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
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