DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS

Transcription

1 CHAPTER Reinforce Concrete Design Fifth Eition DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS A. J. Clark School of Engineering Department of Civil an Environmental Engineering Part I Concrete Design an Analysis 5a FALL 2002 By Dr. Ibrahim. Assakkaf ENCE to Structural Design Department of Civil an Environmental Engineering University of Marylan, College Park CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 1 If the reinforce concrete beam shown in Fig. 1 were constructe using plain roun reinforcing bars, an in aition, if those bars were to be grease or otherwise lubricate before the concrete were poure, the beam woul be as strong as it was mae of plain concrete, without reinforcement. 1

2 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 2 Figure. 1. Bon Stresses ue to Flexure Concrete En slip (a) P Reinforcing bar (c) Bon forces acting on concrete () Bon forces acting on steel (b) Unrestraine slip between concrete an steel CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 3 If a loa is applie as shown Fig. 1b, the bars woul ten to maintain its original length as the beam eflects. The bars woul slip longituinally with respect to ajacent concrete, which woul experience tensile strain ue to flexure. The assumption that the strain in an embee reinforcing bar is the same as that in surrouning concrete, woul 2

3 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 4 Not be vali. In orer for reinforce concrete to behave as intene, it is essential that bon forces be evelope on the interface between concrete an steel, such as to prevent significant slip from occurring at the interface. It is through the action of these interface bon forces that the slip of Fig. 5b is prevente. CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 5 The assumptions for the esign of reinforce concrete inclue: 1. Perfect boning between the concrete an steel exist, an 2. No slippage occur. Base on these assumptions, it follows that some form of bon stress exists at the contact surface between the concrete an steel bars. 3

4 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 6 In beams, this bon stress is cause by the change in bening moment along the length of the beam an the accompanying change in the tensile stress in the bars (flexural bon). The actual istribution of bon stresses along the reinforcing steel is highly complex, ue mainly to the presence of concrete cracks. CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 7 Large local variations in bon stress are cause by flexural an iagonal cracks. High bon stresses have been measure ajacent to these cracks. The high bon stress may result in: Small local slips ajacent to the crack Increase eflection In general, this is harmless as long as failure oes not propagate all along the bar with complete loss of bon. 4

5 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 8 Development Length En anchorage may be consiere reliable if the bar is embee into concrete a prescribe istance known as the evelopment length of the bar. In a beam, if the the actual extene length of the bar is equal or greater than this require evelopment length, then no bon failure will occur. CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 9 Development Length P P Max moment is at a l a Figure 2. Development length l shoul be at least equal to l b 5

6 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 10 Nee for Development Length w Figure 3. Continuous Beam A B C Moment Diagram l l CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 11 Anchorages Versus Development Length If the actual available length is inaequate for full evelopment, special anchorages,such as hooks, must be provie to ensure aequate strength. 6

7 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 12 ACI Coe The provisions of the ACI Coe are irecte towar proviing aequate length of embement, past the location at which the bar is fully stresse, which will ensure evelopment of the full strength of the bar. Therefore, the current metho base on ACI isregar high localize bon stress even though it may result in localize slip between steel an concrete ajacent to the cracks. CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 13 Methos for Determining the Development Length, l The ACI allows the etermination of the evelopment length by two methos: 1. Tabular criteria (ACI Section ) 2. General equation (ACI Section ) In either case, l shall not be less than 12 in. The general equation of the ACI Coe offers a simple approach that allows the user to see the effect of all variables controlling the evelopment length. 7

8 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 14 Methos for Determining the Development Length, l (cont ) This equation (ACI Eq. 12-1) is provie in Section of the ACI Coe, an it is as follows: 3 f y l αβγλ = (1) b 40 f c c k + tr b CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 15 Notations of Eq. 1: (c + k tr )/ b : shall not be taken greater than 2.5 l = evelopment length (in.) f y = yiel strength of nonprestresse reinforcement (psi) f c= compressive strength of concrete (psi); the value of f c shall not excee 100 psi (ACI Coe, Section ) b = nominal iameter of bar or wire (in.) 8

9 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 16 Comments for Eq. 1: 1. α is a reinforcement location factor that accounts for the position of the reinforcement in freshly place concrete. α = 1.3 (ACI Coe, Section ) where horizontal reinforcement is so place that more than 12 in. of fresh concrete is cast in member below the evelopment length or splice. α = 1.0 for other reinforcement. 2. β is a coating factor reflecting the effects of epoxy coating. For epoxy-coate reinforcement having cover less than 3 b or clear spacing between bars less than 6 b, use β = 1.5 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 17 Comments for Eq. 1 (cont ): For all other conitions, use β = 1.2 For uncoate reinforcement, use β = 1.0 The prouct of α an β nee not be taken greater than 1.7 (ACI Coe, Section ) 3. γ is a reinforcement size factor. Where No. 6 an smaller bars are use, γ = 0.8 Where No. 7 an larger bars use, γ = λ is a lightweight-aggregate concrete factor. For lightweight-aggregate concrete when the average splitting tensile strength f ct is not specifie, use λ = 1.3 9

10 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 18 Comments for Eq. 1 (cont ): When f ct is specifie, use f c λ = fct When normal-weight concrete is use, λ = 1.0 (ACI Coe, Section ) 5. c represents a spacing or cover imension (in.) The value of c will be the smaller of either the istance from the center of the bar to the nearest concrete cover (surface) or one-half the center-to-center spacing of the bars being evelope (spacing). CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 19 Comments for Eq. 1 (cont ): The bar spacing will be the actual center-to-center spacing between the bars if ajacent bars are all being evelope at the same location. If, however, an ajacent bar has been evelope at another location, the spacing to be use will be greater than the actual spacing to the ajacent bar. Note in Fig. 4 that the spacing for bars Y may be taken the same as for bars X, since bars Y are evelope in length AB, whereas bars X are evelope at a location other than AB. 10

11 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 20 Comments for Eq. 1 (cont ): A Bars Y B Bars X Figure. 4 s s Single Layer Bars Y Plan View A Bars X (continuous) Bars Y l A Elevation View Section A-A CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 21 Comments for Eq. 1 (cont ): 6. The transverse reinforcement inex K tr is to be calculate from Atr f yt Ktr = 1500sn where A tr = total cross-sectional area of all transverse reinforcement that is within the spacing s an that crosses the potential plane of splitting through the reinforcement being evelope (in 2 ) f yt = yiel strength of transverse reinforcement (psi) 11

12 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 22 Comments for Eq. 1 (cont ): s = maximum center-to-center spacing of transverse reinforcement within the evelopment length l (in.) n = number of bars or wires being evelope along the plane of splitting. CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 23 Reuction in Development Length A reuction in the evelopment length l is permitte where reinforcement is in excess of that require by analysis (except where anchorage or evelopment for f y is specifically require or where the esign inclues provisions for seismic consierations). The reuction factor K ER is given by As require A provie K ER = (2) s 12

13 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No Determine multiplying factors (use 1.0 unless otherwise etermine). a. Use α = 1.3 for top reinforcement, when applicable. b. Coating factor β applies to epoxy-coate bars. Determine cover an clear spacing as multiples of b. Use β = 1.5 if cover < 3 b or clear space < 6 b. Use β = 1.2 otherwise. c. Use γ = 0.8 for No. 6 bars an smaller.. Use λ = 1.3 for lightweight concrete with f ct not specifie. Use fc λ = if fct specifie. f ct CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No Check αβ Determine c, the smaller of cover or half-spacing (both reference to the center of the bar). 4. Calculate Atr f y Ktr =, or use Ktr = 0 (conservative) 1500sn 5. Check c + K b tr

14 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No Calculate K ER if applicable: As K ER = A s require provie 7. Calculate l from Eq. 1 (ACI Coe Eq. 12-1): 3 f y l αβγλ = b 40 f c c k + tr b CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 27 Example 1 Calculate the require evelopment length l into the beam for the negative moment steel shown so as to evelop the tensile strength of the steel at the face of the column. Require A s = 2.75 in 2, f c =4,000 psi, an f y = 60,000 psi. Assume normalweight concrete. 14

15 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 28 Example 1 (cont ) 21 column l 3-#9 bars 21 3-# clear #4 stirrups 14 #4 stirrups full length of beam 1 #4 o.c. spacing 2 for evelopment length l CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 29 Example 1 (cont ) (4) 3 #9 bars : = (2) αβ = ( 1.3)( 1) b (1) α = 1.3, β = 1.0, γ = 1.3 < (3) cover : c = = Dia. #4 stirrup 14 Half -spacing : c = K tr Atr f yt 0.4 = = 1500sn 1500 = 1.0, an λ = 1.0 OK 2( 1.5) 2( 0.5) 2( 2) ( 60,000) = ( 4.5)( 3) From Table 1 Area of 2 #4 stirrups = 2.22 Controls 15

16 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 30 Example 1 (cont ) Table 1. ASTM Stanar - English Reinforcing Bars Bar Designation Diameter in Area in 2 Weight lb/ft #3 [#10] #4 [#13] #5 [#16] #6 [#19] #7 [#22] #8 [#25] #9 [#29] #10 [#32] #11 [#36] #14 [#43] #18 [#57] Note: Metric esignations are in brackets CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 31 Example 1 (cont ) Table 2. Areas of Multiple of Reinforcing Bars (in 2 ) Number of bars #3 #4 $5 #6 Bar number #7 #8 #9 #10 # Table A-2 Textbook 16

17 CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 32 Example 1 (cont ) (5) (6) c + K K ER b tr = As = A s require provie = 3.02 > 2.75 = (7) Calculate the evelopment length l l = 3 40 = f y αβγλ f c c + ktr b 2.5, Therefore, use 2.5 using Eq.1: b CHAPTER 5a. DEVELOPMENT, SPLICES, AND SIMPLE SPAN BAR CUTOFFS Slie No. 33 Example 1 (cont ) Reuction factor 3 l = K ER 40 l = f y αβγλ f c c + ktr b 60, , b ()()() 1 1 ( 1.128) = in. > 12 in OK 17