Title: Lecture 36 Design of TwoWay Floor Slab System
1Lecture 36 - Design of Two-Way Floor Slab System
2Lecture Goals
- Direct Design Method
- Example of DDMs
3Example 1
Using the direct design method, design the
typical interior flat-plate panel. A flat plate
floor system with panels 24 by 20 ft is supported
on 20 in. square columns, 12 ft long. The slab
carries a uniform service live load of 80 psf and
service dead load that consists of 24 psf of
finished in addition to the slab self-weight.
Use fc 4 ksi and fy 60 ksi
4Example 1
The thickness of the slab is found using Table
9.5c
5Example 1
The weight of the slab is given as.
6Example 1
Compute the average depth, d for the slab. Use
an average depth for the shear calculation with a
5 bar (d 0.625 in)
7Example 1
The punch out shear at center column is
Two-way shear.
8Example 1
The punch out shear at center column is
9Example 1
The one-way shear at center column is
10Example 1
Calculate d in both directions. Use 5 for the
reinforcement.
11Example 1
Determine the strip sizes for the column and
middle strip. Use the smaller of l1 or l2 so l2
20 ft
Therefore the column strip b 2( 5 ft) 10 ft
(120 in) The middle strips are
12Example 1
Calculate the strip sizes
13Example 1
Moment Mo for the two directions.
long direction
short direction
14Example 1
Interior panel
15Example 1
The factored components of the moment for the
beam (long).
Negative - Moment Positive Moment
16Example 1
Components on the beam (long).
Column Strip
Negative - Moment Positive Moment
17Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
18Example 1
Computing the reinforcement uses
19Example 1
Compute the reinforcement need for the negative
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 200.6 k-ft
20Example 1
Compute the reinforcement need for the negative
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 200.6 k-ft
21Example 1
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and h 9
in.
22Example 1
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and As
5.98 in2. Use a 5 bar (Ab 0.31 in2 )
Maximum spacing is 2(h) or 18 in. So 6 in lt 18
in. OK
23Example 1
The long direction
24Example 1
Final results in the long direction.
25Example 1
The factored components of the moment for the
beam (short).
Negative - Moment Positive Moment
26Example 1
Components on the beam (short).
Column Strip
Negative - Moment Positive Moment
27Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
28Example 1
The short direction
29Example 1 - Bar development (Fig.
13.3.8)
30Example 1
Final results in short direction.
31Example 2
Using the direct design method, design the
typical exterior flat-plate panel. A flat plate
floor system with panels 24 by 20 ft is supported
on 20 in. square columns, 12 ft long. The slab
carries a uniform service live load of 80 psf and
service dead load that consists of 24 psf of
finished in addition to the slab self-weight.
Use fc 4 ksi and fy 60 ksi
32Example 2
The problem has the same parameters as first
example.
The thickness of the slab is found using
33Example 2
The weight of the slab is given as.
34Example 2
The punch out shear at center column is
35Example 2
The punch out shear at center column is
36Example 2
The one-way shear at center column is
37Example 2
The punch out shear at outside column is
38Example 2
The punch out shear at center column is
39Example 2
The punch out shear at corner column is
40Example 2
The punch out shear at center column is
41Example 2
Moment Mo for the two directions from previous
example.
Mol 411.5 k-ft d 7.9 in. Mos 333
k-ft d 7.3 in
42Example 2
The factored components of the moment for the
beam (long).
43Example 2
Column strip
Components on the beam (long).
44Example 2
Column strip
Components on the beam (long).
45Example 2
Computing the reinforcement uses
46Example 2
Compute the reinforcement need for the internal
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 216.1 k-ft
47Example 2
Compute the reinforcement need for the internal
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 216.1 k-ft
48Example 2
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and h 9
in.
49Example 2
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and As
6.46 in2. Use a 5 bar (Ab 0.31 in2 )
Maximum spacing is 2(h) or 18 in. So 5.5 in lt
18 in. OK
50Example 2
The long direction
51Example 2
The factored components of the moment for the
beam (short) is similar to an interior beam.
52Example 2
Components on the beam (short) interior.
53Example 2
Components on the beam (short) interior.
Middle Strip
Negative - Moment Positive Moment
54Example 2
The short direction
55Example 2
The final results for the interior panel.