1 the following is an example of calculations performed to determine the time dependent camber...
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1
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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 – BT84
Assume 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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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
12
Prestressed Concrete Beam Camber – BT84
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+ Age2pg][1 + 0.7Ψb(tf, ti)]
EciAg Ig
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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
∆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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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 – BT84
ktd = 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 – BT84
ti = 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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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
Ψb(tf, ti) = 1.9 ks khc kf ktd ti-0.118
ti = 1
ks = 1.0
khc = 1.0
kf = 0.625
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Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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
32
Prestressed Concrete Beam Camber – BT84
Concrete 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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Prestressed Concrete Beam Camber – BT84
Kdf = 1 . 1 + EpAps [1+ Ace2
pc][1 + 0.7Ψb(tf, ti)] EciAc Ic
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
34
Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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
36
Prestressed Concrete Beam Camber – BT84
ti = 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
37
Prestressed Concrete Beam Camber – BT84
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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Prestressed Concrete Beam Camber – BT84
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+ Ace2pc][1 + 0.7Ψb(tf, ti)]
EciAc Ic
39
Prestressed Concrete Beam Camber – BT84
Ψb(tf, ti) = 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
khs = 1.02
kf = 0.625
ktd = t = 9855 = 1.00
61 – 4f’c + t 61 – (4)(7.0) + 9855
ti = 90.0
40
Prestressed Concrete Beam Camber – BT84
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
41
Prestressed Concrete Beam Camber – BT84
Ep = 28500 ksi
Eci = 4821 ksi
fcgp = 3.39 ksi
Ψb(tf, ti) = 1.9 ks khc kf ktd ti-0.118
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
42
Prestressed Concrete Beam Camber – BT84
Ψb(td, ti) = (1.90)(1.1)(1.0)(0.625)(0.73)(1.0)-0.118 = 0.95
Ψb(tf, td) = 1.9 ks khc kf ktd ti-0.118
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
43
Prestressed Concrete Beam Camber – BT84
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
44
Prestressed Concrete Beam Camber – BT84
∆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
45
Prestressed Concrete Beam Camber – BT84
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
46
Prestressed Concrete Beam Camber – BT84
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 ks khc kf ktd ti-0.118
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
47
Prestressed Concrete Beam Camber – BT84
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
48
Prestressed Concrete Beam Camber – BT84