Chapter 9: Composite Construction - Steel Design 4300401steeldesign401.webs.com/CN Chap09 -...

9.1

Chapter 9: Composite Construction The following information is taken from “Unified Design of Steel Structures,” Second Edition, Louis F.

Geschwindner, 2012, Chapter 9.

9.1 Introduction

Any structural member in which two or more materials having different stress-

strain relationships are combined and called upon to work as a single member may

be considered a composite member.

• When a concrete slab is supported by steel beams, and there is provision for

shear transfer between the two, the result is a composite section.

AISC Specification Chapter I provides rules for the design of composite members,

including encased beams, composite beams, encased columns, and filled columns.

• Typical composite members are depicted in Figures 9.1 and 9.2 (p. 313) of the

textbook.

In recent decades it has been shown that a strengthening effect can be obtained

by connecting steel beams and reinforced concrete slabs together to act as a unit

in resisting loads.

• Steel beams and concrete slabs joined together compositely can often support

33% to 50% more load than could the steel beams alone in non-composite action.

Composite beams were first used in bridge design in the United States around

1935.

• In 1944 the American Association of State Highway and Transportation

Officials (AASHTO) adopted its first set of specifications for highway bridge

construction.

• The use of composite bridge decks is commonplace today throughout the United

States.

- In these bridges, the longitudinal shears are transferred from the stringers

(beams) to the reinforced-concrete slab with shear connectors, causing the

slab to assist in carrying the loads (ref. Figure 9.3, p. 313 of the textbook).

The first approval for composite building floors was given by the 1952 AISC

Specification.

• Composite building floors, common today, consist of a reinforced-concrete slab

on a steel beam with shear connectors.

- A formed steel deck is used for almost all composite building floors.

9.2

9.2 Advantages and Disadvantages of Composite Beam Construction

The floor slab in composite construction not only acts as a slab for resisting the

live loads, but it also acts as an integral part of the beam.

• The floor slab serves as a large cover plate for the upper flange of the steel

beam, increasing the strength of the beam by providing lateral stability.

An advantage of composite floors is that they make use of the high compressive

strength of the concrete by putting a large portion of the slab in compression.

• At the same time, a large portion of the steel beam is kept in tension, more

than is normally the case for steel-frame structures.

- The result is less steel tonnage required for the same loads and spans.

• Composite sections have greater stiffnesses than non-composite sections.

- As a result, deflections are smaller – possibly only 20% to 30% as large.

- The ability of the structure to take an overload is greater than for a non-

composite structure.

An additional advantage of composite construction is the possibility of having

smaller overall floor depths – of particular importance for tall buildings.

• Smaller floor depths permit reduced building heights, with the consequent

advantage of lower costs for walls, plumbing, wiring, ducts, elevators, and

foundations.

• Reduced beam depths save in fireproofing costs since a coat of fireproofing

material is provided on smaller and shallower steel shapes.

The only disadvantages of composite construction may be the added cost for

shoring and shear connectors.

• The increased strength and reduction in the required steel weight normally

offset the added cost of the shear connectors and this increased cost of the

shear connectors is usually not a true disadvantage.

9.3 Shored Versus Unshored Construction

Two methods of construction are available for composite beams: shored and

unshored construction.

• The difference between the two approaches is how the weight of the wet

concrete is carried.

After the steel beams are placed, the concrete slab is formed and poured.

• The formwork for the concrete, the wet concrete and other construction loads

must be supported by the beams or by temporary shoring.

9.3

- If no shoring is used, the steel beams must be designed to support all of

these loads.

• Most specifications say that after the concrete has gained 75% of its 28-day

strength, the section has become composite and all loads thereafter may be

considered to be supported by the composite section.

When shoring is used, it supports the wet concrete and other temporary

construction loads – but not the weight of the steel beams.

• When the shoring is removed (after the concrete has gained 75% of its 28-day

strength), the weight of the slab is transferred to the composite section, not

just to the steel beams.

• When shoring is used, it is possible to use lighter, less expensive, steel beams.

- However, the savings in steel is not likely to offset the extra cost of the

shoring.

The usual decision is to use heavier steel beams and do without shoring for the

following reasons.

1. Apart from reasons of economy, the use of shoring is a tricky operation,

particularly where settlement of the shoring is possible.

2. The ultimate strengths of composite sections of the same sizes are the same,

whether shoring is used or not.

• If lighter steel beams are selected for a particular span because shoring is

used, the result is a smaller ultimate strength.

3. The use of shoring causes a distribution of stress in the slab and beam that

results in substantial creep.

4. In shored construction, cracks occur over the steel beams requiring the use of

reinforcing steel in the concrete slab.

Shored construction does offer some advantages compared with unshored

construction.

1. Deflections are smaller because they are based on the properties of the

composite section.

2. It is not necessary to make a strength check for the steel beams for the initial

wet concrete load condition.

9.4 Effective Flange

There is no accurate way to know how much of the concrete slab actually acts as

part of the beam.

9.4

• If the beams are closely spaced, the bending stresses in the concrete slab will

be fairly uniform across the compression zone.

• If the beam spacing is large, bending stresses will vary significantly across the

concrete slab.

- The farther a particular part of the concrete slab (a.k.a. the flange) is away

from the steel beam, the smaller will be its bending stresses.

The AISC specifications address this issue by replacing the actual slab with an

“effective slab” that has a constant stress.

• This “equivalent slab” is thought to support the same total compression that is

supported by the actual slab.

• Section I3.1 of the AISC Specification states that the effective width of the

concrete slab is the sum of the effective widths for each side of the beam

centerline, each of which shall not exceed:

1. One-eighth of the beam span, measured center-to-center of supports for

both simple and continuous spans.

2. One-half of the distance from the beam center line to the center line of the

adjacent beam.

3. The distance from the beam center line to the edge of the slab.

• AASHTO requirements for determining effective flange widths are somewhat

different.

9.5 Strength of Composite Beams and Slab

The nominal flexural strength of a composite beam in the positive moment region

may be controlled by any of the following.

• The plastic strength of the steel section.

• The strength of the concrete slab.

• The strength of the shear connectors.

• Web buckling may also limit the nominal strength of the member if the web is

very slender and if a large portion of the web is in compression.

Little research has been done on the subject of web buckling for composite

sections.

• As a result Section I3.2 of the AISC Specification has conservatively applied

the same rules to composite section webs as to plain steel webs.

9.5

• If h/tw ≤ 3.76 (E/Fy)1/2, then there are no slender elements and the positive

nominal flexural strength Mn of a composite section is determined assuming a

plastic stress distribution.

where

h = distance between the web toes of the fillets

= d – 2k

tw = the web thickness

Fyf = the yield stress of the beam flange

and φb = 0.90 (LRFD) and Ωb = 1.67 (ASD)

All of the W, S, and HP shapes in the AISC Manual meet this requirement for

Fy values up to 50 ksi.

• If h/tw > 3.76(E/Fy)1/2, the value of Mn is determined from the superposition

of the elastic stresses with φb = 0.90 and Ωb = 1.67.

- The effects of shoring must be considered for this condition.

The nominal moment capacity Mn of composite sections as determined by load tests

can be estimated very accurately with the plastic theory.

• With this theory, the steel section at failure is assumed to be fully yielded.

• The part of the concrete slab on the compression side of the neutral axis is

assumed to be stressed to 0.85 f’c.

• Any part of the concrete slab on the tensile side of the neutral axis is assumed

to be cracked and incapable of carrying stress.

The plastic neutral axis (PNA) may fall in any one of the following locations.

1. In the concrete slab.

2. In the flange of the steel section.

3. In the web of the steel section.

Fully Composite Beams

Concrete slabs may rest directly on top of the steel beams, or the beams may be

completely encased in concrete for fireproofing purposes.

• Full encasement of the steel beam is very expensive and rarely used.

• When the concrete slab rests on top of the steel beam, mechanical connectors

are used to transfer the load between the beam and the concrete slab.

- Types of shear connectors include stud connectors, spiral connectors, and

channel connectors.

9.6

Round studs that are welded to the top flange of the beam are usually the most

economical shear connectors.

• Studs are available in diameters from ½” to 1” and in lengths from 2” to 8”.

• Section I8.2 of the AISC Specification states that the stud length may not be

less than 4 stud diameters.

• Shop installation of the studs is more economical than field installation, but

there is a tendency to install the studs in the field.

- The studs can be easily damaged during transportation and setting of the

beams.

- The studs are a hindrance to workers walking along the top flanges during

the early stages of construction.

When a composite beam is tested, failure usually occurs with crushing of the

concrete.

• It seems reasonable to assume that the concrete and steel will both have

reached a plastic condition.

- If the plastic neutral axis falls within the concrete slab, the maximum

horizontal shear between the concrete and steel is considered to be equal to

V’ = AsFy.

- If the plastic neutral axis is within the steel section, the maximum

horizontal shear is considered to be equal to V’ = 0.85 f’c Ac, where

Ac = the effective area of the slab

0.85 f’c = average stress at failure in the concrete

Section I3.2d of the AISC Specification says that, for composite action, the total

horizontal shear between the points of maximum positive moment and zero moment

is to be taken as the least of the following.

a. For concrete crushing, if the entire concrete section were in compression

V’ = 0.85 f’c Ac AISC Equation I3-1a

b. For tensile yielding of the steel section, if the entire steel section were in

tension

V’ = Fy As AISC Equation I3-1b

c. For strength of shear connectors, if the shear studs were carrying their full

capacity

V’ = ∑Qn AISC Equation I3-1c

where ∑Qn = the total nominal strength provided by the shear connectors

9.7

Establishing which stress distribution is in effect for a particular combination of

steel and concrete requires calculating the minimum capacity as controlled by the

three components of the composite beam: concrete, steel, and shear connectors.

• Because a fully composite section is assumed, V’ = ∑Qn does not control.

• If Fy As ≤ 0.85 f’c Ac, then only a portion of the concrete section is stressed

and the plastic neutral axis (PNA) is in the concrete slab.

• If Fy As > 0.85 f’c Ac, then the concrete is fully stressed and the steel section

must carry both compression and tension to ensure equilibrium and the plastic

neutral axis (PNA) is in the steel section.

Neutral Axis in Concrete Slab

Although the compression stresses in the concrete slab vary, these stresses are

assumed to be uniform (ref. Figure 9.6, p. 317 of the textbook).

• The compression stress in the concrete is assumed to be 0.85f’c over an area

with a depth “a” and a width “be.”

• The value of “a” can be determined from the following equation where the total

tension in the steel section is set equal to the total compression in the slab.

As Fy = 0.85 f’c a be

a = As Fy/0.85 f’c be

where

a ≤ the slab thickness

be = effective flange width

The nominal or plastic moment capacity of the composite section may be

determined by the following equations.

Compression force: C = 0.85 f’c a be

Tensile force: T = AsFy

Nominal or plastic moment: Mn = Mp = As Fy (d/2 + t – a/2)

9.8

Example Problem – Moment Capacity of Composite Sections

Given: Flat soffit, fully composite beam section shown.

Concrete: f’c = 4 ksi

Steel: Fy = 50 ksi

Find: Moment capacity φbMn

and Mn/Ωb

Solution

W30 x 99 (A = 29.0 in2, d = 29.7”, h/tw = 51.9)

Determine the basis for the positive nominal flexural strength Mn of the composite

section (ref. AISC Specification Section I3.2a).

h/tw = 51.9 < 3.76 (E/Fy)1/2 = 3.76 (29,000/50)1/2 = 90.55

Thus, the positive nominal flexural strength Mn of the composite section is

determined assuming a plastic stress distribution.

Locate the plastic neutral axis (PNA).

a = As Fy/0.85 f’c be = 29.0(50)/0.85(4)(100) = 4.26” < 5”

PNA is in the slab.

Determine the moment capacity assuming a plastic stress distribution.

Mn = Mp = AsFy (d/2 + t – a/2) = 29.0(50)(29.7/2 + 5 – 4.26/2)

= 25,694 kip-inch (2141.2 kip-ft)

LRFD (φb = 0.90): φbMn = 0.90 (2141.2) = 1927.1 kip-ft

ASD (Ωb = 1.67): Mn/Ωb = 2141.2/1.67 = 1282.2 kip-ft

9.9

Partially Composite Beams

A composite section is said to be a partially composite section if the number of

connectors is not sufficient to develop the full flexural strength of the composite

beam.

Composite Beam Design Tables

The moment capacity of a composite section can also be determined by using Table

3-19 of the AISC Manual.

• To use the AISC Manual table, the PNA is assumed to be located either at the

top of the steel flange (TFL) or down in the steel shape.

Y1 = the distance from the PNA to the top of the beam flange (ref. p. 3-14

of the AISC Manual)

= 0 if the PNA is located at the top of the beam flange or in the concrete

slab

Y2 = the distance from the centroid of the effective concrete area to the

top of the beam flange

= t – a/2

The variables used in Table 3-19 of the AISC Manual are defined in Figure 3-3

(p. 3-14 of the AISC Manual).

• The beam is divided into seven PNA locations: five are in the flange of the steel

section and two are in the web of the steel section.

- When the PNA is at the top of the flange (or in the concrete slab), the

entire steel section is in tension.

- As the PNA moves further down, the steel section assumes some of the

compression load and the corresponding strength required by the connectors

is reduced.

The moment capacity of the composite beam in the previous example may be

determined by using Table 3-19 of the AISC Manual as follows.


a = As Fy/0.85 f’c be = 29.0(50)/0.85(4)(100) = 4.26” < 5”

PNA is in the concrete slab.

Y1 = 0 (since the PNA is in the concrete slab)

Y2 = t – a/2 = 5 – 4.26/2 = 2.87”

9.10

From Table 3-19 (p. 3-172) of the AISC Manual, and interpolating between

Y2 = 2.5” and Y2 = 3”.

φbMn = 1927.0 kip-ft

Mn/Ωb = 1282.2 kip-ft

These values compare favorably with the previously calculated values of



Neutral Axis in Top Flange of Steel Beam

If the calculated value of “a” is greater than the slab thickness “t”, then the PNA

will fall within the steel section, either in the flange or below the flange.

• To determine whether the PNA falls within the flange or below the flange,

assume the PNA is located at the base of the flange and determine the total

compressive force C above the PNA and the total tensile force T below the

PNA.

C = 0.85 f’c be t + Fy Af

T = Fy (As – Af)

- If C > T, then the PNA will be in the flange

- If C < T, then the PNA will be below the flange.

When the PNA is in the flange, then the total compressive force and total tensile

force can be equated as follows.

0.85 f’c be t + Fy bf y = Fy As - Fy bf y

From this

y = (Fy As - 0.85 f’c be t)/(2Fy bf)

where

y = the distance to the PNA measured from the top of the top flange

The nominal or plastic moment capacity of the section can be determined from the

following equation (by taking moments about the PNA).

Mp = Mn = 0.85 f’c be t (t/2 + y ) + 2Fy bf y ( y /2) + Fy As (d/2 - y )

9.11

Example Problem – Moment Capacity of Composite Sections

Given: Flat soffit, fully composite beam section shown.

Concrete: f’c = 4 ksi

Steel: Fy = 50 ksi

Find: Moment capacity φbMn

and Mn/Ωb

Solution

W30 x 116 (A = 34.2 in2, d = 30.0”, bf = 10.5”, tf = 0.850”, h/tw = 47.8)

Determine the basis for the positive nominal flexural strength Mn of the composite

section (ref. Specification Section I3.2a).

h/tw = 47.8 < 3.76 (E/Fy)1/2 = 3.76 (29,000/50)1/2 = 90.55

Thus, the positive nominal flexural strength Mn of the composite section is

determined assuming a plastic stress distribution.


a = As Fy/0.85 f’c be = 34.2 (50)/0.85(4) 80 = 6.29” > 4”

PNA is located within the steel flange.

Assume the PNA is at the base of the steel flange.

C = 0.85 f’c be t + Fy Af = 0.85(4)(80)(4) + 50(10.5)(0.850) = 1534 kips

T = Fy (As – Af) = 50 [34.2 – 10.5(0.850)] = 1264 kips

Since C > T, the PNA is within the steel flange and can be located using the

equation that equates the compressive and tensile forces.

y = (Fy As - 0.85 f’c be t)/2Fy bf

= [50 (34.2) – 0.85(4)(80) 4]/2(50) 10.5 = 0.592”

Determine the moment capacity assuming a plastic stress distribution.

Mp = Mn = 0.85 f’c be t (t/2 + y ) + 2Fy bf y ( y /2) + Fy As (d/2 - y )

= 0.85(4)(80)(4)(4/2 + 0.592) + 2(50)(10.5)(0.592)(0.592/2)

+ 50(34.2)(30.0/2 – 0.592)

= 27,642 kip-inch (2303 kip-ft)

LRFD (φb = 0.90): φbMn = 0.90 (2303) = 2073 kip-ft

ASD (Ωb = 1.67): Mn/Ωb = 2303/1.67 = 1379 kip-ft

9.12

The moment capacity of the composite beam in the previous example may be

determined by using Table 3-19 of the AISC Manual as follows.


a = As Fy/0.85f’c be = 34.2(50)/0.85(4)(80) = 6.29” > 4”

PNA is located within the steel flange.

Assume the PNA is at the base of the steel flange.

C = 0.85 f’c be t + Fy Af = 0.85(4)(80) 4 + 50(10.5)(0.850) = 1534 kips

T = Fy (As – Af) = 50 [34.2 – 10.5(0.850)] = 1264 kips

Since C > T, the PNA is within the steel flange and can be located using the

equation that equates the compressive and tensile forces.

y = (Fy As - 0.85 f’c be t)/2Fy bf

= [50(34.2) – 0.85(4)(80) 4]/2(50) 10.5 = 0.592”

Y1 = the distance from the PNA to the top of the beam flange (ref. p. 3-14

of the AISC Manual)

= 0.592”

Y2 = t – a/2 = 4 – 4/2 = 2.00”

From Table 3-19 (p. 3-170) of the AISC Manual, and interpolating between

Y1 = 0.425” and Y1 = 0.638”.



These values compare favorably with the previously calculated values of

φbMn = 2073 kip-ft

Mn/Ωb = 1379 kip-ft

9.13

Neutral Axis in the Steel Section

If the plastic neutral axis is located below the flange, calculations to determine

the moment capacity are similar to the previous calculations.

• The Composite Design tables in Part 3 of the AISC Manual cover most common

cases.

9.6 Strength of Shear Connectors

Sections I8.2a and I8.2b of the AISC Specification provide strength values for

headed steel studs and hot-rolled steel channels, respectively.

Stud Shear Connectors

In Section I8.2a of the AISC Specification, the nominal strength of one stud

shear connector embedded in a solid concrete slab is given by the following

equation.

Qn = 0.5 Asa (f’c Ec)1/2 ≤ Rg Rp Asa Fu AISC Equation I8-1

where

Asa = the cross sectional area of the steel headed stud anchor

f’c = the specified compressive strength of the concrete

Ec = the modulus of elasticity of the concrete

= w 1.5√f’c

w = the unit weight of the concrete

Fu = the specified minimum tensile strength of the steel headed stud anchor

Rg = a coefficient used to account for the group effect of the connectors

(values provided in AISC Specification I8.2a)

Rp = a coefficient used to account for the position effect of the connectors

(values provided in AISC Specification I8.2a)

Values for Qn for stud shear connectors are listed in Table 3-21 of the AISC

Manual.

• These values are given for different stud diameters, for 3 ksi and 4 ksi normal

and lightweight concrete, and for composite sections with or without steel

decking.

Channel Shear Connectors

In Section I8.2b of the AISC Specification, the nominal strength of one channel

shear connector embedded in a solid concrete slab is given by the following

equation.

Qn = 0.3 (tf + 0.5 tw) la (f’c Ec)1/2 AISC Equation I8-2

9.14

where

tf = flange thickness of the channel anchor

tw = web thickness of the channel anchor

la = length of the channel anchor

Number and Placement of Shear Connectors

The number of shear connectors that are to be used between the point of

maximum moment and each adjacent point of zero moment equals the horizontal

force to be resisted V’ divided by the nominal strength of one connector Qn.

Number of shear connectors = V’/Qn

Spacing of Connectors

Section I8.2d of the AISC Specification permits uniform spacings of connectors

on each side of the maximum moment points.

• Tests of composite beams with shear connectors spaced uniformly, and of

composite beams with the same number of connectors spaced in variation with

the shear, show little difference as to ultimate strengths and deflections at

working loads.

• The number of connectors placed between a concentrated load and the nearest

point of zero moment must be sufficient to develop the maximum moment at

the concentrated load.

Maximum and Minimum Spacings

Minimum spacings of shear connectors for composite beams as prescribed by

Section I8.2d of the AISC Specification are as follows (ref. Figure 16.6, p. 572 of

the textbook).

• Except for formed steel decks, the minimum center-to-center spacing along the

longitudinal axes is 6 diameters.

• Except for formed steel decks, the minimum spacing transverse to the

longitudinal axis is 4 diameters.

• Within the ribs of formed steel decks, the minimum permissible spacing is 4

diameters in any direction.

- If the deck ribs are parallel to the axis of the steel beam and more

connectors are required than can be placed within the rib, Section I8.2d of

the AISC Commentary permits the splitting of the deck so that adequate

room is available for the connectors.

• If the flanges of steel beams are narrow, making it difficult to achieve the

minimum transverse spacing, the connectors may be staggered.

9.15

- Figure 16.6 (p. 572 of the textbook) shows possible arrangements for placing

staggered connectors.

The maximum spacing may not exceed 8 times the total slab thickness, or 36”.

Shear connectors must be capable of resisting both horizontal and vertical

movement.

• There is a tendency for the slab and beam to separate vertically, as well as to

slip horizontally.

• The upset heads of stud shear connectors help to prevent vertical separation.

Cover Requirements

Section I8.2d of the AISC Specification requires that shear connectors have at

least 1” of lateral concrete cover in the direction perpendicular to the shear force,

except for connectors installed within the ribs of formed steel decks.

Section I8.1 of the AISC Specification requires the diameter of the studs to be

no greater than 2.5 times the flange thickness of the beam to which they are

welded, unless located directly over the beam web.

• When studs are not placed directly over beam webs, they have a tendency to

tear out of the beam flanges before their full shear capacity is reached.

According to Section I3.2c(1) of the AISC Specification, when a formed steel deck

is used, the steel beam must be connected to the concrete slab with shear

connectors with diameters not larger than ¾”.

• The connectors may be welded through the deck or directly to the steel beam.

- After installation, the connectors must extend at least 1-1/2” above the top

of the steel deck.

• The concrete slab thickness above the steel deck may not be less than 2”.

9.7 Composite Beams with Formed Metal Deck

The combination of formed steel deck and composite design is considered today to

be one of the most economical methods of floor construction.

• The steel deck serves as stay-in-place formwork for the concrete slab.

• Common steel deck profiles are shown in Figure 9.16 (p. 333 of the textbook).

AISC Specification I3.2c provides requirements for composite beams with a

formed steel deck.

• The nominal rib height of the steel deck must be no greater than 3”.

9.16

• The average rib width of the steel deck must be 2” or more.

• The size and length of welded steel headed stud anchors are prescribed.

• The slab thickness above the top of the steel deck must be 2” or more.

• The steel deck must be anchored to the supporting member by steel headed

stud anchors at spacings not to exceed 18”.

Deck Ribs Perpendicular and Parallel to Steel Beam

Table 3-21 of the AISC Manual provides the nominal horizontal shear strengths

for one shear stud (Qn in kips) based upon the stud diameter, the position of the

stud, the profile of the deck, and the orientation of the deck relative to the stud.

The various “deck conditions” (i.e. the orientation of the deck with respect to the

beam) prescribed by Table 3-21 of the AISC Manual are listed as follows.

• “No deck”: There is no formed steel deck.

• “Deck parallel”: The formed steel deck is oriented with ribs parallel to the steel

shape.

- Concrete below the top of the deck can be used in calculating the composite

section properties and must be used in shear stud calculations.

- The nominal horizontal shear strength of the shear stud anchor depends on

the width-to-height ratio of wr/hr ≥ 1.5 or wr/hr < 1.5.

where

hr = nominal rib height

wr = average width of concrete rib or haunch

• “Deck perpendicular”: The formed steel deck is oriented with ribs perpendicular

to the beams with 1, 2 or 3 studs within the same decking rib.

- The space below the top of the rib contains concrete only in the alternating

spaces, so there is no opportunity to transfer force at this level.

- For ribs placed perpendicular to the steel shape, the shear studs must be

placed on one side or the other of the stiffening rib in the middle of the

deck flutes.

- The shear studs are

referred to as either

strong or weak based on

this placement (ref.

AISC Commentary Figure

C-I8.1).

9.17

◦ The strong position is on the side away from the direction from which

the shear is applied.

◦ The weak position is on the same side as the direction from which the

shear is applied.

- Some designers assume conservatively that the studs will always be placed in

the weak positions, with the smaller shear strengths, since it is not an easy

task to make certain that the studs are placed in the strong positions in the

field.

9.8 Fully Encased Steel Beams

For fireproofing purposes, it is possible to encase in concrete the steel beams used

for building floors.

• This practice is not economical because light-weight spray-on fire protection is

much less expensive.

• Furthermore, encased beams may increase the floor system dead load by as

much as 15%.

The nominal flexural strength Mn of concrete-encased members shall be

determined using one of the methods prescribed in Section I3.3 of the AISC

Specification.

9.9 Selecting a Section

The design of a composite beam is somewhat of a trial-and-error procedure that is

simplified using AISC Manual Table 3-19, as illustrated by the following example.

9.18

Example Problem – Design of Composite Beams

Given: Beams 10’ c/c with 36’ simple span are to be selected to support a 4”

lightweight concrete slab on a 3” deep formed steel deck with no shoring.

Steel deck: ribs perpendicular to the beam center lines, with 6” average width

Service loads (along the beams):

wD = 0.78 kip/ft (including the weight of the beam)

wL = 2 kip/ft

Steel: Fy = 50 ksi

Concrete: f’c = 4 ksi, wc = 110 lb/ft3

Find: a) Select the beams.

b) Determine the number of ¾” headed studs required.

c) Compute the service live load deflection.

d) Check the beam shear.

Solution

LRFD

Determine the factored load and moment.

wu = 1.2 D + 1.6 L = 1.2 (0.78) + 1.6 (2) = 4.14 kip/ft

Mu = wuL2/8 = 4.14(36)2/8 = 670.7 kip-ft

Determine the effective flange width (smallest of the following values).

be = 2 (1/8 span) = 2 (1/8) (36 x 12”/’) = 108”

be = 2 (1/2 beam spacing) = 2 (1/2)(10 x 12”/’) = 120”

be = distance from beam center line to the edge of slab (not applicable)

Use be = 108”

Select the W-section

Ycon = distance from top of slab to top of steel flange

= slab thickness + depth of the steel decking = 4 + 3 = 7”

Assume a = 2” < 4” slab thickness (This value is usually quite small, particularly

for relatively light sections.)

Y1 = the distance from PNA to top flange = 0 (This assumes the PNA is located

in the concrete slab.)

9.19

Y2 = distance from the center of gravity of the concrete flange force to the

top flange of the beam = 7 – a/2 = 7 – 2/2 = 6”

Select a W-shape from the composite tables (Table 3-19) of the AISC Manual,

for Y1 = 0, and Y2 = 6”, and Mu = 670.7 kip-ft:

Try W18 x 46 (available flexure strength = 762 kip-ft)

From Table 1-1 of the AISC Manual: A = 13.5 in2, Ix = 712 in4

Notes:

1. For a non-composite beam section, a W24 x 76 beam is required (ref. AISC Table 3-2)

assuming full lateral support - a difference of approximately 6” in depth per floor.

2. A composite section using a W16 x 45 steel beam (φbMn = 701 kip-ft) may have been

considered. However, upon further evaluation, the W16 x 45 composite section is not

adequate for live load deflection.

Check the deflection of the steel beam due to wet concrete plus beam weight.

w = (4”/12)(110 lb/ft3) (10’ tributary width) + 46 = 413 lb/ft (0.413 kip/ft)

Note: The weight of the steel decking and the weight of the concrete within the ribs of the

steel decking are neglected.

M = wL2/8 = 0.413(36)2/8 = 66.9 kip-ft

C1 = 161 (ref. Figure 3-2, p. 3-9 of the AISC Manual)

Deflection: Δ = M L2/C1 Ix (ref. p. 3-8 of the AISC Manual)

= 66.9(36)2/161(712) = 0.756” < 2.5” OK

Verify the available flexural design strength (φbMn) of the composite section.

Determine the controlling shear force:

V’ = AsFy = 13.5(50) = 675 kips (controls)

V’ = 0.85 f’cAc = 0.85 (4) 108.0 (4.0) = 1468.8 kips

Note: For beams supporting the steel

deck with the ribs perpendicular to the

beam, the space below the top of the rib

contains concrete only in the alternating

spaces, so there is no opportunity to

transfer force at this level.

Since AsFy < 0.85 f’cAc the steel controls and the PNA is in the concrete slab.

Required a = AsFy/0.85f’cbe = 675/0.85(4)(108) = 1.84” < 4”


Y2 = t – a/2 = 7.00 – 1.84/2 = 6.08”

9.20

Compute φbMn from Table 3-19 (p. 3-181) of the AISC Manual for the W18 x 46

composite beam for Y1 = 0” and Y2 = 6.08”.

By interpolation, φbMn = 766.0 kip-ft > Mu = 670.7 kip-ft OK

With the same W18 x 46, use the largest value for Y1 that provides a flexural

design strength value (φbMn) no less than Mu = 670.7 kip-ft and Y2 ≈ 6”.

Y1 = 0.303

∑Qn = 492 kips < V’ = 675 kips (a smaller value requiring fewer shear connectors)

Required a = ∑Qn/0.85 f’cbe (ref. p. 3-29 of the AISC Manual)

= 492/0.85(4)(108) = 1.34” < 4”

Y2 = 7.00 – a/2 = 7.00 – 1.34/2 = 6.33”

Recompute φbMn from Table 3-19 (p. 3-181) of the AISC Manual for the

W18 x 46 composite beam for Y1 = 0.303” and Y2 = 6.33”.


Use W18 x 46

Design the studs

Determine the strength (Qn value) for an individual stud.

• From AISC Table 3-21 (p. 3-209), for a ¾” stud, with deck ribs perpendicular to

the beam center line, with the stud placed in the strong position, one stud per

rib, light-weight concrete (f’c = 4 ksi): Qn = 21.2 kips

Determine the total number of connectors on each side of the point of maximum

moment.

∑Qn/Qn = 492/21.2 = 23.2

Use 24 – ¾” studs on each side of the point of maximum moment (i.e. the

centerline of the beam).

Check the spacing of the shear connectors.

Maximum spacing = 8 x slab thickness ≤ 36”

= 8(4) = 32” < 36” (Use 32” maximum spacing)

Minimum spacing = 4 x stud diameter

= 4 (3/4) = 3” (Use 3” minimum spacing)

Actual spacing = (18’ x 12”/’)/24 = 9.0” c/c > 3” OK

< 32” OK

9.21

Compute live load deflections.

Assume the maximum permissible live load deflection = (1/360) span

Δallow = (1/360)(36 x 12) = 1.20”

Compute the live load deflection using the following equation.

Δ = ML L2/C1 Ix (from p. 3-7 of the AISC Manual)


ML = wLL2/8 = 2.0(36)2/8 = 324.0 kip-ft

Ix = lower bound moment of inertia (ref. AISC Table 3-20)

By interpolation (Y1 = 0.303”, Y2 = 6.33”): Ix = 2059 in4

Deflection: Δ = ML L2/C1 Ix

= 324.0(36)2/161(2059) = 1.27” > 1.20” (A little high.)

Check beam shear for the steel section.

Vu = wu L/2 = 4.14(36)/2 = 74.5 kips

φvVn = 195 kips (from Table 3-2 of the AISC Manual) > 74.5 kips OK

Use W18 x 46 with 48 – ¾” headed studs as shear connectors.

ASD

Determine the load combination and moment.

wa = D + L = 0.78 + 2.0 = 2.78 kip/ft

Ma = waL2/8 = 2.78(36)2/8 = 450.4 kip-ft

Determine the effective flange width.

Use be = 108” (as before)









top flange of the beam) = 7 – a/2 = 7 – 2/2 = 6”

9.22


for Y1 = 0, and Y2 = 6”, and Ma = 450.4 kip-ft:



Notes:



2. A composite section using a W16 x 45 steel beam (Mn/Ωb = 466 kip-ft) may have been

initially considered by the author. However, upon further evaluation, the W16 x 45

composite section is not adequate for live load deflection.





M = wL2/8 = 0.413(36)2/8 = 66.9 kip-ft


Deflection: Δ = M L2/C1 Ix (from p. 3-8 of the AISC Manual)

= 66.9(36)2/161(712) = 0.756” < 2.5” OK

Verify the available flexural strength (Mn/Ωb) of the composite section.



V’ = 0.85 f’cAc = 0.85 (4) 108.0 (4.0) = 1468.8 kips









Y1 = 0

Y2 = t – a/2 = 7.00 – 1.84/2 = 6.08”

Compute Mn/Ωb from Table 3-19 (p. 3-181) of the AISC Manual for the

W18 x 46 composite beam for Y1 = 0” and Y2 = 6.08”.

9.23

By interpolation, Mn/Ωb = 509.7 kip-ft > Ma = 450.4 kip-ft OK


design strength value (Mn/Ωb) no less than Ma = 450.4 kip-ft and Y2 ≈ 6”.

Y1 = 0.303


Required a = ∑Qn/0.85f’cbe (ref. p. 3-29 of the AISC Manual)

= 492/0.85(4)(108) = 1.34” < 4”

Y2 = 7.00 – a/2 = 7.00 – 1.34/2 = 6.33”

Recompute Mn/Ωb from Table 3-19 (p. 3-181) of the AISC Manual for the



Use W18 x 46

Design the studs






moment.

∑Qn/Qn = 492/21.2 = 23.2








Actual spacing = (18’ x 12”’/’)/24 = 9.0” c/c > 3” OK

< 32” OK


9.24


Δallow = (1/360)(36 x 12) = 1.20”




ML = wLL2/8 = 2.0(36)2/8 = 324.0 kip-ft




= 324.0(36)2/161(2059) = 1.27” > 1.20” (A little high.)


Va = waL/2 = 2.78(36)/2 = 50.0 kips

Vn/Ωv = 130 kips (from Table 3-2 of the AISC Manual) > 50.0 kips OK


9.25

16.10 Deflections

Deflections for composite beams may be calculated by the same methods used for

other types of beams.

• Deflections for the various types of loads must be considered separately.

- There are deflections due to dead loads and construction loads applied to

the steel section alone if no shoring is used.

- There are deflections due to dead loads applied to the composite section.

- There are deflections due to live loads applied to the composite section.

Concrete beams that are in compression are subjected to long-term creep effects

that cause deflections to increase with time.

• These increases are usually not considered significant for the average

composite beam, unless long spans and large permanent live loads are involved

(ref. Section I3.2.4 of the AISC Commentary).

In calculating deflections for composite sections, a “transformed section moment

of inertia” Itr is used.

• The modulus of elasticity of concrete Ec is used in the calculation for Itr.

Other strategies to compensate for deflections may include the following.

• The steel beams can be cambered for all or some portion of the deflections.

• It may be feasible in some situations to make a floor slab a little thicker in the

middle than on the edges to compensate for deflections.

The designer may want to control vibrations in composite floors subject to

pedestrian traffic or other moving loads.

• This may be the case for large open floor areas with no damping furnished by

partitions (e.g. in shopping malls).

When the AISC Specification is used to select steel beams for composite sections,

the results are often small steel beams and shallow floors.

• Such floors, when unshored, frequently will have large deflections when the

concrete is placed.

- For this reason, designers will often specify cambering of the beams.

- As an alternative, the designer may select larger beams or use shoring.

9.26

• The steel beams selected must have sufficient strength values (φbMn or Mn/Ωb)

to support the beam weight and the weight of the wet concrete during

construction, before the strength of the composite section is achieved.

- The beam sizes are often determined by wet concrete deflections rather

than by flexure considerations.

- It is considered to be good practice to limit these deflections to maximum

values of about 2-1/2”

• An alternate solution for these problems involves the use of partly restrained

(PR) connections (as discussed in Chapter 15).

- Such connections can appreciably reduce mid-span deflections and moments

enabling the use of smaller beams.

- Such connections can also reduce vibrations that are a problem in shallow

composite floors.

16.11 Design of Composite Sections

Composite construction is of particular advantage economically when loads are

heavy, spans are long, and beams are spaced a fairly large intervals.

• For steel building frames, composite construction is economical for spans from

25 to 50 feet with particular advantage in the longer spans.

• Occasionally cover plates are welded to the bottom flanges of the steel beam to

provide an improved economy.

- By adding cover plates to the tensile flange, a slightly better balance in

terms of compressive and tensile areas can be obtained.

Lateral Bracing

After the concrete hardens, it will provide sufficient lateral bracing for the

compression flange of the steel beam.

• During the construction phase before the concrete hardens, lateral bracing may

be insufficient and the design strength of the beam may be reduced.

• When steel-formed decking or concrete forms are attached to the beam’s

compression flange, they usually will provide sufficient lateral bracing.

• Designers must carefully consider lateral bracing for fully encased beams.

Beams with Shoring

If beams are shored during construction, we assume that all loads are resisted by

the composite section after the shoring is removed.

9.27

Beams without Shoring

If temporary shoring is not used during construction, the steel beam alone must be

able to support all the loads before the concrete is sufficiently hardened to

provide composite action.

• Without shoring, the wet concrete loads tend to cause large beam deflections.

- This situation can be countered by cambering the beams.

• Assuming that satisfactory lateral bracing is provided, Section F2 of the AISC

Specification states that the maximum factored moment may not exceed

0.90FyZ.

- The 0.90 in effect limits the maximum factored moment to a value about

equal to the yield moment FyS.

• To calculate the moment to be resisted during construction, the live load should

include the weight of the wet concrete and the live loads due to construction

activities (perhaps 20 psf).

Estimated Steel Beam Weight

It is sometimes useful to estimate the weight of the steel beam.

• Part 5 (p. 5-26) of the third edition of the AISC Manual provided the following

empirical formula for this purpose.

Estimated beam weight = [12Mu/(d/2 + Ycon – a/2) φFy] 3.4

where

Mu = required flexural strength of the composite section

d = nominal steel beam depth

Ycon = distance from the top of the steel beam to the top of the concrete

slab

a = the effective concrete slab thickness (conservatively estimated as 2”)

φ = 0.85

Lower Bound Moment of Inertia

The “lower-bound elastic moment of inertia” of a composite beam can be used to

calculate deflection.

• If calculated deflections using the “lower-bound moment of inertia” are

acceptable, a more complete elastic analysis of the composite section can be

avoided.

To calculate the service load deflections for composite sections, a table of lower

bound moment of inertia values is presented in Part 3 of the AISC Manual (Table

3-20).

9.28

• These values are computed from the area of the steel beam and an equivalent

concrete area of ∑Qn/Fy.

- The remainder of the concrete flange is not used in these calculations.

◦ For partially composite sections, the value of the lower bound moment of

inertia will reflect this situation because ∑Qn will be smaller.

• The lower bound moment of inertia is computed using the following equation

(ref. Figure 16.12).

ILB = Is + As (YENA – d3)2 + (∑Qn/Fy)(2d3 + d1 – YENA)2

AISC Commentary Equation C-I3-1

where

ILB = lower bound moment of inertia

Is = moment of inertia of steel section

As = area of the steel cross section

d1 = the distance from the compression force in the concrete to the top

of the steel section

d3 = distance from the resultant steel tension force for full section

tensile yield to the top of the steel

∑Qn = the sum of the nominal strengths of shear connectors between the

point of maximum positive moment and the point of zero moment to

either side

YENA = the distance from the bottom of the beam to the elastic neutral

axis (ENA)

= [(Asd3 + (∑Qn/Fy)(2d3 + d1))/(As + (∑Qn/Fy))] Equation C-I3-2

Extra Reinforcing

For building design calculations, the beam spans are often considered to be simply

supported, but the steel beams generally do not have perfectly simple ends.

• Some negative moment may occur at the beam ends, with possible cracking of

the concrete slab.

- To prevent or minimize cracking, some extra steel can be placed in the top

of the slab, extending 2 or 3 feet out into the slab.

◦ The amount of steel added is in addition to that needed to meet the

temperature and shrinkage requirements specified by the American

Concrete Institute.

Example Problem – Design of Composite Beams

Example 16-3 (p. 582 of the textbook)

9.29

Given: Beams 10’ c/c with 36’ simple span are to be selected to support a 4”

lightweight concrete slab on a 3” deep formed steel deck with no shoring.

Steel deck: ribs perpendicular to the beam center lines, with 6” average width

Service loads (along the beams): wD = 0.78 kip/ft (including the weight of the

beam), wL = 2 kip/ft

Steel: Fy = 50 ksi

Concrete: f’c = 4 ksi, wc = 110 lb/ft3

Find: a) Select the beams.

b) Determine the number of ¾” headed studs required.

c) Compute the service live load deflection.

d) Check the beam shear.

Solution

LRFD

Determine the factored load and moment.

wu = 1.2 D + 1.6 L = 1.2 (0.78) + 1.6 (2) = 4.14 kip/ft

Mu = wuL2/8 = 4.14(36)2/8 = 670.7 kip-ft

Determine the effective flange width (smallest of the following values).

be = 2 (1/8 span) = 2 (1/8) (36 x 12”/’) = 108”

be = 2 (1/2 beam spacing) = 2 (1/2)(10 x 12”/’) = 120”

be = distance from beam center line to the edge of slab (not applicable)

Use be = 108”








9.30


top flange of the beam = 7 – a/2 = 7 – 2/2 = 6”


for Y1 = 0, and Y2 = 6”, and Mu = 670.7 kip-ft:



Notes:



2. A composite section using a W16 x 45 steel beam (φbMn = 701 kip-ft) may have been







M = wL2/8 = 0.413(36)2/8 = 66.9 kip-ft


Deflection: Δ = M L2/C1 Ix (ref. p. 3-8 of the AISC Manual)

= 66.9(36)2/161(712) = 0.756” < 2.5” OK

Verify the available flexural design strength (φbMn) of the composite section.



V’ = 0.85 f’cAc = 0.85 (4) 108.0 (4.0) = 1468.8 kips










Y2 = t – a/2 = 7.00 – 1.84/2 = 6.08”

9.31

Compute φbMn from Table 3-19 (p. 3-181) of the AISC Manual for the W18 x 46

composite beam for Y1 = 0” and Y2 = 6.08”.



design strength value (φbMn) no less than Mu = 670.7 kip-ft and Y2 ≈ 6”.

Y1 = 0.303


Required a = ∑Qn/0.85 f’cbe (ref. p. 3-29 of the AISC Manual)

= 492/0.85(4)(108) = 1.34” < 4”

Y2 = 7.00 – a/2 = 7.00 – 1.34/2 = 6.33”

Recompute φbMn from Table 3-19 (p. 3-181) of the AISC Manual for the



Use W18 x 46

Design the studs






moment.

∑Qn/Qn = 492/21.2 = 23.2








Actual spacing = (18’ x 12”/’)/24 = 9.0” c/c > 3” OK

< 32” OK

9.32



Δallow = (1/360)(36 x 12) = 1.20”




ML = wLL2/8 = 2.0(36)2/8 = 324.0 kip-ft




= 324.0(36)2/161(2059) = 1.27” > 1.20” (A little high.)


Vu = wu L/2 = 4.14(36)/2 = 74.5 kips

φvVn = 195 kips (from Table 3-2 of the AISC Manual) > 74.5 kips OK


ASD

Determine the load combination and moment.

wa = D + L = 0.78 + 2.0 = 2.78 kip/ft

Ma = waL2/8 = 2.78(36)2/8 = 450.4 kip-ft

Determine the effective flange width.

Use be = 108” (as before)








9.33


top flange of the beam) = 7 – a/2 = 7 – 2/2 = 6”


for Y1 = 0, and Y2 = 6”, and Ma = 450.4 kip-ft:



Notes:



2. A composite section using a W16 x 45 steel beam (Mn/Ωb = 466 kip-ft) may have been







M = wL2/8 = 0.413(36)2/8 = 66.9 kip-ft


Deflection: Δ = M L2/C1 Ix (from p. 3-8 of the AISC Manual)

= 66.9(36)2/161(712) = 0.756” < 2.5” OK

Verify the available flexural strength (Mn/Ωb) of the composite section.



V’ = 0.85 f’cAc = 0.85 (4) 108.0 (4.0) = 1468.8 kips









Y1 = 0

Y2 = t – a/2 = 7.00 – 1.84/2 = 6.08”

9.34

Compute Mn/Ωb from Table 3-19 (p. 3-181) of the AISC Manual for the

W18 x 46 composite beam for Y1 = 0” and Y2 = 6.08”.



design strength value (Mn/Ωb) no less than Ma = 450.4 kip-ft and Y2 ≈ 6”.

Y1 = 0.303


Required a = ∑Qn/0.85f’cbe (ref. p. 3-29 of the AISC Manual)

= 492/0.85(4)(108) = 1.34” < 4”

Y2 = 7.00 – a/2 = 7.00 – 1.34/2 = 6.33”

Recompute Mn/Ωb from Table 3-19 (p. 3-181) of the AISC Manual for the



Use W18 x 46

Design the studs






moment.

∑Qn/Qn = 492/21.2 = 23.2








Actual spacing = (18’ x 12”’/’)/24 = 9.0” c/c > 3” OK

< 32” OK

9.35



Δallow = (1/360)(36 x 12) = 1.20”




ML = wLL2/8 = 2.0(36)2/8 = 324.0 kip-ft




= 324.0(36)2/161(2059) = 1.27” > 1.20” (A little high.)


Va = waL/2 = 2.78(36)/2 = 50.0 kips

Vn/Ωv = 130 kips (from Table 3-2 of the AISC Manual) > 50.0 kips OK


9.36

16.12 Continuous Composite Sections

Section I3.2b of the AISC Specification permits the use of continuous composite

sections.

• The flexural strength of a composite section in a negative moment region may

be considered equal to φbMn for the steel section alone.

- The concrete slab contributes no strength since it is in a tension region.

• Alternatively, the flexural strength may be based on the plastic strength of a

composite section considered to be made up of the steel beam and the

longitudinal reinforcement in the slab if the following conditions are met.

1. The steel section must be compact and adequately braced.

2. The slab must be connected to the steel beams in the negative moment

region with shear connectors.

3. The longitudinal reinforcing in the slab parallel to the steel beam and within

the effective width of the slab must have adequate development lengths.

For a particular beam, the total horizontal shear force between the point of zero

moment and the point of maximum negative moments is to be taken as the smaller

of Fysr Asr and ∑Qn, per AISC Section I3.2d(2).

where

Asr = the cross-sectional area of the properly developed reinforcing steel

Fysr = the yield stress of the reinforcing steel

16.13 Design of Concrete-Encased Sections

For fireproofing purposes, it is possible to encase in concrete the steel beams used

for building floors.

• This practice is not economical because light-weight spray-on fire protection is

much less expensive.

• Furthermore, encased beams may increase the floor system dead load by as

much as 15%.

For the rare situation in which encased beams are used, steel anchors should be

provided to transfer the horizontal shears between the slabs and beams.

The nominal flexural strength Mn of concrete-encased and filled members shall be

determined using one of the methods prescribed in Section I3.3 of the AISC

Specification.

9.37

1. The design strength of the encased section may be based on the superposition

of elastic stresses on the composite section, considering the effects of shoring

for the limit state of yielding where

φb = 0.90 (LRFD) and Ωb = 1.67(ASD)

For unshored construction:

• First, the stresses in the steel section caused by the wet concrete and the

construction loads are determined.

• Then the stresses in the composite section caused by the loads applied after

the concrete hardens are computed and superimposed on the first set of

stresses.

For shored construction:

• All of the loads may be assumed to be supported by the composite section

and the stresses are computed accordingly.

• For stress calculations, the properties of the composite section are

computed by the transformed area method.

- Using the transformed area method, the cross-sectional area of one

material is replaced or transformed by an equivalent area of the other

material.

- It is customary to replace the concrete with an equivalent area of steel

based on a modular ratio n, where n = Ey/Ec, where

Ec = modulus of elasticity for concrete (wc = 90 to 155 lb/ft3)

= wc1.5 33√f’c

- The cross sectional area of the slab (Ac) is replaced with a transformed

or equivalent area of steel equal to Ac/n.

2. The design strength of the encased section may be based on the plastic moment

capacity φbMp or Mp/Ωb of the steel section alone where

φb = 0.90 (LRFD) and Ωb = 1.67 (ASD)

There are no slenderness limitations prescribed by the AISC Specification for

these methods.

• The encasement is effective in preventing both local and lateral buckling.

9.38

Example Problem – Design of Concrete-Encased Sections

Example 16-5 (p. 590 of the textbook)

Given: The encased beam section shown.

No shoring is used.

Simple span = 36’

Service dead load: 0.50 kip/ft before

the concrete hardens plus an

additional 0.25 kip/ft after the

concrete hardens.

Construction live loads: 0.20 kips/ft

Service live load: 1.0 kip/ft after concrete hardens

Effective flange width: be = 60”

Modular ratio: n = 9

Steel: Fy = 50 ksi

Find: Determine the beam adequacy.

Solution

W16 x 45 (A = 13.3 in2, d = 16.1”, Ix = 586 in4)

Calculate the properties of the composite section (neglecting the concrete area

below the slab).

Equivalent area: A = As + Ac/n = 13.3 + 4(60)/9 = 39.97 in2

Locate the neutral axis (referenced from the bottom of the concrete):

yb = (As ys + Ac yc)/A

= {13.3(2.40 + 16.1/2) + [4(60)/9](16 + 4/2)}/39.97 = 15.49”

Compute the moment of inertia for the composite section.

I = Is + As (y’s)2 + Ic + Ac (y’c)

2 = 586 + 13.3(2.40 + 16.1/2 – 15.49)2

+ (1/12)60(4)3/9 + [4(60)/9](16 + 4/2 – 15.49)2

= 586 + 337.84 + 35.56 + 168.0 = 1127.4 in4

9.39

Calculate the stresses before the concrete hardens (assuming the wet concrete is

a live load).

wu = 1.2 D + 1.6 L = 1.2 (0) + 1.6 (0.50 + 0.20) = 1.12 kip/ft

Mu = wuL2/8 = 1.12(36)2/8 = 181.4 kip-ft

All stresses are supported by the steel beam:

c = 16.1/2 = 8.05” (the neutral axis is through the center of the steel

section; top fibers are in compression, bottom fibers are in tension)

ft = Mc/I = [(181.4 x 12”/’) (8.05)]/586 = 29.90 ksi

< φb Fy = 0.90 (50) = 45.0 ksi OK

Calculate the additional stresses after the concrete hardens.

wu = 1.2 D + 1.6 L = 1.2 (0.25) + 1.6 (1.0) = 1.90 kip/ft

Mu = wuL2/8 = 1.90(36)2/8 = 307.8 kip-ft

All stresses are supported by the steel beam:

c1 = yb – 2.40 = 15.49 – 2.40 = 13.09” (from NA to bottom fibers of the steel

section that are in tension)

c2 = 2.40 + 16.10 – 15.49 = 3.01” (from the NA to the top fibers of the steel

section that are in compression)

ft = Mc/I = [(307.8 x 12”/’) (13.09)]/1127.4 = 42.89 ksi

Compute the total tensile stress on the composite section:

ft = 29.90 + 42.89 = 72.82 ksi > φb Fy = 0.90 (50) = 45.0 ksi NG

9.40

For buildings, continuous composite construction with encased sections is

permitted.

• The positive moments are handled exactly as illustrated in the previous

example.

• For negative moments, the transformed

section is taken as shown in Figure 16.15

(p. 592 of the textbook).

- The cross-hatched area represents

the concrete in compression.

- All concrete on the tensile side of the

neutral axis (i.e. above the neutral

axis) is neglected.

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