Aect480 lecture-5
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Aect480 lecture-5 Aect480 lecture-5 Document Transcript

  • Lecture 5 - Page 1 of 9Lecture 5 – T- BeamsConcrete beams are often poured integrally with the slab, forming a much stronger“T” – shaped beam. These beams are very efficient because the slab portioncarries the compressive loads and the reinforcing bars placed at the bottom of thestem carry the tension. A T-beam typically has a narrower stem than an ordinaryrectangular beam. These stems are typically spaced from 4’-0” apart to more than12’-0”. The slab portion above the stem is designed as a one-way slab spanningbetween stems (see Lecture 6).A typical T-beam has the following dimensions and notations:bwbwdOverhangwidthClear distancehf = Slabthicknessb = Effective flange widthNOTE: Stirrups in T-beamare required (not shown inthis sketch)
  • Lecture 5 - Page 2 of 9Assuming T-beams are symmetrical, the following design dimensions are used:Overhang width = smallerb = smallerT-Beam AnalysisT-beams are analyzed similarly to rectangular beams, except thecompression area is a narrow “strip” usually located in the slab.hfa = Effective conc.compressivethickness8hfor½(Clear distance)¼(Beam span)or(2 x overhang width) + bwbwZ = (d -2a)b = Effective flange widthdAc = Shaded area= Effective concretecompression area= (a)(b)As = Total area ofmain tension bars
  • Lecture 5 - Page 3 of 9Mu = Usable moment capacity of T-beam= φTZwhere: φ = 0.9T = Tension force developed in main bars= AsfyAc = Effective concrete compression area=cfT85.0a = Effective concrete compressive thickness=bAcZ = Moment arm distance between center ofcompression to center of tension= d -2a
  • Lecture 5 - Page 4 of 9Example 1GIVEN: A commercial building has T-beams spaced 6’-6” (center-to-center) with a 4”thick concrete slab as shown in the framing plan and cross-section views below. Usethe following information:• Superimposed service floor dead load (NOT including conc. wt.) = 40 PSF• Superimposed service floor live load = 100 PSF• Concrete f’c = 3000 PSI• ASTM A615 Grade 60 barsREQUIRED:1) Determine the maximum factored moment, Mmax, on the T-beam.2) Determine the usable moment capacity, Mu, for the T-beam.AATyp.6’-6”T-beamspan=20’-0”T-beamPerimeter girder ColumnFraming Plan
  • Lecture 5 - Page 5 of 9Step 1 – Determine maximum factored moment, Mmax, on T-beam:Determine area of T-beam = Slab area + Stem area= (6.5’)(0.333’) + (1.1667’)(0.666’)= 2.94 ft2Determine service weight of T-beam = Area of T-beam x Conc. unit wt.= 2.94 ft2(150 lb/ft3)= 441 PLFDet. factored uniform load on T-beam wu = 1.2D + 1.6L= 1.2[(6.5’)(40 PSF) + 441 PLF] + 1.6[(6.5’)(100 PSF)]= 841 PLF + 1040 PLF= 1881 PLF → Use wu = 1.9 KLFDet. Maximum factored moment, Mmax =82Lwu=8)"020)(9.1( 2−KLFMmax = 95 KIP-FT18”Service Dead Load Service Live Load8”16”6’-6”hf = 4”2 - #9 barsSection “A-A” Thru T-Beams(NOTE: stirrups and top bars not shown)
  • Lecture 5 - Page 6 of 9Step 2 – Determine effective concrete slab width “b”:Overhang width = smallerb = smallerStep 3 – Determine effective conc. compression area Ac:T = Tension force developed in main bars= Asfy= 2 bars(1.00 in2per #9 bar)(60 KSI)= 120 KIPSAc = Effective concrete compression area=cfT85.0=)3(85.0120KSIKIPS= 47.1 in2¼(Beam span) = ¼(20’-0” x 12”/ft) = 60” ← USEor(2 x overhang width) + bw = (2 x 32” + 8”) = 72”8hf = 8(4”) = 32” ← USEor½(Clear distance) = ½(78” – 8”) = 35”
  • Lecture 5 - Page 7 of 9Step 4 – Determine usable moment capacity, Mu for the T-beam:a = Effective concrete compressive thickness=bAc="601.47 2ina = 0.79”Z = Moment arm distance between center ofcompression to center of tension= d -2a= 16” -2"79.0Z = 15.6”Mu = φTZ= 0.9(120 KIPS)(15.6”)= 1685 KIP-INMu = 140.4 KIP-FTNOTE: Since Mu = 140.4 KIP-FT > Mmax = 95 KIP-FT,T-beam is ACCEPTABLE.
  • Lecture 5 - Page 8 of 9Heavily-Reinforced T-BeamsT-beams with a lot of tension reinforcement may have a portion of the effectiveconcrete area located within the stem as shown below:The location of the centroid of the effective concrete compression area isfound by methods discussed in AECT 210 – Structural Theory (see Lecture 5).After the location is found, analysis is exactly the same as ordinary T-beams.Similar to ordinary rectangular reinforced concrete beams, the ACI 318 limitsthe amount of tension steel in T-beams so that the steel will yield prior toconcrete compression failure. The maximum area of steel, As is shown in thetable below.Maximum Tensile Steel Permitted in T-BeamsConcrete and Steel Properties: Formula (As = in2)Concrete f’c = 3000 PSISteel fy = 40 KSIAs max = 0.0478[bhf + bw(0.582d – hf)]Concrete f’c = 3000 PSISteel fy = 60 KSIAs max = 0.0319[bhf + bw(0.503d – hf)]Concrete f’c = 4000 PSISteel fy = 40 KSIAs max = 0.0638[bhf + bw(0.582d – hf)]Concrete f’c = 4000 PSISteel fy = 60 KSIAs max = 0.0425[bhf + bw(0.503d – hf)]bwhfbZdAsAc = Shaded area= Effective concretecompression area
  • Lecture 5 - Page 9 of 9Example 2GIVEN: The T-beam from Example 1.REQUIRED: Determine the maximum area of tension steel permitted, As max:Step 1 – Determine As max:From Example 1:Concrete f’c = 3000 PSISteel fy = 60 KSIb = 60”hf = 4”bw = 8”As max = 0.0319[bhf + bw(0.503d – hf)]= 0.0319[(60”)(4”) + 8”(0.503(16”) – 4”)]As max = 8.7 in2NOTE: This area of tension steel As = 8.7 in2is a LOT!! In orderto supply this much steel the beam would require 9 - #9 bars, 15- #7 bars or 20 - #6 bars! It would be far better to change thebeam dimensions than to try to squeeze this many bars into thebeam.