ENCE 710 Design of Steel Structures IV. COMPOSITE STEEL-CONCRET CONSTRUCTION C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland

2 Introduction Following subjects are covered: Composite Action Effective Width Nominal Moment Strength Shear Connectors, Strength and Fatigue Formed Steel Deck Reading: Chapters 16 of Salmon & Johnson AISC LRFD Specification Chapters B (Design Requirements) and I (Design of Composite Members)

3 Composite Action

4 Effective Width AISC-I3 1.Interior B E ≤ L/4 B E ≤ b 0 (for equal beam spacing) 2. Exterior B E ≤ L/8 + (dist from beam center to edge of slab B E ≤ b 0 /2 + (dist from beam center to edge of slab)

5 Nominal Moment Strength Nominal Moment Strength of Fully Composite Section (AISC 14th Edition Art. I3.2a) 1. M n = based on plastic stress distribution on the Composite Section; Φ b = M n = based on superposition of elastic stresses, considering the effect of shoring; Φ b = 0.9

6 Plastic Stress Distribution Case 1 (if a ≤ t s ): S & J Eq. ( to 5) Case 2 (if a > t s ): S & J Eq. ( to 10)

7 Shear Connectors

8 Shear Variation V’ = T max =A s F y V’ = C max = 0.85f c ’b E t s N = C max /Q n or T max /Q n Whichever is smaller

9 Nominal Strength Q n Q n = 1. Headed Steel Stud (AISC Eq. I8-1) 2.Channel Connectors (AISC Eq. I8-2)

10 Nominal Strength Q n

11 Connector Design – Fatigue Strength (AASHTO LRFD Eq b-1) Z r =  d 2  5.5 d 2 /2;(AASHTO LRFD Eq ) where  = 34.5 – 4.28 log N(AASHTO LRFD Eq ) Example:

12 Composite Column Section (rolled steel shape encased in concrete) AISC I2.1. Encased Composite Columns AISC I2.2. Filled Composite Columns (Ref: Separate handout with examples.)

13 Composite Column Section (rolled steel shape encased in concrete) Using Effective Section Properties (I2-4, 5 & 6)

14 Filled Composite Column Example AISC I2-2b (a)Compact (b)Noncompact (c)Slender

15 Filled Composite Column Example For compact sections A c = b f h f +π(r-t) 2 +2b f (r-t)+2h f (r-t) A c = (8.5 in.)(4.5 in.) + π(0.375 in.) 2 + (8.5 in.)(0.375 in.) + 2(4.5 in.)(0.375 in.) = 48.4 in. 2 P 0 = (10.4 in. 2 )(46ksi) (48.4 in. 2 )(5 ksi) = 684kips EI eff = (29,000 kis)(61.8 in. 4 ) + (0.90)(3,900 ksi)(111 in. 4 ) = 2,180,000 kip-in. 2

16 Filled Composite Column Example P e = π 2 (2,180,000 kip-in. 2 )/(1.0(14 ft)(12 in./ft)) 2 = 762 kips P 0 /P e = 684 kips/762 kips = ≤ 2.25 φ c P n = 0.75(470 kips) = 353 kips > 336 kips o.k.