Prestressed Concrete Beam Camber 26 Slab The following

Prestressed Concrete Beam Camber – 26” Slab The following is an example of calculations performed to determine the time dependent camber values for 26 inch precast concrete slab with a span length of 65 feet. The slab has 38 strands with properties as indicated on the Prestressed Slab Design Sheet. 1

Prestressed Concrete Beam Camber – 26” Slab Compute upward deflection at release of strand Assume at this point that the modulus of elasticity is 0. 70 times the final value. ___ EC = (0. 70) 33000 K 1 WC 1. 5 √f’C (AASHTO 5. 4. 2. 4 -1) _____ = (0. 70)(33000)(1. 0)(0. 145)1. 5√(5. 50) = 2991 ksi ES = 28500 (AASHTO 5. 4. 4. 2) I = 63596 in 4 A = 851 in 2 y. C = 5. 36” w = 0. 916 k/ft = 0. 0763 k/in 2

Prestressed Concrete Beam Camber – 26” Slab Compute initial stress loss Total jacking force = (31 kips/strand)(38 strands) = 1178 kips Prestress Moment = (1178)(13. 00 - 5. 36) = 9000 inch kips Maximum moment from beam weight = (0. 0763)(780)2 = 5803 inch kips 8 Net maximum moment = 9000 – 5803 = 3197 inch kips fcgp = Concrete stress at strand cg = 1178 + (3197)(13. 00 - 5. 36) = 1. 77 ksi 851 (63596) 3

Prestressed Concrete Beam Camber – 26” Slab Prestress loss due to elastic shortening = Δfp. ES = Epfcgp Ect Strain in concrete = 1. 77/2991 = 0. 00059177 Stress loss in strand = (28500)(0. 00059177) = 16. 86 ksi Assume 16. 00 ksi will be the loss at release. Revised prestressed force = [ 31. 0 – 16. 00] (0. 153)(38) = 1085 kips 0. 153 Revised prestressed moment = (1085)(13. 00 – 5. 36) = 8289 inch kips fcgp = 1085 + (8289 - 5803)(13. 00 - 5. 36) = 1. 57 ksi 851 (63596) Strain in concrete = 1. 57/2991 = 0. 000524908 Revised stress loss in strand = (28500)(0. 000524908) = 14. 96 ksi 4

Prestressed Concrete Beam Camber – 26” Slab Use a prestress loss of 15. 00 ksi, prestress force = 1091 kips, and strain = 0. 00052632 Prestress after 15. 00 ksi loss = (31)/(0. 153) – 15. 00 = 188 ksi Use the Moment-Area method to determine the upward deflection of the slab at release. Compute the moment in the slab induced by the eccentric prestressed strands. Strand eccentricity = 26 -5. 36 = 7. 64 in 2 Prestress moment in slab = (1091)(7. 64) = 8335 in kips This moment is assumed to be constant over the full length of the slab, neglecting the effect of debonded strands. Use the Moment Area method to determine the prestress 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. 5

Prestressed Concrete Beam Camber – 26” Slab The constant value of M/EI = 8335 = 0. 000043819/in (2991)(63596) The moment of the area between midspan and the end of the beam about the end of the beam = (0. 000043819)(780/2)(780/4) = 3. 33 inches This is the upward deflection caused by the load from the prestress strand not including the weight of the beam. The downward deflection of the beam from self weight = 5 w. L 4 384 EI w = 0. 0763 kips/inch, L = 780 inches, E = 2991 ksi, I = 63596 in 4 ∆ = (5)(0. 0763)(780)4 = 1. 93 inches (384)(2991)(63596) The net upward deflection at midspan at release = 3. 33 -1. 93 = 1. 40 inches 6

Prestressed Concrete Beam Camber – 26” Slab 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 ks = 1. 45 – 0. 13(V/S) ≥ 1. 0 V/S is the volume to surface ratio = (48)(26) = 8. 43 (48)(2) + (26)(2) ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 1+f’c 5 = 0. 77 1 + 5. 5 7

Prestressed Concrete Beam Camber – 26” Slab ktd = t = 90 = 0. 70 61 – 4 f’c + t 61 – (4)(5. 5) + 90 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(0. 70)(1. 0)-0. 118 = 1. 02 Creep deflection = (1. 02)(1. 40) = 1. 43 inches Compute strand stress loss due to creep. Δfp. CR = Ep fcgp Ψ(t, ti) Kid (AASHTO 5. 9. 5. 4. 2 b-1) Eci Kid = 1 1 + Ep. Aps [1+ Age 2 pg][1 + 0. 7Ψb(tf, ti)] Eci. Ag Ig . 8

Prestressed Concrete Beam Camber – 26” Slab Ep = 28500 ksi Aps = (38)(0. 153) = 5. 81 in 2 Eci = 2991 ksi Ag = 851 in 2 epg = 13. 00 – 5. 36 = 7. 64 in Ig = 63596 in 4 Ψb = 1. 02 Kid = 1 1 + (28500)(5. 81) [1+ (851)(7. 64)2][1 + (0. 7)(1. 02)] (2991)(851) 63596 = 0. 834 Δfp. CR = (28500)(1. 57)(1. 02)(0. 834) = 12. 72 ksi (2991) 9

Prestressed Concrete Beam Camber – 26” Slab Compute strand stress loss from shrinkage ∆fp. SR = εbid. Ep. Kid εbid = ks khs kf ktd 0. 48 x 10 -3 (AASHTO 5. 4. 2. 3. 3 -1) khs = 2. 00 – 0. 014 H, H=70, khs = 2. 0 – (0. 014)(70) = 1. 02 εbid = (1. 0)(1. 02)(0. 77)(0. 70)(0. 48 x 10 -3) = 0. 000264 ∆fp. SR = (0. 000264)(28500)(0. 834) = 6. 27 ksi Total stress loss from creep and shrinkage = 12. 72 + 6. 27 = 18. 99 ksi Downward deflection from creep and shrinkage = (3. 33)(18. 99/188) = 0. 34 inches 10

Prestressed Concrete Beam Camber – 26” Slab Strand stress loss from relaxation (AASHTO C 5. 9. 5. 4. 2 c-1) = ∆fp. R 1 = fpt log(24 t) [(fpt/fpy) -0. 55] [1 - 3(Δfp. SR + Δfp. CR) ]Kid K’L log(24 t 1) fpt = stress after transfer = 188 ksi t = 90 days K’L = 45 ti = 1. 0 days fpy = (0. 90)(270) = 243 ksi Δfp. SR = Stress loss in strand from shrinkage = 6. 27 ksi Δfp. CR = Stress loss in strand from creep = 12. 72 ksi Kid = 0. 834 ∆fp. R 1 = (188) log [(24)(90)] [(188/243) – 0. 55] [1 – 3(6. 27+12. 72)](0. 834) = 1. 31 ksi (45) log [(24)(1. 0)] 188 Downward deflection due to relaxation in strand = (0. 98)(3. 33) = 0. 02 inches (188) 11

Prestressed Concrete Beam Camber – 26” Slab Total downward deflection due to creep, shrinkage and relaxation = 0. 34 + 0. 02 = 0. 36” Total upward deflection at 3 months = 1. 40 + 1. 43 – 0. 36 = 2. 47 inches Compute downward deflection due to 40 psf superimposed dead load (90 days) __ ____ EC = 33000 K 1 wc 1. 5 √f’c = (33000)(1. 0)(0. 145)1. 5√ 5. 50 = 4273 ksi w = uniform load/slab = (4. 0)(40) = 160 lbs/ft = 0. 160 kips/ft = 0. 0133 kips/inch ∆ = 5 w. L 4 = (5)(0. 01333)(780)4 = 0. 24” 384 EI (384)(4273)(63596) 12

Prestressed Concrete Beam Camber – 26” Slab Compute concrete stress (tension) at strand cg from this load Moment = w. L 2/8 = (0. 0133)(780)2/8 = 1011 inch kips Stress at strand cg at midspan = My/I = (1011)(7. 64)/(63596) = 0. 12 ksi Strain in concrete = 0. 12/4273 = 0. 00002808 Stress gain in strand = (0. 00002808)(28500) = 0. 80 ksi Compute deflection 5 years after release Upward at release of prestress = 1. 40 inches Compute creep coefficient for 5 years = 1825 days Ψ(t, ti) = 1. 9 ks khc kf ktd ti-0. 118 13

Prestressed Concrete Beam Camber – 26” Slab ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 0. 77 1+f’c 1 + 5. 5 ktd = t = 1825 = 0. 98 61 – 4 f’c + t 61 – (4)(5. 5) + 1825 ti = 1. 0 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(0. 98)(1. 0)-0. 118 = 1. 43 Compute creep coefficient for wearing surface load application (90 days to 5 years) Ψ(t, ti) = 1. 9 ks khc kf ktd ti-0. 118 14

Prestressed Concrete Beam Camber – 26” Slab ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 1+f’c 5 = 0. 77 1 + 5. 5 ktd = t = 1825 = 0. 98 61 – 4 f’c + t 61 – (4)(5. 5) + 1825 ti = 90. 0 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(0. 98)(90)-0. 118 = 0. 84 15

Prestressed Concrete Beam Camber – 26” Slab Upward deflection from release of prestress 5 years after release = (1. 43)(1. 40) + 1. 40 = 3. 40 inches Strain in concrete due to creep = (0. 00052632)(1. 43) = 0. 00075264 Strand stress loss from creep = (28500)(0. 00075264) = 21. 45 ksi Downward deflection from wearing surface 5 years after release = 0. 24 + (0. 84)(0. 24) = 0. 44 inches Strain in concrete due to wearing surface 5 years after release = (0. 84)(0. 00002808) = 0. 00002359 Stress gain in strand from wearing surface creep = (0. 00002359)(28500) = 0. 67 ksi Total strand stress gain from wearing surface = 0. 80 + 0. 67 = 1. 47 ksi 16

Prestressed Concrete Beam Camber – 26” Slab Compute shrinkage coefficient Shrinkage strain (AASHTO 5. 4. 2. 3. 3 -1) = εsh = ks khs kf ktd 0. 48 x 10 -3 ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khs = 2. 00 – 0. 014 H, H=70, khs = 2. 0 – (0. 014)(70) = 1. 02 kf = 5 = 0. 77 1+f’c 1 + 5. 5 ktd = t = 1825 = 0. 98 61 – 4 f’c + t 61 – (4)(5. 5) + 1825 εsh = (1. 0)(1. 02)(0. 77)(0. 98)(0. 48 x 10 -3) = 0. 0003694 Strand stress loss from shrinkage = (28500)(0. 0003694) = 10. 53 ksi Strand stress loss from creep, including wearing surface, = 21. 45 – 1. 47 = 19. 98 ksi Total strand stress loss from creep and shrinkage = 19. 98 + 10. 53 = 30. 51 ksi 17

Prestressed Concrete Beam Camber – 26” Slab Downward deflection due to stress loss = (30. 51)(3. 33) = 0. 54 inches (188) Compute strand stress loss due to relaxation Strand stress loss from relaxation (AASHTO C 5. 9. 5. 4. 2 c-1) = ∆fp. R 1 = fpt log(24 t) [(fpt/fpy) -0. 55] [1 - 3(Δfp. SR + Δfp. CR) ]Kid K’L log(24 t 1) fpt = stress after transfer = 188 ksi t = 1825 days K’L = 45 ti = 1. 0 days fpy = (0. 90)(270) = 243 ksi Δfp. SR = Stress loss in strand from shrinkage = 10. 53 ksi Δfp. CR = Stress loss in strand from creep = 19. 98 ksi 18

Prestressed Concrete Beam Camber – 26” Slab Kid = 1 1 + (28500)(5. 81) [1+ (851)(7. 64)2][1 + (0. 7)(1. 43)] (2991)(851) 63596 = 0. 81 ∆fp. R 1 = (188) log [(24)(1825)] [(188/243) – 0. 55] [1 – 3(10. 53+19. 98)](0. 81) = 1. 31 ksi (45) log [(24)(1. 0)] 188 Downward deflection due to relaxation in strand = (1. 31)(3. 33) =0. 02 inches (188) Total deflection at 5 years = 3. 40 – 0. 44 – 0. 54 – 0. 02 = 2. 40 inches Compute final strand stress loss at 27 years (as on Slab Design Sheet) Compute creep coefficient at 27 years (9855 days) Ψ(t, ti) = 1. 9 ks khc kf ktd ti-0. 118 19

Prestressed Concrete Beam Camber – 26” Slab ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 1+f’c 5 = 0. 77 1 + 5. 5 ktd = t = 9855 = 1. 00 61 – 4 f’c + t 61 – (4)(5. 5) + 9855 ti = 1. 0 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(1. 00)(1. 0)-0. 118 = 1. 46 20

Prestressed Concrete Beam Camber – 26” Slab Compute creep coefficient at 27 years (9855 days) starting at 90 days for wearing surface Ψ(t, ti) = 1. 9 ks khc kf ktd ti-0. 118 ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 0. 77 1+f’c 1 + 5. 5 ktd = t = 9855 = 1. 00 61 – 4 f’c + t 61 – (4)(5. 5) + 9855 ti = 90. 0 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(1. 00)(90)-0. 118 = 0. 86 21

Prestressed Concrete Beam Camber – 26” Slab Strain in concrete due to creep = (0. 00052632)(1. 46) = 0. 00076843 Strand stress loss due to creep = (28500)(0. 00076843) = 21. 90 ksi Strand stress gain due to wearing surface = 0. 80 ksi Strain in concrete due to wearing surface = (0. 00002808)(0. 86) = 0. 000024149 Strand stress gain from wearing surface creep = (0. 000024149)(28500) = 0. 69 ksi Total strand stress loss from creep = 21. 90 – 0. 80 – 0. 69 = 20. 41 ksi Compute shrinkage coefficient at 27 years (9855 days) εsh = ks khs kf ktd 0. 48 x 10 -3 22

Prestressed Concrete Beam Camber – 26” Slab ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khs = 2. 00 – 0. 014 H, H=70, khs = 2. 0 – (0. 014)(70) = 1. 02 kf = 5 = 1+f’c 5 = 0. 77 1 + 5. 5 ktd = t = 9855 = 1. 00 61 – 4 f’c + t 61 – (4)(5. 5) + 9855 εsh = (1. 0)(1. 02)(0. 77)(1. 00)(0. 48 x 10 -3) = 0. 0003770 Strand stress loss from shrinkage = (28500)(0. 0003770) = 10. 74 ksi Total strand loss from creep and shrinkage = 21. 90 - 0. 80 – 0. 69 + 10. 74 = 31. 15 ksi 23

Prestressed Concrete Beam Camber – 26” Slab Compute stress loss from relaxation Kid = 1 1 + (28500)(5. 81) [1+ (851)(7. 64)2][1 + (0. 7)(1. 46)] (2991)(851) 63596 = 0. 81 ∆fp. R 1 = (188) log [(24)(9855)] [(188/243) – 0. 55] [1 – 3(10. 74+20. 41)](0. 81) = 1. 48 ksi (45) log [(24)(1. 0)] 188 Total stress loss = 15. 00 + 31. 15 +1. 48 = 47. 63 ksi Compute shortening 2 weeks after transfer of prestress (as on Slab Design Sheet) Compute creep coefficient at 2 weeks (14 days) Ψ(t, ti) = 1. 9 ks khc kf ktd ti-0. 118 24

Prestressed Concrete Beam Camber – 26” Slab ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khc = 1. 56 – 0. 008 H, H = 70 (From AASHTO Figure 5. 4. 2. 3. 3 -1) khc = 1. 56 – (0. 008)(70) = 1. 0 kf = 5 = 1+f’c 5 = 0. 77 1 + 5. 5 ktd = t = 14 = 0. 26 61 – 4 f’c + t 61 – (4)(5. 5) + 14 ti = 1. 0 Ψ(t, ti) = (1. 90)(1. 0)(0. 77)(0. 26)(1. 00)-0. 118 = 0. 38 Elastic Shortening = Fps L = (1091)(780) = 0. 33” AEci (851)(2991) Shortening from creep = (0. 33)(0. 38) = 0. 13 inches 25

Prestressed Concrete Beam Camber – 26” Slab Compute shrinkage coefficient at 2 weeks (14 days) εsh = ks khs kf ktd 0. 48 x 10 -3 ks = 1. 45 – 0. 13(8. 43) = 0. 354 → 1. 0 khs = 2. 00 – 0. 014 H, H=70, khs = 2. 0 – (0. 014)(70) = 1. 02 kf = 5 = 0. 77 1+f’c 1 + 5. 5 ktd = t = 14 = 0. 26 61 – 4 f’c + t 61 – (4)(5. 5) + 14 εsh = (1. 0)(1. 02)(0. 77)(0. 26)(0. 48 x 10 -3) = 0. 00009802 26

Prestressed Concrete Beam Camber – 26” Slab Shortening from shrinkage = (780)(0. 00009802) = 0. 076 inches Elastic shortening = 0. 33 inches Neglect effects of strand relaxation Total shortening at 2 weeks = 0. 33 + 0. 13 + 0. 076 = 0. 54 inches 27

Prestressed Concrete Beam Camber – 26” Slab 28