Prestressed Concrete Beam Camber

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26 Νοε 2013 (πριν από 3 χρόνια και 11 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
about the end support.

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