Post on 09-Aug-2018
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.
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Sources of building movements
• Deformations due to loads (post-tensioning)• Creepp• Shrinkage• Thermal
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - shrinkage
• Types of shrinkage– Autogenous– Drying– Carbonation– Plastic
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - shrinkage
thth
t )()( εε = ushtsh t)(
35)( εε
+t35+8
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - shrinkage
−610780)( γε ×= shush610780)(
αψλ γγγγγγγγ ⋅⋅⋅⋅⋅⋅= csvscpsh
9
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - shrinkage
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
10
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - shrinkage
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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
• Types of creep– Basic– Drying
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
• 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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
t 6.0t )()( φφ = ut t)(
10)( 6.0 φφ
+
14
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
γφ = 352)( γφ = cu 35.2)(
αψλ γγγγγγγγ ⋅⋅⋅⋅⋅⋅= csvslac
15
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
= −)(25.1 118.0tlalaγ
⎤⎡
−= 0067.027.1 λγ λ For λ > 40
⎥⎦⎤
⎢⎣⎡ ⋅−⋅+= )54.0exp(13.11
32
sv
vsγ ⎥⎦⎢⎣3 s
16
Source: ACI Committee 209
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior - creep
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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
21
ttttt0
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior – graduallyUnrestrained concrete behavior – gradually applied loads
( ) ),(),(1)()(
1),(),( 000 ttdtd
dE
tttt sh
t
c εττφτξσε +⋅+⋅⋅Δ= ∫ )(0
dEt c ττ∫
22
Source: Ghali et al
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Unrestrained concrete behavior – graduallyUnrestrained concrete behavior – gradually applied loads
( ) ),(),(1)(
),(),( 000
0 tttttEtttt shc εχφσε ++
Δ=
)( 0tEc
23
Source: Ghali et al
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Calculation of aging coefficient, χ
80
80
8.00
0),( ttt ≈χ 8.00
0 8.1),(
ttt
+χ
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Calculation of aging coefficient, χ
• 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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Restrained concrete behavior – gradualRestrained concrete behavior – gradual load with relaxation
)()(),( 0
0tEtt csh
φεσ ⋅−
=Δ),(1
),(0
0 ttχφ+
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Age adjusted effective modulus
)()(),( 0
0tEttE c
c φ=
),(1),(
00 ttc χφ+
34
Source: Ghali et al
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – hand calculations of fullyExamples – hand calculations of fully restrained slab
• 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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
39
78.1 +
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
41
78.1 8.0+
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 2 (temporary stress stripExamples – case 2 (temporary stress strip cast at 3 days)
• 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
42
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 2 (temporary stress stripExamples – case 2 (temporary stress strip cast at 3 days)
• Calculate stress relaxation due to creep:
1500creep
312200010550
0001055.0196.23122
150.0=⋅=ε
ksi146.0196.2)572.0(1
31220001055.0)( =+
⋅=∞σ
43
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 2 (temporary stress stripExamples – case 2 (temporary stress strip cast at 3 days)
• 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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
45
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 3 (temporary pour stripExamples – case 3 (temporary pour strip cast at 28 days)
• 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 +
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 3 (temporary pour stripExamples – case 3 (temporary pour strip cast at 28 days)
• 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
47
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – case 3 (temporary pour stripExamples – case 3 (temporary pour strip cast at 28 days)
• 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
48
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – hand calculations of fullyExamples – hand calculations of fully restrained slab
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
49
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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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
50
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• Simplified slab layout
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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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+
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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
54
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
55
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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
56
2008 PTI TECHNICAL CONFERENCE SESSION 2 - DESIGN FOR RESTRAINT IN PT SLABS JONATHAN HIRSCH, P.E.
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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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Examples – computer calculations ofExamples – computer calculations of partially restrained slab
• 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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