6Effective concrete-steel T-Beam The composite beam can be designed as an effective T-Beam, where width of the slab on either side is limited to:1/8 of the beam span½ distance to centerline of adjacent beamThe distance to the end of the slab
7Shoring Temporary shores (supports) during construction are optional. If temporary shores are NOT used, the steel section must have adequate strength to support all loads prior to concrete attaining 75% of f’c
8Shear StrengthDesign shear strength and allowable shear strength of composite beams are based on just the steel section!
9Flexural StrengthPositive Flexural strength fbMn (or Mn/Wb) are determined as follows:fb = (LRFD) and/or Wb = 1.67 (ASD)Mn depends on h/tw as follows:If determine Mn for yield from plastic stress distribution on composite section (flange yield)Else, determine Mn from yielding from superposition of elastic stresses, considering shoring
11Negative momentThe design Negative moment can be based on the steel section alone.Could be based on plastic stress distribution through composite section providedSteel beam is adequately braced compact sectionShear connectors in the negative moment areaSlab reinforcement parallel to steel is properly developed
14Composite beam with formed steel deck Nominal rib height is limited to 3 inches.Width of rib or haunch must be at least 2 inch. For calculations, never more than minimum clear widthMust be connected with shear connectors ¾” or less in diameter. Can be welded through deck or to steel cross-section.Connectors must not extend more than 1.5” above the top of the deck.Must be at least ½” cover
15Composite beam with formed steel deck (cont) Slab thickness must be at least 2”Deck must be anchored to all supporting members at max spacing of 18”.Stud connectors, or a combination of stud connectors and arc spot (puddle) welds may be usedIf ribs are perpendicular to steel, concrete below the steel deck must be neglected for calculation section properties and concrete area
16Composite beam with formed steel deck (cont) For deck ribs parallel to steel beam, concrete below top of steel deck may be included in determining composite section properties and area of concrete.Deck ribs over beams may be split and separated to form concrete haunch.When depth of deck is 1.5” or greater, average width of supported haunch or rib must be at least 2” for the first stud plus four stud diameters for each additional stud.
19Shear Connectors Shear force is transferred by the connectors The total horizontal shear force, V’, between max positive moment and zero moment is the smallest ofConcrete crushing: V’ = 0.85 f’c AcSteel yielding: V’ = As syConnectors fail: V’ = ∑Qn
21Number of shear connectors Number of shear connectors = V’/QnStrength of one shear connectorAsc = x-sectional area of 1 connector,Rg and Rp on next pagessu = tensile strength of connector
22Rg Rg = 1 for Rg = 0.85 for Rg = 0.7 for One stud welded in steel deck rib with deck perpendicular to steel shape;Any number of studs welded in a row through steel deck with deck parallel to steel shape and ratio of rib width to depth ≥ 1.5Rg = 0.85 forTwo studs welded through steel deck rib with deck perpendicular;One stud welded through deck parallel to steel and rib width to depth < 1.5Rg = 0.7 forThree or more studs welded in the deck rib, perpendicular to steel
23Rp Rp = 1.0 for Rp = 0.75 for Rp = 0.6 for Studs welded directly to steel shape (not through steel deck) and having a haunch detail with not more than 50% of the top flange covered by deck or sheet steel closures.Rp = 0.75 forStuds welded in composite slab, deck perpendicular to steel, emid-ht ≥ 2 inchStuds welded through deck, deck parallel to steelRp = 0.6 forStuds welded in composite slab, deck perpendicular to steel and emid-ht < 2 inchemid-ht = distance from edge of stud shank to steel deck web measured at mid height of deck rib in the load bearing direction of the stud (direction of maximum moment)
24ChannelsChannels welded to steel beam may be used as shear connectors.Welds must develop the shear resistance QnEffects of eccentricity must be consideredWhere tf = flange thickness of channel connectortw = web thickness of channel shear connectorLc = length of channel shear connector
25Compressive StrengthConcrete crushing: Cc = 0.85 f’c AcSteel yielding: Ct = As syConnectors fail: Cs = ∑QnSimilar to shear valuesThe location of the plastic neutral axis affects the failure criteria
26Location of Plastic Neutral Axis Case 1: PNA is in the web of the steel. Occurs when concrete compressive force is less than web force, Cc ≤ PywCase 2: PNA is in the thickness of the top flange. Pyw < Cc < CtCase 3: PNA is in the concrete slab. Cc ≥ CtNote: in Case 3, concrete below PNA is neglected!
30Example Composite framing in typical multi-story building 3.25” lightweight concrete, 2” steel deck.Concrete: r = 115 lb/ft2; f’c = 3 ksiAdditional 30% dead load assumed for equipment during constructionDeck is supported on steel beams with stud connectors.¾” diameter, 3.5” longUnshored constructionBeams must support their own weight, weight of concrete before it hardens, deck weight and construction loads.Check floor for vibration with damping ration of 5%.
31Example (p2) Typical beam is 30 ft long. Distance to adjacent beams is 10 ft.Ribs are perpendicular to the beamUniform dead loads on beam are, 500 lb/ft + 30% for equipment loadsSuperimposed loads are 250 lb/ftLive loads (uniform) 500 lb/ft
32Example (p3) Have to pick a beam. Must handle 1.3*0.5 + wt of beam. Using A992 (50 ksi) steel. Assume 22 lb/ft starting estimateW = 1.3* kip/ft = kip/ftFactored load: 1.4*0.672 = 0.941Factored moment: * L2/8 = 0.941*302/8 = kip-ft
33Plastic section modulus Fortunately, a W14x22 has a Z=33.2 in3, I=199 in4, and w=22
34Deflection of the beam The deflection of the beam is given as So camber the beam by 1.6” prior to pouring the concrete. Probably make it 1.5” in drawings.
35Next stepWe know that a W14x22 will handle the unshored loads. We need to consider live loads as well.We can apply the load reduction factor considering our area (30’ x 10’ between beams and supports)R = (A-150) = ( )=0.12So our live load is 0.5*(1-0.12) = 0.44 kip/ft
36Factored load Greater of Factored moment is thus 1.2( ) + 1.6(0.44) = 1.63 kip/ft1.4( ) = kip/ftFactored moment is thusMn = 1.63 * 302/8 = kip-ft
37Concrete compressive force Concrete flange with is lesser ofB = 10x12 = 120” orB = 2 (30 x 12/8) = 90” **Compressive force in concrete is smaller ofCc = 0.85 f’c Ac = 0.85 x 3 x 90 x 3.25 = kipsCt = As sy = 6.49 x 50 = kips **
38Depth of concrete stress block Since Cc > Ct, PNA is in the concrete slab.The distance between the compression and tension forces, e, on the W14x22e = 0.5d – 0.5a= 0.5 x – 0.5*1.414 = inWe are expecting 183.4, so this passes