LRFD-Composite Beam Design

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 "CompositeBeam 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, theadvantages 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 width2. Shored and unshored construction requirements3. Lower bound moment of inertia may be utilized.

A number of papers have been written about these differences and the economies of the LRFDmethod. The reader is referred to the list of references included.

Table 1

S U M M A R Y OF AISC-LRFD SPECIFICATION SECTIONS I3 & I5

SECTION ITEM SUMMARY

I3.1 Effective Width, b = Beam Length/8 (L/8)on each side of beam = Beam Spacing/2 (s/2)(lesser of the 3 values) = Distance to Edge of Slab

I3.5a General hr < 3.0 in. (Height of Rib)Wr > 2. 0 in. (Width of Rib)ds < 3/4 in. (Welded Stud Diameter)Hs = hr + 1 1/2 in. (Minimum Stud Height)

= hr + 3 in. (Maximum Stud Height value for computations)tc > 2.0 in. (Minimum concrete above deck)

15.1 Material Hs > 4ds

I5.2 Horizontal = 0.85f'cAcShear Force = AsFy(lesser of the 3 values) -- • Qn

I5.3 Strength of Stud Qn = 0.5 Asc (f'c Ec) (but not more than Asc Fu)

= 0.5 Asc (f'c wc)3/4 (using E¢ = wcl'5 fx•c in above formula)

I5.6 Shear Connector = 6 ds LongitudinalPlacement and Spacing = 4 ds Transverse (See LRFD Manual Fig. C-I5.1, pg. 6-177)

AISC-LRFD

Table 2

RULES - F O R M E D M E T A L DECK

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

ITEM RIBS PERPENDICULAR RIBS PARALLEL

1. Concrete Area Below Top of Deck NEGLECT INCLUDE

06•wrl, 1} 1.02. Stud Reduction Factor (N•0'85 [•rrj•Wrl{•SrS- 1}-< 1'0 ' [hrrJ [ h r - -<

3. Maximum Stud Spacing 32 in. NOT SPECIFIED

4. Deck Welding 16 in. NOT SPECIFIED

5. Minimum Width of Rib 2 in. NOT SPECIFIED

Steel Tips March 1991

Typical Design Problems

Example 1.

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

i140'

[

- - I .

30' -'

BAt^

^

B

Given:

Loads:

Figure 1

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.

Concrete slab including reinforcing steel

and metal deck 54

Framing 3

Mechanical 4

Ceiling 6

Partition 20

Total D.L. 87 psf

Live Load 100 psf

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

L.L. 20 psf (men & equipment)

For simplicity, the entire 57 psf construction load is

considered as live load during concrete placement.

Solution:

1. Design for construction loads:

a. Strength design

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. duringplacement of concrete.)

Mu - wuL2 - (1'23)(30)2 - 139 kip-ft8 8

Mu (12)(139) _ 51 in.3 (Minimum)Zreq- q•Fy- (0.9)(36)

b. Servicibility design

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

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

wuL2 (2.64)(30)2Mu = T = 8 = 297 kip-ft

For a trial size use formula in LRFD Manual pg. 4-9.12Mu (3.4)

Beam Weight = {d 2}•+Yc - •) Fy

where q• = 0.85 and assume a = 1 in.

d 12Mu (3.4) d a WT Size Z I•+Yc -

Fy(in.) (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 66.5 510

21 396 15.5 26 W21X44 95.4 843

Select W18X35.

NOTE: The original Steel Tips design, based upon ASD,used Grade 50 steel.


' b

I o. o o. %..;;'•, . . . . . .o o ' v * . ' , ) ? o . : .•o° · ° n

i i

d/2 /

! i

1Figure 2

H .,. . . . . . . o ' . . . ..¢' • _ i _ t

Figure 3

tieYc

T

Y•2

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

1

Figure 4

b. Verify flexural strength

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

b = (2)(30)(12) _ 90 in. and8

b = (2)(10)(12) _ 120 in.2

Use 90 in.

Design for full composite action

Tmax= AsFy= (10.3)(36)= 371 k i p s (Governs)Cmax = q• f'c b tc = (0.85)(3.0)(90)(2.5) = 574 kips

Tmax 371a - { f'c b (0.85)(3.0)(90) 1.62 in.

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

Y2 = Yc - • = 5.5 - 0.81 = 4.69 in.

q•Mn= q•AsFy/d + Y21= (0.85)(371)I1---• +4.693

= 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.

= 371 kips (AsFy)

By14.69 - 4.501

*mn= ["•.-.-•.0-•.5-J (364- 351)+ 351

= 356 kip-ft > 297 kip-ft O.K.

kip-ft.

c. Calculate shear studs

For full composite action

•Qn = AsFy= 371 kips

Assume 3/4 inch diameter by 5 inch long studs.

Qn = 0.5Asc(f'c wc)3/4 = (0.5)(0.442) [(3)(145)]3/4= 21.1 kips (<AscFu= 0.442(60)= 26.5 kips)

Check flange thickness

tf= 0.425 > ds/2.5 = 0.3 in. O.K.


Stud Reduction Factor (S.R.F.)0.85 Jwr• •Hs

• / 2 l•J lTrr - 1} < 1.0(Nr)085 6 I5,0_

(mr) 13.0 -

Nr S.R.F. Use

1 1.13 <1.0 1.0

2 0.80 0.80

3 0.65 0.65

Assume 14 stud locations possible per 1/2 beam14 (21.1) = 295 kips

Remaining stud force = 371 - 295= 76 kips

Use twice reduction for doubled stud locations[ (2) (0.8) - 1] 21.1 = 12.7 kips

Total per 1/2 beam = 14 + 76/12.7= 14+6=20

(distributed as shown in Figure 5)

Total = 40 Studs.

II II II II II I l l I I I I I I I

BM Span

Figure 5

d. Design for deflection

Deflection after initial construction deflection

5wL4- (5)(10) [(87 - 57 + 100) ] (30)4 1728A= - -

384E Itr - (384)(29,000,000)Itr

817.-- m.

ItrSee Table 3 for Moment of Inertia, Itr computations.

with Itr for gross area, A = 0.46 in. or L/783with lower bound Itr, A = 0.56 in. or L/643

A DL = 0.13 in.ALL = 0.43 in. or L/837 O.K, (with lower bound Itr = Ilb)

The beam may be cambered for the initialconstruction deflection

( Ireq /Assumeddefiection.A= Iprovided

358A = /5--i-•) 1.0 = 0.70 in.

Camber 3/4 in,

e. Check for shear strength

Vu = [1.2(10)(87) + 1.6(10)(100)]15 / 1000= 39.7 kips

qbVn = •0.6)Fydtw= (0.90)(0.6)(36)(17.7)(0.3)= 103 kips

or from Table on page 3-31 of the LRFD Manualqb Vn = 103 kips

> 39.7 kips O.K.


Table 3

Moment of Inertia Calculation

Type n Y in?AY) y Io Ad2 Itt(in.2) (in.) ( (in.) (in.4) (in.4) (in.4)

1. Gross section 10.3 8.85 91 510 88725.0 21.95 549 36535.3 640 18.13 523 1252 1775

2. Neglecting No tensile concrete for this example. Therefore, Itr is same as for gross section.tensile concrete 1775

3. Reduced concrete area, 10.3 8.85 91 510 473 XJ'On/F,, 10.3 17.7 + 4.69 231 _3_1 473equal to

20.6 322 15.60 511 946 1457

NOTES:

1. Itr = 1457 in.4 is considered the "Lower Bound"

moment of inertia, Iib and may be found directly from

Table on Page 4-49 of the LRFD Manual.

For Y2 = 4.69 in.

W18x35 and

Y1 =

[4.69-4.501Itr = 1430 + l' •--°Y--4-'• 'J (1500-1430) = 1457 in.4

Itr = lib

2. Modular ratio, n = 9

3. Effective concrete width, b = 90 in.

4. Slab thickness, tc = 2.5 in.

5. Ac = 90 (2.5) = 225 in.2

6. Transformed concrete area, A'c = Ac/n = 25.0 in.2

•_.,Ay7. y = - -

•_.,n

8. d=y-y

9. Itr = •_fio + • A d 2

L

Figure 6

NOTE: The purpose and advantage of using the lower bound Itr value found in the LRFD Manual tables is to avoidthe above computations. If the deflections using the lower bound Itr are acceptable, the actual deflectionswill be conservatively less. Lower bound Itr is based upon the area of the beam and an equivalent concrete

area of and is applicable for full as well as partial composite action.


Example 2.

Design a composite interior girder (without cover plate) for anoffice building. See Girder B in Figure 1. The 3-inch deck ribsare oriented paralled to the girder. Girder is assumed loaded asshown in Figure 7.

P P P

I I

L 4oloFigure 7

Given:

Loads:

Span length, L = 40 ft.

Beam spacing, s = 30 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 = 50 ksi

3 inch metal deck, ribs are parallel to girder.

No shoring permitted.

Concrete slab including reinforcing steel

and metal deck 54

Framing 6

Mechanical 4

Ceiling 6

Partition 20

Total D.L. 90 psf

Live 100 psf

Live Load Reduction = 23.1 (1+ D/L)

= 23.1 (1+ 90/100)

= 43.9 %

or = 0.08 (A - 150)

= 0.08 (1200-150)

=84%

or = 40 % maximum

Use 60 psf L.L.

Solution:

1. Design for construction loads:

Assume Framing D.L. = 10 psfConcrete Weight = 50 psf as L.L.Construction L.L. = 20 psf

Pu = (10)(30)[1.2(10) + 1.6(50 + 20)] / 1000 = 37.2 kips

PL_ (37.2)(40)_ 744 kip-ftMu- 2 2

12Mu_ (12)(744)_ 198 in.3 (Minimum)Zreq- •bFy (0.9)(50)

2. Composite Girder Design

a. Trial design for required flexural strength

Pu = (10)(30)[1.2(90) + 1.6(60)]/1000 = 61.2 kips

Mu- PL2_ (61.2)(40)2 - 1224 kip-ft

For a trial size use formula in LRFD Manual pg. 4-9.12Mu (3.4)

Beam Weight = {•-Yc }d - 2 q•Fy

where q) = 0.85 and assume a/2 = 2 in.

d 12Mu (3.4) d WT Size Z I•q-Yc- a

¢ Fy(in.) (in.) (#fit) (in.3) (in.4)

21 1175 14 84 W21X83 196 1830

24 1175 15.5 76 W24X76 200 2100

27 1175 17.0 69 W27X84 244 2850

Select W24x76.

or enter Table on page 4-33 of the LRFD Manual

with Y2 = 3.5 in. (Yc - a/2)PNA = TFL (Full Composite Action)4) Mn = 1230 kip-ft > 1224 kip-ft


I' b/n '--'1C • g• , • " /. :o=:..; •x,.-•'•Xx,•.•\•.: •, ': .•.• ' • , . . • . . . .

d/2 d l 2 + Y o - a / 2 I ' - - " - - : I

2.5b 3.Ob 4.Ob/•c = + =2

Figure 8

4.82" =I ct w

' k ,"1 ' t , , '1

a s s u m e - ! - 6 " _1_ 8" _1_ 8" .!_

Tmax = Cabove + CbelowCabove= (0.85)(3.0)(120)(2.5)= 765 kips

Cbelow = 1120 - 765 = 355 kips

355depth = [(1/2)(120)] (0.85)(3.0) - 2.32 in.

Centroid from top = a/2

a_ (765)(1.25)+355(2.5+2.32/2)2 1120

= 2.01 in.

Assumption of a/2 = 2 in. O.K.

Figure 9

b. Verify flexural strength

Effective concrete widthlesser of,b = (2)(40)(12)/8 = 120 in. andb= (2)(30)(12)/2-= 360in.Use 120 in.

Design for full composite action

Tmax = AsFy = (22.4)(50) = 1120 kips (Governs)

f' =Cmax = 0.85 cAc (0.85)(3.0)[(4.0)(120)] = 1224kips

For Ac see Figure 8.

aY2 = Yc - • = 5.5 - 2.01 = 3.49 in.

qbMn= qbAsFyld+ Y2)= (0.85)(1120)I2-•+ 3.49/12

= 1225 kip-ft > 1224 kip--ft O.K

or from Table page 4-33 for Y2 = 3.5 and TFLOMn = 1230 kip-ft

c. Design for deflection

Initial deflection during construction

19PL3 (19)[(10)(30)(54 + 6)](480)3A=

384Eis (384)(29,000,000)(2100)

= 1.62 in.

Camber 1 1/2 inches.

Composite deflection using Lower Bound Itr (Ilb).

From Table on page 4-46 of LRFD Manual,

with Y2 = 3.5 D.L. = 90 psfPNA = TFL . Construction D.L. = 60 psfIlb = 4780 in4 L.L. = 60 psf

19PL3 (19)[(10)(30)(90 - 60 + 60)1(480)3ATL- 384EI- (384)(29,000,000)(4780)

= 1.07 inches or L/450

ALL= (60/90)(1.07)= 0.71 in. or L/673 O.K.

NOTE: The mooment of inertia using the gross areaequals 5510 in.


d. Shear Connectors

= AsFy For full composite action

= 1120 kips

( ' " ' 1 [ ]Reduction Factor = 0.6 [hr J[ 1 _< 1.0% /

= 0.6 -1 = 0.8

Use 0.8 for stud reduction factor.

Qn = (0.8)(21.1) = 16.9 kips (See Example 1)

1120No.- - - - - - - 67 StudsQn 16.9

67 Studs are required from Zero to Maximum Moment.

Total = 134 $•uds,

Use equal spacing for full length.

e. Check Shear

Vu --- 1.5 (Pu) = 1.5 (61.2) = 92 kips Vn = • (0.6 Fy) d tw = (0.9) (0.6) (50) (23.92) (.44)

= 284 kips > 92 kips Q.K.

NOTE: The original Steel Tips design, based upon ASD,used a W27X94 with 92 studs.

Partial Composite Action

Example 3

Design Beam in Example 1 for p•fial composite action.

SOLUTION:

a. Determine required shear studs

Estimate number of shear studs for partial composite actionusing the following approximate equation

Mu - •Mp '• ,•QnNo. [•Mn - *Mp ) Qn

Where Mu = Moment demand Mp = Steel Beam Capacity with •) = 0.85 Mn = Full Composite Beam Capacity

Mu = 297 kip-ft{Mp = •Fy Z = (0.85) (36) (66.5)/12 = 170 kip-ft{Mn = 356 kip-ft

= AsFy = 371 kips

Qn = 21.1 kips

•= [356-170) •,21.1) 0.47 (17.6)= 8.2

Try 9 studs on each 1/2 beam.

Total = 18 studs.

b. Check flexural strength

•Qn = (9)(21.1) = 190 kips

From Eq. C-I3-4 in commentary of LRFD Manual

190a = 0.85f'cb- (.85)(3.0)(90)- 0.83 in.

Y2= Yc-a/2= 5.5-0.41 = 5.09

From Table on page 4-23 of the LRFD Manual

for W18X35Y2 = 5.0 - 5.09 in.

•Qn = 187 - 190 kips ( PNA = BFL approx.)

•) Mn = 296 kip-ft (approx. equal 297 kip-ft required) O.K.

Therefore, partial composite action with 18 total studs isadequate for the required moment.


c. Check deflection

For deflection computation use the lower bound value givenin the Table on page 4-49 of the LRFD Manual.

For W18x35PNA = BFL +Y2 = 5.0 +_

4Ilb = 1170 in.

A TOTAL = (1775/1170) 0.46 = 0.70 in.ADL = 0.16 in.ALL = 0.54 in. or L/667 O.K.

Obviously any number of studs from 9 (47%) to that for fullcomposite action may be used (per 1/2 Beam Span) with theassociated increase in moment capacity and decrease in de-flection.

Location of. a/2 . effec'ive concreteb

Y2{ •m. •t 1). . • - ' - ' T • I ' - - • : t (pt s)

•...[.. (• Y1(varies - • Sgure below)

I I

Y1 = Distance from top of steel flange to any of the seventabulated PNA locations.

qn (@ point 5) + • qn (@ point 7) qn (@ point 6) =

2

qn (@ point 7) = .25AsFy

Bo$/l{Top Flange

4equ• spaces

I 1 ,, BFL

PNA Flange Locations

Figure 10

DISCUSSION

With the use of the First Edition AISC-LRFD manual,composite beam design can be simplified, particularywith partial composite action. As in the past, AISChas tried to incorporate enough tables and charts tomake repetitive design computations easier. Deter-mining preliminary beam sizes, number of weldedstuds and composite beam deflections is now verystraight forward. With a minimum of assumptions (i.e.location to the compressive force, Y2) preliminarycomparative designs can be done in minutes with theuse of the tables.

The reader is encouraged to read the LRFD ManualPART 4 (Composite Design), PART 6 (Specificationsand Commentary), especially Section I on CompositeMembers, and the other references listed. The numberof articles dealing with LRFD composite membersdesign is growing as designers are becoming morefamiliar with the method and the AISC-LRFD manual.


NOMENCLATURE

Ac

A'cAs

Asc

BFL

C

D.L.E

EcFy

Fu

Hs

IIbIoItrL

L.L.

MnMp

MuNr

P•

PNAQ.

Area of concrete (in.2)

Area of concrete modified by modular ratio (in.2)Area of steel (in.2)

Area of welded stud (in.2)Bottom of flange location

Compressive force (kips)

Dead load (psf)

Modulus of elasticity of steel (29,000,00 psi)

Modulus of elasticity of concrete (ksi)Minimum yield strength of steel (ksi)

Minimum tensile strength of steel (ksi)

Welded stud height (in.)Lower bound moment of inertia (in.4)

Moment of inertia (in.Transformed moment of inertia (in.4)

Span length (ft)

Live load (psf)

Nominal flexural strength 0dp-ft)

Plastic bending moment (kip-fO

Factored Moment (Required flexural strength) (kip-ft)Number of stud connectors in one rib at a beamintersection

Factored point load (kips)Plastic neutral axis

Welded stud shear capacity (kips)

S.R.F.

TTFL

VaVuY1

Y2

Yc

Za

b

d

dsf'c

hrn

tc

tf

twWc

Wr

wu

A

Stud reduction factor

Tensile force (kips)Top of flange locationShear capacity (kips)Shear demand (kips)

Distance from top of beam flange (in.)

Distance from top of beam to concrete flange force (in.)Total thickness of concrete fill and metal deck (in.)

Plastic section modulus (in.3)

Effective concrete flange thickness (in.)Effective concrete flange width (in.)

Depth of steel beam (in.)

Welded stud diameter (in.)

Concrete compressive strength at 28 days. (ksi)Nominal rib height of metal deck (in.)

Modular ratio (E/Ec)

Thickness of concrete above metal deck (in.)

Steel beam flange thickness (in.)

Steel beam web thickness (in.)Unit weight of concrete (lbs./cu. ft)Average metal deck rib width (in.)Factored uniform load (kip/fO

Deflection (in.)Resistance factor

,

2.

3.

4.

5.

6.

7.

REFERENCES

"Manual of Steel Construction, "First Edition, AISC, Chicago, 1986.

STEEL TIPS, "Composite Beam Design with Metal Deck," Steel Committee of California, January 1987.

STEEL TIPS, "The Economies of LRFD in Composite Floor Beams," Steel Committee of California, May 1989.

Smith, J.C., "Structural Steel Design - LRFD Approach," John Wiley & Sons, Inc., N.Y., 1991.

Salmon, C. and Johnson, J., "Steel Structures," Third Edition, Harper & Row, N.Y., 1990.

McCormac, J., "Structural Steel Design - LRFD Method," Harper & Row, N.Y.,1989.

Vinnakota, S., et al., "Design of Partially or Fully Composite Beams, with Ribbed Metal Deck, Using LRFDSpecifications," AISC Engineering Journal, 2nd Quarter, 1988.


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LRFD-Composite Beam Design

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