Lecture 15- Bar Development July 11, 2003 CVEN 444.

Lecture 15- Bar Lecture 15- Bar DevelopmentDevelopment

July 11, 2003CVEN 444

Lecture GoalsLecture Goals

Bar Cut-off PointsSplice Tension Splice Compression Splice

Determining Locations of Determining Locations of Flexural CutoffsFlexural Cutoffs

Given a simply supported beam with a distributed load.


Note:

Total bar length =

Fully effective length

+ Development length


ACI 12.10.3

All longitudinal tension bars must extend a min. distance = d (effective depth of the member) or 12 db (usually larger) past the theoretical cutoff for flexure (Handles uncertainties in loads, design approximations,etc..)


Development of flexural reinforcement in a typical continuous beam.

ACI 318R-02 - 12.10 for flexural reinforcement

Bar Cutoffs - General Bar Cutoffs - General ProcedureProcedure

Determine theoretical flexural cutoff points for envelope of bending moment diagram.

Extract the bars to satisfy detailing rules (from ACI Section 7.13, 12.1, 12.10, 12.11 and 12.12)

Design extra stirrups for points where bars are cutoff in zone of flexural tension (ACI 12.10.5)

1.

2.

3.

Bar Cutoffs - General RulesBar Cutoffs - General Rules

Bars must extend the longer of d or 12db past the flexural cutoff points except at supports or the ends of cantilevers (ACI 12.11.1)

All Bars

Rule 1.

Rule 2. Bars must extend at least ld from the point of maximum bar stress or from the flexural cutoff points of adjacent bars (ACI 12.10.2 12.10.4 and 12.12.2)


Structural Integrity

Simple Supports At least one-third of the positive moment reinforcement must be extend 6 in. into the supports (ACI 12.11.1).

Continuous interior beams with closed stirrups. At least one-fourth of the positive moment reinforcement must extend 6 in. into the support (ACI 12.11.1 and 7.13.2.3)

Positive Moment Bars

Rule 3.



Continuous interior beams without closed stirrups. At least one-fourth of the positive moment reinforcement must be continuous or shall be spliced near the support with a class A tension splice and at non-continuous supports be terminated with a standard hook. (ACI 7.13.2.3).


Rule 3.


Structural IntegrityContinuous perimeter beams. At least one-fourth of the positive moment reinforcement required at midspan shall be made continuous around the perimeter of the building and must be enclosed within closed stirrups or stirrups with 135 degree hooks around top bars. The required continuity of reinforcement may be provided by splicing the bottom reinforcement at or near the support with class A tension splices (ACI 7.13.2.2).


Rule 3.



Beams forming part of a frame that is the primary lateral load resisting system for the building. This reinforcement must be anchored to develop the specified yield strength, fy, at the face of the support (ACI 12.11.2)


Rule 3.


Stirrups

At the positive moment point of inflection and at simple supports, the positive moment reinforcement must be satisfy the following equation for ACI 12.11.3. An increase of 30 % in value of Mn / Vu shall be permitted when the ends of reinforcement are confined by compressive reaction (generally true for simply supports).


Rule 4.



Rule 4.

au

nd V

lM

l


Negative moment reinforcement must be anchored into or through supporting columns or members (ACI Sec. 12.12.1).

Negative Moment Bars

Rule 5.



Interior beams. At least one-third of the negative moment reinforcement must be extended by the greatest of d, 12 db or ( ln / 16 ) past the negative moment point of inflection (ACI Sec. 12.12.3).


Rule 6.



Perimeter beams. In addition to satisfying rule 6a, one-sixth of the negative reinforcement required at the support must be made continuous at mid-span. This can be achieved by means of a class A tension splice at mid-span (ACI 7.13.2.2).


Rule 6.

Moment Resistance Moment Resistance DiagramsDiagrams

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 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 external loading is drawn.

s yn s y

c

where, 2 0.85

A faM A f d a

f b


The ultimate moment resistance of one bar, Mnb is

The intersection of the moment resistance lines with the external bending moment diagram indicates the theoretical points where each bar can be terminated.

nb bs y bs where, area of bar2

aM A f d A


Given a beam with the 4 #8 bars and fc=3 ksi and fy=50 ksi and d = 20 in.


The moment diagram is

Moment Diagram

0

500

1000

1500

2000

2500

3000

0 2 4 6 8 10 12 14 16 18 20

ft

k-in


The moment resistance of one bar is

nb sb y

2s y

c

2nb

ub nb

2

3.16 in 50 ksi5.2 in.

0.85 0.85 3 ksi 12 in.

5.2 in.0.79 in 50 ksi 20 in. 688 k-in.

2

0.9 688 k-in. 620 k-in.

aM A f d

A fa

f b

M

M M


The moment diagram and crossings

Moment Diagram

0

500

1000

1500

2000

2500

3000

0 2 4 6 8 10 12 14 16 18 20

ft

k-in

620 k-in

1240 k-in

1860 k-in

2480 k-in


The ultimate moment resistance is 2480 k-in. 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 termination of the bars.

Moment Diagram

0

500

1000

1500

2000

2500

3000

0 2 4 6 8 10 12 14 16 18 20

ft

k-in

620 k-in

1240 k-in

1860 k-in

2480 k-in

ab

c


Compute the bar development length is

a b

y bd

c

12 or d

12 1.0 in. or 20 in. 20 in.

50000 1.0 in.

20 20 3000

45.6 in. 46 in.

l d

f dl

f

Moment Resistance Moment Resistance DiagramsDiagramsThe ultimate moment resistance is 2480 k-in. 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 termination of the bars.


It is necessary to develop part of the strength of the bar by bond. The ACI 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 12.11.1 specify that 1/3 of positive moment reinforcement must be continuous.


Two bars must extend into the support and moment resistance diagram Mub must enclose the external bending moment diagram.

Example – CutoffExample – Cutoff For the simply supported beam with b=10 in. d =17.5 in., fy=40 ksi and fc=3 ksi with 4 #8 bars. Show where the reinforcing bars can be terminated.

Example – CutoffExample – Cutoff

Determine the moment capacity of the bars.

2s y

c

2nb

3.16 in 40 ksi4.93 in.

0.85 0.85 3 ksi 10 in.

4.93 in.0.79 in 40 ksi 17.5 in.

2

427.7 k-in. 35.64 k-ft.

A fa

f b

M

ExampleExample – Cutoff– CutoffDetermine the location of the bar intersections of moments.

1 bar 35.64 k-ft.

2 bar 71.3 k-ft.

0 M x M mx

3 bar 107 k-ft.

4 bar 142.6 k-ft.


1 bar 35.64 k-ft.

2 bar 71.3 k-ft.

132.5 k-ft. 87.5 k-ft.107 k-ft. 132.5 k-ft.

6 ft.

3.4 ft. 40.8 in. or 41 in.

x

x

3 bar 107 k-ft.

4 bar 142.6 k-ft.


1 bar 35.64 k-ft.

2 bar 71.3 k-ft.

3 bar 107 k-ft.

4 bar 142.6 k-ft.

87.5 k-ft. 0.0 k-ft.71.3 k-ft. 87.5 k-ft.

5 ft.

0.93 ft. 11.1 in. or 11 in.

or 11 in. + 72 in. = 83 in. from center

x

x

Example – CutoffExample – Cutoff

The minimum distance is

a b

y bd

c

12 or d

12 1.0 in. or 17.5 in. 18 in.

40000 1.0 in.

20 20 3000

36.6 in. 37 in.

l d

f dl

f

Example – CutoffExample – CutoffThe minimum amount of bars are As/3 or two bars

ExampleExample – Cutoff– CutoffThe cutoff for the first bar is 41 in. or 3 ft 5 in. and 18 in or 1 ft 6 in. total distance is 41 in.+18 in. = 59 in. or 4 ft 11 in.

Note error it is 4’-11” not 5’-11”

ExampleExample – Cutoff– CutoffThe cutoff for the second bar is 83 in. + 18 in. 101 in. or 8 ft 5 in. (37-in+5-in+18-in+41-in= 101-in.)

Note error it is 4’-11” not 5’-11”

ExampleExample – Cutoff– CutoffThe moment diagram is the blue line and the red line is the envelope which encloses the moment diagram.

Bar SplicesBar Splices

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

Types of Splices

1. Butted &Welded

2. Mechanical Connectors

3. Lay Splices

Must develop 125% of yield strength ACI 12.14.3.2 and ACI 12.14.3.4

Tension Lap SplicesTension Lap Splices

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

Types of Splices

1. Contact Splice

2. Non-Contact Splice (distance between the bars 6” and 1/5 of the splice length ACI 12.14.2.3)

Splice length (development length) is the distance the two bars are overlapped.

Types of SplicesTypes of Splices

Class A Splice (ACI 12.15.2)

When over entire splice length.

and 1/2 or less of total reinforcement is spliced win the req’d lay length.

2

dreq's

provideds A

A

Types of SplicesTypes of Splices

Class B Splice (ACI 12.15.2)

All tension lay splices not meeting requirements of Class A Splices

Tension Lap Splice (ACI Tension Lap Splice (ACI 12.15)12.15)

where As (req’d) = determined for bending

ld = development length for bars (not allowed to use excess reinforcement modification factor)

ld must be greater than or equal to 12 in.

Tension Lap Splice (ACI Tension Lap Splice (ACI 12.15)12.15)

Lap Splices shall not be used for bars larger than No. 11. (ACI 12.14.2)

Lap Splices should be placed in away from regions of high tensile stresses -locate near points of inflection (ACI 12.15.1)

Compression Lap Splice (ACI Compression Lap Splice (ACI 12.16)12.16)

Lap, req’d = 0.0005fy db for fy 60000 psi Lap, req’d = (0.0009fy -24) db for fy > 60000 psi Lap, req’d 12 in

For fc 3000 psi, required lap splice shall be multiply by (4/3) (ACI 12.16.1)

Compression Lap Splice (ACI Compression Lap Splice (ACI 12.17.2)12.17.2)

In tied column splices with effective tie area throughout splice length 0.0015 hs factor = 0.83

In spiral column splices, factor = 0.75

The final splice length must be 12 in.

Example – Splice TensionExample – Splice Tension

Calculate the lap-splice length for 6 #8 tension bottom bars in two rows with clear spacing 2.5 in. and a clear cover, 1.5 in., 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= 5 ksi and fy = 60 ksi

a.

b.

c.


For #8 bars, db =1.0 in and = = = =1.0

yd

b c tr

b

3

40

3 60000 1.042.4 43 in.

1.5 in. 040 50001.0 in.

fl

d f c K

d


The As(provided) /As(required) > 2, class A splice applies; therefore lst = 1.0 ld >12 in., so lst = 43 in. > 12 in. The bars spliced are less than half the number

The As(provided) /As(required) < 2, class B splice applies; therefore lst = 1.3 ld >12 in., so lst = 1.3(42.4 in.) = 55.2 in. use 56 in. > 12 in..

Class B splice applies and lst = 56 in. > 12 in.

Example – Splice Example – Splice CompressionCompression

a) fy = 60 ksi

b) fy = 80 ksi

Calculate the lap splice length for a # 10 compression bar in tied column when fc= 5 ksi and


For #10 bars, db =1.27 in.

ydy

b c

d d

0.020.003

0.02 6000016.97 or 18

5000

18 1.27 in. 22.86 in. 23 in.

flf

d f

l l

Check ls > 0.005 db fy = 38.1 in. So ls = 39 in.


For #10 bars, db =1.27 in. The ld = 23 in.

Check ls > (0.0009 fy –24) db

=(0.0009(80000)-24)(1.27in.) = 61 in.

So use ls = 61 in.

Lecture 15- Bar Development July 11, 2003 CVEN 444.

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Transcript of Lecture 15- Bar Development July 11, 2003 CVEN 444.