Introduction. ENCE 710 Design of Steel Structures IV. COMPOSITE STEEL-CONCRET CONSTRUCTION. Effective Width. Composite Action

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1 ENCE 710 Design of Steel Strutures V. COMPOSTE STEEL-CONCRET CONSTRUCTON C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland ntrodution Following subjets are overed: Composite Ation Effetive Width Nominal Moment Strength Shear Connetors, Strength and Fatigue Formed Steel Dek Reading: Chapters 16 of Salmon & Johnson ASC LRFD Speifiation Chapters B (Design Requirements) and (Design of Composite Members) Composite Ation Effetive Width ASC-3 1. nterior B E L/4 B E b 0 (for equal beam spaing). Exterior B E L/8 + (dist from beam enter to edge of slab B E b 0 / + (dist from beam enter to edge of slab) 3 4

2 Nominal Moment Strength Plasti Stress Distribution Nominal Moment Strength of Fully Composite Setion (ASC 14th Edition Art. 3.a) 1. h / tw p 3.76 / E F yf. M n = based on plasti stress distribution on the Composite Setion; Φ b = 0.9 h / tw p 3.76 / E F yf M n = based on superposition of elasti stresses, onsidering the et of shoring; Φ b = Case 1 (if a t s ): S & J Eq. ( to 5) Case (if a > t s ): S & J Eq. ( to 10) 6 Shear Connetors Shear Variation V = C max = 0.85f b E t s V = T max =A s F y N = C max /Q n or T max /Q n 7 Whihever is smaller 8

3 Metal Dek Nominal Strength Q n Q n = 1. Headed Steel Stud (ASC Eq. 8-1) Q n 0.5A w f ' E R R g p A s F u. Channel Connetors (ASC Eq. 8-) Q n 0.3( t 0.5t ) L f w f ' E 9 10 Nominal Strength Q n Partial Composite ASC Table 3-1 Q n 11 1

4 Partial Composite Partial Composite Why Partial Composite: 1. An oversized steel beam must be seleted from the available rolled beam sizes for arhitetural reasons or ease of fabriation (repeatability) or when defletion ontrols and strength requirements are adequately met by less than fully omposite ation.. When the ribs of the metal dek are perpendiular to the beam, the longitudinal spaing of shear onnetors must be ompatible with the pith of the ribs. Quite often it may not be possible to fit in suffiient number of shear onnetors, for the beam to be designed as a fully omposite beam. (VNNAKOTA, FOLEY and VNNAKOTA) 13 (VNNAKOTA, FOLEY and VNNAKOTA) 14 Partial Composite ASC Composite Table (VNNAKOTA, FOLEY and VNNAKOTA) 15 16

5 ASC Composite Table To utilize ASC Manual Table 3-19, 1. The distane from the ompressive onrete flange fore to beam top flange, Y, must first be determined. Fifty perent omposite ation [ Q n 0.50(AsFy)] is used. The ompression blok depth, a trial = Q n /0.85f b =0.50(A s F y )/0.85f b 3. Y =Y on - a trial / where Y on = distane from top of steel beam to top of slab ASC Composite Table To utilize ASC Manual Table 3-19 (ont.), 5. For the W setion that provides suffiient available strength φ b M n M u 6. The atual value for the ompression blok depth, a, is determined. 7. Defletions for omposite members may be determined using the lower bound moment of inertia provided by Speifiation Commentary Equation C-3-1 and tabulated in ASC Manual Table Enter ASC Manual Table 3-19 with the required strength and Y to selet a plasti neutral axis loation (PNA loation 5 (BFL) with designed Q n ) Connetor Design Fatigue Strength nz r p V Q sr (AASHTO LRFD Eq ) Z r = d 5.5 d /; (AASHTO LRFD Eq & ) (Higher of Fatigue & in the new AASHTO LRFD) Composite Column Setion (rolled steel shape enased in onrete) ASC.1. Enased Composite Columns ASC.. Filled Composite Columns (Ref: Separate handout with examples.) where = log N (AASHTO LRFD Eq ) Example: 19 0

6 Composite Column Setion (rolled steel shape enased in onrete) Filled Composite Column Example Using Effetive Setion Properties ( 4, 5 & 6) P 0 As Fy Asr Fyr 0.85A f ' ASC -b (a) Compat (b) Nonompat () Slender P e1 K L 1 ; Es s 0.5Es se C1 E 1 Filled Composite Column Example For ompat setions A = b f h f +π(r-t) +b f (r-t)+h f (r-t) A = (8.5 in.)(4.5 in.) + π(0.375 in.) + (8.5 in.)(0.375 in.) + (4.5 in.)(0.375 in.) = 48.4 in. b h1 ( b )( h ) 8 ( r t) h 4( r t) 1 ( r t)( ) ( )( ) 111in P 0 As Fy Asr Fyr 0.85A f ' P 0 = (10.4 in. )(46ksi) (48.4 in. )(5 ksi) = 684kips Es s 0.5Es se C 3 = (9,000 kis)(61.8 in. 4 ) + (0.90)(3,900 ksi)(111 in. 4 ) =,180,000 kip-in. 3 4 Filled Composite Column Example P e1 K L 1 P e = π (,180,000 kip-in. )/(1.0(14 ft)(1 in./ft)) = 76 kips P 0 /P e = 684 kips/76 kips = p0 / e kips P (684 ) kips Pn P0 470 φ P n = 0.75(470 kips) = 353 kips > 336 kips o.k. (-) 4