Semi-Integral-Abutment Design Example (USA Unit)

VOL. V - PART 2 DATE: 11May2007 SHEET 1 of 20

INTEGRAL / JOINTLESS BRIDGES SEMI-INTEGRAL ABUTMENTS

GENERAL INFORMATION FILE NO. 20.03-1

GENERAL INFORMATION: This section of the chapter establishes the practices and requirements necessary for the design and detailing of semi-integral abutments. For general requirements and guidelines on the use of semi-integral abutments, see File Nos. 20.01-1 thru -5. Sample design calculations are provided to assist the designer and are intended to correspond to the sample details shown in File Nos. 20.03-2 thru -10. Note that calculations are provided for the backwall and associated structural components only. Plan and elevation views and sections of the semi-integral abutment are provided in this chapter for information on the shape of the backwall in relation to the semi-integral backwall and to illustrate some additional details required on the abutment sheets. “Back of stem” is the reference line on the semi-integral abutment and substructure layout sheets. “End of slab” is the reference line on remaining sheets. Additional sample details are provided to assist the designer in the detailing of semi-integral abutments. These details are provided to show differences in details between steel/concrete stringers, bridges with/without skew and semi-integral abutment layout. Preferred practice on semi-integral abutment layout falls in the following order: 1. Wingwalls oriented transversely to traffic, elephant ears, with the terminal wall on the

superstructure. See File Nos. 20.03-11 thru -14. 2. Wingwalls oriented parallel to traffic, u-back wings, with the terminal wall on the

superstructure. Offset the inside face of wall 3 feet from the face of rail/parapet to allow for dynamic deflection of the attached guardrail. See File Nos. 20.03-15 thru -18.

It is generally desirable to eliminate potential conflicts between superstructure and substructure components. As such, the second layout preference should only be used where right-of-way (R/W), maintenance of traffic (MOT) or design restrictions make the preferred layout not feasible. For design/detailing check list for semi-integral abutments, see File Nos. 20.03-19 and -20.

VOL. V - PART 2

DESIGN OF SEMI-INTEGRAL BRIDGE Given and Assumptions: The calculations provided below do not fully correspond to the details shown in File Nos. 20.03-11 thru -14 but are similar. 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

W = 43.33 Total bridge width Bridge

L = 250.0 ft Bridge length Bridge LThermal = 125.0 ft Length of thermal expansion

H = 6.33 ft Backwall height Backwall

T = 1.58 ft Backwall thickness Backwall

S = 9.33 ft Beam spacing Beam

Overhang = 3.0 ft Slab (and integral backwall) overhang

Cover = 3.5 in Cover over reinforcing steel in backwall

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

f’ = 3,000 psi Compressive strength of wing concrete cWall

fy = 60,000 psi Yield strength of reinforcing steel

θ = 30 deg Bridge skew angle

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

DAS = 1.5 ft Depth of approach slab at backwall

H = 3.0 in Height of bearing Bearing

T = 1 in Thickness of bottom flange bottomflange

T = 15.0 in Wing thickness wing

'c

1.5

sb

fw

En

33= = 8 Modular ratio of concrete to steel for backwall

'cWall

1.5

sw

fw

En

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

DATE: 11May2007 SHEET 2 of 20


SAMPLE DESIGN CALCULATIONS FILE NO. 20.03-2

VOL. V - PART 2




VOL. V - PART 2

Design of Backwall: Determine backwall moments and shears

2Backwallp (HK

21w )γ= Earth pressure resultant per foot

2ft 33lbs)(4)(6. k/1000 1 x pcf 145

21w )(= = 11.6 klf

θ cosSL Beam= 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 = 2 = 0.08(11.6 klf)(10.77 ft) = 107.6 ft-kip Maximum positive moment

2 2

neg 0.10wlM = = 0.10(11.6 klf)(10.77 ft) = 134.6 ft-kip Maximum negative moment

= 0.6(11.6 klf)(10.77 ft) = 75.0 k Maximum shear 0.6wlVmax =

= 1.1(11.6 klf)(10.77 ft) = 137.4 k Maximum reaction at girder 1.1wlRmax = Check to make sure overhang does not govern.

=⎟⎠⎞

⎜⎝⎛=

2

OH cosOverhang0.5wM

θ

2

30 cosft 3.0klf) 0.5(11.6 ⎟

⎠⎞

⎜⎝⎛

o




MOH = 69.6 ft-kip < M Interior support governs neg

=⎟⎠⎞

⎜⎝⎛=

θ cosOverhangwVOH ⎟

⎠⎞

⎜⎝⎛

o30 cosft 3.0klf) (11.6 = 40.2 k < V Interior support governs max

VOL. V - PART 2

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. f




s = 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 = v = 75 psi c.allowable

Allowable shear stress in concrete

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

h = 19.0 in Height of section resisting flexure h = TBackwall

– D = 76.0 in – 18.0 in = 58.0 in Width of section resisting flexure b = HBackwall AS

d1 = h – Cover = 19.0 in – 3.5 in = 15.5 in Depth to first mat of reinforcing steel

d2 = h – 10.0 in = 19.0 in – 10 in = 9.0 in Depth to second mat of reinforcing steel

d’ = 3.5 in Depth to compression steel

Try #6 bars at ~9” spacing. For this backwall height, there are 7 bars in each tension layer.

2As1 = 3.08 in A 2s2 = 3.08 in

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=s2s1

2s21s1

AAdAdA

d Depth to centroid of tension steel

2

22

22

in 12.25in 3.08in 3.08

in) (9.0in 3.08in) (15.5in 3.08d =⎟⎟⎠

⎞⎜⎜⎝

⎛

++

=

2 = 3.08 in 7 bars in compression layer Asc

Performing section analysis (including the compression steel) Y = 3.75 in Distance from compression face to neutral axis NA

Fs1 = 29,800 psi < fs.allowable = 30,000 psi OK In first layer of tension steel

Fc = 1,200 psi < fc.allowable = 2,000 psi OK

VOL. V - PART 2

Shear Design: V = Vmax = 75.0 k

( ) psi 106in) in(12.25 58.0

k) lbs/1 k(1000 75.0bd

Vv max === Actual shear stress in concrete

, Shear reinforcement required v = 75 psi v > vc.allowable c.allowable




Use #4 stirrups, with 2 legs in the shear plane, Av = 0.4 in2. Equation (8-7) from AASHTO Sec. 8.15.5.3

( )( ) in 6.7

in psi)58.0 75psi (106psi) (30,000in 0.4

b vvfAs

2

ec.allowabl

es.allowablv =−

=−

= Stirrup spacing

Check AASHTO Sec. 8.19.1.2 to see if spacing above is OK.

( )y

v.min f50bsA =

Rearranging the above equation, with Av = 0.4 in2:

( )in 8.3

in) 50(58.0psi) (60,000in 0.4

50bfA

s2

yv.min ===

s = 6.7 in controls 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

Zr = 10.15 k

13.5k 10.15k 137.4

ZRn

r

maxstuds ===

7/Therefore, use 7, 8” φ studs on each side of beam web, for a total of 14 studs.

VOL. V - PART 2

Determine reaction at acute corner and wing buttress: See lateral force derivation in files nos. 20.07-4 thru -6.

( )( )k 263.8

30 tanft 250ft 43.331

30 ft)tan klf(43.33 [11.6

θ tanLW

1

θ tan WwR

Bridge

Bridge

Bridgep =

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

=o

o ]

Determine size of rub plates:




Δt = 120 deg.

= 125.0 ft LThermal

tThermalrub ΔL 32ΔL α⎟

⎠⎞

⎜⎝⎛=

in 0.78F)F)(120 deg. per 10ft)](6.5 in/1 ft(12 [125

32ΔL 6

rub =×⎟⎠⎞

⎜⎝⎛= − o

Estimated maximum movement in one direction at abutment. Assume that the temperature increase will only be two-thirds of the total range.

hrp = H - 3 in - 2 in - H – TBackwall Bearing bottomflange Height of rub plates (See note 4 on File No. 20.03-19)

= 6.33 ft(12 in/1 ft) – 3 in – 2 in – 3 in – 1 in = 67 in hrp Fg = 2,000 psi Maximum “galling stress” for ASTM A276 Type 316 steel, of which the rub plates are constructed.

= 0.55 F = 1,100 psi Allowable galling stress fg g

( )grp

pmin fh

Rw = Minimum rub plate width

in 3.58psi) in(1,100 67

k) lbs/1 k(1000 263.8w min ==

VOL. V - PART 2

Ensure the minimum rub plate width is maintained during extremes of the temperature cycle. w = w




min + ΔL = 3.58 in + 0.78 in = 4.36 in rub Use 5 in x 64 in x 0.5 in rub plate Design wing haunch to resist load transferred through the rub plates:

Slope = 1.5 Rate of slop of wing per ft

Assume that the resultant is uniformly distributed along the rub plate

klf 41.7ft 6.33k 263.8

HR

wBackwall

pw ===

( )2

HwM

2Backwallw

s = Moment about seat level

kipft 835.42

ft) klf(6.33 41.7M2

s −==

VOL. V - PART 2

= 263.8 k Shear force V = Rp

= 2.0 ft + H (Slope) Length of shear plane that is resisting RLwing Backwall p

= 2.0 ft + 6.33 ft(1.5) = 11.5 ft Lwing C = 4.5 in Centroid of tension steel cg

= L – C Distance from compression face to cg of dwing wing cg tension steel

= 11.5 ft – (4.5 in(1 ft/12 in) = 11.1 ft dwing

)f125%(0.95v 'cwall wingallowablec =

= 65 psi Allowable shear stress for Group IV vc allowable wing

= 15.0 in Twing

psi 132ft)] in/1 ft(12 in[11.1 15.0

k) lbs/1 k(1000 263.8dT

Vvwingwing

=== Actual shear stress




v > vc allowable, Shear reinforcement required

2Use #4 stirrups, with 2 legs in the shear plane, Av = 0.4 in . Equation (8-7) from AASHTO Sec. 8.15.5.3

( ) wing wingallowablec

allowable sv

T vvfA

s−

= Stirrup spacing

( ) in 11.9in) (15.0psi 65 - psi 132

psi) (30,000in 0.4s2

==

Check AASHTO Sec. 8.19.1.2 to see if spacing above is OK.

( )y

v.min f50bsA =

Rearranging the above equation, with Av = 0.4 in2:

in 32in) 50(15.0

psi) (60,000in 0.450T

fAs

2

wing

yv.min ===

s = 11.9 in controls

VOL. V - PART 2 DATE: 11May2007 SHEET 10 of 20



Design Wing Reinforcement to resist moment due to Rp: Try two rows of #5 bars, with 5 bars per row

d1 = Lwing – Cover Depth to first mat of reinforcing steel d1 = 11.5 ft(12 in/1 ft) – 3.5 in = 134.5 in d2 = d1 – 3 in = 134.5 in – 3 in = 131.5 in Depth to second mat of reinforcing steel As1 = 1.53 in2 As2 = 1.53 in2

Performing section analysis (including the compression steel)

fs1 = 25,940 psi < fs allowable = 30,000 psi OK in first layer of tension steel fc = 350 psi < fc allowable = 2,000 psi OK

Thickness of the EPS layer: Thickness of EPS layer as per File No. 20.06-6: Δt = 120°F ΔL = (LthermalαΔt) Total range of movement at abutment due to temperature ΔL = 125.0 ft(12in/1ft)(6.5x10-6 per deg. F )120°F = 1.17 in EPSt = 10(0.01HBackwall + 0.67ΔL) EPSt = 10[(0.01)(76.0 in)+ (0.67)(1.17 in)] = 15.4 in Therefore, use EPSt = 16 in.

NOTE: DESIGN FOR PRESTRESSED CONCRETE BEAMS IS SIMILAR.

VOL. V - PART 2

CHECK LIST FOR SEMI-INTEGRAL ABUTMENTS

1 Wing haunch at acute corner shall be designed to resist the moment and shear induced by the force resulting from the passive earth pressure and the skew. Rub plates and the additional backwall thickness are only required at the acute corners of skewed bridges. Rub plates to be centered vertically and horizontally over contact area.



CHECK LIST FILE NO. 20.03-19

2 Minimum thickness of the preformed joint filler between the backwall and the wing at the

obtuse corner shall be 1”. This may be increased due to thermal expansion in the transverse direction.

3 Extend wing 6” above finished grade. Not required for bridges without skew or where

terminal wall is on the substructure.

4 Top of rub plate to begin 3” below top of deck. Bottom of rub plate to maintain 2” clear from top of bottom flange for steel superstructures; 3” clear from bottom of beam for concrete. Preformed joint filler to extend as shown.

5 Provide distance from back of stem to break in seat to allow for contraction and creep with 1”

clear.

6 Delete this note if railings are used or slip forming of parapets is not allowed.

7 Bridge plans shall be arranged such that backwall details follow the Deck Plan. For general sheet order, see File No. 01.02-4.

8 Show plan and elevation view of integral backwall at a preferred scale of 3/8” = 1’-0”. The

elevation view should be projected down from the plan view. When bridge is not on skew and where sufficient room is not available in elevation view, plan view is not required.

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

10 “End of slab” shall be used as the reference line for layout of integral backwalls.

11 Label skew angle (if applicable).

12 The minimum width of integral backwall shall be 1’-7” for steel stringers and 1’-10” for

concrete stringers. Clipping flanges is preferable to increases in thickness where required due to skew.

13 All ST series and SV series bars shall be aligned parallel to the beam/girder centerline. The

maximum spacing shall be 12”. ST0602 bars between the backwall and the approach slab (where applicable) are not required outside of the exterior beam/girder.

14 Thickness of backwall shall be increased by 10” at the acute corner of skewed bridges

outside of the exterior stringers. The increase in thickness shall end at the top surface of the bottom flange for steel stringers or 1” above the bottom of beam for concrete stringers.

15 ST0501, ST0602, SV0402 and SV0504 shall be galvanized. All other backwall reinforcing

steel shall be epoxy-coated.

16 Distance between face of integral backwall and back of stem shall be a minimum of 4”.

VOL. V - PART 2

CHECK LIST FOR SEMI-INTEGRAL ABUTMENTS (Continued)

17 The approach slab seat (7”) shall be provided on all integral backwalls regardless of whether the bridge will have an approach slab.



CHECK LIST FILE NO. 20.03-20

18 In case of single span semi-integral bridge, use the temporary blocking note shown.

Otherwise, delete it.

19 Show sections taken through the integral backwall at a preferred scale of 3/4” = 1’-0”. Coordinate the sections to provide the necessary details with repetition only where required.

20 Location and details of holes in the web and the studs should be included with the

beam/girder details.

21 For additional details concerning the use of EPS material and calculations for the required thickness, see File No. 20.06-6.

22 To ensure adequate cover on ST0602 bar, the designer must modify the approach slab

standard.

23 Maximum spacing is 12”.

24 Note not needed for PCBT-53 and larger.

25 The minimum embedment into the backwall is 6” for steel stringers and 9” for concrete stringers.

26 When approach slab is used with concrete superstructure, hook ST0602 bar and embed as

shown.

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

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

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

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Semi-Integral-Abutment Design Example (USA Unit)