Table  A3.1—Design index categories used with the COE slab thickness selection method

360R-68 ACI COMMITTEE REPORT
D. Assume spacing of stiffening beams as sketched in
Fig. A4.1.
A4.1.2 Design for edge lift
A. Calculate approximate depth of stiffening beams where
d = (x)
1.176
and
1) Long direction: L = 120 ft; assume 
β = 10 ft, 6β =
60 ft. Governs
= 0.424 in.
x = 15.20;
d = (15.20)
1.176
; = 24.54 in., say 26 in.
2) Short direction: assume 
β = 10 ft; L = 58 ft < 6β.
Therefore, 58 ft governs.
Δ
allow
= 12(58)/1700
*
= 0.409 in.
x = 12.21
d = (12.21)
1.176
= 18.97 in.
18.97 in. < 26 in. Use 26 in. for trial depth.
B. Check soil bearing pressure under beams
1) Allowable soil pressure
x
L
( )
0.35 
S
( )
0.88 
e
m
( )
0.74
Y
m
( )
0.76
12
Δ P
( )
0.01
---------------------------------------------------------------------
=
Δ
allow
12 60
( )
1700∗
-----------------
=
x
120
(
)
0.35
15.00
(
)
0.88
5.0
(
)
0.74
0.338
(
)
0.76
12 0.424
(
) 2280
(
)
0.01
----------------------------------------------------------------------------------------------
=
x
58
( )
0.35
15
( )
0.88
5.0
(
)
0.74
0.338
(
)
0.76
12 0.409
(
) 2280
(
)
0.01
-----------------------------------------------------------------------------------
=
q
allow
= 3400 lb/ft
2
= 3.40 k/ft
2
2) Applied loading
Slab = 120 × 58 × 0.333 × 0.150 = 347.65 kips
Beams = 9 × 58 × 1.0 × 1.833 × 0.150 = 143.52
Beams = 5 × 111 × 1.0 × 1.833 × 0.150 = 152.59
Perimeter = 2.280 × 35 = 811.68
Live load = 0.040 × 58 × 120 = 278.40
Total: 1733.84 kips
For 1.0 ft wide beams, the assumed spacing on Fig. A4.1
provides 1077 ft
2
of bearing area. The soil bearing pressure
is then: w = 1733.84/1077 = 1.610 kips/ft
2
1.610 < 3.40, so soil bearing pressure is OK for the
assumed beam layout.
C. Calculate section properties for full slab width
Long Short
direction
direction
Beam depth d, in. 26
26
Beam width b, in. 12 12
Number of beams 5 9
Total beam width nb, in. 60 108
Slab thickness t, in. 4 4
Moment of inertia I, in.
4 
208,281 387,791
Section moduli, in.
3
S
T 
33,702 66,861
S
B 
10,509 19,198
Cross sectional area, in.
2 
4104 8136
Depth to neutral axis c
g,
in. –6.18 –5.80
Allowable concrete stress, ksi
Tension 0.329 0.329
Compression 1.350 1.350
Tensile cracking stress, ksi 0.411 0.411
D. Calculate minimum number of tendons required
1) Number of tendons required for minimum average
prestress of 50 psi. Stress in tendons immediately after
anchoring:
f
ps
= 0.7f
pu
= (0.7) (270) = 189 ksi
Stress in tendons after losses: f
ps
= 189 – 30 = 159 ksi
= 8.44
= 16.72
2) Number of tendons required to overcome slab-
subgrade friction on polyethylene sheeting:
N
T
(minimum prestress) area slab
(
)
×
1000 effective tendon stress
(
)
area tendon
(
)
×
×
-------------------------------------------------------------------------------------------------------------------
=
N
T
long
(
)
50 psi
(
) 4103 in.
2
(
)
1000 lb/kip
(
) 159 ksi
(
) 0.153 in.
2
(
)
-----------------------------------------------------------------------------------
=
N
T
short
(
)
50 psi
(
) 8136 in.
2
(
)
1000 lb/kip
(
) 159 ksi
(
) 0.153 in.
2
(
)
-----------------------------------------------------------------------------------
=
Fig. A4.1—Beam layout for apartment house example.
*
The 1700 value is based on experience; refer to 
Chapter 9
 for typical values.

DESIGN OF SLABS-ON-GROUND
360R-69
Weight of beams and slab = 643.76 kips
N
T
= 0.50
N
T
= 0.50
= 9.92 strands (each direction)
3) Total number of tendons
N
T
(long) = 8.44 + 9.92 = 18.36, use 19 tendons
N
T
(short) = 16.72 + 9.92 = 26.64, use 27 tendons
4) Design prestress forces
Because maximum moments occur near the slab perim-
eter, friction losses will be minimal at points of maximum
moments. Therefore, assume total prestressing force effec-
tive for structural calculations.
Long direction: P
r
= (19) × 24.3k = 461.7 kips
Short direction: P
r
= (27) × 24.3k = 656.10 kips
E. Calculate design moments
1) Long direction
M
l
= 
M
l
= 
M
l
= 2.81 ft-kips/ft
2) Short direction
M
s
= (d)
0.35
[(19 + e
m
)/ 57.75](M
l
)
M
s
= (26)
0.35
[(19 + 5.0)/57.75](2.81) = 3.65 ft-kips/ft
A4.1.3 Design for edge lift continued; service moments
compared with design moments
F. Calculate allowable service moments and compare with
design moments
1) Long direction
a) Tension in bottom fiber
(12 × 58)M
t
= S
B
[(P
r
/A) + f
t
] – P
r
e
(12 × 58)M
t
= 10,509[(461.7/4104) + 0.329] –
(461.7)(4.18)
M
t
= 2710 in.-kips/(12 × 58) = 3.89 ft-kips/ft
3.89 > 2.81 OK
b) Compression in top fiber
(12 × 58)M
c
 = S
T
[f
c
 – P
r
/A] – P
r
e
(12 × 58)M
c
= 33,702[1.350 – (461.7/4104)] –
(461.7)(4.18)
M
c
= 39,776 in.-kips/(12 × 58) = 57.15 ft-kips/ft
57.15 > 2.81 OK
2) Short direction
a) Tension in bottom fiber
(12 × 120)M
t
= 19,198 [(656.1/8136) +0.329] –
(656.1)(3.80)
M
t
= 5371 in.-kips/(12 × 120) = 3.73 ft-kips/ft
3.73 > 3.65 OK
b) Compression in top fiber
(12 × 120)M
c
= (66,861) [1.350 – (656.1/8136)] –
(656.1)(3.80)
M
c
= 82,377 in.-kips/(12 × 120) = 57.21 ft-kips/ft
57.21 > 3.65 OK
G. Deflection calculations
1) Long direction
a) Allowable differential deflection
= 11.91 ft
6
β = 66.48 ft < 120 ft, so 6β governs
Δ
allow
= 12 (66.48)/800 = 0.997 in.
b) Expected differential deflection
Δ = 0.296 in. 0.296 < 0.997 OK
2) Short direction
a) Allowable differential deflection
= 12.94 ft
6
Δ = 77.64 > 58 ft, so 58 ft governs
Δ
allow
= 12(58)/800 = 0.870 in.
b) Expected differential deflection
Δ = 0.236 in. 0.236 < 0.870 in. OK
Deflections for edge lift bending are much less than allow-
able in both long and short directions.
μ
( ) W
( )
f
ps
( ) tendon area
(
)
-------------------------------------------
0.75
(
) 643.76
(
)
159
(
) 0.153
(
)
-------------------------------------
S
( )
0.10
de
m
(
)
0.78
Y
m
( )
0.66
7.2 L
( )
0.0065
P
( )
0.04
--------------------------------------------------------
14.50
(
)
0.10
26
5.0
×
(
)
0.78
0.338
(
)
0.66
7.2 120
(
)
0.0065
2280
(
)
0.04
------------------------------------------------------------------------------------
β
1 12
E
c
I
E
s
--------
4
⁄
1 12
1,500,000
208,281
×
1000
-------------------------------------------------
4
⁄
=
=
Δ
L
( )
0.35
S
( )
0.88
e
m
( )
0.74
Y
m
( )
0.76
15.90 d
( )
0.85
P
( )
0.01
---------------------------------------------------------------------
=
Δ
120
(
)
0.35
14.50
(
)
0.88
5.0
(
)
0.74
0.338
(
)
0.76
15.90 26
( )
0.85
2280
(
)
0.01
----------------------------------------------------------------------------------------------
=
β
1 12
1,500,000
387,791
×
1000
-------------------------------------------------
4
⁄
=
Δ
58
( )
0.35
15
( )
0.88
5.0
(
)
0.74
0.338
(
)
0.76
15.90 26
( )
0.85
2280
(
)
0.01
-----------------------------------------------------------------------------------
=

360R-70 ACI COMMITTEE REPORT
H. Shear calculations
1) Long direction
a) Expected shear
V
l
= 
V
l
= 
V
l
= 1.300 kips/ft
b) Permissible shear stress
v
c
= 1.5
= 1.5
= 82.2 psi
c) Total design shear stress
V = = 48.33 psi
48.33 < 82.2 psi OK
2) Short direction
a) Expected shear
V
s
= 
V
s
= 1.235 kips/ft
b) Total design shear stress
V = = 52.78 psi
52.78 < 82.2 psi OK
Shear stresses are OK in both short and long directions.
A4.1.4 Design for center lift
A. Calculate design moments
1) Long direction
M
l
= A
o
[B(e
m
)
1.238
+ C]
A
o
=
1/727[(L)
0.013
(S)
0.306
(d)
0.688
(P)
0.534
(Y
m
)
0.193
]
A
o
=
1/
727[(120)
0.013
(14.50)
0.306
(26)
0.688
(2280)
0.534
(0.384)
0.193
]
A
o
= 1.612
0 
≤ e
m
≤ 5
e
m
= 4.0 B = 1.0 C = 0
M
l
= (1.612)(4.0)
1.238
= 8.97 ft-kips/ft
2) Short direction
M
s
= [(58 + e
m
)/60]M
l
M
s
= [(58 + 4.0)/60]8.97 = 9.27 ft-kips/ft
B. Calculate allowable moments and compare with design
moments
1) Long direction
a) Tension in top fiber
(12 × 58)M
t
 = S
T
[(P
r
/A) + f
t
] + P
r
e
(12 × 58)M
t
= (33,702)[(461.7/4104) + 0.329] +
(461.7)(4.18)
M
t
= 16,809 in.-kips/(12 × 28) = 24.15 ft-kips/ft
24.15 > 8.97 OK
b) Compression in bottom fiber
(12 × 58)M
c
 = S
B
[ f
c
 – (P
r
/A)] + P
r
e
(12 × 58)M
c
= 10,509[1.350 – (461.7/4104)] +
(461.7)(4.18)
M
c
= 14,935 in.-kips/(12 × 58) = 21.46 ft-kips/ft
21.46 > 8.97 OK
2) Short direction
a) Tension in top fiber
(12 × 120)M
t
= 66,861[(656.1/8136) + 0.329] +
(656.1)(3.80)
M
t
 = 29,882 in.-kips/(12 × 120) = 20.75 ft-kips/ft
20.75 > 9.27 ft-kips/ft
b) Compression in bottom fiber
(12 × 120)M
c
= 19,198[1.350 – (656.1/8136)] +
(656.1)(3.80)
M
c
= 26,862 in.-kips/(12 × 120) = 18.65 ft-kips/ft
18.65 > 9.27 OK
Moment capacities exceed expected service moments for
center lift loading in both long and short directions. By
observation, deflection and shear calculations are within
permissible tolerances. For a detailed discussion for slabs on
expansive soils, see Design and Construction of Post-
Tensioned Slabs-on-Ground (PTI 2004).
C. Tendon and beam requirements
1) Long direction: Use nineteen 1/2 in., 270k strands in
slab. Two spaces at 3 ft 2 in. on center and 16 spaces at 3 ft
1-1/4 in. on center beginning 1 ft from each edge. Five beams,
12 in. wide, 26 in. deep, evenly spaced at 14 ft 3 in. on centers.
2) Short direction: Use twenty-seven 1/2 in., 270k
strands in slab. Two spaces at 4 ft 6 in. and 24 spaces at 4
ft 6-1/2 in. on center beginning 1 ft from each edge. Nine
beams, 12 in. wide, 26 in. deep, evenly spaced at 14 ft 10-1/2 in.
on centers.

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