Prestressed Concrete Beam Camber

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26 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)

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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
-
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.

___

E
C

= (0.70) 33000K1W
C
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 in4 A = 851 in2 y
C

= 5.36” w = 0.916 k/ft = 0.0763 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)(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

f
cgp

= Concrete stress at strand cg =
1178
+
(3197)(13.00
-

5.36)
=
1.77 ksi

851 (63596)

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 = 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

f
cgp

=

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

5

Prestressed Concrete Beam Camber

26” Slab

Use 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

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

6

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 =

5wL
4

384EI

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

7

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, t
i
) = 1.9 k
s

k
hc

k
f

k
td

t
i
-
0.118

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) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f

=

5
=

5
=
0.77

1+f’
c

1 + 5.5

8

Prestressed Concrete Beam Camber

26” Slab

k
td

=

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.

Δf
pCR

=
E
p

f
cgp

Ψ(t, ti) K
id
(AASHTO 5.9.5.4.2b
-
1)

E
ci

Kid =

1 .

1 +
E
p
A
ps

[1+
A
g
e
2
pg
][1 + 0.7Ψ
b
(t
f
, t
i
)]

E
ci
A
g

I
g

9

Prestressed Concrete Beam Camber

26” Slab

E
p

= 28500 ksi

A
ps

= (38)(0.153) = 5.81 in
2

E
ci

= 2991 ksi

A
g

= 851 in
2

e
pg

= 13.00

5.36 = 7.64 in

I
g
= 63596 in
4

Ψ
b

= 1.02

K
id

=

1

= 0.834

1 +
(28500)(5.81)

[1+
(851)(7.64)
2][1 + (0.7)(1.02)]

(2991)(851) 63596

Δf
pCR

=
(28500)(1.57)(1.02)(0.834)

=
12.72 ksi

(2991)

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 5.4.2.3.3
-
1)

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
) =
0.000264

∆f
pSR

= (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

11

Prestressed Concrete Beam Camber

26” Slab

Strand stress loss from relaxation (AASHTO C5.9.5.4.2c
-
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 = 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 = 12.72 ksi

K
id

= 0.834

∆f
pR1

=
(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)

12

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

∆ =
5wL
4

=
(5)(0.01333)(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 = 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 =
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, t
i
) = 1.9 k
s

k
hc

k
f
k
td

t
i
-
0.118

14

Prestressed Concrete Beam Camber

26” Slab

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f

=

5
=

5
=
0.77

1+f’
c

1 + 5.5

k
td

=

t
=

1825
=
0.98

61

4f’c + t 61

(4)(5.5) + 1825

t
i

= 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)

Ψ(t, ti) = 1.9 k
s

k
hc

k
f

k
td

t
i
-
0.118

15

Prestressed Concrete Beam Camber

26” Slab

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f
=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

1825
=
0.98

61

4f’c + t 61

(4)(5.5) + 1825

t
i
= 90.0

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

=
0.84

16

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

17

Prestressed Concrete Beam Camber

26” Slab

Compute shrinkage coefficient

Shrinkage strain (AASHTO 5.4.2.3.3
-
1) = ε
sh

= k
s

k
hs

k
f
k
td

0.48x10
-
3

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hs

= 2.00

0.014H, H=70, k
hs

= 2.0

(0.014)(70) =
1.02

k
f
=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

1825
=
0.98

61

4f’
c

+ t 61

(4)(5.5) + 1825

ε
sh

= (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

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 C5.9.5.4.2c
-
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 = 10.53 ksi

Δf
pCR

= Stress loss in strand from creep = 19.98 ksi

19

Prestressed Concrete Beam Camber

26” Slab

K
id

=

1

= 0.81

1 +
(28500)(5.81)

[1+
(851)(7.64)
2
][1 + (0.7)(1.43)]

(2991)(851) 63596

∆f
pR1

=
(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 k
s

k
hc

k
f
k
td

t
i
-
0.118

20

Prestressed Concrete Beam Camber

26” Slab

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f
=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

9855
= 1.00

61

4f’
c

+ t 61

(4)(5.5) + 9855

t
i
= 1.0

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

=
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, t
i
) = 1.9 k
s

k
hc

k
f
k
td

t
i
-
0.118

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f

=

5
=

5
=
0.77

1+f’
c

1 + 5.5

k
td

=

t
=

9855
= 1.00

61

4f’
c

+ t 61

(4)(5.5) + 9855

t
i

= 90.0

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

22

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)

ε
sh

= k
s

k
hs

k
f

k
td

0.48x10
-
3

23

Prestressed Concrete Beam Camber

26” Slab

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hs

= 2.00

0.014H, H=70, k
hs

= 2.0

(0.014)(70) =
1.02

k
f
=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

9855
= 1.00

61

4f’
c

+ t 61

(4)(5.5) + 9855

ε
sh

= (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

24

Prestressed Concrete Beam Camber

26” Slab

Compute stress loss from relaxation

K
id

=

1

= 0.81

1 +
(28500)(5.81)

[1+
(851)(7.64)
2
][1 + (0.7)(1.46)]

(2991)(851) 63596

∆f
pR1

=
(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, t
i
) = 1.9 k
s

k
hc

k
f

k
td

t
i
-
0.118

25

Prestressed Concrete Beam Camber

26” Slab

k
s

= 1.45

0.13(8.43) = 0.354 →
1.0

k
hc

= 1.56

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

k
hc

= 1.56

(0.008)(70) =
1.0

k
f

=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

14
= 0.26

61

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)
-
0.118 = 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

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) = 0.354 →
1.0

k
hs

= 2.00

0.014H, H=70, k
hs

= 2.0

(0.014)(70) =
1.02

k
f

=

5
=

5
=
0.77

1+f’c 1 + 5.5

k
td

=

t
=

14
= 0.26

61

4f’
c

+ t 61

(4)(5.5) + 14

ε
sh

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

27

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

28

Prestressed Concrete Beam Camber

26” Slab