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1 Prestressed Concrete Beam Camber – 26 Slab The following is an example of calculations performed to determine the time dependent camber values for 26-

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Presentation on theme: "1 Prestressed Concrete Beam Camber – 26 Slab The following is an example of calculations performed to determine the time dependent camber values for 26-"— Presentation transcript:

1 1 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.

2 2 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. ___ E C = (0.70) 33000K1W C 1.5 f C (AASHTO ) _____ = (0.70)(33000)(1.0)(0.145)1.5(5.50) = 2991 ksi ES = (AASHTO ) I = in4 A = 851 in2 y C = 5.36 w = k/ft = k/in

3 3 Prestressed Concrete Beam Camber – 26 Slab Compute initial stress loss Total jacking force = (31 kips/strand)(38 strands) = 1178 kips Prestress Moment = (1178)( ) = 9000 inch kips Maximum moment from beam weight = (0.0763)(780) 2 = 5803 inch kips 8 Net maximum moment = 9000 – 5803 = 3197 inch kips f cgp = Concrete stress at strand cg = (3197)( ) = 1.77 ksi 851 (63596)

4 4 Prestressed Concrete Beam Camber – 26 Slab Prestress loss due to elastic shortening = Δf pES = E p f cgp E ct Strain in concrete = 1.77/2991 = Stress loss in strand = (28500)( ) = ksi Assume ksi will be the loss at release. Revised prestressed force = [ 31.0 – 16.00] (0.153)(38) = 1085 kips Revised prestressed moment = (1085)(13.00 – 5.36) = 8289 inch kips f cgp = ( )( ) = 1.57 ksi 851 (63596) Strain in concrete = 1.57/2991 = Revised stress loss in strand = (28500)( ) = ksi

5 5 Prestressed Concrete Beam Camber – 26 Slab Use a prestress loss of ksi, prestress force = 1091 kips, and strain = Prestress stress after ksi loss = (31)/(0.153) – = 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 = = 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.

6 6 Prestressed Concrete Beam Camber – 26 Slab The constant value of M/EI = 8335 = /in (2991)(63596) The moment of the area between midspan and the end of the beam about the end of the beam = ( )(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 = 5wL 4 384EI w = kips/inch, L = 780 inches, E = 2991 ksi, I = in 4 = (5)(0.0763)(780) 4 = 1.93 inches (384)(2991)(63596) The net upward deflection at midspan at release = = 1.40 inches

7 7 Prestressed Concrete Beam Camber – 26 Slab Compute the upward deflection 3 months after release. The creep coefficient is determined from AASHTO formula , Ψ(t, t i ) = 1.9 k s k hc k f k td t i k s = 1.45 – 0.13(V/S) 1.0 V/S is the volume to surface ratio = (48)(26) = 8.43 (48)(2) + (26)(2) k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = f c

8 8 Prestressed Concrete Beam Camber – 26 Slab k td = t = 90 = – 4f c + t 61 – (4)(5.5) + 90 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(0.70)(1.0) = 1.02 Creep deflection = (1.02)(1.40) = 1.43 inches Compute strand stress loss due to creep. Δf pCR = E p f cgp Ψ(t, ti) K id (AASHTO b-1) E ci Kid = E p A ps [1+ A g e 2 pg ][ Ψ b (t f, t i )] E ci A g I g

9 9 Prestressed Concrete Beam Camber – 26 Slab E p = ksi A ps = (38)(0.153) = 5.81 in 2 E ci = 2991 ksi A g = 851 in 2 e pg = – 5.36 = 7.64 in I g = in 4 Ψ b = 1.02 K id = 1 = (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.02)] (2991)(851) Δf pCR = (28500)(1.57)(1.02)(0.834) = ksi (2991)

10 10 Prestressed Concrete Beam Camber – 26 Slab Compute strand stress loss from shrinkage f pSR = ε bid E p K id ε bid = k s k hs k f k td 0.48x10 -3 (AASHTO ) k hs = 2.00 – 0.014H, H=70, k hs = 2.0 – (0.014)(70) = 1.02 ε bid = (1.0)(1.02)(0.77)(0.70)(0.48x10 -3 ) = f pSR = ( )(28500)(0.834) = 6.27 ksi Total stress loss from creep and shrinkage = = ksi Downward deflection from creep and shrinkage = (3.33)(18.99/188) = 0.34 inches

11 11 Prestressed Concrete Beam Camber – 26 Slab Strand stress loss from relaxation (AASHTO C c-1) = f pR1 = f pt log(24t) [(f pt /f py ) -0.55] [1- 3(Δf pSR + Δf pCR ) ]K id KL log(24t 1 ) f pt f pt = stress after transfer = 188ksi t = 90 days K L = 45 t i = 1.0 days f py = (0.90)(270) = 243 ksi Δf pSR = Stress loss in strand from shrinkage = 6.27 ksi Δf pCR = Stress loss in strand from creep = ksi K id = f pR1 = (188) log [(24)(90)] [(188/243) – 0.55] [1 – 3( )](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)

12 12 Prestressed Concrete Beam Camber – 26 Slab Total downward deflection due to creep, shrinkage and relaxation = = 0.36 Total upward deflection at 3 months = – 0.36 = 2.47 inches Compute downward deflection due to 40 psf superimposed dead load (90 days) __ ____ EC = K 1 wc 1.5 fc = (33000)(1.0)(0.145) = 4273 ksi w = uniform load/slab = (4.0)(40) = 160 lbs/ft = kips/ft = kips/inch = 5wL 4 = (5)( )(780)4 = EI (384)(4273)(63596)

13 13 Prestressed Concrete Beam Camber – 26 Slab Compute concrete stress (tension) at strand cg from this load Moment = wL 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 = Stress gain in strand = ( )(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, t i ) = 1.9 k s k hc k f k td t i

14 14 Prestressed Concrete Beam Camber – 26 Slab k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = f c k td = t = 1825 = – 4fc + t 61 – (4)(5.5) t i = 1.0 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(0.98)(1.0) = 1.43 Compute creep coefficient for wearing surface load application (90 days to 5 years) Ψ(t, ti) = 1.9 k s k hc k f k td t i

15 15 Prestressed Concrete Beam Camber – 26 Slab k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = fc k td = t = 1825 = – 4fc + t 61 – (4)(5.5) t i = 90.0 Ψ(t, t i ) = (1.90)(1.0)(1.0)(0.77)(0.98)(90) = 0.84

16 16 Prestressed Concrete Beam Camber – 26 Slab Upward deflection from release of prestress 5 years after release = (1.43)(1.40) = 3.40 inches Strain in concrete due to creep = ( )(1.43) = Strand stress loss from creep = (28500)( ) = ksi Downward deflection from wearing surface 5 years after release = (0.84)(0.24) = 0.44 inches Strain in concrete due to wearing surface 5 years after release = (0.84)( ) = Stress gain in strand from wearing surface creep = ( )(28500) = 0.67 ksi Total strand stress gain from wearing surface = = 1.47 ksi

17 17 Prestressed Concrete Beam Camber – 26 Slab Compute shrinkage coefficient Shrinkage strain (AASHTO ) = ε sh = k s k hs k f k td 0.48x10-3 k s = 1.45 – 0.13(8.43) = k hs = 2.00 – 0.014H, H=70, k hs = 2.0 – (0.014)(70) = 1.02 k f = 5 = 5 = fc k td = t = 1825 = – 4f c + t 61 – (4)(5.5) ε sh = (1.0)(1.02)(0.77)(0.98)(0.48x10-3) = Strand stress loss from shrinkage = (28500)( ) = ksi Strand stress loss from creep, including wearing surface, = – 1.47 = ksi Total strand stress loss from creep and shrinkage = = ksi

18 18 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 c-1) = f pR1 = f pt log(24t) [(f pt /f py ) -0.55] [1- 3(Δf pSR + Δf pCR ) ]K id K L log(24t 1 ) f pt f pt = stress after transfer = 188ksi t = 1825 days K L = 45 t i = 1.0 days f py = (0.90)(270) = 243 ksi Δf pSR = Stress loss in strand from shrinkage = ksi Δf pCR = Stress loss in strand from creep = ksi

19 19 Prestressed Concrete Beam Camber – 26 Slab K id = 1 = (28500)(5.81) [1+ (851)(7.64) 2 ][1 + (0.7)(1.43)] (2991)(851) f pR1 = (188) log [(24)(1825)] [(188/243) – 0.55] [1 – 3( )](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 k s k hc k f k td t i

20 20 Prestressed Concrete Beam Camber – 26 Slab k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = fc k td = t = 9855 = – 4f c + t 61 – (4)(5.5) t i = 1.0 Ψ(t, t i ) = (1.90)(1.0)(1.0)(0.77)(1.00)(1.0) = 1.46

21 21 Prestressed Concrete Beam Camber – 26 Slab Compute creep coefficient at 27 years (9855 days) starting at 90 days for wearing surface Ψ(t, t i ) = 1.9 k s k hc k f k td t i k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = f c k td = t = 9855 = – 4f c + t 61 – (4)(5.5) t i = 90.0 Ψ(t, t i ) = (1.90)(1.0)(1.0)(0.77)(1.00)(90) = 0.86

22 22 Prestressed Concrete Beam Camber – 26 Slab Strain in concrete due to creep = ( )(1.46) = Strand stress loss due to creep = (28500)( ) = ksi Strand stress gain due to wearing surface = 0.80 ksi Strain in concrete due to wearing surface = ( )(0.86) = Strand stress gain from wearing surface creep = ( )(28500) = 0.69 ksi Total strand stress loss from creep = – 0.80 – 0.69 = ksi Compute shrinkage coefficient at 27 years (9855 days) ε sh = k s k hs k f k td 0.48x10 -3

23 23 Prestressed Concrete Beam Camber – 26 Slab k s = 1.45 – 0.13(8.43) = k hs = 2.00 – 0.014H, H=70, k hs = 2.0 – (0.014)(70) = 1.02 k f = 5 = 5 = fc k td = t = 9855 = – 4f c + t 61 – (4)(5.5) ε sh = (1.0)(1.02)(0.77)(1.00)(0.48x10-3) = Strand stress loss from shrinkage = (28500)( ) = ksi Total strand loss from creep and shrinkage = – = ksi

24 24 Prestressed Concrete Beam Camber – 26 Slab Compute stress loss from relaxation K id = 1 = (28500)(5.81) [1+ (851)(7.64) 2 ][1 + (0.7)(1.46)] (2991)(851) f pR1 = (188) log [(24)(9855)] [(188/243) – 0.55] [1 – 3( )](0.81) = 1.48 ksi (45) log [(24)(1.0)] 188 Total stress loss = = ksi Compute shortening 2 weeks after transfer of prestress (as on Slab Design Sheet) Compute creep coefficient at 2 weeks (14 days) Ψ(t, t i ) = 1.9 k s k hc k f k td t i

25 25 Prestressed Concrete Beam Camber – 26 Slab k s = 1.45 – 0.13(8.43) = k hc = 1.56 – 0.008H, H = 70 (From AASHTO Figure ) k hc = 1.56 – (0.008)(70) = 1.0 k f = 5 = 5 = fc k td = t = 14 = – 4f c + t 61 – (4)(5.5) + 14 t i = 1.0 Ψ(t, t i ) = (1.90)(1.0)(1.0)(0.77)(0.26)(1.00) = 0.38 Elastic Shortening = F ps L = (1091)(780) = 0.33 AE ci (851)(2991) Shortening from creep = (0.33)(0.38) = 0.13 inches

26 26 Prestressed Concrete Beam Camber – 26 Slab Compute shrinkage coefficient at 2 weeks (14 days) ε sh = k s k hs k f k td 0.48x10 -3 k s = 1.45 – 0.13(8.43) = k hs = 2.00 – 0.014H, H=70, k hs = 2.0 – (0.014)(70) = 1.02 k f = 5 = 5 = fc k td = t = 14 = – 4f c + t 61 – (4)(5.5) + 14 ε sh = (1.0)(1.02)(0.77)(0.26)(0.48x10-3) =

27 27 Prestressed Concrete Beam Camber – 26 Slab Shortening from shrinkage = (780)( ) = inches Elastic shortening = 0.33 inches Neglect effects of strand relaxation Total shortening at 2 weeks = = 0.54 inches

28 28 Prestressed Concrete Beam Camber – 26 Slab


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