Lecture 17 Bar Development

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Lecture 17- Bar Development

• February 24, 2003
• CVEN 444

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Lecture Goals

• Slab design reinforcement
• Bar Development
• Hook development

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Flexural Reinforcement in Slabs
For a 1 ft strip of slab is designed like a beam
As (reqd) is in units of (in2/ft)
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The table will allow to determine the amount of
steel per each foot of slab.
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Flexural Reinforcement in Slabs
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
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Flexural Reinforcement in Slabs
Maximum Minimum reinforcement requirements

• Thin slabs shrink more rapidly than deeper beams.
• Temperature shrinkage (TS) steel is provided
  perpendicular to restrain cracks parallel to
  span. (Flexural steel restrains cracks
  perpendicular to span)

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Flexural Reinforcement in Slabs
Maximum Minimum reinforcement requirements
TS Reinforcement (perpendicular to span) ACI Sec
7.12.2
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Flexural Reinforcement in Slabs
TS Reinforcement (perpendicular to span) ACI
Sec 7.12.2.2
t thickness of the slab
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Flexural Reinforcement in Slabs
Flexural Reinforcement (parallel to span) ACI
Sec 10.5.4
Smax from reinforced spacing
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
A. Concept of Bond Stress and Rebar Anchorage
Internal Forces in a beam
Forces developed in the beam by loading.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
A. Concept of Bond Stress and Rebar Anchorage
Forces in Rebar
Bond stresses provide mechanism of force transfer
between concrete and reinforcement.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Equilibrium Condition for Rebar
Note Bond stress is zero at cracks
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Sources of Bond Transfer
(1) Adhesion between concrete
reinforcement. (2) Friction Note These
properties are quickly lost for tension.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
(3)Mechanical Interlock.
The edge stress concentration causes cracking to
occur.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
(3) Mechanical Interlock (cont).
Force interaction between the steel and concrete.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Splitting cracks result in loss of bond
transfer. Reinforcement can be used to restrain
these cracks.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Splitting Load is Affected by
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Typical Splitting Failure Surfaces.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
Typical Splitting Failure Surfaces.
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Reinforcement Development Lengths, Bar Cutoffs,
and Continuity Requirements
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Reinforcement Development Lengths
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Reinforcement Development Lengths
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Development Length for Bars in Tension
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Development Length for Bars in Tension
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Development Length for Bars in Tension
Development length, ld ACI 12.2.3
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Factors used in expressions for Development
Length (ACI 12.2.4)
a reinforcement location factor
where ab lt 1.7
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
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Factors used in expressions for Development
Length (ACI 12.2.4)
b coating factor (epoxy prevents adhesion
friction between bar and concrete.)
1.5 1.2 1.0
Epoxy-coated bars or wires with cover less than
3db or clear spacing less than 6db All other
epoxy-coated bars or wires Uncoated reinforcement
where ab lt 1.7
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Factors used in expressions for Development
Length (ACI 12.2.4)
g reinforcement size factor (Reflects more
favorable performance of smaller f
bars)
No.6 and smaller bars and deformed wire No. 7 and
larger bars
0.8 1.0
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Factors used in expressions for Development
Length (ACI 12.2.4)
l 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
not less than When normal weight concrete is used
1.3 1.0 1.0
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Factors used in expressions for Development
Length (ACI 12.2.4)
c spacing or cover dimension, in.
Use the smaller of either (a) the distance from
the center of the bar or wire to the nearest
concrete surface. or (b) one-half the
center-to-center spacing of the bar or wires
being developed.
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Factors used in expressions for Development
Length (ACI 12.2.4)
Ktr transverse reinforcement index (Represents
the contribution of confining reinforcement
across potential splitting planes.)
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Factors used in expressions for Development
Length (ACI 12.2.4)
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Factors used in expressions for Development
Length (ACI 12.2.4)
Note It is permitted to use Ktr 0 as a design
simplification even if transverse reinforcement
is present.
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Excess Flexural Reinforcement Reduction (ACI
12.2.5)
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.
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Development Length for Bars in Compression (ACI
12.3)
Compression development length, ldc ldbc
applicable reduction factors 8 in.
Basic Development Length for Compression, ldbc
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Development Length for Bars in Compression (ACI
12.3)
Reduction Factors (ACI 12.3.3)
- Excessive Reinforcement Factor A( s reqd )
/ A( s provided)
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Example - Development
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
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Example - Development
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. gt db
Clear spacing between bars
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Example - Development
Bars are confined with 3 stirrups. The
conditions are met. Determine the factorsa
1.0 (bottom bars), b 1.0 (no coating) and l
1.0 (normal weight concrete) and 54.8
psi lt 100 psi
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Example - Development
So ld 54.8(1.0 in.) 54.8 in. 55 in.
Using the more general formula Ktr 0.0
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Example - Development
c 1.17 in. controls
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Example - Development
So ld 55 in.
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Example - Development
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
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Example - Development
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Example - Development
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Homework
Problems 10.1 10.2