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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.
Example No.2, page 2-14
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)
Example No.3, page 3-15
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
2.1 Design Conditions
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.
2.8 Prestress losses
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
(c) Creep of concrete
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
(e) Total loss of prestress
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.
2.12 Deflections and Camber
For estimating long-time deflections and camber, use data from PCIDesign Handbook. See Design Example No.1, page 1-12.
(a) Prestress at transfer
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
(c) Growth in storage
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 )
(d) Superimposed dead load
f~ = 5000 psi; Ec = 4.3 x 10° psi
4(Deck) 5xO.413(40x12)
384x12x4.3xl03x22230
5xO.140(40x12)4
(e) Long term dead load
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
3.1 Design Conditions
Simple span of 75 ft x 30 ft width
HS20 live load - 2 lanes
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
Precast concrete: normal weight
f~ 5000 psi
f~i = 4000 psi (AASHTO 9.22)
Cast-in-place concrete: normal weight
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
3.6 Prestressing Strands
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
3.7 Flexural strength
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
(b) M1n1mum steel (AASHTO 9.18.2):
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
(c) Creep of concrete
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
(d) Relaxation of prestressing steel
CR s = 20,000
= 20,000
0.4ES - 0.2 (SH + CR )cO.4x9870 - 0.2(6500 + 10,600) 12,630 psi
(e) Total loss of prestress
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.
3.11 Deflections and Camber
For estimating long-time deflections and camber, use data from PClDesign Handbook. See Design Example No.1, page 1-12.
3-15
(a) Prestress at transfer
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
(d) Superimposed dead load
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'
(e) Long term dead load
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
HS20 live load - 2 lanes
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
Precast concrete: normal weight
f~ = 5000 ps1
f~i = 4000 psi (AASHTO 9.22)
Cast-in-place concrete: normal weight
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
4.6 Prestressing Strands
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
4.7 Flexural strength
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
(b) M1n1mum steel (AASHTO 9.18.2):
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.
4.9 Prestress losses
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 =
(c) Creep of concrete
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
(d) Relaxation of prestressing steel
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
(e) Total loss of prestress
~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)
4 strands for 6 ft length (2 in bottom row + 2 in top row)
4 strands for 10 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.
4.13 Deflections and Camber
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'
(c) Growth in storage
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
NOTES