4. Precast Prestressed Concrete Girder Bridge - Design Example

7/23/2019 4. Precast Prestressed Concrete Girder Bridge - Design Example

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4.DesignExamplePrecast/PrestressedConcreteGirder

DesignSteps:

1. ProblemDefinition

2. GirderSelection

3. SectionProperties

4. MomentsandShears

5. FlexuralDesign

o StrandPattern

o PrestressLosses

o

StressChecks

o FlexuralCapacity

o MinimumReinforcement

6. ShearDesign

o VerticalShear

o InterfaceShear

7. AnchorageZoneReinforcement


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1. ProblemDefinition

SpanData

Overall Girder Length = 106 ft

Design Span = 105 ft

Girder is simply supportedSkew = 0

BridgeCrossSectionData

Number Lanes = 4Number Girders = 6

Girder Spacing = 9.00 ft

Roadway Width = 48.00 ftOverall Width = 51.00 ft

DeckThickness

Actual = 9.00 in

Structural = 8.00 in

GirderType

PCI BT72 (72 in. deep bulbtee)Location: Interior

DeadLoad

Future Wearing Surface = 0.025 ksfBarrier Weight = 0.418 klf

LiveLoad

HL93 Design Truck + Design Lane

GirderConcrete

fc = 6.0 ksifci = 4.5 ksi

wc = 0.150 kcf

DeckConcrete

fc = 4.0 ksiwc = 0.150 kcf

PrestressingSteel

Type: 0.5in Diameter 270 ksi LowRelaxation SevenWire Strand

Eps = 28,500 ksi

Pull: 75%H = 75% (Relative Humidity)

Time to Release = 24 hrs

Profile: 2pt. Depressed

Depression Point: 0.45L = 0.45 (105 ft)

ReinforcingSteel(Non-Prestressed)

fy = 60 ksi

Es = 29,000 ksi

Figure1: Bridge Cross Section: Six PCI BT72 Bulb Tee Girders at 90 Spacing


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2.Analysis

2.1 SectionProperties

2.1.1 Bare Girder:

The LRFD Specsallowtheinclusionoftransformedstrandinthesectionproperties

for a prestressed member (Article 5.9.1.4). For simplicity, the contribution of the strandto the section properties is neglected in this example.

Properties of PCI BT72:

A = 767.0 in2

I = 545,894 in4

h = 72.00 in

yb = 36.60 in

yt = 35.40 in

Sb = 545,894 in4

/ 36.60 in = 14,915 in3

St = 545,894 in4/ 35.40 in = 15,421 in3

2.1.2 Composite Section

Figure2: Cross Section of Single Girder with

Composite Deck


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Note: Any thickening of the slab over the top flange of the girder (i.e., a haunch or build

up) will be neglected in the computation of the section properties of the composite

section. However, if they are detailed in the plans, they should be included as additionaldead load.

Effective deck width: (LRFD 4.6.2.6.1)

Onequarter span length = (105 ft / 4)(12) = 315 in Onehalf flange width + (12)(deck thickness):

(42 in) / 2 + (12) (8 in) = 117 in

Average spacing of adjacent girders = (9 ft) (12) = 108 in (Controls)

Transformed deck width = (n) (effective width) = (0.8165) (108 in) = 88.182 in

n =gc

dc

E

E=

cg

cd

f

f

=

KSI0.6

KSI0.4= 0.8165

Component Area yb A yb A (yb ybc)2

Io Ic

Girder 767.00 36.60 28,072 273,111 545,894 819,005

Effective Deck 705.45 76.00 53,614 297,334 3,762 301,096

Total 1,472.50 81,686 1,120,101

ybcg= (A yb) / A = 81,686 in3

/ 1,472.5 in = 55.47 in

ytcg= h ybc= 72.00 in 55.47 in = 16.53 in

ytcd= hc ybc= 80.00 in 55.47 in = 24.53 in

Sbcg= Ic/ ybc= 1,120,101 in4

/ 55.47 in = 20,193 in3

Stcg= Ic/ ytcg= 1,120,101 in4

/ 16.53 in = 67,762 in3

Stcd= (Ic/ ytcd) / n = (1,120,101 in4

/ 24.53 in) / (0.8165) = 55,925 in3

2.2MomentsandShears

2.2.1 Dead Loads

2.2.1.1 Girder Dead Load at Release

The moments for this condition are computed separately from other moments because

the full length of the girder is used in computing these moments, rather than the design

span (distance from centertocenter of bearings). Thefulllengthisusedbecause,


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when the girder cambers upward in the prestressing bed after release, its only points of

contact with the bed (and therefore its support locations) will be at the ends of the girder.

Locations of interest at release conditions:

1. Transfer point (LRFD 5.8.2.3)

l t = 60 db= 60 (0.5 in) = 30.0 in = 2.5 ft

2. Depression point

x = 0.45 L = 0.45 (106 ft) = 47.7 ft

3. Midspan

x = 0.5 L = 0.5 (106 ft) = 53.0 ft

Girder Dead Load

wgdl = (767 in2

/ 144) (0.150 kcf) = 0.799 klf

Mgdli= ( )xL2

xw

L = 106 ft (overall girder length)

Mgdli= 0.799 (x / 2) (106 x) = 42.35 x 0.400 x 2

2.2.1.2 Girder Dead Load Final

L = 105 ft (bearing to bearing)

Mgdl= 41.95 x 0.400 x2

Vgdl=

x

2

Lw = 41.95 0.799 x

2.2.1.3 Deck Dead Load (Structural Deck)

Structural Deck Thickness = 8.0 in

wddl = ((8 in x 108 in) /144) (0.150 kcf) = 0.900 klf

L = 105 ft

Mddl= 47.25 x 0.450 x2

Vddl= 47.25 0.900 x

2.2.1.4 Additional NonComposite Dead Load (NonStructural Deck)

NonStructural Deck Thickness = 1.0 in

wncdl = ((1 in x 108 in) /144) (0.150 kcf) = 0.1125 klf

L = 105 ft

Mncdl= 5.906 x 0.0563 x2

Vncdl= 5.906 0.1125 x

2.2.1.5 Composite Dead Load Barriers

Barriers: (2) (0.418 klf / barrier) = 0.836 klf

wcdl= 0.836 klf / (6 girders) = 0.1393 klf / girder (LRFD 4.6.2.2.1)

L = 105 ft


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Mcdl= 7.313 x 0.0697 x2

Vcdl= 7.313 0.1393 x

2.2.1.6 Composite Dead Load Future Wearing Surface

Future Wearing Surface: (48.00 ft) (0.025 ksf) = 1.200 klf

wfws= 1.200 klf / (6 girders) = 0.200 klf

L = 105 ft

Mcdl= 10.500 x 0.100 x2

Vcdl= 10.500 0.200 x

2.2.2 Live Loads

2.2.2.1 Distribution Factors (LRFD 4.6.2.2.1)

To use the simplified live load distribution factor formulae, the

following conditions must be met: Width of deck is constant O.K.

Number of girders, Nb, 4 O.K. (Nb= 6)

Girders parallel and same stiffness O.K.

Roadway part of overhang, de, 3.0 ft O.K. (de= 1.25 ft)

Curvature < 4o

O.K. (Curvature = 0o

)

Bridge Type: k (LRFD Table 4.6.2.2.11)

Distribution Factor for Moment(2 or More Lanes loaded):

1.0

3s

g2.06.0

tL0.12

K

L

S

5.9

S075.0DF

+= (LRFD Table 4.2.2.2b1)

Provided that: 3.5 S 16 S = 9.00 ft O.K.4.5 ts12.0 ts= 8.00 in O.K.

20 L < 240 L = 105 ft O.K.Nb4 Nb= 6 O.K.

Kg= n (I + A eg2

) (LRFD Eq. 4.6.2.2.11)

n =dc

gc

E

E

=cd

cg

f

f

= KSI0.4

KSI0.6

= 1.2247 (Note

that

this

is

the

reciprocal

of

the

value

for

n

computedearlier)

eg= yt+ (ts/ 2) = 35.40 in + (8 in / 2) = 39.40 in

Kg= 1.2247 [545,894 in4

+ (767 in2

) (39.40 in)2

] = 2,126,758 in4

( )( )

1.0

3

42.06.0

IN8FT1050.12

IN758,126,2

FT105

FT0.9

5.9

FT0.9075.0DF

+= = 0.74lanes/girder

82.02

1

5.5=x

S


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Distribution Factor for Shear(2 or More Lanes Loaded):

0.2

35

S

12

S2.0DF

+= (LRFD Table 4.6.2.2.3a1)

Provided that the following condition is met in addition to the conditions specified above:10,000 Kg( =2,126,758) 7,000,000 O.K.

0.2

35

FT0.9

12

FT0.92.0DF

+= = 0.88lanes/girder

2.2.2.2 Live Load Effects

Figure3: LRFD Design Truck and Design Lane Load

At Midspan:

Design Truck will govern over Design Tandem for this span.Mtruck= 18L 280 (Maximummomentatmidspan)

Mtruck= (18) (105 ft) 280 = 1,610.0 kipft

Mlane=8

Lw 2

=( )8

FT105KLF64.0 2

= 882.0 kipft

Dynamic Load Allowance (Impact Factor) (LRFD Table 3.6.2.11)

1 + IM = 1 + 0.33 = 1.33 (appliedonlytotruckportionofliveload)

MLL+I = DF [Mlane+ 1.33 (Mtruck)] = (0.7423) [882.0 + (1.33) (1,610.0)] = 2,244.2 kipft

At 5.68 ft from Centerline of Bearing (Critical Section for Shear see Section 3.2.1.1 below):

Mtruck= ( ) LxbPbPbP

LxbP

LxbP

LxbP 332211332211 ++=++

Mtruck= [(32 kip) (99.32 ft) + (32 kip) (85.32 ft) +

(8 kip) (71.32 ft)] (5.68 ft) / 105 ft

Mtruck= 350.5 kipft

Mlane= ( )xL2

xw = (0.64 klf) (5.68 ft) (105 ft 5.68 ft) / 2 = 180.5 kipft

MLL+I,Vmax= DF [Mlane+ 1.33 (Mtruck)] = (0.7423) [180.5 + (1.33) (350.5)] = 480 kipft


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Vtruck= Mtruck/ x = 350.5 kipft / 5.68 ft = 61.7 kip

Vlane=

x

2

Lw = 0.64 klf (105 ft / 2 5.68 ft) = 30 kip

VLL+I= DF [Vlane+ 1.33 Vtruck] = (0.8839) [30 + (1.33) (61.7)] = 99.1 kip

2.3 LoadCombinationsandLoadFactors

2.3.1 Applicable Limit States (LRFD Table 3.4.11)

Service I Service III Strength I

2.3.2 Service I

This load combination is the general combination for Service Limit State stress checksandapplies to all conditions other than Service III.

All load factors are equal to 1.0 for this problem.

For moment at midspan:

Acting on the noncomposite girder, MSLnc:

MSLnc= 1,101.1 + 1,240.3 + 155.0 = 2,496.4 kipft

Girder Deck AddnlDeck

Acting on the composite girder, MSLc:

MSLc= 192.0 + 275.6 + 2,244.2 = 2,711.8 kipft

Barrier FWS LL+I

2.3.3 Service III

This load combination is a special combination for Service Limit State stress checks that applies

only to tension in prestressed concretestructures with the objective of crack control.

All load factors are equal to 1.0 for this problem, except that the live load is reduced by a factor of

0.8.

For moment at midspan:

Acting on the noncomposite girder, MSLnc(same as for Service I).

Acting on the composite girder, MSLc:

MSLc= 192.0 + 275.6 + (0.8)(2,244.2) = 2,263.0 kipft

Barrier FWS LL+I

2.3.4 Fatigue(notrequired)

According to LRFD 5.5.3.1, Fatigue need not be checked for concrete deck slabs in multigirder

applications. Fatigue of the reinforcement need not be checked for fully prestressed

components designed to have extreme fiber tensile stress due to Service III Limit State within thetensile stress limit specified in Table 5.9.4.2.21. Fatigue of concrete is checked indirectly by


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satisfying the compression stress limit of 0.4 '

cf for the load combination specified in LRFD

5.9.4.2.1.

2.3.5 Strength I

This load combination is the general combination for Strength Limit State design. Since

the structure is simply supported, the maximum values for the load factors are usedbecause they produce the greatest effect (see LRFD Table 3.4.12).

No distinction is made between moments and shears applied to the noncomposite orcomposite sections for strength computations. The factored loads are applied to the

composite section.

The following load factors apply:

Dead Load Component and Attachments 1.25 DC

Dead Load Wearing Surface and Utilities 1.50 DW

Vehicular Live Load and Impact 1.75 LL and IM

Mu or Vu= 1.25DC + 1.50DW + 1.75(LL + IM)

For moment at midspan, Mu:

Mu= (1.25) [1,101.1 + 1,240.3 + 155.0 + 192.0] + (1.50) (275.6) + (1.75) (2,244.2)

Girder DeckAddnlDeck Barrier FWS LL+I

Mu= 3,360.5 + 413.4 + 3,927.4 = 7,701 kipft

For shear at the critical section for shear, Vu:

Vu= (1.25) [37.1 + 41.8 + 5.2 + 6.5] +(1.50) (9.3) + (1.75) (99.1)

Girder Deck AddnlDeck Barrier FWS LL+I

Vu= 113.3 + 14.0 + 173.4 = 300.7 kip

For moment at the critical section for shear, Mu:

Mu= (1.25) [239.3 + 269.5 + 33.7 + 41.7] + (1.50) (59.9) + (1.75) (480)

Girder Deck AddnlDeck Barrier FWS LL+I

Mu= 730.3 + 89.9 + 840 = 1660.2 kipft


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SummaryofDeadandLiveLoadEffects

MomentsatRelease(kip-ft)

Component Transfer Pt. (2.5 ft) 0.45L (47.7 ft) Midspan (53.0 ft)

Girder 103.4 1,111 1,122

MomentsandShears

At Critical Section for Shear

(5.68 ft from Center of Support, see

Section 3.2.1.1)

Midspan (52.50 ft)

Component V (kip) M (kipft) M (kipft)

ActingonNon-CompositeGirder:

Girder 37.1 239.3 1,101.1

Deck (Structural) 41.8 269.5 1,240.3

Additional NonComposite 5.2 33.7 155.0

SUBTOTAL 84.1 542.5 2,496.4

ActingonCompositeGirder:

Barriers 6.5 41.7 192.0

Future Wearing Surface 9.3 59.9 275.6

Live Load + Impact 99.1 480.0 2,244.2

SUBTOTAL- ServiceI 114.3 611.2 2,711.8

SUBTOTAL- ServiceIII --- --- 2,263

TOTAL- StrengthI 300.7 1660.2 7,701


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3.Design

3.1 FlexuralDesign

3.1.1 Strand Patterns at End of Girder and at Midspan

A trialanderror procedure is used to determine the strand pattern. For simplicity, the

trial designs that were performed to arrive at the strand patterns shown below are not

included as part of this design example.

The section at midspan is considered first. Generally, strands are added to the section in

pairs, filling the available strand locations from the bottom, until the stress limits and

strength requirements at midspan are satisfied.

The end pattern is then determined by draping strandsas required to satisfy the stresslimits at the end of the girder at release. Other methods, which are not considered in this

example, could also be used to control stresses at the end of the girder.

Draped Strands

No. ofStrands

Dist. fromBottom

2 68 IN

2 66 IN

2 64 IN

2 62 IN

Straight Strands

No. ofStrands

Dist. fromBottom

2 8 IN

6 6 IN

10 4 IN10 2 IN

Strands at Midspan

No. ofStrands

Dist. fromBottom

4 8 IN

8 6 IN

12 4 IN

12 2 IN

4.22in.

Figure4: Strand Pattern at End and Midspan

3.1.2 Properties of Assumed Strand Patterns

For the following computations, all c. g. dimensions are measured from the bottom of the girder.

At Midspan (and between Depression Points):

Depression Point Location:

0.45 L = 0.45 (105 ft) = 47.25 ft from CL bearing = 47.75 ft from end of girder

c.g. @ midspan = [(12 strands)(2 in) + (12)(4 in) +(8)(6 in) + (4)(8 in)] / 36 strands = 4.22in

eccentricity @ midspan = eCL= yb c.g. @ midspan = 36.60 4.22 = 32.38 in

At End of Girder:

c.g. @ end = [(10 strands)(2 in) + (10)(4 in) +(6)(6 in) + (2)(8 in) + (2)(62 in) + (2)(64

in) + (2)(66 in) +(2)(68 in)] / 36 strands = 17.56 in


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eccentricity @ end = eend= yb c.g. @ end = 36.60 17.56 = 19.04 in

At Transfer Length from End of Member (2.5 ft):

l t = 60 db= 60 (0.5 in) = 30 in = 2.5 ft (LRFD 5.8.2.3)

c.g. of strand pattern @ end = 17.56 in

c.g. of strand pattern @ depression point = 4.22 in

c.g. @ transfer point = 17.56 in (2.50 ft / 47.25 ft) (17.56 in 4.22 in) = 16.85 in

eccentricity @ transfer point = etr= yb c.g. @ t.p. = 36.60 16.85 = 19.75 in

At Critical Location for Shear (See section 3.2.1.1, 5.68 ft from CL bearing; 6.18 ft from end ofmember):

c.g. @ 6.18 ft = 17.56 in (6.18 ft / 47.75 ft) (17.56 in 4.22 in) = 15.83 in

eccentricity @ 6.18 ft = ecv= 36.60 15.83 = 20.77 in

Total Area of Prestressing Strands:

Aps= (36 strands) (0.153 in2

) = 5.508 in2

C. G. of Straight Strands:c.g. straight = [(10 strands)(2 in) + (10)(4 in) +(6)(6 in) + (2)(8 in)] / 28 strands =

4.00 in

eccentricity of straight strands = estr= yb c.g. straight = 36.60 4.00 = 32.60 in

Area of Straight Prestressing Strands:

Apss= (28 strands) (0.153 in2

) = 4.284 in2

3.1.3 PrestressLosses

3.1.3.1 Components of Prestress Loss

Total loss of prestress is given by: (LRFD 5.9.5.1)

fpT= fpES+ fpLT

Where,

fpES= sum of all losses or gains due to elastic shortening or extension at the time of application of

prestress and/or external loads (ksi), and

fpLT= losses due to longterm shrinkage and creep of concrete, and relaxation of the steel (ksi)

3.1.3.1.1 Elastic Shortening

cgpci

pESp f

E

Ef = (LRFD Eq. 5.9.5.2.3a1)


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fcgp= Stress at c.g. of strands at release (at midspan)

Scgp= Section modulus of bare girder at c.g. of strand pattern

Scgp= I / eCL= (545,894 in4

) / (32.38 in) = 16,860 in3

Initial value for fcgpcan be calculated assuming the stress in strands after release, fi= 0.90 *

0.75fpu

. (LRFD 5.9.5.2.3a)

Alternatively, fpEScan be calculated using the following closedform equation (LRFD Eq.

C5.9.5.2.3a1):

ps

cigg

g

2

mgps

ggmg

2

mgpips

pES

E

EIA)AeI(A

AMe)AeI(fAf

++

+=

Where,

Eci= cI5.1

c fw000,33 = ( ) ksikcf 5.4146.0000,33 5.1

= 3,905 ksi (LRFD Eq. 5.4.2.41)

wc= 0.145 kcf


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ksi27.3

1.49)-1.20.03.19()1.216.626.55(

ffffffff

dfid

dfpSSpRpCDpSDidpRpCRpSRpLT

=

+++++=

+++++= 21

Note that the total longterm loss calculated by the approximate method is quite close to the one

calculated using the refined method. We will use the approximate loss in the remainder of theexample.

3.1.3.2 Prestress Loss and Effective Prestress at Release

Compute initial prestress loss

ESppi ff =

fpi = 18.1 ksi (8.9%ofinitialprestress,fpj)

Compute effective stress and force after losses

fpi= 202.5 18.1 = 184.4 ksi

Pi= (Aps) (fpi) = (5.508 in2

) (184.4 ksi) = 1,016 kip

3.1.3.3 Prestress Loss and Effective Prestress after All Losses (Final)

Compute final prestress loss

fpT= fpES+ fpLT (LRFD Eq. 5.9.5.11)

fpT= 18.1 + 25.4 = 43.5 ksi (21.5%ofinitialprestress,fpj)

Compute effective stress and force after losses

fpe= 202.5 43.5 = 159.0ksi

Pe= (Aps) (fpe) = (5.508 in2) (159.0 ksi) = 876 kip

Check effective stress: (LRFD Table 5.9.31)

Calculate gain in prestressing steel due to deck weight (gdw), superimposed dead load

(gsdl), and 0.8 times the live load with impact (g0.8LLI).

Ec= cc fw 5.1000,33 = ( ) ksikcf 6146.0000,33 5.1 = 4,509 ksi


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4-15

ksi14.16.21.66.3GainTotal

ksig

ksig

ksinS

Mg

inS

inS

E

En

0.8LLI

sdl

ConcGirderPS

cgPS

weightdeck

dw

3

cgpscomp

3

cgPS

c

p

ConcGirderPS

=++=

==

=+

=

=+

==

===

==

===

2.6856,21

12*)2244(*8.0

6.1856,21

12*)276192(

3.63.6*859,16

12*)1551240(*

856,21)25.5122.447.55(

101,120,1

859,1638.32

894,545

3.6509,4

500,28

Stress in Prestressing steel due to prestress after all losses should be limited to 0.8fpy.

fpe= 159.0 + 14.1 = 144.9 ksi < 0.8 fpy= (0.8) (0.9*270 = 243 ksi) = 194.4 ksi O.K.

3.1.4 Midspan

3.1.4.1 Concrete Stresses Due to Loads

Sign convention for stresses: (+) = Compression

() = Tension

Girder Dead Load At Release with L = 106 ft:

ft=t

sw

S

M=

421,15

122,1x 12 = 0.873 ksi fb=

915,14

122,1x 12 = 0.903 ksi

Girder Dead Load Final with L = 105 ft:

ft=t

sw

S

M=

421,15

1.101,1x 12 = 0.857 ksi fb=

915,14

1.101,1x 12 = 0.886 ksi

Deck (Structural) Dead Load:

ft=421,15

3.240,1x 12 = 0.965 ksi fb=

915,14

3.240,1x 12 = 0.998 ksi

Additional NonComposite Dead Load (NonStructural Deck):

ft=421,15

0.155x 12 = 0.121 ksi fb=

915,14

0.155x 12 = 0.125 ksi

Composite Dead Load Barriers + Future Wearing Surface:

ftg=( )

762,67

6.2750.192 +x 12 = 0.083 ksi fbg=

( )193,20

6.2750.192 +x 12 = 0.278 ksi


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4-16

ftd=( )

925,55

6.2750.192 +x 12 = 0.100 ksi

Live Load I:

ftg=762,67

2.244,2x 12 = 0.397 ksi

ftd=925,55

2.244,2x 12 = 0.482 ksi

Live Load III:

fbg=( )

193,20

2.244,28.0x 12 = 1.067 ksi

3.1.4.2 Concrete Stresses Due to Prestress

At Release

Bottom:

fb=

+

b

CLi

S

e

A

1P =

+

915,14

38.32

767

1016,1 = 3.530 ksi

Top:

ft=

t

CLi

S

e

A

1P =

421,15

38.32

767

1016,1 = 0.809 ksi

After all Losses (Final)

Bottom:

fb=

+

b

CLe

S

e

A

1P =

+

915,14

38.32

767

1876 = 3.044 ksi

Top:

ft= Pe1

A

eCL

St

=

421,15

38.32

767

1876 = 0.697 ksi

3.1.4.3 Concrete Stresses at Service Limit State Before Losses (At Release)

Note: Stressesatdepressionpointwillbemorecriticalatrelease.

Service I:

Bottom of Girder (Compressive Stress):

fb= 3.530 0.903 = 2.627 ksi

PS Girder

Check limiting stress: (LRFD 5.9.4.1.1)

2.627 ksi < 0.60 fci = 0.60 4.50 KSI( )= 2.70 ksi O.K.


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4-17

Top of Girder (Tensile Stress):

ft= 0.809 + 0.873 = 0.064 ksi

PS Girder

Check limiting stress, without bonded auxiliary reinforcement: (LRFD 5.9.4.2.1)

0.064 ksi > cif0948.0 = 50.40948.0 = 0.201 ksi 0.200 ksi O.K.

3.1.4.4 Concrete Stresses at Service Limit States After All Losses (Final)

Service III (Tensile Stress in Bottom of Girder):

fb= 3.044 0.886 0.998 0.125 0.278 1.067 = 0.310 ksi

PS Girder DeckAddnlDeck Barrier+FWS LL+I

Check limiting stress: (LRFD Table 5.9.4.2.21)

0.310 ksi > cf190.0 = 00.6190.0 = 0.465 ksi O.K.

Service I (Compressive Stress in Top of Girder): (LRFD 5.9.4.2.1)

Compressive stress due to the sum of effective prestress and permanent loads:

ft= 0.697 + 0.857 + 0.965 + 0.121 + 0.083 = 1.329 ksi

PS Girder Deck AddnlDeck Barrier+FWS

Check limiting stress:

1.329 ksi < cf45.0 = ( )00.645.0 = 2.70 ksi O.K.

Compressive stress due to the sum of effective prestress, permanent loads and transientloads (full service load):

ft= 0.697 + 0.857 + 0.965 + 0.121 + 0.083 + 0.397 = 1.726 ksi

LL+I


1.726 ksi < cwf60.0 = 0.60(1.0)(6.00) = 3.60 ksi O.K

where w= 1.0 for top flange of girder with composite deck because span/thickness ratioof flange must be less than 15. See LRFD 5.7.4.7.2.

Compressivestressduetoliveloadandone-halfthesumofeffectiveprestressandpermanentloads:

ft= 0.397 + 0.5(2.0260.697=1.329) = 1.062 ksi


1.062 ksi < cf40.0 = ( )00.640.0 = 2.40 ksi O.K

Service I (Compressive Stress in Top of Deck): (LRFD 5.9.4.2.1)

Compressive stress due to the sum of effective prestress and permanent loads:

ft= 0.100 ksi


0.100 ksi < cdf45.0 = ( )00.445.0 = 1.80 ksi O.K.

Compressive stress due to the sum of effective prestress, permanent loads and transientloads (full service load):

ft= 0.100 + 0.482 = 0.582 ksi



18/37

4-18

0.582 ksi < cdwf60.0 = ( )00.460.0 = 2.40 ksi O.K

where w= 1.0 for deck between tips of girder flanges, because

span/thickness = (9.00 ft 3.5 ft) / (8 in / 12 in/ft) = 8.2 < 15. See LRFD 5.7.4.7.2.

Compressive stress due to live load and onehalf the sum of effective prestress andpermanent loads:

ft= 0.482 + 0.5(0.100) = 0.532 ksi


0.532 ksi < cdf40.0 = ( )00.440.0 = 1.60 ksi O.K

3.1.4.5 Strength Limit State (Strength I)

Compute nominal moment capacity, Mn

Check whether section behaves as a rectangular beam or a Tbeam:

+

+=

p

pu

ps1c

ysyspups

d

fAkbf85.0

fAfAfAc = depth of neutral axis (LRFD Eq. 5.7.3.1.14)

=

pu

py

f

f04.12k (LRFD Eq. 5.7.3.1.12)

k = 2 [1.04 (243 ksi / 270 ksi)] = 0.280

dp= h + hf c.g. @ midspan = 72.00 + 8.00 4.22 = 75.78 in [dp= 64.17 In at critical section

for shear]

Since no mild tension or compression reinforcement is being considered, terms are

eliminated. Note that the full effective (not transformed) deck width is used in thiscomputation since the deck concrete strength is used.

( )( )( ) ( ) ( ) ( )( )

+

=

in

ksiininksi

ksiinc

78.75

270508.5280.00.10885.00.485.0

270508.52

2

= 4.682 in

Compute depth of compression block, a:

ca 1=

1 = 0.85 for cf = 4.0 ksi (deck concrete) (LRFD 5.7.2.2)

a = (0.85) (4.682 in) = 3.980 in [a = 3.97 in at critical section for shear]

Sincea=3.98IN


19/37

4-19

=

in

inksifps

78.75

682.4280.01270 = 265.3 ksi [fps= 264.6 ksi at critical section for shear]

Compute nominal moment capacity, Mn:

=

2

adfAM

ppspsn (LRFD Eq. 5.7.3.2.21)

( )( )

=2

980.378.753.265508.5 2

ininksiinMn = 107,827 Kin = 8,986 kipft

Compute factored moment resistance, Mr:

Check tension/compression controlled section (LRFD 5.5.4.2)

Calculate net tensile strain in extreme tension steel at nominal resistance, t

dt= 80 2 = 78 in.

( ) ( )

sectioncontrolled-tensionTherefore

c

cdtt

,

005.0047.0003.0*682.4

682.478003.0* =

=

=

= 1.0 for flexure (LRFD 5.5.4.2)

Mr= Mn= 8,986 kipft

Compare factored moment resistance, Mrto required moment, Mu:

Mr= 8,986 kipft > Mu= 7,701 kipft O.K.

3.1.4.6 Reinforcement Limits

Check Minimum Reinforcement:

Mrthe lesser of 1.2 Mcror 1.33 Mu (LRFD 5.7.3.3.2)

Compute Mcr

( ) ( ) crnccdncccpercr Sf1SSMSffM += (LRFD Eq. 5.7.3.3.21)where

cr f37.0f = = 00.637.0 = 0.91 ksi (LRFD 5.4.2.6)

cpef = compressive stress in concrete due to effective prestress forces only

(after losses) at extreme fiber of section where tensile stress is

caused by externally applied loads

cpef = fbafter losses (see Section 3.1.4.2) = 2.926 ksi

dncM = the noncomposite dead load moment

dncM = 1,101.1 + 1,240.3 + 155.0 = 2,496.4 kipft

cS = bcgS = composite section modulus for the tension face

ncS = noncomposite section modulus for the tension face

( )( ) ( )

+= 1

915,14

193,204.496,212/193,20926.291.0

3

33

in

inft-kipinksiksiMcr


20/37

4-20

Mcr = 6,455 kipft 883 kipft = 5,572 kipft

Mcrmust be taken greater than or equal to frSc= 0.91(20,193)/12 = 1,531 kipft

1.2 Mcr= 1.2 (5,572) = 6,686 kipft GOVERNSsince1.2Mcr 1.2Mcr= 6,686 kipft 0.K.

3.1.5 End and Transfer Point at Release

Stressesonlyneedtobecheckedatreleaseatthislocationsincelosseswithtimewillreduce

theconcretestressesmakingthemlesscritical.

3.1.5.1 Compute Concrete Stresses Due to Loads (Girder Only)

ft=t

sw

S

M=

421,15

4.031x 12 = 0.080 ksi fb=

915,14

4.031x 12 = 0.083 ksi

3.1.5.2 Compute Concrete Stresses Due to Prestress

Stresses due to prestress are equal to zero at the ends.

At Transfer Point:

Bottom:

fb=

+

b

tri

S

e

A

1P =

+

32 915,14

75.19

767

1016,1

in

in

inkip = 2.670 ksi

Top:

ft=

t

tri

S

e

A

1P =

32 421,15

75.19

767

1016,1

in

in

inkip = 0.023 ksi

3.1.5.3 Check Concrete Stresses at Service Limit State Before Losses (At Release)

Service I:


fb= 2.670 0.083 = 2.587 ksi

Check stress limit: (LRFD 5.9.4.1.1)

2.587 ksi < cif60.0 = ( )KSI50.460.0 = 2.70 ksi O.K.


ft= 0.023 + 0.080 = 0.103 ksi

Check stress limit, without bonded auxiliary reinforcement (LRFD 5.9.4.1.2)

0.103 ksi > cif0948.0 = 50.40948.0 = 0.201 ksi 0.200 ksi O.K.

3.1.6 Depression Point (0.45L) at Release

Stressesonlyneedtobecheckedatreleaseatthislocationsincemidspanwillgovernforfinal

stressconditions.

3.1.6.1 Compute Concrete Stresses Due to Loads (Girder Only)


21/37

4-21

ft=421,15

111,1x 12 = 0.865 ksi fb=

915,14

111,1x 12 = 0.894 ksi

3.1.6.2 Compute Concrete Stresses Due to Prestress

Bottom:

fb= 3.530 ksi (same

as

at

Midspan)Top:

ft= 0.809 ksi (sameasatMidspan)

3.1.6.3 Check Concrete Stresses at Service Limit State Before Losses (At Release)

Service I:


fb= 3.530 0.894 = 2.636 ksi

Check stress limit in concrete: (LRFD 5.9.4.1.1)

2.636 ksi < cif60.0 = ( )50.460.0 = 2.70 ksi O.K.


ft= 0.809 + 0.865 = 0.056 ksi

Check stress limit, without bonded auxiliary reinforcement (LRFD 5.9.4.1.2)

0.056 ksi > cif0948.0 = 50.40948.0 = 0.201 ksi 0.200 ksi O.K.

3.2 ShearDesign

3.2.1 Transverse Shear Reinforcement

In this example, the girder will be designed for vertical shear at the critical

section for shear. In a full design, other sections along the length of the

girder would have to be designed as well.

3.2.1.1 Critical Section for Shear

Shear design using the SectionalDesignModelis an iterative process that begins byassuming a value for . To avoid a trial and error iteration process, 0.5 cot in the

equation for x(LRFD 5.8.3.4.2) can be taken as 1.0 (C5.8.3.4.2). This example illustrates

the use of both the trial and error and the simplified procedures.

Critical section for shear is at dvfrom the internal face of support. (LRFD 5.8.3.2)

Compute dv:

dv = Effective shear depth= Distance between resultants of tensile and compressive forces

The depth of the compression block, a, was computed in determining the moment

capacity of the section (see Section 3.1.4.5).

( )2

[email protected]

ationcriticalgchh

add fgev +==

= (72.0 + 8.0 15.83) (3.97/2) = 62.2 in


22/37

4-22

But dvneed not be taken less than the greater of: (LRFD 5.8.2.7)

0.9 de= (0.9) (8015.73) = (0.9) (64.27) = 57.84 in

0.72 h = (0.72) (80) = 57.60 in

Therefore, use dv= 62.2 in

Therefore the critical section for shear is:0.50 ft+ 62.2 in / 12 = 5.68 ft from centerline of support (support assumed to be 1.0 ft wide).

At the critical section for shear, Vu = 300.7 kip

3.2.1.2 Component of Shear Resistance from Prestress, Vp

Pf= 876 kip

Angle of center of gravity of strand profile with respect to horizontal, :

= tan1

[(eCL eend) / (dist. to depression point)]

= tan1

[((32.38 in 19.04 in) / 12) / 47.75 ft] = 1.32o

Vp= Pfsin (1.32o) = (876 kip) [sin (1.32o)] = 20.2 kip

3.2.1.3 Governing Equations for Shear

Vu Vr= Vn (LRFD 5.8.2.12)

= 0.90 for shear (LRFD 5.5.4.2.1)

Vn= Vc+ Vs+ Vp (LRFD 5.8.3.31)

Compute maximum shear capacity of section:

Vn max= pvvc Vdbf25.0 + (LRFD 5.8.3.32)

Vn max= (0.25) (6.00 ksi) (6.00 in) (62.2 in) + 20.2 kip = 580 kip

Vn max

= (0.90) (580) = 522 kip > Vu

= 300.7 kip O.K.

3.2.1.4 Concrete Contribution to Shear Resistance, Vc

vvcc dbfV = 0316.0 (LRFD 5.8.3.33)

To use this equation, the quantity must be determined. This quantity is a factor that

represents the efficiency of shear transfer by concrete.

Note that (ksi)fc0316.0 = ( )psifc0.1 , so a value of 2 would provide a concrete

contribution similar to the familiar simplified value of ( )psifV cc =2 bd.

To obtain , the quantities

cf

v

and are needed, where

cf

v

is a relative shear stress and

is the inclination of the compression field.

( )( ) ( )inin

kipkip

db

VVv

vv

pu

u2.620.69.0

2.209.07.300 =

= = 0.84 ksi (LRFD Eq. 5.8.2.91)

c

u

f

v

= ksiksi

0.6

84.0= 0.140


23/37

4-23

Begin iterations by assuming a value for .

Trial 1: Assume = 24.

( )pspss

popspuu

v

u

x

AEAE2

fAcot)VV(5.0N5.0d

M

+

++= 0.001 (LRFD 5.8.3.4.21)

Mu= 1,660 kipft = 19,920 Kin (seeSummaryofDeadandLiveLoadEffects)

Nu= 0 no applied axial loads

fpo= 0.7 (270) = 189.0 ksi (LRFD C5.8.3.4.2)

Aps= area of prestressing steel on flexural tension side of the member, i.e., the

28 straight strands

Aps= 28 * 0.153 = 4.284 in2

( ) ( ) ( )( )

( )( )[ ]284.4500,282

0.189284.424cot2.207.3005.02.62

920,19+

=x 0.001

( ) 188,2444.174

094,1222

7.8090.3153.320 =

+=x = 0.00071

Because xis negative, use Eq. 5.8.3.4.23:

( )pspsscc

popspuu

v

u

xAEAEAE2

fAcot)VV(5.0N5.0d

M

++

++

= (LRFD 5.8.3.4.23)

Ac= Area of concrete on flexural tension side

= Area of girder below h/2 = 80.00/2 = 40.00 in

= (26)(6) + (4.5)(26+6)/2 + (29.5)(6) = 405 in2

( )( )[ ] 948,047,44.174

094,122405696,42

7.8090.3153.320 =

++

=x = 0.00004 = 0.04x103

From Table 5.8.3.4.21, with x= 0.04x103 and

cf

v

= 0.140, find The assumed value for was 24,

so convergence has been achieved.

To avoid the trial and error procedure, the term 0.5 cot in the equation for xcan be assumed

to be 1.0 (LRFD 5.8.3.4.2). This would result in x= 0.05x103and =24.2 and =2.78

per Table 5.8.3.4.21, the same as for the trial and error procedure.

With these values, the concrete contribution, Vc, can now be computed.

( ) ( )( )ininksiVC 2.620.60.678.20316.0= =80.3kip


24/37

4-24

3.2.1.5 Required Shear Reinforcement, Vs

Required Vs= Vu/ Vc Vp= 300.7 / 0.9 80.3 21.8 = 232 kip

Assuming vertical stirrups,

s

cotdfAV

vyvs

= (LRFD 5.8.3.34)

Compute Avon an in2

/ft basis (s = 12 in):

=

cotdf

V12A

vy

sv

( )( )( ) ( ) ( )

=24cot2.6260

23212

inksi

kipinAv =0.332in

2/ft

Using the CSA method,

Check minimum transverse reinforcement:

y

vcv

f

sbf0316.0A = (LRFD 5.8.2.5)

( ) ( )ksi

ininksiAv

60

1200.600.60316.0= = 0.093 in

2/ft < 0.332 and 0.325 in

2/ft O.K.

Check maximum stirrup spacing: (LRFD 5.8.2.72)

Vu= 300.7 kip > 0.125 fc bvdv= (0.125) (6.00) (6.00) (62.2) = 280 kip

Therefore, maximum stirrup spacing is the smaller of d v/4 = 15.6 in. or 12 in. (Governs)

Use #4 stirrups @ 12 in (Av,provd= 0.40 in2/ft)

3.2.2 Interface Shear Reinforcement

28.7350029

)12s(where,5.19)39(

51

)7501(

8.4

10x0.10-

5.0

xe

3-

=+=

==++

=

=+

++

=

s

xes

ppss

popspuu

v

u

s

s

AEAE

fAVVNd

M

( ) ( ) ( )

kip2.1849.1499.0/7.300/VVRequired

kip149.92.620.60.619.50316.0

us ===

==

c

C

V

ininksiV

( )( )( ) ( ) ( ) ftksi kipinAv /in0.3257.28cotin2.6260 2.18412 2==


25/37

4-25

In this example, the girder will be designed for interface shear at the initial

critical section for shear. In a full design, other sections along the length

of the girder would have to be designed as well.

The governing design equation is:

uini VV (LRFD Eqs. 5.8.4.11 and 5.8.4.12)

Vui= 300.7 kips includes all noncomposite and composite loads.

vvi

uui

db

Vv = (LRFD Eq. 5.8.4.21)

Where,

The width of the shear interfaceis equal to the width of the top flange of

the girder, which is 42.00 in. Therefore, bvi= 42.00 in.

dv= Distance between the centroid of the tension steel and the mid

thickness of the slab

= 72 15.83+ 4 = 60.17 in.

ksix

vui 119.017.6042

7.300==

kips/in.5.00.119(42)in.)bvV viuiui === 1(

(Note: Calculations are performed on per inch basis)

Assume that the top surface of the girder was intentionally roughened toan amplitude of 0.25 in and cleaned prior to placement of the deck

concrete. The requirement for intentional roughening of the top of the

girder should be indicated on the plans.

Compute the nominal interface shear resistance, Vni:

cyvfcvni PfAAcV ++= (LRFD Eq. 5.8.4.13)

where:

c = 0.100 ksi and = 1.000 for an intentionally roughened surface (LRFD 5.8.4.3)

Avf = area of shear reinforcement crossing the shear plane within Acv

= 0.40 in2/ 12 in. = 0.0333 in2/ in. (Vertical shear reinforcement provided is 2 # 4 @ 12

in.)

Acv = bvi(1 in.) = 42 in.

fy = 60 ksi (max) per LRFD 5.8.4.1

Pc = permanent net compressive force normal to the shear plane within Acv

= 0kips/in.Vni 2.6]60*0333.0[000.1)42(100.0 =+=

Check limits on Vni(LRFD 5.8.4.3):

VniK1 fcAcv= 0.3(4)(42) = 50.4 kips/in. (LRFD Eq. 5.8.4.14)

K2Acv= 1.8(42) = 75.6 kips/in. (LRFD Eq. 5.8.4.15)

Vni= 6.2 kips/in. [Governs]


26/37

4-26

Check minimum reinforcement requirement per 5.8.4.4:

The minimum reinforcement requirement is waived since the girder/slab interface is intentionally

roughened to an amplitude of 0.25 in., vui(0.119 ksi) is less than 0.210 ksi, and all vertical shearreinforcement will be extended across the interface and adequately anchored in the slab.

O.K.kips/in.Vkips/in.V uini 0.558.5)2.6(9.0 =>==

3.3 LongitudinalReinforcementRequirement

In this example, the longitudinal reinforcement requirement will be

checked at the inside edge of the bearing. The Specifications require that

this requirement must be satisfied at each section of the girder. Therefore, in a full design, other sections along the length of the girder

would also have to be checked.

3.3.1 Required Longitudinal Force

Required Longitudinal Force at a typical section:

Treqd=

+

+

cotVV5.0

VN5.0

d

Mps

uu

v

u (LRFD Eq. 5.8.3.51)

However, at the inside edge of bearing at the simplysupported ends,

Treqd=

cotVV5.0

Vps

u (LRFD 5.8.3.5)

where:

values of Vu, Vs, Vp, and may be the same as those for the section dvfrom the face of the support.

Vs= shear resistance provided by transverse reinforcement, not to exceed Vu/ .

=s

cotdfA vyv (Usefinalvaluesfromsheardesignabove) (LRFD Eq. C5.8.3.31)

=( )( )( ) ( )

in

inksiin

0.12

24cot2.626040.0 2 = 279.4 kip

Vu/ = 300.7 kip / 0.9 = 334.1 kip, so use the computed quantity for Vs.

Treqd= ( ) ( )

24cot2.204.2795.0

9.0

7.300= (334.1 139.7 20.2) cot (24)

= (174.2) (2.25) = 391 kip

3.3.2 Available Longitudinal Force

The force to resist Treqdmust be supplied by the reinforcement on the flexural tension side

of the member. In this case, the available reinforcement consists of the straight strands.

The available force that can be provided by these strands at the critical section for shear


27/37

4-27

must be determined considering the lack of full development due to the proximity to the

end of the girder.

The location at which T must be provided is where the failure crack assumed for thisanalysis, which radiates from inside face of the support, crosses the centroid of the

straight strands. The angle determined above during shear design at this location is

used here. The inside face of the support is 12 in from the end of the girder.

Figure5: Assumed Failure Crack and Location Where

Crack Crosses Straight Strands

The total effective prestress force for the straight strandsis:

Pes = Apsfpe= 4.284 in2(159.0 ksi) = 681 kip

The distance from the bottom of the girder to the centroid of these strands is:

dg = c.g. straight strands = 4.00 in

Measured from the end of the girder, the crack crosses the centroid of the straight strands

at:

x = + cotdgbl = 12 in + 4.00 in (cot 24) = 21.0 in

This location is within the transfer length l t,so the available stress is less than the

effective prestress force for the straight strands. The available prestress force, Tavail, at x istherefore computed assuming a linear variation in stress from the end of the girder to the

transfer length. The transfer length, l t, is 60 dbor 30 in. (LRFD 5.11.4.1.)

Tavail= Pes ( )tx l = 681 kip (21 in / 30 in) = 477 kip

Since Tavail= 477 kip > Treqd= 391 kip, the straight strands are adequateto resist the

required longitudinal force at this location and no additional reinforcement is required.

If the strands had not been adequate to resist the force, additional mild reinforcement

would have been added to provide the remainder of the required force. Alternatively, thestirrup spacing may be reduced to increase Vswhich in turn reduces Treqd.

3.4AnchorageZoneReinforcement:

3.4.1 Anchorage Zone Reinforcement

Article 5.10.10.1 requires that the factored bursting resistance of a pretensioned

anchorage zone be at least 4.0% of the total prestressing force. This resistance isprovided by vertical reinforcement close to the ends of pretensioned girders.

The factored bursting resistance is given by:


28/37

4-28

Pr= fs As (LRFD Eq. 5.10.10.11)

where:

Pr= (0.04) Po = (0.04) [(0.75) (270 ksi) (5.508 in2

)] = 44.61 kip

Note: ThetotaljackingforcepriortoanylossesisusedasthetotalprestressingforcePo

inthiscalculation:

fsis the working stress in the reinforcement, not to exceed 20 ksi

Solving for the required area of reinforcement, As:

( )ksikip

f

PA

s

rs

20

61.44== = 2.23 in2

Therefore, at least 2.23 in2

of vertical reinforcement must be placed within h/4 = 72 in / 4

= 18.0 in from the end of the member. Stirrups placed for vertical or interface shear can

also be used to satisfy this requirement since this reinforcement is only required to resistforces at release.

3.4.2 Confinement Reinforcement

In accordance with Article 5.10.10.2, confinement reinforcement not less than #3 bars at aspacing of not more than 6.0 in shall be placed within 1.5 d (say 1.5 h = 9.00 ft) from the

end of the girder. These bars shall be shaped to enclose the strands.

CONSPAN


29/37

4-29

Slide 1

Prestress Losses

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 2

2

Needed for

Stress checks

Camber & deflection

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 3

3

Components of P/S Loss

fpT = fpES + fpLT

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________


30/37

4-30

Slide 4

4

Components of Prestress Losses

Stress instrands

TimeStrand Prestress Decktensioning transfer placement

Anchorageseating loss

Jacking

Relaxation andtemperature losses

Creep, shrinkageand relaxation

Elastic gaindue to deck placement

A

BC

D

E

F

SIDL

H

IKElastic gain

due to SIDL

Elastic gain

due to LLJ

LL

G

Elastic shortening

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 5

5

P/S Girder Design ExampleStrands at Midspan

No. of

Strands

Dist. from

Bottom4 8 IN

8 6 IN

12 4 IN

12 2 IN

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 6

6

P/S Girder Design Example

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________


31/37

4-31

Slide 7

7

Calculation of fpES

ps

cigg

g

2

mgps

ggmg

2

mgpips

pES

E

EIA)AeI(A

AMe)AeI(fAf

++

+=

cgpci

pESp f

E

Ef =

ksix

fpES 1.18

28500

3905*894,545*767)767*38.32894,545(508.5

767*12*1122*28.32)767*38.32894,545(5.202508.5

2

2

=

++

+=

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 8

8

2005 LRFD Approximate

Long-Term Loss:

fpLT= 10.0(fpiAps / Ag) h st + 12.0 h st + fpR

h = 1.7 0.01H = 1.7 0.01(75) = 0.95

st = 5 / (1 + fci ) = 5 / (1 + 4.5) = 0.91

fpLT= 10.0 * (202.5*5.508/767) * 0.95 * 0.91+ 12.0 * 0.95 * 0.91 + 2.5

= 25.4 ksi

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

___________________________________

Slide 9

9

Long-Term P/S LossesDetailed Method

fpLT= (fpSR + fpCR + fpR1)id +

(fpSD + fpCD + fpR2 fpSS)df

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Slide 10

10

Revised Equations for:

Modulus of Elasticity

Unit Weight

Creep & Shrinkage

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Slide 11

11

MODULUS OF ELASTICITY, Ec

Ec = 33,000K1 (w)1.5

fc K1 is a function of local aggregates

for fc 5 ksi:

w = 0.140 kcf + 0.001 fc


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Slide 13

13

Creep & Shrinkage

Equations revised to account for highstrength concrete

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Slide 14

14

12SF-7-C

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

0 20 40 60 80 100 120 140 160 180

Time (days)

CreepCoefficient

C5

C6

C7

C8

ACI 209

New Eq.

High Strength vs. Conventional Concrete

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Slide 15

15

12SF-7-S

0

100

200

300

400

500

600

0 25 50 75 100 125

Time (days)

Shrinkage(microstrains)

S1

S2

S3

S4

ACI 209

New Eq.

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Slide 16

16

AASHTO LRFD 2005 Formulas

Creep Coefficient t = fhslatd kkkkk.901

Shrinkage Strain sh = fhsstd kkkk6

10480

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Slide 17

17

Correction Factors

hsk = H014.000.2 hc

= H008.056.1

'

ci

ff1

5k+=

HumidityHumidity

(Creep)

(Shrinkage)

)S/V(128.0448.1ks =

tdk = tf461

t'ci+

lak =118.0

it Loading AgeLoading Age (Creep)

Time DevelopmentTime Development

Concrete StrengthConcrete Strength

VolumeVolume--toto--Surface RatioSurface Ratio

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Slide 18

18

)1a2.4.5.9.5.Eq(KEf idpbidpSR =

( ) )1b2.4.5.9.5.Eq(Kt,tf

E

Ef

ididbcgp

ci

p

pCR =

)1a3.4.5.9.5.Eq(KEf dfpbdfpSD =

( ) ( )[ ] ( ) )1b3.4.5.9.5.Eq(0.0Kt,tf

E

EKt,tt,tf

E

Ef

dfdfbcd

c

p

dfidbifbcgp

ci

p

pCD +=

( )[ ] )1d3.4.5.9.5.Eq(t,t1Kf

E

Ef

dfbdfcdf

c

p

pSS +=

( )[ ] )2d3.4.5.9.5.Eq(

I

ee

A

1

t,t7.01

EAf

c

dpc

cdfd

cddddf

cdf

+

+

=

Loss Equations

ksifpR 2.11=

Initial to deck placement:

Deck placement to final:

ksifpR 2.12=

6.55 ksi

16.62 ksi

3.19 ksi 0.00 ksi

1.49 ksi

Total Long Term Loss = 27.3 ksi

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Slide 19

19

P/S Gains

Deck weight (6.28 ksi)Non-composite girder

SDL (1.62 ksi)Composite girder

0.8LL+I (6.23 ksi)Non-composite girder

Total Gain = 14.1 ksi

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Slide 20

20

Detailed vs. Approximate Method

DetailedMethod

ApproximateMethod

Elastic Loss 18.1 ksi 18.1 ksi

Long-Term Loss 27.3 ksi 25.4 ksi

Total Loss 45.4 ksi 43.5 ksi

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Slide 21

21

www.structuresprograms.unomaha.eduInput Sheet

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Slide 22

22

Material and Section Properties

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Slide 23

23

Prestress Losses and Concrete Stresses

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4. Precast Prestressed Concrete Girder Bridge - Design Example

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Transcript of 4. Precast Prestressed Concrete Girder Bridge - Design Example