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##### Prestressed Concrete Beam Camber – BT84

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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**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**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**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**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**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.**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.**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**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**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**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) Eci Kid = 1 . 1 + EpAps [1+ Age2pg][1 + 0.7Ψb(tf, ti)] EciAg Ig**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**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)**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**Prestressed Concrete Beam Camber – BT84**Downward 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**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**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**Prestressed Concrete Beam Camber – BT84**I = 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)**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**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**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**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**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)**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**Prestressed Concrete Beam Camber – BT84**ti = 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”**Prestressed Concrete Beam Camber – BT84**Compute 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+ Ace2pc][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**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**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**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**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**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**Prestressed Concrete Beam Camber – BT84**Kdf = 1 . 1 + EpAps [1+ Ace2pc][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**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**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**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**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**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**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**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**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**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**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**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**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**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**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