Download presentation

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
**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) 33000K1WC1.5 √f’C (AASHTO ) _____ = (0.70)(33000)(1.0)(0.145)1.5√(5.50) = 2991 ksi ES = (AASHTO ) I = in A = 851 in yC = 5.36” w = k/ft = k/in

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 fcgp = Concrete stress at strand cg = (3197)( ) = 1.77 ksi (63596)

4
**Prestressed Concrete Beam Camber – 26” Slab**

Prestress loss due to elastic shortening = ΔfpES = Epfcgp Ect 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 0.153 Revised prestressed moment = (1085)(13.00 – 5.36) = 8289 inch kips fcgp = ( )( ) = 1.57 ksi (63596) Strain in concrete = 1.57/2991 = Revised stress loss in strand = (28500)( ) = ksi

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
**Prestressed Concrete Beam Camber – 26” Slab**

The constant value of M/EI = = /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 = 5wL4 384EI w = kips/inch, L = 780 inches, E = 2991 ksi, I = in4 ∆ = (5)(0.0763)(780)4 = inches (384)(2991)(63596) The net upward deflection at midspan at release = = 1.40 inches

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, ti) = 1.9 ks khc kf ktd ti 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) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c

8
**Prestressed Concrete Beam Camber – 26” Slab**

ktd = t = = 0.70 61 – 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. ΔfpCR = Ep fcgp Ψ(t, ti) Kid (AASHTO b-1) Eci Kid = 1 + EpAps [1+ Age2pg][ Ψb(tf, ti)] EciAg Ig

9
**Prestressed Concrete Beam Camber – 26” Slab**

Ep = ksi Aps = (38)(0.153) = 5.81 in2 Eci = 2991 ksi Ag = 851 in2 epg = – 5.36 = 7.64 in Ig = in4 Ψb = 1.02 Kid = = 0.834 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.02)] (2991)(851) ΔfpCR = (28500)(1.57)(1.02)(0.834) = ksi (2991)

10
**Prestressed Concrete Beam Camber – 26” Slab**

Compute strand stress loss from shrinkage ∆fpSR = εbidEpKid εbid = ks khs kf ktd 0.48x10-3 (AASHTO ) khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02 εbid = (1.0)(1.02)(0.77)(0.70)(0.48x10-3) = ∆fpSR = ( )(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
**Prestressed Concrete Beam Camber – 26” Slab**

Strand stress loss from relaxation (AASHTO C c-1) = ∆fpR1 = fpt log(24t) [(fpt/fpy) -0.55] [1- 3(ΔfpSR + ΔfpCR) ]Kid K’L log(24t1) fpt fpt = stress after transfer = 188ksi t = 90 days K’L = 45 ti = 1.0 days fpy = (0.90)(270) = 243 ksi ΔfpSR = Stress loss in strand from shrinkage = 6.27 ksi ΔfpCR = Stress loss in strand from creep = ksi Kid = 0.834 ∆fpR1 = (188) log [(24)(90)] [(188/243) – 0.55] [1 – 3( )](0.834) = 1.31 ksi (45) log [(24)(1.0)] Downward deflection due to relaxation in strand = (0.98)(3.33) = 0.02 inches (188)

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 = K1 wc1.5 √f’c = (33000)(1.0)(0.145)1.5√5.50 = 4273 ksi w = uniform load/slab = (4.0)(40) = 160 lbs/ft = kips/ft = kips/inch ∆ = 5wL4 = (5)( )(780)4 = 0.24” 384EI (384)(4273)(63596)

13
**Prestressed Concrete Beam Camber – 26” Slab**

Compute concrete stress (tension) at strand cg from this load Moment = wL2/8 = (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, ti) = 1.9 ks khc kf ktd ti-0.118

14
**Prestressed Concrete Beam Camber – 26” Slab**

ks = 1.45 – 0.13(8.43) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c ktd = t = = 0.98 61 – 4f’c + t 61 – (4)(5.5) ti = 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 ks khc kf ktd ti-0.118

15
**Prestressed Concrete Beam Camber – 26” Slab**

ks = 1.45 – 0.13(8.43) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c ktd = t = = 0.98 61 – 4f’c + t 61 – (4)(5.5) ti = 90.0 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(0.98)(90) = 0.84

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
**Prestressed Concrete Beam Camber – 26” Slab**

Compute shrinkage coefficient Shrinkage strain (AASHTO ) = εsh = ks khs kf ktd 0.48x10-3 ks = 1.45 – 0.13(8.43) = → 1.0 khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02 kf = = = 0.77 1+f’c ktd = t = = 0.98 61 – 4f’c + t – (4)(5.5) εsh = (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
**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) = ∆fpR1 = fpt log(24t) [(fpt/fpy) -0.55] [1- 3(ΔfpSR + ΔfpCR) ]Kid K’L log(24t1) fpt fpt = stress after transfer = 188ksi t = 1825 days K’L = 45 ti = 1.0 days fpy = (0.90)(270) = 243 ksi ΔfpSR = Stress loss in strand from shrinkage = ksi ΔfpCR = Stress loss in strand from creep = ksi

19
**Prestressed Concrete Beam Camber – 26” Slab**

Kid = = 0.81 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.43)] (2991)(851) ∆fpR1 = (188) log [(24)(1825)] [(188/243) – 0.55] [1 – 3( )](0.81) = 1.31 ksi (45) log [(24)(1.0)] 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

20
**Prestressed Concrete Beam Camber – 26” Slab**

ks = 1.45 – 0.13(8.43) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c ktd = t = = 1.00 61 – 4f’c + t 61 – (4)(5.5) ti = 1.0 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(1.00)(1.0) = 1.46

21
**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) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c ktd = t = = 1.00 61 – 4f’c + t – (4)(5.5) ti = 90.0 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(1.00)(90) = 0.86

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) εsh = ks khs kf ktd 0.48x10-3

23
**Prestressed Concrete Beam Camber – 26” Slab**

ks = 1.45 – 0.13(8.43) = → 1.0 khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02 kf = = = 0.77 1+f’c ktd = t = = 1.00 61 – 4f’c + t 61 – (4)(5.5) εsh = (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
**Prestressed Concrete Beam Camber – 26” Slab**

Compute stress loss from relaxation Kid = = 0.81 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.46)] (2991)(851) ∆fpR1 = (188) log [(24)(9855)] [(188/243) – 0.55] [1 – 3( )](0.81) = 1.48 ksi (45) log [(24)(1.0)] 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, ti) = 1.9 ks khc kf ktd ti-0.118

25
**Prestressed Concrete Beam Camber – 26” Slab**

ks = 1.45 – 0.13(8.43) = → 1.0 khc = 1.56 – 0.008H, H = (From AASHTO Figure ) khc = 1.56 – (0.008)(70) = 1.0 kf = = = 0.77 1+f’c ktd = t = = 0.26 61 – 4f’c + t 61 – (4)(5.5) + 14 ti = 1.0 Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(0.26)(1.00) = 0.38 Elastic Shortening = Fps L = (1091)(780) = 0.33” AEci (851)(2991) Shortening from creep = (0.33)(0.38) = 0.13 inches

26
**Prestressed Concrete Beam Camber – 26” Slab**

Compute shrinkage coefficient at 2 weeks (14 days) εsh = ks khs kf ktd 0.48x10-3 ks = 1.45 – 0.13(8.43) = → 1.0 khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02 kf = = = 0.77 1+f’c ktd = t = = 0.26 61 – 4f’c + t – (4)(5.5) + 14 εsh = (1.0)(1.02)(0.77)(0.26)(0.48x10-3) =

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
**Prestressed Concrete Beam Camber – 26” Slab**

Similar presentations

Presentation is loading. Please wait....

OK

Solving a Sudoku Puzzle

Solving a Sudoku Puzzle

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on acute coronary syndrome nursing Converter pub to ppt online training Interfacing keyboard and seven segment display ppt online Convert pdf to ppt online nitro Free download ppt on alternative sources of energy Ppt on switching network connection Ppt on book review of fish tales Ppt on team building training module Ppt on programmable logic array buy Ppt on metro rail in india