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Composite Beams (cont’d). Girder Floor beam L S.

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Presentation on theme: "Composite Beams (cont’d). Girder Floor beam L S."— Presentation transcript:

1 Composite Beams (cont’d)

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4 Girder Floor beam L S

5 b tctc h

6 Effective 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 span – ½ distance to centerline of adjacent beam – The distance to the end of the slab

7 Shoring 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

8 Shear Strength Design shear strength and allowable shear strength of composite beams are based on just the steel section!

9 Flexural Strength Positive Flexural strength  b M n (or M n /  b ) are determined as follows: –  b = 0.90 (LRFD) and/or  b = 1.67 (ASD) M n depends on h/t w as follows: If determine M n for yield from plastic stress distribution on composite section (flange yield) Else, determine M n from yielding from superposition of elastic stresses, considering shoring

10 b tctc h a 0.85 f’ c yy

11 Negative moment The design Negative moment can be based on the steel section alone. Could be based on plastic stress distribution through composite section provided – Steel beam is adequately braced compact section – Shear connectors in the negative moment area – Slab reinforcement parallel to steel is properly developed

12 Shear ConnectorsConcrete SlabRibbed steel deck Steel section

13 Effective width b YcYc twtw tftf bfbf tctc hrhr d

14 Composite 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 width Must 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

15 Composite 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 used If ribs are perpendicular to steel, concrete below the steel deck must be neglected for calculation section properties and concrete area

16 Composite 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.

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19 Shear Connectors Shear force is transferred by the connectors The total horizontal shear force, V’, between max positive moment and zero moment is the smallest of – Concrete crushing: V’ = 0.85 f’ c A c – Steel yielding: V’ = A s  y – Connectors fail: V’ = ∑Q n

20 Shear Connectors For negative moments, concrete cannot withstand tension. Rebar yields – Tensile yielding: V’ = A r  yr – Shear connectors: V’ = ∑ Q n

21 Number of shear connectors Number of shear connectors = V’/Q n Strength of one shear connector A sc = x-sectional area of 1 connector, R g and R p on next pages  u = tensile strength of connector

22 RgRg R g = 1 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 ≥ 1.5 R g = 0.85 for – Two studs welded through steel deck rib with deck perpendicular; – One stud welded through deck parallel to steel and rib width to depth < 1.5 R g = 0.7 for – Three or more studs welded in the deck rib, perpendicular to steel

23 RpRp R p = 1.0 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. R p = 0.75 for – Studs welded in composite slab, deck perpendicular to steel, e mid-ht ≥ 2 inch – Studs welded through deck, deck parallel to steel R p = 0.6 for – Studs welded in composite slab, deck perpendicular to steel and e mid-ht < 2 inch e mid-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)

24 Channels Channels welded to steel beam may be used as shear connectors. Welds must develop the shear resistance Q n Effects of eccentricity must be considered Where t f = flange thickness of channel connector t w = web thickness of channel shear connector L c = length of channel shear connector

25 Compressive Strength – Concrete crushing: C c = 0.85 f’ c A c – Steel yielding: C t = A s  y – Connectors fail: C s = ∑Q n Similar to shear values The location of the plastic neutral axis affects the failure criteria

26 Location of Plastic Neutral Axis Case 1: PNA is in the web of the steel. Occurs when concrete compressive force is less than web force, C c ≤ P yw Case 2: PNA is in the thickness of the top flange. P yw < C c < C t Case 3: PNA is in the concrete slab. C c ≥ C t – Note: in Case 3, concrete below PNA is neglected!

27 Case 1 Eff slab d hrhr d/2 tftf PNA 0.85f’ c a yy yy CcCc e

28 Case 2 Eff slab d hrhr d/2 tftf PNA 0.85f’ c a yy yy CcCc e

29 Case 3 Eff slab d hrhr d/2 tftf PNA 0.85f’ c a yy CcCc e

30 Example Composite framing in typical multi-story building 3.25” lightweight concrete, 2” steel deck. – Concrete:  = 115 lb/ft 2 ; f’ c = 3 ksi – Additional 30% dead load assumed for equipment during construction Deck is supported on steel beams with stud connectors. – ¾” diameter, 3.5” long Unshored construction – Beams 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%.

31 Example (p2) Typical beam is 30 ft long. Distance to adjacent beams is 10 ft. Ribs are perpendicular to the beam Uniform dead loads on beam are, 500 lb/ft + 30% for equipment loads Superimposed loads are 250 lb/ft Live loads (uniform) 500 lb/ft

32 Example (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 estimate W = 1.3* kip/ft = kip/ft Factored load: 1.4*0.672 = Factored moment: * L 2 /8 = 0.941*30 2 /8 = kip-ft

33 Plastic section modulus Fortunately, a W14x22 has a Z=33.2 in 3, I=199 in 4, and w=22

34 Deflection 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.

35 Next step We 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) = ( )=0.12 So our live load is 0.5*(1-0.12) = 0.44 kip/ft

36 Factored load Greater of – 1.2( ) + 1.6(0.44) = 1.63 kip/ft – 1.4( ) = kip/ft Factored moment is thus – M n = 1.63 * 30 2 /8 = kip-ft

37 Concrete compressive force Concrete flange with is lesser of – B = 10x12 = 120” or – B = 2 (30 x 12/8) = 90” ** Compressive force in concrete is smaller of – C c = 0.85 f’ c A c = 0.85 x 3 x 90 x 3.25 = kips – C t = A s  y = 6.49 x 50 = kips **

38 Depth of concrete stress block Since Cc > Ct, PNA is in the concrete slab. The distance between the compression and tension forces, e, on the W14x22 e = 0.5d – 0.5a = 0.5 x – 0.5*1.414 = in We are expecting 183.4, so this passes


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