LRFD Composite Beam Design With Metal Deck

MARCH 1991

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

LRFD-COMPOSITE BEAM DESIGN WITH METAL DECK

INTRODUCTION

This is the com This compan panion ion paper to the "STEEL TIP TIPS" S" dat dated ed Jan Januar uary y 1987 entitle entitled d "Co "Compo mposite site Beam Design with Metal Deck". The original paper used allowable stress design (ASD). This "STEEL "ST EEL TIP TIPS" S" ut util iliz izes es th the e sa same me th thre ree e or orig igin inal al ex exam ampl ples es bu butt de desi sign gned ed by th the e Loa Load d an and d Resistance Resista nce Factor Design (LRFD (LRFD)) Method. The purpose is to show the design procedure, procedure, the advantages of the method, and the ease of using the AISC First Edition (LRFD) for design. Three mai n areas have been revised from the ASD Appr Approac oach: h: 1. Determination Determination o f effective slab width 2. Shored and unshored construction requirements 3. Lower bound m oment o f inertia may be util utilize ized. d. A numb er of papers have been written about these differences and the economies of the LRFD method. The reader is referred to the list of references references included.

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Table 1 S U M M A R Y OF

AISC AI SC-L -LRF RFD D

SPEC SP ECIF IFIC ICAT ATIO ION N

SECTION

I T EM

SUMMARY

I3.1

Effective Width, on each side of beam (les (l esse serr of the 3 val value ues) s)

b = Beam Length/8 (L/8) = Beam Spacing/2 (s/2) = Dis Distan tance ce to Ed Edge ge of Sla Slab b

I3.5a

General

hr < 3 . 0 in. Wr > 2. 2. 0 in. ds < 3/ 3/4 in. Hs = hr + 1 1/2 in. = hr + 3 in. tc > 2.0 in.

15.1

Material

Hs > 4ds

I5.2

Horizontal Shear Force (les (l esse serr of th the e 3 va values)

= 0.85f'cAc = AsFy -- • Qn

I5.3

Strength of Stud

Qn = 0.5 Asc (f'c Ec)

SECT SE CTIO IONS NS I3 & I5

(Height of Rib) (Width o f Rib) (Welded Stud Diameter) (Mi nim umStud Heigh Height) t) (Ma (M axi ximu mum m St Stu ud He Heig ight ht va valu lue e fo forr co computations) (Minimum concrete above deck)

(but not more than Asc Fu)

= 0.5 As Asc c (f'c wc)3 wc)3//4 (using (using E¢ = wc wcl' l'5fx fx•c •c in above formula) f ormula) I5.6

Shear Connector Placement and Spacing

= 6 ds = 4 ds

Longitudinal Transverse

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

Table 2 AISC-LRFD

RULE RU LES S

- F O R M E D M E T A L DE DECK CK

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

ITEM

1. Concrete Area Below Top of Deck 

RIBS PERPENDICULAR NEGLECT

2. Stud Reduction Factor

(N•0'85 [• [•rrj•Wrl{•SrS-

3 Maximum Stud Spacing

32 in

RIBS PARALLEL INCLUDE

1}-< 1'0

' • [whrlr,J [ h r - 1} -< 1.0 06

NOT SPECIFIED

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Typical Design Problems Example Exam ple 1.

Solution:

Design a composite interior floor beam (without cover plate) for an office building. building. See Beam A in Figure 1.

1. Design for construction loads: a. Strength design

-

-

I

.

i1 40'

[

wu= s [1.6 (D.L. + L.L.)] = 10 [1.6 ( 57 + 20 )] )] / 100 0 = 1.23 kip/ft kip/ft (Load (Lo ad fa fact ctor or for D. D.L. L. ass assum umed ed same as fo forr L. L.L. L. dur during ing placement of concrete.)

BAt

B

^

Mu - wuL2 wuL2 - (1 (1'2 '23) 3)(3 (30) 0)2 2 - 139 kipkip-ft ft 8 8

^

Zreq-

Mu (12)(139) (12)(1 39) _ 51 in in.3 (Min (Minim imum um)) q•F •Fy y- (0 (0..9)(3 )(36)

b. Servicibility design

30'

Limit construction deflection to 1 in. (without construction L.L.)

-'

5wL4 5wL (5)[(10 (5)[( 10)(5 )(57) 7)](3 ](30) 0)4(1 4(172 728) 8) Ireq= 384EA- (3 (384) 84)(29 (29,0 ,000 00,0 ,000 00)(1 )(1.0 .0))

Figure 1

- 35 358 8 in. in.4 4 (Mi (Mini nimu mum) m) Given:

Span length, L = 30 ft. ft. Beam spacing, s = 10 ft.

2. Compo Composite site Beam Beam Design:

Slab thickness, tc = 2.5 in. Concrete strength, f'c = 3.0 ksi Concrete Conc rete weight, weight, wc = 145 145 pcf (n = 9)

a. Trial design for required flexural strength wu = 10 [1.2(87)+1.6 [1.2(87)+1.6(100)]/1 (100)]/1000 000 = 2.64 kip/ft kip/ft

Steel yield stress, Fy = 36 ksi 3 inch metal deck, ribs perpendicular

Mu =

to beam (hr = 3 in., wr = 6 in.)

For a trial size use formula in LRFD Manual pg. 4-9. 12Mu (3.4) Beam Weight = {d 2} •+Yc •) Fy

No shoring permitted. Do not reduce live load. Loads:

Concrete slab including reinforcing steel and metal deck

54

Framing

3

Mechanical

4

Ceiling

6

Partition

20

Total D.L.

87 p s f  

Live Load

100 p s f  

wuL2 (2 (2.6 .64) 4)(3 (30) 0)2 T = 8 = 297 k ip - f t

where q• = 0.85 0. 85 and assume a = 1 in. d

1 2 M u(3.4) Fy

(in.)

d

•+Yc

a

WT

Size

Z

I

-

(in.)

(#/ft)

(in.3) ( in .4)

14

396

12.0

33

W14X34 54.6

340

16

396

13.0

31

W16X31 54.0

375

18

396

14.0

28

W18X35 66 66.5

510

21

396

15.5

26

W21X44 95.4

843

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b. Verify flexural strength '

b

I o. o •o °

·

°

o. ' % ..;;'•,

Effective concrete width (AISC I3.1) lesser of,

. .v .* .. .' , ) .?o o .o : .

n

i

i

d/2

/

!

i

1

b = (2 (2)( )(30) 30)(1 (12) 2) _ 90 in in.. an and d 8 b = (2)(10)(12) (2)(10)(12) _ 12 120 0 in. in. 2 Use 90 in. Design for full composite action Tmax= AsFy= AsFy= (1 (10. 0.3) 3)(3 (36) 6)= = 37 371 1k ips (Governs) Cmax = q• f' c b tc = (0.85)(3 (0.85)(3.0)(90)( .0)(90)(2.5) 2.5) = 574 kips

Figure 2

a -

Tmax { f'c f'c b

371 (0.8 (0 .85) 5)(3 (3.0 .0)( )(90 90))

1.62 in.

a/2 = 0.81 in. (larger than 0.5 0.5 in. assumed) a

H

.,.

.

.

.

.

o.

'

.

.

.

.¢'• . _. i _ t

Y2 = Yc - • = 5. 5.5 5 - 0.81 = 4.6 4.69 9 in. in. q•Mn= q• q•As AsFy Fy/d /d + Y21= (0 (0.8 .85) 5)(3 (371 71)I )I11----••

+4.693 +4.69 3

= 4270 kil•2n. = 356 kip-ft or from Table on LRFD Manual pg. 4-23 with Y2 = 4.69 in. PNA = TFL (Top flange location) Y1 =0in.

Figure 3

= 37 371 1 kips kips (A (AsF sFy) y) By

tie

Yc

14.69 - 4.501 *m n= ["•.-.-•.0-•.5-J (364(36 4- 351 )+ 351 351

kip-ft.

Y•2 = 35 356 kip-ft

d / 2 + Yc ' a / 2 d/ 2

T

1

> 297 ki kip-ft

O . K. K.

c. Calculate shear studs For full composite action •Qn = AsF sFy y= 371 kips Assume 3/4 inch diameter by 5 inch long studs. Qn = 0.5Asc(f'c wc)3 wc)3//4 = (0.5)(0.442) [(3) [(3)(145 (145)] )]3 3/4 = 21 21.1 .1 kip kipss (