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Lecture 15- Bar Development July 11, 2003 CVEN 444

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Lecture Goals Bar Cut-off Points Splice Tension Splice Compression Splice

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Determining Locations of Flexural Cutoffs Given a simply supported beam with a distributed load.

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Determining Locations of Flexural Cutoffs Note : Total bar length = Fully effective length + Development length

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Determining Locations of Flexural Cutoffs ACI 12.10.3 All longitudinal tension bars must extend a min. distance = d (effective depth of the member) or 12 d b (usually larger) past the theoretical cutoff for flexure (Handles uncertainties in loads, design approximations,etc..)

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Determining Locations of Flexural Cutoffs Development of flexural reinforcement in a typical continuous beam. ACI 318R-02 - 12.10 for flexural reinforcement

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Bar Cutoffs - General Procedure 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.

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Bar Cutoffs - General Rules Bars must extend the longer of d or 12d b 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 l d 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)

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Bar Cutoffs - General Rules 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.

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Bar Cutoffs - General Rules Structural Integrity 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). Positive Moment Bars Rule 3.

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Bar Cutoffs - General Rules Structural Integrity Continuous 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). Positive Moment Bars Rule 3.

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Bar Cutoffs - General Rules Structural Integrity 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, f y, at the face of the support (ACI 12.11.2) Positive Moment Bars Rule 3.

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Bar Cutoffs - General Rules 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 M n / V u shall be permitted when the ends of reinforcement are confined by compressive reaction (generally true for simply supports). Positive Moment Bars Rule 4.

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Bar Cutoffs - General Rules Positive Moment Bars Rule 4.

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Bar Cutoffs - General Rules Negative moment reinforcement must be anchored into or through supporting columns or members (ACI Sec. 12.12.1). Negative Moment Bars Rule 5.

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Bar Cutoffs - General Rules Structural Integrity Interior beams. At least one-third of the negative moment reinforcement must be extended by the greatest of d, 12 d b or ( l n / 16 ) past the negative moment point of inflection (ACI Sec. 12.12.3). Negative Moment Bars Rule 6.

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Bar Cutoffs - General Rules Structural Integrity 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). Negative Moment Bars Rule 6.

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Moment Resistance Diagrams Moment capacity of a beam is a function of its depth, d, width, b, and area of steel, A s. 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 45 o, to provide tensile reinforcement for the negative moments over the support.

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Moment Resistance Diagrams 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.

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Moment Resistance Diagrams The ultimate moment resistance of one bar, M nb is The intersection of the moment resistance lines with the external bending moment diagram indicates the theoretical points where each bar can be terminated.

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Moment Resistance Diagrams Given a beam with the 4 #8 bars and f c =3 ksi and f y =50 ksi and d = 20 in.

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Moment Resistance Diagrams The moment diagram is

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Moment Resistance Diagrams The moment resistance of one bar is

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Moment Resistance Diagrams The moment diagram and crossings

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Moment Resistance Diagrams 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. a b c

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Moment Resistance Diagrams Compute the bar development length is

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Moment Resistance Diagrams 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.

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Moment Resistance Diagrams 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 12d b, 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.

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Moment Resistance Diagrams Two bars must extend into the support and moment resistance diagram M ub must enclose the external bending moment diagram.

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

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Example – Cutoff Determine the moment capacity of the bars.

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Example– Cutoff Example – Cutoff Determine the location of the bar intersections of moments.

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Example– Cutoff Example – Cutoff Determine the location of the bar intersections of moments.

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Example– Cutoff Example – Cutoff Determine the location of the bar intersections of moments.

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Example – Cutoff The minimum distance is

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Example – Cutoff The minimum amount of bars are A s /3 or two bars

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Example– Cutoff Example – Cutoff The 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”

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Example– Cutoff Example – Cutoff The 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”

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Example– Cutoff Example – Cutoff The moment diagram is the blue line and the red line is the envelope which encloses the 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.Lay Splices Must develop 125% of yield strength ACI 12.14.3.2 and ACI 12.14.3.4

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Tension 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.

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Types 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.

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Types of Splices Class B Splice (ACI 12.15.2) All tension lay splices not meeting requirements of Class A Splices

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Tension Lap Splice (ACI 12.15) whereA s (req’d) = determined for bending l d = development length for bars (not allowed to use excess reinforcement modification factor) l d must be greater than or equal to 12 in.

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Tension Lap Splice (ACI 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)

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Compression Lap Splice (ACI 12.16) Lap, req’d = 0.0005f y d b for f y 60000 psi Lap, req’d = (0.0009f y -24) d b for f y > 60000 psi Lap, req’d 12 in For f c 3000 psi, required lap splice shall be multiply by (4/3) (ACI 12.16.1)

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Compression Lap Splice (ACI 12.17.2) In tied column splices with effective tie area throughout splice length 0.0015 h s factor = 0.83 In spiral column splices, factor = 0.75 The final splice length must be 12 in.

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Example – 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 A s(provided) /A s(required) >2 When 4 bars are spliced and A s(provided) /A s(required) < 2 When all bars are spliced at the same location. f c = 5 ksi and f y = 60 ksi a. b. c.

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Example – Splice Tension For #8 bars, d b =1.0 in and = = = =1.0

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Example – Splice Tension The A s(provided) /A s(required) > 2, class A splice applies; therefore l st = 1.0 l d >12 in., so l st = 43 in. > 12 in. The bars spliced are less than half the number The A s(provided) /A s(required) 12 in., so l st = 1.3(42.4 in.) = 55.2 in. use 56 in. > 12 in.. Class B splice applies and l st = 56 in. > 12 in.

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Example – Splice Compression a)f y = 60 ksi b) f y = 80 ksi Calculate the lap splice length for a # 10 compression bar in tied column when f c = 5 ksi and

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Example – Splice Compression For #10 bars, d b =1.27 in. Check l s > 0.005 d b f y = 38.1 in. So l s = 39 in.

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Example – Splice Compression For #10 bars, d b =1.27 in. The l d = 23 in. Check l s > (0.0009 f y –24) d b =(0.0009(80000)-24)(1.27in.) = 61 in. So use l s = 61 in.

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