Design For Restraint In PT Slabs - Post-Tensioning … Convention/S2-1... · Source: Ghali et al...

SDesign For Restraint In PT SlabsJonathan Hirsch, P.E.

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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.

(C) COPYRIGHT POST-TENSIONING INSTITUTE, ALL RIGHTS RESERVED of 74

Sources of building movements

• Deformations due to loads (post-tensioning)• Creepp• Shrinkage• Thermal

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Sources of restraint to movement

• Vertical supporting elements• Stiff or unstressed regions of the floorg• Temporary formwork• Soil

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Design for restraint – importance

• Prevent premature flexural cracking and excessive crack widths

• Prevent entire slab depth crack widths in regions of low bending

• Durability issues• Owner/occupant perception• Litigation

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Unrestrained concrete behavior - loads

• Concrete is frequently assumed to be a linear elastic material

• Assumed stiffness is based upon the secant modulus

• Strains are proportional to the level of stress on the specimen

5



Unrestrained concrete behavior - loads

5000Concrete Stress-Strain Design Curve

3000

4000

s (p

si) Ec

1000

2000

Stre

s

00 0.001 0.002 0.003 0.004

Strain

6

Strain



Unrestrained concrete behavior - shrinkage

• Types of shrinkage– Autogenous– Drying– Carbonation– Plastic

7




thth

t )()( εε = ushtsh t)(

35)( εε

+t35+8

Source: ACI Committee 209




−610780)( γε ×= shush610780)(

αψλ γγγγγγγγ ⋅⋅⋅⋅⋅⋅= csvscpsh

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75.025.1

ttcure

cp +−≈γ

01.040.17 tcure

−=+

λγ λ for 40 ≤ λ ≤ 80

)120(21

03.000.3v

−= λγ λ for 80 ≤ λ ≤ 100

)12.0exp(2.1svs ⋅−=γ

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1

1.2Fraction of Ultimate Shrinkage

0.6

0.8

1

ε h/ε h

0.2

0.4

0.6εsh/εshu

01 10 100 1000 10000

Time (days)Time (days)

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Unrestrained concrete behavior - creep

• Types of creep– Basic– Drying

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• Only linear for low stress levels• Creep is generally higher in tension than p g y g

compression• Creep strain is assumed to be linear factor of

strain due to instantaneous load

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t 6.0t )()( φφ = ut t)(

10)( 6.0 φφ

+

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γφ = 352)( γφ = cu 35.2)(

αψλ γγγγγγγγ ⋅⋅⋅⋅⋅⋅= csvslac

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= −)(25.1 118.0tlalaγ

⎤⎡

−= 0067.027.1 λγ λ For λ > 40

⎥⎦⎤

⎢⎣⎡ ⋅−⋅+= )54.0exp(13.11

32

sv

vsγ ⎥⎦⎢⎣3 s

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1

1.2Fraction of Ultimate Creep

0.6

0.8

1

φ/φu

0.2

0.4

φ φu

01 10 100 1000 10000

Time (days)

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Analysis assumptions

• Concrete is a homogeneous linear elastic material – presence of reinforcement is ignored

• Only axial stresses/strains are considered –flexural considerations are ignored

• Creep behavior is homogeneous linear elastic. Creep in tension and compression are

d t b id ti lassumed to be identical• Results of analyses can be superimposed

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Unrestrained concrete behavior

• For an instantaneous applied load, the elastic strain is proportional to the elastic modulus

• Creep strain is typically assumed to be a linear factor of the instantaneous strain

( ) )()(1)()( 0 ttttttt c εφσε ++( ) ),(),(1)(

),( 000

00 tttt

tEtt sh

c

cc εφε ++=

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Unrestrained concrete behavior

• For loads applied gradually over a period of time, the resulting strain is affected by– Varying concrete strength– Varying creep response due to application of loading

at different timesat different times– Rate of application of loading

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Unrestrained concrete behavior – graduallyUnrestrained concrete behavior – gradually applied loads

• Variable applied load, shape function ξ1

0 6

0.8

1

Δσ(t,t 0)

0.4

0.6

ctio

n of

Δ

0

0.2Fra

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ttttt0




( ) ),(),(1)()(

1),(),( 000 ttdtd

dE

tttt sh

t

c εττφτξσε +⋅+⋅⋅Δ= ∫ )(0

dEt c ττ∫

22

Source: Ghali et al




( ) ),(),(1)(

),(),( 000

0 tttttEtttt shc εχφσε ++

Δ=

)( 0tEc

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Source: Ghali et al



Calculation of aging coefficient, χ

( ))(

1),(1)()(

1)()(),( 0

0 ttdt

dd

EtttEtt

tc

φττφ

ττξ

τφχ −⋅+⋅⋅= ∫ ),()(),( 00 0

ttdEtt t c φττφ ∫

24

Source: Ghali et al




80

80

8.00

0),( ttt ≈χ 8.00

0 8.1),(

ttt

+χ

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26



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• Relaxation• Numerical procedure for calculation of p

relaxation curve and aging coefficient• “Concrete Structures – Stresses and

Deformation – Third Edition” A Ghali, R Favre, and M Elbadry

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Restrained concrete behavior –Restrained concrete behavior –instantaneous load with relaxation

( ) ( )),(1)(),(),(1

)()(),( 0

00

00 tt

tEtttt

tEtttc χφσφσε +

Δ++=

• If we assume that the external strain for the

)()( 00 tEtE cc

If we assume that the external strain for the entire time period is equal to the initial strain

)(tσ)()(),(

0

00 tE

tttc

cσε =

30

)( 0c



Restrained concrete behavior –Restrained concrete behavior –instantaneous load with relaxation

⋅−Δ

),()()( 00 ttttt φσσ

⎞⎛

+=Δ

),(1),(

0

000 tt

ttχφ

σ

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=)(1

),(1)(),( 000 tt

tttttχφφσσ ⎟

⎠⎜⎝ + ),(1 0ttχφ

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Restrained concrete behavior – gradualRestrained concrete behavior – gradual load with relaxation

( ) ),(),(1),(),( 000

0 tttttttt hεχφσε ++Δ

= ( ) ),(),(1)(

),( 000

0 tttttE

tt shc

c εχφε ++

• If we assume that the external strain is zero

0),( 0 =ttcε32



Restrained concrete behavior – gradualRestrained concrete behavior – gradual load with relaxation

)()(),( 0

0tEtt csh

φεσ ⋅−

=Δ),(1

),(0

0 ttχφ+

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Age adjusted effective modulus

)()(),( 0

0tEttE c

c φ=

),(1),(

00 ttc χφ+

34

Source: Ghali et al



Analysis procedurey p

1. Divide the analysis into time intervals and yloading types (instantaneous or gradual)

2. For each interval, calculate an age adjusted effective modulus for each concrete part

3. For the interval, determine the loads applied to each concrete part during the interval, including creep and shrinkage contribution

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Analysis procedurey p

4. Analyze the interval using the elastic modulus y gfor instantaneous loadings and the age adjusted effective modulus for gradually applied loads

5. Superimpose the results of each concrete part ith lt f i i t l l iwith results from previous interval analysis

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Examples – hand calculations of fullyExamples – hand calculations of fully restrained slab

• Case 1 – no pour strip• Case 2 – temporary stress strip, slab stressed p y p

at 3 days with 150 psi then stress strip cast• Case 3 – temporary pour strip, slab stressed at

3 days with 150 psi and pour strip cast at 28 days

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• For all examples, ultimate shrinkage assumed to be 0.0005. All cases assumed to be moist cured for 7 days.

• Ultimate creep factor assumed to be 2.0• εsh(t) and φ(t,t0) will use functions from ACI 209R-

92 presented earlier and χ(t,t0) will be calculated in d ith th i lifi d tiaccordance with the simplified equation.

• Ec at ages 3, 7, and 28 days assumed to be 3122, 3372 d 3823 k i ti l3372, and 3823 ksi respectively

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Examples – case 1 (no pour strip)

• Because structure is fully restrained, all prestress is diverted to supports. The only stresses are due to shrinkage

• Calculated shrinkage strain @ t = ∞ is 0.0005• Calculate χ(t,t0):

725.078.1

7)7,( 8.0

8.0

=+

≈tχ

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78.1 +



Examples – case 1 (no pour strip)

• Calculated stress due to shrinkage is:

337200050 ⋅ ksi688.00.2)725.0(1

33720005.0)( =+

⋅=∞σ

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Examples – case 2 (temporary stress stripExamples – case 2 (temporary stress strip cast at 3 days)

• Because strip is open when slab is stressed, slab receives entire prestress value of 0.150 ksi

• Calculated shrinkage strain @ t = ∞ is 0.0005• Calculate χ(t,t0):

572.0381

3)3,( 8.0

8.0

=+

≈tχ

725.0781

7)7,(

38.1

8.0

8.0

=+

≈

+

tχ

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78.1 8.0+




• Calculate φ(t,t0):6.0∞ 196.20.2098.1

10)3,(

60

6.0 ≈⋅⋅∞+

∞=∞φ

0.20.20.110

)7,( 6.0

6.0

≈⋅⋅∞+

∞=∞φ

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• Calculate stress relaxation due to creep:

1500creep

312200010550

0001055.0196.23122

150.0=⋅=ε

ksi146.0196.2)572.0(1

31220001055.0)( =+

⋅=∞σ

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• Calculate stress due to shrinkage:

337200050 ⋅ ksi688.00.2)725.0(1

33720005.0)( =+

⋅=∞σ

• Superimposing the stresses, the final stress is:

ksi684.0688.0146.0150.0)( =++−=∞σ

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Examples – case 3 (temporary pour stripExamples – case 3 (temporary pour strip cast at 28 days)

• Because strip is open when slab is stressed, slab receives entire prestress value of 0.150 ksi

• Calculated shrinkage strain @ t = ∞ is 0.0005 –0.0001875 = 0.0003125

• Calculate χ(t,t0):

80

889.0288.1

28)28,( 8.0

8.0

=+

≈tχ

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• Calculate φ(t,t0):6.0

196.20.2098.110

)3,( 6.0

6.0

=⋅⋅∞+

∞=∞φ

688.10.2844.010

)28,( 6.0

6.0

=⋅⋅∞+

∞=∞φ

896.00.2098.12510

25)3,28( 6.0

6.0

=⋅⋅+

=φ

46

2510 +




• Calculate stress relaxation due to creep:

1500creep

38230000620

000062.0)896.0196.2(3122

150.0=−⋅=ε

ksi095.0688.1)889.0(1

3823000062.0)( =+

⋅=∞σ

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• Calculate stress due to shrinkage:

382300031250 ⋅ ksi478.0688.1)889.0(1

38230003125.0)( =+

⋅=∞σ

• Superimposing the stresses, the final stress is:

ksi423.0478.0095.0150.0)( =++−=∞σ

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Case: Resulting Stress (tension positive):

1 (no pour strip) 0.688 ksi

2 (stress strip cast at 3 days) 0.684 ksi

3 (pour strip cast at 28 days) 0.423 ksi

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Examples – computer calculations ofExamples – computer calculations of partially restrained slab

• Use the same parameters as the hand worked problem, but in this case account for the flexibility of the supports, which we assume to be relatively stiff shear walls at each end of the buildingbuilding

• Building dimensions assumed to be 250’ x 75’ with shear wall I = 1500 ft4with shear wall I = 1500 ft4

• Structural model will consider axial slab forces

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• Simplified slab layout

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• Case 1 (no pour strip):• In this case, since the slab ends are not fully y

fixed not all the prestress will be diverted into the supports. In this case, we apply 1080 kips

t Thi lt i l b f f 204prestress. This results in a slab force of -204 kips, or -0.028 ksi (compression)Th lti t i i• The resulting creep strain is 0.028/3122(2.196)=0.0000197

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• Equivalent thermal load for creep = 0.0000197 / 0.0000055 = -3.6° F

• The shrinkage strain is 0.0005• Equivalent thermal load for shrinkage = 0.0005 /

0.0000055 = -91° F• Age adjusted effective modulus for creep:

ksiEc 13841962)5720(1

3122)3,( =+

=∞

53

196.2)572.0(1+




• Age adjusted effective modulus for shrinkage:

k iE 13763372)7( ksiEc 13760.2)725.0(1

)7,( =+

=∞

• Resulting stress due to creep (relaxation) = 0.024 ksi• Resulting stress due to shrinkage = 0 626 ksiResulting stress due to shrinkage 0.626 ksi

ksi622.0626.0024.0028.0)( =++−=∞σ

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• Case 2 (stress strip cast after 3 days):• In this case the slab receives the entire

prestress of -0.150 ksi• The resulting creep strain is

0.150/3122(2.196)=0.0001055• Equivalent thermal load for creep = 0.0001055 /

0.0000055 = -19° F

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• The shrinkage strain is 0.0005• Equivalent thermal load for shrinkage = 0.0005 / q g

0.0000055 = -91° F• Age adjusted effective modulus for creep:

ksiE 13843122)3,( ==∞ ksiEc 1384196.2)572.0(1

)3,(+

∞

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• Age adjusted effective modulus for shrinkage:

k iE 13763372)7( ksiEc 13760.2)725.0(1

)7,( =+

=∞

• Resulting stress due to creep (relaxation) = 0.131 ksi• Resulting stress due to shrinkage = 0 626 ksiResulting stress due to shrinkage 0.626 ksi

ksi606.0626.0131.0150.0)( =++−=∞σ

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• Case 3 (pour strip cast after 28 days):• In this case the slab receives the entire

prestress of -0.150 ksi• Only the creep occurring after the pour strip is

cast at 28 days relaxes the prestress. The resulting creep strain is 0.150/3122(2.196-0 896) 0 0000620.896)=0.000062

• Equivalent thermal load for creep = 0.000062 / 0 0000055 = 11 2° F0.0000055 = -11.2° F

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• The shrinkage strain is 0.0005• Equivalent thermal load for shrinkage = 0.0003125 / q g

0.0000055 = -56.8° F• Age adjusted effective modulus for creep and

shrinkage:

3823 ksiEc 1528688.1)889.0(1

3823)28,( =+

=∞

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• Resulting stress due to creep (relaxation) = 0.084 ksi• Resulting stress due to shrinkage = 0.429 ksig g

ksi363.0429.0084.0150.0)( =++−=∞σ )(

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Examples – comparison of sampleExamples – comparison of sample calculations

Case Hand calculations assuming fully

Computer calculations with realistic sized stiffg y

restrained shear walls

1 (no pour strip) 0.688 ksi 0.622 ksi

2 (3 day stress strip) 0.684 ksi 0.606 ksi

3 (28 day pour strip) 0.423 ksi 0.363 ksi

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Multistory Example

• Same basic parameters as previous examples• 10 story building, fixed basey g• No release strips

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Post-crack behavior

• Restraining force after first cracking is a function of the cracking stress and the reinforcement in the slabreinforcement in the slab

• A low reinforcing ratio (about < 0.003) can result in yielding of the steel upon first crackresult in yielding of the steel upon first crack, preventing formation of further cracks

• Gilbert 3 presents a method for estimating theGilbert presents a method for estimating the number of cracks and corresponding crack width for a given reinforcing ratiog g

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Conclusions

• Slabs of normal length (200’-250’) can become highly restrained by typically sized concrete walls placed at the ends of the slab

• In regular parts of multistory buildings the t i t bl i l i ifi trestraint problem is less significant

• Restraint is still a problem near rigid f d ti d f l ti lfoundations and areas of plan or vertical irregularity

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ConclusionsConclusions

• In restrained slabs, much of the initially applied , y ppprestress gets relaxed– “stress strips” do not significantly improve this

condition– Pour strips left open for 28 days or longer improve

this condition moderatelythis condition moderately• Shrinkage is the major contributing factor

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ConclusionsConclusions

• Consider ways to minimize restraint to yshrinkage– Longer moist cure times– Shrinkage compensating cement– Use pour strips

C l l h d i• Calculate the concrete stresses due to restraint• If cracking is expected, provide a reasonable

t f i f t (ACI 318 i l bamount of reinforcement (ACI 318 min. slab reinforcement is generally NOT sufficient!)

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References

1. ACI Committee 209, “Prediction of Creep, Shrinkage, and Temperature Effects in Concrete Structures”, ACI 209R-92, American Concrete Institute, 1992

2 Gh li A F R d Elb d M2. Ghali, A.; Favre, R.; and Elbadry, M., “Concrete Structures Stresses and Deformations Third Edition” Spon Press NewDeformations Third Edition , Spon Press, New York, NY, 2002, pp. 8-19

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References

3. Gilbert, Ian, “Shrinkage Cracking in Fully Restrained Concrete Members”, ACI Structural Journal, March-April 1992, Vol. 89, No. 2, pp. 141-149

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Jonathan Hirsch, P.E.

[email protected]

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Design For Restraint In PT Slabs - Post-Tensioning … Convention/S2-1... · Source: Ghali et al...

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Transcript of Design For Restraint In PT Slabs - Post-Tensioning … Convention/S2-1... · Source: Ghali et al...