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1 Reinforced Concrete Design-I Lec-07 Bond and Development Length By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt. − The basic assumption of the RCC design is that the strain in concrete and reinforcing steel is the same. If the reinforcing steel slips at its ends, this is not valid. Hence it must be ensured that sufficient bond strength is developed at the interface of steel and concrete to avoid slippage of the steel. − Bond Strength and Development length − Two types of bond failure can be expected in reinforcing bars: − Direct pull out of the steel bars, when ample concrete confinement is provided in the form of large spacing of bars or large concrete cover − Splitting of concrete along the bar when cover confinement or bar spacing is insufficient b. Actual Distribution of Flexural Bond Stress − Pure bending case − Concrete fails to resist tensile stresses only where the actual crack is located. Steel T is maximum and T max = M / jd . − Between cracks , concrete does resist moderate amount of tension introduced by bond. − u is proportional to the rate of change of bar force, and highest where the slope of the steel force curve is greatest. − Very high local bond stress adjacent to the crack. 6 − Beam under transverse loads, − According to simple crack sectional theory, T is proportional to the moment diagram and u is proportional to shear force diagram. − In actual, T is less than the simple analysis prediction everywhere except at the actual cracks. − Similarly, u is equal with simple analysis prediction only at the location where slopes of the steel force diagrams are equals .If the slope is greater than assumed, bond stress is greater; if the slope is less bond stress is less. 7 ULTIMATE BOND STRENGTH AND DEVELOPMENT LENGTH − Types of bond failure − Direct pullout of bars (small diameter bars are used with sufficiently large concrete cover distances and bar spacing) − Splitting of the concrete along the bar (cover or bar spacing is insufficient to resist the lateral concrete tension resulting from the wedging effect of bar deformations) 8 Depar tment a. Ultimate Bond Strength − Direct pull out − For sufficiently confined bar, adhesive bond and friction are overcome as the tensile force on the bar is increased. Concrete eventually crushes locally ahead of the bar deformation and bar pullout results. − When pull out resistance is overcome or when splitting has spread all the way 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 vertical plane − Splitting in horizontal plane: frequently begins at a diagonal crack in connection with dowel action. Shear and bond failures are often interrelated. − Local bond failure − Large local variation of bond stress caused by flexural and diagonal cracks immediately adjacent to cracks leads to this failure below the failure load of the beam. − Results small slip and some widening of cracks and increase of deflections. − Harmless as long as the failure does not propagate all along the bar. − Providing end anchorage, hooks or extended length of straight bar (development length concept) 9 Consider a bar embedded in a mass of concrete P = s * [p*db2/4] P = t*[Lb*p*db] db Lb t = P / [Lb*p*db] < tmax s = P/ [p*db2/4] < smax P < tmax * [Lb*p*db] P < smax * [p*db2/4] To force the bar to be the weak link: tmax * [Lb*p*db] > smax * [p*db2/4] Lb > (smax / tmax)* [db/4] Development Length − Ld = development length − the shortest distance over which a bar can achieve it’s full capacity − The length that it takes a bar to develop its full contribution to the moment capacity, Mn Ld Mn 0 Cc Mn = (C or T)*(dist) Ts Steel Limit, smax − Using the bilinear assumption of ACI 318: smax = + fy Lb > (fy / tmax)* [db/4] Lb > fy * db / (4*tmax) Concrete Bond Limit, tmax − There are lots of things that affect tmax − − − − − − The strength of the concrete, f’c Type of concrete (normal weight or light weight) The amount of concrete below the bar The surface condition of the rebar The concrete cover on the bar The proximity of other bars transferring stress to the concrete − The presence of transverse steel Concrete Strength, f’c − Bond strength, tmax, tends to increase with concrete strength. − Experiments have shown this relationship to be proportional to the square root of f’c. Type of Concrete − Light weight concrete tends to have less bond strength than does normal weight concrete. − ACI 318-08 introduces a lightweight concrete reduction factor, l, on sqrt(f’c) in some equations. − See ACI 318-08, 8.6.1 for details Amount of Concrete Below Bars − The code refers to “top bars” as being any bar which has 12 inches or more of fresh concrete below the bar when the member is poured. − If concrete > 12” then consolidation settlement results in lower bond strength on the bottom side of the bar − See ACI 318-08, 12.2.4(a) Surface Condition of Rebar − All rebar must meet ASTM requirements for deformations that increase pullout strength. − Bars are often surface coated is inhibit corrosion. − Epoxy Coating The major concern! − Galvanizing − Epoxy coating significantly reduces bond strength − See ACI 318-08, 12.2.4(b) Proximity to Surface or Other Bars − The size of the concrete “cylinder” tributary to each bar is used to account for proximity of surfaces or other bars. Presence of Transverse Steel − The bond transfer tends to cause a splitting plane − Transverse steel will increase the strength of the splitting plane. b. Development Length − Development length is the length of embedment necessary to develop the full tensile strength of bar, controlled by either pullout or splitting. − In Fig., let − maximum M at a and zero at support − fs at a T = Ab fs _ − Development length concept total tension force must be transferred from the bar to the concrete in the distance ‘l ‘ by bond stress on the surface. − To fully develop the strength T = Ab fy ld , development length − Safety against bond failure: the length of the bar from any point of given steel stress to its nearby end must be at least equal to its development length. If the length is inadequate, special anchorage can be provided. ACI CODE PROVISION FOR DEVELOPMENT OF TENSION REINFORCEMENT Limit (c + ktr) / db = 2.5 for pullout case √f’c are not to be greater than 100 psi. 21 Depar tment For two cases of practical importance, using (c + ktr) / db = 1.5, 22 Example: 23 Continue: 24 Depar tment Continue: 25 ANCHORAGE OF TENSION BARS BY HOOKS In the event that the desired tensile stress in a bar can not be developed by bond alone, it is necessary to provide special anchorage at the end of the bar. 26 27 b. Development Length and Modification Factors for Hooked Bars 28 29 Department of Example 30 Depar tment ANCHORAGE REQUIREMENTS FOR WEB REINFORCEMENT 31 DEVELOPMENT OF BARS IN COMPRESSION − Reinforcement may be required to develop its compressive strength by embedment under various circumstances. − ACI basic development length in compression ldb = 0.02db fy/√f’c 32 Depar tment Determining Locations of Flexural Cutoffs Given a simply supported beam with a distributed load. Determining Locations of Flexural Cutoffs Note: Total bar length = Fully effective length + Development length 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 db (usually larger) past the theoretical cutoff for flexure (Handles uncertainties in loads, design approximations,etc..) Determining Locations of Flexural Cutoffs Development of flexural reinforcement in a typical continuous beam. ACI 318R-02 - 12.10 for flexural reinforcement Bar Cutoffs - General Procedure 1. Determine theoretical flexural cutoff points for envelope of bending moment diagram. 2. Extract the bars to satisfy detailing rules (from ACI Section 7.13, 12.1, 12.10, 12.11 and 12.12) 3. Design extra stirrups for points where bars are cutoff in zone of flexural tension (ACI 12.10.5) Bar Cutoffs - General Rules All Bars Rule 1. 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) 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) Bar Cutoffs - General Rules Positive Moment Bars Rule 3. 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) Bar Cutoffs - General Rules Positive Moment Bars Rule 3. 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). Bar Cutoffs - General Rules Positive Moment Bars Rule 3. Structural Integrity − Continuous perimeter beams. At least onefourth 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). Bar Cutoffs - General Rules Positive Moment Bars Rule 3. 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, fy, at the face of the support (ACI 12.11.2) Bar Cutoffs - General Rules Positive Moment Bars Rule 4. 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). Bar Cutoffs - General Rules Positive Moment Bars Rule 4. Mn ld la Vu Bar Cutoffs - General Rules Negative Moment Bars Rule 5. − Negative moment reinforcement must be anchored into or through supporting columns or members (ACI Sec. 12.12.1). Bar Cutoffs - General Rules Negative Moment Bars Rule 6. Structural Integrity − 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). Bar Cutoffs - General Rules Negative Moment Bars Rule 6. 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). Moment Resistance Diagrams 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. Moment Resistance Diagrams The nominal moment capacity of an under-reinforced concrete beam is a M n As f y d 2 where, a As f y 0.85 fcb To determine the position of the cutoff or bent point the moment diagram due to external loading is drawn. Moment Resistance Diagrams The ultimate moment resistance of one bar, Mnb is a M nb Abs f y d 2 where, Abs area of bar The intersection of the moment resistance lines with the external bending moment diagram indicates the theoretical points where each bar can be terminated. Moment Resistance Diagrams Given a beam with the 4 #8 bars and fc=3 ksi and fy=50 ksi and d = 20 in. Moment Resistance Diagrams The moment diagram is Moment Diagram 3000 2500 k-in 2000 1500 1000 500 0 0 2 4 6 8 10 ft 12 14 16 18 20 Moment Resistance Diagrams The moment resistance of one bar is a M nb Asb f y d 2 3.16 in 2 50 ksi As f y a 5.2 in. 0.85 f cb 0.85 3 ksi 12 in. M nb M ub 5.2 in. 0.79 in 50 ksi 20 in. 688 k-in. 2 M nb 0.9 688 k-in. 620 k-in. 2 Moment Resistance Diagrams The moment diagram and crossings Moment Diagram 3000 2480 k-in 2500 1860 k-in k-in 2000 1240 k-in 1500 1000 620 k-in 500 0 0 2 4 6 8 10 ft 12 14 16 18 20 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. Moment Diagram 3000 2480 k-in 2500 1860 k-in k-in 2000 a 1500 1240 k-in 1000 620 k-in c 500 b 0 0 2 4 6 8 10 ft 12 14 16 18 20 Moment Resistance Diagrams Compute the bar development length is la 12d b or d 12 1.0 in. or 20 in. 20 in. ld l f y d b 20 f c 50000 1.0 in. 20 3000 45.6 in. 46 in. 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. 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 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. Moment Resistance Diagrams Two bars must extend into the support and moment resistance diagram Mub must enclose the external bending moment diagram. Example – 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 – Cutoff Determine the moment capacity of the bars. a M nb As f y 0.85 f cb 3.16 in 40 ksi 4.93 in. 2 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. 2 Example – Cutoff Determine the location of the bar intersections of moments. 1 bar 35.64 k-ft. 3 bar 107 k-ft. 2 bar 71.3 k-ft. 4 bar 142.6 k-ft. M x M0 mx Example – Cutoff Determine the location of the bar intersections of moments. 1 bar 35.64 k-ft. 3 bar 107 k-ft. 2 bar 71.3 k-ft. 4 bar 142.6 k-ft. 132.5 k-ft. 87.5 k-ft. 107 k-ft. 132.5 k-ft. x 6 ft. x 3.4 ft. 40.8 in. or 41 in. Example – Cutoff Determine the location of the bar intersections of moments. 1 bar 35.64 k-ft. 3 bar 107 k-ft. 2 bar 71.3 k-ft. 4 bar 142.6 k-ft. 87.5 k-ft. 0.0 k-ft. 71.3 k-ft. 87.5 k-ft. x 5 ft. x 0.93 ft. 11.1 in. or 11 in. or 11 in. + 72 in. = 83 in. from center Example – Cutoff The minimum distance is la 12d b or d 12 1.0 in. or 17.5 in. 18 in. ld l f y d b 20 f c 40000 1.0 in. 20 3000 36.6 in. 37 in. Example – Cutoff The minimum amount of bars are As/3 or two bars 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” 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” Example – Cutoff The moment diagram is the blue line and the red line is the envelope which encloses the moment diagram. 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 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. Types of Splices Class A Splice (ACI 12.15.2) Asprovided When 2 over entire splice Asreq'd length. and 1/2 or less of total reinforcement is spliced win the req’d lay length. Types of Splices Class B Splice (ACI 12.15.2) All tension lay splices not meeting requirements of Class A Splices Tension Lap Splice (ACI 12.15) where As (req’d) ld = determined for bending = 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 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 12.16) Lap, req’d = 0.0005fy db Lap, req’d = (0.0009fy -24) db Lap, req’d 12 in for fy 60000 psi for fy > 60000 psi For fc 3000 psi, required lap splice shall be multiply by (4/3) (ACI 12.16.1) Compression Lap Splice (ACI 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 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 a. When 3 bars are spliced and As(provided) /As(required) >2 b. When 4 bars are spliced and As(provided) /As(required) < 2 c. When all bars are spliced at the same location. fc= 5 ksi and fy = 60 ksi Example – Splice Tension For #8 bars, db =1.0 in and = = g = l =1.0 3 fy gl ld d b 40 f c c K tr db 3 60000 1.0 42.4 43 in. 40 5000 1.5 in. 0 1.0 in. Example – Splice Tension 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 Compression Calculate the lap splice length for a # 10 compression bar in tied column when fc= 5 ksi and a) fy = 60 ksi b) fy = 80 ksi Example – Splice Compression For #10 bars, db =1.27 in. ld 0.02 f y 0.003 f y db fc 0.02 60000 16.97 or 18 5000 ld 18 1.27 in. 22.86 in. ld 23 in. Check ls > 0.005 db fy = 38.1 in. So ls = 39 in. Example – Splice Compression 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.