Composite Beams cont .

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Composite Beams (cont’d). Floor beam. Girder. S. L. b. t c. h. 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
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Composite Beams (cont\'d)

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Floor bar Girder S L

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b t c h

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Effective solid steel T-Beam The composite shaft can be planned as a viable T-Beam, where width of the section on either side is restricted to: 1/8 of the pillar traverse ½ separation to centerline of contiguous bar The separation to the end of the chunk

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Shoring Temporary shores (bolsters) amid development are discretionary. In the event that brief shores are NOT utilized, the steel area must have satisfactory quality to bolster all heaps before cement achieving 75% of f\' c

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Shear Strength Design shear quality and passable shear quality of composite shafts depend on simply the steel segment!

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Flexural Strength Positive Flexural quality f b M n (or M n/W b ) are resolved as takes after: f b = 0.90 (LRFD) as well as W b = 1.67 (ASD) M n relies on upon h/t w as takes after: If decide M n for yield from plastic stretch circulation on composite area (rib yield) Else, decide M n from yielding from superposition of versatile hassles, considering shoring

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b 0.85 f\' c t c a h s y

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Negative minute The plan Negative minute can be founded on the steel segment alone. Could be founded on plastic push conveyance through composite segment gave Steel shaft is sufficiently supported conservative segment Shear connectors in the negative minute territory Slab fortification parallel to steel is legitimately created

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Shear Connectors Concrete Slab Ribbed steel deck Steel segment

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Effective width b t c Y c h r t w d t f b f

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Composite bar with shaped steel deck Nominal rib tallness is constrained to 3 inches. Width of rib or rump must be no less than 2 creep. For figurings, never more than least clear width Must be associated with shear connectors ¾" or less in measurement. Can be welded through deck or to steel cross-segment. Connectors must not expand more than 1.5" over the highest point of the deck. Must be at any rate ½" cover

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Composite shaft with framed steel deck (cont) Slab thickness must be no less than 2" Deck must be tied down to every supporting part at max separating of 18". Stud connectors, or a mix of stud connectors and curve spot (puddle) welds might be utilized If ribs are opposite to steel, concrete beneath the steel deck must be disregarded for estimation segment properties and solid territory

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Composite bar with shaped steel deck (cont) For deck ribs parallel to steel bar, concrete underneath top of steel deck might be incorporated into deciding composite segment properties and zone of cement. Deck ribs over bars might be part and isolated to frame solid hindquarters. At the point when profundity of deck is 1.5" or more prominent, normal width of bolstered rump or rib must be no less than 2" for the primary stud in addition to four stud distances across for each extra stud.

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Shear Connectors Shear constrain is exchanged by the connectors The aggregate flat shear drive, V\', between max positive minute and zero minute is the littlest of Concrete pounding: V\' = 0.85 f\' c A c Steel yielding: V\' = A s y Connectors fall flat: V\' = ∑Q n

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Shear Connectors For negative minutes, concrete can\'t withstand pressure. Rebar yields Tensile yielding: V\' = A r s yr Shear connectors: V\' = ∑ Q n

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Number of shear connectors Number of shear connectors = V\'/Q n Strength of one shear connector A sc = x-sectional range of 1 connector, R g and R p on next pages s u = rigidity of connector

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R g R g = 1 for One stud welded in steel deck rib with deck opposite to steel shape; Any number of studs welded in succession through steel deck with deck parallel to steel shape and proportion of rib width to profundity ≥ 1.5 R g = 0.85 for Two studs welded through steel deck rib with deck opposite; One stud welded through deck parallel to steel and rib width to profundity < 1.5 R g = 0.7 for at least three studs welded in the deck rib, opposite to steel

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R p R p = 1.0 for Studs welded straightforwardly to steel shape (not through steel deck) and having a backside detail with not more than half of the top spine secured by deck or sheet steel terminations. R p = 0.75 for Studs welded in composite piece, deck opposite to steel, e mid-ht ≥ 2 crawl Studs welded through deck, deck parallel to steel R p = 0.6 for Studs welded in composite section, deck opposite to steel and e mid-ht < 2 creep e mid-ht = remove from edge of stud shank to steel deck web measured at mid stature of deck rib in the heap bearing course of the stud (heading of most extreme minute)

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Channels welded to steel bar might be utilized as shear connectors. Welds must build up the shear resistance Q n Effects of capriciousness must be viewed as Where t f = rib thickness of channel connector t w = web thickness of channel shear connector L c = length of channel shear connector

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Compressive Strength Concrete pulverizing: c = 0.85 f\' c A c Steel yielding: C t = A s y Connectors come up short: C s = ∑Q n Similar to shear values The area of the plastic nonpartisan hub influences the disappointment criteria

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Location of Plastic Neutral Axis Case 1: PNA is in the web of the steel. Happens when concrete compressive constrain is not as much as web compel, c ≤ P yw Case 2: PNA is in the thickness of the top rib. P yw < c < C t Case 3: PNA is in the solid section. C ≥ C t Note: in Case 3, concrete underneath PNA is dismissed!

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Case 1 0.85f\' c a c Eff section h r e PNA d s y d/2 t f s y

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Case 2 0.85f\' c a c Eff chunk h r PNA s y e d/2 t f s y

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Case 3 0.85f\' c a c Eff piece h r PNA e d/2 t f s y

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Example Composite surrounding in run of the mill multi-story building 3.25" lightweight solid, 2" steel deck. Solid: r = 115 lb/ft 2 ; f\' c = 3 ksi Additional 30% dead load expected for gear amid development Deck is upheld on steel shafts with stud connectors. ¾" distance across, 3.5" since quite a while ago Unshored development Beams must bolster their own particular weight, weight of cement before it solidifies, deck weight and development loads. Check floor for vibration with damping proportion of 5%.

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Example (p2) Typical bar is 30 ft long. Separation to contiguous pillars is 10 ft. Ribs are opposite to the pillar Uniform dead loads on bar are, 500 lb/ft + 30% for hardware loads Superimposed burdens are 250 lb/ft Live loads (uniform) 500 lb/ft

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Example (p3) Have to pick a bar. Must handle 1.3*0.5 + wt of pillar. Utilizing A992 (50 ksi) steel. Accept 22 lb/ft beginning assessment W = 1.3*0.5 + 0.022 kip/ft = 0.672 kip/ft Factored stack: 1.4*0.672 = 0.941 Factored minute: 0.941 * L 2/8 = 0.941*30 2/8 = 105.8 kip-ft

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

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Deflection of the pillar The diversion of the shaft is given as So camber the bar by 1.6" preceding pouring the solid. Most likely make it 1.5" in drawings.

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Next stride We realize that a W14x22 will handle the unshored loads. We have to consider live loads too. We can apply the heap decrease figure considering our zone (30\' x 10\' amongst shafts and backings) R = 0.0008(A-150) = 0.0008(300-150)=0.12 So our live load is 0.5*(1-0.12) = 0.44 kip/ft

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Factored stack Greater of 1.2(0.5+0.25+0.022) + 1.6(0.44) = 1.63 kip/ft 1.4(0.5+0.25+0.22) = 1.081 kip/ft Factored minute is along these lines M n = 1.63 * 30 2/8 = 183.4 kip-ft

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Concrete compressive drive Concrete spine with is lesser of B = 10x12 = 120" or B = 2 (30 x 12/8) = 90" ** Compressive constrain in cement is littler of c = 0.85 f\' c A c = 0.85 x 3 x 90 x 3.25 = 745.9 kips C t = A s y = 6.49 x 50 = 324.5 kips **

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Depth of solid stretch square Since Cc > Ct, PNA is in the solid piece. The separation between the pressure and strain powers, e, on the W14x22 e = 0.5d + 5.25 – 0.5a = 0.5 x 13.7 + 5.25 – 0.5*1.414 = 11.393 in We are expecting 183.4, so this passes

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