short span bridges-pci

Copyright© 1984By Precast/Prestressed Concrete Institute

All rights reserved. This design supplementor any part thereof may not be reproducedin any form without the written permission ofthe Precast/Prestressed Concrete Institute.

First Edition, Second Printing, 1988First Edition, Third Printing, 1991

o

PRECAST/PRESTRESSEDCONCRETE INSTITUTE

175 West Jackson BoulevardChicago, Illinois 60604Phone 312-786-0300Fax 312-786-0353

INTRODUCTION

The PCI manual, "Precas t Prestressed Concrete Short Span Bridges - spans to

100 feet," presents practical design aids for constructing, replacing and

widening bridges for spans up to 100 ft. using precast, prestressed concrete

bridge sections. The manual describes how the use of standard precast con

crete integral deck components create economical new bridges, and replace or

widen deficient old bridges. The manual shows actual bridge applications,

and provides a variety of design aids in the form of charts, graphs and

detail drawings. The design aids are arranged for easy selection of an

appropriate section applicable to immediate and future bridge needs. The

manual is intended to save design time and money for county and municipal

bridge engineers and bridge design consultants.

This "Design Supplement" for the Short Span Bridges Manual presents typical

design calculations for four types of precast, prestressed concrete bridges

using standard sections illustrated in the manual. The four examples include

typical design calculations for the precast sections used in the bridge

superstructure. Step-by-step procedures provide a guide to assist engineers

in proper design for prestressed concrete bridges, including proper applica

tion of the design provisions of the "AASHTO Standard Specifications for

Highway Bridges. 1I

Calculations for the four examples are typical for precast, prestressed con

crete sections, and can be easily adapted to other bridge sections shown in

-i

the Short Span Bridges Manual. In addition, other precast concrete sections

and variations are available, all of wh'ch follow the same deslgn procedures

as presented in the examples. Precast concrete suppliers in your geographi

cal area will be pleased to furnish information on the sections they are

equipped to make most economically.

The des1gns comply with the provisions of the AASHTO Standard Specifications

for Highway Bridges, 13th edition, 1983, with one exception; the method of

design for shear reinforcement presented in the 1979 AASHTO Interim specifi

cations is used for the three simple span bridge designs as an acceptable

alternative to the shear design provisions of the 13th edition. Reference to

speCIfic AASHTO provisions are noted throughout the Supplement .... e.g.

(AASHfO 9.2.2), meaning AASHTO, 13th edition - 1983, Division I-Design,

Section 9.2.2.

Extreme care has been taken to be as accurate as possible with the informa

tioll presented. However, as PCI does not actually prepare engineering plans,

it cannot accept responsibility for any errors or oversights in the use of

the material in this Supplement or in the preparation of engineering plans.

-ii-

CLARI FICATIONS

At the time of the Second Printing of this publication, the following clarifications

needed to be made:

1. The equation, Av (minimum) = 10~ b'S , is found on pages 1-12, 2-14, and 3-15.sy

It is from the 1977 Standard Specifications for Highway Bridges. In the 1980

Interim Specifications, the factor 100 was changed to 50. This reduced the required

minimum reinforcement by one-half. PCI supports this change. Changes affecting

the examples in this pUblication are as follows:

Example No.1, page 1-12

Av (minimum) 0.12 in. 2/ft

#3 @22 in. Av = 0.11x2x12/22 = 0.12 in. 2/ft

However, paragraph 9.20.3.2 requires maximum spacing

not exceed 0.75h 0.75x21 = 15.75 in.

Use #3 @ 15 in.


A (minimum) = 0.125 in. 2/ftv

Use welded wire fabric W4 @6 in. spacing per stem.

(Av = 2xO.08 = 0.16 in. 2/ft, 0.125)


Av (minimum) = 0.08 in. 21f t.

Because of subsequent vertical tie spacing requirements, use

#3 stirrups @ 12 in. centers.

(Av = 2xO.ll = 0.22 in. 2/ft)

iii

2. The following equation is used in the first three design examples and is found on

pages 1- 11, 2- 14. and 3-15:

A _(Vu - 0 Vc )s ( Eq. A)v - 2 0f jdsy

The derivation of this equation follows. From the 1977 AASHTO Standard

Specifications for Highway Bridges (Twelfth Edition), paragraph 1.6.13-SHEAR.

the following expression for area of web reinforcement is found:

(v - V )su c2f 'dsi

solving for (V - V ).u c

2A f jd(V - V )= v sy Vs

u c s

where V - V = V (or, V = V + V )U C S U C S

(Eq. B)

From the 1980 Interim Specifications, paragraph 1.6.13, "members subject to shear

shall be designed so that V ~ ~(V + V )u r c s (Eq. C)

Substitute the expression for Vs (Eq. B) into Eq. C and solve for Av'

PCI recommends Eq. A for use for simple spans where the more complex methods for

design of shear reinforcement found in the 13th Edition are not used.

3. The equation for shear reinforcement on the bottom of page 4-17 is derived much

the same way as described in Item 2 above. The derivation begins with

Equation 9-30 and 9-26 found in the 13th Edition of the AASHTO Standard

Specifications. These equations were first presented in the 1980 Interim

Specifications.

4. In Example No.4, page 4-5, moments and shears are computed using tables found in

Reference 4-1. These tables are based on the 65 ft end spans and result in

moment and shear values for the 80 ft center span which are correct as shown.

iv

DESIGN EXAMPLE NO.1MULTI-BEAM SLAB BRIDGE

1.1 Design Conditions

S1mple span of 45 ft x 30 ft width

HS20 live load - 2 lanes

Use multi-beam precast sections (adjacent units) without wearing

surface.

1.2 Mater1als

Concrete: normal weight

f~ == 5000 psi

f~i == 4000 psi (AASHTO 9.22)

Prestress1ng steel: 1/2 in. diameter 270 ksi stress-relieved strand

Strand area == 0.153 sq. in.6

Es == 28 x 10 ps l

1.3 Precast Beam-type

Us1ng span tables in "Short Span Bridges Manual" as a guide, a voided

slab section can span up to 50 ft for HS20 loading. Select 21 in.

depth x 3 ft. width vo1ded slab section, w1th 10 sections required for

the 30 ft. width bridge. Bridge layout and girder section propert1es

are as follows.

1-1

301-0

10 @31_O u width

SECTION

451-0

ELEVATION

....-Keywa y Section Propert1es

Ac 530 ;n. 2

4I 0: 25,750 in.

St 0: Sb 0: 2450 ;n. 3

Wo ;: 552 plf

=U"l

o,....

31-0

~I-------~VOIDED PRECAST BEAM SECTION

1.4 Design Loads and Moments

(a) Dead load

Beam 0: 552 p1f

Barrier rail 350/2* = 175 plf

727 PIf*Assume each rail distributed to two precast beam sections.

MD 0: Wl2/ B = 0.552 x 45 2/8 + 0.175 x 452/8

139 + 44 0: 183l k/beam

1-2

(b) l1ve load

Llve load d1strlbutlon (AASHTO 3.23.4);

NL = total number of lanes 2

Ng = total number of beams 10

S = (12 NL + 9)/Ng = (12x2 + 9)/10 = 3.3

C = stlffness parameter = KW/L 0.8x30/45 0.533

where W= wldth of brldge

L = span length

NL 2N C 20 5 (3 __L)(l for C < 3+ 10 + 7 - 3")

2 2X2)( 12

= 5 (3 - 0.533) = 6.84+ 10 + 7 ~ 3

Load fractlon to each beam = SID = 3.3/6.84 0.482

Llve load lmpact (AASHTO 3.8):

50 50I = L + 125 = 45 + 125 = 0.294

For 11ve load. use moment tables (AASHTO App. A):

HS20 - 45 ft span - lnterpolate

I k 1 I kML = 538.7 Ilane x 2 (wheel) x 0.482 x 1.294 168 Ibeam

1.5 Prestress1ng Strands

Estimate number of strands requlred based on stresses at service load.

Assume concrete tension 1n bottom f1ber governs. (Negative s1gn

indicates tens10n in concrete.)

1-3

Bottom fiber stress due to design loads:

f b = ("0 + "L)/Sb = (168 + 183)12xl000/2450 = -1719 psi

Allowable tensile stress (AASHTO 9.15.2.2) = 6vf~ = 6v5000 = -424 psi

Required prestress stress in bottom fiber = 1719 - 424 = 1295 psi

Bottom fiber stress due to prestress:P P ·e

f b =i!+~-. Sb

where Pse = effective prestress force after lossese = strand eccentricity = 10.5 - 2 = 8.5 in.

Pse 8.5 Pse1.295 = 530 + 2450

solving, reqiured Pse = 242k

Final prestress per strand, assuming 20% prestress losses:

(0.153 x 0.70 x 270)0.8 = 23.1 kips

Number of strands requ1red = 242/23el = 10.5Try 11 - 1/2 in. 270K strands

1.6 Flexural strength

Using Group I loading combination (AASHTO 3.22); strength required:

MU = 1.3("0 + 1.67ML)= 1.3(183 + 1. 67 x 168) = 603' k

1-4

Use approximate value for stress in prestressed reinforcement

(AASHT 0 9. 17. 4) :

f* f ' (1su s0.5 p* fl/fl)

s C

Mcr

270(1 - 0.5 x 0.00246 x 270/5) = 252 ksi

where p* = A;/bd = 11 x 0.153/36 x 19 = 0.00246

For rectangular sections (AASH10 9.17.2); strength provided:

~M = ~A* f* d(l - 0.6 p* f* If')u s su su c

= 1.0 x 11 x 0.153 x 252 x 19(1 - 0.6 x 0.00246 x 252/5)/12

= 6221k

> 603 OK

Note: For factory produced precast prestressed concrete members, the

strength reduction factor ~ = 1.0 (AASHTO 9.14).

1.7 Max1mum and M1n1mum Steel Percentage

(a) Max1mum steel for rectangular sect10ns (AASHTO 9. 18. 1) :

Reinforcement index = p*f* /f'su c= 0.00246 x 252/5 = 0.124 < 0.3 OK

(b) M1n1mum steel (AASHTO 9.18.2):

Total amount of prestressed reinforcement must be adequate to

satisfy ~M > 1.2 M • Cracking stress for normal weightu crconcrete (AASHTO 9.15.2.3):

f 7.5vf l 7.5v5000 = 530 ps1cr c

Pse PseoeSb(~ + ---S--- + fer)

c b

2450(254 254x8.5 0 530) _ 38,'k:: 12 530 + 2450 +. - u

where Pse = 0.8(0.7x270)11xO.153 = 254k (assuming 20% losses)

622/386 = 1.61 » 1.2 OK

1-5

1.8 Prestress losses

Est'mat1on of loss of prestress w'll be based on the approx'mate

procedure presented 'n AASHTO 9.16.2:

(a) Shr1nkage

SH = 17,000 - 150 RH

Assume m'dwest 10cat'on, RH = 70%

SH = 17,000 - 150 x 70 6500 ps'

(b) Elast'c shortening

f ci r concrete stress at level of prestress'ng steel 'mmediately

after transfer

Assume 10% prestress loss 'mmediately after transfer:

Ps' = 0.9(0.7 x 270) x 11 x 0.153 = 286k

fc'r

286 286(8.5)2= 530 + 25,750

139x12x8.525,150 0.191 ksi

ES = ~ x 7913.8 5780 ps'

1-6

= 0.174 ksl

(c) Creep of concrete

CRe = 12 fclr - 7 feds

fcds = concrete stress at level of prestresslng steel due to

superlmposed dead load

MOe 44x12x8.5= -1- = 25,750

CRe = 12 x 791 - 7 x 174 = 8270 psl

(d) Relaxatlon of prestressing steel

CRS

20,000 - 0.4 ES - 0.2(SH + CRe)

CRs = 20,000 - 0.4 x 5780 - 0.2(6500 + 8270) = 14,730 psi

Note: Loss of prestress due to strand relaxation would be substantially less for low-relaxation strand. Uslng an appropriate

expresslon for low-relaxation strand, and assuming same strand

size and grade:

CR s = 5000 - 0.10 ES - 0.05(SH + CRc)= 5000 - 0.10 x 5780 - 0.05(6500 + 8270)

CR s = 3680 psi « 14,730 for stress-relleved strand

(e) Total loss of prestress

6f s = 6500 + 5780 + 8210 + 14730 = 35,280 psi

or 35.3/0.7 x 270 = 18.6% losses

f se = effective prestress = 0.7 x 270 - 35.3 = 153.7 ksi

1-7

1.9 Concrete stresses

Prestressing:

Psi = 0.9(0.7 x 270)11 x 0.153 = 286k

P = 153.7 x 11 x 0.153 = 259 kse

e = 10.5 - 2 = 8.5 in.

Section Properties

A = 530 in. 2c 3

St = Sb = 2450 in.

Concrete stresses at prestress transfer and at service load (in psi)

are summarized below. With straight strands, only stresses at span end

at prestress transfer and midspan at service load need be evaluated.

Midspan stresses at prestress transfer are not critical with straight

strands.

Span End at Midspan atPrestress Transfer Service Load

P = Psi P = PseLoad

f b f t f b ft

PIAc 540 540 489 489

PelS 992 -992 899 -899

MD/S -- -- -897 897

MLIS -- -- -823 823

TotalStress 1532 -452 -332 1310

Allowable O.6f~i 7.5.ff~i 6v"f ' O.4f I

Stress c c

(AASHTO 2400 -474 -424 20009.15.2) OK OK, but must OK OK

debond since>3v"f~i= -190

Tension (-)

1-8

1.10 Debonded Strands at Span Ends

Reference: uUse of Debonded Strands in Pretensioned Bridge Members."

Horn. Daniel G.• and Preston. H. Kent. Journal, Prestressed

Concrete Institute. Vol. 26, No.4, July-August 1981. pp.

42-50.

Since the top fiber stress 1n tension at span ends exceeds 3vf~1'

bonded reinforcement must be provided to res1st the total tensile force

(AASHTO 9.15.2.1) .... or alternatively, some of the strands can be

debonded (bonding of strand does not extend to end of member) to reduce

the stress level. Debonding technique will be utilized in this exampleto illustrate design procedure.

Transfer length over which force in the strand at release is transferred

to the concrete is taken as f s idb/ 3 = (0.63 x 270)0.5/3 = 28.3 1n.,where f s i = stress in prestressing steel at transfer.

Must reduce tensile stress to 3vf~i = 3v4000 = 190 psi. Distance fromend toward center of span where beam dead load stress is sufficient to

reduce top fiber stress to 190 psi; if x = distance from support:

wx f M wx - x)M= ~(1 - x); =5 or f = 2S (1t t

452 - 190 552x (540 - x)= 2(2450)(12)

solving, x = 57.9 in.

It takes 28.3 in. for strand transfer; at that point the beam dead loadstress is:

wxf O = 25 (1 - x)

t

552(28.3)= 2(2450(12)(540 - 28.3) = 136 psi

1-9

Must reduce stress by 452 - (136 + 190) = 126 psi126Must shield 452 x 11 : 3.06 strands.

Shield 4 strands ... 2 symmetrically on each side of centerline for

57.9 - 28.3 : 29.6 in., say 30 in. from each end of beam.

Check stresses:

60

Concrete stress ~ig;nal Prestress stress

~190 psi T~enSi/- __---- t <190 psi

~~ ~hlelded Stress

~' -'i -. Beam DL Stress

ell Bearing10 20 30 40 50

Distance from End of Beam (in.)

~Shield 4 Strands = 30llJEnd of Beam

III 300III

~ 200+J

V'l 100

___ 500

:;. 400

Graph assumes uniform buildup of transfer stress, with distance

from bearing to end of beam assumed to be small and ignored in

span length computations.

Shield Alternate Strands in Symmetrical Pattern

1-10

(33.75 19.15)32 (5.15)8 _ 39 l K/l45 + 45 + 45 - . ane

39.1 x 0.5 x 0.482 x 1.294 = 12.2K/beam

1.11 Shear Strength

The method for design of shear reinforcement presented 1n the 1979

Interim AASHTO Standard Specifications will be used as an acceptable

alternative to the provisions of the 13th Edition, 1983 Specifications.

Check shear at quarter span:

Vo = w1/4 = 0.727 x 45/4 = a.2K/beam

32k 32kak

+-HS20 Truck

1./4 = 11.25 1" 141

~, 141 ~ 5.15 1

4~ .~

45'-0

LANE LOADING

VL =

VL =

Note: The HS20 truck loading is applied to the full lane long1tudinally

to obtain maximum lane shear at the span quarter point. The lane shear

is then distr1buted to an indiv1dual beam, w1th appropr1ate l1ve load

impact.

Vu = 1.3(VD + 1.67 VL) = 1.3(8.2 + 1.67 x 12.2) = 31.,K/beam

Vc = 0.06f~bljd = 0.06 x 5 x 12 x 0.92 x 19 = 63K

but not greater than 180 b'jd = 0.180 x 12 x 0.92 x 19 = 37.7K

where effect1ve width b' is conservatively taken as 36 - 2(12) = 12 1n.

(V u - ~Vc)s (31.1 - 0.9x37.7)12 0.02 in.2/ftAv = 2~fSyjd = 2xO.9x60xO.92x19 =

where, for shear ~ = 0.9 (AASHTO 9.14)

1-11

A (minimum)v100b's

::f SY

100x12x1260,000 0.24 in. 21f t

Use #3 @ 11 in. (2 legs per beam) Ay :: 0.11x2x12/11 :: 0.24 in. 21f t

Add extra stirrups at beam ends (AASHTO 9.21.3):

4% Psi:: 0.04(286) :: 11 .44k, Ay :: 11.44/20 :: 0.572 in. 2

Use 3 #3 U-stirrups @ 2 in. spacing at each end of beam.

1.12 Deflections and Camber

The following data from "PCI Design Handbook" will be used in

estimating long-time deflections and cambers.

SUGGESTED MULTIPLIERS TO BE USED AS A GUIDE IN ESTIMATINGLONG-TIME CAMBERS AND DEFLECTIONS FOR TYPICAL MEMBERS

Deflection to be Considered

At erection:

(1) Deflection (downward) component - apply to theelastic deflection due to the member weight atrelease of prestress

(2) Camber (upward) component - apply to the elasticcamber due to prestress at the time of releaseof prestress

Final:

(3) Deflection (downward) component - apply to theelastic deflection due to the member weight atrelease of prestress

(4) Camber (upward) component - apply to the elasticcamber due to prestress at the time of releaseof prestress

(5) Deflection (downward) - apply to elastic deflection due to superimposed dead load only

(6) Deflection (downward) - apply to elastic deflection caused by the composite topp1ng

1-12

WithoutCompositeTopping

1.85

1.80

2.70

2.45

3.00

WithCompositeTopping

1.85

1.80

2.40

2.20

3.00

2.30

(a) Prestress at transfer

t

e.g. of section__---J.I_' _

e.g. of strandt

straight Strands

286x8.5x(45x12)2

8x3.8xl03x25.750 = 0.90"1'

(b) Beam dead load

5wt4 5xO.552x(45x121 4

384E1 = 384x12x3.8Xl03x25.750At transfer

(c) Growth in storage

Using suggested mu1t1pliers .... camber (upward) component at

erection: 1.80 x 0.38 = 0.68"1'

(d) Superimposed dead load

Ec = 33(150)1.5~5000 = 4.3 x 106 psi

5xO.175x(45x12)4

384x12x4.3xl03x25,750Net after construction

1-13

= 0.14"01----

(e) Long term dead load

Us1ng appropr1ate mult1p11ers:

0.52"'&' x 2.70

0.90"t x 2.45

0.14",&. x 3.00

Net long term

;:; 1.40"'&'

2.20"1'

_ 0.42"'&'

0.38"t OK

Use span center deflect10n at maximum

moment as approximate maximum

Dist. Factor from lane to beam 10ad1ng:

(Distr1but1on)(Wheel)(Impact)

(0.482)(1/2)(1.294) z: 0.312

(f) live load deflection

32k

~ lane Shear Diagram

50"...1_'k 505 I k

~ ~O'k + Lane Moment Diagram

Us1ng moment-area method:

2(501X~5.5 + 503x7x19) 1728X0:i3l2 .

4.3xl0 x257500.52" or i

1036 OK

S1nce AASHIO does not provide guidance on acceptable 11ve load deflec

t10ns for prestressed concrete br1dges, check cr1teria for steel g\rders

in AASHTO 10.6. live load deflection cr1ter1a for steel girders is

pr1mar1ly based on an empir1cal limitat10n for vibration.

DESIGN EXAMPLE NO.2DOUBLE STEMMED TEE BRIDGE


Simple span of 40 ft x 30 ft width

HS20 live load - 2 lanesUse double stemmed precast sections (adjacent units) with cast-in-place

compos1te deck slab.

2.2 Materials

Precast concrete: normal weight

f~ 5000 ps1f~i = 4000 psi (AASHTO 9.22)

Cast-in-place concrete: normal weight

f~ = 4000 psi

Prestressing steel: 1/2 in. diameter 270 ksi stress-relieved strandStrand area = 0.153 sq. in.E

s= 28 x lOb psi

2.3 Preliminary Selection

Using span table in "Short Span Bridges Manual" as a quide , select a

24 in. depth x 6 ft w1dth "medium ll sect10n for the 40 ft span. w1th 5

sect10ns required for the 30 ft width br1dge. Br1dge layout and girder

2-1

section properties are as follows. Composite properties are based on

f' = 4000 psi deck concrete and f' ; 5000 psi beam concrete. Note:c cStemmed sections vary widely and this precast section is chosen as

representative of common practice. A future wearing surface is not

considered in this example; however, some governing agencies may

require an additional wearing surface.

301_0

~ ~

~ ~v-preca s t

Ba r r i er Ra 11

TT TT IT@6'_0" J (350 pH)

~ 5 Double-Stemmed Sect'ons ~SECTION

40 1-0 ......

I 1

ELEVATION

14 ,440

3.2

Compos ite815

849

46,200

20.8

8.7

2220

5310

Beam419

436

22,230

15.8

8.2

1407

2711

Section Properties

AC

( i n. 2)

wD(plf)I (in. 4)

yb(in.)

Yt3Sb( in. )

St

At top of precast beam:

StYt

36 11 *

*May be 48 in. in some cases

72"

DOUBLE STEMMED BEAM SECTION

--.j f.- 4-1/2"

I ......

lit> l4')U

I ~~ \:

I

~ ...'-- '----:rN

H 8 11 ..,.N

U '--',

2-2

2.4 Design Loads and Moments

(a) Dead load

Beam " 43& P1£

Deck slab 413 p l f

Ba r r i erra11 2x350/5* ;; 140 p1£

989 P1£

*With composite deck, assume barrier rails distributed equally to

the five double stemmed units (AASHTO 3.23.2.3.1.1).

MO ;; Wl2/ 8 " 0.436 x 402/8 + 0.413 x 402/8 + 0.140x402/ 8'k" 87 + 83 + 28 ;; 198 Ibeam

( b) l1 ve load

Use live load distribution for Concrete T-Beams according to

AASHTO Table 3.23.1, considering each stem as one tee for lateral

distribution:

5/6 " 3/6 = 0.5, or 2xO.5 ;; 1.0 for & ft wide double tee.

live load impact (AASHTO 3.8):50 50

I =~25 ;; 40 + 125 ;; 0.303 Use 0.30

For live load, use moment tables (AASHTO App. A):

HS20 - 40 ft span - military load not considered

"k 1 I kML ,,449.8 Ilane x 2(wheel) x 1.0 x 1.30 292 Ibeam

2.5 Prestressing Strands

Estimate number of strands required based on stresses at service load.

Assume concrete tension in bottom fiber governs.

f b " 170x12/1407 + (28+292)12/2220 ;; -3.18 ksi

2-3

Allowable tenslle stress (AASH10 9.15.2.2) = 6~f~ = 6~5000 = -424 psl

Requlred prestress stress In bottom f1ber = 3.18 - 0.42 = 2.76 ks1

Bottom flber stress due to prestress:

Pse P ·ef b

se= -t---

Ac Sb

where P = effect1ve prestress force after lossessee = strand eccentrlclty

Note: stress calculat10ns for prestress are based on beam sectlon

propertles. Est1mate e = 12 In.

Pse Pse)(l~2.76 = 419 t 1407

solvlng, requ1red Pse = 253k

Flnal prestress per strand, assumlng 25% losses:

(0.153 )( 0.70 x 270)0.75 = 21.69K

Number of strands requlred = 253/21.69 = 11.7

,fA

~,

~ ~3.5" to c.g

~~T1-112" cover

Try 12-1/2 In. 270K strands

(6 strands per leg)

Orape (two polnt depress) strands at 4 ft each sldeof span center to provlde 3.5 In. from bottom of leg

to center of strands (see sketch). strand eccen

trlclty w1thln center port1on of span (reglon of

hlghest moment) = 15.8 - 3.5 = 12.3 In.

2-4

2.6 Flexural Strength

Us'ng Group I load'ng comb'nat'on (AASHTO 3.22); strength requ'red:

Mu = 1.3(MO + 1.67ML)

= 1.3(198 + 1.67 x 292) = 8911k

Use approx1mate value for stress 1n prestressed re1nforcement

(AASHTO 9.17.4):

f* = fl (1 - 0.5 p* f l / f 1)su S S C

= 270(1 0.5 x 0.000981 x 270/4) 261 ks i

where p* = A*/bd = 12 x 0.153/72 x 26 =sfl = 4000 ps' for deck slabc

0.000981

For rectangular sect'ons (AASHTO 9.17.2); strength prov'ded:

~Mu = ~A; f;u d(l - 0.6 p*f~u/f~)

= 1.0 x 12 x 0.153 x 261 x 26(1 - 0.6 x 0.000981 x 261/4)/12

= 9981k

> 891 OK

Check "a" as a rectangular sect1on:

a = A* f* 10.85f l bs su c= 12 x 0.153 x 261/0.85 x 4 x 72 1.96 1n. < 5.5 OK

Note: for factory produced precast prestressed members, ~ = 1.0.

2.7 Max1mum and M'n'mum Steel Percentage

(a) Max'mum steel for rectangular sections (AASHTO 9.18.1):

Re1nforcement 1ndex = p*f* If I ~ 0.3su c

0.000981 x 261/5 = 0.051 < 0.3 OK

2-5

(b) Minimum steel (AASHTO 9.18.2):

Total amount of prestressed (and non prestressed) reinforcement must be

adequate to develop a design moment strength at least equal to 1.2

times the cracking moment strength (~Mu ~ 1.2 Mer)' where Mer is determined by summing all the moments that cause a stress in the bottom fiber

equal to the cracking stress f . Referring to the sketch below, forcra prestressed composite member,

P Pse-e MO Ma(~) - (--) + (-) + (-) ;; +f- Ac Sb Sb S crc

Solving for M '" (f +Pse Pse·e

Sc - (MO)Sc

Ac+ --)

Sba cr Sb

Since M '" Mo + Mcr a

Mer ;; (fPse Pse-e

S - MDSc

1)+ - + --) (- -er A Sb c Sbc

Cast-in-place

Member

A*S

~-t--.

CG Precast

Member

STRESS CONOITIONS fOR EVALUATING CRACKING MOMENT STRENGTH

A* area ofsAe

;; area of

Sb '" section

prestressed reinforcement

precast member

modulus for bottom of precast member

2-6

Sc = sect10n modulus for bottom of compos1te memberPse = effect1ve prestress force

e = eccentr1c1ty of prestress force

MO = dead load moment of composite member

Ma = add1tional moment to cause a stress 1n bottom fiber equal tocrack1ng stress fcr

Mcr

= (0.530 + 421690 + 260X12.3) 2220 _ 170(2220 _ 1) = 535'k1407 12 1407

where fer = 7.5vf~ = 7.5v5000 = 530 ps1

Pse = (12xO.153xO.7x270)0.75 = 260k

(For 1n1t1a1 estimate, assume 25% losses.)MO = 87 + 83 = 170'k

(Use beam and slab moments only; rema1n1ng portion will

act on compos1te sect1on.)

998/535 = 1.87 » 1.2 OK

Note: strength ratio sufficiently high; reevaluation with a more exactestimate of prestress losses will not be necessary.


Est1mation of prestress losses will be

procedure presented in AASHTO 9.16.2.

center (critical moment location).

6f s = SH + ES + CR + CRc s

(a) Shrinkage

SH = 17,000 - 150 RH

Assume m1dwest location, RH = 70%

SH = 17,000 - 150 x 70 = 6500 psi

2-7

based on the approx1mate

Compute loss values at span

(b) Elastic shortening

f ci r = concrete stress at level of prestressing steel immediately

after transfer

Assume 10% prestress losses due to elastic shortening and

relaxation at release:

Psi = 0.9(12xO.153xO.7x270) 312k

312 312(12.3)2= 419 + 22,230

87x12x12.322,230 = 2.291 k s t

28ES = 3.8 x 2291 16,750 psi


f d = concrete stress at level of prestressing steel due toe ssuperimposed dead load .... deck slab plus barrier rail

= B3x12x12.3 + 2Bx12(20.B - 3.5) 0.677 ksi22,230 46,200

CRe = 12 x 2291 - 7 x 677 = 22,750 psi

(d) Relaxation of prestressing steel

CRS

= 20,000

CR s = 20,000

0.4 ES - 0.2(SH + CRe)

0.4 x 16,750 - 0.2(6500 + 22,750) = 7450 psi

2-8


6f s = 6500 + 16,750 + 22,750 + 7450 = 53,450 psi

or 53.4/0.7 x 270 = 28.3% losses

f se = effect1ve prestress = 0.7 x 270 - 53.4 = 135.6 ksi

2.9 Concrete Stresses

Prestressing:

Psi = 0.9(0.7 x 270)12 x 0.153 = 312k

Pse = 135.6 x 12 x 0.153 = 249k

Set draped strand pattern to satisfy stress condit10ns at beam ends

(AASHTO 9.15.2.1).

sym.a = 16.0 1

Depress 0.1 spanpo lrrt '\ L.. 4 1-0 ...

~ ~ n.a. of

~=========~....----concrete

i--f;~===========i' i.. 3.5"f

11

~, ee i.~ ~

3.5" to e.g ~

~/ eli ~T1-1/2" cover

Note: Use 0.1 span depress po1nt so that l1near prestress moment

exceeds parabo11c load moments.

Top fiber stress at prestress transfer 1s l1m1ted

transfer strength at 4000 psi, 3v4000 = 190 ps1.

transfer:

to 3vf~. Assum1ngTop f1ber stress at

312 312eef t = 419 - -z7ff = -0.190

solv1ng. ee = 8.12 1n.

2-9

Bottom fiber stress at transfer is limited to 0.60f~1 0: 2400 psi:

312 312ee 2.40f b : 419 +~=

solving, ee :: 7.46 in. say 7.4 1n.

Compression governs; e l :: 12.3 - 7.4 :: 4.9 1n.

Note: For this small amount of drape, the end pattern can eas1ly be

adjusted by the prestressed concrete supplier to fit a part1cular

bulkhead.

Sunmariz1ng: Ps1 :: 312k

Pse :: 249k

ecenter:: 12.3 in.

eend:: 7.4 in.

111

of0

0..~

0

0

0~,

8.4'1 to e.g.of strands

At g1rder end

Loads:

MO(beam) :: 871k

MO(beam) at depress 0: 841k

MO(slab) :: 831k

MO(rai1):: 28

l k

ML;: 292

1

k

Top of BeamSection Properties Beam Composite CompositeA (in. 2) 419 815c 3

1407 2220Sb(1n. )

St 2711 5310 14,440

2-10

Concrete stresses at prestress transfer and at serv1ce load (1n ps1)

are summar1zed as follows:

Span End at Depress Pt. at M1dspan atPrestress Transfer Prestress Transfer Serv1ce Load

P = Ps t P = Ps1 P = PseLoad

f b f t f b f t f b f t OT f t SL

P/Ac 745 745 745 745 594 594 --PelS 1640 -852 2727 -1416 2177 -1130 --

MOblS -- -- -713 370 -742 385 --

Mo/S -- -- -- -- -708 367 --MOrIS -- -- -- -- -152 23 63

ML/S -- -- -- -- -1578 242 660

TotalStress 2385 -107 2760 -301 -409 481 723

All owabl e 0.6f~1 3V'f ' 0.6f~1 7.5vf~1 6vf l o 4f ' 0.4f l

Stress c1 c . c c

(AASHTO 2400 -190 2400 -474 -424 2000 16009.15.2)Beam: OK OK Increase OK, but OK OK OKfl = 4000 release mustc t strength re1nforcef • = 5000 to 2760 s1ncec 0.6 >3vf~1

Slab: 4600 ps1 -190 ps1f • :: 4000c

Tens ton (-)

The s11ght 1ncrease 1n requ1red transfer strength to sat1sfy center

span bottom compress1on 1s not uncommon for th1s type of beam sect1on.

Alternat1vely, a deeper sect10n could be used. W1th m1dspan bottom

tens10n and flexural strength requ1rements both w1th1n l1m1t1ng values,

spec1fy f~ = 4600 ps1. Note: H1gher strengths need to be conf1rmed bylocal prestressed concrete supp11er.

2-11

2.10 Nonprestressed Reinforcement

Since the top factor stress in tension at depress points exceeds 3yf~i'

at prestress transfer, bonded reinforcement must be provided to resist

the total tensile force. Referring to sketch:

Total tensile force: (301;46)2X72 + ~6X2XO.36X8 : 25.1k

Use Grade 60 steel @24,000 psi: -3012As : 25.1/24 = 1.05 in.

2.36] 2111 .~

C6Use 4 #5 bars (As : 1.24 in. 2)

Btm.of flange 2411

"+2760

Extend bars from center of span to span location where tensile stress

equals 3yf~i = 3y4600 = 203 psi. If x = distance from support:

f t: f + fOb; where MOb = ~X(l _ x)

P

-203 -107 - 1~(1416 - 852) (436x)l2 (40 - x)= + 2x2711

solving, x = 10.2 ft

Extend bars 12 ft (including development length) each side of span

center.

Use 4 - #5 X 24 1-0 centered in span

2-12

2.11 Shear Strength

The method of des1gn for shear reinforcement presented 1n the 1979

Interim AASH10 Standard Spec1ficat1ons w1ll be used as an acceptable

alternative to the prov1sions of the 13th Edition, 1983 Specificat1ons.

Check shear at quarter span:

Vo = wl/4 = 0.989 x 40/4 ~ 9.9k/beam

14 1

401-0

14 1~HS20 Truck Load

Lane Loading

30 16 2 k(40 + 40)32 + (40)8 = 37.2 /lane

kVL ~ 37.2 x 0.5 x 1.0 x 1.3 ~ 24.2 /beam

Note: The HS20 truck loading is applied to the full lane long1tudinally

to obtain maximum lane shear at the span quarter point. The lane shear

is then distributed to an individual beam, with appropriate live load

impact .

Vu = 1.3 (V D + 1.67 VL) = 1.3 (9.9 + 1.67 x 24.2) ~ 65.4 k/beam

Using average web width with two webs per tee:

0.06f~bljd = 0.06 x 5 (2 x 6.25) 0.96

but not greater than 180 b'jd ~ 0.180k

= 50.1

2-13

kx 23.2 ,; 83.5

(2 x 6.25) 0.96 x 23.2

Cons1der1ng strand contr1but1on to shear strength:

Vertical component of effective prestress force (e.g. of strand rises

4.9 1n./10 ft.),

Vp 249 (4.9/10 x 12) ~ 0.3k

kV + V ~ 50.1 + 0.3 ~ 50.4c P

(V u - ~Vc)s (05.4 - 0.9 x 50.4}12Av 2~f jd = 2 x 0.9 x 00 x 0.96 x 23.2sy

where ~ = 0.9 for shear (AASHTO 9.14)

2= 0.07 in. 1ft

100b'sf sy

100(2 x 0.25)12= 60.000

Also must prov1de 2-#3@12 in. = 0.22 1n. 2/ f t m1n1mum vert1cal ties

for shear transfer between beam and cast-1n-p1ace deck slab

(AASHTO 9.20.4.4).

Use single-leg stirrup of welded wire fabric Wo.S @ 0 in.2spac1ng per stem. (A

y= 2xO.13 = 0.20 1n. 1ft> 0.25)

Multiple layers of smaller size fabric of equ1valent area could also be

used.

Add additional stirrups at beam ends (AASHTO 9.21.3):

4% p ~ = 0.04(312) ~ 12.48k• A = 12.48/20 = 0.624 1n. 2SlY

Use 1 - #4 U-st1rrup per leg at each end of beam.

For shear transfer between beam and cast-in-place deck slab (AASHTO

9.20.4.2). all st1rrup legs must be extended 1nto deck slab, and top

surface of precast beam must be intentionally roughened. Scor1ng the

2-14

surface with a stiff bristled broom is common practice to satisfy the

1l1ntent1onally roughened II requirement.


For estimating long-time deflections and camber, use data from PCIDesign Handbook. See Design Example No.1, page 1-12.


Ec1 = 33Wl.5vf~i = 33(150)1.5v4600 = 4.1 x 106 psi

I I

a aee r I

. ... ....-------e l

Two point depressed

Psi el2 e 1a2fI (8 - -6-)

312 [12.3(40X12)24.1x103x22230 8

(b) Beam dead load

5w!4 5xO.436(40X12)4384EI = 384X12X4.1xl03x22,230


Using suggested multipliers ...

At transfer

:= 1.10 11 1'

= 0.28 1t+= 0.82 111'

1.80x1.10 - 1.85xO.28

2-15

At erection = 1.46 11 1'

= 0.25",1.

-;: 1.11"1'384x12x4.3x103X46200

After construct1on

(Ra t l inq )


f~ = 5000 psi; Ec = 4.3 x 10° psi

4(Deck) 5xO.413(40x12)

384x12x4.3xl03x22230

5xO.140(40x12)4


Us1ng appropriate mult1pliers ....

Beam 0.28"~ x 2.40

Camber 1.10"1' x 2.20

Deck 0.25",1. x 2.30

Ra111ng 0.04"~ x 3.00

Net long term

0.61"+

2.42"1'-;: 0.58 11 +-;: Q.:.l2"+

1.05"1'

(f) Live load deflect10n

Est1mate max1mum l1ve load def1ect1on at span center w1th heavy

truck axles closely spaced and centered 1n span.

i131

40'-0

D1str1but1on factor from lane to

beam load1ng:

(D1st.)(Lane)(lmpact)

(1.00)(1/2)(1.3) = 0.650

32 k[ __--L- -.--__- I -Lane Shear Diagram

'------ 32k

/ +-Lane Moment Diagram

2-16

Using moment-area method:

(416X132

+ 416x7x16.5) 1728xO.650 = O.3P or Span3 4.7xl03x46,200 1300

OK

2.13 Design Summary

Use 24 in. depth x 72 in. width double-stemmed precast beam section with

12 - 1/2" diameter 270K stress-relieved strand (6 strands per stem).

Double depress strands at 4 ft each side of span centerline. Centroid

of strand pattern to be at 3.5 in. above bottom of leg at depress points

and 8.4 in. above bottom of leg at beam ends. Specified concrete

strength to be f~ = 5000 psi with a release strength of f~i = 4600 psi.Use single-leg stirrups of welded wire fabric W6.5 @£> in. spacing per

stem. Use same stirrup detail full span length.

Note: This design results in a heavily prestressed section. If geom

etry restrictions permit, a deeper section with fewer strands might be

considered. Also. some governing agencies may require a £> in. minimum

C.I.P. deck slab.

2 #5 X 24 1 - 0 @ span center

c.g.~ 8.4"

_ 6-1/2" 270K Strands

6'-0

36"

Welded wire fabricW6.5 @611

'I. ,.

r 5-1/2" C.I.P. Deck 1I - - - - ~ - - - - - - - - - - _""'":::1_ - - - -1I I

~.-- I II

11

'

C • g.i:,_--, __ 3.5" L

-'Ot'N

Strands atspan center

Strands atspan end

REINfORCEMENT DETAILS

2-17

DESIGN EXAMPLE NO.3SINGLE SPAN I-GIRDER BRIDGE


Simple span of 75 ft x 30 ft width


Use 4 PCI Standard I-Girders at 8 ft spacing. Consider composite con

struction with 7-1/2 in. deck slab. Illustrate design for a typical

interior girder only. Design of slab not to be considered in this

example. Allow for 2 in. future wearing surface.

3.2 Materials


f~ 5000 psi

f~i = 4000 psi (AASHTO 9.22)


f~ = 4000 psi

Prestressing steel:

Reinforcing bars:

1/2 in. diameter 270 ksi stress-relieved strand

Strand area = 0.153 sq in.

Es = 28 x 106 psi

f = 60,000 psiy

3-1

3.3 Preliminary Girder Selection

Using span table in "Short Span Br1dges Manual" as a gu1de. both Type

III and Type IV I-Girders are poss1b1l1t1es at a 75 ft span. Cons1der

1ng the wide girder spacing and provision for future wearing surface.

select a Type IV section. Br1dge layout and girder section properties

are as follows.

301-0

1 '-1" 26'-10"

7-1/2 11 (25 psf future wearing surface)

TYPE IVI-GIRDER

3 @8'-0 31-0

PrecastParapet(350 pH)

SECTION

ELEVATION

l ,I

3-2

.....,.Ln •

c.g=

~I~Mr-...,.N

coI-

2611 .

Sect10n Propert1es

Ac == 789 1n. 2

I == 260,730 1n. 4

Yb == 24.73 t n .3Sb ~ 10,543 1n.

St == 8908 1n. 3

wD == 822 p l f

PCI STANDARD TYPE IV I-GIRDER

3.4 Compos1te Sect10n Propert1es

Cast-1n-place Ec == 57.000vf~ == 57,OOOv4000 == 3.o0xlOo ps1

Precast Ec == 57.000v5000 == 4.03X106 ps1

n == 3.60/4.03 == 0.89

[ffect1ve flange w1dth (AASHTO 9.8.1.1):

1/4 span == (75x12)/4 == 225 1n.

D1stance center-to-center of g1rders == 8 ft == 96 1n. (governs)

12 x slab th1ckness plus flange width == 12x7.5 + 20 == 110 1n.

3-3

C.G. of Compos1te Seet10n

Area Y Ay

7.5x85.4 ~ 640 57.75 36,960

789 24.73 19,512

1429 56,472

Yeb ~ 56,472/1429 '" 39.52 1n.

Slab

...,.N

",

_r 0.89(96) '" 85.4" 1~ ~ I Ir-- A~ G1 rder

e g \) e.gB~.~'l. ~composlte

_ A~ N(\') It'lr-

Seet10n Modulus of Compos1te Seet10n

I A (Y-Ycb) A(y-yeb) 2 Ie ~ 263,732 + 385,282

Slab 3,002 640 18.23 212,693 '" 649,000 1n. 4

G1rder 260,730 789 14.79 172,589 Scb649,000 16,420 1n. 3

'" 39.52~

263,732 385,282

Set649,000 44,820 . 3

'" (54-39.52) '" , n.

3.5 Des1gn Loads and Moments

(a) Dead load (1nter1or g1rder)

Non-Compos1te:

Girder '" 822 pH

Slab 150 x 0.625 x 8.0 '" 750 pl f

1572 pH

G1rder "0 ~ 0.822 x 752/ 8 '" 578'1<

G1rder + slab "0~ 1.572 x 752/8

'"1105 I k

3-4

Composlte: Distribute to 4

Future wearing surface 25

Parapet

girders (AASHTO 3.23.2.3.1.1)

x 26.83/4 = 168 plf

2 x 350/4 = 175 plf

343 p1f

2 'kMO = 0.343 x 75 /8 = 241 /girder

Note: The exterior girder may be critical with large slab

overhang and should be checked in a separate design.

(b) Live load (interior girder)

Use live load distribution for Prestressed Concrete Girders

(AASHTO Table 3.23.1):

lanes/girder = 0.5(S/5.5) = 0.5(8.0/5.5) = 0.727

Live load impact (AASHTO 3.8):

50 50I = l + 125 = 75 + 125 = 0.25

For live load, use moment tables (AASHTO App. A):'kHS20 - 75 ft span - lane moment = 1075.1

lkML = 1075.1 x 0.727 x 1.25 = 977 /girder


Determ\ne number of strands required based on stress conditions at

service load.

3-5

Des1gn load stresses at m1dspan:

G1rder fb578x12 -0.658 ks1 f t

578x12 0.778 ks110,543~ ;;

8908~

G1rder + slab f b11 05x12 -1.258 ks1 f t

11 05x12 1.488 ks110,543;; ::

8908 =

Composite f b241x12 -0.176 ks1 f t

241x12 0.065 ks1= 16,420 ~ ::44,820

L1ve load f b977x12 -0.714 ks1 f t

977x12 0.262 I<s1::16,420

;; ::44,820

~

Total stress f b :: -2.148 ks1 f t ~ 1 .815 ks1

(Tension) (Compression)

Allowable stresses (AASHTO 9.15.2):

F1na 1 f b :: 6v'f l;; 6v'5000 :: 424 psi (Tension)c

f t = 0.4f' = 0.4x5000 = 2000 psi (Compress1on)cIn1t1al f b

:: 0.6f~1 :: 0.6x4000 = 2400 psi (Compress1on)

f t 7.5v'f~1 == 7.5v'4000 = 474 pS1* (Tens1on)

*When tens1le stress exceeds 3v'f~1 = 190 ps1, bonded re1nforce

ment must be prov1ded to res1st the total tens1le force (AASHTO

9.18.2.1).

L1m1ts of Prestress:

Bottom f1ber Top F1berF1nal - Total stress 2.148 ks 1 T 1.B15ks1C

Allowable stress .424 ks1 T 2.00 ks1 CReq1d prestress after losses 1.724I<s1 C 0

In1t1al - Girder stress .658 ks1 T .778 ks1 CAllowable stress 2.400 ksi C .190 ks1 T

Allowable initial prestress 3.058 ks1 C .968 ksi T

at midspan

3-6

Final bottom fiber stress controls. Required prestress stress in

bottom fiber (after losses) = 1.724 ksi. Bottom fiber stress due to

prestress:

where Psee

effective prestress force after losses

strand eccentricity. With e.G. of strands approximately

4 in. from bottom of beam, e = 24.73 - 4 = 20.73 in.

Pse 20.73 Pse1.724 = 789 + 10,543

solving, required P = 533Kse

Final prestress per strand, assuming 20% losses:

(O.153xO.70x270)0.80 = 23.1 K

Number of strands required = 533/23.1 = 23.1

Try 24-1/2" 270K strands

strand pattern:

Practices on strand pattern and spacing vary with prestressed concrete

supplier. For this example, use 2 rows of 12 strands each. spaced at 2

in. Locate the rows 3 in. and 5 in. respectively from bottom of girder.

e = 24.73 - 4 = 24.7 in. A straight strand pattern will be used in this

example; however, in some geographic areas, a deflected strand pattern

with the longer spans may be considered more appropriate.

11/ r \• • • • • • •••• • • ~ 12 @ 2"••••••• 2" 12 @ 2"• • • • • =r3 U

3-7


Uslng Group I loadlng comb1natlon (AASHTO 3.22); strength requlred:

MU

;; 1.3 (MO

+ 1.6 7ML)

;; 1.3(1105 + 241 + 1.67x977) ;; 38831k

Use approxlmate value for stress ln prestressed re1nforcement(AASHTO 9. 17 .4) :

f* ;; f ' (l - 0.5 p* fl/fl)su S S C

;; 270(1 - 0.5xO.000665x270/4) = 264 ksl

where p* A;/bd;; 24xO.153/96x57.5 = 0.000665

b ;; flange wldth = 96 In.

d = 54 + 7.5 - 4 ; 57.5 1n.

For rectangular sectlons (AASHTO 9.17.2); strength provlded:

~M ;; ~A* f* d(l - 0.6 p*f* If I)u s su su c

;; 1.Ox24xO.153x264x57.5(1 - O.6xO.000665x264/4)/12

;; 4520 ' k > 3883 OK

Check "a" as a rectangular sectlon:

a = A*f* 10.85f 'bs su c

24xO.153x264/0.85x4x96 ;; 2.96 In. « 7.5 OK

Note: For factory produced precast prestressed members, strength

reduction factor ~ = 1.0 (AASHTO 9.14).

For shear deslgn j = (1 - 0.6xO.000665x264/4) 0.97

3-8

3.8 Max1mum and Mlnlmum Steel Percentage

(a) Maxlmum steel for rectangular sectlons (AASHTO 9.18.1):

Reinforcement 1ndex p*f* ff' ~ 0.3su c'" 0.000665x264f4 0.044 < 0.3 OK


Total amount of prestressed reinforcement must be adequate to

develop a des1gn moment strength at least equal to 1.2 times the

cracking moment strength (~Mu > 1.2 Mer); where, for a prestressedcomposite member (See Deslgn Example No.2, page 2-6):

P P -e SM ( f + ~ + ~) S - M (S - 1)

cr cr Ac Sb c 0 Sb

(0.530 555 + 555X20.7)16,420 _ 1105(16,420 _ 1)+ 789 10,543 12 10,543

2563'k

where = 7.5~5000 '" 530 ps1= (24xO.153xO.7x270)O.80 = 555k (assumlng 20% losses)

'k= 1105 Use glrder + slab moment only; rema1n\ng

port10n (wearlng surface t parapet) will

act on composlte sectlon.

~M > 1.2 Mu - cr4520 > 1.2x2563 '" 3076 OK

3.9 Prestress Losses

Estimatlon of prestress losses w111

cedure presented 1n AASHTO 9.16.2.

(cr1tlcal moment location).

6f s = SH + ES + CRc + CRs

3-9

be based on the approxlmate pro

Compute loss values at span center

(a) Shrinkage

SH = 17.000 - 150 RHAssume east coast location. RH = 10%

SH = 17.000 - 150x70 = 6500 psi

(b) Elastic shortening

ES

fci r = concrete stress at level of prestressing steel immediately

after transfer

Assume 10% losses due to elastic shortening and strand relaxation

at release:

Psi = 0.9(24xO.153xO.7x270) = 625k

_ 625 + 625(20.7)2f c1r - 789 260,730

578x12x20.7260.730 = 1.269 k s t

Ec1 = 57.000vf~i = 57.000v4000 = 3.6 x 106 psi

ES 28= 3.6 x 1269 = 9870 psi


3-10

fcds concrete stress at level of prestressing steel due to

superimposed dead load ... deck slab plus parapet.

Note: the parapet is acting on the composite section.

{1105 - 578)20.7x12260,730

+ 241(39.52 - 4)12649,000 = 0.660 ksi

CRC

12x1269 - 7x660 = 10,600 psi


CR s = 20,000

= 20,000

0.4ES - 0.2 (SH + CR )cO.4x9870 - 0.2(6500 + 10,600) 12,630 psi


Af s = 6500 + 9870 + 10,600 + 12,630 = 39,600 psi

or 39.6/0.7x270 = 20.9% losses

f = effective prestress = 0.7x270-39.6 = 149.4 ksise

3.9 Concrete Stresses

Compression (t)

Tension (-)

(a) Initial stresses at Prestress Transfer

Psi = 0.9(24xO.153xO.7x270) = 62S k

At span center (prestress + girder):

625 625x20.7 578x12f t = 789 --a9OS-- + 8908 = +0.118 ksi < 0.6f~i OK

3-11

625 625x20.7f b = 789 + 10,543

578x11.10.543 +1.361 ks1 < 0.6f~, OK

At span ends (prestress only):

625 625x20.7f t = 789 - 8908 = -0.660 ksi > 7.5-1f~i

Since top fiber stress in tension at span ends exceeds 7.5-1f 1

c1(AASHTO 9.15.2.1), must debond some of the strands. or alterna

tively, use a draped strand pattern to reduce the stress level.

With the longer spans. a draped strand pattern may be considered

more appropriate; however, to illustrate design procedure a

straight strand pattern with debonded strands will be used in thisexample.

Estimate number of debonded strands:

Debonded strandsTotal strands

Oebonded strands ~

(f t - 7.5vf~i)

f t

(660 - 474)24 _ 6 8660 -.

Try 8 debonded strands. Shield 4 strands 4 ft from end in top

row and 4 strands 6 ft from end in bottom row. Varied shield

lengths are recommended to avoid stress concentrations.

Revised stress at span ends:

Psi = 625(16/24) = 417 k

f t417 417x20.7

= -0.440 ksi < 7.5-1f~i= 789 - 8908

f b417 417x20.7 +1.347 ksi < 0.6f~i= 789 + 10.543 =

OK

OK

3-12

With debonding, the top fiber stress is now less than the maximum

permitted; however, some bonded reinforcement must also be provided

since the tensile stress still exceeds 3vf~i::: 190 psi. According

to AASHTO 9.15.2.1, the bonded reinforcement must be designed to

resist the total tensile force. Referring to the sketch:

=....II')

Total Tensile force ~ 0.440x13.3x20/2k::: 58.5

Use Grade 60 steel @24,000 psi:

A '" 58.5/24 2::: 2.44 in.s

Use 6 #6 bars x 12'-0 at each

end of girder 2(As::: 2.64 in. )

1.347

At 8 ft from span ends:

(Assume strand transfer length ::: 50db ~ 24 in.)

MDb::: ~X(l_X) ::: O.8~2X8(75_8) 220'k'"

f t625 625x20.7 + 220x12 -0.363 ksi < 7.5vf~i OK:::789 8908 8908

:::

f b625 725x20.7 220x12 ::: +1.770 ksi < 0.6f~i OK'" 7B9 + 10,543 10,543

(b) Service Load Stresses at Span Center

p ::: A*f ::: (24xO.153)149.4 ::: 549kse s se

f t549 549x20.7 11 05x12 + ( 241 + 977)12 +1.234 ksi < a.4f ' OK:::789 B90B + 8908 44,820

;;

c

f b549 549x20.7 1105x12 (241 + 977)12 -0.374 ks t < 6vf ' OK::: 789 + 1a, 543 10,543 16,422 :::

c

3-13

3.10 Shear Strength

The method of design for shear reinforcement presented in the 1979

Interim AASHTO Standard Specifications will be used as an acceptable

alternative to the provisions of the 13th Edition, 1983 Specifications.

Check shear at span quarter point:

Vd = Wl/4 = (1.572 + 0.343)75/4 = 35.9 K/girder

1./4 ::: 18.75 1 141 141

~P. .t!1U5Vl

(56.25 42.25)32 (28.25)8 45.0 kllane= 75 + 75 + 75 =

VL = 45xO.727x1.25 = 40.9 k/girder

Note: The HS20 truck loading is applied to the full lane longitudin

ally to obtain maximum lane shear at the span quarter point. The lane

shear is then distributed to an individual girder by the live load

distribution (lanes/girder), including live load impact.

Vu 1.3 (VD

+ 1.67Vl

) = 1.3(35.9 + 1.67x40.9) ::: 135k

Vc ::: O.06ftb'jd = O.06x5x8xO.97x57.5 ::: 134K

but not greater than 180b'jd ::: 0.180x8xO.97x57.5 80k

where d 54 + 7.5 - 4.0 = 57.5 in.

j ::: 0.97 (See step 3.7)

3-14

(Vu - ~Vc)s (135 - 0.9x80)12 2A = 2 f jd = 2xO.9x60xO.97x57.5 = 0.126 in. 1ftv ~ sy

but not less than 100b's/f = 100x8x12/60,OOO = 0.16 in. 2 / f tsy

Also must provide 2-#3 @12 in. = 0.22 in. 2/ f t minimum vertical ties

for shear transfer between girder and cast-in-place deck slab

(AASHTO 9.20.4.4).

Check stirrup spacing limitations:

AASHTO 9.20.3.2 - (3/4)54 = 40.5 in.but not greater than 24 in.

AASHTO 9.20.4.4 - 4(7.5) = 30.0 in.

Use #3 U-Stirrups @12 in. full span length

Add additional stirrups at girder ends (AASHTO 9.21.3):

Conservatively, use full Psi

k 24% Psi = 0.04(625) = 25 , Av = 25/20 = 1.25 in.

Use 4 - #4 U-Stirrups @3 in. at each end of girder

For shear transfer between girder and cast-in-place deck slab (AASHTO

9.20.4.2), all stirrup legs must be extended into deck slab, and top

surface of girder must be intentionally roughened. Scoring the surface

with a stiff bristled broom is common practice to satisfy the "intentionally roughened II requirement.


For estimating long-time deflections and camber, use data from PClDesign Handbook. See Design Example No.1, page 1-12.

3-15


For straight strands (See Design Example No.1):

2625X20.7(1SX12)2Psiel

1.40 111'- 3 '"8E! 8x3.6xl0 x260.730

(b) Girder dead load

SWl4 sxO .822(7Sx12) 4 '" 0.62 11 +384EI '" 384x12x3600x260.730

At transfer '" 0.18 11 l'

(e) Growth in storage

1.80xl .40 - 1.85xO.62


At erection '" 1.55" l'

5(750)( 75x12)4(Deck) - -- - '" 0.51 11 +384x12x4.0x106x260.730

5(168 + 175)(75x12)4(Railing and fWS) - -- - 0.09 11 +384x12x4.0x106x649.000

After construction = 0.95" l'


Beam

Camber

Slab

Ra 111ng & FWS

0.62 11 + x 2.40

1.40 11 1' x 2.20

0.51"+ x 2.30

0.09 11+ x 3.00

3-16

'" 1 .49 11 += 3.08 11 1'

= 1.1 P+

'" 0.2P+Net long term '" 0.15 111'

(f) l1ve load deflect10n

Estimate maximum live load deflection at span center with heavy

truck axles closely spaced and centered in span.

.

l.Sk

30.2k

I

t33.8k

30.5' "'-HS20 Truck

(D1stribution factor)(impact)

(0.727(1.25) = 0.91

.-lane Shear D1agram

1031 ' k 10561k

~'-------~ ~lane Moment Diagram

Using moment-area method:

(1031X30.52

+ 1043x7x34) 1728xO.913 4.3X103X649.000

or Span < Span* OK2820 1000

*Since AASHTO does not prov1de gu1dance on acceptable live load

deflect10ns for prestressed concrete bridges. check criteria for

steel girders (AASHTO 10.6). Live load deflection criteria for

steel girders is primarily based on an empirical limitation for

vibration (comfort and rideability).

3-17

3.12 Design Summary

Use pel Standard Type IV I-girder with 24-1/2" d1ameter 210K

stress-re11eved strands. Use stra1ght strand pattern w1th 8 strands

sh1elded at girder ends.

Use '3 U-st1rrups @12 in. full span length w1th 4-'4 U-st1rrups @

3 in. at g1rder ends.

ri==-=m=rr~-- 6 #6 x 121_0

'3 U-Stirrups @12"1..----

4 #4 U-Stirrups @ 3" atg1 rder ends

12 @ 211

11. @ 2"

24-1/2" 210K Strands

Debond 8 Strands at each end of g1rder

4 for 4'-0 in btm. row

4 for 6'-0 in top row

Note: St1rrup deta11s vary with prestressed concrete suppl1ers.

Detail shown here for 1llustration only.

REINFORCEMENT DETAILS

3-18

Re1nforclng bars:

DESIGN EXAMPLE NO.4MULTI-SPAN I-GIRDER BRIDGE

4.1 Oes1gn cond tt tons

Continuous span of 65 ft - 80 ft - 65 ft x 30 ft width


Use 4 PCI Standard precast I-g1rders at 8 ft spac1ng. w1th girders made

cont1nuous for live load and super1mposed dead load (AASHTO 9.7.2).

Cons1der composlte constructlon with 7-1/2 in. deck slab. Illustratedesign of a typlcal lnterior girder only for the center 80 ft span.

Design of slab not to be considered in th1s example. Allow for 2 in.future wearing surface.

4.2 Materials


f~ = 5000 ps1

f~i = 4000 psi (AASHTO 9.22)


f~ = 4000 ps t

Prestress1ng steel: 1/2 1n. diameter 270 ks1 stress-re11eved strand.

Strand area = 0.153 sq. in.Es 28 x 106 psi

fy

=: 60.000 psi

4-1

4.3 Prel'm1nary G1rder Select'on

Us\ng span table in "Short Span Bridges Manual" as a guide, select a

Type IV I-girder section based on the BO ft center span. Bridge layoutand g1rder section propert'es are as follows.

301-0

3 @8 1-0 = 241-0

PrecastParapet(350 pl f )

TYPE IVI-GIRDER

7-1/2" (25 psf future wearing surface)

3'-0

1 '_7 11

SECTION

1

4 651-0

.14 801-0

.14651-0

·1

~ r=9= JII I

ELEVATION r

4-2

2011

r 1\ )~ Section Properties

A = 789 in. 2c

4811 1 = 260,730 in.--.

24.73 in.= Yb =.- • 3U'"J

c.g Sb = 10,543 in.= 8908 in. 3

~,~M St =r-

.- wo = 822 pHN

=CXlII' II' 1

- 26 11

PCI Standard Type IV I-Girder

4.4 Composite Section Properties

Cast-in-place Ec 57.000~f~ = 57,OOO~4000 = 3.60x100 psi

Precast E = 57.000~5000 = 4.03xl00 psic

n = 3.60/4.03 = 0.89

Effective flange width (AASHTO 9.8.1.1):

1/4 span = (80x12)/4 = 240 in.

Distance center-to-center of girders = 8 ft = 9o in. (governs)

12 x slab thickness plus flange width = 12x7.5 + 20 = 110 in.

4-3

Ayy

57.75 3&,9&0

24.73 19,512

5&,472

Area

Ycb = 56,412/1429 = 39.52 1n.

7.5x85.4 = &40

789

1429

C.G. of Compos1te Sect10n

Slab

G1rderI

) ~c.g" Compos1te

4i1'

= 85.4" ~r 0.89(9&)

Ir- ~ T

e.g. "Beam\

='iI'

-In -r-~--------;.

Sect10n Modulus of Compos1te Sect10n

I A (Y-Ycb)2

Ie 263,732 + 385,282A(Y-Ycb) =4Slab 3,002 &40 18.23 212,&93 = &49.000 1n.

G1rder 2&0,730 789 14.79 172.589 Scb &49,000 3= = 1&,4201n.

263,732 385,282 39.52

\t 649,000 3= = 44.820 1n.

(54-39.52)

S(slab)t 649,000 3= = 29,100 1n.

(&1.5-39.52)

4.5 Des1gn Loads and Moments

(a) Dead load (1nter1or g1rder)

Non-Compos1te:

G1rder

Slab 150 x 0.625 x 8.0

= 822 pl f

= 150 p l f

1572 pH

G1rder

G1rder + slab

MOb = 0.822x802/8= 658

l k/g1rder

2 I kMo = 1 .572x80 /8 = 1258 /g'rder

4-4

Composite: 0istributeFuture wearing surfaceParapet

to 4 girders

25x26.83/4

(2x350)/4

(AASHTO 3.23.2.3.1.1);:: If>8plf

;:: 175 plf

343 PIf

The composite girders will be made continuous when the deck is

cast so that the future wearing surface, parapet, and live load

will act on a three span continuous structure. Using Reference4-1 "Moments, Shears and Reactions for Continuous Highway

Bridges;" by interpolation between n ;:: 1.2 and 1.3 for 3

cont1nuous spans w1th a span ratio n = 80/65 = 1.23:

moment coeff. at center of centermoment coeff. at interior support

shear coeff. at interior support

span;:: +0.0635

= -0.1259= 0.615

(+)HO = 0.0635 x 0.343 x 65 2 = 92'k/glrder ~0.1259 x 0.343 x 65 2 'k(-)MO

:: :: l8~ Ig1rder •Vo :: 0.615 x 0.343 x 65 ;:: 13.7 Ig1rder

(b) Live load (interior girder)

For HS20 10ad1ng on a three span cont1nuous structure w1th N ::

1.23; by 1nterpolat1on (Ref. 4-1):

moment at center of center span

moment at 1nter1or supportshear at 1nter1or support

·See CLARIFICATIONS, Item 4, page iv.

4-5

:: 725.B ' k/lane'k:: 620.4 Ilane

:: 64.9k/lane

Use live load distribution for Prestressed Concrete Griders

(AASHTO Table 3.23.1):

lanes/girder; 0.5(S/5.5) :: 0.5(8.0/5.5) :: 0.727

Live load impact (AASHTO

positive moment I

negative moment I

3.8) :

; 50/(125

50/(125

+ 80) :: 0.244

+ 72.5) :: 0.254

lk:: 725.8xO.727xl .244 :: 656 /girder

tk620.4xO. 727xl.254 :: 566 Igirder

:: 64.9xO.727xl .244 :: 58.7 k/ghder


Determine number of strands required based on stress conditions at

service load. Assume concrete tension in bottom fiber governs. Design

load stress at midspan:

f _ 1258x12 92x12 656x12b - 10,543 + 16,420 + 16,420 :: -1.978 ks i

Allowable tensile stress (AASHTO 9.15.2.2) :: 6~f~ :: 0~5000 :: 424 psi

Required prestress stress in bottom fiber:: 1.978 - 0.424 :: 1.554 ksi

Bottom fiber stress due to prestress:

P P ese ~

f b :: Ac

+ Sb

Pse Pse x 20.731.554 :: 789 + 10,543

where e :: strand eccentricity. With e.G. of strands approx. 4 in. from

bottom of girder, e :: 24.73 - 4 :: 20.73 in.

ksolving, required Pse :: 480.0

4-0

Final prestress per strand, assuming 20% losses:

(0.153xO.70x210)0.80 ~ 23.l k

Number of strands required = 480.6/23.1 = 20.8

Try 22-112" 270K strands

(Use even number for symmetry)

strand pattern:

Practices in strand pattern and spacing vary with prestressed concrete

supplier. For this example, use bottom row of 12 strands and top row of

10 strands, with 2 in. spacing between strands. Locate the rows 3 in.

and 5 in., respectively, from bottom of girder. e.G. of strand pattern

; (12x3 + 10x5)/22 = 3.9 in. from bottom. e = 24.73 - 3.9 = 20.83 in.

•••••••••

/ -f

• • • • • • • • • • 10 Strands @ 2"12 Strands @ 211


Using Group I loading combination (AASHTO 3.22); strength required:

Mu 1.3(MO + 1.67ML

)

1.3(1258 + 92 + 1.67x656)

Use approximate value for stress in prestressed reinforcement

(AASHTO 9. 17 .4) :

f* = f ' (1 0.5 p* f'/f')su sse

= 270(1 - 0.5 x 0.000609 x 270/4) = 264 ksi

4-7

where p* = A~/bd = 22xO.153/96x57.6 = 0.000609b = flange w1dth = 96 1n.d = 54 + 7.5 - 3.9 = 57.6 1n.f' = 4000 ps1 for deck slabc

For rectangular sect10ns (AASHTO 9.17.2); strength prov1ded:

~Mu = ~A~f;ud(l - 0.6P*f;u/f~)

= 1.Ox22xO.153x264x57.6(1 - O.6xO.000609x264/4)/12

= 41621k

> 3179 OK

Check "a" as a rectangular sect1on:

a = A;f;u/0.85f~b = 22xO.153x264/0.85x4x96= 2.72 1n. «7.5 OK

Note: For factory produced precast prestressed members. strengthreduct10n factor ~ = 1.0 (AASHTO 9.14).

For shear des1gn. j = (1 - 0.6xO.000609x264/4) = 0.98

4.8 Max1mum and M1n1mum Steel Percentage

(a) Max1mum steel for rectangular sect10ns (AASHTO 9.18.1):

Re1nforcement 1ndex = P*f * If I < 0.3su c-

= 0.000609 x 264/4 = 0.04 < 0.3 OK


Total amount of prestressed re1nforcement must be adequate todevelop a des1gn moment strength at least equal to 1.2 t1mes the

crack1ng moment strength (~Mu ~ 1.2 Mer); where. for a prestressed compos1te member (See Des1gn Example No.2. page 2-6):

4-8

McrP P e Sc

(f + ~ + ~)S - M (-- - 1)cr Ac Sb c 0 Sb

(0.530 + 509 + 509x20.83)16420 _ 1258(16.420 _ 1)789 10,543 12 10t543

= 7.5~5000 = 530 ps1

(22xO.153xO.7x270)O.80 = 509k (assum1ng 20% losses)'k

= 1258 Use g1rder + slab moment only; rema1n1ng

port1on (wear1ng surface & parapet) w1l1

act on compos1te sect1on.

~M > 1.2 Mu - cr4162 > 1.2x2283 = 2740 OK

Note: strength rat10 suff1c1ently h1gh; reevaluat10n w1th a more

exact est1mate of prestress losses w11l not be necessary.


Est1mat1on of loss of prestress w1ll be based on the approx1mate

procedure presented 1n AASHTO 9.16.2. Compute loss values at span

center (cr1t1cal moment 10cat1on).

(a) Shr1nkage

SH = 17.000 - 150 RH

Assume m1dwest 10cat10n, RH = 70%

SH = 17.000 - 150 x 70 = 6500 ps1

4-9

(b) Elast~c shorten~ng

ESEs

=-fEc~ c~r

f ci r = concrete stress

after transfer.

Psi P e2si .

= -- +I

bA

at level of prestressing steel immed'ately

Assume 10% losses due to elast'c shorten'ng and strand relaxationat release:

Psi = 0.9(22xO.153xO.7x270) = 573k

f ci r573 573(20.83)2 65Bx12x20.B3 1.049 ks i= 789 + 260,730 260,730 =

ES 28 1.049 8160 psi= 3.6 x =


f cds = concrete stress at level of prestress'ng steel due to

superimposed dead load ... deck slab plus parapet.

Note: the parapet is acting on the composite sect~on.

(1258 - 658)20.83x12 92(39.52 - 4)12= 260,730 + 649,000 = 0.636 ksi

CRe = 12 x 1049 - 7 x 636 = 8136 psi


CRS

= 20,000 0.4 ES - 0.2(SH + CR e)= 20,000 - 0.4 x 8160 - 0.2(6500 + 8136) = 13,809 psi

4-10


~fs ; 6500 + 8160 + 8136 + 13,809 : 36,605 psi

or 36.6/0.7 x 270 19.4% losses

f se : 0.7 x 270 - 36.6 152.4 ksi

4.10 Concrete stresses

Prestress~ng:

Psi 0.9(0.7 x 270)22 x 0.153 = 573 k

Pse : 152.4 x 22 x 0.153 = 513k

Load s :

MO(girder)

Mo(girder & slab)

Mo(parapet)ML

65S'k

1258'k92' k

: 656'k

Sect~on properties2

A (in. )c 3

Sb (~n. )3St (in. )

Beam

789

10,543

8908

ComposHe

1429

16,420

44,820

Top of girder

ComposHe

29,100

Concrete stresses at prestress transfer and at service load (in ps~)

are summarized on Page 4-12.

Since the top f~ber stress in tension at girder ends due to prestress

at transfer exceeds 7.5vf~, three alternative solutions, or a combina

tion, are possible:

(a) Use draped strands to reduce stress level

(b) Use debonded strands to reduce stress level

(c) Provide bonded reinforcement to resist total tensile force

For this example, use a combination of debonded strands plus bonded

reinforcement. See Design Example No.3, page 3-12 for design

procedure.

4-11

Support at Midspan at Midspan atPrestress Transfer Prestress Transfer Service Load

P = Psi P = Psi P = PseLoad

f b f t f b f t f b ft(girder) ft(slab)

P/Ac 725 725 725 725 650 650 --

PelS 1131 -1338 1131 -1338 1013 -1200 --MOb/S - - -- -748 885 -748 885 --MOs/S -- -- -- - -683 808 --

MOplS -- -- -- - -67 25 38

ML/S -- -- -- - -480 176 271

Total 1856 -613 1053 338 -311 1344 309stress

Allowable 0.6f~i 7.5v'f~i O.6f~i OK 6v'f ' o 4f l o 4f lstress c . c . c

(AASHTO 2400 -474 -424 2000 16009.15.2)Beam: OK Must OK OK OK OKfl = 5000 debondc or drapef ~ i '" 4000 strands

Slab:f~ = 4000

Tension (-)

Debond 12 strands at each end of girder:

4 strands for 3 ft length (2 in bottom row + 2 in top row)



Revised stress at prestress transfer:

Psi = 573(10/22) = 260k

260 260x20.83f t = 789 - 8908 = -0.278 ksi < 7.5 v'f~i

4-12

With 12 debonded strands. the top fiber stress ls less than the maxlmum

permHted; however, some bonded relnforcement must also be prov1ded

slnce the tens1le stress stll1 exceeds 3{f~i = 0.190 ksl.

Use 4 #6 bars x 121-0 at top of each end of glrder

4.11 Cont1nuity Reinforcement at P1ers

(a) For l tve load contlnuHy at lnterlor supports, provide negatlve

moment relnforcement wHhin the cast-in-place deck slab (AASHTO

9.7.2.3). Use Grade 60 deformed relnforclng bars.

strength required:

Mu 1.3(MO t 1.67ML)

= 1.3(182 t 1.61 x 566) = 1465'k

For rectangular sections (AASHTO 8.16.3.2):

1465x12 20.9x60(59 _ 4/2) = 5.71 In.

3.101n.

where ~ = 0.9 for flexure (AASHTO 8.16.1.2.2)

d = 54 t 7.5 - 2.5 = 59 in. (Use 2 In. cover)

a 4 In. (1st trial)

A f 5.71x60= O.~5¥~b = O.85x5x26 =

check a

recalculate As 1465x12 2= 0.9x60(59 _ 3.10/2) = 5.67 In.

Use 13 #6 bars ln deck slab over each glrder at pier2(As = 5. 72 1n. )

4-13

(b) Check max'mum reinforcement (AASHTO 8.16.3.1):

P = As/bd = 5.72/26x59 = 0.0037

Us'ng AASHTO Eq. (8-18) with f~ = 5000, f y = 60,000, and Bl = 0.80;

Pb = 0.0335

Pmax = 0.75 Pb = 0.75(0.0335) = 0.0252 > 0.0037 OK

A more complete design would normally consider support settlement,

effects of creep, shr'nkage and temperature, which all require

foundation data not available here and hence will not be covered.

(c) Effect of initial precompression due to prestress in the girders

at the pier may be neglected in negat've moment strength calcula

t'ons 'f max'mum precompression stress's less than 0.4f~ and

cont'nuity re'nforcement rat,o 'S less than 0.015 (AASHTO

9.7.2.3.2) .

Bottom f'ber stress at support due to in,t'al prestress = 1856 psi.

Reduct'on due to debonded strands, 1856(10/22) = 844 psi < 0.4f' OKc

Continuity reinforcement ratio P = 0.0037 < 0.015 OK

(d) Compressive stress in ends of girders at the pier, at service

loads, due to the addition of negative live load moment must not

exceed O.6f~ = O.6x5000 = 3000 psi (AASHTO 9.7.2.4).

Effective prestress stress:

Pse = 513(10/22} = 233k

233 233x20.83f b = 789 + 10543 = 0.756 ksi

4-14

Cont1nu1ty moments acting at p1er section:II<.

MO + ML : 182 + 566 : 748

Check for cracking at top of slab due to serv1ce loads:

748x12ft: 29103: 0.308 ks1 < 7.5Yf~ ~ 0.474 ks1

Compute concrete stresses due to service loads assuming an

uncracked section:

748x12 .f b = 16420 = 0.547 kSl

fb(total) = 756 + 547 1303 psi < 3000 OK

(e) Positive moment connection at p1er

Positive moments can develop at the pier section due to creep and

shrinkage in the girders and deck slab, and live load in remote

spans. This subject is discussed and a numerical example is pre

sented in Reference 4-2 "Design of Continuous Highway Bridges w1th

Precast, Prestressed Concrete Girders."

The computations required to determ1ne the amount of pos1tive

moment reinforcement, us1ng the procedures of Ref. 4-2, are

involved and time consuming. Research is currently in progress to

further evaluate the validity of such computations. In the

interim, empirical methods seem more practical.

Current practice with several state h1ghway bridge des1gn depart

ments is to provide a standard amount of reinforcement in all but

unusual conditions. A typical standard requires:

4 #6 bars in Type I and Type II I-Girders

6 #6 bars in Type II I and Type IV I-Gi rders.

4-15

The bars are embedded at least one development length 1n the endsof the g1rder. A greater embedment length may be requ1red 1f

strands are debonded. The bars project from the ends of the

g1rder approx1mately 12 bar d1ameters, term1nattng w1th a 900 bend

(wtthtn the cast-tn-place pter d1aphragm). Common pract1ce 1s to

bend the bar after the gtrder ts cast and removed from the forms.Extended strands are also somet1mes used.

Computattons ustng the procedures of Reference 4-2 for th1s des1gn

example y1eld a requ1red steel area of 2.13 1n.2. Us1ng thestandard 6 #6 bars for the Type IV gtrder, As = 2.64 1n.2

Use 6 #6 bars proJect1ng from bottom ofeach end of gtrder for pos1tive moment

connectton with1n pter diaphragm.

4.12 Shear strength

For continuous span brtdges, shear strength should be checkedthroughout the span. To 11lustrate destgn procedure, shear strength

calculations at approxtmately d/2 from the support at the p1er w1ll be

shown. Also, for cont1nuous spans, the method of design for shear re1n

forcement presented in the 1979 Interim Spectfications is generally tooconservative near support locations. For th1s destgn example, the more

exact shear design provis1ons of the 13th Ed1tion, 1983 Spec1f1cationswill be used (AASHTO 9.20).

Compute shear strength at d/2 = 61/2 = 30.5 in. = 2.54 ft from support.

Shear due to girder + slab dead load (simple span):kVo = 1.512(40 - 2.54) = 58.9

4-16

Shear due to parapet + wearing surface (continuous span):kVo = 13.7 - O.343x2.54 = 12.8

Shear due to live load (use shear at support):kVl = 58.7

Factored shear force:kVu = 1.3(Vo + 1.61V

l) = 1.3(58.9 + 12.8 + 1.61x58.1) = 221 Ig1rder

Us1ng the more exact shear design prov1s1ons, shear strength provided by

concrete 1s the lesser of Vc1 and Vcw [AASHTO £qs. (9-27) and (9-29)].

Near supports Vcw generally governs. For this example, only Vcw will be

evaluated; however, in a more complete design, both Vcw and Vc1 should

be evaluated.

Vcw = (3.5~f~ + 0.3 fpc)b'd + Vp

where fpc = compressive stress in concrete (after allowance forprestress losses) at centroid of cross section.

Vp vertical component of prestress force. With stra1ght

strand pattern Vp

= 0

With 12 debonded strands Pse = 513(10/22) = 233k

fpC due to prestress 233 233x20.83(39.S2 - 24.73) 0.020 ks1= 189 - 260730 =

f due to dead load = 1S8x12(39.52 - 24.13) 0.108 ks1pc 260730 = 0.128 ks1

Vcw = (3.5~5000 + O.3x128)8x57.6/1000 = 132K

= (Vu - ~Vc)s _ (221 - O.9x132)12 _ 0 394 1n.2/ftAv ~fsyd - 0.9x60x57.6 -.

but not less than 50b 's/fSY = 50x8x12/60,OOO = 0.08 in. 2/ f t

4-17

Also must prov1de 2 #3 @12 1n. = 0.22 1n. 2/ f t m1n1mum vert1cal t1es

for shear transfer between g1rder and compos1te deck slab (AASHTO

9.20.4.4).

Use #4 U-stirrups @12 in. spac1ng with continuous U portion2of stirrup projecting into slab (A y = 0.40 in. 1ft)

Add addit10nal stirrups at girder ends (AASHTO 9.21.3):

Use full Psi'"

4% Ps1 = 0.04(573) = 22.9 k

2A = 22.9/20 = 1.15 in.y

Use 3 #4 U-stirrups @ 3 in. at each end of girder.

For shear transfer between girder and compos1te deck slab (AASHTO

9.20.4.2), all stirrup legs must be extended into deck slab, and top

surface of g1rder must be intentionally roughened.


For est1mating long-time deflect10ns and camber, use data from PCI

Des'gn Handbook. See Des1gn Example No.1, page 1-12.

(a) Prestress at transfer (ignore effects of debonding)

For straight strands (see Design Example No.1):

573x20.83{80x12)2=

8x3.6xl03X260,730

4-18

1 .46"~

(b) G1rder dead load

5wi,4 5xO.822(80X12)40.81",L.384EI :: 384x12x3600x260,730

::

At transfer 0.55 11 1'


1.46x1.80 - O.81x1.B5 At erection 1.13 11 1'::

(d) Superimposed slab load

5wi,4 5xO.750(80X12)4 0.66 11+384[1 :: 384x12x4000x260,730 ::

(e) Superimposed rail and FWS on composite section

Using influence tables from Reference 4-1:

of moment

W :: 0.343 k/ft

Multiply by wg,2

for moment

NCXlIt'loo+.

o+

o....It'lo

o+

.o

I

IDIDoo

Using moment-area ... sum 1st

diagram.

I----~~~---L.--"'--_I

IelL Span

Use 0.1 span = 8 1

moment about pier end

CXl 0.... l""'-N It'l.... 0.0 0

I I

i 651-0 tPier

-0.1218 x 4 x 2 :: -0.97

-0.0570 x 8 x 8 :: -3.65

-0.0066 x 8 x 16 -0.84

+0.0294 x 8 x 24 5.64+0.0510 x 8 x 32 :: 13.06

+0.0582 x 4 x 38 :: 8.85

t :: 22.09

4-19

22.09xO.343x652x1728

4000x649,OOO

After construction

= 0.02"+

(f) Long term dead load

Beam o.81 11J.. x 2.40 = 1.94"J..

Camber 1.46 11 1' x 2.20 = 3.21"1'

Slab 0.66 11J.. x 2.30 = 1.52"+

Ra 111ng &FWS 0.02"+ x 3.00 = 0.06"+

Net long term ::; 0.31"+Negngible sag - say OK

( g) L1ve load deflection

Estimate maximum live load deflection at span center with heavy

truck axles closely spaced and centered in span. Use influence

coeff1c1ents and moment-area method.

(D1str1but1on)(1mpact)

(0.727)(1.244) ::; 0.90

801 - 0 ...

.It. r

~

.... 1,1') ... 1,1') MN N l"'- I"'- l"'- .. . . . . c .. (x 65 ft) for ft-k1p units\DN,..... 1,1') ~ ,.....

\\-fI + + + +

fI I I ,

I(use end span)

Using moment-area ... sum 1st moment about pier end of moment

d1agram.

4-20

-6.24 x 65 x 4 x 2 == - 3,245

-2.24 x 65 x 8 x 8 :; - 9,318

+1.75 x 65 x 8 x 16 :; 14,560

+5.74 x 65 x 8 x 24 71 ,635

+9.75 x 65 x 8 x 32 162,240+10.37 x 65 x 4 x 38 == 102.240

r == 338,100

338,100xO.90x1728 == 0.20" (Negligible)4000x649,OOO

4.14 Design Summary

Use PCI Standard Type IV I-Girder with 22-1/2" diameter 270K stress

re11eved strands. Use straight strand pattern with 12 strands shielded

at each end of girder. Use 4 #6 bars x 121-0 at top of girder at each

end. Use 6 #6 bars projecting from bottom of girder at each end for

pos1tive moment connection wlthin p1er diaphragm.

Use 13 #6 bars (continuity relnforcement) in deck slab over each girder

at pier. Prov1de 2 In. min1mum cover for continuity re1nforcement.

Use #4 U-st1rrups @12 In. spac1ng (near support), with 3 #4 U-st1rrups

@ 3 1n. at g1rder ends. Note: A complete des1gn would requ1re inves

tigation of shear reinforcement requirements for the full span length.

Selected References

4-1 "Moments, Shears and Reactions for Continuous Highway Bridges," American

Institute of Steel Construct1on, AISC Publication No. T106, 1966.

4-2 "Deslgn of Continuous Highway Bridges w1th Precast, Prestressed Concrete

G1rders," Portland Cement Assoc1at1on, [B014.01E, 1969, 19 pp. Also

PCI Journal, Vol. 14, No.2, April 1969.

4-21

2" Clear

• • • • • • • • • •

13 #6 (Gr. 60) or

equivalent over pier

• •

• ••

Debond Strands at each end:2 in each row for 31-0

2 in each row for 61-0

2 in each row for 101-0

*St1rrup details should follow local standards.Detail shown have for illustration only.

4 #6 X 121-0 at girder ends

U-St1rrups @12 in. (near support*)

U-Stirrups @3 in. at girder ends.

6 #6 (Gr. 60) at girder endsor use extended strands

10 @ 2"

if. @ 2"

22 - 1/2 11 270K strands

REINFORCEMENT DETAILS

4-22

short span bridges-pci

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