1 Prestressed Concrete Beam Camber – 26 Slab The following is an example of calculations performed...

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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.

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Compute upward deflection at release of strandAssume at this point 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√(5.50) = 2991 ksi

ES = 28500 (AASHTO 5.4.4.2)

I = 63596 in4 A = 851 in2 yC = 5.36” w = 0.916 k/ft = 0.0763 k/in

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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 8Net maximum moment = 9000 – 5803 = 3197 inch kips

fcgp = Concrete stress at strand cg = 1178 + (3197)(13.00 - 5.36) = 1.77 ksi 851 (63596)

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Prestress loss due to elastic shortening = ΔfpES = 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)


Revised stress loss in strand = (28500)(0.000524908) = 14.96 ksi

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Prestressed Concrete Beam Camber – 26” SlabUse a prestress loss of 15.00 ksi, prestress force = 1091 kips, and strain = 0.00052632

Prestress stress 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 2Prestress 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.

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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 = 5wL4 384EI

w = 0.0763 kips/inch, L = 780 inches, E = 2991 ksi, I = 63596 in4

∆ = (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

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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 – 0.008H, H = 70 (From AASHTO Figure 5.4.2.3.3-1)

khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

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ktd = t = 90 = 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)-0.118 = 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 5.9.5.4.2b-1)

Eci

Kid = 1 .

1 + EpAps [1+ Age2pg][1 + 0.7Ψb(tf, ti)]

EciAg Ig

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Ep = 28500 ksiAps = (38)(0.153) = 5.81 in2

Eci = 2991 ksiAg = 851 in2

epg = 13.00 – 5.36 = 7.64 inIg = 63596 in4

Ψb = 1.02

Kid = 1 = 0.834 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.02)] (2991)(851) 63596

ΔfpCR = (28500)(1.57)(1.02)(0.834) = 12.72 ksi (2991)

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Compute strand stress loss from shrinkage

∆fpSR = εbidEpKid

εbid = ks khs kf ktd 0.48x10-3 (AASHTO 5.4.2.3.3-1)

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) = 0.000264 ∆fpSR = (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

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Prestressed Concrete Beam Camber – 26” SlabStrand 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 = 188ksit = 90 daysK’L = 45ti = 1.0 daysfpy = (0.90)(270) = 243 ksiΔfpSR = Stress loss in strand from shrinkage = 6.27 ksiΔfpCR = Stress loss in strand from creep = 12.72 ksiKid = 0.834

∆fpR1 = (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)

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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 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 = 0.160 kips/ft = 0.0133 kips/inch

∆ = 5wL4 = (5)(0.01333)(780)4 = 0.24” 384EI (384)(4273)(63596)

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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


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

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ks = 1.45 – 0.13(8.43) = 0.354 → 1.0


khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

ktd = t = 1825 = 0.98 61 – 4f’c + t 61 – (4)(5.5) + 1825

ti = 1.0

Ψ(t, ti) = (1.90)(1.0)(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)


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ks = 1.45 – 0.13(8.43) = 0.354 → 1.0


khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77

1+f’c 1 + 5.5

ktd = t = 1825 = 0.98

61 – 4f’c + t 61 – (4)(5.5) + 1825

ti = 90.0

Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(0.98)(90)-0.118 = 0.84

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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

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Compute shrinkage coefficient

Shrinkage strain (AASHTO 5.4.2.3.3-1) = εsh = ks khs kf ktd 0.48x10-3

ks = 1.45 – 0.13(8.43) = 0.354 → 1.0

khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

ktd = t = 1825 = 0.98 61 – 4f’c + t 61 – (4)(5.5) + 1825

εsh = (1.0)(1.02)(0.77)(0.98)(0.48x10-3) = 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

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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 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 = 188ksit = 1825 daysK’L = 45ti = 1.0 daysfpy = (0.90)(270) = 243 ksiΔfpSR = Stress loss in strand from shrinkage = 10.53 ksiΔfpCR = Stress loss in strand from creep = 19.98 ksi

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Kid = 1 = 0.81 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.43)] (2991)(851) 63596

∆fpR1 = (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)


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ks = 1.45 – 0.13(8.43) = 0.354 → 1.0


khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77

1+f’c 1 + 5.5

ktd = t = 9855 = 1.00

61 – 4f’c + t 61 – (4)(5.5) + 9855

ti = 1.0

Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(1.00)(1.0)-0.118 = 1.46

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Compute creep coefficient at 27 years (9855 days) starting at 90 days for wearing surface


ks = 1.45 – 0.13(8.43) = 0.354 → 1.0


khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

ktd = t = 9855 = 1.00 61 – 4f’c + t 61 – (4)(5.5) + 9855

ti = 90.0

Ψ(t, ti) = (1.90)(1.0)(1.0)(0.77)(1.00)(90)-0.118 = 0.86

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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 0.48x10-3

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ks = 1.45 – 0.13(8.43) = 0.354 → 1.0

khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

ktd = t = 9855 = 1.00 61 – 4f’c + t 61 – (4)(5.5) + 9855

εsh = (1.0)(1.02)(0.77)(1.00)(0.48x10-3) = 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

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Compute stress loss from relaxation

Kid = 1 = 0.81 1 + (28500)(5.81) [1+ (851)(7.64)2][1 + (0.7)(1.46)] (2991)(851) 63596

∆fpR1 = (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)


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ks = 1.45 – 0.13(8.43) = 0.354 → 1.0


khc = 1.56 – (0.008)(70) = 1.0

kf = 5 = 5 = 0.77 1+f’c 1 + 5.5

ktd = t = 14 = 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)-0.118 = 0.38

Elastic Shortening = Fps L = (1091)(780) = 0.33”

AEci (851)(2991)

Shortening from creep = (0.33)(0.38) = 0.13 inches

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Compute shrinkage coefficient at 2 weeks (14 days)

εsh = ks khs kf ktd 0.48x10-3

ks = 1.45 – 0.13(8.43) = 0.354 → 1.0

khs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02

kf = 5 = 5 = 0.77

1+f’c 1 + 5.5

ktd = t = 14 = 0.26

61 – 4f’c + t 61 – (4)(5.5) + 14

εsh = (1.0)(1.02)(0.77)(0.26)(0.48x10-3) = 0.00009802

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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

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1 Prestressed Concrete Beam Camber – 26 Slab The following is an example of calculations performed...

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