toward splitting.

This indicates that shear and bond failure are often intricately

interrelated.


21/52

ACI Code expression for development length for

bars in tension/in compression.

B.

Development Length, ld

Shortest length of bar in which the bar stress can

increase from zero to the yield strength, fy.

( ld used since bond stresses, , vary along a bar ina tension zone)


22/52

Development Length, ld

( ld used since bond

stresses, , vary along abar in a tension zone)


23/52

Development length, ld (simplified expression from ACI 12.2.2)

Development length, ld 12 ACI 12.2.1fc 10000 psi for Ch. 12 provisions for development length in ACI Codes.

Clear spacing of bars being developed or

spliced not less than db, clear cover not less

than db, and stirrups or ties throughout ld notless than the code minimum

or

Clear spacing of bars being developed or

spliced not less than 2db and clear cover not

less than db.

No. 6 and smaller No. 7 andlarger bars and deformed bars

wires

c

y

b

d

25 ff

dl =

c

y

b

d

20 ff

dl =

38 47.5


24/52

Development length, ld 12 ACI 12.2.1fc 10000 psi for Ch. 12 provisions for development length in ACI

Codes.Development length, ld (simplified expression from ACI 12.2.2)

Other cases

No. 6 and smaller No. 7 and largerbars and deformed bars

wires

c

y

b

d

50

3

f

f

d

l =

c

y

b

d

40

3

f

f

d

l =

57 71

fc = 4 ksi fy = 60 ksi , ,, = 1.0


25/52

Development length, ld ACI 12.2.3

2.5 limit to safeguard against pullout type failure.

yd

b c ct

b

ct

b

3

40

where, 2.5

fl

d f c Kd

c K

d

=

+ +


26/52

= reinforcement location factor

Horizontal reinforcement so placed that

more than 12 in of fresh concrete is cast

in the member below the development

length or splice

Other reinforcement

1.3

1.0

where < 1.7


27/52

= coating factor (epoxy prevents adhesion &friction between bar and concrete.)

Epoxy-coated bars or wires with cover lessthan 3db or clear spacing less than 6db

All other epoxy-coated bars or wires

Uncoated reinforcement

1.5

1.2

1.0

where < 1.7


28/52

= reinforcement size factor (Reflects more favorableperformance of smaller

bars)

No.6 and smaller bars and deformed wire

No. 7 and larger bars

0.8

1.0


29/52

= lightweight aggregate concrete factor(Reflects lower tensile strength of lightweight

concrete, & resulting reduction in splitting resistance.)

When lightweight aggregate concrete is used.

However, when fct is specified, shall be

permitted to be taken as but notless than

When normal weight concrete is used

1.3

1.0

1.0

ctc7.6 ff


30/52

c = spacing or cover dimension, in.

Use the smaller of either

(a) the distance from the center of the bar or wire to thenearest concrete surface.

or

(b) one-half the center-to-center spacing of the bar orwires being developed.


31/52

Ktr= transverse reinforcement index (Represents the

contribution of confining reinforcement across

potential splitting planes.)

ns

fAK

**1500

yttr

tr =

The transverse reinforcement is the development

length in a column.


32/52

Total cross-section area of all transverse

reinforcement within the spacing s, which

crosses the potential plane of splitting along

the reinforcement being developed with in thedevelopment length, in2.

Specified yield strength of transverse

reinforcement, psi.

Atr=

fyt =


33/52

maximum center-to-center spacing of

transverse reinforcement within ld in.

number of bars or wires being developedalong the plane of splitting.

s =

n =

NoteNote: It is permitted to use Ktr= 0 as a design

simplification even if transverse reinforcementis present.


34/52

Reduction = (As reqd ) / (As provided )

- Except as required for seismic design (see ACI

21.2.1.4)

- Good practice to ignore this provision, since use

of structure may change over time.

- final ld 12 in.

( )

( ) ( )providedn

u

providedn

dreq'nReduction

M

M

M

M ==


35/52

Compression development length,

ldc = ldbc * applicable reduction factors 8 in.

Basic Development Length for Compression, ldbc

=yb

c

yb

dbc

0003.0

0.02

oflarger

fdf

fd

l


36/52

Reduction Factors (ACI 12.3.3)

- Excessive Reinforcement Factor

= A(sreqd ) / A(sprovided)

- Spiral and Ties

If reinforcement is enclosed with spiral

reinforcement 0.25 in. diameter and

4 in. pitch or within No. 4 tiesaccording to 7.10.5 and spaced 4 in.

on center. Factor = 0.75


37/52

For the cross section of a simply

supported beam reinforced with 4

#8 bars that are confined with #3

stirrup spaced at 6 in. Determine

the development length of the

bars if the beam is made of

normal weight concrete fc

= 3 ksi

and fy= 60 ksi


38/52

Check if conditions for spacing and concrete

cover are met:

For #8 bars, db = 1.0 in.Clear cover = 2.5 in - 0.5 in.= 2.0 in. > db

Clear spacing

between bars

( )

( ) ( )

b

b

2*cover 1

spaces12 in. 2 2.5 in. 3 1.0 in.

3

1.33in. d

b n d =

=

=


39/52

Bars are confined with #3 stirrups. The conditions are

met.

Determine the factors; = 1.0 (bottom bars), =1.0 (no

coating) and = 1.0 (normal weight concrete) and

54.8 psi < 100 psi

yd

b c

(for bars > #7)20

fld f

=

cf =


40/52

So ld = 54.8(1.0 in.) = 54.8 in. 55 in. Using the

more general formula Ktr= 0.0

( ) ( ) ( )d

b

1.0 1.0 1.0 6000054.8

20 3000

l

d= =

yd

b c tr

b

340

fld f c K

d

= +


41/52

= = = = 1.0. Also

smaller of distance from center of bar to the nearest

concrete surface c1 or one-half the center-to-center ofbars spacing c2

c =

2

2*cover0.5

spaces

12 in. 5 in.0.5 1.17 in.

3

bc

=

= =

1 2.5 in.c =

c = 1.17 in. controls


42/52

( )d

b

3 60000 1.054.8

40 1.53000

l

d

= =

If < 1.5 use 1.5. 1.17/1.0 = 1.17 1.5tr

b

c K

d

+

So ld = 55 in.


43/52

If the same beam is made of light

weight aggregate concrete and the

bars are epoxy coated and As

required for analysis is 2.79 in2


44/52

The conditions are met.

Determine the factors; = 1.0 (bottom bars), =1.5 (epoxycoating) and = 1.3 (lightweight aggregate concrete) and

Rs = (As(req) / As(provided) ) = 2.79/3.16 = 0.89. The value ofis 1.5 because the concrete cover is less than 3db = 3 in.

s yd

b c

(for bars > #7)

20

R fl

d f

=


45/52

Check that =1.0(1.5) = 1.5 < 1.7

So ld = 95.1(1.0 in.) = 95.1 in. 96 in.

( ) ( )db

0.89 1.5 1.3 60000 95.120 3000

ld = =


46/52

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.


47/52

Three sections are critical for the

negative moment reinforcement:

Section 1 is at the face of thesupport, where the negative

moment as well as stress are at

maximum value. Two

development lengths, x1 and x2

must be checked.


48/52

Section 2 is the section where part

of the negative reinforcement bar

can be terminated. To develop

full tensile force, the bars shouldextend a distance x2 before they

can be terminated Once part of the

bars are terminated the remaining

bars develop maximum stress.


49/52

Section 3 is a point of inflectionThe bars shall extend a distance x3

beyond section 3: x3 must be equal

to or greater than the effective

depth d, 12db or 1/16 the span,which ever is greater. At least 1/3

of the total reinforcement provided

for negative moment at support

shall extend a distance x3 beyondthe point of inflection.


50/52

Section 4 is that of maximum

positive moment and maximum

stresses. Two development

lengths x1 and x2 have to bechecked. The length x1 is the

development length ld specified by

the ACI Code Section 12.11. The

length x2 is equal to or greater thanthe effective depth d, 12db .


51/52

Section 5 is where part of the positive

reinforcement bars may be terminated. To

develop full tensile force, the bars should

extend a distance x2. The remaining barswill have a maximum stress due to the

termination of part of the bars. At the face

of the support section 1, at least 1/4 of the

positive momentreinforcement in continuous members shall expend along thesame face of the member into the support, according to ACI

12.11.1. For simple members at least 1/3 of the reinforcement

shall extend into the support


52/52

Section 6 is at the points of

inflection limits are according to

section 12.11.3 of the ACI Code.

Documents

Rebar Development Lengths_ppt