APPENDIX 6—DESIGN EXAMPLES FOR

DESIGN OF SLABS-ON-GROUND 360R-73
f
r
= concrete modulus of rupture, psi (MPa)
h = slab thickness, in. (mm)
k = modulus of subgrade reaction, lb/in.
3
(N/mm
3
)
L = radius of effective stiffness, in. (mm)
Mn = negative bending moment capacity of the slab,
tension at top slab surface, in.-k (N-mm);
Mp = positive bending moment capacity of the slab,
tension at bottom slab surface, in.-k (N-mm)
P
ult
= ultimate load capacity of the slab, kips
R
e,3
= residual strength factor (JSCE SF4)
S = slab section modulus, in.
3
/in. (mm
3
/mm)
μ = Poisson’s ratio for concrete (approximately 0.15)
A6.2—Assumptions/design criteria
Slab thickness h ............................................. 6 in. (150 mm)
Concrete compressive strength (cylinder) f
c
′
.............................................................. 4000 psi (27.5 MPa)
Concrete rupture modulus f
r
................... 550 psi (3.79 MPa)
Concrete elastic modulus E ...... 3,600,000 psi (25,000 MPa)
Poisson’s ratio 
μ .............................................................0.15
Modulus of subgrade reaction, k .......100 lb/in
3
(0.027 N/mm
3
)
Storage rack load ..........................................15 kips (67 kN)
Base plate............................................ 4 x 6 in. (10 x 15 cm)
A6.2.1 Calculations for a concentrated load applied a
considerable distance from slab edges
The radius of relative stiffness is given by
L = 
[E × h
3
/(12 (1 – 
μ
2
)k)]
0.25
= [3,600,000 × 6
3
/(12(1 – 0.15
2
)100)]
0.25
= 28.5 in.
The section modulus of the slab is
S = 1 in. × h
2
/6 = 12 × 6
2
/6 = 6 in.
3
/in.
The equivalent contact radius of the concentrated load is
the radius of a circle with area equal to the base plate.
a = (base plate area / 3.14)
0.5
(Eq. 
(10-2))
= (24 / 3.14)
0.5
= 2.8 in.
A concentrated load applied a considerable distance away
from slab edges should not exceed the ultimate load capacity
of the slab:
P
ult
= 6(1 + (2a/L)) × (Mp + Mn) (Eq. (10-3))
where
Mp = f
r
× R
e,3
/100 × S
Mn = f
r
× S
Combining Mp and Mn
Mp + Mn = f
r
× S × (1 + R
e,3
/100) (Eq. (10-4))
A safety factor of 1.5 is selected for this example
Mp + Mn = f
r
× S × (1 + R
e,3
/100)/1.5
Solving Eq. (10-3)
15 = 6 (1 + 2 × 2.8/28.5) × (Mp + Mn)/1.5
The minimum required bending moment capacity of the
slab for the applied load is
3.13 in.-k/in. = Mp + Mn
It is known that stresses due to shrinkage and curling can
be substantial. For the purpose of this example, an amount of
200 psi is selected. This translates into an additional moment
of 1.2 in.-k/in. (6.0 in.
3
/in. × 200 psi) to account for
shrinkage and curling stresses. This stress can vary
depending on the safety factor and other issues, including
mixture proportion, joint spacing, and drying environment.
Using Eq. (10-4) to solve for the required residual strength
factor R
e,3
3.13 in.-k/in. + 1.2 in.-k/in. = f
r
× S × (1 + R
e,3
/100)
R
e,3
≥ [(4.33 × 1000/550/6.0) – 1.0]100
R
e,3
≥ 31
Residual load factors for various fiber types and quantities
are available from steel fiber manufacturers’ literature.
Laboratory testing may be used for quality control to verify
residual strength factors on a project basis. The quantity of
steel fibers to provide the residual strength factor shown in
this example would be in the range of 15 to 33 lb/yd
3
(10 to
20 kg/m
3
), depending on the properties (length, aspect ratio,
tensile strength, and anchorage) of the fiber.
A6.2.2 Calculations for post load applied adjacent to
sawcut contraction joint
Assuming 20% of the load is transferred across the joint
(Meyerhof 1962), the load for a concentrated
load applied
adjacent to a sawcut contraction joint should not exceed
0.80 × P
ult
= 3.5(1 + (3a/L)) × (Mp + Mn)/1.5 (Eq. (10-5))
Solving Eq. (10-5),
0.80 × 15= 3.5(1 + 3 × 2.8/28.5) × (Mp + Mn)/1.5
The minimum required bending moment capacity of the
slab for the applied load is 3.97 in.-k/in. = Mp + Mn.
As in the previous example, an additional moment of
1.2 in.-k/in. is used to account for shrinkage. No curling
stress exists at the edge. Using Eq. (10-4) to solve for the
required residual strength factor R
e,3
3.97 in.-k/in. + 1.2 in.-k/in. = f
r
× S × (1 + R
e,3
/100)
R
e,3
≥ [(5.17 × 1000/550/6.0) – 1.0] × 100
R
e,3
≥ 57
The quantity of steel fibers to provide the residual strength
factor shown in this example would be in the range of 40 to
60 lb/yd
3
(25 to 35 kg/m
3
), depending on the mixture
proportion and all mixture constituents, including fiber type
and quantity.

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