Bob_BT84 Camber Power Point

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The following is an example of calculations performed to determine the time dependent camber values and prestress losses for a BT84 precast concrete beam with a span length of 150 feet. The beam

has 56 strands including 28 deflected strands. The assumed beam spacing is 6’-0”. An 8” deck with 2” of build-up was assumed.

Prestressed Concrete Beam Camber – BT84

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Compute upward deflection at release of strand

Assume that at release that the modulus of elasticity is 0.70 times the final value.

EC = (0.70) 33000K1wC1.5 √f’C (AASHTO 5.4.2.4-1)

= (0.70)(33000)(1.0)(0.145)1.5√(7.00) = 3374 ksi

I = 710000 in4 A = 772 in4 yb = 42.37 yC = 4.68” ye = 22.37

w = 0.06925 k/in

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Compute initial stress loss

Total jacking force = (31 kips/strand)(56 strands) = 1736 kips

Start with 15 ksi loss in prestress at release

Strand stress at release = 31/0.153 – 15.0 = 188 ksi

Prestress force at release = (188)(0.153)(56) = 1611 kips

Prestress moment at midspan at release = (1611)(42.37 – 4.68)

= 60735 inch kips

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Maximum moment from beam weight = (0.06925)(1800)2/8 = 28046 inch kips

Net maximum moment = 60735 – 28046 = 32689 inch kips

fcgp = Concrete stress at strand cg = 1611 + (32689)(42.37 – 4.68) = 3.82 ksi 772 710000

Strain in concrete at strand cg = (3.82)/(3374) = 0.0011322

Stress loss in strand = (0.0011322)(28500) = 32.27 ksi

Recompute initial stress loss in strand assuming 30 ksi initial stress loss

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Prestressed Concrete Beam Camber – BT84Assume 30 ksi loss in prestress at release

Strand stress at release = 31/0.153 – 30.0 = 173 ksi

Prestress force at release = (173)(0.153)(56) = 1482 kips

Prestress moment at midspan at release = (1482)(42.37 – 4.68) = 55856 in. kips

Maximum moment from beam weight = (0.06925)(1800)2/8 = 28046 inch kips

Net maximum moment = 55856 – 28046 = 27810 inch kips

fcgp = Concrete stress at strand cg = 1482 + (27810)(42.37 – 4.68) = 3.39 ksi 772 710000

Strain in concrete at strand cg = (3.39)/(3374) = 0.001005

Stress loss in strand = (0.001005)(28500) = 28.64 ksi

Assume an initial loss of 29 ksi

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Use 174 ksi stress in strand, 1491 kips, and a strain value of 0.001005

Use the Moment-Area method to determine the upward deflection of the beam at release.

Compute the moment in the beam induced by the prestressed strands.

See the following drawings that are used in deflection calculations.

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Use the Moment Area method to determine the deflection at midspan from the prestress force at release. The deflection is equal to the moment of the area of the M/EI diagram between the end of the beam and midspan about the end support.

The moment of the area between midspan and the end of the beam about the end of the beam =

(720)(0.000012448)(720/2) + (360/2)(0.000023459)(720+360/4) + (0.000023459 – 0.000012448)(720/2)(2/3)(720) = 8.55”

This is the upward deflection caused by the load from the prestress strand not including the weight of the beam.

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The downward deflection of the beam from self weight = 5wL4

384EI

w = 0.06925 kips/inch, L = 1800 inches, E = 3374 ksi, I = 710000 in4

∆ = (5)(0.06925)(1800)4 = 3.95 inches (384)(3374)(710000)

The net upward deflection at midspan at release =

8.55-3.95 = 4.60 inches

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Compute the upward deflection 3 months after release.

The creep coefficient is determined from AASHTO formula 5.4.2.3.2-1,

Ψ(t, ti) = 1.9 ks khc kf ktd ti-0.118

k s = 1.45 – 0.13(V/S) ≥ 1.0

V/S is the volume to surface ratio = 772 = 2.67 288.8

ks = 1.45 – 0.13(2.67) = 1.10

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khc = 1.56 – 0.008H, H = 70 (From AASHTO Figure 5.4.2.3.3-1)

khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.625 1+f’c 1 + 7.0

ktd = t = 90 = 0.732 61 – 4f’c + t 61 – (4)(7.0) + 90

ti = 1.0 days

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Ψ(t, ti) = (1.90)(1.10)(1.0)(0.625)(0.732)(1.0)-0.118 = 0.956

Creep deflection = (0.956)(4.60) = 4.40 inches

Compute strand stress loss due to creep.

ΔfpCR = Ep fcgp Ψ(t, ti) Kid (5.9.5.4.2b-1) EciKid = 1 . 1 + EpAps [1+ Age2

pg][1 + 0.7Ψb(tf, ti)] EciAg Ig

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Ep = 28500 ksi

Aps = (56)(0.153) = 8.57 in2

Eci = 3374 ksi

Ag = 772 in2

epg = 37.69 in

Ig = 710000 in4

Ψb = 0.956

fcgp = 3.39 ksi

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Kid = 1 = 0.715 1 + (28500)(8.57) [1+ (772)(37.69)2][1 + (0.7)(0.956)] (3374)(772) 710000

ΔfpCR = (28500)(3.39)(0.956)(0.715) = 19.57 ksi (3374)

Compute strand stress loss from shrinkage

∆fpSR = εbidEpKid

εbid = ks khs kf ktd 0.48x10-3 (AASHTO 5.9.5.4.2a, 5.4.2.3.3-1)

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ks = 1.45 – 0.13(2.67) = 1.10

khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02

kf = 5 = 5 = 0.625 1+f’c 1 + 7.0

ktd = t = 90 = 0.732 61 – 4f’c + t 61 – (4)(7.0) + 90

εbid = (1.10)(1.02)(0.625)(0.732)(0.48x10-3) = 0.0002464

∆fpSR = (0.0002464)(28500)(0.715) = 5.02 ksi

Total prestress loss at 90 days from creep and shrinkage = 19.57 + 5.02 = 24.59 ksi

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Prestressed Concrete Beam Camber – BT84Downward deflection from creep and shrinkage loss = (8.55)(24.59/174) = 1.21 inches

Strand stress loss from relaxation (AASHTO C5.9.5.4.2c-1)= ∆fpR1 = fpt log(24t) [(fpt/fpy) -0.55] [1- 3(ΔfpSR + ΔfpCR) ]Kid

K’L log(24t1) fpt

fpt = stress after transfer = 174ksi

t = 90 days

K’L = 45

ti = 1.0 days

fpy = (0.90)(270) = 243 ksi

ΔfpSR = Stress loss in strand from shrinkage = 5.02 ksi

ΔfpCR = Stress loss in strand from creep = 19.57 ksi

Kid = 0.715

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∆fpR1 = (174) log [(24)(90)] [(174/243) – 0.55] [1 – 3(5.02 + 19.57)](0.715) = 0.639 ksi (45) log [(24)(1.0)] 174

Downward deflection due to relaxation in strand = (0.639)(8.55) = 0.03 inches (174)

Total downward deflection due to strand stress loss = 1.21 + 0.03 = 1.24”

Total upward deflection at 3 months = 4.60 + 4.40 – 1.24 = 7.76 inches

Compute downward deflection due to deck

∆ = 5wL4

384EI

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Assumed beam spacing = 6’-0”

Deck thickness = 8”

Assumed build-up = 2”

Sectional area of deck = (0.667)(6.0) + (0.167)(4.0) = 4.67 ft2

Deck weight per foot = (0.150)(4.67) = 0.700 kips/ft = 0.05833 kips/inch

L = 150 feet = 1800 inches

EC = 33000 K1 wc1.5 √f’c = (33000)(1.0)(0.145)1.5√7.00 = 4821 ksi

w = 0.0583 kips/inch

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Prestressed Concrete Beam Camber – BT84I = 710,000 in4

Δ = (5)(0.0583)(1800)4 = 2.33 inches (384)(4821)(710000)

Compute strand stress gain from the deck load

Moment from deck = (0.0583)(1800)2 = 23611 inch kips (8)

Concrete stress = (23611)(37.69) = 1.253 ksi (710000)

Strain in concrete = 1.253 = 0.0002599 4821

Stress gain in strand = (28500)(0.0002599) = 7.41 ksi

Upward deflection from strand stress gain = (8.55)(7.41) = 0.36 inches (174)

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Determine deflections and strand stress losses after deck is in place

Compute the transformed moment of inertia ( I ) for the beam with the deck in place.

Assume deck concrete strength = 4.00 ksi.

Ebeam = 4821 ksi

Edeck = 33000K1WC1.5 √f’C = (33000)(1.0)(0.145)1.5√(4.00) = 3644 ksi

n = Ebeam = 4821/3644 = 1.32 Edeck

Effective deck width = 6.00/1.32 = 4.55 feet = 54.6 inches

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Distance to composite centroid from bottom of beam =

(772)(42.37) + (54.60)(8.00)(88.00) = 58.86 inches 772 + (54.60)(8.00)

Composite Moment of Inertia = IC =

710,000 + (772)(58.86 – 42.37)2 + (54.60)(8.00)3 + (8.00)(54.60)(88.00 – 58.86)2 = 12 1,293,156 in4

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Total transformed area = 772 + (8)(54.6) = 1209 in2

Strand cg to composite girder cg = 58.86 – 4.68 = 54.18 inches

Compute creep deflection five years beyond release of prestress

The creep factor to be applied to deflection at release = Ψb(tf , ti) - Ψb(td , ti)

The creep factor to be used for deflection from deck = Ψb(tf , td)Kdf

Ψ(tf, ti) = 1.9 ks khc kf ktd ti-0.118

ks = 1.10

khc = 1.0

kf = 0.625

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Prestressed Concrete Beam Camber – BT84ktd = t = 1825 = 0.98

61 – 4f’c + t 61 – (4)(7.0) + 1825

ti = 1

Ψ(tf, ti) = (1.9)(1.10)(1.0)(0.625)(0.98)(1.0)-0.118 = 1.28

Ψ(td, ti) = 1.9 ks khc kf ktd ti-0.118

ks = 1.10

khc = 1.0

kk = 0.625

ktd = t = 90 = 0.73

61 – 4f’c + t 61 – (4)(7.0) + 90

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Prestressed Concrete Beam Camber – BT84ti = 1

Ψ(td, ti) = (1.9)(1.10)(1.0)(0.625)(0.73)(1.0)-0.118 = 0.95

Ψb(tf , ti) - Ψb(td , ti) = 1.28 – 0.95 = 0.33

Compute the creep applied to deck deflection at five years beyond release of prestress. Base this deflection on an elastic deflection assuming a full composite moment of inertia at time of application of deck load.

∆ = 5wL4 384EI

∆ = (5)(0.05833)(1800)4 = 1.28 inches (384)(4821)(1293156)

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Apply creep factor

Ψ(tf, td) = 1.9 ks khc kf ktd ti-0.118

ks = 1.45 – 0.13(V/S)

V = 773 in3/in, S = 289 in2/in, V/S = 773/289 = 2.67

ks = 1.45 – 0.13(2.67) = 1.10

khc = 1.0

kf = 0.625 ktd = t = 1825 = 0.982 61 – 4f’c + t 61 – (4)(7.0) + 1825

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Prestressed Concrete Beam Camber – BT84ti = 90 days

Ψ(tf, td) = (1.90)(1.10)(1.0)(0.625)(0.982)(90)-0.118 = 0.754

Creep deflection from deck = (0.754)(1.28) = 0.96 inches

Compute upward deflection of beam from creep applied to upward at release of prestress. As with deck deflection, base this deflection on an elastic deflection assuming a full composite moment of inertia at time of release.

Deflection at release based on non composite moment of inertia = 4.60”.

Deflection if beam were composite = (4.60)(710000) = 2.52” (1293156)

Creep deflection from release of prestress = (0.33)(2.52) = 0.83”

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Prestressed Concrete Beam Camber – BT84Compute downward deflection from strand stress loss due to creep and shrinkage.

From AASHTO 5.9.5.4.3a-1, the strand stress loss from shrinkage after deck placement =

∆fpSD = εbdfEpKdf

Kdf = 1 . 1 + EpAps [1+ Ace2

pc][1 + 0.7Ψb(tf, ti)] EciAc Ic

Ψb(tf, ti) = 1.9 ks khc kf ktd ti-0.118

ti = 90

k s = 1.10

khs = 1.02

kf = 0.625

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ktd = t = 1825 = 0.982 61 – 4f’c + t 61 – (4)(7.0) + 1825

Ψb(tf, ti) = (1.9)(1.1)(1.02)(0.625)(0.982)(90)-0.118 = 0.77

Kdf = 1 = 0.743 1 + (28500)(8.57) [1+ (1209)(54.18)2][1 + (0.7)(0.77)] (3375)(1209) (1293156)

εbdf = ks khs kf ktd 0.48x10-3 = (1.1)(1.02)(0.625)(0.982)(0.48x10-3) = 0.0003305

∆fpSD = (0.0003305)(28500)(0.743) = 7.00 ksi

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Compute strand stress loss from creep from deck pour to 5 years after release

∆fpCD = Ep fcgp[Ψb(tf, ti) - Ψb(td, ti)]Kdf + Ep ∆fcdΨb(tf, td)Kdf

Eci Ec

Ψb(td, ti) = (1.90)(1.10)(1.0)(0.625)(0.732)(1.0-0.118) = 0.956


ti = 1

ks = 1.0

khc = 1.0

kf = 0.625

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ktd = t = 1825 = 0.982 61 – 4f’c + t 61 – (4)(7.0) + 1825

Ψb(tf, ti) = (1.9)(1.10)(1.0)(0.625)(0.982)(1)-0.118 = 1.283

Ψb(tf, td) = 1.9 ks khc kf ktd ti-0.118

ktd = t = 1735 = 0.981 61 – 4f’c + t 61 – (4)(7.0) + 1735

Ψb(tf, td) = (1.9)(1.1)(1.0)(0.625)(0.981)(90)-0.118 = 0.754

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Ψb(tf, td) = (1.9)(1.1)(1.0)(0.625)(0.981)(90)-0.118 = 0.754

Ep = 28500 ksi

Eci = 3374 ksi

Ec = 4821 ksi

fcgp= 3.39 ksi

∆fcd = Change in concrete stress at strand cg due to losses between transfer and deck placement and stress loss from deck placement

Total strand losses between transfer and deck placement =

19.57 + 5.02 + 0.639 = 25.23 ksi

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Prestressed Concrete Beam Camber – BT84Concrete stress loss from prestress loss =

25.23 + (25.23)(0.153)(56)(42.37-4.68)(42.37-4.68) = 0.465 ksi 772 710000

Stress loss in concrete at strand cg from deck placement = My I

M = wL2 = (0.05833)(1800)2 = 23624 inch kips 8 8

y = 37.69 inches, I = 710000 in4

Stress loss in concrete from deck = (23624)(37.69) = 1.25 ksi (710000)

fcd = 0.465 + 1.25 = 1.71 ksi

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Kdf = 1 . 1 + EpAps [1+ Ace2


Kdf = 1 = 0.701 1 + (28500)(8.57) [1+ (1209)(54.18)2][1 + (0.7)(1.283)] (3374)(1209) 1293156

∆fpCD = (28500) (3.39)[1.283 – 0.956](0.701) + (28500) (1.71)(0.754)(0.701) = 11.91 ksi (3374) (4821)

Total stress loss from deck poor to 5 years = 7.00 + 11.91 = 18.91 ksi

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On the non composite section, this stress loss would result in the following deflection:

∆ = (18.91)(8.55) = 0.93 inches 174

With the composite section, ∆ = (0.93)(710000) = 0.51 inches (1293156)

The total upward deflection between the time of the deck pour and 5 years = -0.96 + 0.83 -0.51 = -0.64 inches

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Compute shortening 2 weeks after transfer of prestress

Compute creep coefficient at 2 weeks (14 days)

Ψ(t, ti) = 1.9 ks khc kf ktd ti-0.118

ks = 1.45 – 0.13(2.67) = 1.10

khc = 1.56 – 0.008H, H = 70 (From AASHTO Figure 5.4.2.3.3-1)

khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.625 1+f’c 1 + 7.0

ktd = t = 14 = 0.30 61 – 4f’c + t 61 – (4)(7.0) + 14

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Prestressed Concrete Beam Camber – BT84ti = 1.0

Ψ(t, ti) = (1.90)(1.10)(1.0)(0.625)(0.30)(1.00)-0.118 = 0.39

Shortening at release = (1491)(1800) = 1.03 inches (772)(3374)

Shortening from creep = (1.03)(0.39) = 0.40 inches

Compute shrinkage coefficient at 2 weeks (14 days)

εsh = ks khs kf ktd 0.48x10-3

ks = 1.45 – 0.13(2.67) = 1.10

khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02

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kf = 5 = 5 = 0.625 1+f’c 1 + 7.0

ktd = t = 14 = 0.30 61 – 4f’c + t 61 – (4)(7.0) + 14

εsh = (1.10)(1.02)(0.625)(0.30)(0.48x10-3) = 0.00010098

Shortening from shrinkage = (1800)(0.00010098) = 0.18 inches

Neglect effects of strand relaxation

Total shortening at 2 weeks = 1.03 + 0.40 + 0.18 = 1.61 inches

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Compute total stress loss at 27 years

The total loss in strands up to the time of deck pour (90 days) =

29 + 19.57 + 5.02 + 0.64 = 54.23 ksi

Strand stress gain from deck pour = 7.41 ksi

Strand stress loss after deck pour = 54.23 – 7.41 = 46.82 ksi

Strand stress loss due to shrinkage between time of deck pour and 27 years =

∆fpSD = εbdfEpKdf

Kdf = 1 . 1 + EpAps [1+ Ace2


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ks = 1.45 – 0.13(V/S)

V = 773 in3/in, S = 289 in2/in, V/S = 773/289 = 2.67

ks = 1.45 – 0.13(2.67) = 1.10

khs = 1.02

kf = 0.625 ktd = t = 9855 = 1.00 61 – 4f’c + t 61 – (4)(7.0) + 9855

ti = 90.0

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Ψb(tf, ti) = (1.90)(1.10)(1.02)(0.625)(1.00)(1)-0.118 = 1.33

Kdf = 1 = 0.805 1 + (28500)(8.57) [1+ (1209)(54.18)2][1 + (0.7)(0.78)] (4821)(1209) 1293156

εbdf = ks khs kf ktd 0.48x10-3

εbdf = (1.1)(1.02)(0.625)(1.00)(0.48x10-3) = 0.000336

∆fpSD = (0.000336)(28500)(0.805) = 7.71 ksi

Compute loss in strand from creep from time of deck pour to 27 years

∆fpCD = Ep fcgp[Ψb(tf, ti) - Ψb(td, ti)]Kdf + Ep ∆fcdΨb(tf, td)Kdf

Eci Ec

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Ep = 28500 ksi

Eci = 4821 ksi

fcgp = 3.39 ksi


Ψb(tf, ti) = (1.90)(1.10)(1.02)(0.625)(1.00)(1)-0.118 = 1.33

Ψb(td, ti) = 1.9 ks khc kf ktd ti-0.118

ktd = t = 90 = 0.73

61 – 4f’c + t 61 – (4)(7.0) + 90

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Ψb(td, ti) = (1.90)(1.1)(1.0)(0.625)(0.73)(1.0)-0.118 = 0.95


ti = 1

ktd = t = 9765 = 1.00 61 – 4f’c + t 61 – (4)(7.0) + 9765

Ψb(tf, td) = (1.90)(1.1)(1.0)(0.625)(1.00)(1)-0.118 = 1.31

Compute ∆fcd

Strand stress loss between time of transfer and deck pour = 46.82 ksi

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Prestress force loss = (46.82)(56)(0.153) = 401 kips

Moment loss = (401)(37.69) = 15114 inch kips

Stress loss in concrete = 401 + (15114)(37.69) = 1.32 ksi 772 (710000)

Moment from deck = 23611 inch kips

Concrete stress loss at strand cg = (23611)(37.69) = 1.25 ksi (710000)

∆fcd = 1.32 + 1.25 = 2.57 ksi

∆fpCD = Ep fcgp[Ψb(tf, ti) - Ψb(td, ti)]Kdf + Ep ∆fcdΨb(tf, td)Kdf Eci Ec

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∆fpCD = (28500)(3.39)[1.33 – 0.95](0.777) + (28500)(2.57)(1.31)(0.805) = 24.78 ksi (3374) (4821)

Total prestress loss = 46.82 + 7.71 + 24.78 = 79.31 ksi

Compute strand stress gain from deck shrinkage

∆fpSS = Ep ∆fcdf Kdf [1+0.7Ψb(tf , td)] Ec

∆fcdf = εddf Ad Ecd ( 1 - epc ed ) [1+0.7 Ψd (tf, td)] Ac Ic

εddf = ks khs kf ktd 0.48x10-3

ks = 1.45 – 0.13(V/S) ≥ 1.0

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V = (72.00)(8.00) = 576.0 in3/in

S = (72.00)(2) + (8.00)(2) = 160.0 in2/in

V/S = 436.80/125.20 = 3.60

ks = 1.45 – 0.13(3.60) = 0.982

khs = 1.02

kf = 5 = 5 = 1.0 1+ f’ci 1 + 4.0

ktd = t = 9765 = 0.67 61 – 4f’c + t 61 – (4)(4.0) + 9765

εddf = (0.982)(1.02)(1.0)(1.0)(0.48x10-3) = 0.000478

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Ad = (72.00)(8.00) = 576 in2

Ecd = 33000K1WC1.5 √f’C = (33000)(1.0)(0.145)1.5√(4.00) = 3644 ksi


Ψb(tf, td) = (1.9)(1.1)(1.0)(0.625)(1.0)(1.0)-0.118 = 1.31

Ψd (tf, td) = (1.90)(0.982)(1.2)(1.0)(1.0)(1.0)-0.118 = 1.90

Ac = 1209 in2

Ic = 1293156 in4

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epc = 54.18 inches

ed = 84 + 8/2 – 58.86 = 29.14 inches

∆fcdf = (0.0004704)(576)(3644)( 1 - (54.18)(29.14) ) = -0.167 [1+(0.7)(1.90)] (1209) (1293156)

∆fpSS = 28500 (-0.167)(0.805) [1+(0.7)(1.31)] = -1.52 ksi 4821

Total Stress Loss @ 27 Years = 79.31-1.52 = 77.79 ksi

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