
Concrete Floor Design:

This design guide is intended to provide guidance for the safe design and economical
construction of suspended concrete floor slabs. This design guide and the corresponding
calculations are based on the requirements of ACI 318 and strength design method
where the capacity of the beam is designed to support factored loads.
Slabs are structural elements whose lengths and widths are large in comparison to
their thicknesses. Unlike beams, shear is generally carried by the concrete
without the aid of shear reinforcement. Longitudinal reinforcement is used
to resist bending moments. The slab thickness is typically governed by deflection
criteria or fire rating requirements.

Design of Oneway Slabs:



1. Determine the required thickness for deflection control in the slab from the
table below:

Minimum Oneway Slab Thickness
(unless deflections are calculated)
(Normal Weight Concrete, fy = 60,000 psi (413.7 MPa))
Construction

Minimum thickness, h
(fraction of span length)

Simply supported

1/20

One end continuous

1/24

Both ends continuous

1/28

Cantilever

1/10


2. Determine the temperature steel required running
normal to the flexural steel. ACI 318 requires that oneway slabs have reinforcement
for temperature and shrinkage stresses running normal to the main flexural steel.
The minimum reinforcement ratio, pt, for shrinkage and temperature control (based
on gross concrete area) is 0.0014 for grade40 steel or less, 0.0020 for grade40
and grade50 steel, and 0.0018 for grade60 steel. For steel exceeding grade60,
the minimum reinforcement ratio, pt, shall be calculated as follows:
[SI] pt = (0.0018)(413.7/fy); units fy is in MPa.
[US] pt = (0.0018)(60,000/fy); units fy is in psi. The maximum spacing of
temperature steel is the smaller of 5 times the slab thickness or 18 in. Also
note while no. 3 bars may be used, for constructability and handling reasons it
is common practice to use no. 4 bars.

3. Calculate the flexural steel area required to support the factored moment on
the unit width wide section of the floor slab. Note also for a continuous
slab of multible spans it will be analyzed as a multiple span continous beam.
This means that as the flexural reinforcement required will be on the top of the
slab and the bottom of the slab corresponding to the positive and negative moments
induced on the slab. To keep the design simple, the floor is sometimes designed
with a double mat of steel, both at the top and bottom of the slab to handle the
maximum moments induced on the top and bottom of the slab. The width of the
beam is essentially set as the unit width of the strip and the depth to steal is
generally taken as the depth or height of the beam, h, minus the steel cover, (minimum
of 3/4 in for floors), minus the diameter of steel, d, divided by 2. An assumption
for d is generally used such as no. 4 bars etc. The procedure then for design
of the flexural steel is the same as the iterative design for a known beam section
and is as follows.

 Calculate the initial area of tension steel, Asi using an estimated value for the
distance from the centroid of the compressed area to extreme compression fiber,
dc, of dc = 0.1d.
 Calculate the
area of compression, Ac using the trial area of tension steel Asi.
 Calculate the
depth of the compression zone, t, and the distance from the centroid to the extreme
compression fiber, dc, from the area of compression, Ac.
 Calculate the
nominal moment capacity, Mn, of the design section using the trial area of tension
steel, Asi, and the calculated factor dc.
 Calculate the
adjusted area of steel, As, from the nominal moment capacity, Mn, and the maximum
moment load, Mu, on the beam.

4. Calculate the effective depth of the slab and calculate the
concrete shear strength of the concrete slab. Check it against the factored
shear load on the slab, 0.85Vc > Vu. If the shear strength is insufficient
increase slab thickness until 0.85Vc > Vu.

5. In summary the design procedures for beams and floor slabs are similar with the
following differences:
(a) Minimum steel cover is 3/4 in (1.91 cm).
(b) The minimum steel ratio for flexure is the same as that for temperature and
shrinkage given above.
