Concrete Decks

7212019 Concrete Decks

httpslidepdfcomreaderfullconcrete-decks 116

573

151 Introduction

Bridge decks not only provide the riding surace or traffic but also support and transer live loads

to the main load-carrying members such as stringer and girders on a bridge superstructure Bridge

decks include cast-in-place (CIP) reinorced concrete precast concrete deck panel prestress con-

crete timber 1047297lled and un1047297lled steel grid and steel orthotropic decks Selection o bridge deck

types depends on locations spans traffic environment maintenance aesthetics and lie cycle costs

among other reasons

Tis chapter ocuses on concrete deck and emphasizes the cast-in-place reinorced concrete deck A

design example o a reinorced concrete bridge deck is provided in accordance with the AASHTO LRFD

Bridge Design Speci1047297cations (AASHO 2012) For more detailed discussion o the concrete deck reer-

ences are made to FHWA (2012) and Barker and Puckett (2007) Steel orthotropic decks are discussed

in Chapter 16

152 Types of Concrete Decks

1521 Cast-in-Place Concrete Deck

Te CIP concrete deck slab is the predominant deck type in highway bridges in the United States

Figure 151 shows a reinorcement layout in a CIP concrete deck on the steel plate girder Figure 152

shows a CIP concrete deck under construction Figure 153 shows a typical CIP concrete deck details Its

main advantages are acceptable skid resistance the easier 1047297eld-adjustment o the roadway pro1047297le during

concrete placement to provide a smooth riding surace commonly available materials and contractors

to do the work However CIP slabs have disadvantages including excessive differential shrinkage with

15Concrete Decks

John ShenCalifornia Department

of Transportation

151 Introduction 573

152 ypes o Concrete Decks 573

Cast-in-Place Concrete Deck bull Precast Concrete Deck 153 Materials 576

General Requirements bull Concrete bull Reinorcement bull Construct ion

Practices

154 Design Considerations 577General Requirements bull Design Limit States bull Analysis Methods

155 Design Example 579Bridge Deck Data bull Design Requirements bull Solution bull Calculate

Factored MomentsmdashStrength Limit State I bull Design or Positive

Flexure Design bull Design or Negative Flexure bull Check Service

Limit State-Cracking Control bull Determine the Slab Reinorcement

Detailing Requirements

Reerences 588



574 Bridge Engineering Handbook Second Edition Superstructure Design

the supporting girders and slow construction the tendency o the deck rebar to corrode due to deicing

salts In order to develop cost-competitive ast to construct and durable alternative systems recent

innovations on CIP decks are ocused on developing mixes and curing methods that produce peror-

mance characteristics such as reeze-thaw resistance high abrasion resistance low stiffness and low

shrinkage rather than high strength

FIGURE 151 Reinorcement layout in a cast-in-place concrete deck

FIGURE 152 A cast-in-place concrete deck under construction

Cont reinf

12 min

Min reinf in top slab4 cont 18

Detaildimension typ

Slabthick

8 min

Add reinf when ldquoSrdquo le 11ndash6

2 clear

1 clear4 bars

5 bars

Equal spacing

Extra 5 bars(tot 2 per bay)

Overhang

Max = Lesser of 05S or 6ndash0

Bottom bars

Truss bars

Top bars

ldquoSrdquoldquoSrdquo Girder C to C spacing L L

FIGURE 153 ypical cast-in-place concrete deck details



575Concrete Decks

1522 Precast Concrete Deck

Tere are two types o precast concrete decks ull-depth precast panels and stay-in-place (SIP) precast

prestressed panels combined with CIP topping

Figure 154 shows u ll-depth precast concrete deck panels under construction Te u ll-depth precast

panels have the advantages o signi1047297cant reduction o shrinkage effects and ast construction speed

and have been used or deck replacement with high traffic volumes NCHRP Report 407 (ardos and

Baishya 1998) proposed a ull-depth panel system with panels pretensioned in the transverse direction

and posttensioned in the longitudinal direction NCHRP Report 584 (Badie and ardos 2008) devel-

oped two ull-depth precast concrete bridge deck panel systems a transversely pretensioned system and

a transversely conventionally reinorced system and proposed guidelines or the design abrication and

construction o ull-depth precast concrete bridge deck panel systems without the use o posttensioning

or overlays and (2) connection details or new deck panel systems

Figure 155 shows a partial depth precast panel or SIP precast prestressed panes combined with

CIP topping Te SIP panels act as orms or the topping concrete and also as part o the structuraldepth o the deck Tis system can signi1047297cantly reduce construction time since 1047297eld orming is only

needed or the exterior girder overhangs It is cost-competitive with CIP decks or new structures and

deck replacement However the SIP panel system suffers re1047298ective cracking over the panel-to-panel

joints A modi1047297ed SIP precast panel system has been developed in NCHRP Report 407 (ardos and

Baishya 1998)

FIGURE 154 Full-depth precast concrete deck panels under construction (Courtesy o FHWA)

Cast-in-place concrete

Stay -in-place precast panel

FIGURE 155 ypical stay-in-place precast panel with cast-in-place concrete deck




153 Materials

1531 General Requirements

Material characteristics in a bridge deck shall behave to reduce concrete distress and reinorcement

corrosion and lead to a long service lie with minimum maintenance Expected concrete deck should

behave with the ollowing characteristics (Russell 2004)

bull Low chloride permeability

bull A top surace that does not deteriorate rom reeze thaw or abrasion damage

bull Cracking that is limited to 1047297ne 1047298exural cracks associated with the structural behavior

bull Smooth rideability with adequate skid resistance

NCHRP Synthesis 333 (Russell 2004) recommended that use o the ollowing materials and practices

enhances the perormance o concrete bridge decks

1532 Concrete

bull ypes I II and IP cements

bull Fly ash up to 35 o the total cementitious materials content

bull Silica ume up to 8 o the total cementitious materials content

bull Ground-granulated blast urnace slag up to 50 o the total cementitious materials content

bull Aggregates with low modulus o elasticity low coefficient o thermal expansion and high thermal

conductivity

bull Largest size aggregate that can be properly placed

bull Water-reducing and high-range water-reducing admixtures

bull Air-void system with a spacing actor no greater than 020 mm (0008 in) speci1047297c surace area

greater than 236 mm2

mm3

(600 in2

in3

) o air-void volume and number o air voids per inch otraverse signi1047297cantly greater than the numerical value o the percentage o air

bull Water-cementitious materials ratio in the range o 040minus045

bull Concrete compressive strength in the range o 28minus41 MPa (4000minus6000 psi)

bull Concrete permeability per AASHO Speci1047297cation 277 in the range o 1500minus2500 coulombs

1533 Reinforcement

bull Epoxy-coated reinorcement in both layers o deck reinorcement

bull Minimum practical transverse bar size and spacing

1534 Construction Practices

bull Use moderate concrete temperatures at time o placement

bull Use windbreaks and ogging equipment when necessary to minimize surace evaporation rom

resh concrete

bull Provide minimum 1047297nishing operations

bull Apply wet curing immediately afer 1047297nishing any portion o the concrete surace and wet cure or

at least seven days

bull Apply a curing compound afer the wet curing period to slow down the shrinkage and enhance

the concrete properties

bull Use a latex-modi1047297ed or dense concrete overlay

bull Implement a warrant requirement or bridge deck perormance

bull Gradually develop perormance-based speci1047297cations



577Concrete Decks

154 Design Considerations


bull Maintain a minimum structural depth o concrete deck o 70 in and a minimum concrete cover

o 25 in (64 mm) f cprime ge40 ksi

bull Use prestressing or depth o slabs less than 120 o the design span

bull Place the primary reinorcement in the direction o the skew or the skew angle o the deck less

than 25deg Otherwise place them perpendicular to the main supporting girders

bull Provide shear connectors between concrete decks and supporting beams

bull Provide edge beams at the lines o discontinuity For the deck supported in the transverse direc-

tion and composed with concrete barriers no additional edge beam is needed

1542 Design Limit States

Concrete decks must be designed or Strength I limit state (AASHO 2012) and are usually designed

as tension-controlled reinorced concrete components Strength II limit state o the permit vehicle

axle load does not typically control deck design Concrete decks are also required to meet the require-

ments or Service I limit state to control excessive deormation and cracking Te deck overhang shall

be designed to meet the requirements or Extreme Event II Concrete decks supported by multi-girder

systems are not required to be investigated or the atigue limit state

1543 Analysis Methods

15431 Approximate Method of Analysis

Approximate method o analysis is traditional ly used to design concrete bridge decks (AASHO 4621)

Te method assumes a concrete deck as transverse slab strips o 1047298exure members supported by the lon-

gitudinal girders Te AASHO speci1047297cations (AASHO 2012) require the maximum positive moment

and the maximum negative moment to apply or all positive moment regions and all negative moment

regions in the deck slab respectively Te width o an equivalent interior strip o a concrete deck is

provided in able 151 (AASHO 2012) For deck overhangs the AASHO Article 36134 may apply

For typical concrete deck supported on different girder arrangements with at least three girders and

the distance between the centerlines o the exterior girders not less than 140 f the maximum live load

moments including multiple presence actors and dynamic load allowance based on the equivalent strip

method are provided in AASHO A-4 (AASHO 2012) and are summarized in able 152

15432 Empirical Method of Analysis

Empirical method o analysis (AASHO 972) is a method o concrete deck slab design based on the

concept o internal arching action within concrete slabs In this method effective length o slab shal l be

taken as (1) or slabs monolithic with supporting members the ace-to-ace distance and (2) or slabs

supported on steel or concrete girders distance between the webs o girders Empirical design may beused only i the ollowing conditions are met

bull Cross-rames o diaphragms are used throughout the girders

bull Spacing o intermediate diaphragms between box beams does not exceed 25 f

bull Te deck is composed with supporting steel or concrete girders

bull Te deck is ully cast-in-place and water cured f cprimege 40 ksi

bull Deck o uniorm depth ge70 in except or hunched at girder 1047298anges and the distance between

extreme layers o reinorcement ge40 in

bull Effective length le135 f 60 le effective lengthdesign depth le180

bull Overhangslab depth ge50 or overhangslab depth ge30 with slab composites with continuous

concrete barrier




15433 Re1047297ned Methods of Analysis

Re1047297ned methods o analysis or concrete deck speci1047297ed in AASHO 4632 (AASHO 2012) usually

consider 1047298exural and torsional deormation without considering vertical shear deormation Tey are

more suitable or a more complex deck slab structure or example the end zones o skewed girder decks

TABLE 152 Maximum Live Load Moment per Foot Width

Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)

Negative M LL+IM (kip-ff)

Distance rom Centerline o Girder to Design Section or Negative Moment (in)

00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702

TABLE 151 Equivalent Strips o Concrete Decks

ype o Concrete Deck Direction o Primary Strip

Relative to raffic Width o Primary Strip (in)

Cast-in-place Overhang 450 + 100 X

Cast-in-place Either parallel or

perpendicular

+ M 260 + 66 X

Cast-in-place with stay-in-place

concrete ormwork

Precast post-tensioned minus M 480 + 30S

S = spacing o supporting components (f)

X = distance rom load to point o support (f)

+ M = positive moment

minus M = negative moment



579Concrete Decks

155 Design Example

1551 Bridge Deck Data

A typical section o a steel-concrete composite plate girder bridge is shown in Figure 156

Concrete f c 4000 psiprime= (276 MPa) Ec = 3625 ksi (250 MPa)

Steel Reinorcement A706 Grade 60

f E y s( ) ( )= =60 ksi 414 MPa 29000 ksi 200000 MPa

n E

Es

c

= = 8

Loads Concrete Barrier weight wbarrier = 0410 kl

3 in Future wearing surace wws = 0140 kc (AASHO able 351-1)Reinorced Concrete unit weight wrc = 0150 kc (AASHO C351)

AASHO HL-93 + dynamic load allowance

1552 Design Requirements

Perorm the ollowing design calculations or concrete deck in accordance with the AASHTO LRFD

Bridge Design Speci1047297cations 2012 Edition

bull Select concrete deck thick ness and cover

bull Calcu late Unactored Dead Load Moments

bull Calcu late Unactored Live Load MomentsmdashEquivalent Strip Method

bull Calcu late Factored MomentsmdashStrength Limit State I

bull Design or Positive Flexure

bull Design or Negative Flexure

bull Check Service Limit State

bull Determine the Slab Reinorcement Detailing Requirements

58ndash0

Concretebarriertype 732(Typ)

2

4 12=48ndash0

structureCL

FIGURE 156 ypical section o composite plate girder bridge




1553 Solution

15531 Select Concrete Deck Thickness and Cover

ry dec2k slab thickness t = 9125 in gt Minimum deck thickness = 70 in

DepthSpan= 9125 (144)= 0064 gt 1 20 = 005 No prestressing needed

Use deck top cover C top = 20 in

Use deck bottom cover C bot = 10 in

15532 Calculate Unfactored Dead Load Moments

Dead load or one oot length o concrete deck is calculated as ollows

Deck concrete weightmdashW DC1mdashdeck concrete weightmdash

W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=

Barrier weight W DC2 (concentrate load applied at 7 in rom the edge o deck)

W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier

Future wearing surace o 3 inmdashW DW

W wthickness o wearing surace 10

3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=

Te dead load moments or the deck slab can be calculated using a continuous beam as shown in

Figure 157able 153 lists unactored dead load moments Only the results or Spans 1 and 2 are shown in the

table since the bridge deck is symmetrical the centerline o the bridge

15533 Calculate Unfactored Live Load Moments

From able 152 unactored live load moments including multiple presence actors and dynamic load

allowance are obtained as ollows

For girder spacing S = 12 f maximum positive live load moments are as

M 801 kip-ttLL IM =+

For negative 1047298exure the design sections are located the ace o the support or monolithic concrete

construction 14 the 1047298ange width rom the centerline o the support or steel girder bridges and 13 the

1047298ange width not exceeding 15 in rom the centerline o the support or precast I-girders or open-boxgirders (AASHO 2012 Article 46216)

85

W DC2 W DC2

W DC1W DW

5ndash0 12ndash0Span 1

12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17

FIGURE 157 Concrete deck under unactored dead loads



581Concrete Decks

For this example assume steel girder 1047298ange width = 18 in the design section is at frac14(18)= 45 in rom

the centerline o steel girder as shown in Figure 158 Te negative moment can be obtained conservatively

as the moment at the centerline o the support or interpolated between moments at 3 in and 6 in

M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+

1554 Calculate Factored MomentsmdashStrength Limit State IFor Strength Limit State I load combination actored moment ollows

( ) ( )= η γ + + γ + γ +DC DC1 DC2 DW DW LL LL IM M M M M M u

η= η η η ge 095D R I

For this example use η = 095 γ =125DC γ = 150DW and γ = 175LL

( )( )= + + + +095 [125 15 175 ]DC1 DC2 DW LL IM M M M M M u

TABLE 153 Unactored Dead Load Moments

Distancerom lef

support X

(f)

Location

XS

Deck Load DC1 M DC1 (kip-ff)

Barrier Load DC2 M DC2 (kip-ff)

Future Wearing Surace DW M DW (kip-ff)

Span 1 Span 2 Span 1 Span 2 Span 1 Span 2

00 00 minus1425 minus1352 minus1760 0496 minus0225 minus0475


24 02 minus0097 minus0044 minus1309 0348 0128 minus0055

36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112

96 08 minus0053 minus0058 0045 minus0097 minus0022 minus0006



Design

sectionC webL

bf = 18

14bf = 45

FIGURE 158 Design section or negative moment




15541 Maximum Positive Factored Moments

From able 153 it is seen that the maximum unactored positive moments due to the concrete deckslab barrier and uture wearing surace is located in Span 2 at a distance o 05 S Te maximum live

load positive moment equals 801 kip-ff Tereore the maximum positive actored moment is

M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =

15542 Maximum Negative Factored Moments

From able 153 it is seen that the maximum unactored negative moments due to the concrete deck

slab barrier and uture wearing surace is located Span 1 at the centerline o exterior girder and can be

obtained conservatively as the moment at the centerline o the exterior support or interpolated between

00S and 01S as ollows

M 0679 12 45

12(1425 0679) 1145 kip-ttDC1 = minus minus minus minus = minus

M 1534

12 45

121760 1534 1675 kip-ttDC2 ( )= minus minus

minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus

Te maximum actored negative moment is as

M u 095 (125)( 1145 1675) (15)( 0149) (175)( 896) 18457 kip-tt[ ]= minus minus + minus + minus = minus

1555 Design for Positive Flexure Design

ry 5 bar size bar area = 031 in2 and bar diameter= 0625 in

Te effective depth d e = total slab thicknessmdashbottom covermdashhal bar diameter

( ) ( )= minus minus = minus minus =

bar diameter

29125 10

0625

27813 inbotd t C e

ry 58 in which is less than the maximum spacing 15t = 18 in = = As

12(031)

80465 in2

For a rectangular section with a width o b = 12 in and depth o t = 9125 inConcrete compression block depth

a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in

Distance rom the extreme compression 1047297ber to the neutral axis

=

β = =c

a 0684

0850801 in

1



583Concrete Decks

ensile strain o rebar is

ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005

Tereore the section is tension controlled resistance actor ϕ = 09

M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin

1563 kip-tt 14561kip-tt

( )( )( )= φ = φ minus

= minus

=

= gt =

1556 Design for Negative Flexurery 5 bar size bar area = 031 in2 and bar diameter= 0625 in

Te effective depth d e = total slab thicknessmdashtop covermdashhal bar diameter

d t C e


bar diameter

29125 20

0625

26813 intop

ry 55 in which is less than maximum spacing 15t = 18 in = =12(031)

50744 in2 As

For a rectangular section with a width o b = 12 in and depth o t = 9125 in

Concrete compression block depth

a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin

2098 kip-tt 18457 kip-tt

( )( )( )= φ = φ minus

= minus

=

= gt =




1557 Check Service Limit State-Cracking Control

Concrete cracking is controlled by the proper distribution o 1047298exure reinorcement at service limit state

AASHO (2012) requires steel reinorcement spacing s o the layer closet to the tension ace to satisy

the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107

where γ e is 075 or Class 2 exposure conditions d c is thickness o concrete cover measured rom extreme

tension 1047297ber to the center o the 1047298exural reinorcement f ss is tensile stress in steel reinorcement at ser-

vice limit state and h is overall thickness o the deck

15571 Service I Load Combination

( )= + + +10 10 10DC1 DC2 DW LL+IM M M M M M s

Maximum positive moment

M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =

Maximum negative moment

M u 10 (10)( 1145 1675) (10)( 0149) (10)( 896) 11929 kip-tt[ ]( )= minus minus + minus + minus = minus

15572 Positive Flexure Cracking Control

= + = + =10bar diameter

210

0625

21313ind c

Assume y is the distance o the neutral axis to extreme compression 1047297ber or the transormed rectangu-lar concrete section with b = 12 in d e = 7813 in = =n

E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b

2 =B nAs = minusC nA d s e



585Concrete Decks

For bottom reinorcement designed or positive 1047298exure As = 0465 in2

= = = A b

2

12

26

= = =B nAs 8(0465) 372

( )= minus = minus times times = minusC nA d s e 8 0754 675 29064

( ) ( )( )( )

( ) ( )=

minus + minus=

minus + minus minus=

4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A

Moment o inertia o cracked or the transormed section I cr is

( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e

ensile stress f ss in the steel reinorcement at service limit state is

( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in

It is obvious that 5 8 in meets cracking control requirement

15573 Negative Flexure Cracking Control

= + = + =d c 25bar diameter

225

0625

22813in

Assume y is the distance o the neutral axis to extreme compression 1047297ber or the transormed rectangular

concrete section with b = 12 in d e = 6813 in = =n EE

s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b





For top reinorcement designed or negative 1047298exure As = 0744 in2

= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952

( )( )( )= minus = minus = minusC nA d s e 8 0744 6813 40551

( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in

ry 545 in or negative moment in the top reinorcement

Use 59 in (truss Bar) and 59 in (straight bar) or both top and bottom reinorcement in the

transverse direction as shown in Figure 159

Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18

115 barsEqual spacing

9125 2 clear

1 clear

FIGURE 159 Bridge deck reinorcement detail



587Concrete Decks

1558 Determine the Slab Reinforcement Detailing Requirements

15581 Top of Slab Shrinkage and Temperature Reinforcement

Te top slab distribution reinorcement is or shrinkage and temperature changes near the surace o the

exposed concrete slab AASHO Article 5108 (AASHO 2012) requires the area o reinorcement in

each direction and each ace As shall meet the ollowing requirements

ge+

A bh

b h f s

y

13

2( )

le le As011 060

where b is the least width o component section h is least thickness o component section f y is speci1047297ed

yield strength o reinorcing bars less than 75 ksiry 418 in bar cross section area = 02 in2

As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2

Using 418 in or longitudinal distribution reinorcement and 545 in or ransverse primary

reinorcement meets this requirement15582 Bottom of Slab Distribution Reinforcement

Te distribution reinorcement on the bottom o the slab is placed in the perpendicular direction to

the primary reinorcement or positive moment and calculated based on whether the primary rein-

orcement is parallel or perpendicular to traffic (AASHO 2012) For this example the primary rein-

orcement is perpendicular to traffic AASHO Article 9732 requires that bottom slab distribution

reinorcement ratio shall be larger than S lt220 67 where S is the effective span length taken as the

distance between the 1047298ange tips plus the 1047298ange overhang For steel girder S is taken as girder spacing

o 12 f conservatively

= = lt

220 220

12635 67

S

Bottom primary reinorcement 545 in As

12(031)

450827 in t2= =

Since bottom distribution reinorcement usually placed within the center hal o the deck span total

required distribution reinorcement area

A ( )= =0635 0827 (6) 315inrequired

2

ry 115 bar ( )( )= = gt =11 031 341 in 315 in2required

2 A As

Figure 159 shows the detailed deck reinorcement or the design example




References

AASHO 2012 AASHTO LRFD Bridge Design Speci1047297cations Customary US Units 2012 American

Association o State Highway and ransportation Officials Washington DC

Badie S S and ardos M K 2008 Full-Depth Precast Concrete Bridge Deck Panel Systems NCHRP

Report 584 ransportation Research Board Washington DC

Barker R M and Puckett J A 2007 Design o Highway Bridges 2nd Edition John Wiley amp Sons Inc

New York NY

FHWA 2012 Concrete Deck Design Example Design Step 2 httpwwwwadotgovbridgelrdus_ds2

htmdesignstep21_0

Russell H G 2004 Concrete Bridge Deck Perormance NCHRP Synthesis 333 ransportation Research

Board Washington DC

ardos M K and Baishya M C 1998 Rapid Replacement o Bridge Decks NCHRP Report 407

ransportation Research Board Washington DC




the supporting girders and slow construction the tendency o the deck rebar to corrode due to deicing

salts In order to develop cost-competitive ast to construct and durable alternative systems recent

innovations on CIP decks are ocused on developing mixes and curing methods that produce peror-

mance characteristics such as reeze-thaw resistance high abrasion resistance low stiffness and low

shrinkage rather than high strength

FIGURE 151 Reinorcement layout in a cast-in-place concrete deck

FIGURE 152 A cast-in-place concrete deck under construction

Cont reinf

12 min

Min reinf in top slab4 cont 18

Detaildimension typ

Slabthick

8 min

Add reinf when ldquoSrdquo le 11ndash6

2 clear

1 clear4 bars

5 bars

Equal spacing


Overhang

Max = Lesser of 05S or 6ndash0

Bottom bars

Truss bars

Top bars

ldquoSrdquoldquoSrdquo Girder C to C spacing L L

FIGURE 153 ypical cast-in-place concrete deck details



575Concrete Decks


















Baishya 1998)








153 Materials











1532 Concrete






conductivity





mm3

(600 in2

in3





1533 Reinforcement






resh concrete



at least seven days








577Concrete Decks











































concrete barrier









Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)



00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702






perpendicular

+ M 260 + 66 X


concrete ormwork








579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





575Concrete Decks


















Baishya 1998)








153 Materials











1532 Concrete






conductivity





mm3

(600 in2

in3





1533 Reinforcement






resh concrete



at least seven days








577Concrete Decks











































concrete barrier









Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)



00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702






perpendicular

+ M 260 + 66 X


concrete ormwork








579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC






153 Materials











1532 Concrete






conductivity





mm3

(600 in2

in3





1533 Reinforcement






resh concrete



at least seven days








577Concrete Decks











































concrete barrier









Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)



00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702






perpendicular

+ M 260 + 66 X


concrete ormwork








579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





577Concrete Decks











































concrete barrier









Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)



00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702






perpendicular

+ M 260 + 66 X


concrete ormwork








579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC











Girder

Spacing

S (f)

Positive

Moment

M LL+IM

(kip-ff)



00 30 60 90 120 180 240

40 468 268 207 174 160 150 134 125

45 463 300 258 210 190 165 132 118

50 465 374 320 266 224 183 126 112

55 471 436 373 311 258 207 130 099

60 483 499 419 350 288 231 139 107

65 500 531 457 384 315 253 150 120

70 521 598 517 436 356 284 163 151

75 544 626 543 461 378 315 188 17280 569 648 565 481 398 343 249 216

85 599 666 582 498 414 361 296 258

90 629 681 597 513 428 371 331 300

95 659 715 631 546 466 404 368 339

100 689 785 699 613 526 441 409 377

105 715 852 764 677 589 502 448 415

110 746 914 826 738 650 562 486 452

115 774 972 884 796 707 719 552 487

120 801 1028 940 851 763 674 556 521

125 828 1081 993 904 816 728 597 554

130 854 1131 1043 955 867 779 638 586

135 878 1179 1091 1003 916 828 679 616

140 902 1224 1137 1050 963 867 718 645

145 925 1267 1181 1094 1008 921 757 672

150 947 1309 1223 1137 1051 965 794 702






perpendicular

+ M 260 + 66 X


concrete ormwork








579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





579Concrete Decks

155 Design Example






n E

Es

c

= = 8















58ndash0


2

4 12=48ndash0

structureCL





1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC






1553 Solution









W t wrc109125

12

(10) (015) 0114 kiptDC1 ( )( )= =

=


W w( ) ( )( )= = =10 10 041 041 kipDC2 barrier



3

1210 014 0035 kiptDC2 ws( )( ) ( )( )= =

=












85

W DC2 W DC2

W DC1W DW


12ndash0Span 2

12ndash0Span 3

12ndash0Span 4

5ndash0

17




581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





581Concrete Decks




M 851

3 15

3940 851 896 kip-ttLL IM ( )minus = +

minusminus =+







Distancerom lef

support X

(f)

Location

XS








36 03 0321 0365 minus1083 0273 0229 0079

48 04 0574 0608 minus0858 0199 0280 0163

60 05 0664 0688 minus0632 0125 0280 0196

72 06 0589 0604 minus0406 0051 0230 0179

84 07 0350 0355 minus0181 minus0023 0129 0112




Design

sectionC webL

bf = 18

14bf = 45








M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC









M u 095 (125)(0688 0125) (15)(0196) (175)(801) 14561kip-tt[ ]= + + + =






M 0679 12 45


M 1534

12 45


minusminus = minus

M 0023

12 45

12(0225 0023) 0149 kip-ttDW = minus +

minusminus = minus







bar diameter

29125 10

0625

27813 inbotd t C e


12(031)

80465 in2


a

A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0465 60

085 40 120684 in


=

β = =c

a 0684

0850801 in

1



583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





583Concrete Decks


ε =

minus=

minus= gt

d c

ct

e (0003)(7813 0801)

0801(0003) 0026 0005


M M A f d a

M

r n s y e

u

209 0465 60 7813

0684

21876 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =



d t C e


bar diameter

29125 20

0625

26813 intop


50744 in2 As



a A f

f b

s y

c

( )( )

( )( )( )=

prime = =

085

0744 60

085 40 121094 in


=β

= =c a 1094

0851287

1


ε =

minus=

minus= gt

d c

ct

e (0003)6813 1287

1287(0003) 0013 0005


M M A f d a

M

r n s y e

u

209 0744 60 6813

1094

225174 kip-inin


( )( )( )= φ = φ minus

= minus

=

= gt =







the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC









the ollowing

s f

d e

s ssc ( )le

γ β

minus700

2 AASHO 5734-1

in which

( )β = +

minusd

h d s

c

c

107







M s 10 (10)(0688 0125) (10)(0196) (10)(801) 9019 kip-tt[ ]= + + + =





210

0625

21313ind c


E

Es

c

8 we have

+ minus =b

y nA y nA d s s e2

02

= minus + minus

y B B AC

A

4

2

2

in which = A b




585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





585Concrete Decks


= = = A b

2

12

26

= = =B nAs 8(0465) 372


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

372 372 4 6 29064

2 61912 in

2 2

y B B AC

A


( )

( )( )( )( )( )= + minus = + minus =

3

12 1912

38 0465 7813 1912 157496 incr

32

32 4I

by nA d y s e


( ) ( )( )( )( )

= minus

= minus

=8 9019 12 7813 1912

213463244 ksi

cr

f nM d y

I ss

s e

( ) ( )β = +

minus = +

minus =

d

h d s

c

c

107

11313

0 07 (9125 1313)1240

( )=

γ β

minus = minus =s f

d e

s ss

c

7002

700 (075)

(1240)(3244)(2)(1313) 1043in




225

0625

22813in



s

c

8 we have

+ minus =b

y nA y nA d s s e2

02

=

minus + minus y

B B AC

A

4

2

2

in which = A b






= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC







= = = A b

2

12

26

( )( )= = =B nAs 8 0744 5952


( ) ( )( )( )

( ) ( )=

minus + minus=


4

2

5952 5952 4 6 40551

2 62151in

2 2 2

y B B AC

A


I by

nA d y s e( ) ( )( )

( )( )( )= + minus = + minus =3

12 2151

38 0744 6813 2151 16917 incr

32

32 4


( ) ( )( )( )( )=

minus=

minus=

8 11929 12 6813 2151

1691731561ksi

cr

f nM d y

I ss

s e

( ) ( )( )

β = +minus

= +minus

=d

h d s

c

c

107

125

07 9125 28131637

( )= γ

β minus = minus =s

f d e

s ss

c700 2 700 (075)

(1637)(31561)(2)(2813) 454 in




Bottom bars 59

Top bars

Truss bars

59

59


2ndash0 typ

12ndash0

4 cont 18


9125 2 clear

1 clear




587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC





587Concrete Decks






ge+

A bh

b h f s

y

13

2( )

le le As011 060



As

12(02)

180133in t2= =

A

bh

b h f s

y

13

2( )

13(12)(9125)

2(12 9125)(60)0056in t2gt

+ =

+ =

As ge 011in t2










= = lt

220 220

12635 67

S


12(031)

450827 in t2= =



A ( )= =0635 0827 (6) 315inrequired

2


2 A As





References






New York NY


htmdesignstep21_0


Board Washington DC






References






New York NY


htmdesignstep21_0


Board Washington DC



Documents

Concrete Decks