DESIGN OF FULL INTEGRAL BRIDGE

VOL. V - PART 2 DATE: 11May2007 SHEET 1 of 28

INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS

GENERAL INFORMATION FILE NO. 20.02-1

GENERAL INFORMATION: This section of the chapter establishes the practices and requirements necessary for the design and detailing of integral abutments. For general requirements and guidelines on the use of integral abutments, see File Nos. 20.01-1 thru -5. Included are sample design calculations and details to assist the designer. The bridge shall be within the limits for length, skew and thermal movement as stated in File No. 20.01-1:

maximum span length (single span) = 180 feet maximum skew = 30º

Use a single row of vertical steel H-piles (Preferably HP10x42) and orient piles for weak axis bending whenever possible. The web of the H-pile shall be perpendicular to the centerline of the beams/girders regardless of the skew. This will facilitate the bending about the weak axis of the pile. Minimum spacing of the piles is 4’-0”. Maximum pile spacing shall not exceed beam/girder spacing. Avoid high abutment walls except for short spans where anticipated movements are small and can be easily tolerated. Limit the total abutment height to a maximum of 17 feet from finished grade. Wingwalls supported by the abutment shall be limited to 6 feet for straight wings (elephant ear length beyond the 6 feet extension of footing to support wingwall shall be considered). Length of U-wings shall be limited to 8 feet. The portion of the wall beyond shall be designed as a free-standing retaining wall. If not, an independent retaining wall system, which does not move, shall be considered. For design/detailing check lists for integral abutments, see File Nos. 20.02-23 thru -24 and File Nos. 20.02-27 thru -28.

VOL. V - PART 2

DESIGN OF FULL INTEGRAL BRIDGE

Given and Assumptions:

γ = 145 pcf Unit weight of soil (select backfill material) (See Manual of S&B Division Vol. V – Part 2, file no. 17.102-2)

Kp = 4 Assumes the use of EPS material behind backwall

WBridge = 43.33 ft Total bridge width

LBridge = 150.0 ft Bridge length

LThermal = 75.0 ft Length of thermal expansion

HBackwall = 6.33 ft Backwall height

TBackwall = 2.5 ft Backwall thickness

SBeam = 9.33 ft Girder spacing

Overhang = 3.0 ft Slab (and integral backwall) overhang

Cover = 3.5 in Cover over reinforcing steel in backwall

CS = 0.0208 Crown rate (cross slope) of bridge deck

Δh = 2CS(SBeam) Change in height of footing

Hftg = 3.0 ft Height of footing

f’c = 4,000 psi Compressive strength of backwall and wing concrete

f’cf = 3,000 psi Compressive strength of footing concrete

fy = 60,000 psi Yield strength of reinforcing steel

θ = 30 deg Bridge skew angle

α = 6.5 x 10-6 per deg F Coefficient of thermal expansion

DAS = 1.5 ft Depth of approach slab at backwall

Tbottomflange = 1 in Thickness of bottom flange

'c

1.5

sb

fw

En33

= = 8 Modular ratio of concrete to steel for backwall and wing

'cf

1.5

scf

fw

En

33= = 9 Modular ratio of concrete to steel for footing

DATE: 11May2007 SHEET 2 of 28


SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-2

VOL. V - PART 2

Δh = 4.66 in = 0.39 ft

Hftg = 3.0 ft

Ph = γ Kp (HBackwall + Δh) Passive soil pressure at hinge level Ph =4(145 pcf)(6.33 ft + 0.39 ft) = 3.9 ksf

Pf = γ Kp (HBackwall + Hftg) Passive soil pressure at bottom of footing

Pf = 4(145pcf)(6.33 ft + 3.0 ft) = 5.4 ksf




VOL. V - PART 2

Determine Backwall Moments and Shears: 2

Backwallp h(HK 21w )Δ+= γ Earth pressure resultant per foot

(above hinge)

2ft 0.39ft 33lbs)(4)(6. k/1000 1 x pcf 14521w )( += = 13.1 klf

θ cosS

L Beam= Girder spacing along skew

ft 10.7730 cos

ft 9.33L ==o

For simplicity, use the following equations to determine moments, shear, and reaction.

2pos 0.08wlM = = 0.08(13.1 klf)(10.77 ft)2 = 121.6 ft-kip Maximum positive moment

2

neg 0.10wlM = = 0.10(13.1 klf)(10.77 ft)2 = 152.0 ft-kip Maximum negative moment

0.6wlVmax = = 0.6(13.1 klf)(10.77 ft) = 84.7 k Maximum shear

1.1wlRmax = = 1.1(13.1 klf)(10.77 ft) = 155.2 k Maximum reaction at girder Check to make sure overhang does not govern.

=⎟⎠⎞

⎜⎝⎛=

2

OH cosOverhang0.5wM

θ

2

30 cosft 3.0klf) 0.5(13.1 ⎟

⎠⎞

⎜⎝⎛

o

MOH = 78.6 ft-kip < Mneg Interior support governs

=⎟⎠⎞

⎜⎝⎛=

θ cosOverhangwVOH ⎟

⎠⎞

⎜⎝⎛

o30 cosft 3.0klf) (13.1 = 45.4 k < Vmax Interior support governs

Design Integral Backwall:

Group IV load combination controls. Group IV allowable overstress is 125%. ** For this example, Group IV loading controls the design. It shall be the responsibility of the designer to verify which load case controls, and design accordingly. fs = 125%(0.4Fy) fs = 30,000 psi Allowable stress of steel

fc = 125%(0.4f’c) fc = 2,000 psi Allowable of stress of concrete

)f125%(0.95v 'cec.allowabl = vc.allowable = 75 psi Allowable shear stress in

concrete




VOL. V - PART 2

Flexure design using negative moment: n = 8 Modular ratio of backwall concrete to steel

h = TBackwall h = 30.0 in Height of section resisting flexure

b = HBackwall – DAS = 76 in – 18 in = 58.0 in Width of section resisting flexure

d1 = h - Cover = 30 in - 3.5 in = 26.5 in Depth to first mat of reinforcing steel

d2 = h - (2 x Cover) = 30 in - 2(3.5 in) = 23.0 in Depth to second mat of reinforcing steel

d’ = 3.5 in Depth to compression steel

Try minimum reinforcing, # 6 bars at ~10” spacing. For this backwall height, there are 6 bars in each tension layer.

As1 = 2.64 in2 As2 = 2.64 in2

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=s2s1

2s21s1

AAdAdA

d

⎟⎟⎠

⎞⎜⎜⎝

⎛++

= 22

22

in 2.64in 2.64in) (23.0in 2.64in) (26.5in 2.64d = 24.75 in Depth to centroid of tension steel

Asc = 2.64 in2 6 bars in compression layer Performing section analysis (including the compression steel) Fs = 15,040 psi < fs.allowable = 30,000 psi OK In first layer of tension steel

Fc = 510 psi < fc.allowable = 2,000 psi OK

Shear Design: V = Vmax V = 84.7 k

( ) psi 59in) in(24.75 58.0

k) lbs/1 k(1000 84.7bd

Vv max === Actual shear stress in concrete

v < vc.allowable, Shear reinforcement not required vc.allowable = 75 psi Use stirrup spacing of 12”




VOL. V - PART 2

Shear Stud Design at Girder Ends: Zr = 8.12 k For 7/8” φ AASHTO Sec. 10.38.5.1.1 Zr = 125%(8.12 k) Horizontal shear capacity per stud, with 25% overstress for Group IV loading Zr = 10.15 k

15.3k 10.15k 155.2

ZRn

r

maxstuds ===

Therefore, use 8, 7/8” φ studs on each side of beam web, for a total of 16 studs. Design Overhang and Wing:




Check thinner wing portion only: Lw = 7.0 ft True length of wing [ ] klf 6.5ft) 0.75(6.33pcf)4 (145

21)H x (0.75K γ

21w 22

Backwallpw ===

MOH = 0.5ww(Lw)2 = 0.5(6.5 klf)(7.0 ft)2 = 159.3 ft-kip VOH = ww x Lw = 6.54 klf(7.0 ft) = 45.8 k

b = HBackwall b = 76.0 in Width of section resisting flexure

tw = 16.0 in Thickness of wing

d = tw - cover = 16.0 in – 3.5 in = 12.5 in Depth to first mat of reinforcing steel

VOL. V - PART 2




ry 7 # 6 bars at 10" Spacing

s

s i > fs.allowable = 30,000 psi Redesign

ased on trial and error using 8 # 8 bars at 9” spacing

As = 6.32 in

Fs = 26,800 psi < fs.allowable = 30,000 psi OK

Check shear in wing:

T

= 3.08 in2 A

= 53,600 psF

B

2

Fc = 1,300 psi < fc.allowable = 2,000 psi OK

psi f0.95v '

ca = va = 60 psi

( ) psi 48in) in)(12.5 (76.0

lbs 45,800v =bdVOH == tirrups required

*Since the As required for this section exceeds that calculated for the integral backwall, additional

Allowable shear stress in concrete

< va No s

reinforcing steel will be required for the wing portion of the integral backwall.

VOL. V - PART 2




nalyze Pile Group:

A

ft 50.03ft 43.33W

L Bridge === 30 cos cosftg oθ

ftg = TBackwall Wftg = 2.5 ft Width of footing

Depth of footing

.0 k Live load reaction per truck/per

R = Lftg x Hftg x Wftg x 150 pcf Reaction of footing

Rftg = 50.03 ft(3.0 ft)(2.5 ft)(150 pcf x 1 k/1000 lbs) = 56.3 k

Dneat = 6.33 ft – (8.5 in x 1 ft/12 in) = 5.62 ft Depth of neatwork

Rneat = Lftg x Wftg x Dneat x 150 pcf Reaction of abutment neatwork

Rneat = 50.03 ft(2.5 ft)(5.62 ft)(150 pcf x 1 k/1000 lbs) = 105.4 k

Rss = 604 k Superstructure dead load

ss + Rftg + Rneat Total dead load reaction for

RDL = 604 k+ 56.3 k + 105.4 k = 766 k

gth of footing Len

W Hftg = 3.0 ft VLL = 64 lane ftg

RDL = R footing

VOL. V - PART 2




σp = 9000 psi Maximum pile stress

p = 12.4 in Pile area

APpile = Ap σn Pile capacity

APpile = 111.6 k

2 A

C C

3.6ft 12

ft 43.33ft 12

WN Bridge

L === NL = 3 Number of lanes

LL = 0.75(NL VLL ) = 0.75(3)(64.0 k) = 144.0 k Total live load

R = R + RDL = 144 k + 766 k = 910 k Total load on piles

R ∑ LL

8.2k) (111.6

k 910Cap

RNpile

Piles ==Σ

= Therefore, use 9 piles

etermine Lateral Loading on Piles:

f = 0.5(Ph + P )Dftg = 0.5(3.9 ksf + 5.4 ksf)(3.0 ft) = 14.0 klf

= Rf + Rbw = 14.0 klf + 11.6 klf = 25.6 klf Total resultant along integral

p =

D R f q abutment

k 739.530 cos

)30 ft)(tan klf(43.33 25.6 costanθqWBridge ==

o

o

θF

p is the theoretical force required to restrain all lateral movement. Lateral restraint should be

Fconsidered when transverse displacement can interfere with performance, or with adjacent structures. The designer may wish to employ software such as COM624 or L-Pile to determine the actual forces in the pile. If the force must be restrained (due to proximity to adjacent structures, utilities, retaining/MSE walls, etc. or any facility which may be damaged by lateral displacements), the designer shall provide sufficient restraint in the structure (either through the piles or other supplementary means) to mobilize the lateral force.

VOL. V - PART 2




Design Footing for Bending in the Vertical Plane:

Point loads are girder loads

Design as continuous beam with a combination of both point loads and distributed loads:

oh

= 1.37 ft Cantilever length for distributed L

loads

klf 3.2ft 50.03

k 105.4k 56.3L

RRw

ftg

neatftg =+

=+

= Distributed loads

Sp = 5.0 Pile spacing

For Distributed Loads:

( ) kip-ft 3.02

ft) klf(1.37 3.22

LwM22

ohoh === Overhang moments

intP = 0.08w(Sp )2 =0.08(3.2 klf)(5.0 ft)2 = 6.4 ft-kip Positive moments

intN = 0.1w(Sp )2 = 0.1(3.2 klf)(5.0 ft)2 = 8.0 ft-kip Negative moments

egative moment controls

ftg p

M M N

= 0.6wS = 0.6(3.2 klf)(5.0 ft) = 9.6 k V

VOL. V - PART 2




or Point/Girder Loads:

rom superstructure analysis: P3 = P2

F FP1 = 185.5 k P2 = 241.8 k

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡= 2

p

2p

1girder1 S in 17.5-Sin 17.5

0.1PM

( )

⎥⎦

⎤⎢⎣

⎡= 2

2

girder1 ft) (5.0 n) 1ft/12i x in (17.5-ft 5.0in) ft/12 1 x in (17.5k) 0.1(185.5 M = 13.6 ft-kip

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡= 2

p

2p

2girder2 S in 8.75-Sin 8.75

0.1PM

( )

⎥⎦

⎤⎢⎣

⎡= 2

2

girder2 ft) (5.0 in) ft/12 1 x in (8.75-ft 5.0in) ft/12 1 x in (8.75k) 0.1(241.8 M = 12.9 ft-kip

girder = 13.6 ft-kip Controlling girder moment

M = MintN + Mgirder = 8.0 ft-kip + 13.6 ft-kip = 21.6 ft-kip

esign Integral Abutment Footing (Vertical Plane):

= H h = 36.0 in Height of section resisting flexure

= Wftg b = 30.0 in Width of section resisting flexure

= h – cover d = 32.5 in Depth to first mat of reinforcing

= 9

ry 4 # 6 bars As = 1.76 in2

s = 3,680 psi < fs.allowable = 30,000 psi OK

heck Shear in Footing:

M ∑ D h ftg b

d steel n T F

Fc = 81 psi < fc.allowable = 1,500 psi OK C psi f0.95v '

ca = va = 52 psi

Allowable shear stress in concrete

( ) ==bdV

v ftg [ ]in) in(32.5 30.0

k) lbs/1 (1000 k 9.6 v = 10 psi a ps required

< v No stirru




esign footing for bending in the horizontal plane:

D

f = 0.5(Ph + P )Hf = 0.5(3.9 ksf + 5.4 ksf)(3.0 ft) = 14.0 klf Resultant on footing in

p = 5.0 ft Loh = 1.37 ft

R f horizontal plane S

( )2

LRM

2ohf

oh = = ( )=

2 ft 1.37klf 14.0 2

13.1 ft-kip Overhang moments

intP = 0.08R (Sp )2 = 0.08(14.0 klf )(5.0 ft)2 = 28.0 ft-kip Positive moments

intN = 0.01R (Sp )2 = 0.1(14.0 klf)(5.0 ft)2 = 35.0 ft-kip Negative moments

egative moment controls

ftg = 0.6R Sp = 0.6(14.0 klf )(5.0 ft) = 42.0 k Shear in footing

M f M f N

V f

VOL. V - PART 2




esign Integral Abutment Footing (Horizontal Plane):

= 2.5 ft Height of section resisting flexure

= 3.0 ft Width of section resisting flexure

= h – cover d = 26.5 in Depth to first mat of reinforcing

ry 5 # 6 bars (Minimum reinforcement for integral abutment)

s = 2.2 in

s = 5310 psi < fs.allowable = 30,000 psi OK

:

D h b

d steel T

2A F

Fc = 133 psi < fc.allowable = 1,500 psi OK Check Shear in Footing

psi f0.95v 'ca = v = 52 psi Allowable shear stress in concrete

a

( )bdVftg

( )in) in(26.5 36.0k) lbs/1 (1000 k 42.0v = = = 44 psi < va No stirrups required

etermine Number of Dowels Required:

D

φ = 0.875 in Diameter of dowels

y = 36,000 psi Yield strength of dowels

shear = 0.4Fy Allowable shear in dowels

shear = 14.4 ksi

F F

F

VOL. V - PART 2

4 Adowel

2φπ= Adowel = 0.60 in2 Area of dowel

APdowel = Adowel Fshear = 0.60 in2 (14.4 ksi) = 8.7 k Capacity of one dowel

f = 14.0 klf Passive reaction of soil against

olve for Pile Reaction:

p Effective pile length

perature in one direction

LT = 0.75(α120°LThermal)

T = 0.75(6.5x10 g. F)120°F[75.0 ft(12 in/ 1ft)] = 0.53 in

C

R footing S

= 20.0 ft L

= 71.7 in4 Moment of inertia of piles Ip

ΔLT = Change in length due to tem Δ

-6 ΔL per de

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡= 5.1

3ER

Δ3

p

pTp L

I L Pile reaction at top of pile

( )

( ) ⎥⎦

⎤⎢⎣

⎡= 3

4

p ft in/1 12 x ft 20.0 in 71.7in 0.53ksi) 3(29,000R 5.1 = 0.4 k

olve for Dowel Spacing:

S

> Min spacing of 4φ in 7.4 ft 0.62

ft 5.0k 0.4klf 14.0

k 8.7==

⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

p

pf

doweldowel

SR

R

CAPS - OK




VOL. V - PART 2




TEMPORARY SUPPORT BOLT DETAIL

heck Load on Temporary Support Bolts:

Rsteel = 9.0 k Reaction of steel beam, diaphragms, etc.

CT = 4.0 k Reaction of construction tolerances and other

Rbeam = Rsteel + RCT Rbeam = 13.0 k Total reaction

H = 13.66 in Unsupported height of temporary support

F = 55 ksi

C

R

loads

bolt

Bolt

y

φ bolt = 0.875 in Temporary support bolt diameter

=⎟⎠⎞⎛

4boltφ

⎜⎝=

42 I bolt

π4

4in 0.875 ⎞⎛πin 0.03

42 =

⎟⎠

⎜⎝ Temporary support bolt moment of inertia

==4

2

boltboltA

φπ ( ) 20.60in

4in 0.875

=⎟⎟⎠

⎞⎜⎜⎝

⎛ 2

π Temporary support bolt area

== bolt

boltIr

boltAin 0.22

in 0.6 2

4

=⎟⎟⎠

⎞⎜⎝

Radius of gyration of temporary support bolt

= 1.0 k factor for bolt acting as column

Lr Factor =

in 0.03⎜⎛

k

= ⎟⎠⎞

⎜⎝⎛

in 0.22in 13.661.0

⎟⎟⎠

⎞⎜⎜⎝

⎛

bolt

bolt

rH

kK = 62.1

Hbolt = 9 in + Δh

VOL. V - PART 2

Axial compression in temporary

bolt

beama 2A

R=

)in 2(0.6k) lbs/1 (1000 k 13.0

2=σ = 10,800 psi

support bolt

==y

2

c FE2

Cπ

102ksi 55

ksi) (29,0002 2

=π

Since KLr Factor < Cc, then allowable compression in temporary support bolt:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

EFFactor KLrF

Comp y2

yallow 44

1122 π.

psi 25,000psi) 0(29,000,00psi) (55,000(62.1)psi 55,000 2

=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−= 44

1122 π.

Since σ mpAllow Bolt diameter is sufficient

Check Bending in Shortest Temporary Support Bolt:

Hbolt = 9.0 in

a < co

3

4

bolt

boltbolt in 0.07

2in 0.875

in 0.03

2

IS =⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

=φ

Section modulus of bolt

s = 0.6 Coefficient of static friction

ind P-Load associated with ΔT induced at top of bolt and associated moment:

p = 9 Number of piles

b = 5 Number of beams

LT = Change in length due to temperature

LT = 0.75(α120°LThermal) = 0.75(6.5x10 per deg. F)120°F(75.0 ftx12 in/1ft) = 0.53 in

µbetween the beam flange and the steel bearing plates

F n n Δ

-6 Δ




2RF s

beamfμ

k 3.92

0.6k 13.0 =⎟⎠⎞

⎜⎝⎛==

Friction force of superstructure loading per bolt

VOL. V - PART 2




⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛= TEΔL

P

pp

bolt2p

boltb

3bolt

pp

3p

b

I2nHL

I3nH

I3nL

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

)in 2(9)(71.7in) (9.0ft)] in/1 ft(12 [20.0

in 3(5)(0.03in) (9.0

in 3(9)(71.7ft) in/1 ft(12 (20.0

ksi) in(29,000 0.53P

4

2

4

3

4

3b

))

b = 1.7 k Resultant load at top of bolt

b = Pb Hbolt =1.7 k[9.0 in(1 ft/12 in)] = 1.3 ft-kip Moment induced into bolt

P M

( )bolt b

bb S2n

M=σ ( )3in 2(5)(0.07

k) lbs/1 ft)(1000 in/1 kip(12-ft 1.3= Bending stress

= 22, 300 psi

∑σ = σa + σb = 10,800 psi + 22,300 psi = 33,100 psi Sum of axial compression

wable = 125%(0.55Fy) σallowable = 37,800 psi Allowable bolt stress

∑σ < σallowable OK

va = 0.4Fy va = 22 ksi

bσ

and bending stresses σallo

vb

bolt

b

2AP

= ksi 1.4)in 2(0.60

k 1.72 == < Allowable shear OK

heck Shear in Temporary Support Bolts: C

VOL. V - PART 2




esign Plate A:

D

bf = 12.0 in

de = 1.5 in Clearance between the plate

ends and temporary support bolt centerline

in 12.130 cos

in 1.5 - in (12.0==

)30 cos d-(b e f

e)

=L

initial plate size: (use minimum values)

p = 5 in Plate width

= 0.5 in Minimum plate thickness

A = tp = 2.5 in Area of plate

Assume an w tp

p wp

12

3pp

p

twI = 4

3

in 0.05in) in(0.5 5.0==

12 Moment of inertia of plate

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2tI

Sp

pp

34

in 0.20

2in 0.50in 0.05

=⎟⎠⎞

⎜⎝⎛

Section modulus of plate

Beam/girder flange width

Span between temporary support bolts

VOL. V - PART 2




heck Plate Stresses: C

plf 13,000ft 1.0

k) lbs/1 (1,000 k 13.0b

Rw

f

beam === Loading along length of Plate A

k 6.62

in) ft/12 1 x in klf(12.1 13.02

wLv ep ===

Shear force at temporary support bolt

psi 2,640in 2.5

k) lbs/1 k(1,000 6.6Av

p

p ===τ

τ < 12,000 psi – OK

kip-ft 1.78

in) ft/12 1 x klf(12.1in 13.08

wLM22

ep === Bending moment in plate

psi 102,0000.20in

ft) k)(12in/1 lbs/1 kip(1,000-ft 1.7SM

3p

pb ===σ

bσ > 27

Redesign:

= 5.0 in Revised plate width

tp = 1.0 in Revised plate thickness

p= 5 in2 Area of plate

,000 psi-Increase plate size wp

A

12

3pp

p

twI = 4

3

in 0.4in) in(1.0 5.0==

12 Moment of inertia of plate

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2tI

Sp

pp

34

in 0.8

2in 1.0in 0.4

=⎟⎠⎞

⎜⎝⎛

Section modulus of plate

kip-ft 1.78

in) ft/12 1 x in klf(12.1 13.08

wLM22

ep === Bending moment in plate

psi 23,900in 0.8

ft) in/1 k)(12 lbs/1 kip(1,000-ft 1.7SM

3p

pb ===σ

bσ < 27,000 psi - OK

Check Plate Stresses:




ote: Plate B will function as a plate washer and no structural design is necessary.

hickness of the EPS layer:

File No. 20.06-6:

L thermalαΔt) = 75.0ft(12in/1ft)(6.5x10-6 per deg. F )120°F = 0.7in Total range of move-

t Backwall L

.7 in)] = 12.3 in

t

NOTE: DESIGN FOR PRESTRESSED CONCRETE BEAMS IS SIMILAR

N T

hickness of EPS layer as per T

t = 120°F Δ

= (LΔ ment at abutment due to temperature

PS = 10(0.01H + 0.67Δ ) E

PSE t = 10[(0.01)(76.0 in)+ (0.67)(0

herefore, use EPS = 13 in. T

VOL. V - PART 2

CHECK LIST FOR FULL INTEGRAL – STEEL BEAMS / GIRDERS

1 Show plan and elevation views of integral abutment and footing plan at a preferred scale of



CHECK LIST – STEEL BEAMS / GIRDERS FILE NO. 20.02-23

3/8” = 1’-0”. Regardless of skew, the elevation view should be projected down from the plan view. Sections taken through the integral abutment shall be shown at 3/4” = 1’-0” min.

2 Label the location centerline/baseline as shown on the title sheet.

3 “End of slab” shall be used as the reference line for layout of integral abutments. Any

stations and elevations required shall be referenced to this line.

4 Label the skew of abutment in relation to the end of the slab and the skew of the piles in relation to the centerline of integral abutment.

5 The minimum width of integral abutment shall be 2’-6”. If this width is not sufficient, a 3’-0”

width may be used.

6 All ST series and SV series bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of these bars shall be 12”. ST0602 bars between the backwall and the approach slab (where applicable) are not required outside of the exterior beam / girder.

7 Dowels shall be a minimum of 7/8” ø and shall be spaced at a maximum of 12” center-to-

center along the centerline of integral abutment. Locate first dowel 6” beyond the edge of bottom flange and dimension from the centerline of beam/girder. Dowels shall be embedded 12” into footing.

8 The approach slab seat (7”) shall be provided on all integral abutments regardless of whether

or not the bridge will have an approach slab.

9 Show elevation at bottom of beam/girder along centerline of integral abutment. Show elevation at bottom of footing.

10 For details of when to slope the abutment footing or to detail the abutment footing level, see

File No. 20.06-8.

11 The wing shall project a minimum of 12” below the top of footing. For alternate wingwall details, see File No. 20.06-7.

12 AF04 bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of

these bars shall be 12”.

13 The footing shall be designed with a minimum height of 3’-0” and a maximum height of 4’-0”.

14 A 4” clearance between the end of the beam/girder and the end of slab shall be provided. The flanges may be clipped to provide this clearance. In the case of a minimum width abutment and a skew approaching the maximum of 30° the entire end of the beam/girder may need to be clipped to provide this clearance and to allow the anchor bolts to be located on the centerline of the integral abutment.

VOL. V - PART 2

CHECK LIST FOR FULL INTEGRAL – STEEL BEAMS / GIRDERS (Continued)

15 A series of 1



CHECK LIST – STEEL BEAMS / GIRDERS FILE NO. 20.02-24

1/2” ø holes shall be provided for the SH series bars to pass through. These holes shall be located a minimum of 3” and a maximum of 6” from the flanges and shall be spaced at a maximum spacing of 12”. See File Nos. 20.05-1 thru -4.

16 A distance of 9” minimum to 2’-0” maximum shall be provided between bottom of flange and

top of footing.

17 The fill in front of the integral abutment shall be set at a minimum 9” distance below the top of footing with the edge of berm set at a minimum 3’-0” distance from the face of abutment.

18 Temporary support bolts shall be a minimum 7/8” diameter and shall be embedded a

minimum of 1’-0” into footing.

19 Plate A shall be a minimum of 1/2” thick and 5” wide.

20 The minimum width of plate B is 3” and the minimum length is 5”. The plate shall be 1/2” thick in all cases.

21 For determining the thickness of EPS material, see File No. 20.06-6.

22 For instructions on completing the title block, see File No. 03.03.

23 For instructions on completing the project block, see File No. 03.02.

24 For instructions on developing the CADD sheet number, see File Nos. 01.01-7 and 01.14-4.

25 ST0501, ST0602 and SV0402 shall be galvanized. All other backwall reinforcing steel shall

be epoxy-coated.

26 Approach slabs shall be modified to provide a 10” wide seat.

VOL. V - PART 2

CHECK LIST FOR FULL INTEGRAL – PRESTRESSED CONCRETE BEAMS

1 Show plan and elevation views of integral abutment and footing plan at a preferred scale of



CHECK LIST – PRESTRESSED CONCRETE BEAMS FILE NO. 20.02.27

3/8” = 1’-0”. Regardless of skew, the elevation view should be projected down from the plan view. Sections taken through the integral abutment shall be shown at 3/4” = 1’-0”.

2 Label the location centerline/baseline as shown on the title sheet.

3 “End of slab” shall be used as the reference line for layout of integral abutments. Any

stations and elevations required shall be referenced to this line.

4 Label the skew of abutment in relation to the end of the slab and the skew of the piles in relation to the centerline of integral abutment.

5 The minimum width of integral abutment shall be 2’-6”. If this width is not sufficient, a 3’-0”

width may be used.

6 All ST series and SV series bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of these bars shall be 12”. ST0602 bars between the backwall and the approach slab (where applicable) are not required outside of the exterior beam / girder.

7 Dowels shall be a minimum of 7/8” ø and shall be spaced at a maximum of 12” center-to-

center along the centerline of integral abutment. Locate first dowel 6” beyond the edge of bottom flange and dimension from the centerline of beam/girder. Dowels shall be embedded 12” into footing.

8 The approach slab seat (7”) shall be provided on all integral abutments regardless of whether

or not the bridge will have an approach slab.

9 Show elevation at bottom of beam/girder along face of integral abutment. Show elevation at bottom of footing.

10 For details of when to slope the abutment footing or to detail the abutment footing level, see

File No. 20.06-8.

11 The wing shall project a minimum of 12” below the top of footing. For alternate wingwall details, see File No. 20.06-7.

12 AF04 bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of

these bars shall be 12”.

13 The footing shall be designed with a minimum height of 3’-0” and a maximum height of 4’-0”.

14 A 6” clearance between the end of the beam/girder and the end of slab shall be provided. This distance is to be shown on the prestressed beam standard.

15 Modify the prestressed beam standard to show the location of 11/2” ø holes in beam web.

The four holes shown are for a PCBT-53. Increase (or decrease) the number of holes by one for every beam depth increment above (or below) the 53” deep section shown. The additional holes shall be placed in 8” increments below the bottom hole shown. Depending on slab thickness, bolster thickness and strand pattern, the designer may need to adjust the hole locations so that adequate cover below the approach slab seat is maintained. Holes must lie within the web.

VOL. V - PART 2

CHECK LIST FOR FULL INTEGRAL – PRESTRESSED CONCRETE BEAMS

(Continued)

16 A distance of 9” minimum to 2’-0” maximum shall be provided between bottom of flange and top of footing.



CHECK LIST – PRESTRESSED CONCRETE BEAMS FILE NO. 20.02.28

17 The fill in front of the integral abutment shall be set at a minimum 9” distance below the top of

footing with the edge of berm set at a minimum 3’-0” distance from the face of abutment.

18 Top BC0501 shall extend 3” beyond end of beam. Bottom BC0501 and BC0602 shall extend 10”.

19 For determining the thickness of EPS material, see File No. 20.06-6.

20 The insert plate shall be eliminated from the prestressed beam standard.

21 For instructions on completing the title block, see File No. 03.03.

22 For instructions on completing the project block, see File No. 03.02.

23 For instructions on developing the CADD sheet number, see File Nos. 01.01-7 and 01.14-4.

24 ST0501, ST0602 and SV0402 shall be galvanized. All other backwall reinforcing steel shall

be epoxy-coated.

25 Approach slabs shall be modified to provide a 10” wide seat.

DESIGN OF FULL INTEGRAL BRIDGE

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