Composite Steel Deck

MARCH 1991

by Ron Vogel, Computers and Structures, Inc. March, 1991

LRFD-COMPOSITE BEAM DESIGN WITH METAL DECK

INTRODUCTION
This is the companion paper to the "STEEL TIPS" dated January 1987 entitled "Composite Beam Design with Metal Deck". The original paper used allowable stress design (ASD). This "STEEL TIPS" utilizes the same three original examples but designed by the Load andResistance Factor Design (LRFD) Method. The purpose is to show the design procedure, the advantages of the method, and the ease of using the AISC First Edition (LRFD) for design. Three main areas have been revised from the ASD Approach: 1. Determination of effective slab width 2. Shored and unshored construction requirements 3. Lower bound moment of inertia may be utilized. A number of papers havebeen written about these differences and the economies of the LRFD method. The reader is referred to the list of references included.

Table 1 S U M M A R Y OF SECTION I3.1 ITEM Effective Width, on each side of beam (lesser of the 3 values) General AISC-LRFD SUMMARY b = Beam Length/8 (L/8) = Beam Spacing/2 (s/2) = Distance to Edge of Slab hr < Wr > ds < Hs = = tc > 3.0 in. 2. 0 in. 3/4 in. hr + 11/2 in. hr + 3 in. 2.0 in. (Height of Rib) (Width of Rib) (Welded Stud Diameter) (MinimumStud Height) (Maximum Stud Height value for computations) (Minimum concrete above deck) SPECIFICATION SECTIONS I3 & I5

I3.5a

15.1 I5.2

Material Horizontal Shear Force (lesser of the 3 values) Strength of Stud

Hs > 4ds = 0.85f'cAc = AsFy -- • Qn Qn = 0.5 Asc (f'c Ec) (but not more than Asc Fu) = 0.5Asc (f'c wc)3/4 (using E¢ = wcl'5fx•c in above formula) = 6 ds = 4 ds Longitudinal Transverse

I5.3

I5.6

Shear Connector Placement and Spacing

(See LRFD Manual Fig. C-I5.1, pg. 6-177)

Table 2 AISC-LRFD RULES - F O R M E D M E T A L DECK

(Sections I3.5b and I3.5c)

I E T M

RIBS PERPENDICULAR NEGLECT (N•0'85 [rjWl • r •r• r{SS 1}-< 1'0

RIBS PARALLEL INCLUDE ' [hrrJ <06•wrl, [ h r - 1} - 1.0

1. Concrete Area Below Top of Deck 2. Stud Reduction Factor

3. Maximum Stud Spacing 4. Deck Welding 5. Minimum Width of Rib

32 in. 16 in. 2 in.

NOT SPECIFIED NOT SPECIFIED NOT SPECIFIED

Page 2 Steel Tips March 1991

Typical Design Problems
Example 1. Design a composite interior floor beam (without cover plate) for an office building. See Beam A in Figure 1.Solution: 1. Design for construction loads: a. Strength design

- - I .

i1
40'

[

BAt
^ ^
30' -'

wu= s [1.6 (D.L. + L.L.)] = 10 [1.6 ( 57 + 20 )] / 1000 = 1.23 kip/ft (Load factor for D.L. assumed same as for L.L. during placement of concrete.) B Mu - wuL2 - (1'23)(30)2 - 139 kip-ft 8 8 Mu (12)(139) _ 51 in.3 (Minimum) Zreq- q•Fy- (0.9)(36)

b. Servicibility design Limit constructiondeflection to 1 in. (without construction L.L.)

Figure 1
Given: Span length, L = 30 ft. Beam spacing, s = 10 ft. Slab thickness, tc = 2.5 in. Concrete strength, f'c = 3.0 ksi Concrete weight, wc = 145 pcf (n = 9) Steel yield stress, Fy = 36 ksi 3 inch metal deck, ribs perpendicular to beam (hr = 3 in., wr = 6 in.) No shoring permitted. Do not reduce live load. Loads: Concrete slab includingreinforcing steel and metal deck 54 Framing 3 Mechanical 4 Ceiling 6 Partition 20 Total D.L. Live Load 87 psf 100 psf

5wL4 (5)[(10)(57)](30)4(1728) Ireq= 384EA- (384)(29,000,000)(1.0) - 358 in.4 (Minimum) 2. Composite Beam Design: a. Trial design for required flexural strength wu = 10 [1.2(87)+1.6(100)]/1000 = 2.64 kip/ft Mu = wuL2 (2.64)(30)2 T = 8 = 297 kip-ft

For a trial size use formulain LRFD Manual pg. 4-9. 12Mu (3.4) Beam Weight = {d 2} •+Yc • Fy ) where q = 0.85 and assume a = 1 in. • d (in.) 14 16 18 21 396 396 396 396 12Mu(3.4) Fy d •+Yc a
-

WT (#/ft) 33 31

Size

Z

I

(in.) 12.0 13.0 14.0 15.5

(in.3) (in.4) W14X34 54.6 W16X31 54.0 340 375 510 843

28 W18X35 66.5 26 W21X44 95.4

Construction Loads, D.L. 57 psf (concrete & framing) L.L. 20 psf (men &...