Aect480 lecture-11-development-of-reinforcement-splices

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Aect480 lecture-11-development-of-reinforcement-splices

  1. 1. Lecture 11 – Page 1 of 9Lecture 11 – Development of Reinforcement, Splices, HooksReinforcing bars must be embedded a minimum distance into the concrete inorder to achieve the full tensile capacity, T of the bar. This length is referred toas “Development Length”, Ld.The development length is based upon the BOND between the rebar and theconcrete. Factors affecting this bond include the following:• Type of ribbing on the bar• Presence of epoxy (or other ) coating• Concrete quality• Distance between bar and edge of concrete• Type of end anchorage into the concreteConcreteLdRebarT = Asfy
  2. 2. Lecture 11 – Page 2 of 9Determining Ld for Tension Bars:a) #6 and smaller bars:cybdffdL25αβλ=b) #7 and larger bars:cybdffdL20αβλ=where:db = diameter of barfy = yield strength of bar, PSIf’c = specified concrete compressive strength, PSIα = alpha= Bar location factor= 1.3 for top reinforcement= 1.0 for all other locationsβ = beta= Coating factor= 1.5 for epoxy coated bars= 1.0 for uncoated barsλ = lambda= Lightweight aggregate factor= 1.3 for lightweight aggregate= 1.0 for normal weight aggregate
  3. 3. Lecture 11 – Page 3 of 9Example 1GIVEN: A #6 rebar under tension force. Assume the following conditions:• Concrete f’c = 4000 PSI• Normal weight concrete (λ = 1.0)• ASTM A615 Grade 60 rebar• #6 rebar (γ = 0.8)• Uncoated bar (β = 1.0)• Bar location is bottom of beam (α = 1.0)REQUIRED: Determine the development length, Ld to achieve full tensilestrength of the bar.Step 1 – Use the formula above to determine Ld:cybdffdL25αβλ=PSIPSILd400025)0.1)(0.1)(0.1)(000,60("86⎟⎠⎞⎜⎝⎛=Ld = 28.5”ConcreteLd = 28.5”#6 Rebar
  4. 4. Lecture 11 – Page 4 of 9Assuming “normal” conditions, the following table may be used to determinedevelopment lengths of bars in tension:Development Length Ld of Grade 60 bottom bars in normal weight concreteCondition Concrete f’c No. 6 and smallerbarsNo. 7 and largerbars3000 PSI 44db 55db4000 PSI 38db 47dbClear spacing ofbars > db, clearcover > db 5000 PSI 34db 42db3000 PSI 66db 82db4000 PSI 57db 71dbAll other cases5000 PSI 51db 64dbExample 2GIVEN: The same information as Example 1.REQUIRED: Using the table above, determine the Ld for a #6 bar.Step 1 – Use table above to determine Ld:Concrete f’c = 4000 PSIClear spacing of bars > db, clear cover > dbLd = 38db= 38 ⎟⎠⎞⎜⎝⎛"86Ld = 28.5”
  5. 5. Lecture 11 – Page 5 of 9Determining Ld for Compression Bars:The development length of bars in compression is not as large as thedevelopment length in tension because of the absence of tension crackingin the concrete.Ldc = Development length in compression= larger ofExample 3GIVEN: A #6 bar in compression. Use f’c = 4000 PSI and Grade 60 bars.REQUIRED: Determine the Ldc for the bar.Ldc = Development length in compressionLdc = 14.2”Ldc = Larger ofLdc =cybffd02.0orLdc = 0.0003dbfyLdc =cybffd02.0 =PSIPSI4000)000,60("8602.0⎟⎠⎞⎜⎝⎛= 14.2” ← UseorLdc = 0.0003dbfy = 0.0003 ⎟⎠⎞⎜⎝⎛"86(60,000 PSI) = 13.5”
  6. 6. Lecture 11 – Page 6 of 9Lap Splices of BarsBars are generally fabricated to lengths of about 60’-0”, but transportation,workability and other concerns often require bars to be less than about40’-0” long. For long walls, beams, slabs and other situations requiringlong lengths of bars, lap splicing is commonly used. It is good practice toplace laps at regions of small tension, i.e., low moment.Ls = Length of lap splice= 1.0Ld for “Class A” splice if the area of reinforcementprovided through the splice > twice that required by analysisand not more than 50% of the total reinforcement is splicedwithin the lap length= 1.3Ld for “Class B” splice if reinforcement does notmeet Class A requirementsConcreteLs
  7. 7. Lecture 11 – Page 7 of 9Hooked and Bent BarsHooks are used in concrete members where there is not sufficient straightlength to achieve the full development length Ld.The following is a diagram showing the required lengths of bends andhooks:Ldh = Lhbλ
  8. 8. Lecture 11 – Page 8 of 9Where: Lhb = Basic development length of hook in tension=cbfd1200λ = 1.0 unless otherwise specified below:=000,60yfif using other than Grade 60 bars= 0.7 if side concrete cover > 2½” or end cover > 2”= 0.8 if ties or stirrups spacing < 3db= 1.3 if lightweight concrete=ssAovidedAquired_Pr_ReExample 4GIVEN: A #5 Grade 40 bar is in tension as shown below. Use LIGHTWEIGHTconcrete with f’c = 4000 PSI.REQUIRED: Determine the min. required hook dimensions “X”, “Y” and “Z”.Step 1 – Determine dimension “X”:X = 12db= 12 ⎟⎠⎞⎜⎝⎛"85X = 7½”End cover = 1½”XYZ = LdhCritical sectionSide cover = 1½”
  9. 9. Lecture 11 – Page 9 of 9Step 2 – Determine dimension “Y”:Y = 4db since it is a #6 bar= 4 ⎟⎠⎞⎜⎝⎛"85Y = 2½”Step 3 – Determine length of hooked bar, Lhb:Lhb =cbfd1200=PSI4000"851200⎟⎠⎞⎜⎝⎛= 11.9”Step 4 – Determine total development length, Z = Ldh:Ldh = LhbλWhere: λ = 1.0 since side cover = 1½” < 2½”= 1.3 since lightweight concrete=PSIfy60000=PSIPSI6000040000= 0.67Ldh = Lhbλ= 11.9”(1.0)(1.3)(0.67)Ldh = 10.4”

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