1

Bridge Deck Design

4.1

g g

Interior region of deck

1. Empirical method - described in LRFD 9.7.22. Strip method - loads described in LRFD 3.6.1.3.3

analysis described in LRFD 4.6.23. Refined methods - Finite Element Method, etc, using

loads described in LRFD Chap. 3

DECK SLAB DESIGN METHODS 4.2

Sn equal spans @ S

Overhangs: LRFD 9.7.2.21. Empirical method not applicable to overhangs. 2. Strip method or refined method must be used in the cantilever

overhang to account for wheel loads between the railing and the exterior grider, and for collision loads applied to the railing.

REINFORCING DEFINITIONS

Railing reinforcement

Longitudinal slabreinforcement

[ 2 layers ]

Slab ( "Deck" )Horizontal

•

•

•

•

•

•

•

•

•

•

Vertical

4.3

Transverse slabreinforcement

[ 2 layers ]

Interior region

Overhang

•

•

Railing

2

LRFD DESIGN CRITERIA

Slab transverse reinforcement (interior region):

Service limit stateStrength limit state

Slab transverse reinforcement (overhang):

Service limit stateStrength limit state

gravity loads

gravity loads

4.4

Strength limit stateExtreme event limit state collision (rail) plus

reduced gravity loads

Slab longitudinal reinforcement (all)

Secondary flexure; temperature/shrinkage control

Rail horizontal and vertical reinforcement:

Extreme event limit state collision (rail)

Live Load: HL-93

Deck Concretef’c = 4 ksiwc =150 pcf

Mild Steel (Non-Prestressed)f = 60 ksi

Problem Definition4.5

fy 60 ksiEs =29,000 ksi

DimensionsThickness = 8.0 in. (LRFD 9.7.11 & 13.7.3.1.2)Cover = 2.5 in. (Top) (LRFD 5.12.3)

= 1.0 in. (Bottom)

Future Wearing Surface Allowance: FWS = 30 psf

Deck Design4.6

3

Sn spans @ S

Interior region of deck

Definitions used in empirical design method4.7

Le

S

bf

bw

S - bf

Effective length: Le

The effective length is the distance between the flange tips (S− bf) plusthe flange overhang ½ (bf − bw):

Le = S − ½ (bf + bw)

Empirical MethodBased on extensive lab testsLoad resistance mechanisms» Flexure» Arching Action

FEM verification

4.8

FEM verificationFactor of safetyNo analysis requiredIsotropic reinforcementNot applicable to overhang design

Empirical Method – Design ConditionsDiaphragms at lines of support

Concrete and/or steel girders

CIP and water cured deck

Uniform depth

Effective length to depth ratio

4.9

Effective length to depth ratio

– 6 to 18

Effective length

– 13.5 ft., maximum

4

Empirical Method – Design Conditions

Core depth – 4.0 in., minimum

Slab thickness – 7.0 in., minimum

Minimum overhang to depth ratio

» 5

4.10

» 3, if barrier is composite

fc’ - 4 ksi, minimum

Deck is composite

Empirical Method – DesignBottom Layer, each way: 0.27 in.2 / ft.

(#5 bars @ 13.5 in. spacing – As,prov’d = 0.276 in.2 / ft.)

Top Layer, each way: 0.18 in.2 / ft.(#4 bars @ 13 in. spacing – As,prov’d = 0.185 in.2 / ft.)

G d 60 t l

4.11

Grade 60 steel

Outermost bar in direction of effective length

Maximum spacing – 18 in. o.c.

Reinforcement doubled in end zone if skew exceeds 25°

Steel placement:

Reinforcement shall be provided in each face of the slab with theoutermost layers placed in thedirection of the effective length (i.e., transverse steel outside)

•••

•

transverse steellongitudinal steel

Empirical design method details4.12

Design depth: ts(excluding wearingsurface)

Core depth: tc••

••

Top cover

Bottom cover

Wearing surface

transverse steel

longitudinal steel

Slab thickness:

5

Empirical Method – Final Design4.13

Strip MethodContinuous beam loaded with truck axle loadsEquivalent strip widths – interior, exterior, and overhangDL moments on a per foot width basisLL moments:

M i l d l i

4.14

» Moving load analysis– Truck axles moved laterally– Multiple presence factors– Dynamic load allowance– Total moment divided strip width

» LRFD Table A4.1-1 (used in this design example)

STRIP METHODApplicability: The strip method (also called the approximate method in the LRFD Specification)

provides for selecting all transverse and longitudinal reinforcing in the slab, andthe reinforcing required to connect the railing to the slab.

The deck is subdivided into 1-ft-wide strips perpendicular to the supporting girders. The strips aretreated as continuous beams. The span length is taken as the center-to-center distance betweenthe girders. The girders are assumed to provide rigid point supports for the continuous-beam strip.

4.15

1’ width

n spans @ SS

6

railing (parapet)thickened slab in overhang

main slab wearing surface

S

wearing surface

Dead loads on slab strip4.16

wearing surface

added slab thickness main slab thickness

railing weight per foot length

S

Live loads on slab stripLive load used in deck design is that caused only by truck axle loads ( lane load not included ). Recall the lane width is 10 feet.

6'

16k 16k

8' 6' 4' 6'

16k 16k 16k 16k

4.17

One truck Two adjacent trucks

spacing varies

Two trucks in alternate spans(use influence lines for placement)

6' 6'

16k 16k 16k 16k

Live loads on slab strip ( continued )

6'

16k 16k

8' 6' 4' 6'

16k16k 16k16kOne truck120% of the moment caused by one truck:

m = 1.20

Two trucks100% of the moment caused by two trucks:

m = 1. 0

Use whichever produces the greatest moment at each point across width of deck:

4.18

6’

16 kips16 kips

vary position andspacing of truck(s)

1’ width

7

Live loads on slab strip ( continued )Examples of critical wheel load positions:

6’

1’

16 kMultiply moment produced by one truck by m = 1.2

Maximum negative moment at exterior support ( wheel load in cantilever ):

4.19

Maximum positve moment in end segment:

Position of inner wheel relative tosupport depends on grider spacing.

≈ 0.4 S Approximate location of maximum positivemoment caused by dead loads

6’16 k

Multiply moment produced by one truck by m = 1.2

S

16 k

Live loads on slab strip ( continued )Examples of critical wheel load positions ( continued ):

S < 10 feet

16k 6’ 16k

S > 10 feet

16k 6’ 16k 16k 6’ 16k

4’

Maximum negative moment at first interior support :

4.20

S < 10 feet

Multiply moment produced by one truck by m = 1.2

S > 10 feet

Multiply moment produced by two trucks by m = 1.0

In all cases, the truck load moments must be increased by the impact factor: MLL = MTruck ( 1 + IM ) = MTruck ( 1 + 0.33 )

Live loads on slab strip ( continued )The moments computed using the truck load patterns just described are distributed to the1’-wide deck strip by dividing by each moment value (foot-kips) by equivalent strip widthsthat are defined in LRFD Table 4.6.2.1.3-1:

Equivalent strip width: SW

4.21

1’ widthMdesign

)feet(SW)kipsfoot(M)foot/kipsfoot(M LL

design−

=−

8

Live loads on slab strip ( continued )

Computer analyses are required to consider all possible live load combinations and positions.AASHTO LRFD Table A4.1-1 (shown on the next page) gives values of maximum live loadmoments per foot width that may be used in designing the interior slab segments in slab-girder bridges. The moments given in this table account for:

1. Grider spacing

2. Placement of trucks to produce the maximum positive moment beween girders, and the maximum negative at the interior girders ( m = 1.2 applied to moments caused by one truck; m = 1.0 applied to moments caused by two trucks)

4.22

3. The dynamic (impact) factor of 1.33

These moments may be combined directly with the dead load moments to determine the the critical moments (per foot width) to be used in designing the deck reinforcing in the

interior deck segments.

The live load moments at the exterior girder must be computed separately. The criticalload at the exterior girder is generally that associated with truck collision into the rail.

Negative Moment , k-ft/ftDistance from CL of girder to design section for negative moment

GirderSpacing

ft

PositiveMoment

k-ft/ft 0 in. 3 in. 6 in. 9 in. 12 in. 18 in. 24 in.4’-0” 4.68 2.68 2.07 1.74 1.60 1.50 1.34 1.254’-3” 4.66 2.73 2.25 1.95 1.74 1.57 1.33 1.204’-6” 4.63 3.00 2.58 2.19 1.90 1.65 1.32 1.184’-9” 4.64 3.38 2.90 2.43 2.07 1.74 1.29 1.20

Table may be used for all positive LL moments

Table may be used for negative LL moments at all interior supports

Must compute negative LL moment in overhang

AASHTO LRFD Table A4.1-1

Live loads on slab strip ( continued )4.23

5’-0” 4.65 3.74 3.20 2.66 2.24 1.83 1.26 1.125’-3” 4.67 4.04 3.47 2.89 2.41 1.95 1.28 0.985’-6” 4.71 4.36 3.73 3.11 2.58 2.07 1.30 0.995’-9” 4.77 4.63 3.97 3.31 2.73 2.19 1.32 1.026’-0” 4.83 4.88 4.19 3.50 2.88 2.31 1.39 1.07

13’-0” 8.54 11.31 10.43 9.55 8.67 7.79 6.38 5.8613’-3” 8.66 11.55 10.67 9.80 8.92 8.04 6.59 6.0113’-6” 8.78 11.79 10.91 10.03 9.16 8.28 6.79 6.1613’-9” 8.90 12.02 11.14 10.27 9.40 8.52 6.99 6.3014’-0” 9.02 12.24 11.37 10.50 9.63 8.76 7.18 6.4514’-3” 9.14 12.46 11.59 10.72 9.85 8.99 7.38 6.5814’-6” 9.25 12.67 11.81 10.94 10.08 9.21 7.57 6.7214’-9” 9.36 12.88 12.02 11.16 10.30 9.44 7.76 6.8615’-0” 9.47 13.09 12.23 11.37 10.51 9.65 7.94 7.02

Treating the slab strip as a continuous beam with rigid point supports results in overly-conservative negative moments at the centerline of each girder. The girder flangeshelp resist the negative moment at the girders, thus reducing the negative moment forwhich the slab must be designed

Live loads on slab strip ( continued )4.24

dc

Maximum negative momentin slab when girder is treatedas a point support.

Actual negative momentexperienced by slab

Moment computed at dcwhen girder is treatedas a point support

9

From LRFD 4.6.2.1.6:

The design section for negative moments in precast I-shaped and T-shaped concrete beamsmay be taken at one-third the flange width but not exceeding 15 0 inches from the centerline

dc

Live loads on slab strip ( continued )4.25

may be taken at one-third the flange width, but not exceeding 15.0 inches, from the centerlineof the point support assumed at the centerline of the girder.

"0.153bd f

c ≤=

Table A4.1-1 gives values of negative live load moment at distances dc = 3, 6, 9, 12, 18 and 24”from the girder centerline. Moments at other dc locations can be found by interpolating momentvalues at adjacent tabulated distances.

From LRFD Commentary 4.6.2.1.6: This reduction in negative moment replaces the use of reduced span length to compute moments as done in the standard specification.

Strip MethodLimit states» Service: crack control» Fatigue: need not be checked» Strength: factored moments» Extreme event: vehicular collision

4.26

Strip Method – DL Moments

DL / LL ratioC = 10 or 12Self weight = 8(150)/12

= 100 psf = 0.1 ksf

cwl

M2

=

4.27

.ft/.ftkip 81.010

9x1.0M2

DL −==

.ft/.ftkip 24.010

9x03.0M2

FWS −==

Future wearing surface = 30 psf = 0.3 ksf

10

Strip Method – LL MomentsTable A4-1Span = 9 ft.Critical section for negative moment» (1/3) bf = 14 in. (governs) ≤ 15 in.» Use 12 in (conservative)

4.28

» Use 12 in. (conservative)

.ft/.ftkip29.6M posLLI -=

.ft/.ftkip71.3M negLLI -=

Strip Method – Service LS Moments

Service Limit State:

» Negative Interior Moment:

Mneg = -(0.81+0.24+3.71) = -4.76 kip-ft. / ft.

4.29

» Positive Moment:

Mpos = (0.81+0.24+6.29) = 7.34 kip-ft. / ft.

Strip Method – Strength LS Moments

Strength Limit State

» Negative Interior Moment:

Mneg,str Mneg,str = -(1.25x0.81 + 1.5x0.24 + 1.75x3.71)

= -7.87 kip-ft. / ft.

4.30

p

» Positive Moment:

Mpos,str = 1.25x0.81 + 1.5x0.24 + 1.75x6.29

= 12.38 kip-ft. / ft.

11

Strip Method – Flexure DesignMneg,str = -7.87 kip-ft. / ft.

» Try No. 5 at 10 in. o.c.

» As = (12/10)(0.31 in.2/bar)

= 0.372 in.2 / ft.

⎟⎠⎞

⎜⎝⎛ −=

2ad

bfA

M ysn φφ

bf85.0fA

a 'c

ys=

4.31

in. 0.65 ===85.0

547.085.0ac.in 547.0a ==

12)(0.85)(4)()(0.372)(60

( ) ( )

flexure for 0.9 section controlled-tension Therefore

ccd

section controlled mpressionTension/Co Check

tt

=

≥=−

=−

=

φ

ε

,

005.0021.0003.0*65.0

65.019.5003.0*

Strip Method – Flexure Design

⎟⎠⎞

⎜⎝⎛ −=

2ad

bfA

M ysn φφ

ft /ftki238)547.0195()60)(372.0)(90.0(M ⎟⎞

⎜⎛

4.32

ft./ft.-kip 23.8)2

19.5()12(

))()((Mφ n =⎟⎠⎞

⎜⎝⎛ −=

ϕMn = 8.23 kip-ft. / ft. > Mneg,str = 7.87 kip-ft. / ft. O.K.

Strip Method – Crack ControlMaximum spacing of tension reinforcement

» LRFD Article 5.7.3.4 applies if fMneg > 0.8fr

ksi38.0424.0*8.0f24.0*8.0f8.0 'cr ===

12*764

4.33

Therefore, Article 5.7.3.4 applies

ksi 38.0ksi 45.0

68*1212*76.4f2Mneg

>=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

12

Strip Method – Crack ControlMaximum spacing of tension reinforcement

css

e d2fβγ700

s −≤

exposure2 Class for 75.0γ,where

e =

4.34

dc = cover – extreme tension fiber to center of extreme reinf.

= 2.5” (clear cover) + 0.625 (diameter of No. 5 bar)/2

= 2.81 in.

77.1)81.28(7.0

81.21)dh(7.0

d1β

c

cs =+=+=

--

fs = Stress in reinf. based on cracked section analysis

bεc fc

NeutralAxis

kds

13kds

C

Strip Method – Crack ControlCalculate fs

4.35

M

d s

sε fs

Axis

jd = (1 - )dk3s s

Elevation Section Strain Stress Resultant Forces

T

Figure 3: Reinforced concrete rectangular beam section at service load

Strip Method – Crack Controlwhere:

M = -4.76 kip-ft./ft.As = No. 5 at 10” o.c. = 0.31/10*12 = 0.372 in.2/ ft.ds = 8 – 2.5 – 0.625/2 = 5.19 in.

sss jdA

Mf =

n = modular ratio = Es / Ec = 29,000 / 3,830 = 7.57. Use 8 ≥ 6 OK(LRFD 5.7.1) ksi 830,30.4)150.0)(000,33(fw000,33E 5.1'

c5.1

cc ===

4.36

bdA

ρ s=

00597.0)19.5)(12(

372.0ρ ==

( ) nρnρnρ2k 2 -+=

[ ] 265.0)8)(00597.0()8)(00597.0()8)(00597.0)(2(k 2 =+= -

3k1j -= 912.03/265.01j =−= ksi 4.32

)19.5)(912.0)(372.0()12*76.4(fs ==

13

Strip Method – Crack Control

Provided No. 5 at 10 in. o.c. > 3.53 in. o.c. N.G.

Reduce spacing to 7 in. o.c.R i d i i 7 2 i O K

.in53.381.2*24.32*77.1

75.0*700d2

fβγ700

s css

e ==≤

4.37

Revised maximum spacing = 7.2 in. O.K.

Therefore, for negative interior moments: Provide #5 @ 7 in. o.c. (As prov'd = 0.53 in.2 / ft.)Mneg prov’d = 11.5 kip-ft./ft.

Similar calculations for Mpositive suggestNo. 5 at 8 in. o.c. are adequate (As, prov’d = 0.465 in2/ft.)Mpos prov’d = 13.3 kip-ft./ft.

LONGITUDINAL STEELSecondary longitudinal reinforcement (parallel to the girders ) in the bottom of the slab is required as a percentage of the primary positive-moment transverse reinforcement perpendicular to the girders:

Longitudinal steel percentage of primary reinforcement = eL

220 % ≤ 67%

where:

Le = effective length (feet) is the distance between the flange tips plus the flange overhang:

Le = S − (bf + bw) / 24

4.38

S

° ° ° ° ° °• • • • • • • • • •

Le

bf

bw

S - bf

longitudinal steel (bottom of slab)SbL

temperature steel

Strip Method – Distribution Reinforcement(LRFD 9.7.3.2)

At bottomIn secondary directionPercent of reinforcement for Mpositive

ft. 8.5 in.102 6 - 108 S where %,67S

220===≤

4.39

Governs 67% ,%67%755.8

220>=

As = 0.67(0.47 in.2 / ft.) = 0.31 in.2 / ft.

Provide #5 @ 12 in. o.c. (As prov'd = 0.310 in.2 / ft.)

14

Strip Method – Shrinkage & Temp. Reinf.

28.10.5.60.011.0

18.10.5.)(2

3.1

−≤≤

−+

=

EqA

Eqfhb

bhA

s

ys

11.0085.060*)8510(*2

8*510*3.1::1

==+

= ss AthereforeA

deckofwidthfullConsiderMethod

,

42’ – 6” = 510 in

8 in

12:2 = indeckofwidthunitConsiderMethod

4.40

Maximum spacing: 3*8=24 in or 18 in (governs)

Provide No. 4 @ 18 in. o.c. (As prov'd = 0.27 in2 / ft.)

11.0,087.060*2496*3.1

24)012(*2968*12

12:22

===

=+===

=

ss AthereforeA

inperimeterDryinginArea

indeckofwidthunitConsiderMethod

12 in

8 in

Strip Method – Minimum Reinforcement

ft./ft.-kip 7.9 in.-kip 94.7 )8*12(*74.0M2

cr ===

Mr ≥ lesser of 1.2 Mcr or 1.33 Mu (LRFD 5.7.3.3.2)

crcr SfM =

ksi 74.0437.0f37.0f cr ==′=

4.41

pp)6

(cr

Mpos prov’d = 13.3 kip-ft/ft and Mneg prov’d = 11.5 kip-ft/ft> 9.5 kip-ft/ft OK

ft/ft-kip 1.33Mft/ft-kip 1.33M

(governs) ft/ft-kip M

strpos,

strneg,

cr

47.1638.12*33.147.1087.7*33.1

5.99.7*2.12.1

==

==

==

Empirical vs. Traditional

Total reinforcement per square foot of deck:

Empirical method:2[0.276 + 0.185] = 0.922 in.2 / ft. (- 41%)

4.42

Traditional method:0.53 + 0.465 + 0.310 + 0.27 = 1.575 in.2 / ft. (+ 71%)

15

DECK CANTILEVER / RAILING ANALYSIS

The deck cantilever is designed for the negative moment at the exterior girder causedby whichever of the two load conditions shown below produces the greatest value ofthis moment:

Dead load moment: railing, deck, FWS

PLUS

4.43

Full truck live load(gravity load)

Critical momentlocation

Reduced truck gravity load plus lateral load applied to railing

Critical momentlocation

OR

Deck Cantilever / Railing Analysis ( continued )

Truck gravity load:The truck live load is applied as a uniformly distributed load WL located 1.0 foot

from the inside face of the railing:

WL = uniform load distributed longitudinally

4.44

1.0 ftdc

Two methods for determining WL:Simplified method - may only be used if railing is continuousStrip method - must be used if railing is discontinuous;

may be use if railing is continuous

Deck Cantilever / Railing Analysis ( continued )

Truck gravity load - simplified method:

continuous railing

WL

4.45

1.0 ft

WL = 1.0 kip/ft

16

Truck gravity load - strip method:

Deck Cantilever / Railing Analysis ( continued )

1.0 ft

WL16k

16k

discontinuous rail

gap

4.46

SW 1.0 ft1.0 ft

SWX

SWOverhang (inches) = 45 + 10.0 • X(feet)

Overhangw

LSW

PW =

RAILING DESIGNTruck load applied to railing:

The primary purpose of the railing (also called the barrier or parapet) is to contain andredirect vehicles using the structure in order to: (1) protect the occupants of a vehiclethat collides with the railing, (2) protect other vehicles on the structure at the time of thecollision, (3) protect persons and vehicles on roadways and other areas under the bridge.

The railing and deck overhang are designed to survive a vehicle collision force whoseintensity depends on the nature of the highway and the traffic it carries. Designparameters are grouped into six (6) Test Levels in AASHTO 13.7.2 . Four of the mostcommon Test Levels are:

4.47

TL-1 - Test Level One: Applicable to roads with low speeds and traffic volumes

TL-2 - Test Level Two: Applicable to local and collector roads with favorable site conditions, reduced posted speeds,and a small number of heavy vehicles

TL-3 - Test Level Three: Applicable to high-speed arterial highways with favorable site conditions, and very low mixturesof heavy vehicles

TL-4 - Test Level Four: Applicable to high-speed highways, freeways, expressways and interstate highways with a mixtureof trucks and heavy vehicles

Railing Design ( continued ) Fv

Critical section for negative moment in deck overhang

H ( to top of wall )

FL

Lt Ft

4.48

Design Forces and Railing Test Levels Designations TL-1 TL-2 TL-3 TL-4 TL-5A TL-5B TL-6

Ft Transverse ( KIP ) 13.5 27 54 54 116 124 175

FL Longitudinal ( KIP ) 4.5 9 18 18 39 41 58

Fv Vertical Down ( KIP ) 4.5 4.5 4.5 18 50 80 80

Lt ( IN ) [ Given in FT in table ] 48 48 48 42 96 96 96

Minimum Rail Height H ( IN ) 27 27 27 32 40 54 90

Table A13.2-1

17

The minimum design rail collision load is based on the conditions (the highway type traffic volumeand site features) for which bridge will be used (LRFD Table 13.7.1.1). The moments produced atthe base of the rail by the load that causes the rail to fail are applied to the deck at the outer edgeof the cantilever overhang.

Rw is force required to “break” wall - it must equalor exceed design load Ft given in Table A13.2-1

H

Rw

Railing Design ( continued )

Vehicle loads used in rail collision Extreme Event

4.49

MCTVCT

TCT = VCT

1’ width deck strip

MCT

collision Extreme Event Limit State analysis:

Reduced wheel loadis applied with Rw

TCT = VCT

MCT

Rw

Forces per foot deck width causedby horizontal force Rw

MSLAB

Railing Design ( continued )Railing design model - Case 1 ( interior ) :

45°

Rw

Lc

H

Rw

Lc = c

w2

ttM

MH822

L2

L+⎟⎟

⎠

⎞⎜⎜⎝

⎛+ ( in )

Rw = ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

−

2HL

MM8LL2

H2 ccw

tc ( k )

TCT = VCT = H2LR

c

w+ ( k/in )

4.50

MCT

VCT

H2Lc +

MCT = H2LHR

c

w+ ( in-k/in )

H = height of the railing wall ( inches )Lt = length along which impact force is assumed distributed to wall ( inches )

Mw = average unit ultimate moment resistance of wall about vertical axis ( in-k/in )Mc = average unit ultimate moment resistance of wall about horizontal axis ( in-k/in )

Lc = width of failure mechanism that offers the least Rw ( inches )

Railing Design ( continued )Railing design model - Case 2 ( end ) :

45°

Rw

Lc

HLc = c

w2

ttM

MH822

L2

L+⎟⎟

⎠

⎞⎜⎜⎝

⎛+ ( in )

Rw = ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

−

2HL

MM8LL2

H2 ccw

tc ( k )

R

4.51

Rw

MCT

VCT

TCT = VCT = HLR

c

w+ ( k/in )

MCT = HLHR

c

w+ ( in-k/in )

18

Overhang DesignDesign Case 1: DL and trans. & long. Vehicle impact forces

Load & Resistance Factors = 1.0. – extreme event limit state

Design Case 2: DL & vert. vehicle impact forces

Load & Resistance Factors = 1 0 extreme event limit state

4.52

Load & Resistance Factors = 1.0. – extreme event limit state

Typically does not govern for concrete barriers

Design Case 3: Strength I Limit State

1.25DC + 1.5 DW + 1.75 (LL+IM)

Vehicle Impact ForcesExtreme Event Test Vehicle – TL4(LRFD 13.7.2)

Design Forces and Designations

Ft Transverse Force 54 KIPFL Longitudinal Force 18 KIP

4.53

FL Longitudinal Force 18 KIPFv Vertical Force Down 18 KIPLt and LL 3.5 FTLv 18 FTHe min (Height of impact above deck) 32 INH Minimum Height of Barrier 32 IN

Safety BarrierStrength of Barrier: ILDOT F-Shape Concrete Barrier

4.54

19

Strength of Barrier – Yield Line Case 1 4.55

Strength of Barrier

( )++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

MMMHLLL

:1 Case Line Yield

c

wbttc

2 822

4.56

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=HLMMM

LLR cc

wbtc

w

2

1 882

2

Strength of Barrier – Yield Line Case 2 4.57

20

Strength of Barrier

HLMMM

LLR

MMMHLLL

:2 Case Line Yield

ccwbw

c

wbttc

⎟⎟⎞

⎜⎜⎛

++⎟⎟⎞

⎜⎜⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

2

2

2

22

22

4.58

HLL wbtc

w ⎟⎠

⎜⎝⎟⎠

⎜⎝ −2 2

Strength of Barrier 4.59

Strength of Barrier

( )

kip 134.4

ft

:1Case Line Yield

=⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

=+

+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

HLMMM

LLR

MMMHLLL

ccwb

tcw

c

wbttc

2

1

2

882

2

7.13822

:2CaseLineYield

4.60

Rw for barrier = 61.6 kip

61.6 kip > Ft = 54 kip OK

controls kip, 61.6

ft6.3

:2Case Line Yield

=⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

HLMMM

LLR

MMMHLLL

ccwb

tcw

c

wbttc

2

2

2

22

22

21

Flexural Design of Deck

P

M

h d

4.61

M

Distribution of Mc and T

» At the inside of barrier– M over Lc

Yield Line Case 1

4.62

– T over Lc+2H

Distribution of Mc and T

– T over Lc+ H

» At the inside of barrier– M over Lc

Yield Line Case 2

4.63

22

Flexural Design of DeckAt the inside face of the barrier:MDC = (8/12)*(0.150)*(1.5)2 / 2 = 0.06 kip-ft. / ft.Mbarrier = (0.450)*(1.5)2 / 2 = 0.34 kip-ft. / ft.Mc = 13.9 kip-ft. / ft. (Flexural strength of barrier about hor. axis)

Design forces for deck (at inside face of barrier):M = MDC + Mbarrier + Mc

= 0 06 + 0 34 + 13 9

4.64

= 0.06 + 0.34 + 13.9= 14.30 kip-ft. / ft.

P = T1 (yield line case 1) = 6.94 kip / ft., at centroid of deck

P

M

h d

Reinforcement at Top of Deck

P

M C

T+P

h da

Strains Stresses Forces

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −=

22222 1ahPadThdPadCMn

4.65

As = 0.185 in.2 / ft. (Empirical design – No. 4 @ 13” o.c.)T1 = T + PT1 = T + P = 0.185x60 = 11.1 kip / ft.C = 11.1 - 6.94 = 4.16 kip / ft.a = 4.16/(0.85x12x4) = 0.10 in.c = a/0.85 = 0.10/0.85 = 0.12 in.de = 8 – 2.5 – 0.5/2 = 5.25 in.

⎠⎝⎠⎝⎠⎝⎠⎝

Reinforcement at Top of Deck

M = 2 53 < M = 14 30 kip-ft / ft NG

P

M C

T+P

h da

Strains Stresses Forces

.ft/.ftkip53.2.ft/.inkip3.30210.0

2894.6

210.025.51.11Mn −=−=⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=

4.66

Mn = 2.53 < M = 14.30 kip-ft. / ft. NG

.ft/.ftkip0.15.ft/.inkip7.179292.0

2894.6

292.006.54.44Mn −=−=⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=

Mn = 15.0 > M = 14.30 kip-ft. / ft. OK

Provide additional No. 7 at 13 in. o.c.alternating with No. 4 at 13 o.c.» As = (0.20+0.60)/13*(12) = 0.74 in.2 / ft.» T = 0.74x60 = 44.4 kip / ft.

23

» Away from barrier:– Dispersion at 30 to 45 deg

4.67