pretensioned beam example

8/13/2019 Pretensioned Beam Example

1/15

CE437/537,Spring2011 PretensionedBeamExample 1/15

Pretensionedbeamsaretypicallymanufacturedbyavendorinaprestressingyard. Structuralengineers

selectappropriatebeams(forexamplehollowcoreslabsectionsanddoubleTbeams)forbuildingsfrom

loadtablesprovidedbythevendor. Pretensionedbridgegirders(e.g.AASHTOandbulbTgirders)can

bedesignedbythevendorusingspecialcomputersoftware.

Studentscangainanunderstandingofthebehaviorofpretensionedbeamsbyanalyzingtheresponseof

atypical

pretensioned

beam

at

each

stage

of

its

life.

Example: SelectapretensionedDoubleTbeamfromthePCImanualandcheckitagainstcriteriainACI

31808. Span=52ft,SDL=0,LL=60psf.

1. Selectashapeandprestressinglayout. FromthePCIloadtableshowninFigure1,selecta10DT26with10diameter270ksilowrelaxationstrandswithoneharppoint(atmidspan).

Figure1. SpanloadtablefromPCIManual(5th

Edition)for10DT26.


2/15


Usethefollowingsectionandmaterialproperties. Assumethatthetendonsarejackedto70%oftheir

tensilestrength. Alsoassumethatattransfer,10%ofthestressislostduetoseatingoftendon

anchorages. Calculatethetendoneccentricitiesatthecriticalsections(transferpoint=strand

developmentlengthfromendofbeam,0.4span,andmidspan).

sp a nlength L 52 ft SDL 0 psf

LL 60 psf Concrete:

SectionProperties Mat'lProperties

Shape 10DT26 f'c 5,000 psi strength@28days

A 689 in2

1 0.80

bf 120 in UW 150 pcf unitweight

tf 4 in SW 718 plf selfweight

I 30,716 in4

yt 5.71 in f'ci/f'c 80% strengthlevel at transfer

yb 20.29 in f'ci 4,000 psi strengthattransfer

sectionmodulus St 5,379 in3

=I/yt Ec 4287 ksi mod.ofelasticity

Sb 1,514 in3

=I/yb Eci 3834 ksi

=0.033*UW^1.5*SQRT(f'ci)

SteelStrands:SectionProperties Mat'lProperties

numberofstrands Nstrands 10 fpu 270 ksi tensilestrength

stranddiameter ps 0.5 in Eps 28,500 ksi modulus

area ofstrand Astrand 0.153 in2

pu 0.045 max.rec'dstrain

area ofal l strands Aps 1.53 in2

frombeambottom ys_end 10 in

ys_mid 3 in

Jacking&Release

ja cki ng stress level fjacking/fpu 75%

seatingloss ftrans/fjacking 90%

fpo 182 ksi =f pu*fjacking/fpu*ftrans/fjacking

tress force a ttransfer Po 279 k =fpo*Aps

developmentlength Ldevel 30.38 in =fpo*ps/3

TendonProfile

Beam

e n d ra n sf e r

Pt 0.4

S pa n Mi d

Spaneccentricity e 10.29 10.97 15.89 17.29 in =yb ys_mid

Moments

TransferPt 0.4Span MidSpan

x/L 0.0487 0.400 0.500

MCoef 0.0232 0.120 0.125 =0.5*(x/L x/L^2)

omentdu etose l f wt MSW

44.9 233 243 kft =SW/1000*L^2*MCoef

mentdu etolive load ML 37.6 194.7 203 kft =LL/1000*bf/12*L^2*MCoef


3/15


2. Calculatethelossofprestressduetoelasticshortening,creepandshrinkageoftheconcreteandrelaxationoftheprestressingstrands. SeePrestressLossesontheclasswebsiteforan

explanationoftheconcretestressesduetoprestressing.

3. Calculatethestressesintheconcreteattransferandatservice. TheallowablestressesfromACI31808areshownonFigureA2.

LossofPrestress (@0.4L)PSforcea ttransfer P

o 279 k

e 15.89 in

fci_CGS 1.251 ksi =Po/A + Po*e^2/I MSW

*12*e/I

elas ticshortening ES 9.30 ksi =fci_CGS/Eci*Eps

creep CR 16.64 ksi =2*ES*Eci/Ec

l tosurfacearea ratio V/S 2.05

relativehumidity RH 75 %

shrinkage SH 5.12 ksi =0.0000082*Eps*(1 0.06*V/S)*(100 RH)

C 1.0

relaxation RE 3.76 ksi =f'c/1000 0.04*(ES+CR+SH)*C

Total_Loss 34.8 ksi

fpe 168 ksi =fpu*fs_jacking/fpu Total_Loss

ectiveprestressforce Pe 257 k =fpe*Aps

StressesatTransfer(allstressesinpsi)Transfer

Pt 0.4

Span Mid

Span

Allowable

concrete stressesat: compression tension

top ofbeam ft 64 2800 379

bottomofbeam fb 2069 =0.7*f'ci =6*SQRT(f'ci)

ft 101 50 2400 190

fb 1486 1666 =0.6*f'ci =3*SQRT(f'ci)

where:

ft =(Po/A+Po*e/St MSW

*12/St)*1000

fb =(Po/A

Po*e/Sb

+

M

SW

*12/Sb)*

1000


4/15


4. Checktheflexurestrengthunderoverload(atultimateconditions).Thestressstraingraphforprestressingstrands(seeFigureA3)isnotbilinear,asassumedfor

rebar. Thestressinthestrandscanbecalculatedasafunctionofthetotalstrainintheprestressing

usingtheequationsatthebottomofthefigure.

The

total

strain

in

the

tendons

is

the

sum

of

the

strain

due

to

the

effective

prestress

force

(Pe)

plus

thestrainintheconcreteattheCGSduetothefailureloads(seeFigureA4). ThestrainattheCGS

duetothefailureloadsismosteasilycalculatedbyfirstcalculatingthestrainrequiredto

decompresstheconcrete,thencalculatingthetensilestrainintheconcreteatultimate(similarto

anormallyreinforcedconcretebeam).

Theinternalforcesmustbecalculatediteratively,sincetheforceintheprestressingisafunctionof

thestrainintheprestressing,andthestraindistributionisafunctionoftheinternalforces. When

checkingtheflexurestrengthusinghandcalculations,itsconvenienttostartwithanassumedvalue

ofthestressintheprestressingthatisclosetotheultimatetendonstrength,saywithin5ksi.

Theavailableflexurestrength,Mn,mustbegreaterthanthemomentduetofactoredloads,Mu,andmustbegreaterthan1.2xthecrackingmoment,Mcr. Thislaststipulationistoensureaductile

failure: iftheflexurestrength(Mn whichisbasedontheassumptionthattheconcreteinthetensilezonehascracked)islessthantheuncrackedstrengthofthebeam,thenwhenthe

overloadedbeamdoescrackitwillfailsuddenly.

StressesatService(allstressesinpsi) Allowable

TransferPt 0.4Span MidSpan compression tension

cretestresses du e to:

sustainedloads ftsustained

51 134 89 2250 849

=0.45*f'c =12*SQRT(f'c)

a l l service loads ftall

33 568 541 3000 849

=0.6*f'c =12*SQRT(f'c)

a l l service loads fball

1578 324 228 3000 849

=0.6*f'c =12*SQRT(f'c)

where:

ftsustained

=(Pe/A+Pe*e/St MSW

*12/St)*1000

ftall

=(Pe/A

+

Pe*e/St

M

SW

*12/St

ML

*12/St)*

1000

fball

=(Pe/A Pe*e/Sb+MSW

*12/Sb+ML*12/Sb)*1000


5/15


5. CheckDeflectionsTheengineermustcalculatethecamberofthebeamwhenitiserectedandthelongtermcamber

ofthebeamsothatthefinishedstructureperformsasintended. Deflectionsduetoliveloadsare

alsochecked

against

the

ACI

allowable

deflections

listed

at

the

bottom

of

Figure

A

2.

FlexureStrength0.4Span MidSpan

e 15.89 17.29 in

du e toprestressing sPe

0.00588 0.0058836 =Pe/(Aps*Eps)

decompression

i nconcrete a tCGS CGSPe

0.00058 0.00067 =Pe/(A *Ec) Pe*e^2/(I*Ec)

effective depthtoPS dp 21.60 23.00 in =yt+e

depthto neutral axis c 1.01 1.01 in mustbe


6/15


DeflectionduetoPrestressing. Equationstocalculatethedeflectionduetoprestressingcanbe

derivedfromthemomentdistributionscausedbytheprestressingforces.

= dxEIM

PS

Forabeam

with

asingle

depression

point,

the

moment

diagram

due

to

prestressing

is

as

shown

below:

Doubleintegrationoftheequationabove:

+= dxLx

ePdxePIE oendoci2/

')(

12

'

8

22L

IE

ePL

IE

eP

ci

o

ci

endomidspan +=

eend

emide

Poeend

Poe

TendonProfile

MomentsduetoPS

x

L/2 L/2


7/15


CrackedMomentofInertia. Ifthebeamwillcrackunderserviceloads,thenthecrackedmomentof

intertia(Icr)mustbecalculated. Thetransformedsectionshownbelowisconstructedinwhichthe

areaoftheprestressingismultipliedbythemodularratioofsteeltoconcrete,n=Eps/Ec. the

concretebelowtheneutralaxis(NA)isneglectedbecauseitisintensionandcracked. Inthefigure

below,K=curvature

=strain

/(distance

from

NA).

Thefirsttaskistocalculatethelocationoftheneutralaxis,xinchesbelowthetopoftheflange,

forwhichthecompressionforcesarebalancedbythetensionforces.

02

)(2

1

)(2

1

)(

,2

1

2

2

__

_

=+

=

=

==

==

=

=

ppspsf

ppsf

cppsfc

cppspsps

cctopctopc

pspsftopc

dAnxAnx

b

xdAnxb

EnxdKAxbExK

EnxdKEf

ExKEf

fAxbf

TC

Usingthequadraticequationtosolveforxgives(andlettingnAps=A,bf=b,anddp=d)

bdAbAAxx '2'',

2

21 +=

Thecrackedmomentofinertiaisthen

23

)('3

xdAxb

Icr +=

bf

x

nAps

dp

flange

NA

strains concrete

stress

K


8/15


EffectiveMomentofInteria. Thebeammaycracknearmidspan(at0.4Lforbeamswithasingle

depressionpoint)butitwillnotcrackunderserviceloadsoverthewholelength. Equation98inACI

31808canbeusedtocalculateaweightedaverageofthecrackedmomentofinertiaandthe

uncracked(grossmomentofinertia,Ig)dependingontherelativemagnitudeofthecracking

momentMcr andthemomentduetoappliedloads,Ma. Theeffectivemomentofinertiaforthe

entire

span,

Ie

is

cra

crg

a

cre I

M

MI

M

MI

+

=

33

1

Thesituationiscomplicatedbythepresenceofprestressingforces. Forpretensionedbeams,

Ma=theliveloadmoment

Mcr=theportionoftheliveloadmomentnecessarytocausecracking

Crackingis

imminent

when

at

the

bottom

of

the

beam,

the

stresses

due

to

dead

load

plus

apercent

oftheliveloadequalthedecompressionstressplusthetensilestrengthoftheconcrete

riondecompress

bcrackingcausetoL

bDb ffff +=+

%

Dbr

Pb

crackingcausetoLb ffff

e +=% , wherethedecompressionstrain= +vestrainduetoPe

Writingtheequationsaboveintermsofmoments

bLb

bDbr

Pb

a

cr

Sf

Sfff

M

M e )( +

=

Definethetotalstressatthebottomofthebeamas

Lb

Db

Pb

Tb ffff

e ++=

then

Db

Pb

Tb

Lb ffff

e =

and

Lb

rT

b

a

cr

Lb

rT

bLb

a

cr

f

ff

M

M

f

fff

M

M =

+= 1,


9/15


Toillustratethecalculationprocedureforabeamthatcracksunderliveload,increasetheliveload

to80psfforthisexample.

Deflection

ImmediateDeflectionsduePSandSW:

eend 10.29 in

emid 17.29 in

e' 7.00 in =emid eend

deflection a ttransfer transf 1.72 i n

=Po*eend*(L*12)^2/(8*Eci*I)+ Po*e'*(L*12)^2/(12*Eci*I)

lection du e tos e l f wt SW

1.00 in =5*SW/1000*L^4*1728/(384*Eci*I)

LL

Icr:

n 6.65 =Eps/Ec

A 1.53 in2

=Aps

nA 10.17 in2

=n*A

d 21.60 in =dp

b 120.00 in =bf

sqrt 229.9 in2

=SQRT(nA^2+2*b*nA*d)

x1 1.83 in =

(

nA

+

sqrt

)

/

bx2 2.00 in =(nA sqrt)/b

fromtop fibertoNA x 1.83 in

i a fo rcrackedsection Icr 4,221 in4

=b*x^3/3 + nA *(d x)^2

Ieff:

s du e a ll l oads a t0.4L fbT

838 psi

fbL

2058 psi =ML*12000/Sb

Mcr/Ma 0.850 =1 (fbT fr)/fb

L

Ieff 20,508 in4

=(Mcr/Ma)^3

*

I

+(1

(Mcr/Ma)^3)

*

Icr

deflection du e toLL LL

1.50 in =5*LL/1000*bf/12*L^4*1728/(384*Ec*Ieff)

owable deflectdu e LL LL

max 1.73 in =L*12/360


10/15


11/15


FigureA1. Stressesinconcreteduetopretensioning.


12/15


PrestressingSteel: ThefollowingcriteriaarespecifiedbyACIfortheprestressingsteel(Section18.5.1,

pg287):

Maxstressduetojackingforce=min(0.94fpy,0.80fpu)

Maxstressattransfer=min(0.82fpy ,0.74fpu)

Stage DesignCriteria

1. Concrete

stressesat

transferofPT

forceto

concrete

(ACI18.4.1,

pg284)

atends elsewhere

maxtension '6ic

f '3ic

f

maxcomp.

'7.0ic

f

'6.0ic

f

2. Concrete

stressesunder

serviceloads

(ACITable

R18.3.3,

pg284)

Sustained

loadsAllloads

maxtension '12ic

f

maxcomp. '45.0ic

f

'60.0ic

f

3. Flexure

strength) 290)pg18.8.2,(ACIcrn

un

MM

MM

2.1

4. Deflections

(ACITable

9.5b,pg124)

240max

360max

L

L

erectionafter

LL

=

=

FigureA2. RelevantdesigncriteriainACI31808

PP

wSW+SDL+LL

PP

wSW


13/15


FigureA3. Stressvsstrainforprestressingtendon.


14/15


FigureA4. Straininprestressingsteelatultimateflexurestrengthofbeam

straininsteelduetodecompressing

theconcreteattheCGS:

straininsteelduePe:

+

PePe

psps

ePs

EA

Pe =

1. Aftertransfer&losses

(butnogravityloads)2. Gravityoverloadsareapplied

tobeamuntilfailure

straindistribution

in

concreteduePe

dp

straindistribution

in

concreteduetooverloads

=

cc

e

cc

eP

EI

eP

EA

Pe

CGS

2

+=

strainin

theCGSg

Pe

CG


15/15


FigureA5. Deflectionmultipliersforestimatinglongtermdeflections.

pretensioned beam example

Documents