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Lecture 7 - Page 1 of 10

Lecture 7 Composite Steel Beams

Steel beams are usually used to carry a metal deck-supported concrete slab. Innon-composite construction, the beam does NOTinteract structurally with theslab the slab is simply dead weight. This is because the slab is not adequately

bonded to the beam.

Non-Composite Construction

The word composite means 2 or more different materials. In compositeconstruction, the slab is adequately bonded to the steel beam by means ofheaded shear studs resulting in a composite beam. The concrete acts like alarge flange in compression, while a much greater portion of the steel beam

acts in tension. The result is a VERY efficient beam as much as 40% to 60%lighter weight steel than non-composite.

Composite Construction

Headed shear studswelded thru metal deck tobeam flan e


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Notice that the neutral axis (N.A.) in the non-composite beam is located in themiddle of the section. This indicates that half of the beam section is in tensionand half is in compression.

In the composite section, the compression is carried ENTIRELY by the concrete,while the tension is carried by the beam.

Composite Design per LRFD

AISC Spec Ch. I3 (p. 16.1-83) AISC Part 3 Table 3-19 (p. 3-156 thru 189) AISC Part 3 Table 3-21 (p. 3-207)

From a side view, a composite beam looks like the following:

Ycon

Ten.

Comp.

N.A

Non-Composite Section Composite Section

N.A

Ten.

Comp.

d

Shear stud

ConcreteMetal Deck

Beam


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For LRFD composite design, the following terms are used:

where: b = Effective slab width (from AISC Spec. p. 16.1-83)

8

BeamSpan

= smaller of2

gBeamSpacin

Distance to edge of slab

a = Effective concrete thickness

2 for initial trial size

=bf

FA

c

ys

'85.0

As= Cross-sectional area of beam

= from properties

fc= Specified concrete compressive strength= 4 KSI (usually)

Y2 = Ycon-2

a

Y2

b = Effective slab width

a Ycon


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The best way to see how the design of a composite beam is accomplished is thruan example problem:

Example 1 (LRFD)GIVEN: A floor structure using A992 steel beams and the following

superimposed SERVICE loads:

Service live load = 75 PSF Service dead load (not incl. beam weight) = 63 PSF

In addition, use diameter shear studs, 4 concrete over 1 metaldeck and normal weight concrete fc= 4000 PSI.

REQUIRED: Design the lightest weight W14 beam A using compositeconstruction.

Step 1 Determine factored maximum moment Mu:

wu= 1.2D + 1.6L= 1.2[8(63 PSF + 30 PLF)] + 1.6[8(75 PSF)]= 1601 PLF= 1.6 KLF

Mu=8

2Lwu

=8

)'30(6.1 2KLF

= 180 KIP-FT

Beam A

30-0

3@8-0=24-0

Assumed beam weight


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Step 2 Determine Trial approx. beam weight:

Approx. Beam weight =

+ 2285.0

)/"12(4.3

a

Y

d

F

ftM

cony

u

Assume d = 14 (since using W14)a = 2Ycon= 4

Approx. Beam weight =

+

2

"2"4

2

"14)50(85.0

)/"12(1804.3

KSI

ftFTKIP

= 17.28 PLF

From AISC Table 3-19 p. 3-184 TRY W14x22

Step 3 Determine concrete flange force, Qn:

Qn= AsFy

where: As= Area of beam

= 6.49 in2

(from properties for W14x22)

Qn= (6.49 in2)(50 KSI)

= 324.5 KIPS

(NOTE: From AISC Table 3-19 p. 3-184 the value of Qncanbe found looking at Y1 = 0.000 Qn = 325 KIPS)


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Step 4 Determine effective concrete slab width, b:

8

BeamSpan=

8

)/"12('30 ft= 45 USE

b = smaller of2

gBeamSpacin=

2

)/"12('8 ft= 48

Distance to edge of slab = N/A

Step 5 Determine Y2 for usage in AISC Table 3-19 p. 3-184:

Y2 = Ycon-2

a

where a =bf

FA

c

ys

'85.0

=)"45)(4(85.0

325

KSI

KIPS

= 2.12

Y2 = 4 - 2

"12.2

= 2.94

Use Y2 = 3

Step 6 Determine required beam size from AISC Table 3-19:

W14x22Y2 = 3

Qn= 325 KIPS

Design strength in flexure = 240 KIP-FT > 180 KIP-FT

See Table 3-19 Below


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Page 3-184


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Step 7 Determine number of shear studs required:

Number of studs required =)(

2

studn

n

Q

Q

where: Qn= AsFy= 325 KIPS

Qn(stud)= Nominal horz. shear strength of stud= From AISC Table 3-21 p. 3-207

= 17.2 KIPS

Number of studs required = studperKIPS

KIPS

__2.17

)325(2

= 37.8 studs

Use 38 dia. studs

Step 8 Check beam shear at coped end:

Assume beam is coped 1

Factored beam end reaction =2

)'30(6.1 KLF

= 24 KIPS

13.7

1

W14x22 beam

Girder

Normal wt. conc. fc= 4 KSIDeck perpendicular1 weak stud per rib


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Set up a ratio of vVnfor the full W14x22 beam section to thereduced beam section:

depthduced

V

depthFull

V nvnv

_Re_

=

From AISC p. 3-67 vVn= 94.8 KIPS for full depth

)"5.1"7.13("7.13

8.94

= nv

VKIPS

vVn= 84.4 KIPS > 24 KIPS for reduced section OK

Step 9 Draw summary sketch:

30-0 span

W14x22 A992 beam

38 Dia. shear studs weldedthru metal deck along center ofbeam spaced evenly along lengthof beam


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Example 2GIVEN: The beam from Example 1. All loads and other conditions arethe same.REQUIRED: Design lightest weight W14 Beam A using NON-COMPOSITEconstruction.

Step 1 Determine Mu:

Mu= 180 KIP-FT

Step 2 Design lightest weight W14 beam:

From AISC ZxTable p. 3-18:

Use W14x34 bMpx= 205 KIP-FT > 180 KIP-FT

(Note: The W14x34 non-composite beam is 55% heavierthan a W14x22 beam that is used as a composite beam.)

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