CE 241 - 01b Composite Beams

1

Professor Thomas A. Sabol

CEE 241– Advanced Steel Design

Composite Floor and Roof Slab Systems

Professor Thomas A. SabolDepartment of Civil and Environmental Engineering



COMPOSITE CONSTRUCTION

Welding shear connectors

(headed studs)

Headed shear stud prior to

weldingMetal deck spanning

perpendicular to beam

Almost all steel framed floors and topped roofs (i.e. those with concrete vs. those with only metal deck) are composite systems

Deck flutes – the deck spans

parallel to the flutesHeaded shear stud welded

through deck to the steel beam

Metal deck spanning

perpendicular to beam

2




Thickness of concrete slab a function of:•Structural requirements (often not a governing requirement)•Fire-resistive requirements (often governs)•Type of concrete (lightweight vs. normal weight)•Vibration requirements (benefit of added mass)•Seismic and foundation demands (need to reduce weight)

Metal deck thickness and typical spans

Concrete topping

thickness

Deck thickness

3” most common for floors. 2”

sometimes used. 1-1/2” usually

limited to non-composite roofs




Thickness of metal deck a function of:•Structural requirements (spans between beams)•Vibration requirements•Fire rating

Flute

3




Fire rating of a metal deck and concrete fill:

• Hourly fire ratings are used as a measure of the ability of the composite deck and slab to contain a fire and keep it from spreading from floor to floor.

• The “fire” is defined in ASTM E119 – it is a laboratory standard, not a fire in a real building





• For the duration of the fire test, the floor must carry the design load, not allow a 250o

temperature rise through the slab, and not permit flames or hot gasses to penetrate the assembly.

• The building code controls the number of hours required (see IBC Table 601)

• Concrete cover is often controlled by fire rating required rather than structural requirements.

4





Usually, deck fireproofing

isn’t considered economical

110 to 115

pcf

145 to 150 pcf

Very common system for Type I-A

(the most fire-resistive

construction) or I-B construction



Section I.3 of the AISC Specification addresses composite flexural members

Advantages of Composite Construction: Uses material efficiently: concrete in compression

& steel in tension (web & flange)

Reduces weight and increases stiffness compared to non-composite: deflection only 20-30% of non-composite for the same size beam


5



Advantages of Composite Construction: Reduces structural depth

Efficiently supports high live loads or high “post-composite” dead loads because these are applied after the steel beam has become composite with the concrete – takes advantage of the added strength of the composite section

Provides greater reliability because stability limit states are usually not critical




“Phases” of Composite Beams

• Pre-composite: Bare steel beam supports uncured concrete. Design of beam would be just as you learned in undergraduate steel class.

• Composite:

Beam and concrete slab work together. A shear transfer mechanism must be present.

Concrete must be sufficiently cured to resist the loads applied after composite action is available.

Composite Construction

6



“Types” of Loads

• Pre-composite - loads present before composite action can be developed

Dead (e.g., weight of uncured concrete, beam and deck self-weight)

Live (e.g., temporary construction storage)

• Composite – after composite action is available:

Dead (e.g., superimposed loads such as ceilings, floor finishes, landscaping, etc.)

Live (e.g., the building code prescribed live-load).

Composite Construction



Prior to development of composite action, total weight of uncured (i.e. wet) concrete must be carried by steel beam

Sometimes, the beam is not strong or stiff enough and it must be shored so it does not fail or it doesn’t deflect too much

SHORED CONSTRUCTION

Shoring – exact

placement varies

Beam

Metal deck and

concrete fill

7



Shoring supports the steel beam (i.e., shored beam carries “no” load) because shoring supports weight of wet concrete until composite action is available. This usually results in a smaller beam.

SHORED CONSTRUCTION

Shoring post

Note that the supporting beam

must be strong enough to

support weight of shored beam

Beam being

shored



If unshored, steel beam alone must have adequate strength (i.e., Mn) to resist all pre-composite loads (i.e. Mu) applied prior to concrete reaching 75% of its design strength, f’c

UNSHORED CONSTRUCTION

Beam

Metal deck and

concrete fill

Beam must carry uncured concrete

as a non-composite steel member

until concrete reaches 75% of f’c

8



Reasons not to shore:

UNSHORED CONSTRUCTION

High expense and difficult construction logistics make shoring almost always more expensive than potential savings from reduced beam weight in shored construction

Creep - Concrete slab always in compression which can lead to greater long-term deflections

Therefore, shored steel frame construction is almost never used. Unshored construction is usually preferred.



In many cases, the size of beam is governed by the “pre-composite” condition, but it may not be stiff enough (i.e., too much deflection).

The composite section usually possesses more than enough strength to support the superimposed composite loads.

So, to improve efficiency, we “camber” the beam (pre-deflect the beam upwards)

CAMBER

Beam

Camber – deform beam in opposite direction

of anticipated dead load deflection

9



Camber allows us to use a more efficient section

Generally, it is recommended that beams be cambered if 80% of pre-composite deflection, based on pre-composite dead load, is > 1/2" to 3/4"

Camber in 0.25 in. increments and limit to about 2.5 in. max. for 30 to 40 ft. long beams (longer beams may have more camber)

Beam

Camber – deform beam in opposite direction

of anticipated dead load deflection

CAMBER



Goal of using camber: obtain a level or nearly level slab once the pre-composite dead load has been applied and the composite action develops

CAMBER

Camber in pre-composite beam

Level slab after the concrete has been placed

10



Use a consistent topping thickness rather than trying to obtain a level beam during placement. “Level” beams run the risk of excessive deflection due to “ponding” of concrete

CAMBER



“Level” pre-composite beam

Composite beam with excessive concrete to maintain level slab



Do not over-camber

Sometimes the camber “doesn’t come out” because of fixity and over-estimation of pre-composite loads

CAMBER



In most situations, highly restrained

connections should not be cambered

Pinned connections allow the rotation

required to flatten-out a cambered beam

11



Problem locations: interior columns adjacent to longer spans, roof beams with large camber

Area around interior column won’t deflect. If camber in adjacent beam doesn’t come out, an unintended low spot at the column may result

If camber doesn’t come out of roof beams, roof drainage can be impacted unfavorably.

CAMBER

Long-span beam with excessive

camber (potential high spot)

Area around column

(potential low spot)

Roof beam with

excessive camber

(potential high

spot that

interferes with

roof drainage)



If camber doesn’t come out:

Shear connector can protrude above slab

Concrete cover may be unequal or insufficient to obtain required fire rating

CAMBER



Over-cambered beam with protruding shear connectors

Over-cambered beam with uneven concrete thickness

12



General recommendations:

Camber ≤ 0.8Δpre-composite (considers restraint)

Δpre-composite should only include those loads present before composite condition (e.g., beam self-weight, concrete slab)

When evaluating camber for cantilevered beams with and without backspan, consider actual conditions

CAMBER

Does this beam rotate?

Is this a column with significant stiffness?

Relative length of cantilever and backspan

will change apparent deformed shape

Cantilever is unlikely to exhibit

significant composite action

(concrete is in tension)

Based on “pre-composite” dead loads or

other loads you are sure will be present



Cambering uses hydraulic rams or heat. Operator has to be careful not to buckle the beam. Cambering is a trial and error process

Beam

Beam is pushed beyond elastic limit to achieve a

permanent set. When hydraulic rams are released,

the beam springs back by the elastic deformation.

The residual set is the camber.

Hydraulic ramHydraulic ram

Reference line

(a string)

CAMBER

Cambering with

hydraulic rams

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CAMBER

Heat is applied to create camber on

the opposite site of the desired

crown (i.e., upward camber).

Heat marks go on bottom side (beam’s

crown (camber) is to the top)

Cambering with heat



Pre-Composite Condition (Positive Moment)

If decking is perpendicular to the beam, it is usually assumed that the unbraced length is 0 ft.

If decking is parallel to the beam (often applies to girders) the deck isn’t oriented in its strongest direction. The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).

Composite Condition (Positive Moment)

Concrete slab braces beam; therefore, unbraced length is 0 ft.

UNBRACED LENGTH OF BEAM

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Pre-Composite Condition (Negative Moment)

Bottom flange is in compression (often applies to cantilevers). The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).

Composite Condition (Negative Moment)

Concrete is in tension and bottom flange is in compression (often applies to cantilevers). The unbraced length is then based on conventional bracing considerations (e.g., location of perpendicular beams).

UNBRACED LENGTH OF BEAM



How much slab acts as part of the composite beam? (See Section I3.1a)

Concrete stress decreases with perpendicular distance from beam. Concrete over the beam flange is under greatest stress. Nevertheless, AISC Specification assumes constant stress over assumed (effective) width with reasonable accuracy.

Concept of effective width is similar to that used for “t-beams” by ACI 318.

EFFECTIVE FLANGE WIDTHRefers to AISC 360 – Specification

for Structural Steel Buildings

Actual stress distribution

in concrete slab

Effective width with uniform

stress distribution

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How much slab acts as part of the composite beam?

be is taken on each side of the beam center line. Effective width is sum of the be values.

t is concrete topping thickness

EFFECTIVE FLANGE WIDTH

t

be1

Effective Flange Width

be2

Slab edge

(where occurs)

Bottom of

steel deck

Top of

steel deckbe2

be1

be1

be2 Slab edge

Effective

flange width

Effective

flange width



EFFECTIVE FLANGE WIDTH

1 span of beam

81

distance from centerline of beam to beam Use smallest value2

of beam to slab edge

eb

Centerline

Effective flange width = be1 + be2

t

be1

Effective Flange Width

be2

Slab edge

(where occurs)

Bottom of metal

deck

Top of

steel deck

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SHEAR TRANSFER

Headed shear stud

prior to welding

For the metal deck/concrete and the steel beam to work together effectively, adequate shear transfer must be provided

Shear connectors – studs, headed (“Nelson”) studs –most common

3/4" most common diameter

Length > 4 stud

Beam flange

Beam web

Shear

connectorShear

demand Q



SHEAR TRANSFER

Although not directly related to composite strength calculations, the deck must also be attached to the steel framing

Decking usually attached to steel beam using puddle welds or power-actuated fasteners

Puddle welds used to attach

decking to steel framing

Powder-actuated fasteners

fired through deck into steel

attach decking

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SHEAR TRANSFER

Most failures occur with slab crushing so we usually assume plastic behavior in both steel and concrete (i.e., we can fully develop Mp in the beam)

Required strength of shear connectors (V’), between point of M+

max and M = 0, is the least of the following (Section I3.2d):

V’ = 0.85 f'cAc Compressive strength of concreteV’ = AsFy Tensile strength of steel beamV’ = Qn Total nominal strength of shear studs

where Ac = area of concrete within effective slab width (in2)

As = area of steel cross section (in2)

Ac = Effective width x slab

topping thickness



SHEAR TRANSFER

Beam

Uniform load

M+maxn* = required number

of shear studs

from M+max to M0

*For most beams (e.g. those

with symmetric uniform

loads), total number of studs

required for a beam is 2n and

they are evenly distributed

along the length of the beam.

M0

If two or more studs per

flute are required, the

additional studs are added

starting at the supports

and working toward the

middle

Usually install one

stud per flute (i.e., 12”

on center)

18



SHEAR TRANSFER

Beam (Girder)

Point loads

M0

M+max

n = required number

of shear studs

from M+max to M0

If the shear diagram shows no (or very little) shear

along a portion of beam, theoretically, no shear

connectors are required. Nevertheless, minimum

spacing of shear connectors governs in this region

Shear diagram

Example of Beam with Point Loads



No -factor for strength of studs.

Nominal strength of shear connector is:

where

Asa = Area of shear connector shank (in.2)

Ec = Young's modulus of concrete (ksi)

Fu = Specified minimum tensile strength (ksi)

Rg = Varies between 1.0 and 0.75 depending on number of studs and direction of deck flutes

Rp = Varies between 1.0 to 0.6 depending on how many studs are welded in a given flute

SHEAR CONNECTOR STRENGTH (Qn)

0 5n sa c c g p sa uQ . A f ' E R R A F (Eq. I8-1)

19



“Parallel” usually

refers to girders

“Perpendicular”

usually refers to

beams


Condition Rg RP

No decking 1.0 0.75

Decking oriented parallel to the steel shape

wr/hr≥ 1.5

wr/hr < 1.5

1.0

0.85

0.75

0.75

Decking oriented perpendicular to the steel shape

Number of studs occupying the same decking rib

1

2

3 or more

1.0

0.85

0.7

0.6

0.6

0.6

hr = nominal rib height, in.wr = average width of concrete rib or haunch

(as defined in Section I3.2c), in.

Decking oriented perpendicular

to steel shape

(See User Note on page 16.1-98 of the Steel Manual for common values of Rg and Rp)




f'c Qn

(ksi) (pcf) (kips)3.0 110 17.7 3/4" A36 studs

145 17.7

4.0 110 17.7145 17.7

Most common stud diameter and concrete strength

SHEAR CONNECTOR STRENGTH (Qn) for one stud per rib in the “weak” position (see next slide)

Table I3-21 of the

Specification gives values

for other conditions

20



Stiffening rib in

deck flute

Weak and Strong Stud Positions

Direction of

shear

Direction of

shear

Stud in the

“strong” position

Stud in the

“weak” position

Deck flutes usually have a stiffening rib that requires stud be located to one side or the other

Stud strength can be a function of position“Weak” position typically assumed unless otherwise

indicated (since specific placement not usually specified)

“More” concrete

to support stud



2” min

3” max

1.5” min

Ignore concrete in the flute (rib) unless

the rib is parallel to beam (e.g. a girder

condition -- typically about 50% of the

concrete is available in this case)

Metal deck flutes (ribs)

perpendicular to beams

0.5” min


parallel to beam (girder)

Partial Framing Plan

Concrete Cover and Thickness


perpendicular to beams

21



For uniform loads, a uniform spacing of studs shows similar performance compared with a spacing that follows the statical shear distribution (i.e. VQ ÷ Ib) …therefore we use uniform spacing for uniform loads

For Point Loads - Number of connectors between load and nearest point of zero moment must be sufficient to develop M+

max at point load.

NUMBER, SPACING AND COVER

Moment Diagram M+max

Provide required number of

connectors over this distance

P Key concept

for girder

design

M = 0



In areas of zero (low) shear, the maximum spacing of studs would usually govern


M = 0

M+max

Provide required number of

connectors over this distance

P

Key concept

for girder

design

Moment Diagram

Shear Diagram

P

Provide shear connectors no

further apart than 8 x slab

thickness

22



Minimum center to center spacing along beam is 6(where is stud diameter) and minimum center to center transverse spacing along beam is 4. If “formed slab” (i.e., no steel deck): 4 either direction.

Studs require min. 1" lateral concrete cover (except where deck is used)

of stud < 2.5 tf if not over webs


6min

4min

Studs

Top of beam flange

May stagger studs if beam is

too narrow – this can squeeze

in a few more studs over a

given distance.Multiple Studs in a Row (or Flute)



When composite action is small or a beam has a point of “zero” shear (e.g., the middle segment of a girder loaded at its third points), provide shear connectors spacing governed by maximum spacing

Maximum spacing < 8 x total slab thickness (36 in. max.) to prevent vertical separation between slab and beam flange when slab goes into compression


6 (min) -

4min

Studs

Top of beam flange8 x toal slab

thickness

(max)

23



When decking runs perpendicular to the beam, studs are installed based on the flute spacing in the deck, typically 12 in. on center

When decking runs parallel to the beam (e.g., a girder), the stud spacing is independent of the deck flute spacing and 6 x stud diameter would govern


If flutes are perpendicular

to beam, flute spacing

establishes stud spacing

If flutes are parallel to beam,

stud spacing is independent of

flute spacing



If fully composite section has moment strength Mn*, but you only need Mn* where < 1.0, you canprovide enough studs just to reach Mn*

A partially composite beam will have fewer than the maximum number of studs required to develop the full composite strength of the beam and slab system: same beam and slab but fewer studs = lower moment strength

Partially composite beams are the most common in actual building design

It is recommended that > 25%

PARTIALLY COMPOSITE BEAMS

24



Many designers will often specify beam studs at 12 in. on center, even if design requires fewer, because:

• Reduces construction errors by keeping things consistent

• Provides added flexural strength at a nominal cost (i.e. studs are less expensive than strengthening the beam later)

PARTIALLY COMPOSITE BEAMS



Flexural capacity of composite section may be governed by:• Tensile strength of beam, • Compressive strength of concrete slab or • Strength of shear connectors.

The least of these values establishes the limit state that governs the design

Composite section cannot transfer more shear than can be developed by the shear connectors – if the shear connectors have less strength than the concrete slab or the beam, then this will establish the limit on the flexural state of the composite section.

FLEXURAL CAPACITY Key

concept

Usually, when it is a

partially composite

beam

25



Plastic stress

distribution signifies

section can achieve Mp

Wide flange sections

with Fy ≤ 50 ksi satisfy

this, so you may assume

entire section yields

hc = distance between toes of fillet d-2k

yfw

c

FE

th

76.3

FLEXURAL CAPACITY

If web is slender and in compression, it may buckle.

Use plastic stress distribution in the composite section if:

h c

The flange and the web do not

meet at a right angle due to the

rolling process that creates the

wide flange shape. There is a

small radius at the intersection

called a “fillet.”k

k

d



Use elastic stress distribution if:

Superimpose elastic stresses and effect of shoring must be considered.

The elastic stress distribution requirement applies most frequently when built-up members are used (e.g. plate girders) and relatively thin plates are used.

3 76 where 0 9c

yfw

h E. ; .Ft

FLEXURAL CAPACITY Not a common condition

-- you must calculate

stress distribution

based on level of strain

26



0.85f’c

Fy

Calculating the moment capacity of a composite section is a function of plastic neutral axis (PNA) location.

Possible plastic neutral axis locations:•Neutral axis in the concrete slab •Neutral axis in top flange of beam•Neutral axis in web of steel beam

T (from steel beam)

C (from concrete)

b = effective

widthConcrete in tension

(doesn’t contribute

to flexural strength

but does influence

internal moment arm

dimension)

PNA

FLEXURAL CAPACITY

PNA in slab

Steel beam

weaker than

concrete slab

Stress

Distribution

t = total slab

thickness

Internal

moment arm



Plastic Neutral Axis in Concrete Slab:

(Note: If a > t, then PNA is not in slab and you need to revise your assumptions and recalculate the flexural capacity.)

2 2

0 9

n p s y

d aM M A F t

.

t

d

a

d/2 + t - a/2

a/2

d/2

T = AsFy

C = 0.85f’cab

b = effective widthConcrete in

tension (doesn’t

contribute to

flexural strength

but does influence

internal moment

arm dimension)

PNA

FLEXURAL CAPACITY

T must equal

C to satisfy

equilibrium

0.85f’c

Fy

Internal

moment arm

27



If deck is used, the methodology to calculate the flexural capacity is the same except the slab is elevated above the beam flange by the height of the deck (hr). Adjust geometry.

FLEXURAL CAPACITY

(Note: If a > hc, then PNA is not in slab and you need to limit the compression strength to C = 0.85f’cbhc. If PNA is no lower than t, you can assume PNA is “in the slab” to calculate Mn.)

td

a

d/2 + t - a/2

a/2

d/2

T = AsFy

C = 0.85f’cabPNA

0.85f’c

Fy

hc

hr

Concrete in

compression

Concrete in tension

(below PNA)

Deck



Fy

0.85f’c

Fy

Neutral Axis in Top Flange of Beam:

As a first guess, we assume neutral axis is at base of flange:

FLEXURAL CAPACITY

0 85 c f y

s f y

C . f' b t AF

T A A F

t C

T

0.85f’cbt

b

y

d-y

FybfybfN.A.

Fy(As - bfy)

y is depth of

steel in

compression

Not all of

the steel is

in tension

Cconc

Csteel

Tsteel

Adjust “t” as required

for concrete topping

over steel deck

28



Fy

0.85f’c

FyTsteel

Plastic Neutral Axis in Top Flange of Beam:

C = Cconc + Csteel = Tsteel = T

But if C ≠ T with assumed PNA at base of flange, then our assumption is wrong and PNA is not at base of flange.

FLEXURAL CAPACITY

t C

T

0.85f’cbt

b

y

d-y

FybfybfP.N.A.

Fy(As - bfy)

Cconc Csteel

Note: T = C to

satisfy equilibrium



Fy

0.85f’c

FyTsteel

Plastic Neutral Axis in Top Flange of Beam:

Setting C = T and solving for Mn:

FLEXURAL CAPACITY

0 85

0 85

2

0 9 0 85 22 2 2

c y f y s y f

y s c

y f

n p c y f y s

. f ' b t F b y F A F b y

F A . f' bty

F b

t y dM M . . f' bt y F b y F A y

C

T

t C

T

0.85f’cbt

b

y

d-y

FybfybfP.N.A.

Fy(As - bfy)

Cconc

Csteel

29



Neutral Axis in Web of Steel Beam

Where Cs = Fy[(bftf) + tw(d-2tf-x)]

Using equilibrium and summing moments about the PNA, one can solve for Mn (derivation not shown).

FLEXURAL CAPACITY

0.85f’c

Fy

Fy

t

C

T

b

bf

0.85f’cbt = Cconc

Fybftf = Cflange

Fy(PNA - tf)tw = Tweb

Fytw(d-2tf-x) = Cweb

d

x

tw

tf

PNAFybftf = Tflange



0.85f’c

Fy

Fy

Neutral Axis in Web of Steel Beam

Note that ΣQn = Cconc since this is the force that must be transferred across the slab(deck)/beam interface

Note: PNA = 0.5(AsFy+Cconc) = Fy[As/2 +tw(PNA-d/2)]

FLEXURAL CAPACITY

t

C

T

b

bf

0.85f’cbt = Cconc

Fybftf = Cflange

Fy(PNA - tf)tw = Tweb

Fytw(d-2tf-x) = Cweb

d

x

tw

tf

PNAFybftf = Tflange

30



“Inadequate slab” means that

strength of concrete slab is less

than tensile strength of steel

Limit States of Composite Beams Based on Relative Strength of Components

FLEXURAL CAPACITY



(Assuming Beams Without Shoring)

The bare steel section must have enough strength to resist weight of uncured concrete and other pre-composite loads (i.e., pre-composite dead and live loads).

If partially cured composite section will support incidental construction loads, it is recommended that an allowance for 10 – 20 psf be included as a “pre-composite” live load.

FLEXURAL CAPACITY

31



If the decking is judged not provide bracing for the compression flange, Lb must be considered in designing the beam. Otherwise, Lb = 0.

To avoid yielding the beam, it is recommended that for “pre-composite” condition:

Mu < FyZ

where:

Mu = Required strength (i.e., factored moment) due to loads from concrete + weight of other pre-composite loads

= 0.9

FLEXURAL CAPACITY

This is Z for the

steel beam alone



“Transformed moment of

inertia” substitutes steel for

concrete at the ratio of n =

(Es/Ec) ≈ 10 and then calculates

Itr using parallel axis theorem

Lower Bound Moment Of Inertia

Moment of inertia will vary with the applied moment and location of the neutral axis because of the amount of uncracked concrete

Use of a transformed moment of inertia (Itr)using elastic theory will underestimate deflections by 15% to 30%

DEFLECTION ESTIMATE

32



Lower bound moment of inertia (ILB) is moment of inertia at required strength level (i.e., it uses only the provided shear transfer (Qn) and enough of the slab to balance Qn)

AISC recommends that effective moment of inertia (Ieff) be about 0.75Itr

DEFLECTION ESTIMATE



Transformed Moment Of Inertia

DEFLECTION ESTIMATE

Effective width of concrete is

transformed (reduced) to equivalent

width of steel based on ratio of n = Es/Ec

(i.e., beff/n)

Transformed moment of inertia (Itr) of the composite

section can be calculated using parallel axis theorem

using the transformed properties

If some of the concrete in the decking

is in compression, an effective width

of 50% may be used when doing the

transformation (only applies when

flutes are parallel to the beam (e.g.,

girders)

beff

33



Lower Bound Moment Of Inertia

Full elastic analysis involves many assumptions, so, as an alternative, AISC offers the use of Lower Bound Moment of Inertia

DEFLECTION ESTIMATE

2 23 3 12n

LB s s ENA ENAy

QI I A Y d d d Y

F

d

d1

Equivalent concrete area =Qn/Fy

YENA

ENA

a/2

Location of effective

Concrete flange force Qn

C

T d3

Distance from resultant

tension force (for full

tension yield) to top of

steel (usually, d/2)



YENA = distance from bottom of beam to elastic neutral axis (ENA)

=

DEFLECTION ESTIMATE

3 3 12ns

y

ns

y

QA d d d

F

QA

F

d

d1

Equivalent concrete area =Qn/Fy

YENA

ENA

a/2

Location of effective

Concrete flange force Qn

C

T d3

34



Design Method Summary

1. Determine pre-composite and composite loads for strength evaluation (factored loads) and serviceability evaluation (unfactored loads)

2. Determine required strength of beam supporting pre-composite loads (using factored loads)

3. Check deflection of pre-composite beam using unfactored dead loads and establish initial camber recommendation (at most, 80% of DL deflection)

4. Check beam under composite loads (factored) and determine required number of studs

5. Check deflection under composite loads (unfactored)

6. Summarize design recommendations



Center of beam spacing

is 10’/2 = 5’

Design a composite beam to satisfy the following requirements.

Assume beams at 10' center to center, L = 36’, 4" lightweight concrete + 3" deck (t = 7”). Steel is ASTM A992 (i.e. Fy = 50 ksi)

DL = 0.78klf, LL = 1.2klf , Fy = 50 ksi, f'c = 3 ksi

EXAMPLE 1

klfuw 862216178021 .).(.).(.

kftklf

uM 74628

36862 2

.)(.

Factored Loads:

Moment:

Governs2 36 12

1088

2 5 12 120

( )b "

b ( ) "

Effective Flange Width: L = 36’

35



EXAMPLE 1

Try a W18 x 35 As = 10.3 in2

Where is plastic neutral axis? Assume that PNA is in slab.

.in87.1

.)in108()ksi3(85.0)ksi50(in.3.10

'85.02

bf

FAa

c

ys

t

d

ad/2 + t - a/2

a/2

d/2

T = AsFy

C = 0.85f’cba

fy

b = effective

widthConcrete in

tension

N.A

.

Since a < t, the PNA is in concrete slab

Note: really

it is because

a < t-3”h

ch

r

Deck

Concrete in

compression



Composite section may be oversized because Mn is much greater than Mu. We may be able to reduce the steel section, but we still have to check its ability to support the pre-composite loads (i.e. the loads prior to development of composite action).

What is available strength assuming full composite action?

EXAMPLE 1

nM2 2

17 7 1 87 10 9 10 3 50 7

2 2 12

576 1 462 7

p s y

ft k ft k

d aM A F t

. .( . ) . ( )

. . O.K .

36



Governs (assuming you

use only the minimum

number of studs – no

longer assuming full

composite action)

Design of Studs

Use Qn = 17.7k for f'c = 3 ksi (weak position) lightweight concrete and =3/4"

What is magnitude of force used to establish the number of studs for shear transfer? Select minimum of:

(a) 0.85 f'cab = 0.85 (3) (1.87) (108) = 515k

or

(b) AsFy = 10.3 (50) = 515k

or

EXAMPLE 1

kx

Qinin

ftinkft

n 3372

2871

72717

127462.

..

. /

(c)22a

td

Mu

This equation only

applies if the PNA is

above the top of the

beam flange!



This beam is an example of a partially composite section. A fully composite section would require sufficient studs to transfer at least 515k but only 372k is required to transfer shear due to design loads. Amount of composite action is about 72% (i.e. 372/515 = 0.72)

Minimum number of required studs on each side of beam centerline is:

Say, 22 studs each side of beam centerline or a total of 44 studs

Note that actual Qn is 22(17.7k) = 378.4kThis value is used later in the deflection calculations.

EXAMPLE 1

372 3

21 6517 7min

. kn .

. k

37



Compare strength of W18x35 against loads present before composite section is developed (i.e. pre-composite condition) :

2

52 10 0 52

0 78 0 52 0 26

20 101 2 0 52 1 6

1000

0 94

0 94 36152

8249 152

psf klfconc

klf

psfklf

u

klf

klfft k

u

ft k ft kp

w ' .

DL' . . .

x 'w . ( . ) .

.

. ( ')M

M

OK

A W18x35 is OK for non-composite loads

Assumed construction

live load

EXAMPLE 1

Weight of concrete slab

Superimposed

dead load (not

used in this check

because it is

applied when the

composite

strength is

available)



Typically, anything

below L/240 for DL

only is considered too

flexible

Service load deflection before composite section forms:

4

0.52

5(0.52)(36) 17281.29"

384(29000)(510)

36' 12325 240

1.29 325

klfw

x LOK

A W18x35 has significant pre-composite dead load deflection, but we can camber out most of this (or we could upsize the beam).

EXAMPLE 1 Camber is based

on DL only in

this case

38



EXAMPLE 1

Beam

Camber 1.0”

How much camber is required?Normally, we camber a section if the pre-composite deflection is greater than about 0.5 to 0.75 in. And we camber in 0.25in. increments.

Therefore, camber in this case is at most 1.25 in. (Δpre-composite =1.29 in. initial deflection).

AISC recommends camber be based on 0.8 Δpre-composite = 1.03”, so use 1 in. of camber



2 2

3 3 1

3 3 1

2

2

17 7 378 4 1 8710 3 17 7 7

2 50 2378 4

10 350

15 17

nLB s s ENA ENA

y

ns

y

ENA

ns

y

QI I A Y d d d Y

F

QA d d d

FY

QA

F

( . ) . .. .

..

. "

d3

= d

/2

d1

Equivalent concrete

area =Qn/Fy

YE

NA

ENA

a/2

Service Live Load Deflection After Composite Section

Lower Bound I:

EXAMPLE 1

As = 10.3 in2

Is = 510 in4

d = 17.7 in.

ΣQn = 378.4 k

d1 = tslab – a/2

tslab = 7 in.

a = 1.87 in.

a

t sla

b

39



22

4

4

17 7 378 4 1 87510 10 3 15 17 17 7 7 15 17

2 50 2

1481

5 1 2 36 1728384 29000 1481

36 121 06 1 2

360 36036 12

408 3601 06

LB

LL

. . .I . . . .

in.

( . )( )( )( )

L x. " Max.Deflection . "

xO.K .

.

Substitute YENA into equation for Lower Bound Moment of Intertia:

EXAMPLE 1

Building code

typically limits LL

deflection to

L/360

AISC recommends

limiting deflection

to less than 1” for

typical spans



Final design recommendations are:

W18x35 with 44 - 3/4”diameter studs. Camber beam 1.0”

Note: For a complete design, you must list the beam size, the total number of studs, and the amount of camber.

EXAMPLE 1

This is

important!

40



Design composite beam for the following loads using AISC Manual

L = 45' Beams spaced 10’o.c.Fy = 50 ksi f'c = 4.0 ksi3" deck + 4.5" hardrock ( = 145 pcf)

DL slab= 0.075ksf

DL beam weight = 0.008ksf (assumed)DL all other = 0.010ksf (ceiling, floor, MEP, etc.)

LL = 0.1ksf (unreduced)

Line loads on beamTotal dead load = 0.093ksf (10 ft) = 0.93klf

Total live load = 0.10ksf (10 ft) = 1.0klf

EXAMPLE 2



Construction loadsConstruction dead load = 0.083ksf (10 ft) = 0.83klf

Construction live load = 0.020ksf (10 ft) = 0.20klf

Determine required flexural strength

EXAMPLE 2

2

1.2(0.93 ) 1.6(1.0 ) 2.72

2.72 (45 )688

8

klf klf klfu

klf ftft k

u

w

M

Determine beffThe effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline

Concrete slab

plus beam

weight

41



Note use of “2” to

account for beff on

both sides of beam

EXAMPLE 2

Determine beff(1) One-eighth of the beam span

(2) One-half the distance to centerline of adjacent beam

(3) One-half the distance to edge of slab (N.A.)

45(2) 11.3

8

ftft

10(2) 10.0 controls

2

ftft



Using the figures on pages 3-30 and 3-31 and the tables starting on page 3-156 of the AISC Steel Manual, we can significantly simplify the effort required to design the beam.

EXAMPLE 2

Different locations for PNA

in beam top flange

Different locations for PNA

in beam web

TFL assumes

compression

force is in

concrete only

Y2 = tslab – a/2

42



Using the figures on pages 3-14 and 3-15 and the tables starting on page 3-158 of the AISC Steel Manual, we can significantly simplify the effort required to design the beam.

EXAMPLE 2

…



Calculate moment arm for concrete force measured from top of steel shape (Y2):

Assume a = 1” (this is often a reasonable first guess)

Enter Manual Table 3-19 with the required strength and Y2 = 7.0in. Select a beam and neutral axis location that indicates sufficient available strength

Select a W21x50 as a trial beam.

When PNA Location 5 (BFL), this composite shape has an available strength of

Mn = 769ft-k > 688ft-k OK

EXAMPLE 2

12 7.5" 7.0"

2 2slab

aY t

This is our required

strength for the

composite beam

43



From Table 3-19:

EXAMPLE 2

Y2 = 7.0 in.



EXAMPLE 2

Check the beam deflections and available strength

Check the deflection of the beam under pre-composite conditions, considering only the pre-composite dead loads as contributing to the deflection (i.e. no pre-composite live load)

Limit Δpre-composite to a maximum of 2.5 in. to facilitate concrete placement

From Manual Table 3-20, a W21x50 has Ix = 984in4, so the member does not satisfy the deflection criterion under construction loads

3 3

44 4 /5 5(0.83 )(45 ) (1728 )

1060384 384(29000 )2.5

klf ft in ftinDL

req ksi in

wI

E

44



EXAMPLE 2

Check the beam deflections and available strength

Ix of beam

(only)

(Table 3-20)



EXAMPLE 2

Using Manual Table 3-20, revise the trial member selection to a W21x55, which has Ix = 1140in4

Check selected member strength as an unshored beam under construction loads assuming adequate lateral bracing through the deck attachment to the beam flange.

Calculate the required strength

2

1.4 1.4(0.83 ) 1.16

1.2 1.6 1.2(0.83 ) 1.6(0.2 ) 1.32

1.32 (45 )331

8

klf klf

klf klf klf

klf ftft k

u

DL

DL LL

M

Governs

45



EXAMPLE 2

The design strength for a W21x55 is 473ft-k > 331ft-k OK

For a W21x55 with Y2 = 7.0in, the member has sufficient available strength when the PNA is at Location 6 and ΣQn = 292kips

Mn = 767ft-k > 688 ft-k OK

Mp

Y2 = 7.0 in.

ΣQn

(Table 3-19)

(Table 3-19)



Check a (were we right to assume a = 1 in?)

0.716in < 1.0in OK (we are close and required concrete compression area is less than we assumed)

Check live load deflectionΔLL= ℓ/360 = [(45ft)12in/ft]/360 = 1.5in

Lower bound moment of inertia for composite beams is tabulated in Manual Table 3-20.

In some situations

you need to go back

and recalculate

strength

EXAMPLE 2

' /

2920.716

0.85 0.85(4 )10 (12 )

kipsinn

ksi ft in ftc

Qa

f b

46



EXAMPLE 2

For a W21x55 with Y2 = 7.0in and the PNA at Location 6, ILB = 2440in4

1.3in < 1.5in OK

3 3

4

4 4 /5 5(1.0 )(45 ) (1728 )1.3

384 384(29000 )2440

klf ft in ftinLL

LL ksi inLB

wEI

(Table 3-20)

Y2 = 7.0 in.

Note that PNA is in the web



EXAMPLE 2

Determine if beam has sufficient shear strength

Determine the required number of shear stud connectors

Using perpendicular deck with one ¾ in. diameter weak stud per rib (per foot) in normal weight 4ksi concrete, Qn = 17.7k/stud

Total number of shear connectors = 2(17 studs) = 34 studs

45(2.72 ) 61.2

2234 61.2 OK

ftklf k

u

k kn

V

V

/

29217 studs on each side of the beam

17.7

kn

k studn

QQ

47



EXAMPLE 2

Check spacing of shear connectors

Since each flute is 12in, use one stud every flute, starting at each support, and proceed for 17 studs on each end of the span (checking to make sure maximum spacing is satisfied).

(Usually, the designer would simply require one stud per foot without trying to save the 4 or 5 extra studs for a 22.5 ft half span.)

6dstud < 12in < 8tslab, therefore shear stud spacing requirements are met

The studs are to be 5in. long, so they will extend a minimum of 1.5in. into the slab



EXAMPLE 2

Final design:

Use W21x55 beam with minimum of 34 studs and camber 1.75 in.

(Note:

(So, use camber of 1.75 in.)

4

4max21 55

10600.8(2.5 ) 1.86

1140

inreq in in

inW x

I

I

Documents

CE 241 - 01b Composite Beams